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https://math.stackexchange.com/questions/3073471/what-is-the-probability-that-a-player-does-not-have-at-least-1-card-of-each-suit
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# What is the probability that a player does not have at least 1 card of each suit with a 52-card deck?
There are 52 cards in a deck, with 4 suits, each with 13 cards. You distribute the 52 cards, at random, to each of four players so that each has 13 in their hand. What is the probability that at least one player does not have at least 1 card of each suit?
I found this post relevant, though it only focuses on the first 13 cards.Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present?. My question is after all 52 cards have been dealt. I've used this as a starting point, but am currently stuck as to how to approach this.
• Very often these questions are easier if you look at them the other way round. How would you approach it if the question was "What's the probability that some player has at least one card from each suit?" and can you see how that's relevant to your question? – postmortes Jan 14 at 17:22
• The post I linked to is the answer I'm looking for, I'm dumb. Thanks for your reply, you helped me to figure this out. – John Matthesson Jan 14 at 18:15
• John, I the post you linked does not have the answer you are looking for. Call the players A,B,C,D. The post you linked is the probability that A is missing some suit. Your current question is the probability that at least one of A, B, C or D is missing some suit. – Mike Earnest Jan 14 at 18:20
Perhaps there's some smart way to count the number of permutations of the 52 cards such that each player gets at least one card of each suite, which is the complement of what is asked for. But sometimes we do not need the exact solution and a simulation can give a pretty good approximation. In this case, with $$10^6$$ samples, we can estimate the probability being asked for is about $$0.184171$$. I bet this is pretty close the exact result.
Here's the code in Mathematica
In[2]:= n = 52
Out[2]= 52
In[13]:= n0 = n/4
Out[13]= 13
In[3]:= l = Range[n]
Out[3]= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, \
18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, \
35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52}
In[14]:= ll = Partition[l, n0]
Out[14]= {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, {14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26}, {27, 28, 29, 30, 31, 32,
33, 34, 35, 36, 37, 38, 39}, {40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52}}
In[39]:= containAll[sample_] := Module[{i, sl, crossl},
sl = Partition[sample, n0];
crossl = Tuples[{sl, ll}];
If[And @@ Map[ContainsAny[#[[1]], #[[2]]] &, crossl], 1, 0]
]
In[53]:= s0 =
Total[Parallelize[
Map[(Permute[l, #] // containAll) &,
RandomPermutation[n, 1000000]]]] // AbsoluteTiming
Out[53]= {191.22, 815829}
In[59]:= 1 - 815829/1000000 // N
Out[59]= 0.184171
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Discover a lot of information on the number 3306: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 3306
Is 3306 a prime number? No
Is 3306 a perfect number? No
Number of divisors 16
List of dividers 1, 2, 3, 6, 19, 29, 38, 57, 58, 87, 114, 174, 551, 1102, 1653, 3306
Sum of divisors 7200
## How to write / spell 3306 in letters?
In letters, the number 3306 is written as: Three thousand three hundred and six. And in other languages? how does it spell?
3306 in other languages
Write 3306 in english Three thousand three hundred and six
Write 3306 in french Trois mille trois cent six
Write 3306 in spanish Tres mil trescientos seis
Write 3306 in portuguese TrĂªs mil trezentos seis
## Decomposition of the number 3306
The number 3306 is composed of:
2 iterations of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
Other ways to write 3306
In letter Three thousand three hundred and six
In roman numeral MMMCCCVI
In binary 110011101010
In octal 6352
In US dollars USD 3,306.00 (\$)
In euros 3 306,00 EUR (€)
Some related numbers
Previous number 3305
Next number 3307
Next prime number 3307
## Mathematical operations
Operations and solutions
3306*2 = 6612 The double of 3306 is 6612
3306*3 = 9918 The triple of 3306 is 9918
3306/2 = 1653 The half of 3306 is 1653.000000
3306/3 = 1102 The third of 3306 is 1102.000000
33062 = 10929636 The square of 3306 is 10929636.000000
33063 = 36133376616 The cube of 3306 is 36133376616.000000
√3306 = 57.49782604586 The square root of 3306 is 57.497826
log(3306) = 8.103494278381 The natural (Neperian) logarithm of 3306 is 8.103494
log10(3306) = 3.5193028492354 The decimal logarithm (base 10) of 3306 is 3.519303
sin(3306) = 0.86468775665388 The sine of 3306 is 0.864688
cos(3306) = 0.5023097485545 The cosine of 3306 is 0.502310
tan(3306) = 1.7214234028748 The tangent of 3306 is 1.721423
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## Compute Expected Value, Pass GO, Collect \$200
Expected Value – such a great time to talk about games, probability, and decision making! Today’s lesson started with a Monopoly board in the center of the room. I had populated the “high end” and brown properties with houses and hotels. Here’s the challenge:
When I play Monopoly, my strategy is often to buy and build on the cheaper properties. This leaves me somewhat scared when I head towards the “high rent” area if my opponents built there. It is now my turn to roll the dice. Taking a look at the board, and assuming that my opponents own all of the houses and hotels you see, what would be the WORST square for me to be on right now? What would be the BEST square?
For this question, we assumed that my current location is between the B&O and the Short Line Railroads. The conversation quickly went into overdrive – students debating their ideas, talking about strategy, and also helping explain the scenario to students not as familiar with the game (thankfully, it seems our tech-savvy kids still play Monopoly!). Many students noted not only the awfulness of landing on Park Place or Boardwalk, but also how some common sums with two dice would make landing on undesirable squares more likely.
ANALYZING THE GAME
After our initial debates, I led students through an analysis, which eventually led to the introduction of Expected Value as a useful statistic to summarize the game. Students could start on any square they wanted, and I challenged groups to each select a different square to analyze. Here are the steps we followed.
First, we listed all the possible sums with 2 dice, from 2 to 12.
Next, we listed the Monopoly Board space each die roll would causes us to land on (abbreviated to make it easier).
Next, we looked at the dollar “value” of each space. For example, landing on Boardwalk with a hotel has a value of -\$2,000. For convenience, we made squares like Chance worth \$0. Luxury Tax is worth -\$100. We agreed to make Railroads worth -\$100 as an average. Landing on Go was our only profitable outcome, worth +\$200. Finally, “Go to Jail” was deemed worth \$0, mostly out of convenience.
Finally, we listed the probability of each roll from 2 to 12.
Now for the tricky computations. I moved away from Monopoly for a moment to introduce a basic example to support the computation of expected value.
I roll a die – if it comes out “6” you get 10 Jolly Ranchers, otherwise, you get 1. What’s the average number of candies I give out each roll?
This was sufficient to develop need for multiplying in our Monopoly table – multiply each value by its probability, find the sum of these and we’ll have something called Expected Value. For each initial square, students verified their solutions and we shared them on a class Monopoly board.
The meaning of these numbers then held importance in the context of the problem – “I may land on Park Place, I may roll and hit nothing, but on average I will lose \$588 from this position”.
HOMEWORK CHALLENGE: since this went so well as a lesson today, I held to the theme in providing an additional assignment:
Imagine my opponent starts on Free Parking. I own all 3 yellow properties, but can only afford to purchase 8 houses total. How should I arrange the houses in order to inflict the highest potential damage to my opponent?
I’m looking forward to interesting work when we get back to school!
## Pulling In To the Station
My school isn’t 1-1 with technology yet, though there are rumblings we will get there next year….or the year after….or 2031…anyway, it’s time to get techy! My new classroom features 4 computer stations in the back – nice to have, but not super-helpful with classes of about 24 each. Station-model classroom structure has been super-helpful in my pre-calculus class in the first month. Besides the chance for all students to participate in rich technology-based activities, I’ve had the opportunity to carve out valuable small-group time with students. Here’s an example:
In our first pre-calc unit, we review functions and their shirts, folding in new ideas like the step function, piecewise and even/odd functions. My objective for the class was for students to consider functions in varied forms. As students entered class, playing cards were drawn to establish their groupings, so there were 3 groups of 7 or 8. With 15 minutes on the classroom clock, students started on their first station:
1. Group 1 gathered in a small group with me in a circle of desks, where we worked through proving functions even or odd, and sketching their graphs.
2. Group 2 worked at the computer stations on a Desmos Marbleslides featuring quadratic functions, with many students pairing up to work together. If you have never tried a Marbleslides, run and play now – we’ll wait for you to come back…
3. Group 3 worked out in the courtyard (hey, my new classroom leads outside – which is nice) on a group task involving a piecewise function.
After groups had rotated through all 3 activities, we had time to recap / share and assess our learning over the hour. Here’s why I need to do this more:
• The small group station let me touch base with every student, assess strengths, find out what we need to work on, and provide feedback to everyone.
• Marbleslides is sneaky awesome! When students begin to obsess over function shifts and how to restrict domains and don’t want to peel away from their computer, you know something is going right.
• Class went fast! It felt like the mixed practice from Let It Stick was now becoming part of my classroom culture.
• My pre-calc is mostly 11th and 12th graders, who have had a pretty traditional classroom experience in their math lives. I can sense they appreciate that something difference is happening.
• All students are responsible for their learning. Even the least-active task, the piecewise function, was used the next class for sharing out and a jumping-off point.
## A Bulleted Assemblage of Items for the New School Year (but not a list)
The “list” article is a popular device, and one which often draws the eyeballs. Lists are also, often, a cop-out – a way to express many ideas without having to dig too deeply. I hate lists….
As I start my new school year tomorrow, I give you this bulleted assemblage of items which are on my mind as I look forward to our first day.
• Fawn Nguyen’s 7 Deadly Sins of Teaching Math is required reading for all professionals. In particular, I strive to pay more attention to my (teacher talking / student talking ) ratio. I like to think I am strong in this area, but I need to do better. Before the end of the last school year, our district screened the movie “Most Likely to Succeed” to all professional staff. In an opening scene, the teacher provides first-day freshmen with an opening day task – and then leaves the room. The students struggle, the teacher eventually intervenes, but a powerful classroom culture is established. I want to provide more tasks to my students where I’m simply not needed.
• I have used a number of opening-day activities for AP Statistics over 14 years. Distracted Driving and the Henrico hiring case are two I used most often. But I think Doug Tyson’s Smelling Parkinson’s activity could be my new favorite. It’s a powerful premise which gets kids talking about the possible vs the plausible on day 1, with a hint of simulation thrown in for good measure. I show the video below to the class and right away the statistical importance of what we do for the entire school year is established.
• Desmos Activity Builder will take on a much bigger role in my classroom. I’ve created activities for both my Pre-Calculus and my freshman Prob/Stat class to review their understanding, and also to serve as my “getting to know you” opportunity. Look forward to sharing out how it goes.
• Shoes. I hate new shoes. They’re tight and often rip apart the back of my ankle until I break them in. If we can have pre-washed jeans, then we can have pre-worn shoes. We need our best people on this.
• Who knew a cute Pythagorean triple generator could be of interest to so many. After I posted about an interesting share from Ken Sullins at the PCTM summer conference, so many folks chimed in with their ideas. Thanks especially to Joel Bezaire who shared additional ideas from Twitter Math Camp. I’m using this in my pre-calc class on day 1.
• I’ve given the same probability problem to my freshmen for the last few years. I love everything about this problem on day 1: it gets kids talking, it gets kids struggling, and it tells me much about their problem solving background.
OK, maybe this was a list after all. I need to do some last-minute ironing.
## Who Assessed it Better? AP Stats Inference Edition
Free-response questions and exam information in this post freely available on the College Board – AP Statistics website
Today I am stealing a concept from Dan Meyer’s task comparison series “Who Wore it Best”, and bringing it to the AP Statistics exam world. In the series of 6 free-response questions on the AP Stats exam, it is not unusual for one question to focus solely on inference. Compare these two questions, which each deal with inference for proportions.
From 2012:
From 2016:
I read (graded) the question from 2012 as an AP Exam reader, and observed a variety of approaches. I find that while many students understand the structure of a hypothesis test, it’s the nuance – the rationales for steps – which are often lost in the communication. In the 2012 question, students were expected to do the following:
1. Identify appropriate parameters
2. State null and alternate hypotheses
3. Identify conditions
• Independent, random samples and normality of sampling distribution
4. Name the correct hypothesis procedure
5. Compute / communicate test statistic and p-value
6. Compare the p-value to an alpha level
7. Make an appropriate conclusion in context of the problem
It’s quite a list. And given that individual AP exam problems are worth a total of 4 points, steps are often combined into one scoring element. Here, naming the test and checking conditions were bundled – as such, precision in providing a rationale for conditions was often forgiven. For example, if students identified the large sample sizes as a necessary condition, this was sufficient, even if there was no recognition of a link to normality of sampling distributions. Understanding the structure of a hypothesis test – with appropriate communication – was clearly the star of the show. While inference is one the “big ideas” in AP Statistics, my view is that questions like this from 2012 encourage cookbook statistics, where memorized structure take the place of deeper understanding.
So, it was with much excitement that I saw question 5 from 2016. Here, the interpretation of a confidence interval was preserved. But I appreciate the work of the test development committee in parts b and c; rather than have students simply list and confirm conditions for inference, the exam challenges students to be quite specific about their rationales. With parts b and c, students certainly struggled more than with the conditions in 2012, but I hope their inclusion causes statistics teachers to consider their approach to hypothesis conditions. The mean scores for each question speak to struggles students on this question, compared to traditional hypothesis testing structure.
• 2012: 1.56 (out of 4)
• 2016: 1.27 (out of 4)
The inclusion of part b of question 5 this year, where students were asked to defend the np > 10 condition, was perfect timing for my classes. This year, I tried a new approach to help develop student understanding of the binomial distribution / sampling distribution relationship. I found that while many students will continue to resort to the “short cut” – memorizing conditions – a higher proportion of students were able to provide clear communication of this inference condition.
The AP Statistics reading features “Best Practices Night” – where classroom ideas are shared. You can find resources from the last few years at Jason Molesky’s APStatsMonkey site. I shared my np > 10 ideas with the group, and received many positive comments about it. Enjoy my slides here, and feel free to contact me with questions regarding this lesson:
Finally, I can’t express how wonderfully rich a professional-development experience the AP reading is. I always find myself with a basket of new classroom ideas and contacts to share with – it’s stats-geek Christmas. For me, 2016 is the year the #MTBoS started to make its mark at the Stats reading – I met so many folks from Twitter, and we held our first-ever tweet-up!
Also, the vibrant Philadelphia-area stats community was active as always. We meet as a group a few times a year to share ideas and lessons; seeing so many from this area participate in the reading makes us all better with what we do for our students.
## The People in My Math Neighborhood
Oh, who are the people in your neighborhood?
Say, who are the people in your neighborhood?
The people that you meet each day
I work with an awesome group of people at a high school outside of Philadelphia. They are my colleagues, the people I share ideas with on a daily basis, and some of my closest friends.
But in recent years, my math neighborhood has grown considerably. I suppose I discovered the power of the online neighborhood 4 or 5 years ago, developing and growing a wonderful network of professional colleagues through the #MTBoS. And my relationship with this neighborhood has grown from a mechanism for sharing ideas, to a source of inspiration, positive thought, discussion and reflection.
We are now 3 weeks after the NCTM Annual Conference in San Francisco. It’s easy to forgot the little things which occur in a big conference, and I hopefully will find time to reflect and utilize new ideas later. NCTM this year has done a wonderful job of providing a means to continue reflections and growth outside of the conference, along with archiving session resources. Here, I highlight 4 sources of inspiration, and friends in my math neighborhood, as I look back on my San Francisco experiences.
GRAHAM FLETCHER – Graham, an elementary specialist from Georgia (or is he Canadian? such a chameleon), challenged teachers to consider the mathematical story we share with students in his ShadowCon talk. How is your story different than the one being told by your colleague teaching the same material just across the hall?
High school teachers may be intrigued by Graham’s discussion of fractions, reducing and equivalence and the role of “simplifying”. His talk has caused me to think about the many odd restrictions we place on student work: i.e. “write the equation of your line in standard form”, and their necessity in my math story. Graham’s call to action – challenging teachers to identify their own “simplifying fractions” (something they teach not currently in the standards) – is an appropriate task for all grades.
ROBERT KAPLINKSY – Robert was featured on the MathEd Out podcast last summer, and I recall taking a walk, listening last year when it occurred to me that Robert’s path to becoming a math teacher was eerily similar to mine. His ShadowCon talk, “Empower”, reminds me that no matter how top-down our education world may feel, we all have a role to empower others and become influential in our math neighborhoods. I appreciate the multiple mechanisms Robert suggests for fostering empowerment, and his call to action that we thank a colleague who helped us feel empowered is a wonderful way to close out a school year – and look forward to new things.
PEG CAGLE – I have admired Peg’s ideas for some time now, and was thrilled to meet and chat with her last year at Twitter Math Camp. Even though I rarely teach geometry, I felt pulled to Peg’s session “Paper Cup + Gust of Wind”, and was awed by the simplicity, engagement, theme-building in this simple task. By rolling a paper Dixie cup along a surface, Peg develops a lesson which extends through the school year, building complexity each time.
Day 1:Explain what happens when we roll out the cup
Day 40: Convince a skeptic of the shape it makes. Find its area.
Day 105: find area of shape based on dimensions
Day 140: How can you build a cup from a single sheet (with base) of 8.5 x 11 paper to trace out the maximum area as it rolls?
Day 175 (after trig ratios): how do you NOW find the area of the shape, given its dimensions
This session has caused me to think about other simple tasks which could become full-course themes. Peg’s inspiration came from a cup blowing in the breeze – you never know where the next fun math idea will come from!
CHRISTINE FRANKLIN – Why was I so nervous and awe-struck to meet Christine at the AP Stats forum in San Francisco? Because she is so awesome – and was the inspiration for my NCTM talk on Variability and Inference, geared towards the middle school community. It was at Professional Night at the AP Stats reading 2 years ago where Christine diagrammed the historical path stats has taken in K-12 curriculum, and the parallels between AP and middle school descriptions. Christine was recently named the K-12 statistical ambassador by the ASA, and a sweeter person could not fill the job.
Hoping I never move out of the neighborhood!
## Activity Builder Reflections
We’re now about 9 months into the Desmos Activity Builder Era (9 AAB – after activity-builder). It’s an exciting time to be a math teacher, and I have learned a great deal from peeling apart activities and conversing with my #MTBoS friends (run to teacher.desmos.com to start peeling on your own – we’ll wait…). In the last few weeks, I have used Activities multiple times with my 9th graders. To assess the “success” of these activities, I want to go back to 2 questions I posed in my previous post on classroom design considerations, specifically:
• What path do I want them (students) to take to get there?
• How does this improve upon my usual delivery?
AN INTRODUCTION TO ARITHMETIC SERIES (click here to check out the activity)
My unit or arithmetic sequences and series often became buried near the end of the year, at the mercy of “do we have time for this” and featuring weird notation and formulas which confused the kids. I never felt quite satisfied by what I was doing here. I ripped apart my approach this year, hoping to leverage what students knew about linear functions to develop an experience which made sense. After a draft activity which still left me cold, awesome advice by Bowen Kerins and Nathan Kraft inspired some positive edits.
In the activity, students first consider seats in a theater, which leads to a review of linear function ideas. Vocabulary for arithmetic sequences is introduced, followed by a formal function for finding terms in a sequence. It’s this last piece, moving to a general rule, which worried me the most. Was this too fast? Was I beating kids over the head with a formula they weren’t ready for? Would the notation scare them off?
The path – having students move from a context, to prediction, to generalization, to application – was navigated cleanly by most of my students. The important role of the common difference in building equations was evident in the conversations, and many were able to complete my final application challenge. The next day, students were able to quickly generate functions which represent arithmetic sequences, and with less notational confusion than the past. It certainly wasn’t all a smooth ride, but the improvement, and lack of tooth-pulling, made this a vast improvement over my previous delivery.
DID IT HIT THE HOOP? (check out the activity)
Dan Meyer’s “Did It Hit the Hoop” 3-act Activity probably sits on the Mount Rushmore of math goodness, and Dan’s recent share of an Activity Builder makes it all the more easy to engage your classes with this premise. In class, we are working through polynomial operations, with factoring looming large on the horizon. My 9th graders have little experience with anything non-linear, so this seemed a perfect time to toss them into the deep end of the pool. The students worked in partnerships, and kept track of their shot predictions with dry-erase markers on their desks. Conversations regarding parabola behavior were abundant, and I kept mental notes to work their ideas into our formal conversations the next day. What I appreciate most about this activity is that students explore quadratic functions, but don’t need to know a lick about them to have fun with it – nor do we scare them off by demanding high-level language or intimidating equations right away.
The next day, we explored parabolas more before factoring, and developed links between standard form of a quadratic and its factored form. Specifically, what information does one form provide which the other doesn’t, and why do we care? The path here feels less intimidating, and we always have the chance to circle back to Dan’s shots if we need to re-center discussion. And while the jury is out on whether this improves my unit as a whole, not one person has complained about “why”…yet.
MORE ACTIVITY BUILDER GOODNESS
Last night, the Global Math Department hosted a well-attended webinar featuring Shelley Carranza, who is the newest Desmos Teaching Faculty member (congrats Shelley!). It was an exciting night of sharing – if you missed it, you can replay the session on the Bigmarker GMD site.
## When Binomial Distributions Appear Normal
We’re working through binomial and geometric distributions this week in AP Stats, and there are many, many seeds which get planted in this chapter which we hope will yield bumper crops down the road. In particular, normal estimates of a binomial distribution – which later become conditions in hypothesis testing – are valuable to think about now and tuck firmly into our toolkit. This year, a Desmos exploration provided rich discussion and hopefully helped students make sense of these “rules of thumb”.
Each group was equipped with a netbook, and some students chose to use their phones. A Desmos binomial distribution explorer I had pre-made was linked on Edmodo. The explorer allows students to set the paremeters of a binomial distribution, n and p, and view the resulting probability distribution. After a few minutes of playing, I asked students what they noticed about these distributions.
A lot of them look normal.
Yup. And now the hook has been cast. Which of these distributions “appear” normal, and under what conditions? In their teams, students adjusted the parameters and assessed the normality. In the expressions, the normal overlay provides a theoretical normal curve, based on the binomial mean and standard deviation, along with error dots. This provides more evidence as students debate normal-looking conditions.
Each group was then asked to summarize their findings:
• Provide 2 settings (n and p) which provide firm normality.
• Provide 2 settings (n and p) which provide a clearly non-normal distribution.
• Optional: provide settings which have you “on the fence”
My student volunteer (I pay in Jolly Ranchers) recorded our “yes, it’s normal!” data, using a second Desmos parameter tracker. What do we see in these results?
Students quickly agreed that higher sample sizes were more likely to associate with a normal approximation. Now let’s add in some clearly non-normal data dots. After a few dots were contributed, I gave an additional challenge – provide parameters with a larger sample size which seem anti-normal. Hers’s what we saw:
The discussion became quite spirited: we want larger sample sizes, but extreme p’s are problematic – we need to consider sample size and probability of success together! Yes, we are there! The rules of thumb for a normal approximation to a binomial had been given in a flipped video lecture given earlier, but now the interplay between sample size and probability of success was clear:
And what happens when we overlay these two inequalities over our observations?
Awesomeness! And having our high sample sizes clearly outside of the solution region made this all the more effective.
Really looking forward to bringing this graph back when we discuss hypothesis testing for proportions.
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Not an official ACM page
[Problem 2 | 1997 East-Central problem set | Ed's programming contest problem archive | my home page]
## ACM East Central Region 1997 Regional Programming Contest
### Problem 1 - Polygon Puzzler
We define a simple polygon as an area enclosed by endpoint-connected line segments such that no line segment intersects another (except for adjoining segments at their endpoints). A simple polygon can thus be defined by an ordered list of its vertices (the endpoints of the enclosing line segments). A planar polygon is a polygon whose vertices all lie in the same plane.
For this problem you are asked to compute the area of a simple planar polygon oriented in three space. That is, although the vertices of the polygon lie in some two-dimensional plane, the vertices are specified in three-dimensional Cartesian coordinates.
### Input
The input will consist of an ordered sequence of coordinates for the vertices of the polygon. Each line of the input will contain the three-dimensional Cartesian coordinates for a single vertex in the order x, y, z. The values for the x, y, z components will be separated by a single space. Input values should be considered to be double precision floating point and may be positive or negative. The coordinates of the final line of input input will be the same as the coordinates on the first line of input. No polygon will have more than 1024 vertices.
### Output
The output should be the area of the polygon specified by the input and should be rounded to the nearest 1/1000 (i.e., three places after the decimal point should be printed).
### Sample Input
```1.401117996399998e+00 1.509291958378880e-01 1.186959898555237e-01
1.918738650437130e-01 1.067473024933127e+00 9.075713530920345e-01
1.401117996399998e+00 -1.509291958378880e-01 -1.186959898555237e-01
-1.918738650437130e-01 -1.067473024933127e+00 -9.075713530920345e-01
-1.401117996399998e+00 1.509291958378880e-01 1.186959898555237e-01
```
```4.000
```
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Exponential Properties Involving Products
## Add exponents to multiply exponents by other exponents
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Exponential Properties Involving Products
by CK-12 //basic
Learn how to use basic exponent rules.
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## Exponential Properties Involving Products
Write repeated products in exponential form and simplify expressions involving exponents.
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## Chapter 9--Concept 9.1 (Lesson): Exponential Properties Involving Products
Learn how to use basic exponent rules.
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## Recognize and Apply the Power of a Product Property
Learn to recognize and apply the power of a product property when multiplying monomials
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## Product Rules for Exponents
This concept explores the product rules for exponents.
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## Multiplying Monomials - Power of a Product Property
by Laura Warden //basic
Recognize and apply the power of a product property when multiplying monomials.
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## Product Rules for Exponents
This concept explores the product rules for exponents.
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## Exponential Properties Involving Products
Write repeated products in exponential form and simplify expressions involving exponents.
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## Exponential Properties Involving Products
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## Multiplication and Powers
This concept explores the product rules for exponents.
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## Exponential Properties Involving Products
Write repeated products in exponential form and simplify expressions involving exponents.
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## Exponential Properties Involving Products
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## Exponential Properties Involving Products
by Susan Sudberry //basic
Learn how to use basic exponent rules.
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## Multiplication Properties of Exponents
by EPISD Algebra 1 Team //at grade
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## Exponential Properties Involving Products - CCSS Extended
Write repeated products in exponential form and simplify expressions involving exponents.
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## Multiplying Powers with the Same Base: Powers with the Same Base
by SFDR Algebra I //at grade
Write repeated products in exponential form and simplify expressions involving exponents.
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## Exponential Properties Involving Products
by Barbara Zingg //basic
Learn how to use basic exponent rules.
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• PLIX
## Exponential Properties Involving Products: Slider Exploration
Exponential Properties Involving Products Interactive
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## Product of Powers Property - Overview
by CK-12 //basic
Overview
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## Multiple Exponent Properties Involving Products - Overview
by CK-12 //basic
Overview
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• Video
## Multiplying Monomials
by CK-12 //basic
Learn how to multiply monomials.
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• Video
## Exponential Notation
How to represent expressions in exponential notation, as explained by James Sousa.
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## Exponent Properties Involving Products
by CK-12 //basic
Shows examples of how to use exponent properties involving products to solve problems.
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## Product of Powers: A Sample Application
This video demonstrates a sample use of the product of powers.
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## Product of Powers Property - Example 1
by CK-12 //basic
Simplifying Expressions Using the Product of Powers Property
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## Products of Powers: An Explanation of the Concept
This video provides an explanation of the concept of product of powers.
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## Multiple Exponent Properties Involving Products - Example 1
by CK-12 //basic
Simplifying Expressions Using Multiple Exponent Properties Involving Products
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## Multiple Exponent Properties Involving Products - Example 2
by CK-12 //basic
Solving Word Problems Using Exponent Properties Involving Products
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## Product of Powers: An Explanation of the Concept
This video provides an explanation of the concept of product of powers.
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## Exponential Properties Involving Products Quiz
Quiz for Exponential Properties Involving Products.
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• Critical Thinking
## Exponential Properties Involving Products Discussion Questions
A list of student-submitted discussion questions for Exponential Properties Involving Products.
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• Real World Application
## Bats
Using exponents to figure out the wingspan of bats.
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• Study Guide
## Properties of Exponents Study Guide
This study guide reviews properties of exponents and the exponential form: product rule, quotient rule, power rule for exponents, power rule for quotients. It also looks at how to evaluate zero, negative, and fractional exponents.
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• Study Guide
## Exponential Properties Involving Products
by Kevin Cui //
Learn to solve multiplication problems involving different kinds of exponents.
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## Powers, Roots and Indices (Exponents)
by Kyle Bardman //basic
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• Flashcards
## Exponential Properties Involving Products Flashcards
by CK-12 //basic
These flashcards help you study important terms and vocabulary from Exponential Properties Involving Products.
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• Flashcards
## Recognize and Apply the Power of a Product Property
by Kyle Bardman //basic
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Applications of trigonometry
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1 Applications of trigonometry This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms and conditions of this license allow for free copying, distribution, and/or modification of all licensed works by the general public. 1
2 Table of Sine, Cosine and Tangent functions ANGLE Sine Cosine Tangent ANGLE Sine Cosine Tangent
3 Question 1 Questions The Pythagorean Theorem is a mathematical relationship between the lengths of the sides of a right triangle. Sometimes this relationship is represented graphically: Explain how this geometrical illustration relates to the Pythagorean Theorem (c 2 = a 2 + b 2 ). file i02676 Question 2 Geometry may be used to elegantly prove the Pythagorean Theorem. Shown here are two squares of equal area, each with four right triangles embedded within, occupying area of their own: a b a b b a c c a Area = b b Area = c 2 b 2 a c b b Area = c a a a 2 b a a b b a Explain how these two figures prove the Pythagorean Theorem, where a 2 + b 2 = c 2. file i
4 Question 3 Use the Pythagorean Theorem to determine the unknown lengths of the following (right) triangles:? 10 6? ? 6 file i
5 Question 4 Examine this right triangle (a triangle with one 90 o angle), and identify which side is the opposite, the adjacent, and the hypotenuse as viewed from two different angles (a and b): b X Y a 90 o Z As viewed from angle a: Opposite side = Adjacent side = Hypotenuse side = As viewed from angle b: Opposite side = Adjacent side = Hypotenuse side = file i
6 Question 5 Use trigonometric functions (sine, cosine, tangent) to determine all unknown lengths and angles in these right triangles: Y X 4 a b d e c 8 4 Z f A 37 o 7.5 g B 5 D h 59 o C o E j F file i02067 Question 6 Calculate the angles θ and Φ in this right triangle: Φ x 14 θ 14 file i
7 Question 7 Suppose you were laying out the forms for a house foundation, and had to ensure the corners were square (exactly 90 o ): All corner angles should be 90 o How could you ensure this, using trigonometry? How would you ensure this if you had no calculator with you to calculate trig functions, and no protractor or framing square with you to measure angles? Hint: remember that the lengths happen to form a right triangle! file i02678 Question 8 A model rocket enthusiast wants to approximately measure the peak altitude of her rocket by measuring the angle from horizontal (θ) sighting the rocket at its apogee and the distance from the launch pad to the point of observation. Which trigonometric function will she need to use in order to calculate the rocket s height? Rocket θ Launch pad (A) Tangent (B) Cosine (C) Cosecant (D) Sine (E) Secant file i
8 Question 9 We may analyze the horizontal and vertical components of an angled force by sketching a right triangle and calculating the lengths of that triangle s sides. The angled force vector length will be the magnitude of the applied force, while the opposite and adjacent side lengths will represent the horizontal and vertical components of that force. In this example, we see a person pushing a lawnmower. The handle of this lawnmower transmits a force of 20 pounds from the person s body to the lanwmower frame, at an angle of 32 degrees from horizontal: F = 20 lb force F x =? F y =? 32 o Lawnmower Use trigonometry to calculate how much of this angled force translates into forward (horizontal) force to drive the lawnmower forward against the friction of the grass, and how much of the applied force translates to downward (vertical) force pressing the lawnmower wheels against the ground. F forward = F x = F down = F y = lbs lbs If the person doing the lawnmowing wishes to apply their force more efficiently to the mower (i.e. more of their force going into the forward component and less pushing downward), should they hold the lawnmower handle at a shallower angle or at a steeper angle? Suggestions for Socratic discussion A technique highly recommended for word-problems is to sketch a picture of the problem and label elements of that picture with the given information. Do this, and compare your sketch with those of your classmates. How, specifically, does this aid your problem-solving? file i
9 Question 10 A truck with a winch on the front is pulling a car up a steep hill. The tension (pulling force) in the cable is 1000 pounds. How many pounds is the horizontal component of this force that the truck has to resist with its brakes? How many pounds is the vertical component of this force that the truck has to resist with its suspension (springs)? Assume that the angle of the hill is 35 o from level: F = 1000 lb Force on suspension Force on brakes 35 o F brakes = lbs F suspension = lbs Suggestions for Socratic discussion A technique highly recommended for word-problems is to sketch a picture of the problem and label elements of that picture with the given information. Do this, and compare your sketch with those of your classmates. How, specifically, does this aid your problem-solving? file i
10 Question 11 In the act of turning a wind turbine s rotor, wind also exerts a horizontal force on the supporting tower. To stabilize the wind turbine tower, guy wires are attached near the top of the tower to anchor points on the ground: Wind Wind turbine 800 N of horizontal force created by cable tension Concrete anchor "Guy" wire 40 o cable tension =??? 30 m Suppose that a wind is exerting 800 Newtons of horizontal force on the tower, measured at the point of connection between the guy wire and the tower. This connection point is 30 meters from ground level, and the angle of the guy wire to the ground is 40 o. Assuming that the single guy wire is bearing the full load of this wind force, how many Newtons of tension will this guy wire experience? F tension = lbs Supposing we are interested in minimizing the amount of tension the guy wire must bear, should we re-locate the concrete anchor closer to the base of the wind turbine tower or farther away from the base of the wind turbine tower? file i
11 Question 12 A house has a 3:12 roof slope. Supposing that one of the triangular truss structures in the roof bears 500 pounds of snow load (equivalent to 500 pounds of force applied to the peak of the triangle), how much tensile (pulling) force will there be in the horizontal beam? 500 lb 12 ft 3 ft tensile force??? House F beam = lbs file i02577 Question 13 A 150 pound weight is suspended by a hinged strut, which is kept at an angle of 45 o by the tension in a cable. Determine the compressive force within the strut as well as the tension within the cable. Assume that the cable is perfectly horizontal and that the strut is weightless: cable strut 45 o weight (150 lb) Hinge F strut = F cable = lbs lbs What will happen to the forces on the strut and on the cable as the strut is angled closer to vertical? file i
12 Question 14 A 200 newton weight is suspended by a hinged strut, which is kept at an angle of 30 o by the tension in a cable. Determine the compressive force within the strut as well as the tension within the cable. Assume that the cable is perfectly horizontal and that the strut is weightless: cable strut 30 o Weight (200 N) Hinge F strut = F cable = N N file i
13 Question 15 If the rear wheel of this bicycle supports 70% of the rider s weight, determine the compressive force within the seat stays (the struts leading from the seat of the bicycle to the center of the rear wheel, as opposed to the chain stays connecting the crank bracket to the wheel center) if the rider weighs 190 pounds. Give your answer in the form of a total amount of compressive force for the two seat stays (there are two of them, one on each side of the rear wheel). Assume that the chain stays are level with the ground: seat stays 65 o wheel chain stays 70% of rider s weight F seatstay = lbs file i
14 Question 16 A 15 pound weight hangs from the ceiling, suspended from two equal-length strings. The angle of each string is 50 o from horizontal: ceiling string string 50 o 50 o weight (15 lb) Calculate the amount of tension in each string. Hint: because the strings are equal length and have equal angles with respect to horizontal, the tension in each string will likewise be equal. F tension = lbs Suggestions for Socratic discussion Perform a thought experiment where you imagine the angles in this problem approaching 90 o (i.e. vertical strings). What happens to the tension within each string as they become more vertical? file i
15 Question 17 A 50 pound weight hangs by two strings, string A pulling at an angle and string B pulling horizontally. The angle of string A is 60 o from level (i.e. 30 o from vertical). Calculate the tension within each string: string A 60 o string B weight (50 lb) F tensiona = F tensionb = lbs lbs file i
16 Question 18 A process called delayed coking is used in the oil refining industry to convert heavy oils and tars into higher-valued petroleum products. A process vessel called a coke drum has a removable lid held down by a series of bolts, and alternatively by a hydraulic ram. When it comes time to open up the coke drum, the hydraulic ram is pressurized to maintain adequate force on the coke drum lid, the bolts are removed, and then the ram s fluid pressure is reduced until the lid springs open from the force of the gas pressure inside the coke drum: Hydraulic hose Ram Lid 38 o Hinge Coker deck Top of coke drum (contains 5 PSI gas) Calculate the hydraulic pressure necessary to hold down the lid on the coke drum when the gas pressure inside the drum is 5 PSI and all hold-down bolts have been removed from the lid. Assume a lid diameter of 30 inches, and a ram piston diameter of 4 inches. Hint: sketch a right triangle, representing forces as side lengths on the triangle the ram s diagonal force will translate into both a horizontal force on the lid (which you may ignore) and a vertical force on the lid (which is what we re interested in here). Hydraulic P = PSI Suggestions for Socratic discussion Which direction will the horizontal force component be exerted on the lid? Identify the potential hazards of a hydraulic oil leak in this system. Compare the effects of a slow leak (e.g. a leaky fitting connecting the hose to the ram) versus a catastrophic leak (e.g. the hose bursting from excess pressure). file i
17 Answer 1 Answers This illustration represents the Pythagorean Theorem in geometric form: the squares of the lengths of the opposite and adjacent sides, when added, equal the square of the length of the hypotenuse. 100 squares 36 squares 64 squares The side lengths of this triangle are 6, 8, and 10. Their corresponding areas are 36, 64, and 100. Answer 2 In both squares, the total area is equal because the squares have identical dimensions. In both squares, the area occupied by the four triangles is also equal, because the triangles all share the same dimensions. Therefore, the total shaded area in each square must also be equal. As you can see, the left square s shaded area is equal to c 2, while the right square s shaded area is equal to the sum of a 2 + b 2, thereby proving the Pythagorean Theorem. 17
19 Answer 4 As viewed from angle a: Opposite side = Y Adjacent side = Z Hypotenuse side = X As viewed from angle b: Opposite side = Z Adjacent side = Y Hypotenuse side = X Imagine the triangle being a room, viewed from above, with you standing at the location of the angle in question. The hypotenuse is always the longest of the three sides in a right triangle, regardless of which angle is being considered. The adjacent side is the non-hypotenuse side that you can reach from where you stand (i.e. the side that is adjacent to or next to the angle). The opposite side is the non-hypotenuse side that you can see on the other side of the room from your location (i.e. the side that is opposite the angle). As viewed from angle "a" As viewed from angle "b" Hypotenuse (Next to the corner) b Adjacent Hypotenuse Opposite (On the opposite side of the room) 90 o Adjacent a (Next to the corner) 90 o Opposite (On the opposite side of the room) 19
20 Answer 5 a = 45 o c = o e = o b = 45 o d = o f = o X = Y = Z = Y X 4 a b d e c 8 4 Z f A 37 o 7.5 g B 5 D h 59 o C o E j F g = 53 o h = 31 o j = 59 o A = C = E = B = D = F = Answer 6 θ = Φ = 45 o 20
21 Answer 7 Using two adjacent sides of the foundation form, measure the diagonal and calculate whether its length is equal to the length of the hypotenuse for a right triangle. To do this with no calculator and with no angle-measuring tools, you could mark lengths along the two sides of the form equal to multiples of 3 and 4, then measure a diagonal length between those end points to see if it was equal to the same multiple of 5: Multiple of 4 Multiple of 5? Multiple of 3 Answer 8 The tangent function is the proper one to use, since the rocket s height is the side opposite the angle and the distance between the enthusiast and the launch pad is the side adjacent to the angle. tan θ = Opposite side length Adjacent side length In this particular example, the rocket s height (h) may be calculated from the angle (θ), the launch-pad distance (d) and the person s height (x) as such: h = x + d tan θ Answer 9 F forward = F x = (20 lb) (cos 32 o ) = lbs F down = F y = (20 lb) (sin 32 o ) = lbs Maximum pushing efficiency will be realized when the person shrinks to the height of the lanwmower deck and is pushing 100% horizontally (0 o angle from the ground). Answer 10 The truck s brakes must resist pounds of horizontal force, to keep it from rolling in the direction of the slope. The truck s suspension must bear pounds of vertical force (in addition to the truck s own front-end weight). 21
22 Answer 11 F tension = N The concrete guy wire anchor should be relocated farther away from the tower in order to minimize wire tension. A farther distance will decrease the angle to ground. As this angle approaches zero degrees (level with the ground), the guy wire tension will approach the minimum value of 800 Newtons. Answer 12 It might be easier to approach this problem if we consider one-half of the truss at a time. The 500 pounds snow load will be evenly borne by each side of the truss, such that the tension in each half of the horizontal beam handles half of the force necessary to create the 500 pounds of upward force resisting the snow. Our pitch ratio of 3:12 (or 1:4) tells us the horizontal beam tension will be four times the snow load. If each half of the truss bears 250 pounds of snow, then each half of the horizontal beam will experience 1000 pounds of tension. The two beam halves tensions translate into 250 pounds of vertical force (each), making 500 pounds of vertical force. The 1000 pounds of tension in each half of the beam do not add up to 2000 pounds because the tension in each beam half is pointed in opposite directions in order for the horizontal forces to cancel (otherwise, the truss would be accelerating horizontally!): 250 lb 250 lb Forces are from the snow, acting on the truss assembly 12 ft 3 ft 3 ft 12 ft 1000 lb 1000 lb Answer 13 F strut = lbs F cable = 150 lbs As the strut angle becomes more vertical, the strut experiences less compressive force and the cable also experiences less tension. Answer 14 F strut = 400 N F cable = N Answer 15 Combined seat stay compression = lbs. Ideally, this will equate to pounds of compressive force in each of the two seat stays. Answer 16 F tension = lbs (in each string) 22
23 Answer 17 In this system of forces, string A exerts both a vertical and a horizontal force at the junction point of the three strings, just above the weight. String B only exerts a horizontal force at that point: A x A A y 60 o B (no B y ) B x weight (50 lb) Because string A is the only one with a vertical force component, it must be doing all the work in suspending the 50 pound weight. String B merely pulls laterally, and does not contribute to the effort of suspending the weight. Therefore, if since we know A y is the only force component working to oppose the 50 pound vertical force of the weight, we can solve for A using trigonometry: A y = 50 lb A 60 o (50 lb) A x We have a right triangle with an opposite side length of 50 and an angle of 60 o. To solve for the length of the hypotenuse: F tensiona = (50 lb)/(sin 60 o ) = lbs So, string A must have a tension of pounds. That this figure is greater than 50 makes sense, because the angle of string A with respect to vertical makes it less efficient at opposing the force of the weight. If the string were perfectly vertical, its tension would only have to be 50 pounds to suspend the weight, but since it is not perfectly vertical its tension must be more than that. Being angled, string A also exerts a horizontal force. Since nothing in this system is moving, it must be in a condition of translational equilibrium, so the horizontal component of tension A (known as A x ) must be opposed by another force. This other force is the tension of string B. We may calculate A x (and, consequently, B as well), by using trigonometry again: F tensionb = A x = ( lb)(cos 60 o ) = lbs 23
24 Answer 18 With a piston diameter of 4 inches, a hydraulic pressure of PSI is necessary to generate pounds. This is a minimum pressure, for safety reasons. More than PSI won t do any harm, but less than this amount will fail to hold down the lid! 24
Unit 2: Right Triangle Trigonometry RIGHT TRIANGLE RELATIONSHIPS
Unit 2: Right Triangle Trigonometry This unit investigates the properties of right triangles. The trigonometric ratios sine, cosine, and tangent along with the Pythagorean Theorem are used to solve right
Week 11, Lesson 1 1. Warm Up 2. Notes Sine, Cosine, Tangent 3. ICA Triangles
Week 11, Lesson 1 1. Warm Up 2. Notes Sine, Cosine, Tangent 3. ICA Triangles HOW CAN WE FIND THE SIDE LENGTHS OF RIGHT TRIANGLES? Essential Question Essential Question Essential Question Essential Question
Right-angled triangles and trigonometry
Right-angled triangles and trigonometry 5 syllabusref Strand: Applied geometry eferenceence Core topic: Elements of applied geometry In this cha 5A 5B 5C 5D 5E 5F chapter Pythagoras theorem Shadow sticks
Name Date PD. Pythagorean Theorem
Name Date PD Pythagorean Theorem Vocabulary: Hypotenuse the side across from the right angle, it will be the longest side Legs are the sides adjacent to the right angle His theorem states: a b c In any
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Load and Motor Control Contents SERIES CAVITY DESCRIPTION FLOW PRESSURE PAGE NO. LPM/GPM BAR/PSI STANDARD PILOT ASSISTED CB101... C10-3... Load Control Cartridge Valve...45/12... 380/5500... 5-6 MHC-010-S***
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https://em.emedu.org.tw/elec_sim/5-2.html
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Module 5.2 Bouncing Signal in T. Line w/ Impulse Source
Module 5.2 Bouncing Signal in T. Line w/ Impulse Source
Formula:
1. $\Gamma_S$: 訊號源端的反射係數
$\Gamma_S = \frac{Z_S-Z_0}{Z_S+Z_0}$
2. $\Gamma_L$ : 負載端的反射係數
$\Gamma_L = \frac{Z_L-Z_0}{Z_L+Z_0}$
3. $h(t)$ : heavyside function
$\left\{ \begin{array}{rcl} h(t) = 0 \mathrm{,\quad for}\hspace{2mm} t < 0 \\ h(t) = 1 \mathrm{, \quad for}\hspace{2mm} t\geq 0 \end{array}\right.$
4. $V(z,t)$ :
$= \sum\limits_{n=0}^{\infty}{V_+ \left( \Gamma_S \Gamma_L \right)^n \{[h(-2nL+v_pt-z) - h(-2nL+v_pt-W-z)] \\ \quad + \Gamma_L \times [h(v_pt+z-(2n+2)L)-h(v_pt+z-(2n+2)L-W)]\}}$
Parameters:
1. $V_0$ : Amplitude of DC voltage (V)
2. $Z_S$ : Source impedance (Ω)
3. $Z_L$ : Load impendance (Ω)
4. $Z_0$ : Characteristic impendance of the transmission line (Ω)
5. $L$ : Length of the transmission line (m)
6. $v_p$ : The speed of electromagnetic wave in transmission line (m/s)
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# Is the metric topology determined by its convergent sequences?
I am aware that in a first countable space (and thus any metric space) is completely determined by its convergent sequences and their limits, i.e.,
If $$\tau_1$$ and $$\tau_2$$ are two first countable topologies on a set $$X$$ such that $$x_i\to c$$ in $$\tau_1$$ iff $$x_i\to c$$ in $$\tau_2$$, then $$\tau_1 = \tau_2$$.
However, this raises the following question: If two metrics on a space have the same convergent sequences, will they have the same limits as well (and thus the same topology)?
• maybe more accurate to say "two metrics on a space" rather than "two metric spaces", since problem doesn't make much sense if the underlying set isn't the same.
– M W
Feb 18 at 5:29
• @MW agreed and edited.
– Atom
Feb 18 at 8:41
• math.stackexchange.com/q/4554239/688539 Feb 18 at 11:06
I believe this is true -- we can recover what the sequences converge to.
Say $$(a_n)$$ is a sequence in $$X$$ that we know converges, but we don't know what it converges to. There will be a unique $$x \in X$$ so that the new sequence $$(a_1, x, a_2, x, a_3, x, a_4, x, \ldots)$$ is convergent (which we can detect), and this $$x$$ is necessarily the limit of the sequence $$(a_n)$$.
Of course, this process is not "computable" in any sense. I have no idea how one would find such an $$x$$ in practice (though maybe it's doable in case $$X$$ is compact...) but, in the abstract, this shows that the convergent sequences remember limits.
I hope this helps ^_^
• Maybe I am missing something, but is this an answer to OP's question? There are two metric spaces, for which say $a_n\to x_1$ (in $X_1$) and $a_n\to x_2$ (in $X_2$). Feb 18 at 3:15
• @WishYouTheBest As the answer noted, the $x$ in the answer is unique. That is to say, if $x_1 \neq x_2$, then $(a_1, x_1, a_2, x_1, a_3, \cdots)$ is convergent in $X_1$ but not in $X_2$, so the two spaces do not share convergent sequences in that case. Feb 18 at 3:46
• @DavidGao Thanks, I couldn't figure out your point! Feb 19 at 11:21
This is just to expand a bit on HallaSurvivor's answer.
It turns out that limits of convergent sequences can be encoded in sequences themselves: In any $$T_1$$ space, $$a_i\to x$$ iff the sequence $$a_1, x, a_2, x, a_3, x, a_4, \ldots$$ converges. Hence the topology of a first countable $$T_1$$ space is completely determined by its convergent sequences.
• Worth noting that $T_1$ is absolutely necessary here, since for $X=\{0,1\}$ the distinct topologies $\tau_i=\{\emptyset,\{i\},X\}$, for $i=0,1$, each have the property that every sequence converges (in $\tau_0$ they all converge to $1$, and in $\tau_1$ they all converge to $0$). As a sillier example, the trivial topology has this property as well.
– M W
Feb 18 at 4:57
• @MW I was just going to add that! :)
– Atom
Feb 18 at 5:05
• Here is an example to see that first countable is also necessary: the weak and norm topology on $\ell^1$ have the same convergent sequences (with the same limits) but are very different topologies Feb 18 at 8:45
• @AlessandroCodenotti A more accessible example would be cocountable and discrete topologies on an uncountable set. :)
– Atom
Feb 18 at 8:49
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# geometry
posted by .
E And C are mid points od line AD and line DB; line AD is congruent to line DB and angle A is congruent to angle 1. What do you do to prove that ABCE is a isosceles trapezod ??
• geometry -
The non-parallel sides of an isosceles
trapezoid are equal.
The corresponding angles formed by these sides are also equal.
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0
# How much is 100 minuts in a day?
Updated: 9/24/2023
Wiki User
9y ago
Best Answer
1 hr 40 mins
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9y ago
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6 hours 35 minutes
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First we convert 40% to a decimal. we get 40/100 or 0.40 then we multiply it by 60 0.4*60 = 24 so 40% of 60 minuts is 24 minuts
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There are: 36*60 = 2160 minutes
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Approx 253360 minutes
1 1/2 hours.
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77
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he does waste time because he day dreams and he day dreams for about 30 minuts to an hour
185 minuts
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hHow many minuts long was the you have a dream speach?
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100 dollars a day
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Shake the machine, but there's not much else you can do.
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## Exercises
1. Define a semi-algebraic model that removes a triangular nose'' from the region shown in Figure 3.4.
2. For distinct values of yaw, pitch, and roll, it is possible to generate the same rotation. In other words, for some cases in which at least , , or . Characterize the sets of angles for which this occurs.
3. Using rotation matrices, prove that 2D rotation is commutative but 3D rotation is not.
4. An alternative to the yaw-pitch-roll formulation from Section 3.2.3 is considered here. Consider the following Euler angle representation of rotation (there are many other variants). The first rotation is , which is just with replaced by . The next two rotations are identical to the yaw-pitch-roll formulation: is applied, followed by . This yields .
1. Determine the matrix .
2. Show that .
3. Suppose that a rotation matrix is given as shown in (3.43). Show that the Euler angles are
(3.85)
(3.86)
and
(3.87)
5. There are 12 different variants of yaw-pitch-roll (or Euler angles), depending on which axes are used and the order of these axes. Determine all of the possibilities, using only notation such as for each one. Give brief arguments that support why or why not specific combinations of rotations are included in your list of 12.
6. Let be a unit disc, centered at the origin, and . Assume that is represented by a single, algebraic primitive, . Show that the transformed primitive is unchanged after any rotation is applied.
7. Consider the articulated chain of bodies shown in Figure 3.29. There are three identical rectangular bars in the plane, called . Each bar has width 2 and length 12. The distance between the two points of attachment is 10. The first bar, , is attached to the origin. The second bar, , is attached to , and is attached to . Each bar is allowed to rotate about its point of attachment. The configuration of the chain can be expressed with three angles, . The first angle, , represents the angle between the segment drawn between the two points of attachment of and the -axis. The second angle, , represents the angle between and ( when they are parallel). The third angle, , represents the angle between and . Suppose the configuration is .
1. Use the homogeneous transformation matrices to determine the locations of points , , and .
2. Characterize the set of all configurations for which the final point of attachment (near the end of ) is at (you should be able to figure this out without using the matrices).
8. A three-link chain of bodies that moves in a 2D world is shown Figure 3.30. The first link, , is attached at but can rotate. Each remaining link is attached to another link with a revolute joint. The second link, , is a rigid ring, and the other two links are rectangular bars.
Assume that the structure is shown in the zero configuration. Suppose that the linkage is moved to the configuration , in which is the angle of , is the angle of with respect to , and is the angle of with respect to . Using homogeneous transformation matrices, compute the position of the point at in Figure 3.30, when the linkage is at configuration (the point is attached to ).
9. Approximate a spherical joint as a chain of three short, perpendicular links that are attached by revolute joints and give the sequence of transformation matrices. Show that as the link lengths approach zero, the resulting sequence of transformation matrices converges to exactly representing the freedom of a spherical joint. Compare this approach to directly using a full rotation matrix, (3.42), to represent the joint in the homogeneous transformation matrix.
10. Figure 3.12 showed six different ways in which 2D surfaces can slide with respect to each other to produce a joint.
1. Suppose that two bodies contact each other along a one-dimensional curve. Characterize as many different kinds of joints'' as possible, and indicate the degrees of freedom of each.
2. Suppose that the two bodies contact each other at a point. Indicate the types of rolling and sliding that are possible, and their corresponding degrees of freedom.
11. Suppose that two bodies form a screw joint in which the axis of the central axis of the screw aligns with the -axis of the first body. Determine an appropriate homogeneous transformation matrix to use in place of the DH matrix. Define the matrix with the screw radius, , and displacement-per-revolution, , as parameters.
12. Recall Example 3.6. How should the transformations be modified so that the links are in the positions shown in Figure 3.25 at the zero configuration ( for every revolute joint whose angle can be independently chosen)?
13. Generalize the shearing transformation of (3.84) to enable shearing of 3D models.
Implementations
14. Develop and implement a kinematic model for 2D linkages. Enable the user to display the arrangement of links in the plane.
15. Implement the kinematics of molecules that do not have loops and show them graphically as a ball and stick'' model. The user should be able to input the atomic radii, bond connections, bond lengths, and rotation ranges for each bond.
16. Design and implement a software system in which the user can interactively attach various links to make linkages that resemble those possible from using Tinkertoys (or another popular construction set that allows pieces to move). There are several rods of various lengths, which fit into holes in the center and around the edge of several coin-shaped pieces. Assume that all joints are revolute. The user should be allowed to change parameters and see the resulting positions of all of the links.
17. Construct a model of the human body as a tree of links in a 3D world. For simplicity, the geometric model may be limited to spheres and cylinders. Design and implement a system that displays the virtual human and allows the user to click on joints of the body to enable them to rotate.
18. Develop a simulator with 3D graphics for the Puma 560 model shown in Figure 3.4.
Steven M LaValle 2012-04-20
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Discussion and 2 replies | Statistics homework help
Case Problem Par, Inc.
Par, Inc., is a major manufacturer of golf equipment. Management believes that Par’s market share could be increased with the introduction of a cut-resistant, longer-lasting golf ball. Therefore, the research group at Par has been investigating a new golf ball coating designed to resist cuts and provide a more durable ball. The tests with the coating have been promising.
One of the researchers voiced concern about the effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances comparable to those of the current-model golf ball. To compare the driving distances for the two balls, 40 balls of both the new and current models were subjected to distance tests. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models. Theresults of the tests, with distances measured to the nearest yard, follow. These data are available in the file Golf.
Do:
In a managerial report, use the methods of hypothesis testing to
• Provide descriptive statistical summaries of the data for each model, written in complete sentences.
• Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
• Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?
• Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
Discuss
Post by classmate 1
Hello everyone,
Here are my findings and managerial report after using the methods of hypothesis testing:
After reviewing the data for both current and new golf ball driving distance, a t-Test was performed to determine if the current golf balls travel a longer distance than the new ones in a sample size of 40. Assuming that the data is normally distributed, the t-Test displayed a sample mean for the current golf ball of 270.28 and a sample mean for the new golf ball of 267.5. The difference between the means is not significant. The variance for the current golf ball is 76.61, and the variance for the new golf ball is 97.95. The p-value is 0.094. The standard deviation for the current golf ball is 8.75, and the standard deviation for the new golf ball is 9.9. The “t Critical one-tail” is 1.66, and the “t Critical two-tail” is 1.99.
The hypothesis test was needed to determine if there is enough evidence to support that the current golf balls travel a greater distance than the new ones proposed for Par, Inc (null hypothesis) or if, in fact, they are the same (alternative hypothesis).
The p-value is a random variable derived from the distribution and is a measure of evidence against the null hypothesis (Hung et al., 1997). In this case, the p-value is 0.094, which is less than 0.05, indicates strong evidence against the null hypothesis as there is less than a 5% probability that the null would be correct; hence, the p-value is greater than the significance level, and one should fail to reject the null hypothesis (McLeod, 2019). There appears no evidence to support that the current golf balls travel a more significant distance than the new ones. Given the existing data, there is no evidence to support that the alternative hypothesis is correct, as one would need more data.
Based on my hypothesis testing conclusion, my recommendations for Par, Inc would be to utilize a more extensive-sized data sampling for current and new golf balls. An increased sample size would decrease the standard error of associated sampling distributions (Anderson et al., 2021).
References
Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning
H. M. James Hung, O’Neill, R. T., Bauer, P., & Kohne, K. (1997). The Behavior of the P-Value When the Alternative Hypothesis is True. Biometrics53(1), 11–22. https://doi.org/10.2307/2533093 (Links to an external site.)
McLeod, S. (2019). What a p-value tells you about statistical significance. P. https://www.simplypsychology.org/p-value.html
Comment by instructor
To write a report on hypothesis test, the very first step is to clearly state the null and alternative hypothesis. Can you add that part in?
Post by classmate 2
Wk2 Discussion
• Provide descriptive statistical summaries of the data for each model, written in complete sentences.
• From the data that was given we compared two different types of golf balls to see if there is a difference in the current ones being used and the new durable cut resistant ones. The data compares 40 different golf ball drives with each test being for each of the golf balls. When comparing the statistical data given, The current golf ball had an average distance of 270.28 versus the new golf balls having an average distance of 267.48 which is a drop of 1.04% for averages. For highest drive distance per choice was the same at 289, while the low had a difference of 255 for the old ones versus 250 for the newer gold balls, this also had a decrease of 1.96%. The median between the two golf balls was similar to the average at 270 for the current and 265 for the newer ones. On average the change between the current golf balls to the new, more durable golf balls was each of the drives lost an average of 2.8. The p-value is 0.094, T-Critical one-tail is 1.66, and the T-Critical two-tail is 1.99.
• Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
• Null hypothesis: Current golf balls that Par uses travel a greater distance than the more durable, cut resistant golf balls.
• Alternate hypothesis: New golf balls travel the same distance as current ones.
• Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test?
• With the p-value (0.094) being more than the (0.05) 5% probability making the null hypothesis not proven and would lean more towards the alternate hypothesis. This means that the data is not showing a lower than 5% accuracy making the null hypothesis true.
• What is your recommendation for Par, Inc.? Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
• With a test of this size showing almost identical results with little differences, a large scale would give a more accurate account. Based on there really only having such a small change in distance, many variable could have been at play to skew these results. If all the hits were the same order, golfers fatigue, and wind speed being some of these variables. If the scale was out of 100 than the results would supply more accurate result since it would give them an out of 100% answer they would need to confirm this theory. They would have to test the on almost identical condition days with the golfer hitting the ball in one order the first time and opposite order the second time. This would eliminate a lot of the “what if” variables that affect the overall numbers on this (mythbusters logic).
• Reference:
• Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning
Comment by instructor
like how this post is broken into subsections. It’s very easy to read.
Your alternative hypothesis is incorrect. Can you fix it?
Post by classmate 3
rovide descriptive statistical summaries of the data for each model, written in complete sentences.
The sample mean for the current golf ball is 270.28. The sample mean for the new golf ball is 267.5. On average the current golf ball had a 2.775-yard advantage. You get this number by subtracting the distance between the two means.
The value of the test statistic is 1.3284.
Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
Par is looking into designing a new golf ball that will be cut resistant and offer driving distances comparable to those of the current model golf balls. To compare the driving distance of the two balls, 40 balls of both new and current models are subject to distance tests. A one tailed hypothesis test is to be conducted to see whether that mean driving distance for the current golf balls is shorter than the new gold ball.
Letting:
M1= Population mean driving distance for the current golf ball.
M2= Population mean driving distance for the new golf ball.
Hypothesis test:
H0: M1-M2 (<=) 0
H2= M1-M2 (>=) 0
Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?
The P-value of the test is 0.0940. If the P-value is less than or equal to the level of significance 0.05, the null hypothesis can be rejected. Therefore, in this test the null hypothesis cannot be rejected.
Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
With more data, par should have a good idea of the difference between the means of the two golf balls. Continued study for this case should be very easy using a mechanical hitting machine that can hit several hundred golf balls without much trouble and would help to come a good data-based conclusion.
The data does not provide statistical evidence that the new golf ball has a lower mean driving distance than the current golf ball. Failing to reject H0 means the research finding is inconclusive. The data did not show the new golf ball with a significant lower mean driving distance, which means the researcher should not be ready to conclude the mean distance for the new golf ball is equal to or better than the current golf ball. A potential for a type 2 error exists with this conclusion.
Post by classmate 4
Provide descriptive statistical summaries of the data for each model, written in complete sentences.
The descriptive statistical summaries from the data for each model using the T Test: Two-Sample Assuming Unequal Variances lead me to the current values; Current Mean comes out to 270.28, Current Variance is 76.61, Current Observation is 40, DF is 76, T Stat is 1.33, P(T<=t) one 0.09, T Critical one 1.66, P(T<=t) two 0.19 and lastly T Critical two being at 1.99. New Mean is 267.5, New Variance is 97.95 and the New Observation is also at a 40.
Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
It is not always obvious how the null and alternative hypotheses should be formulated. Care must be taken to structure the hypotheses appropriately so that the hypothesis testing conclusion provides the information the researcher or decision maker wants.” (Ch. 9.1)
Ho: u < 270 Ha: u > 270
Ho: u < 76.61 Ha: u > 76.61
From the current to the new mean the value between is 2.78 in favor of the current mean and the current variance versus the new variance is -21.34 in favor of the new variance.
Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?
The p- value is listed at 0.09 and is currently 0.05% in value which in turn means the null hypothesis cannot be rejected.
Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
With a larger sample size of testing more golf balls, there will be a larger margain of error as well.
References:
Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning
Post by classmate 5
Provide descriptive statistical summaries of the data for each model, written in complete sentences.
Par, Inc. wants to know if their current golf balls travel a greater distance than their new golf balls. A sample size of 40 golf balls from both the current and new golf balls. After running a t-test these are the figures I got. The mean for the current golf balls is 270.275 and for the new golf balls it is 267.5. The variance of the current golf balls is 76.61 and the variance for the new ones is 97.94. The t critical one tail is 1.66, while the t critical two tail is at 1.99.
Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
A hypothesis test will have to be done to determine which golf balls travel a greater distance.
The null hypothesis is that the current golf balls travel a greater distance than the new golf balls.
The alternative hypothesis is that the new golf balls travel a greater distance than the current ones
Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?
The p value for the one tailed test is 0.094. The p value is more than the level of significance of 0.05 which means that the null hypothesis is not rejected.
Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
I would suggest that the sample size does increase. This will provide more information about both golf balls and cut down on possible errors.
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Cornerstones of Financial Accounting (2nd Edition) View more editions Solutions for Chapter AP3
• 1264 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 1
Compound interest:
Compound interest refers to the method of calculating the time value of money. Under this method interest is calculated based on the previous balances of the principal and undistributed interest. That is interest for the current period is added to the previous balance to calculate the interest for the next period.
Calculate quarterly interest rate:
Annual interest rate is 12%
Calculate the future value at the end of the 5year:
Particulars Amount (in \$) Initial balance 6,000.00 1st quarter interest (6000 X 3%) 180.00 Balance at the end of 1st quarter 6,180.00 2nd quarter interest (6180 X 3%) 185.40 Balance at the end of 2nd quarter 6,365.40 3rd quarter interest (6365.40 X 3%) 190.96 Balance at the end of 3rd quarter 6,556.36 4th quarter interest (6556.36 X 3%) 196.69 Balance at the end of 4th quarter 6,753.05 Balance at the end of 1 year 5th quarter interest (6753.05 X 3%) 202.59 Balance at the end of 5th quarter 6,955.64 6th quarter interest (6955.64 X 3%) 208.67 Balance at the end of 6th quarter 7,164.31 7th quarter interest (7164.31 X 3%) 214.93 Balance at the end of 7th quarter 7,379.24 8th quarter interest (7379.24 X 3%) 221.38 Balance at the end of 8th quarter 7,600.62 Balance at the end of 2 year 9th quarter interest (7600.62 X 3%) 228.02 Balance at the end of 9th quarter 7,828.64 10th quarter interest (7828.64 X 3%) 234.86 Balance at the end of 10th quarter 8,063.50 11th quarter interest (8063.50 X 3%) 241.90 Balance at the end of 11th quarter 8,305.40 12th quarter interest (8305.4 X 3%) 249.16 Balance at the end of 12th quarter 8,554.57 Balance at the end of 3 year 13th quarter interest (8554.57 X 3%) 256.64 Balance at the end of 13th quarter 8,811.20 14th quarter interest (8811.20 X 3%) 264.34 Balance at the end of 14th quarter 9,075.54 15th quarter interest (9075.54 X 3%) 272.27 Balance at the end of 15th quarter 9,347.80 16th quarter interest (9347.80 X 3%) 280.43 Balance at the end of 16th quarter 9,628.24 Balance at the end of 4 year 17th quarter interest (9628.24 X 3%) 288.85 Balance at the end of 17th quarter 9,917.09 18th quarter interest (9917.09 X 3%) 297.51 Balance at the end of 18th quarter 10,214.60 19th quarter interest (10214.60 X 3%) 306.44 Balance at the end of 19th quarter 10,521.04 20th quarter interest (10521.04 X 3%) 315.63 Balance at the end of 20th quarter 10,836.67 Balance at the end of 5 year
Thus, the balance at the end of 5 year is \$ 10,836.67
Corresponding Textbook
Cornerstones of Financial Accounting | 2nd Edition
9780538473453ISBN-13: 0538473452ISBN: Authors:
Alternate ISBN: 9781133012535, 9781133012559, 9781133292029, 9781133419549, 9781467214070
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The sum of the measures of the interior angles of a triangle equals 180º.
Examples:
1 Find m∠B and m∠C.
Solution: mA + mB + mC = 180
38 + x + (x + 2) = 180
40 + 2x = 180
2x = 140
x = 70 = m∠B
x + 2 = 72 = m∠C
2. The angles in a triangle are represented by (4x - 6)º, (2x + 1)º and (x + 3)º. Is this a right triangle?
Solution:
(4x - 6) + (2x + 1) + (x + 3) = 180
7x - 2 = 180
7x = 182
x = 26
(4x - 6)º = 98º
(2x + 1)º = 53º
(x + 3)º = 29º
No. The triangle is obtuse.
3 m∠ABC=m∠BCD Find m∠ACD.
Solution: mBCD = 56º.
In ΔABC, 85º + 56º + mBCA = 180
mBCA = 39º
m
ACD = 56º - 39º = 17º
4. The angles in a triangle are in the ratio of 1 : 2 : 3. Find the measure of the angles in the triangle.
Solution:
x + 2x + 3x = 180
6x = 180
x = 30
The angles are 30º, 60º, and 90º.
In your previous studies of this theorem, you most likely saw how a triangle's angles can be cut off and rearranged to form a straight angle of 180º (as shown below).
We now need to use more sophisticated ideas to establish that this theorem is actually true. Throughout history, several different methods of proof of this theorem have appeared.
Let's take a look at a few of these methods of proof.
Transformational Proofs:
Transformational Proof Using Translation
[Also written as m∠A + m∠ABC + m∠C = 180.]
Proof:
• Translate ΔABC so C' coincides with B forming a straight line from point C, through point B, to point B' (along vector ).
• Since translations are rigid transformations, we know that angle measure is preserved making m∠C = m∠A'BB'.
• A straight angle, ∠B'BC, whose measure by definition is 180º, exists giving
m∠A'BB' + m∠A'BA + m∠ABC
= 180.
• Angles ∠C and ∠A'BB' are congruent corresponding angles making . It can also be said that these sides are parallel because in a translation the corresponding segment sides of the pre-image and image are parallel.
• As a result, ∠A and ∠A'BA are congruent alternate interior angles. If 2 parallel lines are cut by a transversal, the alternate interior angles are congruent.
• By substitutions, m∠C + m∠A + m∠ABC = 180.
Transformational Proof Using Rotation
[Or m∠ABC + m∠ACB + m∠BAC = 180.]
Proof:
• Rotate ΔABC about the midpoint of .
• Then rotate ΔA'B'C' about the midpoint of .
• Since rotations are rigid transformations, angle measure is preserved and m∠ABC = m∠A'B'C' and m∠B'A'C' = m∠B''A''C''.
• These alternate interior angles will be congruent, making
If 2 lines are cut by a transversal and the alternate interior angles are congruent, the lines are parallel.
Since both and pass through point B and are parallel to , we know the segment from A to B'' is straight (there is only one line through B parallel to - Parallel Postulate).
• A straight angle is an angle whose rays form a straight line, making ∠ABB'' a straight angle with a measure of 180º. Now, m∠ABC + m∠A'BB' + m∠B''BA' = 180.
• Rotations preserve angle measure, making m∠A'BB' = m∠ACB and m∠BAC = m∠BA'C = m∠A'BB''.
• By substitution, m∠ABC + m∠ACB + m∠BAC = 180.
Proof Using An Auxiliary Parallel Line
Statements Reasons 1. ΔABC 1. Given 2. Draw auxiliary line through B parallel to . 2. Through a point not on a line, only one line may be drawn parallel to a given line. 3. ∠DBE is a straight angle. 3. A straight line forms a straight angle. 4. m∠DBE = 180º 4. A straight angle has a measure of 180º. 5. m∠1 + m∠2 + m∠3 = m∠DBE 5. Angle Addition Postulate (whole quantity) 6. 6. If 2 parallel lines are cut by a transversal, the alternate interior angles are congruent. 7. m∠1 = m∠A; m∠3 = m∠C 7. Congruent angles are angles of equal measure. 8. m∠A + m∠2 + m∠C = 180º 8. Substitution
Slightly different version:
Proof Using An Auxiliary Parallel Line with an Extension
Statements Reasons 1. ΔABC 1. Given 2. Extend through C to E, and draw an auxiliary line through C parallel to . 2. Through a point not on a line, only one line may be drawn parallel to a given line. 3. ∠ACE is a straight angle. 3. A straight line forms a straight angle. 4. m∠ACE = 180º 4. A straight angle has a measure of 180º. 5. m∠1 + m∠2 + m∠3 = m∠ACE 5. Angle Addition Postulate (whole quantity) 6. 6. If 2 parallel lines are cut by a transversal, the alternate interior angles are congruent. 7. 7. If 2 parallel lines are cut by a transversal, the corresponding angles are congruent. 7. m∠B = m∠2; m∠3 = m∠A 7. Congruent angles are angles of equal measure. 8. m∠1 + m∠B + m∠A = 180º 8. Substitution
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## Official Guide Explanation:Data Sufficiency #81
Background
This is just one of hundreds of free explanations I've created to the quantitative questions in The Official Guide for GMAT Review (12th ed.). Click the links on the question number, difficulty level, and categories to find explanations for other problems.
These are the same explanations that are featured in my "Guides to the Official Guide" PDF booklets. However, because of the limitations of HTML and cross-browser compatibility, some mathematical concepts, such as fractions and roots, do not display as clearly online.
Question: 81
Page: 279
Difficulty: 4 (Moderately Easy)
Category 1: Arithmetic > Descriptive Statistics > other
Explanation: Statement (1) tells us nothing: Between the list of numbers and k < n, there's no way to find the value of n.
Statement (2) is also insufficient. Given a list of 5 (or any odd number of) numbers, the median must be one of the numbers in the set. Since the median of this list is 10 and 10 isn't one of the numbers listed, either k = 10 or n = 10. If we put the numbers in order, it looks like this:
6,(k/n),(k/n),12,17
Either k or n is the median--whichever is larger.
Taken together, the statements are sufficient. Since (1) tells us that n is larger than k, the order of the numbers is:
6,k,n,12,17
The middle number is the median, so n = 10. Choice (C) is correct.
You should follow me on Twitter. While you're at it, take a moment to subscribe to GMAT Hacks via RSS or Email.
Total GMAT Math The comprehensive guide to the GMAT Quant section. It's "far and away the best study material available," including over 300 realistic practice questions and more than 500 exercises! Click to read more.
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# Module One Post Eight (Dance Related to Math)
http://mathrelationstodance.blogspot.ca/
This website talks about angles, musicality, and balance in dance. Dancers must know how high to lift their leg, this is determined in degrees. Their legs are lifted at 45 degrees, 90 degrees and higher. Dancers must also know how to transfer their weight in order to maintain balance. Dancers must also have the ability to memorize movements and counts, as well as on what counts to perform. This site also gives great explanations as to why they think these forms of math are useful to dancers.
This site was useful because not only did it have good information, it also gave great explanations as to why they think these forms of math are useful to dancers. This site also explained angles which was helpful since other sites didn’t give much information on it.
# Module Two Post Three (2 point perspective)
In this website I found, I am going to mostly talk about how two point perspective fits in with math. To state the obvious, this type of perspective is used a lot in geometry. As each shape is drawn, it will be either perpendicular, or parallel to each other. every line that is drawn will make its way to one of the vanishing points, creating angles as it goes.
You can think of the process of drawing two point perspective like a formula for math. It has multiple steps that has to go in a certain order to make the answer (drawing) correct.
These are the two drawings that are made in the website I have explained. As you can see, you can see there are multiple 3D prisms that are used to create the scenes.
2 Point Perspective Drawing: Step by Step Guide for Beginners
# Module Two Post Two (Going deeper into techniques of 2 point persepcetive)
Once again, I am not using a website, as I felt that I needed to write a new post to go deeper about the techniques I breifly mentioned in post sixteen. My art teacher, Ms.Lehtonen did a whole unit about perspective drawing. In that unit she taught us how to draw in that way and how the mathematical techniques that are used.
I already mentioned that you always need to use a ruler, but that is not all. There are only certain amount of lines that are used. these lines are vertical, and diagonal pointing to either of the vanishing points. These lines will create angles which will help you know how big the other corners will be. The angle made in one corner of ( a building ), will be the same as the all the corners that look like they are in that position.
This image greatly describes how each corner is the same as if a mirror is placed between each horizontal line, the other corner will be equal.
# Module One Post Fourteen (math in art)
This time I substituted the website for a video, and it clearly explains drawing perspective in math. You need geometry, and angles to make sure everything is in proportion.
In one point perspective, when you are drawing the object like a cube, the lines has to create a 90 degree angle. After, the lines that finish up the cube have to be parallel to each other. All of the lines have to be straight too. (unless you are drawing 0 point perspective.)
As you watch the video, you can see that first, the basic shape that defines the soon to be cube is 2d. When you form the cube, it turns into a 3d shape.
# Module 1 Post 15 (The Geometry of Ballet)
This last prezi https://prezi.com/28tsc8jiwg71/the-geometry-of-ballet/ didn’t have much information on rotation, translation, and reflection. Although it did talk about the angles. While performing an arabesque a dancer has to have her leg at a 90-135 degree angle. While performing a grand jete a dancer has to have her legs at 180 degrees so basically a straight line.
This was useful because normally when I view sites with the title geometry of ballet, I normally think it’ll be about geometry but this site was very different and had different examples. This will be useful because I know that angles are even involved in the studio.
# Module 1 Post 12 (Mathematics Of Ballet)
https://mathofballet.weebly.com/blog
This site was like many others I found. It was someone else’s blog but they did find really good information. There is a variety of different examples of why math is involved in ballet. For example spacial formations, angles, symmetry and counting the music. All of this is important to a ballerina. When a professional company performs they need apply all of these (especially spacial formation).
This was helpful because there was a lot of examples and I have multiple different forms of math to chose from. The one I found the most useful was the spacial formations because that is very important to a ballerina.
# Module One Post Fifteen (following through with a pass)
This website isn’t associated with lacrosse or math specifically, but has useful information.
Even though it’s not directly associated with any of these things it can still relate to my topic. Following through with a pass associates with almost all sports, meaning if you don’t follow through with a pass it may not go where you want it to. It relates to math because passes relate to angles.
https://www.teamsnap.com/community/skills-drills/football/football-passing/695-finishing-the-throw-follow-through
# Module One Post ten (Hockey math)
Cite used: http://spinnakers.org/mathonice.htm
In this cite it displays many key examples of how math is implied in the game of hockey. For instance, if you are shorthanded the fraction would be 5/4, meaning the opposing team has five and you have four. A great demonstration of fractions which is used practically every game. Another example is when angling someone to the outside of the boards. You want to cut them off from an angle so that they run out of space, this is a key aspect for defensemen like me. We are told and use this strategy on all times when the opposing team is rushing in. Lastly, the triangle, this shape is commonly used as a direction of offence, players (forwards) will shape up in this formation in the offensive zone creating good passing lanes. In all i found this extremely useful, as it is filled with a lot of information and examples. Definitely recommend to anyone!
# Module One Post Seven [Goal tending in Soccer]
I think goal tending is one of the most important and hard positions in soccer because you are guarding such a big net. I think that angles has a lot of play when it comes to goal tending. The goaltender has to follow the line of shot when a player is shooting on him. In my opinion the person who is shooting against the goalie is always looking for an angle but the goalie can narrow his or her’s shooting angle by creeping out of his box https://www.mathedpage.org/conics/soccer/
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## Welcome to H3 Maths
Blog Support for Growing Mathematicians
### Speed Mathematics – from the horrors of a Concentration Camp
August28
Can you multiply 5132437201 by 452736502785 in seventy seconds without using a calculator?
A Russian Jew, Jakow Trachtenberg, developed his system of “speed mathematics” while he was in Hitler’s concentration camps as a political prisoner during World War II. Read more about this amazing Mathematician here. You can download a copy of his work at this link.
by posted under Uncategorized | Comments Off on Speed Mathematics – from the horrors of a Concentration Camp
#### Post Support
10 x 9 x 8 + (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020
NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]
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# Search results
1. ### Solution to general IVP
Bump...any suggestions?
2. ### Solution to general IVP
Homework Statement Find all solutions of the IVP y'' + a(t)y' + b(t)y = 0, y(t0) = 0, y'(t0) = 0 where t0 is any fixed point on the t-axis and the coefficients are continuous. The Attempt at a Solution I know this has to do with the Existence and Uniqueness theorem. How would I apply...
3. ### Differential Equations - Exponential Decay
These are what I think I'm not sure how to find graph solutions from my equation: y = 4 + Ce^(-.5t) Thanks for your help!
4. ### Differential Equations - Exponential Decay
So...first interval - exponentially decreasing graph when y = R/k, it's a constant, straight line along R/k? and when y is big, it's increasing?
5. ### Differential Equations - Exponential Decay
Homework Statement We have the ODE y' = -ky + R for a population y(t) where death rate exceeds birth rate, counteracted by a constant restocking rate. I'm assuming k is the decay constant and R is the restocking rate The population at time t0 = 0 is y0, and I have to find a formula for...
6. ### Line Integral
Ok thanks! But what if it's something like F(x,y,z)=(xy, x/z, y/z) ? Then if you plug in 0, you get a denominator of zero...
7. ### Line Integral
Homework Statement What is the line integral of F(x,y,z) = (xy, x, xyz) over the unit circle c(t) = (cost, sint) t E (0,2pi) ? Homework Equations integral= (f(c(t))*c'(t))dt) The Attempt at a Solution Ok, so I tried solving this like I would any other line integral using the given...
8. ### Integration of a function.
ohh....but how is the triple integral from 0 to 4-x^2-y^2 ? I know it's the function, but graphically, I don't understand. Oh well, my homework was due 10 minutes ago and I just turned it in (leaving this question blank) lol. Thanks for the answer though!
9. ### Integration of a function.
Homework Statement Hey guys, I have one question: how can I integrate the function f(x,y,z)=x + y + z over the region between the paraboloid 4-x^2-y^2 and the xy-plane? Homework Equations For the paraboloid region, I used polar coordinates and found the volume of the region to be 8pi...
10. ### Volume of region R between paraboloid and xy-plane
Alright, I used polar coordinates, i think, I got 8pi.
11. ### Volume of region R between paraboloid and xy-plane
Would it be plus/minus sqrt(4-x^2) ?
12. ### Volume of region R between paraboloid and xy-plane
Ok, for my limits, i got x is between 0 and 2, and y is between 0 and 2...so i did a double integral and got 16/3. Is that right?
14. ### 4-ball B^4
Homework Statement What is the x-simple description of a 4-ball = { ||x|| \leq r } Homework Equations It's a 4-ball, so isn't the equation x^{2} + y^2 +z^2 +w^2 ? The Attempt at a Solution For my limits, I got x is between \pm \sqrt{1-z^2} and y is between -1 and 1, and z is...
15. ### Volume of region R between paraboloid and xy-plane
Umm, isn't the equation just z=4-x^2-y^2 ? Ok, thanks, I'll try a double integral, so for the limits, I would just set z=0, right? and solve for x and y?
16. ### Volume of region R between paraboloid and xy-plane
Homework Statement So my question is: what is the volume of the region R between the paraboloid 4-x^2-y^2 and the xy-plane? Homework Equations I know how to solve it, it is a triple integral, but how do you find the limits of integration? The Attempt at a Solution Do I set x=0...
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1. ## Determinants
Hi, Thanks for the help before.
Can you help me with this new problem.
Define the following determinants of the matrices whose elements come from the ring F[x]:
δ1=1
δ2= | 0 -1|
|-1 x|
δ3= | 0 -1 0|
| 0 x -1|
|-1 0 x|
δ4= | 0 -1 0 0|
| 0 x -1 -1|
| 0 0 x -1|
|-1 0 0 x|
Continue in this fashion. δn is the determinant of an nxn matrix each of whose entries is either 0 or -1 or x according to the following rule. The diagonal entries are all equal to x except for the first diagonal entry which is 0. Each entry along the super diagonal, (i.e. just above the main diagonal) equals -1, as does the entry in the lower left-hand corner. All other entries are 0. Evaluate δn.
Thank you!
2. Originally Posted by mivanova
Hi, Thanks for the help before.
Can you help me with this new problem.
Define the following determinants of the matrices whose elements come from the ring F[x]:
δ1=1
δ2= | 0 -1|
|-1 x|
δ3= | 0 -1 0|
| 0 x -1|
|-1 0 x|
δ4= | 0 -1 0 0|
| 0 x -1 -1|
| 0 0 x -1|
|-1 0 0 x|
Continue in this fashion. δn is the determinant of an nxn matrix each of whose entries is either 0 or -1 or x according to the following rule. The diagonal entries are all equal to x except for the first diagonal entry which is 0. Each entry along the super diagonal, (i.e. just above the main diagonal) equals -1, as does the entry in the lower left-hand corner. All other entries are 0. Evaluate δn.
Thank you!
we have $\delta_1=1.$ suppose $n>1.$ so i guess this is how $\delta_n$ is defined: $\delta_n=\begin{vmatrix}0 & -1 & 0 & 0 & \cdots & 0 \\ 0 & x & -1 & 0 & \cdots & 0 \\ 0 & 0 & x & -1 & \cdots & 0 \\ . & . & . & . & \cdots & . \\
. & . & . & . & \cdots & . \\ . & . & . & . & \cdots & . \\ -1 & 0 & 0 & 0 & \cdots & x \end{vmatrix}.$
what you need to do is to multiply the first row by $x$ and add the result to the
second row to get rid of $x$ in the second row. then multiply the second row by $x$ and add the result to the third row to get rid of $x$ in the third row. continue until all $x$ are gone. thus:
$\delta_n=\begin{vmatrix}0 & -1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & -1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & -1 & \cdots & 0 \\ . & . & . & . & \cdots & . \\
. & . & . & . & \cdots & . \\ . & . & . & . & \cdots & . \\ -1 & 0 & 0 & 0 & \cdots & 0 \end{vmatrix}$
$=(-1)^n \begin{vmatrix}-1 & 0 & 0 & \cdots & 0 \\ 0 & -1 & 0 & \cdots & 0 \\ 0 & 0 & -1 & \cdots & 0 \\ . & . & . & \cdots & . \\
. & . & . & \cdots & . \\ . & . & . & \cdots & . \\ 0 & 0 & 0 & \cdots & -1 \end{vmatrix}=(-1)^{2n-1}=-1$
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# Multiplication As Repeated Addition Worksheets
### About These 15 Worksheets
These worksheets are like magical maps that will guide you through the land of numbers and help you discover the secrets of multiplication. Understanding the link between multiplication and addition is like getting a golden key to the world of math. Once you have it, so many things just click into place. It’s like finding out the secret behind a magic trick.
### What is Multiplication as Repeated Addition?
Imagine you have a bunch of candies, and you want to know how many candies you have in total. You could count them one by one, but that might take a long time. Instead, you can use multiplication as repeated addition to find the answer faster!
Let’s say you have 5 candies in each row, and you have 3 rows of candies. Instead of counting each candy, you can add 5 + 5 + 5, which gives you 15 candies in total. This is like doing 3 times 5, which is another way of saying 3 groups of 5.
### Exploring Different Types of Problems
Now that we understand the magic behind multiplication as repeated addition, let’s dive into the different types of problems you might find on these worksheets:
Basic Repeated Addition – In this type of problem, you might be given a simple multiplication expression and asked to write it as repeated addition. For example, if you see 4 x 3, you can turn it into 3 + 3 + 3 + 3 or 4 groups of 3.
Multiplication Arrays – You might encounter grids of objects, like boxes of pencils arranged in rows and columns. Your task will be to count the total number of pencils by adding the objects in each row. For instance, if there are 2 rows of 5 pencils each, you’d have 5 + 5 = 10 pencils. This helps you see how multiplication is like arranging things in rows and columns.
Matching Equations – Here, you’ll see a set of multiplication equations and their corresponding repeated addition expressions. Your job is to match the multiplication problem with the right repeated addition. For example, you’d match 2 x 4 with 4 + 4.
Word Problems – Get ready for some real-world adventures! You’ll encounter problems that involve situations you might face in everyday life. For instance, “Sarah has 3 bags, and each bag has 6 marbles. How many marbles does she have in total?” You’ll need to recognize that this problem is all about 3 groups of 6 marbles.
Creating Repeated Addition – Flip the previous problems around! You might see repeated addition expressions and need to write them as multiplication equations. If you read “5 + 5 + 5,” you can turn it into 3 x 5, showing that there are 3 groups of 5.
Grouping Objects – Get ready to group! You’ll see groups of objects, like apples in baskets. Your task is to figure out how many objects there are in all by using multiplication as repeated addition. If you have 4 baskets with 2 apples each, you can add 2 + 2 + 2 + 2 or simply say 4 x 2, showing 4 groups of 2.
Drawing Arrays – Time to unleash your artistic skills! You might need to draw an array of objects given a multiplication problem. If you have 3 x 3, you can draw 3 rows with 3 objects in each row. Counting the objects will give you 9, which is the answer to the multiplication problem.
Reinforcing with Repeats – In some worksheets, you’ll practice the same multiplication fact over and over again. This is like building your multiplication muscles! For example, you might see 2 x 6, 2 x 6, 2 x 6… and so on.
### Why These Worksheets Matter?
Multiplication as repeated addition worksheets might seem like fun puzzles, but they’re more than that. They’re tools that help you understand multiplication in a deep and meaningful way. By seeing how multiplication connects to adding things repeatedly, you’re building a strong foundation for math adventures to come.
As you solve these problems, you’re becoming a math explorer, discovering the treasures hidden within numbers. So, grab your pencils and dive into these worksheets with excitement because you’re on a journey to become a multiplication master!
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# Calculating Rotational Inertia (A Bit of Confusion)
1. Dec 4, 2013
### Lemniscates
Hey all,
Perhaps this is a bit stupid...
I'm familiar with the normal procedure of calculating rotational inertia (using integration, parallel axis theorem, etc.). However, I had a confusing thought: if the center of mass of a body is the point at which you can treat as all of the mass being concentrated there and also all the external forces, then why can't I say something like this:
If a uniform rod of length L is swinging around a pivot, and the center of mass of the rod is a distance L/2 away from the pivot, then why can't we treat that center of mass as the place where all the mass is concentrated and use the rotational inertia formula for a particle (MR^2) and subsequently get M(L/2)^2? I know it's ML^2/3, but what is the problem here?
Last edited: Dec 4, 2013
2. Dec 4, 2013
### Zag
Hello Lemniscates,
You have to be careful when you use qualitative statements such as: "the center of mass of a body is the point at which you can treat as all of the mass being concentrated there". In fact, this is only true for translational motion! If you follow the mathematical derivation of this statement, you will notice that one of the underlying assumptions is that the motion under consideration is translational. Therefore, it makes no sense and it is mathematically incorrect to apply this statement to a system undergoing rotational motion.
Best,
Zag
3. Dec 4, 2013
### Lemniscates
Ah. That's interesting. But what about problems involving physical pendulum calculations? From what I've read and have been taught, when you calculate the torque induced by gravity in a physical pendulum, you use the distance from the center of mass to the pivot as the length of the moment arm, essentially saying that all of the gravity acts at the center of mass (and is that not a use of the center of mass/external forces statement?). So how is that justified if the statement about centers of mass can only be applied to bodies in translational motion?
Last edited: Dec 4, 2013
4. Dec 4, 2013
### SteamKing
Staff Emeritus
It's like swinging a sledgehammer: if you grasp the end of the handle and swing the hammer, it takes a certain amount of effort to swing the head. However, if you turn the hammer around and swing the handle by grasping the head, it takes less effort than the first instance. The c.g. of the hammer has not changed, but its location from the axis of rotation has. Since you are not swinging the heavy head in as large a radius of arc in the second instance as in the first instance, it takes less effort.
5. Dec 5, 2013
### cjl
In effect, you somewhat can. You are correctly accounting for the moment of inertia due to the rotation of the center of mass around the pivot, but in that motion, the rod is also rotating about its center of mass. Your formulation would be correct if the rod were somehow maintaining its orientation while revolving around the pivot (I'm sure such an apparatus could be constructed). However, since the rod is rotating, you can do the following.
You can consider your situation as the superposition of two situations:
1) The rotation of the center of mass of the rod around the pivot (which, as you surmised, gives ML2/4)
2) The rotation of the rod about its own center of mass (which gives a moment of inertia of ML2/12)
If you sum these two factors, you get that the total effective moment of inertia is ML2/4 + ML2/12 = 4ML2/12 = ML2/3. This is a useful way to find the overall moment of inertia of a system, since all you need to know is the moment of inertia of the individual parts of the system about their own centers of mass, and the distance each center of mass is from the pivot of the overall system. This is known as the Parallel Axis Theorem
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# A random meeting with a formal ergodic theorist
Most of the people, say more than 99.999999% of the population, has no idea of what Ergodic theory is. Therefore, if you meet someone randomly, you may be sure that he/she does not have idea at all of what it does mean. Among many other examples you can tell in order to introduce the concept of ergodicity, today, Jana Rodríguez Hertz told us one that was quite nice and that I would not like to keep to myself.
The example Jana told us was the following:
Suppose you want to know the number of fish in a lake, say $n.$ It is extremely difficult to count them, however you can estimate $n$. In order to do this, you can take $p$ fish and mark them, then allow them to return to the lake. After some time, you take $m$ fish and you count how many are marked. If you assume that after enough time the proportion of marked fish is the same at “every scale”, i.e., if you take any amount $k$ of fish you will always obtain a number $r(k)$ of marked ones so that $\frac{r(k)}{k}=\frac{p}{n},$ then you can estimate $n$ by $\frac{p m}{r(m)},$ where $r(m)$ is the number of marked fish you obtained when you took in total $m$ fish. Mathematically the validity of this statement is justified if you have the mixing property. This is a strong condition of the system.
Another strategy for estimating $n$ is to capture $m$ fish each $10$ days, count how many are marked, and then allow them to return to the lake. Suppose the $i$-th time you captured fish you obtained $r_i$ marked ones. If you do this many times, say $N,$ then you can estimate $n$ by $\frac{Np}{\sum_{i=1}^N\frac{r_i}{m}}= \frac{Npm}{\sum_{i=1}^N r_i}.$ Mathematically the validity of this statement is justified by the ergodic property. This is weaker than the mixing condition of the system needed in the previous estimation strategy.
In formal terms you may introduce the concept of ergodicity as the behavior of a system where the second estimation approach does work. Sadly or luckily, who knows, ergodicity is unlikely to occur in most of the dynamical systems in nature :).
# Maths does not need colours
If the world would be in black and white maths was exactly the same that it is.
There is indeed no need of colours in maths at all. The basic fact that justifies this is that we can code maths with only zeros and ones, or in other words, in black and white. In this sense, maths is a science that do not “see in colours”.
In practical terms, if we could create a mathematical machine to simulate a human being it should be unable to see colours, however it could distinguish among them. Let me explain this. It is easy to write a program in order to distinguish colours, and surely it will work well in most of the cases. However, distinguish is different from see. If we take a look to what the machine really sees whenever it watches a colour, we should realise that it is indeed an image in black and white of what a colour is. The reason is that the machine understands in mathematical terms. This makes a huge difference between our perception of the word and the one of a mathematically made machine.
Paradoxically, it is usually said that teaching mathematics needs colours. This is because we are not mathematically programmed machines, moreover, colours allows us to incorporate, understand, remember and differentiate mathematical objects (surely a phycologist could say more about this). More in general, we are used to perceive our surrounding using colours, so, it is not very surprising that representing maths with colours do help us to understand them better (this is an interesting fact, that many maths professors could exemplify vasty).
It is a bit sad that a colour is indeed a not definable object in mathematics, so a proper definition of something so trivial is far beyond what we (mathematicians) can define. Defining a colour is a problem that more successfully than we, physicists, artists, phycologists, physicians and writers have accomplished, however, its fully comprehension looks at least intriguing.
# Fire simulation
I was playing the nice video game Ni no Kuni. To be more specific, I was playing a stage called The Mountain of Fire, about a volcano that is going to erupt (in three minutes!). In the scene there are molten rocks, lava and fire. A natural question at this point is: how does the game simulate fire?
A short answer by a mathematician is using the Navier-Stokes Equation, and the one by a ”wikipedian” is using the paper by Jos StamHowever, this problem is far more fascinating than this. My goal here is to explain in a few simple words why. I will start exhibiting the Navier-Stokes Equation, then I will comment about how to simulate fire and I will exhibit some simulations. Finally, I will enumerate current applications and possible ones.
The Navier-Stokes Equation
The Navier-Stokes equation is a basic model for incompresible fluids, like the water. We start by considering a velocity field
\begin{aligned} u:&\mathbb R\times \mathbb R^3& \to & \mathbb R^3\\ &(t,x)&\mapsto& u(t,x)=(u_1(t,x),u_2(t,x),u_3(t,x)) \end{aligned}
and the pressure
\begin{aligned} p:&\mathbb R\times \mathbb R^3& \to & \mathbb R\\ &(t,x)&\mapsto& p(t,x) \end{aligned}
that depend on the time $t$ and position $x.$ Each fluid has a viscosity constant $\nu>0.$ The Navier-Stokes equation is given by the system of equations
\begin{aligned} &\frac{\partial}{\partial t} u+ (u \cdot\nabla)u=\nu \Delta u-\nabla p,\\ &\nabla \cdot u=0,\\ &u(0,x)=u_0(x). \end{aligned}
Given a smooth initial velocity field $u_0(x),$ a global smooth solution of the Navier-Stokes equation is a smooth velocity field $u:[0,+\infty)\times\mathbb{R}^3\to\mathbb{R}^3$ and a pressure function $p:[0,+\infty)\times\mathbb{R}^3\to\mathbb{R}$ satisfying the system of equations.
The justification of the Navier-Stokes equation will not be discussed here, a note on its more important points can be found here, for example.
A solution $u$ of the Navier-Stokes equation is said to blow up if $\sup_{x\in\mathbb{R}^3}|u(t,x)|\to +\infty$ as $t\to +\infty,$ i.e., at some time there is flow with arbitrarily high velocity. Whether is theoretically possible to always find global solutions that do not blow up is an open problem. You can read an explanation of why it is a complicated problem in Terry weblog.
How to simulate Fire
There is a well developed research field called Computational Fluid dynamics, that in particular, studies algorithms to simulate solutions of the Navier-Stokes equation. Numerical simulations using real data suggest, indeed, that there is always a global solution that do not blow up (as it is expected). If you are interested in understanding how to simulate solutions of the Navier-Stokes equation you can read this paper, for example.
Simulations
The Simulations showed here were made using some examples provided in the software Fire Dynamics Simulator (FDS) and Smokeview (SMV).
Applications
As I motivated this post, the Navier-Stokes equation can be applied to simulate fire and smoke in modern video games. However, there are many other applications. I can enumerate some natural examples:
1. Weather forecasting.
2. Chemical and Energy industries.
3. Evacuation routes and placement of fire alarms in case of smoke or fire in cities, forest, buildings, etc. See for example
[RUSHMEIER, H. 1994. Rendering Participating Media: Problems and Solutions from Application Areas. In Proceedings of the 5th Eurographics Workshop on Rendering, 35–56].
4. Ventilation design of the ocean, cities, buildings, miners, computers, etc.
5. Design of planes, trains, ships, cars, bikes, etc. More in general, the design of any device that interact with a fluid in order to accomplish a specific function, like, high stability, high velocity, etc.
6. Design of anti-tsunami systems.
7. Mathematical Cardiology, i.e. the modelling and simulation of the circulatory system.
8. Increasing the velocity of transmission of information about certain properties that depend on fluids.
Artistic representation of fluids
The representation of fluids in arts is typical of an artistic movement called Expressionism. Expressionist artists combine together the representation of certain specific physical phenomena and the representation of daily life situations, as a results, they create an emotional response to the viewer. Usually, the expressionism is studied from the point of view of the psychology, however, from a mathematical physic point of view we are interested in the specific physical phenomena that the artists use. We show examples of artistic representation of fluids by expressionist artists.
###### The Scream (or The Cry) 1893, Edvard Munch.
The Cry is a representation of different fluids interacting together: a human being, a river and the atmosphere.
###### The Starry Night 1889, Vincent van Gogh.
The Starry Night has been studied by many people, I will limite myself here to mention that the astronomer Charles A. Whitney studied the painting from the astronomic point of view. In particular, the brightest star is Venus and the moon was waning gibbous. From my point of view van Gogh is a step away (in physical terms) from the Navier-Stokes equation model of incompressible flows, because he is more likely representing the superfluidity of the lights. Even though, you can experimentally check (by watching here) that The Starry Night is closer related to the Navier Stokes equation than you may had thought.
# Why maths and life are two different things?
Of course I do not intend to even try to answer this question, rather, I will explain an example that happened to me yesterday.
Imagine there are three lifts on the same wall. You are in a rush (the stairs are far away not to be considered) and you need to stand up somewhere in order to take the lift as quick as possible. Where should you stand?
##### |
Mathematically the answer is simple if you consider that each lift has the same probability of arriving (all are available, have the same speed, arrive to every floor, etc… ). In this case you should locate in front of the lift in the middle.
In the real life, however, this is far from being optimal. First of all, it is not very polite. Secondly, you may not been able to see if one of the other lifts arrive before. So a better answer is to consider the smallest distance from the lift in the middle that allows you to see the other two lifts. This answer uses the mathematical answer, but it improves it accordingly the real word.
This example clearly shows that maths and life are two different things, however, maths can help you to improve a decision in real life if you use it carefully.
Esta es la primera vez que escribo sobre la matemática en la filosofía. El tema que vamos a tratar esta basado en el libro de Byung-Chul Han, llamado La sociedad del cansancio (Müdigkeitsgesellschaft).
Vamos a remontarnos a la matemática en los años 40. El matemático Estadounidense George Dantzig es contratado por la USAF como consultor matemático. En 1947, propone el problema de programación lineal junto a un método para resolverlo. Esto da inicio a una de las áreas más importantes de la matemática aplicada actual, el de programación lineal y el desarrollo de métodos para resolver problemas con muchas variables. El método propuesto por Dantzing resulta no ser muy eficiente, ya que puede tomar tiempo exponencial (Klee y Minty, 1972). Sin embargo, hoy en día es sabido que el problema puede ser resuelto en tiempo polinomial usando el método del elipsoide (Khachiyan, 1979). A pesar de esto, el método del elipsoide sin modificaciones es en la práctica completamente inútil.
Desde los años 80 en adelante la optimización ha tomado un rol preponderante en la industria. El estudio e implementación de modos eficientes para resolver problemas de la vida real, junto con el avance de los computadores, han permitido desarrollar en el último tiempo tecnología impensada hace un par de años.
Que efectos tiene la matemática, y en especial, el concepto de optimización en la sociedad?
Esta es la pregunta que se plantea Byung-Chul Han en su libro La sociedad del cansancio. Según el autor, la enfermedad de la sociedad postmoderna es el estado patológico de exceso de positividad y del sentimiento que todo es posible. La sociedad, se ha transformado en una sociedad de rendimiento, donde abundan los gimnasios, las torres de oficina, los bancos, los grandes centros comerciales y los laboratorios genéticos. Los sujetos no son como antes sujetos de obediencia, pero de rendimiento, emprendedores de sí mismos, explotadores de sí mismos. El ser humano en su conjunto se convierte en una máquina de rendimiento cuyo objetivo consiste en funcionar sin alteraciones y sin negatividad, ya que estas ralentizan el proceso de optimización.
Entre las consecuencias de este exceso de positividad se encuentra la violencia, que se despliega en una sociedad permisiva y pacífica. Una violencia que no es privativa, sino que exhaustiva. Este tipo de violencia que sacia y nos hace obeso, no nos da la oportunidad de estar aburridos, sino que al contrario, nos hace estar todo el tiempo pendiente de muchas variables, de muchos datos, y todos simultáneamente. Según el autor, los animales con mayor capacidad para llevar a cabo muchas actividades simultáneamente son los más agresivos y los más salvajes. (Byung-Chul Han)
Otra arista de la violencia producida por el exceso de positividad se ve reflejada en la supresión del esquema negativo de la prohibición. Con el fin de aumentar la productividad se sustituye la disciplina por el rendimiento. El rendimiento es determinado por el nivel de producción. La prohibición tiene un efecto negativo, ya que bloquea e impide un crecimiento ulterior. La positividad del poder hacer es mucho más eficiente que la negatividad del deber. (Byung-Chul Han)
En cuánto al individuo, el exceso de positividad genera un sentimiento de depresión y fracaso. Surge el concepto de ”nada es posible”, ”nada me resulta” que sigue la aceptación del de ”todo es posible”. (Byung-Chul Han)
El resultado final del exceso de positividad en la sociedad es de cansancio y agotamiento excesivo. Es un tipo de cansancio que no solo no permite ver al otro, sino que vé en el otro al mismo tiempo el YO. Esto conlleva finalmente a una sociedad del dopaje, donde los individuos son ápaticos y muy violentos. (Byung-Chul Han)
Antes de explicar la solución que Byung-Chul Han propone al exceso de positividad, y por consiguiente, una solución a la actual enfermedad de la sociedad postmoderna, producto del áfan por la optimización; vamos a abogar por las matemáticas y en particular por la optimización.
En las báses del desarrollo de las matemáticas se encuentra un proceso contemplativo y un sentimiento intenso de impotencia y fracaso. Es muy acertado en este proceso lo que escribe Byung-Chul Han, la pura agitación no genera nada nuevo, solo reproduce y acelera lo existente. Por ejemplo, existe un sitio web especializado para hacer preguntas en matemáticas http://mathoverflow.net/, donde muchos matemáticos agitados hacen preguntas públicas de modo que otros matemáticos puedan responderle de manera casi instántanea. El resultado es un sitio lleno de preguntas (muy interesantes), pero que no aporta al desarrollo de las matemáticas, ya que es más bien un caos acelerado de toda la matemática ya existente.
No es la matemática, ni la optimización que están detrás de la enfermedad del exceso de positividad de la sociedad que plantea Byung-Chul Han, sino que es resultado natural en la sociedad de la disponibilidad de tecnologías nuevas.
Las soluciones esbozadas en ”La sociedad del cansancio” son: rescatar la importancia del sosiego, de la vida contemplativa, de la rabia y del miedo. (Byung-Chul Han)
El sosiego, en el sentido de Nietzsche:
Por falta de sosiego, nuestra civilización desemboca en una barbarie. En ninguna época, se han cotizado más los activos, es decir, los desasosesgados. Cuéntase, por tanto, entre las correcciones necesarias que deben hacérsele al carácter de la humanidad el fortalecimiento en amplia medida del elemento contemplativo.
La vida contemplativa, en el sentido de Catón:
Nunca está nadie más activo que cuando no hace nada, nunca está menos solo que cuando está consigo mismo
La rabia y el miedo, en el sentido de Byung-Chul Han:
Lo interesante de la propuesta de Byung-Chul Han en relación a la incorporación en la sociedad del concepto de optimización, es el hecho de constituir una visión postmoderna de la simbiosis Matemática-Sociedad, con la que tendremos que aprender a convivir del modo más humano posible.
Referencias: Recibí recientemente como regalo de cumpleaños el libro ”La sociedad del cansancio” de parte de mi amigo Angello Estefane.
# Bingo winning strategy
I will write about a problem by my friend Edgardo Roldán Pensado. He told me it long time ago. If you are a mathematician the solution may be trivial, however, I personally admire the way he faced the situation and came out with this nice problem.
I won’t bore you with a longer introduction, let start the problem.
Suppose you bought two days ago (in amazon.com for example) a Royal Bingo Supplies Wooden Bingo Game (like the one in the picture)
• Wooden Bingo Game Set with instructions.
• Perfect for old-fashioned fun with a nostalgic twist.
• Includes with 18 Bingo cards, 150 Bingo chips, a Bingo board, brass cage and 75 wooden balls.
• Great for parties, barbeques or family game nights.
• Recommended for ages 3 and up.
Today you received your game and you invite seven other friends to play together. Each player takes a bingo card and you start playing. Five hours later, when everybody is already tired of playing, you decide to count the times each player has won. Everybody is very surprised how lucky a single player was, who won many more times than anybody else.
Is there a Bingo winning strategy? Or do you have a very lucky friend?
Well…Everybody has a very lucky friend, so no discussion about this. However, there is also a Bingo winning strategy, so, there is a chance the winner was a Bingo´s tactician (with a bit of luck).
How does the Bingo´s tactician play?
This guy chose his bingo card at the end, after analysing the bingo cards of the other player. Each card was (hopefully, I do not know if it is true) created with a uniform random distribution (for each square you choose a number from 1 to 75 with probability $1/75.$). He considered a metric on the set of bingo cards, for example, given two bingo cards $a=(x_1,\ldots,x_{24})$ and $b=(y_1,\ldots,y_{24}),$ $d(a,b)=\#\{i:x_i\neq y_i\},$ with the convention that $\# \emptyset :=0.$ He was a bit lucky enough to be able to find a bingo card that maximises the $d$ distance with respect to the bingo cards of the other player.
Why is this a winning strategy?
Suppose that the $7$ bingo cards chosen by the opponents of the winner were very close with respect to the distance $d,$ moreover, suppose that all the $7$ bingo cards were exactly the same. On the other hand, the “lucky one” chose a different bingo card. So there is $\frac{1}{2}$ probability one of the other seven players (and then all) wins, and $\frac{1}{2}$ probability the “lucky one” wins. Now, suppose, the $7$ bingo cards chosen by the opponents are very close (with respect to $d$), but all different. Then the “lucky one” player wins with probability close to $\frac{1}{2},$ whilst and the other players with probability close to $\frac{1}{14}.$
# Explaining maths without using them
At the time I was finishing my thesis some non mathematician friends asked me to explain them what I was actually doing. In this post I write the explanation I elaborated for them. In one sentence, the first problem one need to know how to solve is the following.
How do you explain a mathematical result without using maths at all?
The answer is well known to everybody teaching maths, however, I will elaborate a bit on this. The first step is to understand the historical roots of the problem you solved, together with the main motivations that people had to need to develop maths in such a determined direction that your problems appeared naturally. Once, you have understood completely this part, you need to associate each problem you solved to something that the people you want to explain the problem to can understand well. This means to associate each result to something completely different but that behaves exactly the same. Finally, you need to somehow transport all this information and deliver it to your audience in a comprehensive way. The last part of the process is well performed by an ”artistic drawing” (whatever that means) together with an explanation of each element. The hardest part of the communication is the explanation of the proof of each result, it happened indeed, that this becomes an issue of time.
In what follows I will answer a more specific question.
Problem:
Explain your thesis to your friends with time constrains conditions.
My thesis can be coded by the following drawing.
The drawing has 5 main elements: a black disk at the top left, a black disconnected curve at the top and bottom, a red apple shape at the left, a blue swirl at the bottom and a green tree shape in the centre and right of the drawing. I will explain each of the five element in what follows.
1. Black disk: If one goes back in time from the questions I faced in my thesis, we arrive to the three body problem. This is represented by the three white bodies inside the black disk. I can suggest you some nice reference to read about, like [Wikipedia, Celestial Encounters by F. Diacu and P. Holmes]. In a few words: The n-body problem consists in determining the position of $n$ planet at any time $t$, given the initial conditions of known position and velocity at time $t=0.$ For $n=2$ the problem can be easily solved, one can prove indeed, that the trajectory of one planet with respect to the other always lies along a conic section. The problem for $n=3$ was addressed by Henri Poincaré and literally he found the chaos in it. He was not able to find a solution, but instead, introduces qualitative methods to understand the solution, like, periodic solutions, recurrence theorem, non existence of uniform integrals, asymptotic solutions, dependence of the solutions with respect to a parameter, homoclinic orbits, first return map, invariant curves and homoclinic tangles. These qualitative methods are still our tools to understand dynamic systems that are too complicated to be understand in a deterministic way, i.e. those dynamical systems which equations can be solved.
2. Black disconnected curve: This curve represents the continuous trajectory of a particle that moves because of some physical laws acting on it. We suppose that there is a finite measure on the drawn system, we assume further that it is invariant under the physical laws. Under these assumptions, by the Poincare recurrence theorem, almost every particle will enter in a finite amount of time into a set $A,$ providing the measure of this set is strictly positive. The set $A$ in the picture corresponds to the black disk, and the black disconnected curve eventually enters $A$ after some (unknown) finite amount of time. The question that we face is how can we estimate the time that takes the particle to enter for first time $A.$ We solved this problem under very strong hypothesis on the physical laws. Estimations for this problem under our hypothesis are known, however, we refined the existing bound.
3. Blue swirl: This represents chaos in the classic sense of the diagram of phase of ODE´s (Ordinary differential equation). Recall that when we have an ODE $x'=Ax$ in $\mathbb{R}^2,$ the diagram of phase corresponds to the plot of the vectors $(y_1,y_2)$ in the $(x_1,x_2)$-plane, where we draw at the point $x= (x_1,x_2)$ the vector $(y_1,y_2)=Ax.$ This represent the velocity field of the solutions of the ODE. We assume $A$ is not singular, so that the equilibrium is $x=0.$ The diagram of phase will be determined by the eigenvalues and eigenvectors $\lambda_1,\lambda_2$ of the matrix $A.$ There are 7 cases for the equilibrium state. The equilibrium state is:
1. Stable if $\lambda_1<\lambda_2<0.$
2. Unstable if $0<\lambda_1<\lambda_2.$
3. Saddle if $\lambda_1<0<\lambda_2.$
4. Degenerate if $\lambda_1=\lambda_2\in \mathbb{R}.$
5. Center if $\lambda_1=i \beta, \beta\in\mathbb{R}\setminus \{0\}.$
6. Stable spiral if $\lambda_1=\alpha+i \beta, \alpha<0, \beta\in\mathbb{R}\setminus \{0\}.$
7. Unstable spiral if $\lambda_1=\alpha+i \beta, \alpha>0, \beta\in\mathbb{R}\setminus \{0\}.$ In the drawing we represented this case.
4. Red apple shape: This represents a seed that gives birth to a tree after a process that allows to ”build” a complex structure from iterations of a simpler one. This is analogous to the system build by a simple iterated function scheme that consists of two disjoint contractions on the unit interval, these contractions have associated a unique non-empty closed invariant set. This sets comes from a structure similar to the one of a tree, where one branch divided into to smaller, and each smaller one into two smaller and so on.
5. Green tree shape: The green structure represent a tree that we consider analogous to the invariant set of an iterated function system. The problem that we face in the last chapter is: What happens if we perturb a little bit the seed of the tree, how different is going to be our tree? The analogy with this is: perturbing a little bit both maps of our iterative function and study how smoothly that affect the invariant set. A mathematical way to do this is by considering the for example the ”measure” (called Hausdorff dimension) of the invariant set.
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# Algebra 2 URGENT
Simplify:
1. 0
2. 0
3. 4
1. Simplify 2+4-5
1. 0
2. 0
posted by Simplify
## Similar Questions
1. ### math
I am not sure how to do these. 1. Simplify 4√6/√30 by rationalizing the denominator. Show your work. 2. Simplify (2√5 + 3√7)^2 Show your work. Justify each step.
2. ### math
2. Simplify. Please be sure to show all of your work. -3(-9) – |-5 – 3 3. Simplify. Please show all of your work 9c^3+7c-(3c^3-12+c) 4. Solve 4x- 3(5x-8) =23-9(x+2) . Please show all of your work 5. Solve the following
3. ### Algebra 1
Simplify using long division. (remember to show all your work.) 3. (72-8x^2+4x^3-36x)/(x-3) 4. (8b^3-6)/(2b-1) simplify the expression, if possible. state the excluded values, if any. 5. 10x^2-25X+15^5x-5 6.2w^2-18^w^2+6w+9
4. ### Math
Simplify the expressions in each pair. (2 marks each) (Show your work) i) (x^2-x-2)/(x+1),g(x)=(x-2) My work on simplifying the functions: (x-1)(x+2)/(1) (x-1)(x+2) ii) (x^2-x-12)/(x+3), g(x)=(x^2-6x+8)/(x-2) My work on
5. ### Math
Consider the equation below. 1/5(x+2)+2x = 6x-10 Part A: Which property can be used to simplify the expression 1/5(x+2)? Answer _______________ Part B: Move all x-terms to one side of the equation and simplify. Show your work.
6. ### algebra
simplify and show all work 11+12^2-(-11)*12 is this right11+144-(-11)*12 155-(-11)*12 144*12 1,928 simplify -(8m)-(-7m+2) 15m-2 according to order of operation what should be done first when evaluating this expression do not solve
8. ### algebra
Let f(x)= 2x^2 +7 and g(x)= x – 6. (a) Find the composite function (f*g)(x) and simplify. Show work. (b) Find ( f*g)(1). Show work.
9. ### Calculus
Let f (x) = 5x^2 – 9 and g(x) = x – 3. (a) Find the composite function ( f o g)(x) and simplify. Show work. (b) Find ( f o g)(2) . Show work.
10. ### ALGEBRA
SIMPLIFY AND SHOW WORK 12-1/9-4/3+(1/2)^2
11. ### Algebra 2 URGENT
Simplify: 1/u^2-2u-1/u^2-4 Please Show All Work!
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# math
posted by on .
factor each of the following:
a)x^2+10x+25
b)x^2+16x+28
c)x^2-9x+14
• math - ,
x^2+10x+25=(x+5)*(x-5)
Becouse:
(x+5)*(x-5)=x*x+5*x-5*x-25=x^2+5x-5x+25=x^2+25
x^2+16x+28=(x+2)*(x+14)
Becouse:
(x+2)*(x+14)=x*x+2*x+14x+28=x^2+2x+14x+28=x^2+16x+28
x^2-9x+14=(x-7)*(x-2)
Becouse:
(x-7)*(x-2)=x*x-7x-2x+2*7=x^2-9x+14
• math - ,
i don't realy get it still like i do get it but at the same time i don't :( sorry im being an thik head lol
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# Learning Electronics
Learn to build electronic circuits
# Logarithms for analog circuits
### Question 1:
The concept of a mathematical power is familiar to most students of algebra. For instance, ten to the third power means this:
103 = 10 ×10 ×10 = 1000
. . . and eight to the seventh power means this:
87 = 8 ×8 ×8 ×8 ×8 ×8 ×8 = 2,097,152
Just as subtraction is the inverse function of addition, and division is the inverse function of multiplication (because with inverse functions, one ündoes" the other), there is also an inverse function for a power and we call it the logarithm.
Re-write the expression 103 = 1000 so that it uses the same quantities (10, 3, and 1000) in the context of a logarithm instead of a power, just as the subtraction is shown here to be the inverse of addition, and division is shown to be the inverse of multiplication in the following examples:
3 + 8 = 11 (+ and - are inverse functions) 11 - 3 = 8
2 ×7 = 14 (× and �are inverse functions) 14 �2 = 7
103 = 1000 (powers and logs are inverse functions) log10 ??? = ???
103 = 1000 (powers and logs are inverse functions) log10 1000 = 3
Notes:
In my experience, most American students are woefully underprepared for the subject of logarithms when they study with me. Admittedly, logarithms do not see as much use in everyday life as powers do (and that is very little for most people as it is!). Logarithms used to be common fare for secondary school and college students, as they were essential for the operation of a slide rule, an elegant mechanical analog computing device popular decades ago.
The purpose of this question is to twofold: to get students to realize what a logarithm is, and also to remind them of the concept of inverse functions, which become very important in analog computational circuits.
### Question 2:
Given the following mathematical expression, write another one defining a logarithm using the same variables:
If: xy = z Then: log? ? = ?
If: xy = z Then: logx z = y
Notes:
Nothing special here. Indeed, the answer to this question may be derived from any algebra textbook.
### Question 3:
Electronic calculators with logarithm capability have at least two different types of logarithms: common logarithm and natural logarithm, symbolized as "log" and "ln", respectively. Explain what the difference is between these two types of logarithms.
The common logarithm function assumes a "base" value of ten, whereas the natural logarithm assumes a base value of e (Euler's constant).
Follow-up question: what is the approximate value of e? How can you get your calculator to give you the answer (rather than looking it up in a math book?
Notes:
Some calculators, of course, allow you to extract the logarithm of any number to any base. Here, I simply want students to become familiar with the two logarithm functions available on the most basic scientific calculators.
Note that some calculators will show just enough digits of e to give the false impression that they repeat (ten digits: e = 2.718281828). If anyone suggests that e is a (rational) repeating decimal number, correct this misunderstanding by telling them it is irrational just like p.
### Question 4:
Note the following logarithmic identities, using the "common" (base 10) logarithm:
log10 = 1
log100 = 2
log1000 = 3
log10000 = 4
In the first equation, the numbers 10 and 1 were related together by the log function. In the second equation, the numbers 100 and 2 were related together by the same log function, and so on.
Rewrite the four equations together in such a way that the same numbers are related to each other, but without writing "log". In other words, represent the same mathematical relationships using some mathematical function other than the common logarithm function.
101 = 10
102 = 100
103 = 1000
104 = 10000
Notes:
An illustration like this helps students comprehend what the "log" function actually does.
### Question 5:
Note the following logarithmic identities, using the "common" (base 10) logarithm:
log0.1 = -1
log0.01 = -2
log0.001 = -3
log0.0001 = -4
In the first equation, the numbers 0.1 and 1 were related together by the log function. In the second equation, the numbers 0.01 and 2 were related together by the same log function, and so on.
Rewrite the four equations together in such a way that the same numbers are related to each other, but without writing "log". In other words, represent the same mathematical relationships using some mathematical function other than the common logarithm function.
10-1 = 0.1
10-2 = 0.01
10-3 = 0.001
10-4 = 0.0001
Notes:
An illustration like this helps students comprehend what the "log" function actually does.
### Question 6:
Examine the following progression of mathematical statements:
(102)(103) = 100000
102+3 = 100000
105 = 100000
What does this pattern indicate? What principle of algebra is illustrated by these three equations?
Next, examine this progression of mathematical statements:
log105 = log100000 = 5
log102+3 = log100000 = 5
log102 + log103 = log100000 = 5
What does this pattern indicate? What principle of algebra is illustrated by these three equations?
First pattern:
The product of two base numbers with different exponents is equal to that base number raised to the power of the exponents' sum.
Second pattern:
The sum of two logarithms is equal to the logarithm of those two numbers' product.
Notes:
In this question, I want students to begin to see how logarithms relate multiplication to addition, and how powers relate addition to multiplication. This is an initial step to students recognizing logarithms as transform functions: a means to transform one type of mathematical problem into a simpler type of mathematical problem.
### Question 7:
Examine this progression of mathematical statements:
(100)(1000) = 100000
(100)(1000) = 105
log[(100)(1000)] = log105
log100 + log1000 = log105
log102 + log103 = log105
2 + 3 = 5
What began as a multiplication problem ended up as an addition problem, through the application of logarithms. What does this tell you about the utility of logarithms as an arithmetic tool?
That logarithms can reduce the complexity of an equation from multiplication, down to addition, indicates its usefulness as a tool to simplify arithmetic problems. Specifically, the logarithm of a product is equal to the sum of the logarithms of the two numbers being multiplied.
Notes:
In mathematics, any procedure that reduces a complex type of problem into a simpler type of problem is called a transform function, and logarithms are one of the simplest types of transform functions in existence.
### Question 8:
Suppose you owned a scientific calculator with two broken buttons: the multiply (×) and divide (). Demonstrate how you could solve this simple multiplication problem using only logarithms, addition, and antilogarithms (powers):
7 ×5 = ???
The answer to this problem was easy enough for you to figure out without a calculator at all, so here are some more practice problems for you to try:
23 ×35 =
781 ×92 =
19.4 ×60 =
0.019 ×2.6 =
Here I will show you the steps to using logarithms to solve the first multiplication problem:
7 ×5 = ???
7 ×5 = 10log7 + log5
7 ×5 = 100.8451 + 0.6990
7 ×5 = 101.5441
7 ×5 = 35
Since the others are easy enough for you to check (with your non-broken calculator!), I'll leave their solutions in your capable hands.
Notes:
Incidentally, there is nothing special about the common logarithm to warrant its exclusive use in this problem. We could have just as easily applied the natural logarithm function with the same (final) result:
7 ×5 = ???
7 ×5 = eln7 + ln5
7 ×5 = e1.9459 + 1.6094
7 ×5 = e3.5553
7 ×5 = 35
### Question 9:
Examine this progression of mathematical statements:
1000 100 = 10
1000 100 = 101
log �� 1000 100 �� = log101
log1000 - log100 = log101
log103 - log102 = log101
3 - 2 = 1
What began as a division problem ended up as a subtraction problem, through the application of logarithms. What does this tell you about the utility of logarithms as an arithmetic tool?
That logarithms can reduce the complexity of an equation from division, down to subtraction, indicates its usefulness as a tool to simplify arithmetic problems. Specifically, the logarithm of a quotient is equal to the difference between the logarithms of the two numbers being divided.
Notes:
In mathematics, any procedure that reduces a complex type of problem into a simpler type of problem is called a transform function, and logarithms are one of the simplest types of transform functions in existence.
### Question 10:
Suppose you owned a scientific calculator with two broken buttons: the multiply (×) and divide (). Demonstrate how you could solve this simple multiplication problem using only logarithms, addition, and antilogarithms (powers):
12 �3 = ???
The answer to this problem was easy enough for you to figure out without a calculator at all, so here are some more practice problems for you to try:
122 35 =
781 92 =
19.4 60 =
3.5 0.21 =
Here I will show you the steps to using logarithms to solve the first multiplication problem:
12 �3 = ???
12 �3 = 10log12 - log3
12 �3 = 101.0792 - 0.4771
12 �3 = 100.6021
12 �3 = 4
Since the others are easy enough for you to check (with your non-broken calculator!), I'll leave their solutions in your capable hands.
Notes:
Incidentally, there is nothing special about the common logarithm to warrant its exclusive use in this problem. We could have just as easily applied the natural logarithm function with the same (final) result:
12 �3 = ???
12 �3 = eln12 - ln3
12 �3 = e2.4849 - 1.0986
12 �3 = e1.3863
12 �3 = 4
### Question 11:
Examine this progression of mathematical statements:
(1000)2 = 1000000
(1000)2 = 106
log[(1000)2] = log106
(2)(log1000) = log106
(2)(log103) = log106
(2)(3) = 6
What began as an exponential problem ended up as a multiplication problem, through the application of logarithms. What does this tell you about the utility of logarithms as an arithmetic tool?
That logarithms can reduce the complexity of an equation from exponentiation, down to multiplication, indicates its usefulness as a tool to simplify arithmetic problems. Specifically, the logarithm of a number raised to a power is equal to that power multiplied by the logarithm of the number.
Notes:
In mathematics, any procedure that reduces a complex type of problem into a simpler type of problem is called a transform function, and logarithms are one of the simplest types of transform functions in existence.
### Question 12:
Suppose you owned a scientific calculator with two broken buttons: the power (yx) and root (x {y}). Demonstrate how you could solve this simple power problem using only logarithms, multiplication, and antilogarithms (powers):
34 = ???
The answer to this problem was easy enough for you to figure out without a calculator at all, so here are some more practice problems for you to try:
256 =
5643 =
0.2242 =
410.3 =
Here I will show you the steps to using logarithms to solve the first multiplication problem:
34 = ???
34 = 10(4 log3)
34 = 10(4)(0.4771)
34 = 101.9085
34 = 81
Since the others are easy enough for you to check (with your non-broken calculator!), I'll leave their solutions in your capable hands.
Notes:
Incidentally, there is nothing special about the common logarithm to warrant its exclusive use in this problem. We could have just as easily applied the natural logarithm function with the same (final) result:
34 = ???
34 = e(4 ln3)
34 = e(4)(1.0986)
34 = e4.3944
34 = 81
### Question 13:
Examine this progression of mathematical statements:
� 1000 = 101.5
log � 1000 = log( 101.5 )
log( 10001/2 ) = log( 101.5 )
1 2 (log1000) = log( 101.5 )
1 2 (log103) = log( 101.5 )
3 2 (log10) = log( 101.5 )
3 2 (1) = log( 101.5 )
3 2 = log( 101.5 )
3 2 = 1.5
What began as a fractional exponent problem ended up as a simple fraction, through the application of logarithms. What does this tell you about the utility of logarithms as an arithmetic tool?
That logarithms can reduce the complexity of an equation from fractional exponentiation, down to simple fractions, indicates its usefulness as a tool to simplify arithmetic problems. Specifically, the logarithm of a root of a number is equal to the logarithm of that number divided by the root index.
Notes:
In mathematics, any procedure that reduces a complex type of problem into a simpler type of problem is called a transform function, and logarithms are one of the simplest types of transform functions in existence.
### Question 14:
Suppose you owned a scientific calculator with two broken buttons: the power (yx) and root (x {y}). Demonstrate how you could solve this simple root problem using only logarithms, division, and antilogarithms (powers):
3 � 8 = ???
The answer to this problem was easy enough for you to figure out without a calculator at all, so here are some more practice problems for you to try:
4 {13} =
5 {209} =
2.5 {9935} =
9.2 {0.15} =
Here I will show you the steps to using logarithms to solve the first multiplication problem:
3 � 8 = ???
3 � 8 = 10( 1/3 log8 )
3 � 8 = 10( 1/3 (0.9031) )
3 � 8 = 100.3010
3 � 8 = 2
Since the others are easy enough for you to check (with your non-broken calculator!), I'll leave their solutions in your capable hands.
Notes:
Incidentally, there is nothing special about the common logarithm to warrant its exclusive use in this problem. We could have just as easily applied the natural logarithm function with the same (final) result:
3 � 8 = ???
3 � 8 = e( 1/3 ln8 )
3 � 8 = e( 1/3 (2.0794) )
3 � 8 = e0.6931
3 � 8 = 2
### Question 15:
You may be wondering why anyone would bother using logarithms to solve arithmetic problems for which we have perfectly good and effective digital electronic calculator functions at our disposal. For example, why would anyone do this:
10log7 + log5
. . . when they could just do the following on the same calculator?
7 ×5
The quick answer to this very good question is, "when it is more difficult to directly multiply two numbers." The trouble is, most people have a difficult time imagining when it would ever be easier to take two logarithms, add them together, and raise ten to that power than it would be to simply multiply the original two numbers together.
The answer to that mystery is found in operational amplifier circuitry. As it turns out, it is much easier to build single opamp circuits that add, subtract, exponentiate, or take logarithms than it is to build one that directly multiplies or divides two quantities (analog voltages) together.
We may think of these opamp functions as "blocks" which may be interconnected to perform composite arithmetic functions:
Using this model of specific math-function "blocks," show how the following set of analog math function blocks may be connected together to multiply two analog voltages together:
Notes:
The purpose of this question is simple: to provide a practical application for logarithms as computational aids in an age of cheap, ubiquitous, digital computing devices.
### Question 16:
Logarithms have interesting properties, which we may exploit in electronic circuits to perform certain complex operations. In this question, I recommend you use a hand calculator to explore these properties.
Calculate the following:
10log3 =
log(108) =
eln3 =
ln(e8) =
10(log3 + log5) =
e(ln3 + ln5) =
10(log2.2 + log4) =
e(ln2.2 + ln4) =
10(log12 - log4) =
e(ln12 - ln4) =
10(2 log3) =
e(2 ln3) =
10([log25/2]) =
e([ln25/2]) =
10log3 = 3
log(108) = 8
eln3 = 3
ln(e8) = 8
10(log3 + log5) = 15
e(ln3 + ln5) = 15
10(log2.2 + log4) = 8.8
e(ln2.2 + ln4) = 8.8
10(log12 - log4) = 3
e(ln12 - ln4) = 3
10(2 log3) = 9
e(2 ln3) = 9
10([log25/2]) = 5
e([ln25/2]) = 5
Notes:
Discuss what mathematical operations are being done with the constants in these equations, by using logarithms. What patterns do your students notice? Also, discuss the terms "log" and äntilog," and relate them to opamp circuits they've seen.
Ask your students whether or not they think it matters what "base" of logarithm is used in these equations. Can they think of any other arithmetic operations to try using logarithms in this manner?
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# Confusing Derivation of torque acting on disk question
• Bonbon32
In summary, when the disk is on the tabletop, it exerts a force on the table surface. This force is given by the product of the disk's mass and the pressure it produces.f
#### Bonbon32
Homework Statement
A disk of mass M and angular speed w falls into table top. The kenetic coffienet of friction from table top is meu. Create an equation for dt=dr *F. Using this information.
Relevant Equations
Iaplha = torque
Fr = torque
Not sure how they obtained an answer of (2meukenetic *Mg/R^2)*r^2 dr = dt
When they say disk I am assuming it's not ring shaped. So since F * Dr= DT where I guess F acting on a small point on the disc is constant from the table top. And Dr is the variable distance from a random small point from the axis which summed together times force give total torque. I plug in F friction using mgmeu. Not to sure where to go from there.
#### Attachments
• IMG_20191224_002338.jpg
80 KB · Views: 148
Last edited by a moderator:
The disk as a whole, lying on the tabletop, exerts a force ##M g## on the table surface, right? Assuming that the weight is evenly distributed over the entire disk-table interface, that means that the disk presents a pressure of ##Mg/\pi R^2##.
A given circular element of the disk will have an area associated with it (##\pi r^2 dr##). What is the force associated with that area element (you've got pressure and area)? Work out the ##d\tau## that that circular element will provide.
The disk as a whole, lying on the tabletop, exerts a force ##M g## on the table surface, right? Assuming that the weight is evenly distributed over the entire disk-table interface, that means that the disk presents a pressure of ##Mg/\pi R^2##.
A given circular element of the disk will have an area associated with it (##\pi r^2 dr##). What is the force associated with that area element (you've got pressure and area)? Work out the ##d\tau## that that circular element will provide.
Makes sense so (meu mg/R^2 )* r^2 *Dr which gives me a torque, not sure where they got the 2 from? Which is on the numerator next to the meu value. Finally got one more question if you can bear with me. Would the meu mg the friction be dF instead as we are taking only a small circular position about the rotating axis? So it becomes meu*dm*g ? If we aren't assuming anything.
Makes sense so (meu mg/R^2 )* r^2 *Dr which gives me a torque, not sure where they got the 2 from?
What's the area of the circular element of radius r and width dr?
Would the meu mg the friction be dF instead as we are taking only a small circular position about the rotating axis? So it becomes meu*dm*g ?
Force is pressure times area. so the frictional force contribution for a given area element becomes ##dF = P \times dA##, where dA is the differential area element.
By the way, you can find a table of mathematical letters and symbols in the edit window top bar menu if you click on the ##\sqrt{x}## icon. There you will find, for example, the Greek letter μ. You don't need to call it "meu", if you can use the actual letter.
Added an image to clarify what a "circular element" refers to.
Last edited:
Bonbon32
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Upon reaching the age of sixty, the president of the board : GMAT Sentence Correction (SC)
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# Upon reaching the age of sixty, the president of the board
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Upon reaching the age of sixty, the president of the board of education decided to retire over searching for another job.
A)Upon reaching the age of sixty, the president of the board of education decided to retire over searching for another job.
B)After reaching the age of sixty, the president of the board of education decided retirement than to search for another job.
C)When he reached the age of sixty, the president of the board of education decided to retire rather than to search for another job.
D)When reaching the age of sixty, the president of the board of education decided to retire instead of searching for another job.
E)When he reached the age of sixty, the president of the board of education decided to retire rather than searching for another job.
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29 Aug 2005, 10:46
I will go with C as it is parallel ( to retire ....to search)
A - i find "retire over" to be incorrect
B - not parallel
D - When reaching is incorrect
E - not parallel
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29 Aug 2005, 22:08
Even I would go for C.
Reason:
A : Improper usage of parallelism
B : "decided...than to" is again wrong
C : Correct one
D : "When reaching ..." sounds odd
E : "to retire...than searching" Improper parallelism
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30 Aug 2005, 01:04
C,
parallellism here to retire rather than to search
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30 Aug 2005, 05:36
A)Upon reaching the age of sixty, the president of the board of education decided to retire over searching for another job.
- Not Parallel
B)After reaching the age of sixty, the president of the board of education decided retirement than to search for another job.
- Not parallel
C)When he reached the age of sixty, the president of the board of education decided to retire rather than to search for another job.
D)When reaching the age of sixty, the president of the board of education decided to retire instead of searching for another job.
- Awkward and not parallel
E)When he reached the age of sixty, the president of the board of education decided to retire rather than searching for another job.
- Awkward and not parallel
C is the best choice.
30 Aug 2005, 05:36
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http://poj.org/problem?id=1923
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Fourier's Lines
Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 2809 Accepted: 919
Description
Joseph Fourier was a great mathematician and physicist and is well known for his mathematic series. Among all the nineteen children in his family, Joseph was the youngest and the smartest. He began to show his interest in mathematics when he was very young. After he grew up, he often corresponded with C. Bonard (a professor of mathematics at Auxerre) by exchanging letters.
In one letter written to Bonard, Fourier asked a question: how to draw 17 lines on a plane to make exactly 101 crossings, where each crossing belongs to exactly two lines. Obviously, this is an easy problem, and Figure-1 is a solution that satisfies his requirement. Now the problem for you is a universal one. Can we draw N lines on a plane to make exactly M crossings, where each crossing belongs to exactly two lines? If we can, how many pieces, at most, can these lines cut the plane into?
Input
The input may have several sets of test data. Each set is one line containing two integers N and M (1 <= N <= 100, 0 <= M <= 10000), separated by a space. The test data is followed by a line containing two zeros, which indicates the end of input and should not be processed as a set of data.
Output
Output one line for each set of input in the following format:
Case i: N lines cannot make exactly M crossings.
if the drawing of these lines is impossible;
or:
Case i: N lines with exactly M crossings can cut the plane into K pieces at most.
Note: Even if N or M equals to one, you should use the words "lines" and "crossings" in your output.
Sample Input
```4 3
4 6
4 2
5 11
17 101
0 0
```
Sample Output
```Case 1: 4 lines with exactly 3 crossings can cut the plane into 8 pieces at most.
Case 2: 4 lines with exactly 6 crossings can cut the plane into 11 pieces at most.
Case 3: 4 lines cannot make exactly 2 crossings.
Case 4: 5 lines cannot make exactly 11 crossings.
Case 5: 17 lines with exactly 101 crossings can cut the plane into 119 pieces at most.
```
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# v= (L) di/dt assistance
• posted
First of all, I took calculus and learned derviatives as: X^3 d/dx = 3X^2.
The voltage across an inductor is given by: v= (L) di/dt
I'm unsure how to take the derivative of that formula because there is a di in the numerator.
I understand it can be said as delta i / delta t. But this would be in the case of t approaching 0 or a linear change in current; i.e. a linear ramp.
What if I had a poteniometer that was changing at an acceleration rate instead of a linear rate? Then my i would not be linear, it would be curved.
What if I wanted to know the current at exactly X seconds not using a delta assumption?
Any calculus assistance will be appreciated.
p.s. I am aware this formula can be used in Diff EQ and a formula will show the voltage at every point in time.
• posted
Certainly a better way of writing this is: d/dX (X^3) = 3x^2.
Your misunderstanding is mostly of the physics. The mathematical aspect is not so important. In uyour context, I presume, you need to have a specific functional form for i. Not to be original, suppose i(t) = t^3 and L = 2 henries, what would v be as a function of time?
Bill
• posted
Not sure why you want the derivative of it in the first place. That would give you dv/dt on left and d^2i/dt (second derivative of i) on the right.
To understand the voltage/current relationship, suppose the applied voltage is v = ksin(wt). Substitution would give you...
ksin(wt) = (L) di/dt k/L sin(wt) = di/dt
Now, *integrate* both sides....
k/L ((1/w)-cos(wt)) = i (ignoring the integration constant for now...)
Note how i is a negative cosine function while we started with v as a sin function. This shows the current is 90 degrees out of phase with the voltage (current lagging voltage).
If the applied voltage is something more complex than a simple sinusoid function, the integration can get a bit more difficult, but the principle is the same.
Or, if you want you can use a current source driving current through the inductor. Let us say the current source is a sin function such that i = ksin(wt). We find the derivative of i...
di/dt = d(ksin(wt) / dt = kw cos(wt)
Substituting in the formula for the inductor....
v=(L) di/dt = (L) kw cos(wt)
Again, the shift from sin to cosine shows the voltage out of phase by 90 degrees from the current (voltage leading current). And again, if the injected current is something other than a sinusoid, the principle is the same, but you may have more work to do to arrange so that i can be reduced to di/dt.
Hope this helps...
daestrom
• posted
Question about your answer. When you took the integral of both sides, does the i come out as a constant because d/dt disappeared.
That example is fine if I'm applying a sine wave, but what if I have a DC source? Using deltas will be a good approximation, however, it's not exact because the current is di/dt and not delta i / delta t.
I believe I'm confusing this too much. :)
• posted
Your questions may require more math than I can muster. But you can only use di/dt and its integrals on smooth continuous functions, not step functions. So for DC with a step change on/off, I don't think you can use smooth integrals.
Where.... k/L sin(wt) = di/dt
integrated to
k/L ((1/w)-cos(wt)) = i
No, 'i' is not a constant . The right hand side is the instantaneous current, which happens to be a function of time (say, f(t) = i ). Reversing things and taking the derivative...
k/L ((1/w)-cos(wt)) = f(t) = i
k/l sin(wt) = f'(t) = di/dt
The sine function works out nicely in part because the function is continuous and we can ignore any transient parts by only looking at the 'steady-state'. If you replace the applied voltage with something like a step change at time 0, where at all times less than zero the voltage and current are zero and then voltage steps to some fixed positive value, then you have....
v=0, i=0 (t0)
Notice that there is no 'upper limit' to current, it just keeps growing forever (at the rate di/dt = v/L). This is because there is no resistance in this 'theoretical' circuit. Of course a real circuit has a resistance, which means we apply Kirchoff's law to get....
0=v - Ri - Ldi/dt (t>0)
Solving this R-L circuit can be found in many calculus textbooks. If you ignore the transient part, obviously i=v/R for the steady-state where di/dt = 0 (kind of the definition of 'steady-state' in this case). But the transient part can be 'interesting' and will show the current rise from zero up to the steady-state is a function of e^(-R/Lt). Hence the 'time constant' for such a circuit is TC=R/L
Does that help at all??
daestrom
• posted
Well yes, a switch would be consider linear. I'm thinking more of say a poteniometer where I'm changing the rate of current from 1 amp / second,
2 amps / second, then 4 amps / second, then 8 amps / second.
Of course this would be difficult to achieve but in that example a linear ramp would clearly not be accurate. So using delta i / delta t would be incorrect.
Now using that example, the current would be changing on a curve.
I think I was pampered with d/dx (x^3) = 3x^2.
Now I'm having problems trying to solve (basic) electronic equations.
• posted
Principles are the same, just the math can get ugly.
Suppose you have....
di/dt = t^2
Integrating (assuming i = 0 at t=0)...
i = 1/3 t^3 + 0
Obviously....
v = L di/dt = L t^2
t=0 di/dt = 1A/s t=1 di/dt = 2 A/s t=2 di/dt = 4 A/s t=3 di/dt = 8 A/s
This is a 'doubling' function so it's exponential. This one I can see from inspection....
di/dt = 2^(t)
So...
v = L (2^(t))
Integrating this one is a bit harder...
di/dt = 2^(t) di/dt = e^(t*ln2)
i = int(e^(t*ln2)) dt
Now, I *think* that you can integrate this term by setting u=(t*ln2) and du=ln2dt. Then knowing int(e^u du) = e^u
i=1/ln2*int(e^u du) i=1/ln2*e^u i= 1/ln2 *e^(t*ln2)
-or- i = 1/ln2 * 2^(t)
But I could be mistaken here.
Well, if it was *easy*, you wouldn't have to go to school to learn it now, would ya? :-)
The 'trick' is if you can find a function that describes di/dt in the first place. The more complicated that function is, the harder it is to integrate it to find 'i'.
daestrom
• posted
You take the derivative of i with respect to t because i is actually some function of t.
In other words, i changes as t changes (or di/dt = 0).
In order to make use of the above relationship, you have to know what i(t) is or know what v(t) is and work backwards to find i(t)
• posted
First, instead of X^3 d/dx = 3X^2, the correct form is d/dX X^3 = 3X^2. That is, you take the derivative with respect to X 'of' X^3. This dictates the order of the operator in the equation.
Second, the function v = L di/dt can simply be interpretted as saying that the voltage across an inductor is proportional to the rate of change of current through the inductor. (The constant of proportionality is L.) So, if the current changes faster the voltage developed across the inductor will be larger. If a current that drives an inductor is suddenly switched off, for example, then di/dt will be very large, so v will be very large. This accounts for the term "inductive kickback."
Let's take the case you describe that causes the variation of the current to accelerate. For example, let i(t) = K t^2 where K is a constant.
Since v = L di/dt, then v = L d(K t^2)/dt = 2 L K t
What does the result mean? It means that the voltage across the inductor continues to increase linearly with time. The "scale factor" is 2LK. So at t = 0, v = 0. At t = 1, v = 2LK. At t = 2, v = 4LK. And so on.
• posted
After reading your explainations and playing with some circuits in MultiSim, I believe I have a better understanding.
I was trying to understand the formula (v = L di/dt) without understanding what I was doing or inputting a function.
If I understand everything correctly:
• Any linear change such as a continous increase (or decrease) in current over time will be linear and thus making the formula delta i / delta t.
• If I'm applying some function to a circuit, then I need to solve for it such as what you did.
I did have some issues when simulating. I was unable to obtain a higher voltage than my input. Adjusting the input frequency only caused the voltage across the inductor to equal my input voltage but only for a different time period (obviously equal to the inductor formula).
It would have been nice to generate 100 volts with 5 volts input. Most online HV circuits I found online contained step-up transformers and any inductor sites beats the inductor formula to death without explaining it.
Another problem I saw: I had a 1H in series with a 100 ohm resistor and opened the circuit (via a switch) and the voltage across the inductor was a triangle wave (above and below ground) that appeared to never hault.
• posted
This may be a limit of multisim (never used it), or the way you set it up, don't know.
Well 20:1 is asking a bit much, but it certainly is possible to use an inductor to 'pump' to a higher voltage. One simple circuit is to arrange the DC supply to feed through the inductor and return via a FET transistor. Between the inductor and FET, tap off with a diode to feed a capacitor. Now arrange a 'suitable oscillator' to turn on the FET so the inductor basically shorts the DC supply. When current builds up, suddenly switch off the FET. The inductor magnetic field will collapse and create a voltage spike that will momentarily forward bias the diode and push some charge into the capacitor. Once inductor current drops near zero, turn on the FET and start over again. This uses the strong magnetic field built up in the inductor when it's 'shorting' the power supply to 'pump' charge the capacitor. But as with all things, there are limits to how much load you can draw off the high-voltage capacitor because if you draw more charge off than the 'pump' circuit can put in, voltage on the capacitor drops.
(I'm sure someone here could devise a better circuit or tell me where I may have made a mistake with this, but the principle is used in a lot of DC-DC converters that are used for step-up)
That certainly is a multisim issue. What should happen is you get a very high voltage 'spike', but it has a very short duration. A 'real' switch would end up arcing momentarily from the high voltage spike and the energy in the inductor would be dissipated in the arc.
daestrom
• posted
--------- Part of the problem is just what circuit you are trying to simulate and the source involved (voltage or current) and it may be that the simulation software simply doesn't like the initial conditions given or that the model is not realistic (e.g. switches in series with an ideal current source would be a serious problem- a non-ideal source can be converted to a voltage source behind the switch). The circuits involved don't have the complexity to need a simulator so you should be able to get a good handle on it without one.
• posted
"Don Kelly" wrote in news:EAmCj.86204\$w94.33585@pd7urf2no:
MultiSim allows you to set the initial conditions. Typically I leave it at the default (let the simulator determine initial conditons), however, this time I have it set to 'zero'.
I have a 1H inductor directly to ground; the 1H makes the math easier. The switch allows for a linear ramp instead of the FET; it was causing non-linearity most likely because of the internal capacitance.
I do, however, get double the voltage across the inductor than I calculate. As an example, my switch takes 9.9us to open (as measured in the program) and it's telling me the current is 7.4mA. That should be [7.4mA / 9.9us] * 1H = 747v. but my measured voltage is 1492v. I'm not sure if this is a fault of the program since I have an inductor directly to ground. The circuit turns into a Diff EQ if I add a resistor and makes the math tricky to deal with while "learning" the basics.
• posted
I have no idea of what multisim is doing but it appears that you are calculating an average voltage over the switch opening. I do have a problem with the "measured" value -if it is the multisim result, it isn't actually measured -however, it does agree with the maximum value assuming such a ramp down. I don't know enough about multisim (having never used it) to say what it is actually doing. Is it trying to represent an ideal current source with a switch in series? If so, problems occur. If you are representing an ideal current source that ramps down from 7.4 to 0 in 9.9 microseconds- that is another animal and your value of 750V is good. Otherwise??? Again, exactly what circuit is it that you are trying to represent? A great deal of the problem appears to be that you are trying to model an unrealistic system. Resistance always exists so there is no such thing as a pure inductance. I would suggest that for true learning purposes, you ignore multisim which is a tool for those that have a firm hand on basics. It is not a substitute for the work involved in learning the basics (as you have found) , including the dread DEQ (1st and second order are sufficient at this stage and the solutions of such are very straightforward and simple - by the book .
Don Kelly [email protected] remove the X to answer
• posted
yes, i know this is late, nevertheless...
Steve wrote: [...]
that is accurate, and also true for non-linear change in current. the issue is, you cannot develop a voltage across a pure inductance without a change in current through the inductance.
use a current source, not a voltage source.
the time constant for L-R circuits is, naturally, L/R. your description suggests that you simulated an unrealistic condition -- you open switch, current in your inductor must drop to zero, but if you don't add another conductive path somewhere the delta-i has no place to "go". your simulator then does whatever it can do, and the result isn't necessarily realistic.
hth, hand.
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.
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## Saturday, November 23, 2013
### Thinking Flexibly About Exponential Form
I had a great discussion with my Calculus class about how the exponential sequence 2^x is
1, 2, 4, 8, 16, ... but that there is nothing special about the base 2. There has to be a number "a", such that (3^a)^x = 2^x, because there has to be an exponent "a" such that 3^a = 2. Together they found that number, and we said that g(x) = 3^(0.631x) is therefore an approximation of f(x) = 2^x. Similarly, there must be an exponent k such that (e^k)^x = 2^x. So, the kids took a minute or so to find that k = ln(2), so h(x) = e^(0.693x) must also be an approximation of f(x) = 2^x. We discussed how h is easily differentiable now that we are Chain Rule pros.
I then set the kids loose on an activity that asks them to think flexibly about the exponential form. The reason why I like this activity is because the kids distinguish between an exact form that is "natural" for the situation versus the usual e^(kx) approximation. (The kids know that they can substitute k with an exact expression without losing precision, however.) For example, for #1 in the worksheet, where the kids are looking at bacteria that double every 6 minutes, both the kids and I think that it's natural to think of the sequence as {10, 20, 40, 80, ...} and to observe the general form y = 10(2)^x. To fix the problem that we want it to double only ONCE by x = 6 minutes, not doubling 6 times, we divide our exponent by 6 to get y = 10(2)^(x/6). For me, this is the natural way of writing the equation and testing it initially to make sure that it fits the bill. If they then want to differentiate it, they then turn it into base e, where e^k = 2^(1/6), so that y = 10(e)^(kx) replaces y = 10(2)^(x/6). The kids figure out that k = (1/6)*ln(2) or around k = 0.116. So, y = 10(e)^(0.116x) is an approximation of our exact function, and it has the benefit of being easily differentiable.
In thinking about exponential form as being fluid, the kids can consider equivalent compound-interest scenarios. I gave them a couple of scenarios to play with and to explain, in order to get at that idea. I am pretty happy with the level of understanding they have with this concept, seeing that it's the second time this year we've seen exponential compounding. After that, they didn't seem to have much trouble working through our practice quiz on the exponential topic. Overall, I am pretty happy with the way our differentiation technique unit has gone. We're moving a little bit slower than I had hoped, but their understanding of the connections between concepts has been really great!!! Both they and I are still excited to walk into this class everyday, and that's a good feeling. I anticipate that by January, we'll be wrapped up with all the differentiation techniques (including related rates problems, which I've been sprinkling into the mix periodically), and the kids will be ready to start thinking backwards and/or to do a differentiation project.
Stay tuned!
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# Sound Seems Amplified Over Water
by Ron Kurtus (revised 18 November 2012)
If you are sitting in a boat, a sound coming from the shore will seem louder than the same sound heard by a person on land. Sound seems to be amplified when it travels over water.
The reason is that the water cools the air above its surface, which then slows down the sound waves near the surface. This causes refraction or bending of the sound wave, such that more sound reaches the boat passenger.
Sound waves skimming the surface of the water can add to the amplification effect, if the water is calm.
Questions you may have include:
• Why does sound lose its loudness with distance?
• What happens to the speed of sound with temperature?
• How does the amplitude increase?
This lesson will answer those questions. Useful tool: Units Conversion
## Amplitude decreases with distance
As any waveform expands or spreads out from its source, its amplitude decreases. You can see this with water waves in a pond. With sound, the amplitude of the waves relates to the loudness of the sound that we perceive.
Amplitude of waves decreasing as they spread
## Speed and temperature
When the air temperature is 24° C (75° F), the speed of sound is 346 meters per second (m/s) or 775 miles per hour (mph). The relationship between the speed of sound and the air temperature is approximated from the equation:
v = 331.4 + 0.6Tc m/s
where Tc is the Celsius temperature.
If the water was at 15.5° C (60° F), the air just above the surface would be close to the same temperature. The air temperature would then increase at further distances from the water until it reached the normal air temperature.
The follow chart is the air temperature and speed of sound at different heights above the water.
#### miles/hour
91.5 cm (3 ft) or more 24 75 346 775
61 cm (2 ft) 21 70 344 771
30.5 cm (1 ft) 18.3 65 343 767
Near water level 15.5 60 341 764
Speed of sound decreases closer to cooler water
## Effect of refraction and reflection
Refraction of the sound waves in cooler air and the reflection off the surface of the water effect the sound that is heard.
### Waves bent by refraction
When a wave strikes a material in which it travels slower, its direction is changed slightly if it strikes the material at an angle. This effect is called refraction. You have seen refraction when light passes through a pane of glass at an angle. The image is displaced, because the light rays were refracted.
(See Refraction of Light for more information on this subject.)
The same thing happens when sound enters a material in which its speed is slower than in normal air. The direction of the sound waves change slightly.
Since the temperature of the water in a lake or ocean is usually cooler than the normal air temperature, the air just above the water level is cooled by the water. The temperature varies according to the distance from the surface of the water. This gradient of speeds would result in a lens effect due to refraction of sound. That means sound would tend to focus and thus increase its apparent loudness.
Cool air bends sound and thus increases amplitude
It is a strange effect that follows the principles of sound and wave motion.
### Reflection off surface of water
If the water is smooth or calm, the sound waves skim the surface of the water and are reflected toward the observer in the boat, adding to the amplification. However, if the water is choppy, the sound is randomly reflected and make no contribution to the amplitude of the sound.
## Summary
A sound coming from the shore will sound louder to a person sitting in a boat in the water than the same sound heard by a person on land.
Sound seems to be amplified when it travels over water, because the water cools the air above its surface. Cool air slows down the sound waves near the surface, causing refraction or bending of the sound wave. Then more sound reaches the boat passenger.
Sound waves skimming the surface of the water can add to the amplification effect, if the water is calm.
Know the principles of science
## Resources and references
Ron Kurtus' Credentials
### Websites
Speed of Sound - Hyperphysics site
Physics Resources
## Questions and comments
Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.
## Students and researchers
www.school-for-champions.com/science/
sound_amplified_over_water.htm
Please include it as a link on your website or as a reference in your report, document, or thesis.
## Where are you now?
School for Champions
Physics topics
## Also see
### Let's make the world a better place
Be the best that you can be.
Use your knowledge and skills to help others succeed.
Don't be wasteful; protect our environment.
### Live Your Life as a Champion:
Take care of your health
Seek knowledge and gain skills
Do excellent work
Be valuable to others
Have utmost character
#### Be a Champion!
The School for Champions helps you become the type of person who can be called a Champion.
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# Variable Cost Ratio
A company’s variable production costs expressed as a percentage of its net sales
## What is the Variable Cost Ratio?
The variable cost ratio is a cost accounting tool used to express a company’s variable production costs as a percentage of its net sales. The ratio is calculated by dividing the variable costs by the net revenues of the company. The company’s net revenue includes the sum of its returns, allowances, and discounts subtracted from the total sales.
The primary motive for calculating the variable cost ratio is to consider costs that may be subject to variations with changes in levels of production and compare them to the amount of revenues generated by the sales of that particular cycle of production.
In calculating the ratio, fixed costs, which are the expenses that remain constant regardless of variations in production levels, are excluded. Examples of fixed costs include building lease, employee salaries, etc.
### Summary
• The variable cost ratio is a cost accounting tool used to express a company’s variable production costs as a percentage of its net sales.
• The primary motive of calculating the ratio is to consider costs that may be subject to variations with the changes in production levels and compare them to the amount of revenues generated by the sales of that particular cycle of production.
• It is an important factor in determining the overall profitability of a company.
### How to Calculate the Variable Cost Ratio
The formula for the calculation of the variable cost ratio is as follows:
##### Variable Cost Ratio = Variable Costs / Net Sales
An alternate formula is given below:
##### Variable Cost Ratio = 1 – Contribution Margin
The contribution margin is a quantitative expression of the difference between the company’s total sales revenue and the total variable costs of production of goods that were sold. The contribution margin is expressed in percentage points.
### Sample Calculation
There are several ways in which the variable cost ratio can be calculated. Under the first method, the mathematical calculation is performed on a per-unit basis. In such a situation, consider a product with a per-unit variable cost of \$10 and a per-unit sales price of \$100. It gives a variable cost ratio of 0.1 or 10%.
The calculation can also be done by utilizing totals over a given period of time. Consider a situation wherein the total variable costs of production are \$1,000 per month, and the total revenues generated per month are \$10,000. The variable cost ratio, in this situation, is 0.1 or 10%.
### Significance
The variable cost ratio is an important factor in determining the overall profitability of a company. It indicates whether the business can achieve a desirable balance of revenue streams such that a rise in revenues is faster than that of expenses. It is used to express in quantitative terms the relationship shared by the sales of a company and the particular costs of production that are associated with the revenues being considered.
The ratio is a useful evaluation metric for a company’s management to determine necessary break-even or minimum profit margins, make profit projections, and identify the optimal sales price for its products. In a situation where a company incurs high variable costs as a percentage of its net sales made, it is most likely that it does not need to cover a lot of fixed costs per month.
It means that the company will need to generate enough revenue to cover the fixed costs involved in the production process. It just enables the company to stay in business without generating any substantial profits from sales.
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# Showing $\sum c_n = \sum a_n \sum b_n$ if $\sum c_n x^n=\sum a_n x^n \sum b_n x^n$ on $[0, 1)$ [closed]
Given that $$\sum c_n x^n=\sum a_n x^n \sum b_n x^n$$ on $$[0, 1)$$ and all partial sums $$\sum c_n x^n, \sum a_n x^n, \sum b_n x^n$$ converges uniformly on $$[0, 1]$$, is it true that $$\sum c_n = \sum a_n \sum b_n$$?
## closed as off-topic by Carl Mummert, Nosrati, Scientifica, Delta-u, MicahSep 26 '18 at 16:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Nosrati, Scientifica, Delta-u, Micah
If this question can be reworded to fit the rules in the help center, please edit the question.
Yes. In the conditions you mentioned the sum $$\sum c_n$$ is equal to the Abel sum which is $$\lim_{x\to{1^{-}}}\sum c_n x^n$$, and same can be said about the other two sums. So then:
$$\sum c_n=\lim_{x\to{1^{-}}}\sum c_n x^n=\lim_{x\to{1^{-}}}\sum a_n x^n\sum b_n x^n=\lim_{x\to{1^{-}}}\sum a_n x^n\lim_{x\to{1^{-}}}\sum b_n x^n=$$
$$=\sum a_n\sum b_n$$
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Objects > Functions
# Functions
A function is a rule that turns input values into output values. For example, the function f(x) = 2·x + 3 is a rule that says “To get an output value, start with the input value, multiply it by 2, and then add 3 to the result.” To evaluate a function you follow the rule for a particular input value. For example, f(5) = 2(5) + 3 = 13.
There are three kinds of functions in Sketchpad:
A symbolically defined function is defined by an algebraic rule.
A derivative function is defined by finding the derivative of another function.
A data-defined function is defined by the top stroke of a drawing or the top edge of a picture.
Once you've created a function, see Using Functions for information on the many ways you can use a function.
Create a Symbolically-Defined Function
To define a new function symbolically, for example, f(x) = 2·sin(x), choose This command opens Sketchpad’s Calculator to allow you to define how the function is calculated. To create a new function and plot it immediately, choose To determine whether the function appears in y= notation or f(x) notation, choose the desired notation from the Calculator’s Equation pop-up menu. See also: How to Use Signum to Construct a Piecewise Function
Create a Derivative Function
To create a derivative function, select the function you want to differentiate and choose
The derivative of a symbolically-defined function is usually defined symbolically, as shown on the left. In rare cases of exceptionally complex symbolically-defined functions, the derivative may be defined approximately.
The derivative of a data-defined function (defined by a drawing or picture) is always defined approximately, as shown on the right.
Symbolic Derivative Approximate Derivative
To determine whether the function appears in y= notation or f(x) notation, set the desired notation for the parent function.
When Sketchpad creates a derivative function, it does not keep track of domain restrictions. For instance, f(x) = ln(x ) has a domain restriction: it's defined for x > 0. Sketchpad shows the correct derivative, f '(x ) = 1/x, but without the domain restriction: the derivative is defined for x 0.
Create a Data-Defined Function
To create a data-defined function based on a drawing or picture, select the drawing or picture and choose The function is defined by the top edge of the drawing or picture. To determine how much smoothing is applied to the function, choose Edit | Properties | Function. An unsmoothed function follows every pixel in the top edge of the drawing or picture; a smoothed function reduces the small variations. To determine whether the function appears in y= notation or f(x) notation, choose
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# G-sharp major 7th suspended 4th chord
The Solution below shows the G-sharp major 7th suspended 4th chord in root position, 1st, 2nd, and 3rd inversions, on the piano, treble clef and bass clef.
The Lesson steps then explain how to construct this 7th chord using the 3rd, 5th and 7th note intervals, then finally how to construct the inverted chord variations.
For a quick summary of this topic, have a look at Seventh chord.
Key C C# Db D D# Eb E E# Fb F F# Gb G [G#] Ab A A# Bb B B# Cb
## Solution - 4 parts
### 1. G-sharp major 7th suspended 4th chord
This step shows the G-sharp major 7th suspended 4th chord in root position on the piano, treble clef and bass clef.
The G-sharp major 7th suspended 4th chord contains 4 notes: G#, C#, D#, F##.
The chord spelling / formula relative to the G# major scale is: 1 4 5 7.
G-sharp major 7th suspended 4th chord note names
Note no.Note intervalSpelling
/ formula
Note name#Semitones
from root
1root1The 1st note of the G-sharp major 7th suspended 4th chord is G#0
2G#-perf-4th4The 2nd note of the G-sharp major 7th suspended 4th chord is C#5
3G#-perf-5th5The 3rd note of the G-sharp major 7th suspended 4th chord is D#7
4G#-maj-7th7The 4th note of the G-sharp major 7th suspended 4th chord is F##11
Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 7/5/4.
The staff diagrams and audio files contain each note individually, ascending from the root, followed by the chord containing all 3 notes.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 2. G-sharp major 7th suspended 4th 1st inversion
This step shows the G-sharp major 7th suspended 4th 1st inversion on the piano, treble clef and bass clef.
The G-sharp major 7th suspended 4th 1st inversion contains 4 notes: C#, D#, F##, G#.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 5/4/2, so the chord is said to be in five-four-two position.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 3. G-sharp major 7th suspended 4th 2nd inversion
This step shows the G-sharp major 7th suspended 4th 2nd inversion on the piano, treble clef and bass clef.
The G-sharp major 7th suspended 4th 2nd inversion contains 4 notes: D#, F##, G#, C#.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 7/4/3, so the chord is said to be in seven-four-three position.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 4. G-sharp major 7th suspended 4th 3rd inversion
This step shows the G-sharp major 7th suspended 4th 3rd inversion on the piano, treble clef and bass clef.
The G-sharp major 7th suspended 4th 3rd inversion contains 4 notes: F##, G#, C#, D#.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 6/5/2, so the chord is said to be in six-five-two position.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
## Lesson steps
### 1. Piano key note names
This step shows the white and black note names on a piano keyboard so that the note names are familiar for later steps, and to show that the note names start repeating themselves after 12 notes.
The white keys are named using the alphabetic letters A, B, C, D, E, F, and G, which is a pattern that repeats up the piano keyboard.
Every white or black key could have a flat(b) or sharp(#) accidental name, depending on how that note is used. In a later step, if sharp or flat notes are used, the exact accidental names will be chosen.
The audio files below play every note shown on the piano above, so middle C (marked with an orange line at the bottom) is the 2nd note heard.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 2. G-sharp tonic note and one octave of notes
This step shows 1 octave of notes starting from note G#, to identify the start and end notes of the scale used to build this chord.
The numbered notes are those that might be used when building this chord.
Note 1 is the root note - the starting note of the chord - G#, and note 13 is the same note name but one octave higher.
No. Note 1 2 3 4 5 6 7 8 9 10 11 12 13 G# A A# / Bb B C C# / Db D D# / Eb E F F# / Gb G G#
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 3. G-sharp major scale note interval positions
This step describes the G# major scale , whose note intervals are used to define the chord in a later step.
The major scale uses the W-W-H-W-W-W-H note counting rule to identify the scale note positions.
To count up a Whole tone, count up by two physical piano keys, either white or black.
To count up a Half-tone (semitone), count up from the last note up by one physical piano key, either white or black.
The tonic note (shown as *) is the starting point and is always the 1st note in the major scale.
Again, the final 8th note is the octave note, having the same name as the tonic note.
No. Note 1 2 3 4 5 6 7 8 G# A# / Bb C C# / Db D# / Eb F G G#
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 4. G-sharp major scale note interval numbers
This step identifies the note interval numbers of each scale note, which are used to calculate the chord note names in a later step.
To identify the note interval numbers for this major scale, just assign each note position from the previous step, with numbers ascending from 1 to 8.
No. Note 1 2 3 4 5 6 7 8 G# A# B# C# D# E# F## G#
To understand why the note names of this major scale have these specific sharp and flat names, have a look at the G# major scale page.
Both the note interval numbers and note names from the piano diagram above will be used in later steps to calculate the chord note names.
### 5. 7th chord qualities
This step defines a seventh chord, names the 7th chord qualities and identifies the notes that vary between them.
#### 7th chord definition
Whereas a triad chord contains 3 notes, a 7th chord contains 4 notes that are played together or overlapping.
#### 7th chord qualities
7th chords exist in eight different chord qualities, which are diminished, half-diminished, minor, minor-major , dominant, major, augmented, and augmented-major.
Each chord quality name is the name of the entire chord as a whole, not its individual notes (which will be covered later).
#### Triad chord qualities using the 1st, 3rd, 5th and 7th scale notes
All of these 7th chord qualities are based on the 1st, 3rd, 5th and 7th notes of the major scale piano diagram above.
Depending on the chord quality, the 3rd, 5th and 7th scale note names of the major scale above might need to be adjusted up or down by one or more half-notes / semitones / piano keys.
It is these variations of the 3rd, 5th and 7th notes that give each one a distinctive sound for any given key (eg. C-flat, E etc).
In fact, these 7th chords are based on triad chords - the first 3 notes of any 7th chord are identical to a specific triad chord quality, with one extra note added to make it a 7th chord.
#### Suspended 7th chords - using the 2nd or 4th scale notes
A suspended chord is known in music theory as an altered chord because it takes one of the above chord qualities and modifies it in some way.
Unlike all of the above qualities, Suspended triad chords do not use the 3rd note of the major scale (at all) to build the chord.
The 3rd note is suspended, ie. removed completely, and replaced by either the 2nd note of the major scale - a suspended 2nd, or more commonly by the 4th note of the major scale - a suspended 4th.
Musically, this is interesting, since it is usually the 3rd note of the scale that defines the overall character of the chord as being major (typically described as 'happy') or minor ('sad').
Without this 3rd note, suspended chords tend to have an open and ambiguous sound.
The steps below will detail the major 7th suspended 4th triad chord quality in the key of G#.
### 6. 7th chord note intervals
This step defines the note intervals for each chord quality, including the intervals for the G-sharp major 7th suspended 4th 7th chord. It also shows how the 7th chord qualities are related to the triad chord qualities they are based on.
Each individual note in a 7th chord can be represented in music theory using a note interval, which is used to express the relationship between the first note of the chord (the root note), and the note in question.
The root note is always the 1st note (note interval 1 in the above diagram) of the major scale diagram above. ie. the tonic of the major scale.
Then there is one note interval to describe the 2nd note, and another to describe the 3rd note of the chord, and finally another interval for the 4th chord note.
In the same way that the entire chord itself has a chord quality, the intervals representing the individual notes within that chord each have their own quality.
These note interval qualities are diminished, minor, major, perfect and augmented.
Below is a table showing the note interval qualities for all 7th chords, together with the interval short names / abbrevations in brackets.
The final column shows the triad chord quality that the 7th chord is based on, so the 2nd and 3rd note quality columns are the same as the triad table for the same key.
seventh chord note interval qualities
7th chord quality2nd note quality3rd note quality4th note qualityBased on triad quality
diminishedminor (m3)diminished (d5)diminished (d7)diminished
half-diminishedminor (m3)diminished (d5)minor (m7)diminished
minorminor (m3)perfect (P5)minor (m7)minor
minor majorminor (m3)perfect (P5)major (M7)minor
dominantmajor (M3)perfect (P5)minor (m7)major
majormajor (M3)perfect (P5)major (M7)major
augmentedmajor (M3)augmented (A5)minor (m7)augmented
augmented-majormajor (M3)augmented (A5)major (M7)augmented
major suspended
(2nd/4th)
major (M2) or
perfect (P4)
perfect (P5)major (M7)suspended
(2nd/4th)
dominant
suspended 4th
perfect (P4)perfect (P5)minor (m7)suspended 4th
The numbers in brackets are the note interval number (ie the scale note number) shown in the previous step.
Looking at the table above, the note intervals for the chord quality we are interested in (major 7th suspended 4th), in the key of G# are G#-perf-4th, G#-perf-5th, and G#-maj-7th.
The links above explain in detail the meaning of these qualities, the short abbrevations in brackets, and how to calculate the interval note names based on the scale note names from the previous step.
### 7. G-sharp major 7th suspended 4th chord in root position
This step shows the G-sharp major 7th suspended 4th chord note interval names and note positions on a piano diagram.
Each note interval quality (diminished, minor, major, perfect, augmented) expresses a possible adjustment ie. a possible increase or decrease in the note pitch from the major scale notes in step 4.
If an adjustment in the pitch occurs, the note name given in the major scale in step 4 is modified, so that sharp or flat accidentals will be added or removed.
But crucially, for all interval qualities, the starting point from which accidentals need to be added or removed are the major scale note names in step 4.
For this chord, this is explained in detail in G#-perf-4th, G#-perf-5th and G#-maj-7th, but the relevant adjustments for this major 7th suspended 4th chord quality are shown below:
G#-4th: Since the 4th note quality of the major scale is perfect, and the note interval quality needed is perfect also, no adjustment needs to be made. The 4th note name - C#, is used, and the chord note spelling is 4.
G#-5th: Since the 5th note quality of the major scale is perfect, and the note interval quality needed is perfect also, no adjustment needs to be made. The 5th note name - D#, is used, and the chord note spelling is 5.
G#-7th: Since the 7th note quality of the major scale is major, and the note interval quality needed is major also, no adjustment needs to be made. The 7th note name - F## is used, and the chord note spelling is 7.
If it is still not clear why the interval qualities are organised / related as they are, please refer to each of the interval links above.
#### G-sharp major 7th suspended 4th seventh chord note names
The final chord note names and note interval links are shown in the table below.
Note Interval No. Interval def Spelling 1 4 5 7 G# C# D# F## root G#-perf-4th G#-perf-5th G#-maj-7th 1 4 5 7 0 5 7 11
The piano diagram below shows the interval short names, the note positions and the final note names of this triad chord.
In music theory, this 7th chord as it stands is said to be in root position because the root of the chord - note G#, is the note with the lowest pitch of all the chord notes.
The note order of this chord can also be changed, so that the root is no longer the lowest note, in which case the chord is no longer in root position, and will be called an inverted 7th chord instead.
For 7th chords, there are 3 possible inverted variations as described below.
#### Figured bass notation
The figured bass notation for a 7th chord in root position is 7/5/3, with the 7 placed above the 5, and the 5 above the 3.
These numbers represent the interval between the lowest note of the chord and the note in question.
So another name for this inversion would be G-sharp major 7th suspended 4th triad in seven-five-three position.
For example, the 7 represents note F##, from the G#-7th interval, since the chord root, G#, is the lowest note of the chord (as it is not inverted). .
In the same way, the figured bass 5 symbol represents note D#, from the G#-5th interval, and the 4 symbol represents note C#, from the G#-4th interval
Since figured bass notation works within the context of a key, we don't need to indicate in the figured bass symbols whether eg. the 3rd is a major, minor etc. The key is assumed from the key signature.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 8. G-sharp major 7th suspended 4th 1st inversion
This step shows the first inversion of the G-sharp major 7th suspended 4th.
To invert a chord, simply take the first note of the chord to be inverted (the lowest in pitch) and move it up an octave to the end of the chord.
So for a 1st inversion, take the root of the 7th chord in root position from the step above - note G#, and move it up one octave (12 notes) so it is the last (highest) note in the chord.
The second note of the original 7th chord (in root position) - note C# is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this chord in 1st inversion is 5/4/2, with the 5 placed above the 4, and the 4 placed above the 2 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be G-sharp major 7th suspended 4th triad in five-four-two position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original chord root!), and the note in question.
For example, the 5 represents note G#, from the C#-5th interval, since the lowest (bass) note of the chord - now inverted, is C#.
In the same way, the figured bass 4 symbol represents note F##, from the C#-4th interval, and the 2 symbol represents note D#, from the C#-2nd interval
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 9. G-sharp major 7th suspended 4th 2nd inversion
This step shows the second inversion of the G-sharp major 7th suspended 4th.
For a 2nd inversion, take the first note of the 1st inversion above - C#, and move it to the end of the chord.
So the second note of the 1st inversion - note D# is now the note with the lowest pitch for the 2nd inversion.
Or put another way, the third note of the original 7th chord (in root position) is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this chord in 2nd inversion is 7/4/3, with the 7 placed above the 4, and the 4 placed above the 3 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be G-sharp major 7th suspended 4th triad in seven-four-three position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original chord root!), and the note in question.
For example, the 7 represents note C#, from the D#-7th interval, since the lowest (bass) note of the chord - now inverted, is D#.
In the same way, the figured bass 4 symbol represents note G#, from the D#-4th interval, and the 3 symbol represents note F##, from the D#-3rd interval
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 10. G-sharp major 7th suspended 4th 3rd inversion
This step shows the third inversion of the G-sharp major 7th suspended 4th.
For a 3rd inversion, take the first note of the 2nd inversion above - D#, and move it to the end of the chord.
So the second note of the 2nd inversion - note F## is now the note with the lowest pitch for the 3rd inversion.
Or put another way, the fourth note of the original 7th chord (in root position) is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this chord in 3rd inversion is 6/5/2, with the 6 placed above the 5, and the 5 placed above the 2 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be G-sharp major 7th suspended 4th triad in six-five-two position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original chord root!), and the note in question.
For example, the 6 represents note D#, from the F##-6th interval, since the lowest (bass) note of the chord - now inverted, is F##.
In the same way, the figured bass 5 symbol represents note C#, from the F##-5th interval, and the 2 symbol represents note G#, from the F##-2nd interval
In 3rd inversion, often the 6 symbol is not shown at all, as it is assumed.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
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# Thread: Solving differential equations by inspection!
1. ## Solving differential equations by inspection!
I have this question 1. Solve the following DE’s, by inspection.
y’= -sin x
I Know that y=cos x and y'= -sin x. Then is y=cos x the answer? What does it mean when solve by inspection???
2. ## Re: Solving differential equations by inspection!
Solving by inspection involves exactly what you did (well, almost, you forgot the constant of integration)...you know that:
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\cos(x)+C)=-\sin(x)$ and so we must have:
$\displaystyle y=\cos(x)+C$
3. ## Re: Solving differential equations by inspection!
Oh thanks for the reply!
Then let's say for y'' = 3 and I did. y'=3x+c
y= (3/2)x^2 +C
Then is y= (3/2)x^2 + C answer for y"=3?
Thank you.
4. ## Re: Solving differential equations by inspection!
No, for the second example, you should observe that:
$\displaystyle \frac{d^2}{dx^2}\left(\frac{3}{2}x^2+c_1x+c_2 \right)=3$ hence:
$\displaystyle y=\frac{3}{2}x^2+c_1x+c_2$
,
,
### how to solve differential equations by inspection
Click on a term to search for related topics.
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https://socratic.org/questions/what-is-the-net-area-between-f-x-x-sinx-and-the-x-axis-over-x-in-0-3pi
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# What is the net area between f(x) = x-sinx and the x-axis over x in [0, 3pi ]?
May 7, 2018
int_0^(3π)(x-sinx)dx=((9π^2)/2-2) m^2
#### Explanation:
$f \left(x\right) = x - \sin x$ , $x$$\in$$\left[0 , 3 \pi\right]$
$f \left(x\right) = 0$ $\iff$ $x = \sin x$ $\iff$ $\left(x = 0\right)$
(Note: $| \sin x | \le | x |$ , $\forall$$x$$\in$$\mathbb{R}$ and the $=$ is true only for $x = 0$)
• $x > 0$ $\iff$ $x - \sin x > 0$ $\iff$ $f \left(x\right) > 0$
So when $x$$\in$$\left[0 , 3 \pi\right]$ , $f \left(x\right) \ge 0$
Graphical help
The area we are looking for since $f \left(x\right) \ge 0$ ,$x$$\in$$\left[0 , 3 \pi\right]$
is given by int_0^(3π)(x-sinx)dx $=$
int_0^(3π)xdx - int_0^(3π)sinxdx $=$
[x^2/2]_0^(3π)+[cosx]_0^(3π) $=$
(9π^2)/2+cos(3π)-cos0 $=$
((9π^2)/2-2) ${m}^{2}$
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-1-equations-and-inequalities-1-1-linear-equations-1-1-exercises-page-93/34
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## Precalculus (6th Edition)
Published by Pearson
# Chapter 1 - Equations and Inequalities - 1.1 Linear Equations - 1.1 Exercises - Page 93: 34
#### Answer
$x = \frac{19}{5} = 3.8$
#### Work Step by Step
$-8(x+3) = -8x - 5(x+1)$ $-8x - 24 = - 8x - 5x -5$ $-8x-24=-13x-5$ $-8x-24+13x=-13x-5+13x$ $5x-24=-5$ $5x-24+24=-5+24$ $5x=19$ $x=\frac{19}{5}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Vector cross Product
1. Jan 26, 2010
### xcgirl
1. The problem statement, all variables and given/known data
C= B|A| + A|B|
D= A|B|-B|A|
C and D are orthogonal
Find a third vector perpendicular to both C and D
2. Relevant equations
[AxB] = |A||B|sin(theta)
3. The attempt at a solution
I know that to find the answer I need to find the cross product of C and D. I have done similar problems, but the unit components (i,j,k) have always been given. I can't figure out a way to do this without having those.
Thanks for any help, even a hint!
2. Jan 26, 2010
### rl.bhat
CXD = (B|A| + A|B|)X( A|B|-B|A|)
Do the cross multiplication of right hand side. Note that AXA = BXB = 0 And AXB = -BXA
You can multiply the magnitudes of A and B directly.
3. Jan 26, 2010
### xcgirl
so I can do...
BxA|A||B| + BxB|A||A| + AxA|B||B| - AxB|B||A|
BxA|A||B| - AxB|B||A|
so would that be the final answer? thanks for the help, this all just seems a little weird to me. I didnt know that you could just essentially "foil" it like that
4. Jan 26, 2010
### rl.bhat
Last step
CXD = 2(AB)*BXA, because AXB = BXB = 0
5. Jan 26, 2010
### xcgirl
thanks, you're a lifesaver! i totally get it now
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# Electric Potential and Potential Difference
The definition of electrical potential is that the ability of charged particles to try to do the work. The unit of electrical potential is volts.
When 2 equally charged particles are brought close to, they fight to repel one another whereas dissimilar charges attract one another. This means, each charged particle incorporates a tendency to try to do work.
The electrical potential at a purpose owing to a charge "is one potential unit if one joule of work is done in delivery a unit positive charge i.e. positive charge of 1 coulomb from time thereto purpose,
Mathematically it's expressed as,
Electric potential = Work done / charge
= W/Q
In electrical circuits flow of current is usually from higher electric potential to lower electric potential. I.e. The distinction between the electrical potential at any 2 offer points in a very circuit is thought as potential. This can be known as between the 2 points and measured in volts. It’s denoted as V.
For illustration, let the electric potential of a charged particle A is say V1 whereas the electric potential of a charged particle B is say V2. Then the potential between the 2 particles A and B is V1-V2. If V1-V2 is positive we are saying that A is at higher potential than B whereas if V1-V2 is negative we are saying that B is at higher potential than A.
Consider 2 points having potential of V volts between them, as shown within the fig. The point A is at upper potential than B. As per the definition of potential unit, the V joule of work is to be performed to maneuver unit charge from point B to point A.
Thus, once such 2 points, that are at totally different potentials are joined in conjunction with the assistance of wire, the electrical current flows from higher potential to lower potential i.e. the electrons begin flowing from lower potential to higher potential thus, to take care of the flow of electrons i.e. flow of electric current, there should exist a potential between the 2 points.
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By: Doris (offline) Tuesday, October 26 2010 @ 04:40 AM CDT (Read 2563 times)
Doris
Pls can some one help to provide the working and answer for my Pri. 5 boy math problem sum :
There were two identical flights of steps. For the first Flight of steps, Albus walked up some steps and ran 4 steps, and took a total of 75 seconds. For the second flight of steps, Albus walked up some steps and ran 11 steps, and took a total of 40 seconds. How long will Albus take if he had walked up BOTH flights of steps? Leave answer in seconds.
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By: CanCan (offline) Tuesday, October 26 2010 @ 11:19 AM CDT
CanCan
[/img]
First you draw 2 equal models to represent the number of identical flight of stairs. Note: this means that there are equal number of steps in each flight of stairs. For the top model, draw 4 run steps (shown in red) and name the remainding as steps that were walked (in pink and yellow). The bottom model, draw 11 steps (shown in red) and name the remainding as steps that were walked (in pink).
Since running steps will take the same time, you can cancel the 4 red units at the top with the bottom 4 red unit. Similiarly, since the walking steps take the same time, you can cancel the yellow units on top with the bottom one. The yellow unit represents a certain unknown number but they have to be the same number.
Thus when you take 75 - 40 = 35
This leaves you with the difference caused by running the 7 steps in red versus the walking steps in pink. In other words, walking the 7 steps take 35 seconds more than running the 7 steps. Note: the pink units also represents 7 steps as you have begin with the same number of stairs
35/7=5 Walking each step takes 5 sec more than running each step.
To find out how long he take to finish walking one flight of steps, change the 4 running steps to walking by adding a 5 sec each. Then add it to the 75 sec.
5x4= 20
20+75=95
2 flights of stairs = 95x2=190sec
Full time tutor,
CanCan
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# Search by Topic
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Broad Topics > Algebraic expressions, equations and formulae > Quadratic equations
### Proof Sorter - Quadratic Equation
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### Target Six
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Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
### Square Mean
##### Age 14 to 16 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
##### Age 16 to 18 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
### Continued Fractions II
##### Age 16 to 18
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
### Interactive Number Patterns
##### Age 14 to 16 Challenge Level:
How good are you at finding the formula for a number pattern ?
### Golden Eggs
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Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
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Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
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### Always Two
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### Two Cubes
##### Age 14 to 16 Challenge Level:
Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to. . . .
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### In Between
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##### Age 14 to 16 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
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### Golden Mathematics
##### Age 16 to 18
A voyage of discovery through a sequence of challenges exploring properties of the Golden Ratio and Fibonacci numbers.
### Golden Construction
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Draw a square and an arc of a circle and construct the Golden rectangle. Find the value of the Golden Ratio.
### Golden Fibs
##### Age 16 to 18 Challenge Level:
When is a Fibonacci sequence also a geometric sequence? When the ratio of successive terms is the golden ratio!
### Xtra
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Find the sides of an equilateral triangle ABC where a trapezium BCPQ is drawn with BP=CQ=2 , PQ=1 and AP+AQ=sqrt7 . Note: there are 2 possible interpretations.
### Darts and Kites
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### Good Approximations
##### Age 16 to 18 Challenge Level:
Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.
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Two perpendicular lines are tangential to two identical circles that touch. What is the largest circle that can be placed in between the two lines and the two circles and how would you construct it?
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A bag contains red and blue balls. You are told the probabilities of drawing certain combinations of balls. Find how many red and how many blue balls there are in the bag.
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Sketch the graphs for this implicitly defined family of functions.
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##### Age 14 to 16 Challenge Level:
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
### Pentakite
##### Age 14 to 18 Challenge Level:
ABCDE is a regular pentagon of side length one unit. BC produced meets ED produced at F. Show that triangle CDF is congruent to triangle EDB. Find the length of BE.
### Implicitly
##### Age 16 to 18 Challenge Level:
Can you find the maximum value of the curve defined by this expression?
### Halving the Triangle
##### Age 16 to 18 Challenge Level:
Draw any triangle PQR. Find points A, B and C, one on each side of the triangle, such that the area of triangle ABC is a given fraction of the area of triangle PQR.
### Golden Ratio
##### Age 16 to 18 Challenge Level:
Solve an equation involving the Golden Ratio phi where the unknown occurs as a power of phi.
##### Age 16 to 18 Short Challenge Level:
Can you solve this problem involving powers and quadratics?
### A Third of the Area
##### Age 14 to 16 Short Challenge Level:
The area of the small square is $\frac13$ of the area of the large square. What is $\frac xy$?
### Resistance
##### Age 16 to 18 Challenge Level:
Find the equation from which to calculate the resistance of an infinite network of resistances.
### Bird-brained
##### Age 16 to 18 Challenge Level:
How many eggs should a bird lay to maximise the number of chicks that will hatch? An introduction to optimisation.
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# .
In mathematics, given a local field K, such as the fields of reals or p-adic numbers, whose multiplicative group of non-zero elements is K×, the Hilbert symbol is an algebraic construction, extracted from reciprocity laws, and important in the formulation of local class field theory. As the name suggests, it was in some sense introduced by David Hilbert, although it would be anachronistic to say that of the local field formulation.
Explicitly, it is the function (–, –) from K× × K× to {−1,1} defined by
$$(a,b)=\begin{cases}1,&\mbox{ if }z^2=ax^2+by^2\mbox{ has a non-zero solution }(x,y,z)\in K^3;\\-1,&\mbox{ if not.}\end{cases}$$
Contents
Properties
The following three properties follow directly from the definition, by choosing suitable solutions of the diophantine equation above:
• If a is a square, then (a, b) = 1 for all b.
• For all a,b in K×, (a, b) = (b, a).
• For any a in K× such that a−1 is also in K×, we have (a, 1−a) = 1.
The (bi)multiplicativity, i.e.,
(a, b1b2) = (a, b1)·(a, b2)
for any a, b1 and b2 in K× is, however, more difficult to prove, and requires the development of local class field theory.
The third property ensures that the Hilbert symbol factors over the second Milnor K-group $$K^M_2 (K)$$, which is by definition
K×K× / (a ⊗ 1−a, aK× \ {1})
By the first property it even factors over $$K^M_2 (K) / 2$$. This is the first step towards the Milnor conjecture.
Interpretation as an algebra
The Hilbert symbol can also be used to denote the central simple algebra over K with basis 1,i,j,k and multiplication rules $$i^2=a, j^2=b, ij=-ji=k$$. In this case the algebra represents an element of order 2 in the Brauer group of K, which is identified with -1 if it is a division algebra and +1 if it is isomorphic to the algebra of 2 by 2 matrices.
Hilbert symbols over the rationals
For a place v of the rational number field and rational numbers a, b we let (a, b)v denote the value of the Hilbert symbol in the corresponding completion Qv. As usual, if v is the valuation attached to a prime number p then the corresponding completion is the p-adic field and if v is the infinite place then the completion is the real number field.
Over the reals, (a, b) is +1 if at least one of a or b is positive, and −1 if both are negative.
Over the p-adics with p odd, writing $$a = p^{\alpha} u$$ and $$b = p^{\beta} v$$, where u and v are integers coprime to p, we have
$$(a,b)_p = (-1)^{\alpha\beta\epsilon(p)} \left(\frac{u}{p}\right)^\beta \left(\frac{v}{p}\right)^\alpha$$, where $$\epsilon(p) = (p-1)/2$$
and the expression involves two Legendre symbols.
Over the 2-adics, again writing $$a = 2^\alpha u and b = 2^\beta v$$, where u and v are odd numbers, we have
$$(a,b)_2 = (-1)^{\epsilon(u)\epsilon(v) + \alpha\omega(v) + \beta\omega(u)}$$ , where $$\omega(x) = (x^2-1)/8.$$
It is known that if v ranges over all places, (a, b)v is 1 for almost all places. Therefore the following product formula
$$\prod_v (a,b)_v = 1$$
makes sense. It is equivalent to the law of quadratic reciprocity.
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Accuracy is up to 5 decimal places for meters to feet and to the nearest 16th of an inch for meters to feet and inches. The conversion factor from meters to feet is 3.2808398950131, which means that 1 meter is equal to 3.2808398950131 feet: 1 m = 3.2808398950131 ft. To convert 5.2 meters into feet we have to multiply 5.2 by the conversion factor in order to get the length amount from meters to feet. How many candles are on a Hanukkah menorah? 5.3 centimeter in inch. Who is the longest reigning WWE Champion of all time? Step 2: Convert the decimal feet to inches The total distance d in inches (in) is equal to the distance d in meters (cm) divided by 0.0254:. Once this is very close to 3.28 feet, you will almost always want to use the simpler number to make the math easier. The meter is the base unit in the International System of Units and is equal to the distance travelled by light through a vacuum in 1/299,792,458 seconds. Copyright © 2020 Multiply Media, LLC. feet to meters formula. How much is 3 meters in feet and inches and centimeters? m * 100 cm 1 m * 1 in 2.54 cm * 1 meter is equal to 3.28084 feet: 1 m = (1/0.3048) ft = 3.28084 ft. 7.5 meters equal 24.6062992126 feet (7.5m = 24.6062992126ft). 1 meters to feet = 3.28084 feet. In this case we should multiply 5 Feet by 0.3048 to get the equivalent result in Meters: 5 Feet x 0.3048 = 1.524 Meters. Simply use our calculator above, or apply the formula to change the weight 5.5 … To find the inch value, multiply the fractional part by 12. How many candles are on a Hanukkah menorah? Meters to feet conversion table 5.5 meters equal 18.0446194226 feet (5.5m = 18.0446194226ft). The distance d in feet (ft) is equal to the floor value of the distance d in inches (in) divided by 12: How far? 5 Meters to Feet Conversion Meters to Feet - Distance and Length - Conversion. The distance d in feet (ft) is equal to the distance d in meters (m) divided by 0.3048: d (ft) = d (m) / 0.3048. How short? How big is 5'5 in other units? How big? In other words, 5 meters is 0.5970 times the length of a London bus, and the length of a London bus is 1.680 times that amount. A meter , or metre, is the fundamental unit of length in the metric system, from which all other length units are based. In this case we should multiply 5 Meters by 3.2808398950131 to get the equivalent result in Feet: 5 Meters x 3.2808398950131 = 16.404199475066 Feet 5 Meters is equivalent to 16.404199475066 Feet. One meter is a length measurement, equal to 3.28 feet. Easily convert Meters to feet, with formula, conversion chart, auto conversion to common lengths, more. One meter is, therefore, slightly longer than a yardstick. Why don't libraries smell like bookstores? The distance d in feet (ft) is equal to the floor value of the distance d in inches (in) divided by 12: 5.3 centimeter in feet. Meters to Feet formula If you stretched out your large intestine, it would be about as long as the width of a queen size bed. Area of a path, decking or patio to help estimate how long it will take to pressure wash. 5 Feet is equivalent to 1.524 Meters. swap units ↺ Amount. cm: in: m: mm: 5feet5 and a quarter inch in cm: 165.735 cms: 5 foot 5 and a half inch in cm: 166.37 centimeters: 5ft5 and three quarters of an inch in cm: 167.005 centimeters: 5foot5 in meters: 1.65735 meters Today we're going to look at the conversion of meters to feet.Both of these are units of length. 1 foot is exactly 0.3048 meters. 1 meter is equal to 3.2808 feet: 1 m = 3.2808 ft. 1 meter is equal to 39.37 inches: 1 m = 39.37 in. When did organ music become associated with baseball? The large intestine is about 5 feet (1.5 meters) long. Data should be separated by coma (,), space ( ), tab, or in separated lines. 1 m is equivalent to 1.0936 yards, or 39.370 inches. 1 meter is equal to 3.2808 feet: 1 m = 3.2808 ft. 1 meter is equal to 39.37 inches: 1 m = 39.37 in. The total distance d in inches (in) is equal to the distance d in meters (cm) divided by 0.0254:. If you need to be super precise, you can use one meter = 3.2808398950131 feet. All Rights Reserved. One meter is a length measurement and equals approximately 3.28 feet. Definitions. It is equal to 100 centimeters, 1/1000th of a kilometer, or about 39.37 inches. Combinations. Convert 3 meters to centimeters 3 = 300 cm Convert 3 meters to feet and inches 3 meters = 9 feet and 10.11 inches Convert 3 meters to inches 3 = 118.11 inches Convert 3 meters to feet The following equation allows the simple conversion between feet and meters: meter = feet / 3.2808. 2.5 meters equals 8.202 feet because 2.5 times 3.281 (the conversion factor) = 8.202 All In One Unit Converter Please, choose a physical quantity, two units, then type a value in any of the boxes above. How long is 27 meters? Common building codes generally suggest that the nosing have a minimum length of 0.75 inches (1.9 cm) and a maximum length of 1.25 inches (3.2 cm). 27 m to ft conversion. Convert 5 meters to centimeters 5 = 500 cm Convert 5 meters to feet and inches 5 meters = 16 feet and 4.85 inches Convert 5 meters to inches 5 = 196.85 inches Convert 5 meters to feet To convert 5.5 Meters to Feet you have to multiply 5.5 by 3.2808398950131, since 1 Meter is 3.2808398950131 Feet. cm: in: m: mm: 5feet10 and a quarter inch in cm: 178.435 cms: 5 foot 10 and a half inch in cm: 179.07 centimeters: 5ft10 and three quarters of an inch in cm: 179.705 centimeters: 5foot10 in meters: 1.78435 meters d (in) total = d (m) / 0.0254. How many inches in 5 meters? How to convert 5.5 Meters to Feet. You are currently converting Distance and Length units from Meters to Feet. How far is 27 meters in feet? How far is 5 meters in feet? 5.5 meters equals 18.04 feet because 5.5 times 3.281 (the conversion factor) = 18.04 5.3 meters is how many feet. 1 Meter is equal to 3.2808398950131 Feet. 10 meters to feet = 32.8084 feet. Convert 5 Feet to Meters. Converting 7.5 m to ft is easy. ›› Quick conversion chart of meters to feet. Convert Feet to Meters. To calculate 5 Feet to the corresponding value in Meters, multiply the quantity in Feet by 0.3048 (conversion factor). The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. What are the release dates for The Wonder Pets - 2006 Save the Ladybug? Copyright © 2020 Multiply Media, LLC. To convert feet to meters, multiply the foot value by 0.3048 or divide by 3.280839895. Why don't libraries smell like bookstores? 5 meters x 3.28 = 16.4 feet; 2.7 meters x 3.28 = 8.856 feet ... A cubic meter is an amount of volume equal to a cube one meter long, one meter wide, and one meter tall. Meters: Feet and Inches (No Input): Note: Fill in one box to get results in the other box by clicking "Calculate" button. Accuracy is up to 5 decimal places for meters to feet and to the nearest 16th of an inch for meters to feet and inches. 5 meter sticks 1 meter, is 3.28,so that time 5 about 3 five foot people stand head to toe sio its 15 feet long Direct Conversion Formula5 m* 1 ft 0.3048 m = 16.40419948 ft How long will the footprints on the moon last? If you need to be super precise, you can use one meter = 3.2808398950131 feet. X (ft) = Y (m) ÷ 0.3048 How to convert from Feet to Meters How big is 5'10 in other units? Convert cm, km, miles, yds, ft, in, mm, m. 5 m to cm: 5 m to feet: 5 m to in: 5 m to km: 5 m to miles: 5 m to mm: 5 m to yd: How much is 5 meters in feet? The foot is the standard unit of length in the system of measurement used primarily in the United States. 1 Meter (m) is equal to 3 feet and 3.3700787 inches. What are some samples of opening remarks for a Christmas party? We can also form a simple proportion to calculate the result: The conversion factor from meters to feet is 3.2808398950131, which means that 1 meter is equal to 3.2808398950131 feet: 1 m = 3.2808398950131 ft To convert 50.5 meters into feet we have to multiply 50.5 by the conversion factor in order to get the length amount from meters to feet. What does contingent mean in real estate? The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. One meter equals to the length of the path that a light travels in vacuum for the time of 1/299,792,458 second. The integer part of the result is the foot value. One foot equals 12 inches exactly. A meter, or metre, is the fundamental unit of length in the metric system, from which all other length units are based. How big? How far is 5 meters? How long is 5 feet 5 inches in meters? One foot equals 12 inches exactly. A meter is a SI unit scientifically accepted as the base unit of distance and length. Note that rounding errors may occur, so always check the results. What does contingent mean in real estate? Step 1: Convert from meters to feet. Who is the longest reigning WWE Champion of all time? Meter Conversion: 1 meter = 100 centimeters → meters to centimeters; 1 meter = 3.28 feet → meters to feet and inches; 1 meter = 39.37 inches → meters to inches; 1 meter = 0.001 kilometer 1 meter = 1000000000 nanometers → meters to nanometers; 1 meter = 1.0936 yards → meters to yards; Mile Conversion: 1 mile = 5280 feet → miles to feet It gives the conversion results of meters to feet based on a range of 0.01m to 100m. Convert 2 meters to feet: d (ft) = 2m / 0.3048 = 6.5617ft. A meter, or metre, is the fundamental unit of length in the metric system, from which all other length units are based. 5 m to ft conversion. How long is it? 1 Foot = 0.3048 Meter. 20 meters to feet = 65.6168 feet. How much is 5 meters in feet and inches and centimeters? Lay the meter stick on the ground, and place the rulers end-to-end next to it. (Routemaster Double-Decker, RM standard specification) The Routemaster Double-Decker buses, well-known as icons of London, measure 8.38 m in length. people stand head to toe sio its 15 feet long. It is equal to 100 centimeters, 1/1000th of a kilometer, or about 39.37 inches. All Rights Reserved. 2. Learn that one meter equals 3.28 feet. 1 meter = 3.28 x feet, so, 0.5 x 1 meter = 0.5 x 3.28 feet, or. 5 Meters (m) = 16.4042 Feet (ft) Meters : The meter (symbol m) is the fundamental unit of length in the International System of Units (SI). A meter is a length unit in the metric system defined as the distance travelled by light in 1/299792458 seconds. A foot is a unit of length equal to exactly 12 inches or 0.3048 … How to convert meters to feet+inches. 5 meters = 196.850394 inches12 inches = 1 footTherefore,196.85 divided by 12 =16 feet 4 inches (rounded) Full wave loop (feet) = 1005 / frequency in Mhz Full wave loop (meters) = 306.32 / frequency in Mhz Cut wire slightly longer for connecting insulators and pruning. Along with other units like a kilometer or an inch, a meter is one of the fundamental units in SI. Type in your own numbers in the form to convert the units! Meters. 5 meters to feet = 16.4042 feet. To calculate 5 Meters to the corresponding value in Feet, multiply the quantity in Meters by 3.2808398950131 (conversion factor). 1 m = 3.2808398950131 ft The main purpose of a nosing is to improve safety by providing extra space on which a person can place their feet. How wide? What is 5 meters in inches, feet, meters, km, miles, mm, yards, etc? There are 3.2808398950131 feet in a meter. To convert directly between meters and feet or meters and inches (plus many other units of length, distance and height), please use the length and distance converter. Example. 1 metre is equal to 1 meters, or 3.2808398950131 feet. You will also find a converter for centimeters to feet and inches here . Number of building plots of a length and width are possible on an area of land in hectares or acres. 5.3 centimeter in meter. d (in) total = d (m) / 0.0254. One meter equals 39.37 inches, which is approximately 3 feet 3 inches. meter = feet * 0.3048. meter = feet / 3.280839895. 1 meter to feet is 3.28084 ft = 3 ft, 3 3/8 in 2 meters to feet is 6.56168 ft = 6 ft, 6 3/4 in Take a look at the articles section for a feature on how many feet are in a meter. Once this is very close to 3.28 feet, you will almost always want to use the simpler number to make the math easier. How tall is 5 meters? 40 meters to feet = 131.2336 feet. Converting 5.5 m to ft is easy. To. For example, to calculate how many meters is 50 feet, multiply 50 by 0.3048, that makes 15.24 meters is 50 feet. 5.3 centimeter in centimeter. From To Link to this page. Meters to Feet and Inches Conversion in Batch. From. 5 meters is 16 feet and 4.85 inches. 5 feet 5 inch = 1.651 metres How long is 1.58m in feet? When did organ music become associated with baseball? 5.3 centimeter in mile. What is 5 meters in feet? How wide? Three rulers (3 feet) will almost be as long as the meter stick. meters (m) feet (ft) Swap == > 1 m = 3.2808399 ft : 1 ft = 0.3048 m: Algebraic Steps / Dimensional Analysis Formula. A foot is a unit of length equal to exactly 12 inches or 0.3048 meters. How short? What is a sample Christmas party welcome address? Similarly, a cubic foot (ft 3) is equal to a cube one foot long, one foot wide, and one foot tall. 5 meter sticks 1 meter, is 3.28,so that time 5 about 3 five foot Since 1983, the metre has been officially defined as the length of the path travelled by light in a vacuum during a time interval of 1/299,792,458 of a second. 25 meters to feet = 82.021 feet. How narrow? The result is the following: 5.5 m × 3.2808398950131 = 18.045 ft. 5.5 m = 18.045 ft. We conclude that five point five Meters is equivalent to eighteen point zero four five Feet: 27 Meters = 88.582677 Feet (rounded to 8 digits) Display result as. How long? How to convert meters to feet. To convert meters to feet and inches, first multiply the meter value by 3.2808399 to convert into feet. 1 meter to feet is 3.28084 ft = 3 ft, 3 3/8 in 2 meters to feet is 6.56168 ft = 6 ft, 6 3/4 in To convert meters to feet, divide your figure by 0.3048. ››. 0.5 meter = 1.64 feet. 15 meters to feet = 49.2126 feet. How long? 2.5 meters to feet can also be determined using the Meters to Feet conversion table. How tall is 3 meters? 5.3 centimeter in millimeter. Find additional conversions: Meters to feet converter. Floor, wall or ceiling area of a reception room, lounge, kitchen, living room, dining room, hallway, staircase, bedroom, bathroom, toilet, study, tv room, garage cellar or a basement storage room. 50 meters to feet = 164.04199 feet. 30 meters to feet = 98.4252 feet. Simply use our calculator above, or apply the formula to change the weight 7.5 m to ft. You can test this using a meter stick and 1 foot (12 inch) rulers. 5.3 centimeter in kilometer. 5.3 centimeter in yard. Use this page to learn how to convert between metres and feet. How to convert meters to feet+inches. How far? How narrow? 15.24 meters is 50 feet is about 5 feet 5 inches in meters ( cm ) by! Close to 3.28 feet, you will almost always want to use the simpler number make... 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You have to multiply 5.5 by 3.2808398950131, since 1 meter = 3.28 x feet, multiply foot... - conversion meter ( m ) is equal to 100 centimeters, 1/1000th of a nosing is to safety.
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1.3 Order of Operations ( ) + X - 4343 . The Order of Operations tells us how to do a math problem with more than one operation, in the correct order.
Presentation on theme: "1.3 Order of Operations ( ) + X - 4343 . The Order of Operations tells us how to do a math problem with more than one operation, in the correct order."— Presentation transcript:
1.3 Order of Operations ( ) + X - 4343
The Order of Operations tells us how to do a math problem with more than one operation, in the correct order.
P lease E xcuse M y D ear A unt S ally This will help to you to remember the order of operations.
Add + Subtract - Multiply x Divide Please Excuse My Dear Aunt Sally P E MDMD ASAS Parentheses ( ) Exponents 4 3
Please Excuse My Dear Aunt Sally Parentheses ( ) Always do parentheses 1 st.
Please Excuse My Dear Aunt Sally Exponents 4 3 Always do Exponents 2 nd.
Multiply x Divide Please Excuse My Dear Aunt Sally Do multiplication and division 3 rd, from left to right.
Add + Subtract - Please Excuse My Dear Aunt Sally Do addition and subtraction 4th, from left to right.
Let’s Try Some Problems!
PEMDAS 3+2 3 – (9+1) 3+2 3 – 10 3+8-10 11-10 1
PEMDAS 3 (9+1) + 6 2 3(10)+6 2 3(10)+36 30+36 66
PEMDAS 4+5 (6-2) 4+5 (4) 4+20 24
PEMDAS 4+ 10 ( 2 3 )-16 4+10(8) -16 4+ 80 -16 84-16 68
PEMDAS 21 + 10 2 10 21+100 10 21 + 10 31
PEMDAS 10+7 2 -2 (5) 10+49–2(5) 10+49- 10 59 - 10 49
PEMDAS 64 [9(3)–19] 64 (27 –19) 64 8 8
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# Basic Maths with Buttons
Coloured buttons or counters are perfect to use for hands on Maths practice. Whether for counting, adding, subtracting or dividing it’s much easier for most children to work out and understand these concepts with something tangible in front of them. They are also ideal for sorting games.
We’re taking part in the Manipulatives series over at School Time Snippets where 21 bloggers have joined together to show how maths can be made fun and hands on using simple supplies found in the home.
The possibilities are endless for Maths games with manipulatives like this. First we used the buttons to practice sorting by attributes of colour and shape. You could also sort by size, or other differing features.
Although this is a simple task, children can expand on the activity by estimating beforehand which will be the largest group, or even making an estimate of how many they think will be in each group and then checking their answer by counting. When estimating you can talk about rounding up and down accurately, and talk about the difference between an estimate and a guess.
It’s interesting for children to consider the different ways of categorising each button too, whether by it’s shape, colour or size. You could take this further by drawing out a venn diagram and laying the buttons out on top, to look at how each one can fit into overlapping categories.
Using buttons or similar manipulatives is also great for getting hands on with simple addition, subtraction and division as I mentioned. My boys enjoyed working together on this, with each one setting problems for the others. My younger boys definitely find it easier to work out the simple sums when using tools like these and it helps to build their confidence as they see themselves as able to add together much larger numbers compared to what they would manage just with mental calculations.
With such an open ended material, the children can adapt the maths problems to their level and work with the tools in the way that best suits them, which is great.
The plastic buttons that we used for this activity are very similar to these on Amazon. I like this type since they are quite chunky and have a good variety of shapes.
For more hands on Maths ideas with manipulatives, do check out the Manipulatives series over at School Time Snippets, and follow our STEM board on Pinterest for more Maths (along with the other STEM topics)
Follow Anna – In The Playroom’s board STEM : Science Technology Math and Engineering for Kids on Pinterest.
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## Father’s Day Handprint Craft
### 2 thoughts on “Basic Maths with Buttons”
1. I never thought to use buttons but what a great idea so many to choose from to make it fun and inexpensive, thank you for the inspiration!
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# CLASS-2CIRCLE THE SMALLEST NUMBER
CIRCLE THE SMALLEST NUMBER
1) 164, 234, 216
Ans.) In above given number series the numbers are given 164, 234, 216. If we compare hundred place of three given numbers then we can find 1 from number 164 is smaller than 2 (1 < 2) from 234 and 216 respectively. So, here we can conclude that 164 is the smallest number between three given numbers.
2) 345, 567, 322
Ans.) In above given number series the numbers are given 345, 567, 322. If we compare hundred place of three given numbers then we can find 3 from number 345 and 322 is smaller than 5 (3 < 5) from 567. We can understand here 345 & 322 are small than 567. Now we will compare between tens place of the numbers 345 & 322. 4 from 345 is greater than 2 from 322. So, here we can conclude that 322 is smallest number between three given numbers.
3) 451, 532, 231
Ans.) In above given number series the numbers are given 451, 532, 231. If we compare hundred place of three given numbers then we can find 2 from number 231 is smaller than 4 (2 < 4) and 5 (2 < 5) from 451 & 532 respectively. So, here we can conclude that 231 is smallest number between three given numbers.
4) 425, 124, 130
Ans.) In above given number series the numbers are given 425, 124, 130. If we compare hundred place of three given numbers then we can find 1 from numbers 124 & 130 is smaller than 4 (1 < 4) from 425. There are two numbers 124 & 130 where 1 is in hundreds place so, very difficult to identify by hundreds place. Now we will consider tens place where 2 from 124 is smaller than 3 (2 < 3) from 130, so 124 is smaller than 130. Here we can conclude that 124 is smallest number between three given numbers.
5) 234, 389, 127
Ans.) In above given number series the numbers are given 234, 234, 127. If we compare hundred place of three given numbers then we can find 1 from number 127 is smaller than 2 (1 < 2) from 234 & and also smaller than 3 (1 < 3) from 389. Here we can conclude that 127 is smallest number between three given numbers.
6) 357, 259, 453
Ans.) In above given number series the numbers are given 357, 259, 453. If we compare hundred place of three given numbers then we can find 2 from number 259 is smaller than 3 (2 < 3) from 357 & and also smaller than 4 (2 < 4) from 453. Here we can conclude that 259 is smallest number between three given numbers.
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# Shark Attack Internal Analysis
3648 Words15 Pages
Mathematics Internal Assessment - Exploration Investigating the trends or patterns in shark attack activity over the past 13 years (from 2001 to 2013) in the United States of America and looking at the possible outcomes of the the activity increasing or decreasing in the next few years. Author: Cheyenne Hylton SUBJECT: MATHEMATICS TEACHER’S NAME: MS. STEELE CANDIDATE NAME: CHEYENNE HYLTON CANDIDATE NUMBER: 003311 0013 MATHEMATICS LEVEL: STANDARD LEVEL (SL) Introduction On this planet there are a handful of species that are at the top of the food chain, but which species are really at the top? Sharks and human beings both exist at the top of the food chain but both species of mammals find it difficult to coexist, more specifically in a…show more content…
So, if it greater than 40 it has to be 41. The Median The median is the middle value of an ordered data set. If there are n values, ‘n’ being the number of values in a set of data, listed in order from smallest to largest, the median is the ((n+1)/2)th the data value. Step 1: Use the formula to figure out the median of data. Median formula = ((n+1)/2) n = the number of values Therefore (13+1)/2 = 7 The 7th values when put in a numerical increasing order is 38, thus the median is 38. Mode The mode is the number that occurs most frequently in the data. In figure 1.5, the table has the total attacks rounded off to the nearest tens is done for simplicity and can assist in finding the mode. The mode is the most frequently occuring value in the data set. The max mode using the total amount of attacks rounded off to the nearest tens is 50 and 40. The min mode using the total amount of attacks rounded off to the nearest tens is 30. So the mode at: The mode at 50 = occurs 5 times The mode at 40 = occurs 5 times The mode at 30 = occurs 3 times A Bar Graph Illustrating the Mode of the data (Figure
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1
JEE Advanced 2022 Paper 1 Online
+3
-1
Let $$p, q, r$$ be nonzero real numbers that are, respectively, the $$10^{\text {th }}, 100^{\text {th }}$$ and $$1000^{\text {th }}$$ terms of a harmonic progression. Consider the system of linear equations
$$\begin{gathered} x+y+z=1 \\ 10 x+100 y+1000 z=0 \\ q r x+p r y+p q z=0 \end{gathered}$$\$
List-I List-II
(I) If $$\frac{q}{r}=10$$, then the system of linear equations has (P) $$x=0, \quad y=\frac{10}{9}, z=-\frac{1}{9}$$ as a solution
(II) If $$\frac{p}{r} \neq 100$$, then the system of linear equations has (Q) $$x=\frac{10}{9}, y=-\frac{1}{9}, z=0$$ as a solution
(III) If $$\frac{p}{q} \neq 10$$, then the system of linear equations has (R) infinitely many solutions
(IV) If $$\frac{p}{q}=10$$, then the system of linear equations has (S) no solution
(T) at least one solution
The correct option is:
A
(I) $$\rightarrow$$ (T); (II) $$\rightarrow$$ (R); (III) $$\rightarrow$$ (S); (IV) $$\rightarrow$$ (T)
B
(I) $$\rightarrow$$ (Q); (II) $$\rightarrow$$ (S); (III) $$\rightarrow$$ (S); (IV) $$\rightarrow$$ (R)
C
(I) $$\rightarrow(\mathrm{Q})$$; (II) $$\rightarrow$$ (R); (III) $$\rightarrow(\mathrm{P})$$; (IV) $$\rightarrow$$ (R)
D
(I) $$\rightarrow$$ (T); (II) $$\rightarrow$$ (S); (III) $$\rightarrow$$ (P); (IV) $$\rightarrow$$ (T)
2
JEE Advanced 2019 Paper 1 Offline
+3
-1
Let $$M = \left[ {\matrix{ {{{\sin }^4}\theta } \cr {1 + {{\cos }^2}\theta } \cr } \matrix{ { - 1 - {{\sin }^2}\theta } \cr {{{\cos }^4}\theta } \cr } } \right] = \alpha I + \beta {M^{ - 1}}$$,
where $$\alpha$$ = $$\alpha$$($$\theta$$) and $$\beta$$ = $$\beta$$($$\theta$$) are real numbers, and I is the 2 $$\times$$ 2 identity matrix. If $$\alpha$$* is the minimum of the set {$$\alpha$$($$\theta$$) : $$\theta$$ $$\in$$ [0, 2$$\pi$$)} and {$$\beta$$($$\theta$$) : $$\theta$$ $$\in$$ [0, 2$$\pi$$)}, then the value of $$\alpha$$* + $$\beta$$* is
A
$$- {{17} \over {16}}$$
B
$$- {{31} \over {16}}$$
C
$$- {{37} \over {16}}$$
D
$$- {{29} \over {16}}$$
3
JEE Advanced 2017 Paper 2 Offline
+3
-1
How many 3 $$\times$$ 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries of MTM is 5?
A
198
B
162
C
126
D
135
4
JEE Advanced 2016 Paper 2 Offline
+3
-1
Let $$P = \left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {16} & 4 & 1 \cr } } \right]$$ and I be the identity matrix of order 3. If $$Q = [{q_{ij}}]$$ is a matrix such that $${P^{50}} - Q = I$$ and $${{{q_{31}} + {q_{32}}} \over {{q_{21}}}}$$ equals
A
52
B
103
C
201
D
205
EXAM MAP
Medical
NEET
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Use these Maths activities for the UK Election in 2017 bring purpose and context to Key Stage 2 investigations for primary school mathematics; third in the summer series of blogs linking classroom KS2 Maths lessons and resources to topical, calendar-based events.
Though a general election wasn't due until 2020, Prime Minister Theresa May has called for one to take place on Thursday 8th of June. The country has scrambled to prepare for the election and campaigning is now underway.
Here at Third Space Learning, we've scrambled to prepare for the election in our own special way – via a topical Maths post of course! The general election is a fantastic opportunity to help pupils understand the mechanisms of government, as well as how voting works.
What better way to show pupils how to use their power as citizens of the UK? Or improve their wider social awareness? With the activities below you can help your pupils prepare for life outside of a school, all while conducting a great Maths lesson!
#### UK Election Maths activity 1: 'The Ice Cream Ballot'
Follow these instructions to hold an ice cream ballot for your class:
1. Print out copies of the ballot card above and give one to each of your pupils.
2. Ask pupils to rank the four flavours in the column to the right (in ranking order best to worst with values 1-4).
3. Then as a class (or group) collect them in and tally which received the most votes using the relative majority system (which ice cream had the most ones next to it).
4. Then tally votes using the absolute majority system. This is a little more complex, so the steps for ordering are broken down here:
• Order flavours by the number of ones.
• Then take the flavour with the lowest number of ones and redistribute to the other three flavours, using the twos on the ballot papers.
• Now do the same for the new lowest group.
• Stop when one flavour has more than 50% of the votes.
Discuss: Did the same nominee win using both categories? If not, why? Which do you think is the fairest system?
#### UK Election activity 2: 'Maths Majority'
Use the image above as an instigator for the following discussion:
This image is of how many people voted in the 2010 general election, by political party. Use it to work out how many people voted overall. Now work out how many none voters would need to vote for the Labour Party, the Liberal Democrats, etc. to have more voters than the Conservative party.
Finally, work out what fraction/percentage of people did not vote in the 2010 election.
#### UK Election Maths activity 3: 'Election Estimations'
The bar graph shows the number of votes each of the main political parties received in the last general election. Estimate the number of votes received by each party.
Extension: As a class, decide on the closest estimation for each party. Using those estimations, work individually to create a pie chart showing how many people voted for each political party.
Good luck with the election Maths, don't forget to tweet us @thirdspacetweet and tell us what your class or school finds our election Maths activities!
This post is the third of our new topical Maths investigation series for summer-term 2017 which links classroom KS2 Maths lessons and resources to calendar-based events. Follow the links below to each post:
Or follow the links here for our hugely popular spring-term topical Maths series:
To find out more about our 1-to-1 Maths specialist intervention for your KS2 pupils, visit our Year 6 SATs foundation page, and for more information on earlier years, see our Year 3, 4, & 5 catch up programme
Put two of your pupils on our
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# Spark Starter Guide 2.5: Hypothesis Testing
## Introduction
In this section, we are going to cover the Spark Pearson’s Chi-squared ( χ2) statistic . We will also introduce Spark’s ML Pipelines and a new transformer: the StringIndexer.
The purpose of data, in general, is to use the data to help make effective decisions. But how do we know with any certainty that our analysis will lead to better decisions? We rarely, if ever, have all the desired data about a business problem, academic problem, or any problem for that matter. Therefore, it is impossible to know with absolute certainty whether our analysis is correct. This is where the science of statistics tries to provide insight into things that are unknowable. Statistics deals with samples of data that represent the entire population of data and tries to make reasonable assertions with sample data.
The purpose of data is to use the data to help make effective decisions
Hypothesis testing is a statistical test on two samples of data to determine if we can accept or reject the null hypothesis. To do this, hypothesis testing, tests whether the result occurred by chance using statistical operations. There are only two possible outcomes: a statistically significant result or a change that can only be attributed to chance. Hypothesis testing works using two statistical hypothesis’: the null and alternative hypothesis’. The null hypothesis is the default hypothesis and means nothing has changed or the result is purely from chance. The other option is the alternative hypothesis and means some other cause is impacting the results. For example, if we wanted to evaluate whether a coin is a fair coin then the null hypothesis would be that half the flips land heads and the other half land tails. The alternative hypothesis would be that the ratio of heads and tails does not result in fifty percent split.
Statistics deals with samples of data that represent the entire population of data and tries to make reasonable assertions with sample data
The Pearson’s Chi-squared test is a type of hypothesis testing that tests feature columns against the label column (outcome column) for independence. Every feature column in the test is paired with the label column and the Chi-squared statistic is computed. With the Chi-squared statistic, every column, both features and label, must be a categorical column. Which means it cannot use continuous columns. Continuous columns represent number data that can change continuously or the disticnt values cannot be counted because they are infinite. An example would be a column that represented by numbers: weight, revenue, sales in dollars, and so on.
## Getting Data
The data set for this section is the Adult data set from data.world. Citation: Dua, D. and Graff, C. (2019). UCI Machine Learning Repository [http://archive.ics.uci.edu/ml]. Irvine, CA: University of California, School of Information and Computer Science. To access the data on data.world requires a free account. After signing up, download the file “`adult.data.csv`” (once downloaded change the file name to “`adult_data.csv`“). This data set consists of categorical columns of various personal characteristics with the aim of trying to predict whether adults make more than \$50,000 per year. This data set of categorical columns will be a good fit for Hypothesis Testing. Upload the CSV to your Hadoop cluster and save it in HDFS. Once in HDFS make a DataFrame called `adult_df` from the CSV file. This CSV doesn’t have a header so be sure to set header equal to false. We can specify the columns with the `toDF()` method. The final code will look like:
PySpark:
```adult_df = spark.read.format("csv").load("hdfs://…/adult/adult_data.csv"
, sep = ","
, inferSchema = "true"
, header = "false").toDF("age", "workclass", "fnlwgt", "education", "education-num", "marital-status", "occupation", "relationship", "race", "sex", "capital-gain", "capital-loss", "hours-per-week", "native-country", "class")
```
Spark Scala:
```val adult_df = spark.read.format("csv")
.option("sep", ",")
.option("inferSchema", "true")
```
This DataFrame has several columns we do not want. We ultimately want a DataFrame of only categorical columns. So we will drop the four columns we do not want and call the DataFrame “`adult_cat_df`“.
PySpark:
```adult_cat_df = adult_df.drop("fnlwgt", "education-num", "capital-gain", "capital-loss")
```
Spark Scala:
```val adult_cat_df = adult_df.drop("fnlwgt", "education-num", "capital-gain", "capital-loss")
```
## StringIndexer
Since we are dealing only with categorical columns we need a new transformer for our ML pipeline: the StringIndexer. The purpose of the StringIndexer is to convert categorical columns (either strings or numbers) into indices by numbering every unique entry starting from zero up to the total number of unique entries in the column. In Spark the StringIndexer can only be used on one column at a time. So, to use the StringIndexer on multiple columns, which we will see in the exercise below, we simply make more StringIndexers. But the ML Pipeline makes packaging them together very easy. The StringIndexer works by creating a new column in the DataFrame.
Start by calling the method StringIndexer with two parameters. The first parameter is the input column and is the categorical column you would like converted into an index. The second parameter is the output column and is the new column name you will be creating. Normally, the output column is the input column name concatenated with “_index”, but you may name the new string index column however you like. A PySpark StringIndexer would look like:
```from pyspark.ml.feature import StringIndexer
category_indexer = StringIndexer(
inputCol = "category"
, outputCol = "category_index"
)
```
A Spark Scala StringIndexer example would be:
```import org.apache.spark.ml.feature.StringIndexer
val category _indexer = new StringIndexer()
.setInputCol("category")
.setOutputCol("category_index")
```
## ML Pipelines
The purpose of Spark’s ML Pipeline API is to make it very easy to combine or chain algorithms together into one workflow. All the elements of a ML Pipeline are either transformers or estimators. A transformer usually transforms one DataFrame into another by adding one or more columns to the original DataFrame. A transformer also implements the method `transform()`. An estimator is an algorithm that trains on data and implements the method `fit()`. An estimator is used on a DataFrame and outputs a model, which is itself a transformer.
To use Spark’s ML Pipelines use the imports: `from pyspark.ml import Pipeline` for PySpark and `import org.apache.spark.ml.Pipeline` for Scala. Proceed to use any indexers, vector assemblers, or any machine learning transformers and estimators. Once you have all the pieces you can assemble them in a pipeline. Call the Pipeline function which will have “`stages`” which are the variable names for the individual transformers and estimators in the pipeline. The stages are called in order from left to right in the pipeline. So make sure your stages follow a natural progression of transforming and processing the DataFrame. An hypothetical example using a DataFrame named “`df`” with made up columns “`category`“, “`sales_range`“, and “`prediction`” would look like the following:
PySpark:
```from pyspark.ml import Pipeline
pipeline = Pipeline(stages = [category_indexer, sales_range_indexer, prediction_indexer, assembler])
```
Spark Scala:
```import org.apache.spark.ml.Pipeline
val pipeline = new Pipeline()
.setStages(Array(category_indexer, sales_range_indexer, prediction_indexer, assembler))
```
Next we create a “`model`” by calling the `fit()` method on our pipeline and passing a DataFrame. In the example below the variable “`model`” is a `PipelineModel` which accepts new data and applies the data to the stages in our pipeline.
PySpark
```model = pipeline.fit(df)
```
Spark Scala:
```val model = pipeline.fit(df)
```
Lastly, we transform the newly created model and produce our final DataFrame that has the output or prediction column. In our hypothetical example it would look like:
PySpark
```transformed = model.transform(df)
```
Spark Scala:
```val transformed = model.transform(df)
```
Let’s go through a detailed exercise and see Spark’s ML Pipeline in action.
## Exercise 11: Spark’s Pearson’s Chi-squared test statistic
In this exercise we will use ML Pipelines to calculate Pearson’s Chi-squared test statistic.
1. Import the Python and Scala imports needed for the ML Pipelines and the Chi-squared statistic
PySpark
```from pyspark.ml import Pipeline
from pyspark.ml.feature import VectorAssembler
from pyspark.ml.feature import StringIndexer
from pyspark.ml.stat import ChiSquareTest
```
Spark Scala:
```import org.apache.spark.ml.linalg.Vector
import org.apache.spark.ml.Pipeline
import org.apache.spark.ml.feature.VectorAssembler
import org.apache.spark.ml.feature.StringIndexer
import org.apache.spark.ml.stat.ChiSquareTest
```
1. Create `StringIndexer`s for the two feature columns, `"age"` & `"education"`, and create a `StringIndexer` for the label column `"class"`. These three columns will be converted to numerical columns so they can be used by the ML Pipeline.
PySpark
```age_indexer = StringIndexer(
inputCol = "age"
, outputCol = "age_index"
)
education_indexer = StringIndexer(
inputCol = "education"
, outputCol = "education_index"
)
class_indexer = StringIndexer(
inputCol = "class"
, outputCol = "class_index"
)
```
Spark Scala:
```val age_indexer = new StringIndexer()
.setInputCol("age")
.setOutputCol("age_index")
val education_indexer = new StringIndexer()
.setInputCol("education")
.setOutputCol("education_index")
val class_indexer = new StringIndexer()
.setInputCol("class")
.setOutputCol("class_index")
```
These three columns are converted to indexes because the ML Pipeline and the Chi-squared test only work on numbers. Since these columns are strings we have transformed them into numbers.
1. Use the `VectorAssembler` on the `"age_index"` and the `"education_index"` numerical columns to create a new column of vectors called `"features"`.
PySpark
```assembler = VectorAssembler(
inputCols = ["age_index", "education_index"]
, outputCol = "features"
)
```
Spark Scala:
```val assembler = new VectorAssembler()
.setInputCols(Array("age_index", "education_index"))
.setOutputCol("features")
```
1. Construct a ML Pipeline called “`pipeline`” with stages of the three indexers and the assembler.
PySpark
```pipeline = Pipeline(stages = [education_indexer, age_indexer, class_indexer, assembler])
```
Spark Scala:
```val pipeline = new Pipeline()
.setStages(Array(education_indexer, age_indexer, class_indexer, assembler))
```
1. Use the pipeline to apply a model to the “`adult_cat_df`” DataFrame that returns a new DataFrame that we call “`model`“. Then apply that model to the “`adult_cat_df`” DataFrame to transform the DataFrame with the four new columns: “`age_index`“, “`education_index`“, “`class_index`“, and “`features`“. Name the final DataFrame “`transformed`“.
PySpark
```model = pipeline.fit(adult_cat_df)
```
Spark Scala:
```val model = pipeline.fit(adult_cat_df)
```
1. We are now ready to perform our Chi-squared test. Call the `ChiSquareTest.test()` with `"transformed"` as the first parameter, the second parameter `"features"` which is a column of combined vectors, and the third parameter `"class_index"` which is the name of the column label we are testing against. Display the final DataFrame.
PySpark
```chi_test = ChiSquareTest.test(transformed, "features", "class_index")
chi_test.show(truncate=False)
```
Spark Scala:
```val chi_test = ChiSquareTest.test(transformed, "features", "class_index")
chi_test.show(truncate=false)
```
Output:
```+---------+----------------+--------------------------------------+
|pValues |degreesOfFreedom|statistics |
+---------+----------------+--------------------------------------+
|[0.0,0.0]|[72, 15] |[3502.0364642305994,4429.653302288618]|
+---------+----------------+--------------------------------------+
```
1. Lastly, we can extract the columns `pValues``degreesOfFreedom`, and the `statistics` from the DataFrame and print the data in a nicer fashion. Follow the PySpark and Spark Scala code below:
PySpark
```chi = chi_test.head()
print("pValues: " + str(chi.pValues))
print("degreesOfFreedom: " + str(chi.degreesOfFreedom))
print("statistics: " + str(chi.statistics))
```
Spark Scala:
```val chi = chi_test.head
println(s"pValues: \${chi.getAs[Vector](0)}")
println(s"degreesOfFreedom: \${chi.getSeq[Int](1).mkString("[", ",", "]")}")
println(s"statistics: \${chi.getAs[Vector](2)}")
```
Output:
```pValues: [0.0,0.0]
degreesOfFreedom: [72,15]
statistics: [3502.0364642305994,4429.653302288618]
```
The full code example is below.
PySpark
```from pyspark.ml import Pipeline
from pyspark.ml.feature import VectorAssembler
from pyspark.ml.feature import StringIndexer
from pyspark.ml.stat import ChiSquareTest
age_indexer = StringIndexer(
inputCol = "age"
, outputCol = "age_index"
)
education_indexer = StringIndexer(
inputCol = "education"
, outputCol = "education_index"
)
class_indexer = StringIndexer(
inputCol = "class"
, outputCol = "class_index"
)
assembler = VectorAssembler(
inputCols = ["age_index", "education_index"]
, outputCol = "features"
)
pipeline = Pipeline(stages = [education_indexer, age_indexer, class_indexer, assembler])
chi_test = ChiSquareTest.test(transformed, "features", "class_index")
chi_test.show(truncate=False)
print("pValues: " + str(chi.pValues))
print("degreesOfFreedom: " + str(chi.degreesOfFreedom))
print("statistics: " + str(chi.statistics))
```
Spark Scala:
```import org.apache.spark.ml.linalg.Vector
import org.apache.spark.ml.Pipeline
import org.apache.spark.ml.feature.VectorAssembler
import org.apache.spark.ml.feature.StringIndexer
import org.apache.spark.ml.stat.ChiSquareTest
val age_indexer = new StringIndexer()
.setInputCol("age")
.setOutputCol("age_index")
val education_indexer = new StringIndexer()
.setInputCol("education")
.setOutputCol("education_index")
val class_indexer = new StringIndexer()
.setInputCol("class")
.setOutputCol("class_index")
val assembler = new VectorAssembler()
.setInputCols(Array("age_index", "education_index"))
.setOutputCol("features")
val pipeline = new Pipeline()
.setStages(Array(education_indexer, age_indexer, class_indexer, assembler))
val chi_test = ChiSquareTest.test(transformed, "features", "class_index")
chi_test.show(truncate=false)
println(s"pValues: \${chi.getAs[Vector](0)}")
println(s"degreesOfFreedom: \${chi.getSeq[Int](1).mkString("[", ",", "]")}")
println(s"statistics: \${chi.getAs[Vector](2)}")
```
```+---------+----------------+--------------------------------------+
|pValues |degreesOfFreedom|statistics |
+---------+----------------+--------------------------------------+
|[0.0,0.0]|[72, 15] |[3502.0364642305994,4429.653302288618]|
+---------+----------------+--------------------------------------+
pValues: [0.0,0.0]
degreesOfFreedom: [72, 15]
statistics: [3502.0364642305994,4429.653302288618]
```
Spark ML Pipelines make transforming data and applying machine learning algorithms very easy. We didn’t use any machine learning algorithms in this pipeline but we could have. A machine learning algorithm is simply an estimator that would get added right into the pipeline.
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1. Hi, I was doing part (d) of this question and was wondering why you are allowed to sub in t = 4 and x = 32 to work out the integration constant for the equation of t is greater then 4.
2. (Original post by Glavien)
Hi, I was doing part (d) of this question and was wondering why you are allowed to sub in t = 4 and x = 32 to work out the integration constant for the equation of t is greater then 4.
Draw a sketch of both graphs on the same coordinate axes. It should make sense from there on. I choose to use triangles to find the distance for the second half of the graph because it's simpler than integrating it twice.
3. (Original post by Glavien)
Hi, I was doing part (d) of this question and was wondering why you are allowed to sub in t = 4 and x = 32 to work out the integration constant for the equation of t is greater then 4.
Why are we doing integration constants? It's a definite integral, is it not? Displacement is the area under the velocity graph.
4. (Original post by Zacken)
Why are we doing integration constants? It's a definite integral, is it not? Displacement is the area under the velocity graph.
This was my solution after looking at the mark scheme. The bit I don't get is why they substituted t = 4 and x = 32 to work out the constant. Surely you can't as t=4 is not in the domain of the function.
5. (Original post by Glavien)
This was my solution after looking at the mark scheme. The bit I don't get is why they substituted t = 4 and x = 32 to work out the constant. Surely you can't as t=4 is not in the domain of the function.
I understand your concerns but at A-Level this is pretty much expected without justification. What you're really doing is saying so .
For what it's worth, I'd have used areas right away.
6. (Original post by Zacken)
I understand your concerns but at A-Level this is pretty much expected without justification. What you're really doing is saying so .
For what it's worth, I'd have used areas right away.
Thanks for the help, I prefer your method of definite integrals.
7. (Original post by aymanzayedmannan)
Draw a sketch of both graphs on the same coordinate axes. It should make sense from there on. I choose to use triangles to find the distance for the second half of the graph because it's simpler than integrating it twice.
Thanks, I'll try that.
8. (Original post by Glavien)
Thanks for the help, I prefer your method of definite integrals.
No problem!
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# FAQ: Iterables and Iterators - Finite Iterator: Chain
This community-built FAQ covers the “Finite Iterator: Chain” exercise from the lesson “Iterables and Iterators”.
Paths and Courses
This exercise can be found in the following Codecademy content:
## FAQs on the exercise Finite Iterator: Chain
There are currently no frequently asked questions associated with this exercise – that’s where you come in! You can contribute to this section by offering your own questions, answers, or clarifications on this exercise. Ask or answer a question by clicking reply () below.
If you’ve had an “aha” moment about the concepts, formatting, syntax, or anything else with this exercise, consider sharing those insights! Teaching others and answering their questions is one of the best ways to learn and stay sharp.
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import itertools
odd = [5, 7, 9]
even = {6, 8, 10}
all_numbers = list(itertools.chain(odd, even))
print(all_numbers)
• Print the result which will be:
``````[5, 7, 9, 8, 10, 6]
``````
I’m probably missing something obvious here, but why does the 6 appear at the end of the list and not at index position 3? I know it must be to do with the different data types of “odd” and “even” but still don’t see why it would put the first value of “even” at the end of the returned list.
1 Like
Yes you’ve got the right idea, it’s the because sets aren’t ordered. Unlike dictionaries which have been insertion ordered since 3.7, you should never rely on the order of a set.
In CPython at least the hashing function used may give you the illusion of order when you iterate through a set (the implementation relies on a hash function which means there is an order, technically). It’s bad form to rely on this though and the bits used for hashing even change depending on the length of the set and all sorts of clever stuff. It’s all very interesting but it’s the kind of implementation detail you should not be basing your program on . If you need the output of the set to be ordered then create a new ordered type before using it further.
1 Like
What does SKU mean in this exercise?
“stock-keeping unit.” it was defined earlier in the unit, so easy to miss, but it basically means like the barcode values on products at stores.
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# sum of interior angles of a 7-point star.
Printable View
• Apr 4th 2011, 03:50 AM
kingman
sum of interior angles of a 7-point star.
Dear sir ,
I would appreciate if anyone can show me the solution to the below question.
thanks
Kingman
What is the value of the angles A+B+C+D+E+F+G in the beolw figure ?
• Apr 4th 2011, 06:44 AM
Soroban
Hello, kingman!
I have a very primitive solution.
Quote:
$\text{What is the sum of the angles }A\!+\!B\!+\!C\!+\!D\!+\!E\!+\!F\!+\!G\text{ in the figure?}$
Place a pencil on $\,AC$, the eraser at $\,A$, the point at $\,C.$
Rotate the pencil about $\,C$ so that the eraser is at $\,E$.
Rotate the pencil about $\,E$ so that the point is at $\,G.$
Rotate the pencil about $\,G$ so that the eraser is at $\,B.$
Rotate the pencil about $\,B$ so that the point is at $\,D.$
Rotate the pencil about $\,D$ so that the eraser is at $\,F.$
Rotate the pencil about $\,F$ so that the point is at $\,A.$
Rotate the pencil about $\,A$ so that the eraser is at $\,C.$
We find that the pencil has undergone $1\tfrac{1}{2}$ rotations.
Therefore: . $A\!+\!B\!+\!C\!+\!D\!+\!E\!+\!F\!+\!G \:=\:540^o$
• Apr 4th 2011, 07:01 AM
mathaddict
Quote:
Originally Posted by kingman
Dear sir ,
I would appreciate if anyone can show me the solution to the below question.
thanks
Kingman
What is the value of the angles A+B+C+D+E+F+G in the beolw figure ?
Assuming that the star is uniform, you can try this method.
Lets begin with the triangle with angle A as one of its angle, label the two remaining angles with 1 and 2 (your preference). Then, label the opposite angles in the neighbouring triangle as well. Introduce some new angles, and repeat the process until you have all the angles in the 7 exterior triangles labelled. Now, label the heptagon inside, ie 180 - 1, 180 -2 ....
A + 1 + 2 = 180
B + 2 + 3 = 180
Do for all the rest and get 7 equations and sum them all up:
A+...+G + 2 (1 + 2 + ... + 7) =1260
Then set up another equation by summing up all the angles in the inner heptagon.
Yeah, just follow these boring steps if it comes out in a test, else go for Soroban's more interesting way of solving.(Wink)
• Apr 4th 2011, 07:21 AM
kingman
Thanks very much Soroban for the graphical approach to the question but I wonder whether it can done in the similar fashion iwhen solving the above problem if we take a 5 -point star instead and taking advantage of fact that the exterior angle of a triangle equals to sum of the interior opposite angles.
Thanks
kingman
• Apr 4th 2011, 09:02 AM
bjhopper
another method
mark intersection of GE and DF, H.StraigtenAGEF to form a rectangle.Let AGD be equilateral and GHD isosceles 30-30-120. Revised figure contains 4 90's and 3 60's= 540
• Apr 4th 2011, 06:54 PM
kingman
thanks very much for the solution but would appreciate very much if you can draw out how the final figure looks like and how a rectangle can be formed by straigtening AGEF .Finally can you explain how AGD becomes equilateral and GHD becomes isosceles .
thanks
• Apr 4th 2011, 09:10 PM
bjhopper
sum of interior angles
Hi kingman,
if you look at your diagram with my added pointH you can see that BGHD is shaped like akite and ACEF can be straitened to a rectanglewhich I will call akite with tails at F and E.THe rearrangement does not change the sum of the interior angles but makes it easy to find the sum.Draw the kite as previously described. Extend GH and DH .Draw AC parallel to GD.Drop perpendiculars from A and C meeting DH extended at Fand GH extended at E. Count the angles AC must be equal to FE
bjh
• Apr 4th 2011, 10:41 PM
kingman
Thanks very much for the response and would appreciate if you can use Paintbrush and make a sketch
of the diagram you have just described. Really expeiencing difficulty in visually something without a diagram.
thanks
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Labware - MA35 Multivariable Calculus - Three Variable Calculus
MA35 Labs 3 » Three Variable Calculus Contents3.3 More Derivatives 3.3.2 Second Partial Derivative 3.3.3 Hessian Matrices 3.3.4 Taylor Series 3.4 Integration Search
Second Partial Derivative
Text
A second partial derivative is a partial derivative of a function which is itself a partial derivative of another function.
There are nine types of second partial derivatives for functions of three variables.
1. fxx(x,y,z) = Partial derivative of fx(x,y,z) with respect to x.
2. fyx(x,y,z) = Partial derivative of fx(x,y,z) with respect to y.
3. fzx(x,y,z) = Partial derivative of fx(x,y,z) with respect to z.
4. fxy(x,y,z) = Partial derivative of fy(x,y,z) with respect to x.
5. fyy(x,y,z) = Partial derivative of fy(x,y,z) with respect to y.
6. fzy(x,y,z) = Partial derivative of fy(x,y,z) with respect to z.
7. fxz(x,y,z) = Partial derivative of fz(x,y,z) with respect to x.
8. fyz(x,y,z) = Partial derivative of fz(x,y,z) with respect to y.
9. fzz(x,y,z) = Partial derivative of fz(x,y,z) with respect to z.
Third, fourth, fifth, and in general, nth partial derivatives for any positive integer n, exist as well.
Demos
Second Partial Derivatives This demo shows graphs of f(x, y, z), one first partial derivative, and one second partial derivative. Red indicates a multiple of 5 (0 in this example). A yellow region next to a red one has a value greater than that multiple of 5, while a magenta region has a lower value. The demo starts out with the example fzx(x, y, z). To change the first derivative, change g(x, y, z) to f_x(x, y, z), f_y(x, y, z), or f_z(x, y, z). The second derivative, h(x, y, z), will be some derivative of the first derivative with respect to x, y, or z, chosen the same way as before. For example, to see fxy(x, y, z), change g(x, y, z) to f_y(x, y, z) and h(x, y, z) to g_x(x, y, z). Try changing x0, y0, and z0 to see different layers of these graphs and also to see how their values change as x, y, and z change. (For example, if the second derivative shown is fxy, notice how this second derivative graph describes the changes in fy as you change x0.)
Exercises
• 1. Find the following second partial derivatives as functions of x, y, and z:
• fxz(x, y, z), where f(x, y, z) = x2 + xy + xz
• fyy(x, y, z), where f(x, y, z) = x2yz + xy2z + xyz2
• fzy(x, y, z), where f(x, y, z) = zey
• fxy(x, y, z), where f(x, y, z) = 1/(xyz)
• 2. Suppose the demo above did not provide graphs of the first and second partial derivatives. How could you use the graph of f(x, y, z) to determine what the graph of fxy(x, y, z) would look like?
• 3. Consider the function f(x, y, z) = Axy + Bxz + Cyz. Describe the relationships between the mixed second partial derivatives. Will these relationships exist for any function of three variables f(x, y, z) or are there exceptions?
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# Units of Measure for Gases
## Presentation on theme: "Units of Measure for Gases"— Presentation transcript:
Units of Measure for Gases
1 atm.= 760 torr = 760 mmHg =101,325Pa, or ~1 x 105 Pa
Unlike liquids and solids, gases take up the volume of their container.
Atmospheric pressure is measured with a barometer. Atmospheric pressure is the measure of the weight of air. (not mass). It is the measure of the gravitational pull on air. At higher altitudes atm. Pressure is lower because air is thinner. Think of it as the heavier molecules sink, or there are less of them at higher altitudes.
P = F/A A is area, F is Force = ma, (is mass x acceleration),. a = 9
P = F/A A is area, F is Force = ma, (is mass x acceleration), a = 9.8 m/s2) Since Force = (mass x a), then Kg m / s2 = units for force = 1 Newton (N) And since Area, A = m then P = N / m2 which is 1 Pascal (Pa)
Gas Laws Boyle’s Law – shows that pressure and volume are inversely proportional, at constant temperature This means that PV = constant, i.e., as pressure goes up volume goes down and their product is a constant. From this we get the equation: P1V1 = P2V2 This relationship is only true under low pressure conditions. Under these conditions a gas is referred to as an IDEAL GAS. At high pressures gasses deviate from ideal behavior and are referred to as REAL GASSES, and PV = constant.
Charles’ Law – Volume is proportional to temperature at constant pressure
From this relationship we arrive at the equation: V1 / T1 = V2 / T2
Avagadro’s Law At constant temperature and pressure the volume of a gas is directly proportional to the # of moles of gas i.e., at the same temperature and pressure, equal volumes of gasses have the same number of moles. From this relationship the following equation is derived: V1 / n1 = V2 / n2
The Ideal Gas Law From Boyle’s Law : P V = constant
Charles’Law: V / T = constant Avogadro’s Law: V / n = constant The equation PV = nRT can be derived. PV = nRT R = the universal gas constant = L atm. / mol K
The Combined Gas Law P1 V1 / T1 = P2 V2 / T2
This equation is derived from the Ideal Gas Law: P1 V1 / T1 = P2 V2 / T2 For all gas laws the temperature must be in Kelvins.
Examples: In the laboratory an instrument called a manometer is used to measure pressure. Problem 9, page 224 deals with a manometer. The difference in height of the liquid in the U-tube is added to the atmospheric pressure to determine the pressure in the flask, if the pressure in the flask is greater. It is subtracted if the pressure in the flask is less than the outside pressure.
Molar Mass of a Gas The Ideal gas law can also be used to calculate the molar mass of a gas as follows: g R T PV or d R T P
Gas Stoichiometry Gas stoichiometry is the same, except to determine volume at STP, we can use the relationship L / mole. If the conditions are NOT STP then we must use the relationship ngas = PV/ RT and V = nRT / P See sample exercise 5.13 in text
Dalton’s Law of Partial Pressures
The total pressure of a mixture of different gasses is the sum of the pressures that each gas would exert if it were alone. Ptotal = P1 + P2 + P3 +…= n1RT + n2RT + n3RT +… V V V …and since RT and V are constant: Ptotal = ntotal (RT / V)
Mole Fraction of a Gas The mole fraction is the ratio of the moles of a particular gas in a mixture to the total number of moles of gas in the mixture. It can be calculated using the following equation: = Pi / P total Xi = n i / n total Therefore the partial pressure of a gas, Pi can be calculated: Pi = Xi P total
The Kinetic Theory of Gases
This theory is used to explain the similarities in the behavior of all gases. Gases consist of particles (atoms or molecules) in continuous, random motion. They undergo frequent collisions with each other and with the walls of the container. The pressure associated with a gas is due to the forces from these wall collisions.
Collisions between gas particles are elastic
Collisions between gas particles are elastic. That is, when two particles collide, no energy is lost, nor converted to heat. One particle may speed up while the other slows down, but no energy is lost. The average Energy of Translational Motion of a gas particle is directly proportional to temperature. That is, the energy of motion of a particle from one place to another is related to its speed by: Et = mu2 / 2, where E = kinetic energy of translation, m = mass of particle, u = velocity.(equation from KE=1mv2/2) Volume of gas particles is negligible compared to the container it is in. Attractive forces between gas particles are neglected. Note: KE of a gas is directly proportional to its Kelvin Temp.
Gas Pressure Relationships
At constant V,T – Pressure increases as n increases (more particles = more collisions) At constant n,T – pressure increases as volume decreases (smaller volume = particles striking walls more often) At constant n,V - Pressure increases as Temperature increases (higher temperature = faster movement of particles, thus more collisions and with greater force)
The Root Mean Square Velocity for a gas particle
U = (3RT / M)1/2 units = m/s , cm/s, etc. M = molar mass of the gas Derivation for this equation is in Appendix 2 and on pages Note: look at your table for R values and you will see that for cm/s, R = 8.31 x 107 g cm2/s2 mol K
U =( (3(8.31 x 107g cm2/s2 mol K)x 298K))1/2
Example: Find the average speed of an oxygen molecule in air at room temperature, (25oC). 25oC = T = 298 K, MM O2 = 32.0 g/mol U = (3RT/MM)1/2 U =( (3(8.31 x 107g cm2/s2 mol K)x 298K))1/2 32.0 g / mol U = 4.82 x 104 cm / s = 482 m / s Convert this velocity to miles / hour
Graham’s Law of Effusion
Effusion is the flow of gas particles through a small opening or pinhole. Rate of effusion is directly proportional to the average velocity of the particles, so: Rate of effusion of A = uA Rate of effusion of B uB
Therefore, uA/ uB = (3RT/MA)1/2 / (3RT/MB)1/2
And so rate of effusion of A = MB1/2 = time B rate of effusion of B MA time A Note: Heavy molecules take longer to effuse.
Example: In an effusion experiment it required 45 s for an unknown gas, X, to pass through a small opening into a vacuum. Under the same conditions, it took 28 s for the same number of moles of Ar to effuse. Find the molar mass of the unknown gas. Time B / Time A = (MMB / MMA)1/2 28 s / 45 s = (39.9 g/mol / MMx)1/2 (28 s / 45 s)2 =39.9 g/mol / MMx MMx = 1.0 x 102 g / mol
Real Gases and the van der Waals Equation
Real gases approach “ideal” behavior at low pressures and high temperatures. The van der Waals equation allows us to calculate for real gas conditions. As a partial derivation, we start with the Ideal Gas Law. We then have to make some corrections.
To get the Real Volume, we have to realize that the volume of the container is not truly the volume of the gas. The gas takes up some of that volume. The correction factor for Volume is nb, where n = # of moles of gas and b is a constant that has been calculated from experimental results for individual gases. To make the correction for Volume: V-nb is used instead of V
The pressure must also have a correction factor to allow for small attractive forces that occur in a real gas. Pobserved = P- a (n/V)2 where n/V = moles of gas particles / Liter and ‘a’ is a proportionality constant. This correction of ‘n / V’ is because pressure depends on the concentration of the particles, i.e., moles / volume.
The final corrected equation is:
P = nRT– a n 2 V-nb V There is actually a longer version of this equation but it is not addressed in your book.
Example: # 55 Calculate the pressure exerted by mol N2 in a L container at 25 oC. aN= 1.39 atm L2 / mol bN = L / mol T = 298 K n = mol V = 1.000L
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# Basic Multiplication Level 7
## Test your multiplication skills with this self-marking exercise to be completed without a calculator.
##### Level 1Level 2Level 3Level 4Level 5Level 6Level 7Level 8DescriptionHelpMore Arithmetic
Take your time, do your best. If you get stuck on one, just try the rest!
971 x 31 = 728 x 13 = 107 x 21 = 678 x 42 = 957 x 44 = 237 x 65 = 7910 x 41 = 9942 x 93 = 2210 x 90 = 8018 x 67 = 4247 x 56 = 8004 x 35 =
Check
This is Basic Multiplication Level 7. You can also try:
Level 1 Level 2 Level 3 Level 4 Level 5 Level 6 Level 8
## Instructions
Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.
When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
## Transum.org
This web site contains over a thousand free mathematical activities for teachers and pupils. Click here to go to the main page which links to all of the resources available.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 19 October 'Starter of the Day' page by E Pollard, Huddersfield:
"I used this with my bottom set in year 9. To engage them I used their name and favorite football team (or pop group) instead of the school name. For homework, I asked each student to find a definition for the key words they had been given (once they had fun trying to guess the answer) and they presented their findings to the rest of the class the following day. They felt really special because the key words came from their own personal information."
Comment recorded on the 10 April 'Starter of the Day' page by Mike Sendrove, Salt Grammar School, UK.:
"A really useful set of resources - thanks. Is the collection available on CD? Are solutions available?"
#### Beat The Clock
It is a race against the clock to answer 30 mental arithmetic questions. There are nine levels to choose from to suit pupils of different abilities.
There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.
A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.
Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.
Subscribe
## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.
## Maths Map
Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All:
## Description of Levels
Close
Level 1 - Multiplying single digit numbers up to and including 5
Level 2 - Multiplying numbers up to and including 10
Level 3 - Multiplying numbers up to and including 12
Level 4 - Multiplying a two digit number by a number up to and including 10 (first six no bridging)
Level 5 - Multiplying two digit numbers up to and including 19
Level 6 - Multiplying two digit numbers up to and including 99
Level 7 - Multiplying three and four digit numbers by two digit numbers
Level 8 - Mixed multiplication questions that get very hard indeed
More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
## Curriculum Reference
See the National Curriculum page for links to related online activities and resources.
## Multiplication
This video shows how to multiply numbers using column method (also referred to as traditional or long multiplication) and is from Corbettmaths.
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# Ways to Make a Number Using Subtraction Part 2
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Consider the following options:
8 – 2
9 – 4
7 – 5
8 – 4
Let's identify all the correct ways to make number six.
First check the difference of all the given options.
8 – 2 = 6
9 – 4 = 5
7 – 5 = 2
8 – 4 = 4
Out of all the options, 8 – 2 is the only combination that makes the number six.
So, 8 – 2 makes the number six.
ds
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
### Help
##### Remember :
The smallest number is the one that comes first while counting.
##### Solution :
To arrange the given numbers in order from smallest to greatest, find the smallest number among all the given numbers.
21,27,23
21 is the smallest number.
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# 5 Students Selected
Here it is:Suppose 5 students are randomly selected in a class of 42. How many selections are possible if 5 students are selected with the first winning $100, second$75, third $50, fourth$25, fifth $10. A student can win more than one.I don't even know where to start. Permutations? Combination? I feel like this has multiple steps, but of what I do not know. Thanks! #### soroban ##### Elite Member Hello, SadAtMath! Evidently, you are over-thinking the problem. Suppose 5 students are randomly selected in a class of 42. How many selections are possible if 5 students are selected with the first winning$100, second $75, third$50, fourth $25, fifth$10.
A student can win more than one prize.
I don't even know where to start. Permutations? Combination?
I feel like this has multiple steps, but of what I do not know. Thanks!
There are 42 choices for the winner of the 1st prize.
There are 42 choices for the winner of the 2nd prize.
There are 42 choices for the winner of the 3rd prize.
There are 42 choices for the winner of the 4th prize.
There are 42 choices for the winner of the 5th prize.
Get it?
There are: .$$\displaystyle 42^5 \:=\:130,\!691,\!232$$ possible selections.
##### New member
Really?
Are you trying to tell me that there are 130691232 possible selections? That is, 42^5? I felt like with the various prizes available it would ve more difficult. Sheesh, probability really stumps me. Thanks!
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### Part 1 Instructions - Page 1 of 7
• Experiment
This is an experiment in 2 parts.
• Rounds
This part of the experiment consists of 10 rounds.
• Introduction
In each round, you will have to decide whether or not to buy a commodity. The price of the commodity is the same for everyone taking part in the experiment, whereas its value to each person is different.
• Price
The price of the commodity (for everyone) is £2.40 throughout this part of the experiment.
• Value
The value of the commodity to you in a given round is determined by two things: its maximum value to you and the number of people other than you that buy it.
• Maximum Value
The maximum value of the commodity to you is chosen randomly at the start of each round, with all values in the range £0.00 to £10.00 being equally likely. This maximum value is private to you and it is not shown to anyone else. Similarly, you cannot see the others' maximum values, which are different to yours.
• Actual Value
The actual value of the commodity to you is equal to its maximum value multiplied by the proportion of people other than you that buy it. So the actual value is always lower than the maximum value unless everyone else buys the commodity. The actual value is zero if no one else buys the commodity.
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Question concerning the relativity of length
1. Oct 30, 2006
myoho.renge.kyo
let K' = a stationary rigid rod. let its length be AB (the length 299,792,458 meters between the two ends A and B of K'). the axis of K' is lying along the x-axis of a stationary system of coordinates K.
let there be a stationary red clock and an observer stationed at A.
let there be a stationary blue clock and another observer at the origin of K.
let the red clock and the blue clock synchronize.
let t = the time values of events as measured by the observer stationed at the origin of K using the blue clock.
let T = the time values of events as measured by the observer stationed at A using the red clock.
let a ray of light depart from A at t0 = T0 in the direction of B, and let the ray of light be reflected at B at t1 = T1.
the observer stationed at A measures the time (T1 - T0 = 1 s) required by the ray of light to travel from A to B.
the observer stationed at the origin of K measures the time (t1 - t0 = 1 s) required by the ray of light to travel from A to B.
c = AB/(T1 - T0) = AB/(t1 - t0).
c is the velocity of the ray of light as measured by the observer stationed at the origin of K, whether K' is stationary or moving at a constant speed v in the direction of increasing x.
let a constant speed v in the direction of increasing x be imparted to K'.
let a ray of light depart again from A at t0 = T0 in the direction of B, and let the ray of light be reflected at B at T1 and at t1.
the observer stationed at A measures again the time (T1 - T0 = 1 s) required by the ray of light to travel from A to B.
according to the relativity of time, the observer stationed at the origin of K is expected to measure the time [t1 - t0 = (1 s)/sqrt(1 - v^2/c^2)] required by the ray of light to travel from A to B. therefore, T1 is not equal to t1.
since c is the velocity of the ray of light as measured by the observer stationed at the origin of K, whether K' is stationary or moving at a constant speed v in the direction of increasing x, c = rAB/[(1 s)/sqrt(1 - v^2/c^2)], where rAB is not equal to AB.
What is the value of rAB, such that c = AB/(T1 -T0) = rAB/[(1 s)/sqrt(1 - v^2/c^2)]? thanks!
Last edited: Oct 30, 2006
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# E&M question
1. Apr 2, 2006
### matt85
The distance between an object and its image is fixed at 40.0 cm. A converging lens of focal length f = 3.62 cm forms a sharp image for two positions of the lens. What is the distance a between these two positions?
Determine:
The distance between these two positions a = ____cm
Could someone help me with the formula i want to be using?
2. Apr 3, 2006
### andrevdh
The thin lens formula gives
$$\frac{1}{f}=\frac{1}{o}+\frac{1}{i}$$
with i the image distance and o the object distance. Now say you find a position between the object and fixed screen (image position) such that you get a sharp image on the screen. The object image is now o and the image distance i while the sum of these two being 40 cm. But according to the formula if you make the object distance i the image distance should be o! Which gives you the other position of the lens where you should get a sharp image on the screen. In the second case the object distance will therefore be i and the image distance will be o.
Last edited: Apr 3, 2006
3. Apr 3, 2006
### andrevdh
I would guess that with a mech advantage of 2 you would have problems lifting your lecturer! Three would give you a fighting chance!
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# 6th Grade Standard Math Final Exam Review
### 30 terms by westberry
#### Study only
Flashcards Flashcards
Scatter Scatter
Scatter Scatter
## Create a new folder
Helping kids review each and every concept tested on our final exam!
### What is the Commutative Property?
In addition or multiplication, switching the order will still result in the same answer.
Example: a + b = b + a
### What is the Distributive Property?
Multiply a number by the sum of other numbers. Example: 5 (4 + 6) = 5x4 + 5x6 = 20 + 30 = 50!
### Which property is shown in this example? p (q + r) = pq + pr
The Distributive Property.
### Which property is shown in the two examples below? 5 x 1 = 5 5 x 0 = 0
The Identity Property.
### How do you change a percent into a fraction? Example: 90%
Take away the % sign, and put the 90 over 100. Then, simplify the fraction. 90/100 = 9/10
### How would you find a certain percent of a number? Example: What number is 24% of 75?
Change the percent to a decimal (remove the % sign and move the decimal two spaces to the left), then multiply it by the number you're finding the % of. Example: 0.24 x 75 = 18
### Explain how to change a decimal to a percent! Examples: 8.45 and 0.005
Move the decimal two spaces to the right & add a % sign. Examples: 845% and 0.5%
### How do you change a percent to a decimal? Examples: 8.2% and 45%
Remove the percent sign, then move the decimal two spaces to the left. Examples: 0.082 and 0.45
### How would you change a fraction into a percent? Examples: 4/5 and 1/23
Two ways will work...either change the fraction into a new fraction with 100 as a denominator (if your denominator is a factor of 100), OR change the fraction into a decimal & move the decimal two spaces left & add a % sign. Examples: 4/5 = 80/100 = 80% and 1/23 = 0.0435 = 4.35%
### If you want to change a fraction into a decimal, it's easy. You just.... Example: 1/12
Put the numerator inside the division box, and divide it by the denominator! Example: 0.8333
### How many feet are in one yard?
There are 3 feet in one yard. A good estimate of one yard is about the size of a classroom door's width.
### Let's say you are taking your final exam & you can't remember how many yards there are in a mile, or how many cups there are in one gallon...WHAT can you DO!?
Look at the conversion charts on your reference sheet :). All of that info is there.
### When you're converting units with the metric system (like...change 200 kg to grams?), what sentence can help you remember how many spaces to move the decimal?
King Henry Doesn't usually drink chocolate milk.
### What (besides the funny sentence) do each of these letters REALLY stand for?
The commonly used prefixes in the metric system: Kilo, Hecto, Deka, unit, deci, centi, and milli. The first three are capitalized because they are LARGER than a regular unit (meter, liter or gram). The others are smaller than a unit without a prefix.
After the 6.
### Use KHDudcm to answer: Change 95.1 g to mg. 95.1g = ________ mg?
95.1 g = 95,100 kg (move the decimal three spaces to the right, since moving from u to m on the acronym is moving three letters to the right.
### To convert Fahrenheit to Celcius and the other way around, you have to use formulas. Do you have to memorize them?
NO! They're on your reference sheet, at the bottom! If you're looking for a Fahrenheit answer, use the formula that has f = blahblahblah. If you're looking for Celcius, use the C = blahblahblah formula!
### To find the perimeter of a figure...what do you do?
You add up the lengths of all of the sides!
### How would you find the perimeter of a circle, since it doesn't have sides?
Find the circumference. Sing to yourself, "If I need a circumference, I'll just use pi D!" pi D means that to find the circumference of a circle, you multiply its diameter by pi. Units are plain, not squared, since you are measuring distance and not area.
3
### What's the formula for finding the area of a parallelogram?
A = bh (Area = base times height). A common mistake kids make is then taking half of that number. Don't divide by two unless you're finding the area of a triangle. :)
### Units for area are always...
SQUARED! That's the little two. Area is two-dimensional (2D), that's why the units are squared.
### Units for volume are always...
CUBED! That's the little three. Volume has three dimensions (3D), that's why the units are cubed.
### When you're graphing an inequality, when do you use an open dot on the number line?
When x is < > some number. For example: x < 4 would have an open dot on 4, and the line would go to the left (since x stands for all of the numbers LESS than four).
### When you're graphing an inequality on the number line, when would you use a closed dot? (hey if you got the last one...you've got this one!)
When x is ≥ ≤ some number. For example: x ≥ 4 would have a closed dot on 4, and the line would go to the right (since x stands for all of the numbers GREATER THAN or equal to four.)
### To solve equations you have to use:
Opposite operations.
### When you're graphing the point (2,1) - describe how you would "move" on the coordinate plane to graph it.
You'd go OVER 2 (to the right, since 2 is positive), and then you'd go UP 1 (since 1 is positive also).
Line up the dot!
7!
Example:
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# Which coordinate is cyclic in this case
1. Oct 12, 2012
### aaaa202
Consider a simple two particle system with two point masses of mass m at x1 and x2 with a potential energy relative to each other which depends on the difference in their coordinates V = V(x1-x2)
The lagrangian is:
L = ½m(x1')2 + ½m(x2')2 + V(x1-x2)
Obviously their total momentum is conserved d/dt(mx1' + mx2') = 0, which can be verified by plugging in to the lagrangian. But there is no cyclic coordinates in the lagrangian. Is it possible to put it in a form where this hidden cyclic coordinate is shown?
2. Oct 12, 2012
### Staff: Mentor
Hint: Try to express your lagrangian in terms of
y1=x1+x2
y2=x1-x2
3. Oct 12, 2012
### aaaa202
ahh nice.
So you get:
1/4my12 + 1/4my22 - V(y2) = 0
Is it possible to transform to a situation with all coordinates cyclic?
4. Oct 13, 2012
### raopeng
It seems to be the motion in an inertial frame of reference. Well in that case x1 - x2 must be constant I think, so potential energy can be omitted as a constant. What is clear is that potential energy is a function of generalised coordinates, so it has to be Cartesian coordinates as well.
5. Oct 13, 2012
### vanhees71
First of all the Lagrangian in the new coordinates
$$L=\frac{\mu}{2}(\dot{y}_1^2+\dot{y}_2^2)-V(y_2)$$
with $\mu=m/2$.
To answer the question, whether you can find a set of coordinates, which all are cyclic, you should read about the Hamilton-Jacobi partial differential equation and action-angle variables.
In your case there is for sure another conserved quantity! Think which that might be!
6. Oct 13, 2012
### aaaa202
well that's the energy but that has nothing to do with cyclic coordinates.
Also I did read Hamilton-Jacobi theory but that takes its basis in the hamiltonian formulation. So I guess it's not really possible to transform to a frame with all coordinates cyclic UNLESS you use the hamiltonian formulation with more freedom to vary your conjugate variables?
7. Oct 13, 2012
### Staff: Mentor
To get a second cyclic variable, you would need some parameter which describes the time-evolution of your (y2-)system with fixed energy. If V is quadratic (or at least gives oscillations in some way), this would be the phase of the oscillation, for example.
I doubt that you can do this transformation in an explicit way with a general V.
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interpret the practical meaning of the derivative
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40) The value of a certain automobile purchased in 1997 can be approximated by the function v(t)=25(0.85)^t , where t is the time in years, from the date of purchase, and v is the value, in thousands of dollars.
(a) Evaluate and interpret v(4).
(b) Find an expression for v1(t) including units.
(c) Evaluate and interpret v1(4).
(d) Use v(t), v1(t) and any other considerations you think are relevant to write a paragraph in support of or in opposition to the following statement: From a momentary point of view, it is best to keep this vehicle as long as possible".
56) Let f(v) be the gas consumption ( in liters/km ) of a car going at velocity v( in km/hr). In other words , f(v) tells you how many liters of gas the car uses to go one kilometer , if it is going at a velocity v, you are told that
f(80)=0.05 and f1(80)=0.0005.
a) Let g(v) be the distance the same car goes on one liter of gas at velocity v. what is the relationship between f(v) and g(v) ? find g(80) and g1(80).
b) Let h(v) be the gas consumption in liters per hour. In other words, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. what is the relationship between h(v) and f(v)?
Find h(80) and h1(80).
c) How would you explain the practical meaning of the values of these functions and their derivatives to a driver who knows no calculus?
© BrainMass Inc. brainmass.com October 15, 2018, 6:20 pm ad1c9bdddf - https://brainmass.com/math/calculus-and-analysis/interpret-the-practical-meaning-of-the-derivative-124983
Solution Summary
The solution evaluates the functions and calculates and interprets the meaning of its derivative in the contexts.
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Further Inference in the Multiple Regression Model Prepared by Vera Tabakova, East Carolina University.
Presentation on theme: "Further Inference in the Multiple Regression Model Prepared by Vera Tabakova, East Carolina University."— Presentation transcript:
Further Inference in the Multiple Regression Model Prepared by Vera Tabakova, East Carolina University
6.1 The F-Test 6.2 Testing the Significance of the Model 6.3 An Extended Model 6.4 Testing Some Economic Hypotheses 6.5 The Use of Nonsample Information 6.6 Model Specification 6.7 Poor Data, Collinearity and Insignificance 6.8 Prediction
If the null hypothesis is not true, then the difference between SSE R and SSE U becomes large, implying that the constraints placed on the model by the null hypothesis have a large effect on the ability of the model to fit the data.
Hypothesis testing steps: 1. Specify the null and alternative hypotheses: 2. Specify the test statistic and its distribution if the null hypothesis is true 3. Set and determine the rejection region Using α=.05, the critical value from the -distribution is. Thus, H 0 is rejected if.
4. Calculate the sample value of the test statistic and, if desired, the p-value 5. State your conclusion Since, we reject the null hypothesis and conclude that price does have a significant effect on sales revenue. Alternatively, we reject H 0 because.
The elements of an F-test : 1. The null hypothesis consists of one or more equality restrictions J. The null hypothesis may not include any ‘greater than or equal to’ or ‘less than or equal to’ hypotheses. 2. The alternative hypothesis states that one or more of the equalities in the null hypothesis is not true. The alternative hypothesis may not include any ‘greater than’ or ‘less than’ options. 3. The test statistic is the F-statistic
4. If the null hypothesis is true, F has the F-distribution with J numerator degrees of freedom and N-K denominator degrees of freedom. The null hypothesis is rejected if. 5. When testing a single equality null hypothesis it is perfectly correct to use either the t- or F-test procedure; they are equivalent.
Example: Big Andy’s sales revenue 1. 2. If the null is true 3. H 0 is rejected if 4. 5. Since 29.95>3.12 we reject the null and conclude that price or advertising expenditure or both have an influence on sales.
Figure 6.1 A Model Where Sales Exhibits Diminishing Returns to Advertising Expenditure Slide 6-12Principles of Econometrics, 3rd Edition
6.4.1 The Significance of Advertising
Since, we reject the null hypothesis and conclude that advertising does have a significant effect upon sales revenue.
Economic theory tells us that we should undertake all those actions for which the marginal benefit is greater than the marginal cost. This optimizing principle applies to Big Andy’s Burger Barn as it attempts to choose the optimal level of advertising expenditure.
Big Andy has been spending \$1,900 per month on advertising. He wants to know whether this amount could be optimal. The null and alternative hypotheses for this test are
Because, we cannot reject the null hypothesis that the optimal level of advertising is \$1,900 per month. There is insufficient evidence to suggest Andy should change his advertising strategy.
Reject H 0 if t ≥ 1.667. t =.9676 Because.9676 < 1.667, we do not reject H 0. There is not enough evidence in the data to suggest the optimal level of advertising expenditure is greater than \$1900.
6.6.1 Omitted Variables
1. Choose variables and a functional form on the basis of your theoretical and general understanding of the relationship. 2. If an estimated equation has coefficients with unexpected signs, or unrealistic magnitudes, they could be caused by a misspecification such as the omission of an important variable.
3. One method for assessing whether a variable or a group of variables should be included in an equation is to perform significance tests. That is, t-tests for hypotheses such as or F-tests for hypotheses such as. Failure to reject hypotheses such as these can be an indication that the variable(s) are irrelevant. 4. The adequacy of a model can be tested using a general specification test known as RESET.
6.7.1 The Consequences of Collinearity
The effects of imprecise information: 1. When estimator standard errors are large, it is likely that the usual t- tests will lead to the conclusion that parameter estimates are not significantly different from zero. This outcome occurs despite possibly high or F-values indicating significant explanatory power of the model as a whole.
2. The estimators may be very sensitive to the addition or deletion of a few observations, or the deletion of an apparently insignificant variable. 3. Despite the difficulties in isolating the effects of individual variables from such a sample, accurate forecasts may still be possible if the nature of the collinear relationship remains the same within the new (future) sample observations.
MPG = miles per gallon CYL = number of cylinders ENG = engine displacement in cubic inches WGT = vehicle weight in pounds
Identifying Collinearity 1. Examining pairwise correlations. 2. Using auxiliary regression If the R 2 from this artificial model is high, above.80 say, the implication is that a large portion of the variation in is explained by variation in the other explanatory variables.
Mitigating Collinearity 1. Obtain more information and include it in the analysis. 2. Introduce nonsample information in the form of restrictions on the parameters.
Slide 6-48Principles of Econometrics, 3rd Edition a single null hypothesis with more than one parameter auxiliary regressions collinearity F-test irrelevant variable nonsample information omitted variable omitted variable bias overall significance of a regression model regression specification error test (RESET) restricted least squares restricted sum of squared errors single and joint null hypotheses unrestricted sum of squared errors
Slide 6-49Principles of Econometrics, 3rd Edition
Slide 6-50Principles of Econometrics, 3rd Edition (6A.1) (6A.3) (6A.2) (6A.4)
Slide 6-51Principles of Econometrics, 3rd Edition (6A.5)
Slide 6-52Principles of Econometrics, 3rd Edition
Slide 6-53Principles of Econometrics, 3rd Edition
Slide 6-54Principles of Econometrics, 3rd Edition
Slide 6-55Principles of Econometrics, 3rd Edition (6B.1)
Slide 6-56Principles of Econometrics, 3rd Edition
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Factor 3x2 - 15x - 42.
Question
Updated 7/1/2016 8:11:59 AM
Edited by jeifunk [7/1/2016 8:11:57 AM], Confirmed by jeifunk [7/1/2016 8:11:59 AM]
s
Original conversation
User: Factor 3x2 - 15x - 42.
Weegy: 3x^2 - 15x - 42 = 3(x^2 - 5x - 14) = 3(x - 7)(x + 2)
User: Factor 9b2 - 4.
Weegy: 9b^2 - 4
= (3b)^2 - 2^2
= (3b - 2)(3b + 2)
User: Factor k2 + 13k + 12.
Weegy: k^2 + 13k + 12 = (k + 12)(k + 1)
User: Factor 360t + 10t3 - 120t2.
Weegy: 360t + 10t^3 - 120t^2
= 10t(36 + t^2 - 12t)
= 10t(t - 6)^2
= 10t(t - 6)(t - 6)
User: Simplify m - {n + [p - (m + n - p)]}.
Question
Updated 7/1/2016 8:11:59 AM
Edited by jeifunk [7/1/2016 8:11:57 AM], Confirmed by jeifunk [7/1/2016 8:11:59 AM]
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m - {n + [p - (m + n - p)]};
= m - {n + (p - m - n + p)};
= m - {n + 2p - m - n}
= m - {2p - m};
= m + m - 2p;
= 2m - 2p
Confirmed by jeifunk [7/1/2016 8:12:07 AM]
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# Question: Triangles with Intersection
## Comment on Triangles with Intersection
### When two triangles are
When two triangles are similar and we are taking the ratio of the sides. Is it the bigger triangle over smaller or vice versa?
My solution:
x/12 = 5/10
x = 6
### It doesn't matter whether you
It doesn't matter whether you compare the bigger triangle over smaller or vice versa; the outcome will be the same in both cases. However, when comparing sides, you must compare CORRESPONDING sides. You have not done so.
The side with length 12 (in the larger triangle) does not correspond to the side with length x (in the smaller triangle).The side with length 12 is between the angles denoted with a check mark and a DOT, whereas the side with length x is between the angles denoted with a check mark and a HEART. As such, those two sides are not CORRESPONDING, which explains why your solution is incorrect.
Does that help?
### Thankyou :-) Awesome videos
Thankyou :-) Awesome videos helping me a lot during my prep
### Where are the 1200 + practice
Where are the 1200 + practice questions located ?
### Most of the 1200 practice
Most of the 1200 practice questions can be found in the Related Resources box beneath most video lessons (e.g., https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...)
These questions help to reinforce the concepts/strategies covered in that particular lesson. You will find that I have personally answered most of those linked questions, using the strategies covered in the video lessons.
It's also important to note that most of those practice questions are official GRE questions, which are (of course) the most representative of what you will see on test day.
### Awesome. Nice explanation,
Awesome. Nice explanation, thanks.
### I got the similar result as
I got the similar result as Sneha but I have a slightly different explanation. Can we make the smaller triangle CED flip upside down? If we imagine that then the triangle CED will look exactly similar to triangle AEB and also the side CE corresponds to AE and similary DE corresponds to BE. When I take the ratio CE/AE= DE/BE, I too get the same result of 6. So as per your explanation, a proper correspondency only happens when the two sides and the angle within them match and are equal? So we should not only see the sides but also the angles within them. Is this understanding correct? One more question- Is this theory pointing to the SAS,AAS,SSA, etc. these properties? Please clarify.
### Hi Deepak,
Hi Deepak,
Be careful; side CE does NOT correspond to side AE.
To determine which sides correspond, we must examine where each side lies in relationship to the given angles.
At 1:00 in the video, I have used dots, hearts and check-marks to label all of the angles and show which angles the two triangles have in common.
Notice that side CE lies between the angles denoted with a dot and a check-mark.
So, the side that corresponds to side CE will be side on the big triangle that also lies between the angles denoted with a dot and a check-mark. That side is EB
So, side CE corresponds to side EB (not AE, as you suggest)
ASIDE: The rules (SAS, ASA and SSS) apply to CONGRUENT triangles (i.e., triangles that are EXACTLY the same).
The key concept is this question is the concept of SIMILAR triangles (triangles that share the same angles, but are different sizes)
Does that help?
Cheers,
Brent
### Yes, that really helps. I
Yes, that really helps. I understand the difference of congruency and similar triangles now. :) Thank you so much. Also, do we need to know the congruency properties as well for GRE?
### No, you don't need to know
No, you don't need to know the congruency properties.
Thank you :)
### https://greprepclub.com/forum
https://greprepclub.com/forum/in-the-figure-above-a-is-the-center-of-the-circle-12255.html
In this question did you mean both triangles have angle B or E because i can't seem to understand how both triangles have angle B
I'll direct you to my solution at https://greprepclub.com/forum/in-the-figure-above-a-is-the-center-of-the...
Let's first focus on ∆EAF
If we let ∠E = star, and we let ∠F = face, then ∆EAF has the following 3 angles: star, face and 90°
This means star + face + 90° = 180° (angles in a triangle add to 180°)
Now let's focus on ∆EBC
∠E is the same as in ∆EAF. So, ∠E = star
Also notice that ∠C = 90°
KEY: We can already see that ∆EBC shares 2 of its angles with ∆EAF
This means the 3rd angle (∠B) must also be the same as in ∆EAF
In other words, ∠B = ∠F = face
Does that help?
Cheers,
Brent
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > upgr2wlk Structured version Visualization version GIF version
Theorem upgr2wlk 26433
Description: Properties of a pair of functions to be a walk of length 2 in a pseudograph. Note that the vertices need not to be distinct and the edges can be loops or multiedges. (Contributed by Alexander van der Vekens, 16-Feb-2018.) (Revised by AV, 3-Jan-2021.) (Revised by AV, 28-Oct-2021.)
Hypotheses
Ref Expression
upgr2wlk.v 𝑉 = (Vtx‘𝐺)
upgr2wlk.i 𝐼 = (iEdg‘𝐺)
Assertion
Ref Expression
upgr2wlk (𝐺 ∈ UPGraph → ((𝐹(Walks‘𝐺)𝑃 ∧ (#‘𝐹) = 2) ↔ (𝐹:(0..^2)⟶dom 𝐼𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
Proof of Theorem upgr2wlk
Dummy variable 𝑘 is distinct from all other variables.
StepHypRef Expression
1 upgr2wlk.v . . . 4 𝑉 = (Vtx‘𝐺)
2 upgr2wlk.i . . . 4 𝐼 = (iEdg‘𝐺)
31, 2upgriswlk 26406 . . 3 (𝐺 ∈ UPGraph → (𝐹(Walks‘𝐺)𝑃 ↔ (𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))})))
43anbi1d 740 . 2 (𝐺 ∈ UPGraph → ((𝐹(Walks‘𝐺)𝑃 ∧ (#‘𝐹) = 2) ↔ ((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ∧ (#‘𝐹) = 2)))
5 iswrdb 13250 . . . . . . . . 9 (𝐹 ∈ Word dom 𝐼𝐹:(0..^(#‘𝐹))⟶dom 𝐼)
6 oveq2 6612 . . . . . . . . . 10 ((#‘𝐹) = 2 → (0..^(#‘𝐹)) = (0..^2))
76feq2d 5988 . . . . . . . . 9 ((#‘𝐹) = 2 → (𝐹:(0..^(#‘𝐹))⟶dom 𝐼𝐹:(0..^2)⟶dom 𝐼))
85, 7syl5bb 272 . . . . . . . 8 ((#‘𝐹) = 2 → (𝐹 ∈ Word dom 𝐼𝐹:(0..^2)⟶dom 𝐼))
9 oveq2 6612 . . . . . . . . 9 ((#‘𝐹) = 2 → (0...(#‘𝐹)) = (0...2))
109feq2d 5988 . . . . . . . 8 ((#‘𝐹) = 2 → (𝑃:(0...(#‘𝐹))⟶𝑉𝑃:(0...2)⟶𝑉))
11 fzo0to2pr 12494 . . . . . . . . . . 11 (0..^2) = {0, 1}
126, 11syl6eq 2671 . . . . . . . . . 10 ((#‘𝐹) = 2 → (0..^(#‘𝐹)) = {0, 1})
1312raleqdv 3133 . . . . . . . . 9 ((#‘𝐹) = 2 → (∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))} ↔ ∀𝑘 ∈ {0, 1} (𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}))
14 2wlklem 26432 . . . . . . . . 9 (∀𝑘 ∈ {0, 1} (𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))} ↔ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))
1513, 14syl6bb 276 . . . . . . . 8 ((#‘𝐹) = 2 → (∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))} ↔ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)})))
168, 10, 153anbi123d 1396 . . . . . . 7 ((#‘𝐹) = 2 → ((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ↔ (𝐹:(0..^2)⟶dom 𝐼𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
1716adantl 482 . . . . . 6 ((𝐺 ∈ UPGraph ∧ (#‘𝐹) = 2) → ((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ↔ (𝐹:(0..^2)⟶dom 𝐼𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
18 3anass 1040 . . . . . 6 ((𝐹:(0..^2)⟶dom 𝐼𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)})) ↔ (𝐹:(0..^2)⟶dom 𝐼 ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
1917, 18syl6bb 276 . . . . 5 ((𝐺 ∈ UPGraph ∧ (#‘𝐹) = 2) → ((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ↔ (𝐹:(0..^2)⟶dom 𝐼 ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)})))))
2019ex 450 . . . 4 (𝐺 ∈ UPGraph → ((#‘𝐹) = 2 → ((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ↔ (𝐹:(0..^2)⟶dom 𝐼 ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))))
2120pm5.32rd 671 . . 3 (𝐺 ∈ UPGraph → (((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ∧ (#‘𝐹) = 2) ↔ ((𝐹:(0..^2)⟶dom 𝐼 ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))) ∧ (#‘𝐹) = 2)))
22 3anass 1040 . . . 4 (((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ∧ 𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)})) ↔ ((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
23 an32 838 . . . 4 (((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))) ↔ ((𝐹:(0..^2)⟶dom 𝐼 ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))) ∧ (#‘𝐹) = 2))
2422, 23bitri 264 . . 3 (((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ∧ 𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)})) ↔ ((𝐹:(0..^2)⟶dom 𝐼 ∧ (𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))) ∧ (#‘𝐹) = 2))
2521, 24syl6bbr 278 . 2 (𝐺 ∈ UPGraph → (((𝐹 ∈ Word dom 𝐼𝑃:(0...(#‘𝐹))⟶𝑉 ∧ ∀𝑘 ∈ (0..^(#‘𝐹))(𝐼‘(𝐹𝑘)) = {(𝑃𝑘), (𝑃‘(𝑘 + 1))}) ∧ (#‘𝐹) = 2) ↔ ((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ∧ 𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
26 2nn0 11253 . . . . . . 7 2 ∈ ℕ0
27 fnfzo0hash 13172 . . . . . . 7 ((2 ∈ ℕ0𝐹:(0..^2)⟶dom 𝐼) → (#‘𝐹) = 2)
2826, 27mpan 705 . . . . . 6 (𝐹:(0..^2)⟶dom 𝐼 → (#‘𝐹) = 2)
2928pm4.71i 663 . . . . 5 (𝐹:(0..^2)⟶dom 𝐼 ↔ (𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2))
3029bicomi 214 . . . 4 ((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ↔ 𝐹:(0..^2)⟶dom 𝐼)
3130a1i 11 . . 3 (𝐺 ∈ UPGraph → ((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ↔ 𝐹:(0..^2)⟶dom 𝐼))
32313anbi1d 1400 . 2 (𝐺 ∈ UPGraph → (((𝐹:(0..^2)⟶dom 𝐼 ∧ (#‘𝐹) = 2) ∧ 𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)})) ↔ (𝐹:(0..^2)⟶dom 𝐼𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
334, 25, 323bitrd 294 1 (𝐺 ∈ UPGraph → ((𝐹(Walks‘𝐺)𝑃 ∧ (#‘𝐹) = 2) ↔ (𝐹:(0..^2)⟶dom 𝐼𝑃:(0...2)⟶𝑉 ∧ ((𝐼‘(𝐹‘0)) = {(𝑃‘0), (𝑃‘1)} ∧ (𝐼‘(𝐹‘1)) = {(𝑃‘1), (𝑃‘2)}))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 384 ∧ w3a 1036 = wceq 1480 ∈ wcel 1987 ∀wral 2907 {cpr 4150 class class class wbr 4613 dom cdm 5074 ⟶wf 5843 ‘cfv 5847 (class class class)co 6604 0cc0 9880 1c1 9881 + caddc 9883 2c2 11014 ℕ0cn0 11236 ...cfz 12268 ..^cfzo 12406 #chash 13057 Word cword 13230 Vtxcvtx 25774 iEdgciedg 25775 UPGraph cupgr 25871 Walkscwlks 26362 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-8 1989 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 ax-rep 4731 ax-sep 4741 ax-nul 4749 ax-pow 4803 ax-pr 4867 ax-un 6902 ax-cnex 9936 ax-resscn 9937 ax-1cn 9938 ax-icn 9939 ax-addcl 9940 ax-addrcl 9941 ax-mulcl 9942 ax-mulrcl 9943 ax-mulcom 9944 ax-addass 9945 ax-mulass 9946 ax-distr 9947 ax-i2m1 9948 ax-1ne0 9949 ax-1rid 9950 ax-rnegex 9951 ax-rrecex 9952 ax-cnre 9953 ax-pre-lttri 9954 ax-pre-lttrn 9955 ax-pre-ltadd 9956 ax-pre-mulgt0 9957 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-ifp 1012 df-3or 1037 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-eu 2473 df-mo 2474 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ne 2791 df-nel 2894 df-ral 2912 df-rex 2913 df-reu 2914 df-rmo 2915 df-rab 2916 df-v 3188 df-sbc 3418 df-csb 3515 df-dif 3558 df-un 3560 df-in 3562 df-ss 3569 df-pss 3571 df-nul 3892 df-if 4059 df-pw 4132 df-sn 4149 df-pr 4151 df-tp 4153 df-op 4155 df-uni 4403 df-int 4441 df-iun 4487 df-br 4614 df-opab 4674 df-mpt 4675 df-tr 4713 df-eprel 4985 df-id 4989 df-po 4995 df-so 4996 df-fr 5033 df-we 5035 df-xp 5080 df-rel 5081 df-cnv 5082 df-co 5083 df-dm 5084 df-rn 5085 df-res 5086 df-ima 5087 df-pred 5639 df-ord 5685 df-on 5686 df-lim 5687 df-suc 5688 df-iota 5810 df-fun 5849 df-fn 5850 df-f 5851 df-f1 5852 df-fo 5853 df-f1o 5854 df-fv 5855 df-riota 6565 df-ov 6607 df-oprab 6608 df-mpt2 6609 df-om 7013 df-1st 7113 df-2nd 7114 df-wrecs 7352 df-recs 7413 df-rdg 7451 df-1o 7505 df-2o 7506 df-oadd 7509 df-er 7687 df-map 7804 df-pm 7805 df-en 7900 df-dom 7901 df-sdom 7902 df-fin 7903 df-card 8709 df-cda 8934 df-pnf 10020 df-mnf 10021 df-xr 10022 df-ltxr 10023 df-le 10024 df-sub 10212 df-neg 10213 df-nn 10965 df-2 11023 df-n0 11237 df-xnn0 11308 df-z 11322 df-uz 11632 df-fz 12269 df-fzo 12407 df-hash 13058 df-word 13238 df-edg 25840 df-uhgr 25849 df-upgr 25873 df-wlks 26365 This theorem is referenced by: umgrwwlks2on 26719
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# hmwk_6_sol (1) - Discrete Structures Assignment 6 Due at...
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Discrete Structures Sept 26 2011 Assignment 6: Due at beginning of class Wednesday, Oct 12 Prof. Hopcroft *******Note: this homework is subject to the style guide on the website. Points will be deducted for homeworks not following the guidelines.******** *******Please print out and staple the grade sheet to the back of your homework!****** 1. If, for a given a there exists x such that ax 1 mod m does that imply that a and m are relatively prime? Prove why or why not. Proof. This implies that x is a ’s inverse in mod m . We have a theorem, that says if a and m are relatively prime, then x exists (Euclid’s Extended Algorithm). Hence, we have the intuition that we should try to prove a and m are relatively prime. But note, we can not directly use Euclid’s Algorithm, as it assumes what we want to prove. (Another way to phrase it, is we want to prove the reverse direction.) There are a few proofs that work, I present the most straight forward one: Using the given let us manipulate it a little bit: ax 1 mod m ax - km = 1 for some integer k ax = 1 + km Now, let us look at the gcd gcd( ax, m ) = gcd( km + 1 , m ) . It may seem obvious from here, but if we add a little bit of argumentation, our proof will be mathe- matically sound. Let d be the gcd( km + 1 , m ). Because d divides m and d divides km + 1, this implies that d divides 1. Which, can only be the case if d = 1. Hence, we have shown gcd( ax, m ) = 1, which implies no number that divides m divides either a or x . Hence, there is no number (other than 1) that divides both m and a .
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Quick Answer: What Percentage Is 50 Minutes Of An Hour?
What is .50 of an hour?
For example, 50 percent of an hour equals 30 minutes, because 0.50 * 60 equals 30..
What jobs pay \$500 an hour?
What Salary Equals \$500/Hour?ProfessionWageLawyers\$59.11Physician Assistants\$53.97Actuaries\$52.09Mathematicians\$50.5038 more rows
What is 6.75 hours in hours and minutes?
6.75 hours with the decimal point is 6.75 hours in terms of hours. 6:75 with the colon is 6 hours and 75 minutes. . 75 = fractional hours.
What is .15 of an hour?
Option 2: Use our minutes conversion chartMinutesDecimal HoursDecimal Hours9.15.8210.17.8311.18.8512.20.8716 more rows
What is .8 of an hour?
Billing Increment Chart—Minutes to Tenths of an HourMinutesTime25-30.531-36.637-42.743-48.86 more rows
How much is 500 an hour?
In this case, you can quickly compute the annual salary by multiplying the hourly wage by 2000. Your hourly pay of 500 dollars is then equivalent to an average annual income of \$1,000,000 per year.
Is \$24 an hour good?
Assuming all things equal, \$24 per hour would be slightly above the median household income in the US. It is also worthwhile to look at sites like Glassdoor to see what others make in your field. You can filter by company, location, job, etc.
What is percentage formula?
If you have to turn a percentage into a decimal, just divide by 100. For example, 25% = 25/100 = 0.25. To change a decimal into a percentage, multiply by 100. So 0.3 = 0.3 × 100 =30% .
What decimal is 50 minutes of an hour?
0.833Minutes to Decimal Hours CalculatorMinutesDecimal Hours470.783480.800490.817500.83356 more rows
How do you calculate percentage of hours?
Divide the number of working hours by the number of hours in a full-time work week, and multiply this by 100. For example: a full-time work week is in this case 40 hours, but Corinne will be working 34 hours. The calculation will then be: (34/40) * 100 = 85%. This number should be filled in at percentage employment.
What jobs make \$100 an hour?
Here is the list of the top jobs that pay over \$100 an hour:Life coach.Underwater welder.Freelance photographer.Political speechwriter.Tattoo artist.Massage therapist.Interior designer.Commercial pilot.More items…•
Is 100 minutes an hour and 40 minutes?
This conversion of 100 minutes to hours has been calculated by multiplying 100 minutes by 0.0166 and the result is 1.6666 hours.
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## Thursday, November 26, 2009
### The power of proportion!
I was walking to school, when I saw a UFO land on my school. I ran inside and saw that the aliens were holding Mr.Harbeck hostage, then he yelled. "Noel, save me with your proportion skills." Then I replied, "Don't worry Mr.Harbeck, Noel is here!" So then i grabbed a text book and yelled out the following question:
Delia was paid \$35 for 5 h of babysitting. How much should she receive for 3 h? Use a unit rate to find the answer.
This is what I did:
I got to the unit rate per hour, that looked like this:
5h/5=1h
\$35/5=\$7
So then I got \$7/hour.
I then had to get the \$ for 3 hours:
1h x 3 = 3hours
\$7 x 3 = \$21
I saw that one of the tenticles that were holding Harbeck got loose, and it only had 2 tenticles left. So I hit him with another question:
David can saw a log into three pieces in 7 minutes. If he continues sawing at a constant rate, how long will it take him to saw a similar log into 6 pieces?
This is my answers:
To get from 3 pieces to 6 pieces you would have to double it, so I just doubled 7 minutes to get 14 minutes.
Then I noticed a liquid falling onto Harbeck's head, it was the aliens eyeballs. They melted so I decided to give another question a shot and it would die. Here was the last question:
Two quarters has the same value as ten nickels.
What is the values of five quarters in nickels?
Here's what I did:
I used a unit rate to find the answer to this question.
2 quarters / 2 = 1 quarter
10 nickels / 2 = 5 nickels
then I followed up with this;
1 x 5 = 5 quarters
5 x 5 = 25 nickels
Once I finished that sentence the alien exploded and the room was covered in green goo. Harbeck then thanked me and pressed on the intercom; "Office, Can you get Moses down here to clean this up."
THE END!
#### 6 comments:
Jex (Juan Exequiel)8-17 said...
Good job Noel. I like how you explained the questions clearly and put in the answers. Maybe next time you should put some pictures in so it wont look that plain.
Good job again.
Next time do the blog earlier.
Ralph817 said...
Good Job. I liked how you explained everything and used colours. Nect time you should add pictures and do your blog on time.
×Jusŧıŋ£8-17× said...
good job, you finally did that ., but then its late .
Celdrick--Petilla--8-17 said...
GOOG JOB! but its late..well better late than never..... but add pics...next time do it on time
Brendan817 said...
Good job Noel ! Maybe next time you can post your alien story on time. I like your colors because they stand out . I also like your storyline .
Layton 8-17 said...
great job noel! i liked ow you used alot of colour it made it very exiting but you should have used more pictures. but anyways great job!!
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Can't prove these identities?
• Nov 20th 2010, 06:16 AM
doodofnood
Can't prove these identities?
1) sin(x+y)sin(x-y) = cos^2 y - cos^2 x
(sin x cos y + cos x sin y)(sin x cos y - cos x sin y) = cos^2 y - cos^2 x
From there I don't know where to go...
2) tan(pi/4 + x) + tan(pi/4 - x) = 2sec2x
For this one I don't even know where to start, I can't even find a formula that I could use for the LS.
All and any help is very much welcomed!
• Nov 20th 2010, 06:24 AM
Unknown008
Quote:
1) sin(x+y)sin(x-y) = cos^2 y - cos^2 x
(sin x cos y + cos x sin y)(sin x cos y - cos x sin y) = cos^2 y - cos^2 x
From there I don't know where to go...
From there, use the difference of 2 squares.
Then, use the Pythagorean identity to substitute the sin by cos and you'll get it. (Smile)
2. Start by using the compound angle identity for tan!
• Nov 20th 2010, 06:33 AM
TheCoffeeMachine
Quote:
Originally Posted by doodofnood
1) sin(x+y)sin(x-y) = cos^2 y - cos^2 x
(sin x cos y + cos x sin y)(sin x cos y - cos x sin y) = cos^2 y - cos^2 x
$\displaystyle (a+b)(a-b) = a^2-b^2$ (difference of two squares).
Quote:
2) tan(pi/4 + x) + tan(pi/4 - x) = 2sec2x
For this one I don't even know where to start, I can't even find a formula that I could use for the LS.
$\displaystyle \displaystyle \tan(x+y) = \frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$
Oh, late. Ah, well... :)
• Nov 20th 2010, 06:46 AM
doodofnood
so for the first 1 i got down to
sin^2xcos^2y - cos^2xsin^2y
= (1-cos^2x)cos^y - cos^2x(1-cos^2y)
now what...?
• Nov 20th 2010, 06:51 AM
TheCoffeeMachine
Quote:
Originally Posted by doodofnood
now what...?
Well, why did you stop? Keep going, expand the little barackets.
• Nov 20th 2010, 06:58 AM
doodofnood
so if i expand i get..
cos^2y - cos^4xy - cos^2x - cos^4xy
then cos^4xy cancels out?
:O
• Nov 20th 2010, 07:07 AM
harish21
Quote:
Originally Posted by doodofnood
so if i expand i get..
cos^2y - cos^4xy - cos^2x - cos^4xy
then cos^4xy cancels out?
:O
$\displaystyle (1-cos^2x)cos^2y - cos^2x(1-cos^2y)$
$\displaystyle = cos^2y -cos^4y-cos^2x+cos^4y$
now the $\displaystyle cos^4y$ cancel out and you are left with..
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# WebGL2Fundamentals.org
Fix, Fork, Contribute
# WebGL2 3D - Spot Lighting
In the last article we covered point lighting where for every point on the surface of our object we compute the direction from the light to that point on the surface. We then do the same thing we did for directional lighting which is we took the dot product of the surface normal (the direction the surface is facing) and the light direction. This gave us a value of 1 if the two directions matched and should therefore be fully lit. 0 if the two directions were perpendicular and -1 if they were opposite. We used that value directly to multiply the color of the surface which gave us lighting.
Spot lighting is only a very small change. In fact if you think creatively about the stuff we've done so far you might be able to derive your own solution.
You can imagine a point light as a point with light going in all directions from that point. To make a spot light all we need to do is choose a direction from that point, this is the direction of our spotlight. Then, for every direction the light is going we could take the dot product of that direction with our chosen spotlight direction. We'd pick some arbitrary limit and decide if we're within that limit we light. If we're not within that limit we don't light.
In the diagram above we can see a light with rays going in all directions and printed on them is their dot product relative to the direction. We then have a specific direction that is the direction of the spotlight. We choose a limit (above it's in degrees). From the limit we compute a dot limit, we just take the cosine of the limit. If the dot product of our chosen direction of the spotlight to the direction of each ray of light is above the dot limit then we do the lighting. Otherwise no lighting.
To say this another way, let's say the limit is 20 degrees. We can convert that to radians and from that to a value for -1 to 1 by taking the cosine. Let's call that dot space. In other words here's a small table for limit values
`````` limits in
degrees | radians | dot space
--------+---------+----------
0 | 0.0 | 1.0
22 | .38 | .93
45 | .79 | .71
67 | 1.17 | .39
90 | 1.57 | 0.0
180 | 3.14 | -1.0
``````
Then we can the just check
``````dotFromDirection = dot(surfaceToLight, -lightDirection)
if (dotFromDirection >= limitInDotSpace) {
// do the lighting
}
``````
Let's do that
First let's modify our fragment shader from the last article.
``````#version 300 es
precision highp float;
// Passed in from the vertex shader.
in vec3 v_normal;
in vec3 v_surfaceToLight;
in vec3 v_surfaceToView;
uniform vec4 u_color;
uniform float u_shininess;
+uniform vec3 u_lightDirection;
+uniform float u_limit; // in dot space
// we need to declare an output for the fragment shader
out vec4 outColor;
void main() {
// because v_normal is a varying it's interpolated
// so it will not be a unit vector. Normalizing it
// will make it a unit vector again
vec3 normal = normalize(v_normal);
vec3 surfaceToLightDirection = normalize(v_surfaceToLight);
vec3 surfaceToViewDirection = normalize(v_surfaceToView);
vec3 halfVector = normalize(surfaceToLightDirection + surfaceToViewDirection);
- float light = dot(normal, surfaceToLightDirection);
+ float light = 0.0;
float specular = 0.0;
+ float dotFromDirection = dot(surfaceToLightDirection,
+ -u_lightDirection);
+ if (dotFromDirection >= u_limit) {
* light = dot(normal, surfaceToLightDirection);
* if (light > 0.0) {
* specular = pow(dot(normal, halfVector), u_shininess);
* }
+ }
outColor = u_color;
// Lets multiply just the color portion (not the alpha)
// by the light
outColor.rgb *= light;
// Just add in the specular
outColor.rgb += specular;
}
``````
Of course we need to look up the locations of the uniforms we just added.
`````` var lightDirection = [?, ?, ?];
...
var lightDirectionLocation = gl.getUniformLocation(program, "u_lightDirection");
var limitLocation = gl.getUniformLocation(program, "u_limit");
``````
and we need to set them
`````` gl.uniform3fv(lightDirectionLocation, lightDirection);
gl.uniform1f(limitLocation, Math.cos(limit));
``````
And here it is
A few things to note: One is we're negating `u_lightDirection` above. That's a six of one, half dozen of another type of thing. We want the 2 directions we're comparing to point in the same direction when they match. That means we need to compare the surfaceToLightDirection to the opposite of the spotlight direction. We could do this in many different ways. We could pass the negative direction when setting the uniform. This would be my 1st choice but I thought it would be less confusing to call the uniform `u_lightDirection` instead of `u_reverseLightDirection` or `u_negativeLightDirection`
Another thing, and maybe this is just a personal preference, I don't like to use conditionals in shaders if possible. I think the reason is it used to be that shaders didn't actually have conditionals. If you added a conditional the shader compiler would expand the code with lots of multiply by 0 and 1 here and there to make it so there were not any actual conditionals in the code. That meant adding conditionals could make your code explode into combinatorial expansions. I'm not sure that's true anymore but let's get rid of the conditionals anyway just to show some techniques. You can decide yourself whether or not to use them.
There is a GLSL function called `step`. It takes 2 values and if the second value is greater than or equal the first it returns 1.0. Otherwise it returns 0. You could write it like this in JavaScript
``````function step(a, b) {
if (b >= a) {
return 1;
} else {
return 0;
}
}
``````
Let's use `step` to get rid of the conditions
`````` float dotFromDirection = dot(surfaceToLightDirection,
-u_lightDirection);
// inLight will be 1 if we're inside the spotlight and 0 if not
float inLight = step(u_limit, dotFromDirection);
float light = inLight * dot(normal, surfaceToLightDirection);
float specular = inLight * pow(dot(normal, halfVector), u_shininess);
``````
Nothing changes visually but here is that
One other thing is right now the spotlight is super harsh. We're either inside the spotlight or not and things just turn black.
To fix this we could use 2 limits instead of one, an inner limit and an outer limit. If we're inside the inner limit then use 1.0. If we're outside the outer limit then use 0.0. If we're between the inner limit and the outer limit then lerp between 1.0 and 0.0.
Here's one way we could do this
``````-uniform float u_limit; // in dot space
+uniform float u_innerLimit; // in dot space
+uniform float u_outerLimit; // in dot space
...
float dotFromDirection = dot(surfaceToLightDirection,
-u_lightDirection);
- float inLight = step(u_limit, dotFromDirection);
+ float limitRange = u_innerLimit - u_outerLimit;
+ float inLight = clamp((dotFromDirection - u_outerLimit) / limitRange, 0.0, 1.0);
float light = inLight * dot(normal, surfaceToLightDirection);
float specular = inLight * pow(dot(normal, halfVector), u_shininess);
``````
And that works
Now we're getting something that looks more like a spotlight!
One thing to be aware of is if `u_innerLimit` is equal to `u_outerLimit` then `limitRange` will be 0.0. We divide by `limitRange` and dividing by zero is bad/undefined. There's nothing to do in the shader here we just need to make sure in our JavaScript that `u_innerLimit` is never equal to `u_outerLimit`. (note: the example code does not do this).
GLSL also has a function we could use to slightly simplify this. It's called `smoothstep` and like `step` it returns a value from 0 to 1 but it takes both an lower and upper bound and lerps between 0 and 1 between those bounds.
`````` smoothstep(lowerBound, upperBound, value)
``````
Let's do that
`````` float dotFromDirection = dot(surfaceToLightDirection,
-u_lightDirection);
- float limitRange = u_innerLimit - u_outerLimit;
- float inLight = clamp((dotFromDirection - u_outerLimit) / limitRange, 0.0, 1.0);
float inLight = smoothstep(u_outerLimit, u_innerLimit, dotFromDirection);
float light = inLight * dot(normal, surfaceToLightDirection);
float specular = inLight * pow(dot(normal, halfVector), u_shininess);
``````
That works too
The difference is `smoothstep` uses a hermite interpolation instead of a linear interpolation. That means between `lowerBound` and `upperBound` it interpolates like the image below on the right whereas a linear interpolation is like the image on the left.
It's up to you if you think the difference matters.
One other thing to be aware is the `smoothstep` function has undefined results if the `lowerBound` is greater than or equal to `upperBound`. Having them be equal is the same issue we had above. The added issue of not being defined if `lowerBound` is greater than `upperBound` is new but for the purpose of a spotlight that should never be true.
### Be aware of undefined behavior in GLSL
Several functions in GLSL are undefined for certain values. Trying to raise a negative number to a power with `pow` is one example since the result would be an imaginary number. We went over another example above with `smoothstep`.
You need to try to be aware of these or else your shaders will get different results on different machines. The spec, in section 8 lists all the built in functions, what they do, and if there is any undefined behavior.
Here's a list of undefined behaviors. Note `genType` means `float`, `vec2`, `vec3`, or `vec4`.
``genType asin (genType x)``
Arc sine. Returns an angle whose sine is x. The range of values returned by this function is [−π/2, π/2] Results are undefined if ∣x∣ > 1.
``genType acos (genType x)``
Arc cosine. Returns an angle whose cosine is x. The range of values returned by this function is [0, π]. Results are undefined if ∣x∣ > 1.
``genType atan (genType y, genType x)``
Arc tangent. Returns an angle whose tangent is y/x. The signs of x and y are used to determine what quadrant the angle is in. The range of values returned by this function is [−π,π]. Results are undefined if x and y are both 0.
``genType acosh (genType x)``
Arc hyperbolic cosine; returns the non-negative inverse of cosh. Results are undefined if x < 1.
``genType atanh (genType x)``
Arc hyperbolic tangent; returns the inverse of tanh. Results are undefined if ∣x∣≥1.
``genType pow (genType x, genType y)``
Returns x raised to the y power, i.e., xy. Results are undefined if x < 0. Results are undefined if x = 0 and y <= 0.
``genType log (genType x)``
Returns the natural logarithm of x, i.e., returns the value y which satisfies the equation x = ey. Results are undefined if x <= 0.
``genType log2 (genType x)``
Returns the base 2 logarithm of x, i.e., returns the value y which satisfies the equation x=2y. Results are undefined if x <= 0.
``genType sqrt (genType x)``
Returns √x . Results are undefined if x < 0.
``genType inversesqrt (genType x)``
Returns 1/√x. Results are undefined if x <= 0.
``````genType clamp (genType x, genType minVal, genType maxVal)
genType clamp (genType x, float minVal, float maxVal)``````
Returns min (max (x, minVal), maxVal). Results are undefined if minVal > maxVal
``````genType smoothstep (genType edge0, genType edge1, genType x)
genType smoothstep (float edge0, float edge1, genType x)``````
Returns 0.0 if x <= edge0 and 1.0 if x >= edge1 and performs smooth Hermite interpolation between 0 and 1 when edge0 < x < edge1. This is useful in cases where you would want a threshold function with a smooth transition. This is equivalent to:
``` genType t;
t = clamp ((x – edge0) / (edge1 – edge0), 0, 1);
return t * t * (3 – 2 * t);
```
Results are undefined if edge0 >= edge1.
``````mat2 inverse(mat2 m)
mat3 inverse(mat3 m)
mat4 inverse(mat4 m)``````
Returns a matrix that is the inverse of m. The input matrix m is not modified. The values in the returned matrix are undefined if m is singular or poorly conditioned (nearly singular).
• Fundamentals
• WebGL2 vs WebGL1
• Image Processing
• 2D translation, rotation, scale, matrix math
• 3D
• Lighting
• Structure and Organization
• Geometry
• Textures
• Rendering To A Texture
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https://www.esaral.com/q/find-the-perimeter-of-the-shaded-region-in-the-figure-70963
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# Find the perimeter of the shaded region in the figure,
Question:
Find the perimeter of the shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles.
Solution:
Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC
$=\frac{1}{2} \times 2 \pi r+\frac{1}{2} \times 2 \pi r+14+14$
$=2 \pi r+28$
$=2 \times \frac{22}{7} \times \frac{14}{2}+28$
$=72 \mathrm{~cm}$
Hence, the perimeter of the shaded region is 72 cm.
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https://pyret.org/docs/latest/op-precedence.html
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2.4.1 But why not use precedence?
#### 2.4Combining Multiple Operators
Pyret has only one rule for using multiple operators in a single expression: different operators must be explicitly grouped by parentheses, and evaluation always proceeds from left to right.
Therefore, the following expressions are not allowed:
```1 + 1 - 1
1 + 1 > 1
1 + 1 == 2
(3 * 4 / 2)
(3 * 4) / 1 + 1
3 * (4 / 2) + 1```
And will raise an error like:
Pyret disallows mixing operators without clearly defining the operator precedence using parentheses. Conversely, any number of identical operators can be grouped without pairwise parentheses. These expressions are all valid in Pyret:
```1 + (1 - 1)
(1 + 1) > 1
1 + 1 + 1
1 - 1 - 1
(1 + 1) == 2
3 * (4 / 2)
(3 * (4 / 2))
(3 * 4) / (1 + 1)
(3 * (4 / 2)) + 1```
##### 2.4.1But why not use precedence?
Pyret does not use implicit operator precedence or the order of operations that you learned in math class. “Please Excuse My Dear Aunt Sally” does not apply here.
Note for non-American readers: if you’ve never heard of dear Aunt Sally, it’s a mnemonic often used to memorize a standard order-of-operations.
Implicit operator precedence is a common source of errors among even experienced developers, so getting in the habit of explicitly defining precedence using parentheses is a good idea even when using languages that support implicit precedence.
Pyret has many operators, besides just the arithmetic ones: comparison and logical operators, equality operators, testing operators, and others. While it’s possible to memorize the precedence among just a few of these operations, it’s both tedious and unenlightening to memorize precedences among every possible combination of operators. Instead, parentheses make the programmer’s intent explicit.
To make life easier, as said above, Pyret allows you to group multiple uses of the same operator without parentheses: instead of having to write (1 + (2 + 3)) + 4, you can simply write 1 + 2 + 3 + 4. The astute reader may immediately object that while this seems fine for addition, it may be confusing for subtraction: after all, 1 - (2 - 3) produces a different result than (1 - 2) - 3, because subtraction is not associative. Even this is more subtle than it may seem at first: roughnums are not associative even for addition!
```check:
(~100000000000000 + ~-100000000000000) + ~0.0001 is-roughly ~0.0001
~100000000000000 + (~-100000000000000 + ~0.0001) is-roughly ~0
end```
A similarly nuanced problem occurs with comparison operators: writing 1 < 2 < 3 is legal, but will produce an error at runtime, because true cannot be compared to 3. Likewise, 1 == 1 == 1 will produce false, because 1 == 1 evaluates to true, which is not equal to 1. Some other languages attempt to make these expressions “work” and evaluate to true, but by doing so they’ve effectively created new operators that take in three arguments, since their behavior cannot be expressed in terms of any pairwise usage of a binary operator.
Pyret takes the firm stance that since every operator has its own quirks, it does not make sense to create a complex, hard-to-predict set of rules for how different operators interact. Instead, it uses just one single rule, with the easy use of parentheses to resolve any unintended behaviors.
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https://fr.maplesoft.com/support/help/view.aspx?path=NumberTheory%2FNumberOfPrimeFactors
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Number of Prime Factors - Maple Help
Home : Support : Online Help : Mathematics : Number Theory : Number of Prime Factors
NumberTheory
NumberOfPrimeFactors
number of prime factors counted with multiplicity
Calling Sequence
NumberOfPrimeFactors(n)
Omega(n) $\mathrm{\Omega }\left(n\right)$
Parameters
n - integer
Description
• The NumberOfPrimeFactors(n) command computes the number of prime factors of the integer n counted with multiplicity.
• Every prime number divides 0 evenly, so 0 has infinitely many prime factors. However, for consistency with, for example, the Divisors command, NumberOfPrimeFactors(0) returns an error.
• To determine the number of distinct prime divisors of n (that is, without respect to multiplicity), use the distinct = true (or just distinct) option.
• Omega and $\mathrm{\Omega }$ are aliases of NumberOfPrimeFactors.
• You can enter the command Omega using either the 1-D or 2-D calling sequence. For example, Omega(8) is equivalent to $\mathrm{\Omega }\left(8\right)$.
Examples
> $\mathrm{with}\left(\mathrm{NumberTheory}\right):$
> $\mathrm{NumberOfPrimeFactors}\left(5\right)$
${1}$ (1)
> $\mathrm{NumberOfPrimeFactors}\left(-9\right)$
${2}$ (2)
> $\mathrm{NumberOfPrimeFactors}\left(12\right)$
${3}$ (3)
> $\mathrm{NumberOfPrimeFactors}\left(12,'\mathrm{distinct}'\right)$
${2}$ (4)
> $\mathrm{\Omega }\left(57\right)$
${2}$ (5)
> $S≔\mathrm{sum}\left(\mathrm{\Omega }\left(f\left(i\right)\right),i=1..n\right)$
${S}{≔}{\sum }_{{i}{=}{1}}^{{n}}{}{\mathrm{\Omega }}{}\left({f}{}\left({i}\right)\right)$ (6)
> $\mathrm{eval}\left(S,\left[\mathrm{=}\left(f,k↦2\cdot k+1\right),n=15\right]\right)$
${21}$ (7)
> $\mathrm{NumberOfPrimeFactors}\left(0\right)$
Compatibility
• The NumberTheory[NumberOfPrimeFactors] command was introduced in Maple 2016.
• For more information on Maple 2016 changes, see Updates in Maple 2016.
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https://dsp.stackexchange.com/questions/48785/wikipedia-equation-for-dft-seems-to-be-bad/48787
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Wikipedia equation for DFT seems to be bad?
I was writing a simple fourier transform implementation and looked at the DFT equation on wikipedia for reference, when I noticed that I was doing something differently, and after thinking about it felt that the wikipedia version must be wrong because it's very simple to think of a signal that when fourier transformed (with that equation) will return an incorrect spectrum: Because the equation wraps the signal around the complex plane only once (due to the $n/N$ with $0<n<N-1$), any signal that is periodic an even number of times (while wrapping the complex plane) will have no spectrum as the usual peaks (while going around the unit circle) that would appear during a DFT will cancel each other out (when an even number of them appear).
To check this I wrote some code which produced the following image, which seems to confirm what my thoughts.
"Time using equation" uses the equation $$X_f = \sum_{n=0}^{N-1} x_n (\cos(2\pi f t_n) - i \sin(2\pi f t_n))$$ with $t$ a vector of time (so the time $t_n$ at which $x_n$ was sampled for example). It can be found in the function ft below.
The wikipedia equation, linked above, is copied here for reference: $$X_f = \sum_{n=0}^{N-1} x_n \left(\cos\left(2\pi f \frac{n}{N}\right) - i\sin\left(2\pi f \frac{n}{N}\right)\right)$$ It can be found in the function ft2.
import numpy as np
import matplotlib.pyplot as plt
plt.style.use('ggplot')
def ft(t, s, fs):
freq_step = fs / len(s)
freqs = np.arange(0, fs/2 + freq_step, freq_step)
S = []
for freq in freqs:
real = np.sum(s * np.cos(2*np.pi*freq * t))
compl = np.sum(- s * np.sin(2*np.pi*freq * t))
tmpsum = (real**2 + compl**2) ** 0.5
S.append(tmpsum)
return S, freqs
def ft2(s, fs): # Using wikipedia equation
nump=len(s)
freq_step = fs / nump
freqs = np.arange(0, fs/2 + freq_step, freq_step)
S = []
for i, freq in enumerate(freqs):
real = np.sum(s * np.cos(2*np.pi*freq * i/nump))
compl = np.sum(- s * np.sin(2*np.pi*freq * i/nump))
tmpsum = (real**2 + compl**2) ** 0.5
S.append(tmpsum)
return S, freqs
def main():
f = 5
fs = 100
t = np.linspace(0, 2, 200)
y = np.sin(2*np.pi*f*t) + np.cos(2*np.pi*f*2*t)
fig = plt.figure()
ax.set_title('Signal in time domain')
ax.set_xlabel('t')
ax.plot(t, y)
S, freqs = ft(t, y, fs)
ax.set_xticks(np.arange(0, freqs[-1], 2))
ax.set_title('Time using equation')
ax.set_xlabel('frequency')
ax.plot(freqs, S)
S, freqs = ft2(y, fs)
ax.set_title('Using Wiki equation')
ax.set_xlabel('frequency')
ax.set_xticks(np.arange(0, freqs[-1], 2))
ax.plot(freqs, S)
plt.tight_layout()
plt.show()
main()
Obviously it seems rather unlikely that I would have randomly found an error on such a high profile wiki page. But I can't see a mistake in what I've done?
• To get a deeper understanding of the meaning of a DFT, I recommend you read my first two blog articles: "The Exponential Nature of the Complex Unit Circle" (dsprelated.com/showarticle/754.php) and "DFT Graphical Interpretation: Centroids of Weighted Roots of Unity" (dsprelated.com/showarticle/768.php). – Cedron Dawg Apr 27 '18 at 13:57
• Thanks I'll take a look. I'm honestly very surprised at the attention this got when it's all due to a very silly bug in my code. – Nimitz14 Apr 27 '18 at 18:20
• I'm surprised too. The continuous vs discrete thing is a big deal though. My blog is all about the discrete case without any reference to the continuous case which is different than teaching the discrete case as a sampled version of the continuous case. – Cedron Dawg Apr 27 '18 at 18:44
You have a bug in ft2. You are incrementing i, and freq together. That's not how you want your summation to work. I messed around with fixing it, but it got messy. I decided to rewrite it from a discrete perspective instead of trying to use the continuous terminology. In the DFT, the sampling rate is irrelevant. What matters is how many samples are used (N). The bin numbers (k) then correspond to frequency in units of cycles per frame. I tried to keep you code as intact as possible so it would remain easily comprehensible to you. I also unfurled the DFT calculation loops to hopefully reveal their nature a little bit better.
Hope this helps.
Ced
import numpy as np
import matplotlib.pyplot as plt
def ft(t, s, fs):
freq_step = fs / len(s)
freqs = np.arange(0, fs/2, freq_step)
S = []
for freq in freqs:
real = np.sum(s * np.cos(2*np.pi*freq * t))
compl = np.sum(- s * np.sin(2*np.pi*freq * t))
tmpsum = (real**2 + compl**2) ** 0.5
S.append(tmpsum)
return S, freqs
def ft3(s, N): # More efficient form of wikipedia equation
S = []
slice = 0.0
sliver = 2*np.pi/float(N)
for k in range(N/2):
sum_real = 0.0
sum_imag = 0.0
angle = 0.0
for n in range(N):
sum_real += s[n] * np.cos(angle)
sum_imag += -s[n] * np.sin(angle)
angle += slice
slice += sliver
tmpsum = (sum_real**2 + sum_imag**2) ** 0.5
S.append(tmpsum)
return S
def ft4(s, N): # Using wikipedia equation
S = []
for k in range(N/2):
sum_real = 0.0
sum_imag = 0.0
for n in range(N):
sum_real += s[n] * np.cos(2*np.pi*k*n/float(N))
sum_imag += -s[n] * np.sin(2*np.pi*k*n/float(N))
tmpsum = (sum_real**2 + sum_imag**2) ** 0.5
S.append(tmpsum)
return S
def ft5(s, N): # Roots of Unity Weighted Sum
sliver = 2 * np.pi / float( N )
root_real = np.zeros( N )
root_imag = np.zeros( N )
angle = 0.0
for r in range(N):
root_real[r] = np.cos( angle )
root_imag[r] = -np.sin( angle )
angle += sliver
S = []
for k in range( N/2 ):
sum_real = 0.0
sum_imag = 0.0
r = 0
for n in range( N ):
sum_real += s[n] * root_real[r]
sum_imag += s[n] * root_imag[r]
r += k
if r >= N : r -= N
tmpsum = np.sqrt( sum_real*sum_real + sum_imag*sum_imag )
S.append( tmpsum )
return S
def main():
N = 200
fs = 100.0
time_step = 1.0 / fs
t = np.arange(0, N * time_step, time_step)
f = 5.0
y = np.sin(2*np.pi*f*t) + np.cos(2*np.pi*f*2*t)
fig = plt.figure()
ax.set_title('Signal in time domain')
ax.set_xlabel('t')
ax.plot(t, y)
S, freqs = ft(t, y, fs)
ax.set_xticks(np.arange(0, freqs[-1], 2))
ax.set_title('Time using equation')
ax.set_xlabel('frequency')
ax.plot(freqs, S)
S = ft3(y, N)
ax.set_title('Using Wiki equation')
ax.set_xlabel('frequency')
ax.set_xticks(np.arange(0, freqs[-1], 2))
print len(S), len(freqs)
ax.plot(freqs, S)
plt.tight_layout()
plt.show()
main()
• btw you were probably having problems because my code assumed python3 ;) – Nimitz14 Apr 27 '18 at 18:33
• @Nimitz14, Not a big deal. I added the "float()" and a bunch of ".0"s on the numbers. Your code ran fine, the only thing I had to remove was the "plt.style.use('ggplot')" statement. – Cedron Dawg Apr 27 '18 at 18:46
• @Nimitz14, I forgot to mention, I added a ft5 routine to the code that pre-calculates the roots of unity values and really shows how the DFT is calculated using the same roots for each bin. – Cedron Dawg Apr 27 '18 at 21:19
i am not gonna look through your code. the wikipedia page looks okay, but it is a good example of the "format war" or "notation war" or "style war" between mathematicians and electrical engineers. some of it, i think the math people are right. EEs should have never adopted "$j$" for the imaginary unit. that said, this is a better expression of the DFT and inverse is:
DFT: $$X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j 2 \pi nk/N}$$
iDFT: $$x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] \, e^{j 2 \pi nk/N}$$
because electrical engineers doing DSP like to use $x[n]$ as the a sequence of samples in "time" and $X[k]$ as the sequence of discrete samples in "frequency". mathematicians might like this better:
DFT: $$X_k = \sum\limits_{n=0}^{N-1} x_n \, e^{-i 2 \pi nk/N}$$
iDFT: $$x_n = \frac{1}{N} \sum\limits_{k=0}^{N-1} X_k \, e^{i 2 \pi nk/N}$$
and that is the same as the wikipedia page.
you might need to pay more attention to the use of $+$ or $-$ in the exponent and how that translates to $+$ or $-$ against the $\sin(\cdot)$ term.
• If we used i instead of j, we couldn’t say ELI the ICE man. ELJ the JCE man doesn’t have the same ring. Civilization would be threatened – Stanley Pawlukiewicz Apr 26 '18 at 3:30
• elijah the juice man? – robert bristow-johnson Apr 26 '18 at 9:59
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# Blum integer
In mathematics, a natural number n is a Blum integer if n = p × q is a semiprime for which p and q are distinct prime numbers congruent to 3 mod 4.[1] That is, p and q must be of the form 4t + 3, for some integer t. Integers of this form are referred to as Blum primes.[2] This means that the factors of a Blum integer are Gaussian primes with no imaginary part. The first few Blum integers are
21, 33, 57, 69, 77, 93, 129, 133, 141, 161, 177, 201, 209, 213, 217, 237, 249, 253, 301, 309, 321, 329, 341, 381, 393, 413, 417, 437, 453, 469, 473, 489, 497, ... (sequence A016105 in the OEIS)
The integers were named for computer scientist Manuel Blum.[3]
## Properties
Given n = p × q a Blum integer, Qn the set of all quadratic residues modulo n and coprime to n and aQn. Then:[2]
• a has four square roots modulo n, exactly one of which is also in Qn
• The unique square root of a in Qn is called the principal square root of a modulo n
• The function f : QnQn defined by f(x) = x2 mod n is a permutation. The inverse function of f is: f−1(x) = x((p−1)(q−1)+4)/8 mod n.[4]
• For every Blum integer n, −1 has a Jacobi symbol mod n of +1, although −1 is not a quadratic residue of n:
${\displaystyle \left({\frac {-1}{n}}\right)=\left({\frac {-1}{p}}\right)\left({\frac {-1}{q}}\right)=(-1)^{2}=1}$
No Blum integer is the sum of two squares.
## History
Before modern factoring algorithms, such as MPQS and NFS, were developed, it was thought to be useful to select Blum integers as RSA moduli. This is no longer regarded as a useful precaution, since MPQS and NFS are able to factor Blum integers with the same ease as RSA moduli constructed from randomly selected primes.[citation needed]
## References
1. ^ Joe Hurd, Blum Integers (1997), retrieved 17 Jan, 2011 from http://www.gilith.com/research/talks/cambridge1997.pdf
2. ^ a b Goldwasser, S. and Bellare, M. "Lecture Notes on Cryptography" Archived 2012-04-21 at the Wayback Machine. Summer course on cryptography, MIT, 1996-2001
3. ^ Sloane, N. J. A. (ed.). "Sequence A016105 (Blum integers: numbers of the form p * q where p and q are distinct primes congruent to 3 (mod 4))". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
4. ^ Menezes, Alfred; van Oorschot, Paul; Vanstone, Scott (1997). Handbook of applied cryptography. Boca Raton: CRC Press. p. 102. ISBN 0849385237. OCLC 35292671.
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# Solving a complicated equation for approximate analytical Solution using Mathematica
• Mathematica
• djymndl07
djymndl07
TL;DR Summary
Solving a complicated equation for approximate analytical Solution
Hello there, I am trying to solve the Following equation for r,
$$2 a Q^4+5 r^4 \left(3 c (\omega +1) r^{1-3 \omega }-2 r (r-3 M)-4 Q^2\right)=0$$
Clearly this is unsolvable. But if we substitute a=0 and c=0 we get one of the solution, ##r=\frac{1}{2} \left(\sqrt{9 M^2-8 Q^2}+3 M\right)##. Can I obtain approximate analytical solution of the above equation which gives the same value when substitutions a=0 and c=0 are applied. If yes, then how? I have tried AsymptoticSolve, but got no answer.
Are a, c, Q, M, and ω all constants (i.e. not functions of r)? I take it you want r(a,c,Q,M,ω). If you put in values for a,c,Q,M,ω, you could get numerical solutions that might help guide you. What's the magnitude of ω compared to 1? Could you do an expansion if ω is much larger or smaller than 1?
a,c,Q,M, ##\omega## Are arbitrary constants. ##\omega## lies between -1 and -3. Other constants may take any positive value.
Last edited by a moderator:
Then I am afraid you're doomed. Techniques exist if you know something about these parameters (depending on what it is that you know) but if they can literally be anything, you need the general solution. Which does not exist.
I assume this equation came from some physical problem. So maybe you know some possible values of the constants. Then you could put in those constants and then solve numerically for r as a function of ω, for example. Is that a possible approach?
DeBangis21 and djymndl07
djymndl07 said:
Can I obtain approximate analytical solution
phyzguy said:
solve numerically
I don't think that's what he wants.
phyzguy said:
I assume this equation came from some physical problem. So maybe you know some possible values of the constants. Then you could put in those constants and then solve numerically for r as a function of ω, for example. Is that a possible approach?
Yes, numerically I can do that. but some analytical solution, even if it is an approximate one would be better.
djymndl07 said:
Yes, numerically I can do that. but some analytical solution, even if it is an approximate one would be better.
As @Vanadium 50 said, I don't think that's possible.
How do approximations work? You have a big piece plus a small piece, and you neglect the small piece. What is the small piece here? You tell us you cannot tell - it could be anything. OK, that's fair, but it also means you can't approximate.
I got some way to do that in mathematica. Thank you everyone for the reply. One can see the link Here if interested.
Note the continued questioning about the size of the parameters. Also note that the expression you got assumes a and c are small (which you told us was not something we can assume).
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This site is supported by donations to The OEIS Foundation.
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A096587 Triangle read by rows: T(n,k)=number of Catalan knight paths in Quadrant I from (0,0) to (n,k). A Catalan knight moves (1 right and 2 up) or (1 right and 2 down) or (2 right and 1 up) or (2 right and 1 down). 6
1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 2, 2, 0, 0, 1, 3, 3, 1, 2, 3, 3, 0, 0, 1, 2, 4, 9, 8, 3, 3, 4, 4, 0, 0, 1, 12, 12, 10, 11, 18, 15, 6, 4, 5, 5, 0, 0, 1, 14, 22, 42, 39, 27, 22, 30, 24, 10, 5, 6, 6, 0, 0, 1, 54, 61, 64, 72, 98, 87, 56, 38, 45, 35, 15, 6, 7, 7, 0, 0, 1, 86, 128, 213, 217, 181, 167 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,12 COMMENTS A005220=Column 1; A005201=Column 2. LINKS FORMULA T(0, 0)=1; T(1, 2)=1; for n>=2, T(n, 0)=T(n-2, 1)+T(n-1, 2), T(n, 1)=T(n-2, 0)+T(n-2, 2)+T(n-1, 3); for k>=2, T(n, k)=T(n-2, k-1)+T(n-2, k+1)+T(n-1, k-2)+T(n-1, k+2). EXAMPLE Rows: 1 0 0 1 1 1 0 0 1 0 1 2 2 0 0 1 T(3,2) counts these paths: (0,0)-(1,2)-(2,0)-(3,2) and (0,0)-(1,2)-(2,4)-(3,2). CROSSREFS Cf. A005220, A005201, A096588. Sequence in context: A225927 A029392 A035379 * A136438 A059848 A036865 Adjacent sequences: A096584 A096585 A096586 * A096588 A096589 A096590 KEYWORD nonn,tabl AUTHOR Clark Kimberling, Jun 28 2004 STATUS approved
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Last modified May 25 03:00 EDT 2013. Contains 225634 sequences.
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# Math
posted by .
Sam ate 1/4 of the pizza. If the pizza has 8 slices, how many slices are left?
• Math -
jerome is making gift baskets using a ribbon that is 98 centimeters long he cuts 12 cenontimeter pieces of ribbon to tie on each basket what was the length of the ribbon after jerome cut off 5 pieces
• Math -
8 - 1/4
8/1 - 1/4
8*4/1*4 - 1/4
32/4 - 1/4
31/4
• Math -
Sam ate 1/4 of 8 is 1/4 x 8 = 2 slices
There are 6 slices left.
• Math -
For the ribbon problem
98 - 12(5) = length of ribbon.
• Math -
6
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Home > Standard Error > How To Find Standard Error
# How To Find Standard Error
## Contents
The researchers report that candidate A is expected to receive 52% of the final vote, with a margin of error of 2%. For each sample, the mean age of the 16 runners in the sample can be calculated. Advertisement Autoplay When autoplay is enabled, a suggested video will automatically play next. For a value that is sampled with an unbiased normally distributed error, the above depicts the proportion of samples that would fall between 0, 1, 2, and 3 standard deviations above navigate here
Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some n is the size (number of observations) of the sample. Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s. doi:10.2307/2340569. https://en.wikipedia.org/wiki/Standard_error
## How To Find Standard Error
ADDITIONAL INFO Links About FAQ Terms Privacy Policy Contact Site Map Explorable App Like Explorable? Statistic Standard Deviation Sample mean, x σx = σ / sqrt( n ) Sample proportion, p σp = sqrt [ P(1 - P) / n ] Difference between means, x1 - Data distributionsSummarizing spread of distributionsInterquartile range (IQR)Practice: Interquartile range (IQR)Measures of spread: range, variance & standard deviationComparing range and interquartile range (IQR)The idea of spread and standard deviationCalculating standard deviation step In this scenario, the 400 patients are a sample of all patients who may be treated with the drug.
Very slow. If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. Figure 1. How To Find Se In Statistics Journal of the Royal Statistical Society.
The calculations involved are somewhat complex, and the risk of making a mistake is high. How To Get Standard Error From Standard Deviation Regressions differing in accuracy of prediction. Despite the small difference in equations for the standard deviation and the standard error, this small difference changes the meaning of what is being reported from a description of the variation https://en.wikipedia.org/wiki/Standard_error The standard error of the mean (SEM) (i.e., of using the sample mean as a method of estimating the population mean) is the standard deviation of those sample means over all
This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called Calculating Standard Error In Excel Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. AP Statistics Tutorial Exploring Data ▸ The basics ▾ Variables ▾ Population vs sample ▾ Central tendency ▾ Variability ▾ Position ▸ Charts and graphs ▾ Patterns in data ▾ Dotplots They report that, in a sample of 400 patients, the new drug lowers cholesterol by an average of 20 units (mg/dL).
## How To Get Standard Error From Standard Deviation
As will be shown, the mean of all possible sample means is equal to the population mean. http://vassarstats.net/dist.html National Center for Health Statistics typically does not report an estimated mean if its relative standard error exceeds 30%. (NCHS also typically requires at least 30 observations – if not more How To Find Standard Error Comments View the discussion thread. . Standard Error Example Because the age of the runners have a larger standard deviation (9.27 years) than does the age at first marriage (4.72 years), the standard error of the mean is larger for
Next, consider all possible samples of 16 runners from the population of 9,732 runners. http://xvisionx.com/standard-error/find-the-standard-error-of-the-estimated-mean-difference.html It is rare that the true population standard deviation is known. Compare the true standard error of the mean to the standard error estimated using this sample. The graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16. Std Error Calculation
The data set is ageAtMar, also from the R package openintro from the textbook by Dietz et al.[4] For the purpose of this example, the 5,534 women are the entire population The ages in one such sample are 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. American Statistical Association. 25 (4): 30–32. http://xvisionx.com/standard-error/how-to-find-standard-error-in-excel.html doi:10.4103/2229-3485.100662. ^ Isserlis, L. (1918). "On the value of a mean as calculated from a sample".
For the purpose of this example, the 9,732 runners who completed the 2012 run are the entire population of interest. Calculating Standard Error Of Proportion It is useful to compare the standard error of the mean for the age of the runners versus the age at first marriage, as in the graph. Correction for finite population The formula given above for the standard error assumes that the sample size is much smaller than the population size, so that the population can be considered
## Test Your Understanding Problem 1 Which of the following statements is true.
Close Yeah, keep it Undo Close This video is unavailable. Hyattsville, MD: U.S. Standard Error of the Estimate Author(s) David M. How Do You Calculate The Standard Error Solution The correct answer is (A).
The table below shows how to compute the standard error for simple random samples, assuming the population size is at least 20 times larger than the sample size. Home > Research > Statistics > Standard Error of the Mean . . . Because of random variation in sampling, the proportion or mean calculated using the sample will usually differ from the true proportion or mean in the entire population. http://xvisionx.com/standard-error/find-standard-error-of-mean-calculator.html For the purpose of hypothesis testing or estimating confidence intervals, the standard error is primarily of use when the sampling distribution is normally distributed, or approximately normally distributed.
Standard error of the mean This section will focus on the standard error of the mean. Loading... The graph shows the ages for the 16 runners in the sample, plotted on the distribution of ages for all 9,732 runners. The confidence interval of 18 to 22 is a quantitative measure of the uncertainty – the possible difference between the true average effect of the drug and the estimate of 20mg/dL.
In an example above, n=16 runners were selected at random from the 9,732 runners. If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the Naturally, the value of a statistic may vary from one sample to the next. Working...
National Center for Health Statistics (24). Watch QueueQueueWatch QueueQueue Remove allDisconnect Loading... It will be shown that the standard deviation of all possible sample means of size n=16 is equal to the population standard deviation, σ, divided by the square root of the Loading...
The standard deviation of the age for the 16 runners is 10.23, which is somewhat greater than the true population standard deviation σ = 9.27 years. The standard deviation of the age was 9.27 years. The margin of error of 2% is a quantitative measure of the uncertainty – the possible difference between the true proportion who will vote for candidate A and the estimate of Sign in to make your opinion count.
Notation The following notation is helpful, when we talk about the standard deviation and the standard error. Search this site: Leave this field blank: . The distribution of the mean age in all possible samples is called the sampling distribution of the mean. Sign in 8 Loading...
Despite the small difference in equations for the standard deviation and the standard error, this small difference changes the meaning of what is being reported from a description of the variation
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2 Bronze
## How to calculate disk work load?
Can someone tell me the calculation, how to calculate?
An application created 4500 I/Os at peak workload with a peak work loads with read/write ratio of 2:1. What is disk load at peak activity for a RAID 6 configuration?
1 Solution
Accepted Solutions
4 Tellurium
## Re: How to calculate disk work load?
Hi Arslan,
I ran your question by Sagar Patil, who manages the ISM curriculum. Here is his answer:
Can someone tell me the calculation, how to calculate?
An application created 4500 I/Os at peak workload with a peak work loads with read/write ratio of 2:1. What is disk load at peak activity for a RAID 6 configuration?
Therefore, reads = 2/3 * 4500 = 3000 I/Os
And writes = 1/3 * 4500 = 1500 I/Os
For RAID 6, write penalty is 6.
Therefore, Disk load = (3000 * 1) + (1500 *6) = 3000 + 9000 = 12000 I/Os
5 Replies
Highlighted
4 Tellurium
## Re: How to calculate disk work load?
Hi Arslan,
Can you show how you are currently calculating the answer to this question?
Highlighted
2 Bronze
Hi Kate,
Highlighted
2 Bronze
## Re: How to calculate disk work load?
Hy kate,
I found the solution and formula is "read+(write*write penalty)".
So above calculation will be
reads are 66.66% and writes are 33.33%
RAID 6 penalty is 6
Hence 66.66 + (6*33.33) = 66.66 + 199.98
Highlighted
2 Bronze
## Re: How to calculate disk work load?
Sorry for above calculation, as i did not calculate with 4500
66.66% reads of 4500 = 2997
33.33% Writes of 4500 = 1499.8
4 Tellurium
## Re: How to calculate disk work load?
Hi Arslan,
I ran your question by Sagar Patil, who manages the ISM curriculum. Here is his answer:
Can someone tell me the calculation, how to calculate?
An application created 4500 I/Os at peak workload with a peak work loads with read/write ratio of 2:1. What is disk load at peak activity for a RAID 6 configuration?
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# Series of functions: convergence interval type vs. monotony of the function and calculation of supx∈[r,∞)|fn(x)|\mathrm{sup}_{x \in [r,\infty)}|f_n(x)|
I have two doubts regarding the normal convergence of a function series.
Consider a real functions series fn(x):A⊂R→Rf_n(x): A \subset \mathbb{R} \to \mathbb{R} of wich I want to study the normal convergence.
∑n≥0fn(x)\sum_{n \geq 0} f_n(x)
Suppose that I find that the series
∑n≥0|fn(x)|\sum_{n \geq 0} |f_n(x)|
converges pointwise in a open interval I=(a,+∞)I=(a,+ \infty).
Since normal convergence is stronger than pointwise convergence of ∑n≥0|fn(x)|\sum_{n \geq 0} |f_n(x)|, only x∈Ix \in I should be considered. Besides that, proving that series converges in a open interval (as II is) would imply that it converges normally also in the corresponding closed interval, that is I′=[a,∞)I’=[a,\infty), but this would also imply that the series ∑n≥0|fn(x)|\sum_{n \geq 0} |f_n(x)| converges pointwise in x=ax=a which is not true. Therefore I will look in “smaller” intervals inside II, so for x∈Jr=[r,∞)x \in J_r=[r,\infty) with r>ar>a.
The normal convergence requires to find some constants MnM_n such that
supx∈[r,∞)|fn(x)|
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# Convert 4 meters per minute to inches per hour (4 m/min to in/h conversion)
## 4 meters per minute is equal to how many inches per hour?
4 meters per minute is equal to 9448.8 inches per hour.
• 4 meters per minute = 9448.8 inches per hour
• 4.1 meters per minute = 9685.02 inches per hour
• 4.2 meters per minute = 9921.24 inches per hour
• 4.3 meters per minute = 10157.46 inches per hour
• 4.4 meters per minute = 10393.68 inches per hour
• 4.5 meters per minute = 10629.9 inches per hour
• 4.6 meters per minute = 10866.12 inches per hour
• 4.7 meters per minute = 11102.34 inches per hour
• 4.8 meters per minute = 11338.56 inches per hour
• 4.9 meters per minute = 11574.78 inches per hour
## How to convert 4 meters per minute to inches per hour?
To convert 4 meters per minute to inches per hour, multiply the value in meters per minute by 2362.2.
### What is the formula to convert 4 meters per minute to inches per hour?
The conversion formula to convert 4 meters per minute to inches per hour is :
inches per hour = meters per minute × 2362.2
### What is the conversion factor to convert 4 meters per minute to inches per hour?
The conversion factor to convert 4 meters per minute to inches per hour is 2362.2
### Examples to convert m/min to in/h
#### Example 1
Convert 4.2 m/min to in/h.
Solution:
Converting from meters per minute to inches per hour is very easy.
We know that 1 m/min = 2362.2 in/h.
So, to convert 4.2 m/min to in/h, multiply 4.2 m/min by 2362.2 in/h.
4.2 m/min = 4.2 × 2362.2 in/h
4.2 m/min = 9921.24 in/h
Therefore, 4.2 meters per minute converted to inches per hour is equal to 9921.24 in/h.
#### Example 2
Convert 4.8 m/min to in/h.
Solution:
1 m/min = 2362.2 in/h
So, 4.8 m/min = 4.8 × 2362.2 in/h
4.8 m/min = 11338.56 in/h
Therefore, 4.8 m/min converted to in/h is equal to 11338.56 in/h.
If you don't want to do the calculation from 4 meters per minute to inches per hour manually, you can simply use our 4 meters per minute to inches per hour calculator.
### How to use the 4 meters per minute to inches per hour converter?
To use the 4 meters per minute to inches per hour converter, follow these steps:
1. Enter the value in meters per minute that you want to convert into the "meters per minute" text box.
2. Click on the "Convert" button.
3. The conversion result in inches per hour will be displayed in the "inches per hour" text box.
4. The step by step conversion process will be displayed in the "Conversion steps" text box.
5. To copy the meters per minute to inches per hour conversion steps, click on the "Copy" button.
6. To report an incorrect conversion, click on the "Report incorrect conversion" button.
7. To reset the converter, click on the "Reset" button.
8. To inverse the conversion and convert 4 Inches per hour to Meters per minute, click on the "Swap" button.
9. To convert between other units of speed, you can choose the units from the "From" and "To" drop-down menus.
## Meters per minute to inches per hour conversion table
The meters per minute to inches per hour conversion chart below shows a list of various meters per minute values converted to inches per hour
Meters per minute (m/min) Inches per hour (in/h)
4 m/min 9448.8 in/h
4.05 m/min 9566.91 in/h
4.1 m/min 9685.02 in/h
4.15 m/min 9803.13 in/h
4.2 m/min 9921.24 in/h
4.25 m/min 10039.35 in/h
4.3 m/min 10157.46 in/h
4.35 m/min 10275.57 in/h
4.4 m/min 10393.68 in/h
4.45 m/min 10511.79 in/h
4.5 m/min 10629.9 in/h
4.55 m/min 10748.01 in/h
4.6 m/min 10866.12 in/h
4.65 m/min 10984.23 in/h
4.7 m/min 11102.34 in/h
4.75 m/min 11220.45 in/h
4.8 m/min 11338.56 in/h
4.85 m/min 11456.67 in/h
4.9 m/min 11574.78 in/h
4.95 m/min 11692.89 in/h
## Meters per minute (m/min)
### 4 meters per minute equivalents in other speed units
• 4 meters per minute = 576000 centimeters per day
• 4 meters per minute = 24000 centimeters per hour
• 4 meters per minute = 400 centimeters per minute
• 4 meters per minute = 6.68 centimeters per second
• 4 meters per minute = 18897.64 feet per day
• 4 meters per minute = 787.4 feet per hour
• 4 meters per minute = 13.12 feet per minute
• 4 meters per minute = 0.2186668 feet per second
• 4 meters per minute = 226771.6 inches per day
• 4 meters per minute = 9448.8 inches per hour
• 4 meters per minute = 157.48 inches per minute
• 4 meters per minute = 2.624668 inches per second
• 4 meters per minute = 5.76 kilometers per day
• 4 meters per minute = 0.24 kilometers per hour
• 4 meters per minute = 0.004 kilometers per minute
• 4 meters per minute = 6.66668 × 10-5 kilometers per second
• 4 meters per minute = 0.1293332 knots
• 4 meters per minute = 2.223732 × 10-10 light speeds
• 4 meters per minute = 0.0001943632 mach
• 4 meters per minute = 5760 meters per day
• 4 meters per minute = 240 meters per hour
• 4 meters per minute = 0.0666668 meters per second
• 4 meters per minute = 5760000000 microns per day
• 4 meters per minute = 240000000 microns per hour
• 4 meters per minute = 666666.68 microns per minute
• 4 meters per minute = 66666.8 microns per second
• 4 meters per minute = 3.579332 miles per day
• 4 meters per minute = 0.1493332 miles per hour
• 4 meters per minute = 0.002666668 miles per minute
• 4 meters per minute = 4.14248 × 10-5 miles per second
• 4 meters per minute = 5760000 millimeters per day
• 4 meters per minute = 240000 millimeters per hour
• 4 meters per minute = 4000 millimeters per minute
• 4 meters per minute = 66.68 millimeters per second
• 4 meters per minute = 6299.2 yards per day
• 4 meters per minute = 262.48 yards per hour
• 4 meters per minute = 4.36 yards per minute
• 4 meters per minute = 0.0726668 yards per second
## Inches per hour (in/h)
### 4 inches per hour to meters per minute conversion (in/h to m/min)
• 4 inches per hour = 0.001693332 meters per minute
• 4.1 inches per hour = 0.0017356653 meters per minute
• 4.2 inches per hour = 0.0017779986 meters per minute
• 4.3 inches per hour = 0.0018203319 meters per minute
• 4.4 inches per hour = 0.0018626652 meters per minute
• 4.5 inches per hour = 0.0019049985 meters per minute
• 4.6 inches per hour = 0.0019473318 meters per minute
• 4.7 inches per hour = 0.0019896651 meters per minute
• 4.8 inches per hour = 0.0020319984 meters per minute
• 4.9 inches per hour = 0.0020743317 meters per minute
### 4 inches per hour equivalents in other speed units
• 4 inches per hour = 243.84 centimeters per day
• 4 inches per hour = 10.16 centimeters per hour
• 4 inches per hour = 0.1693332 centimeters per minute
• 4 inches per hour = 0.002822224 centimeters per second
• 4 inches per hour = 8 feet per day
• 4 inches per hour = 0.3333332 feet per hour
• 4 inches per hour = 0.00555556 feet per minute
• 4 inches per hour = 9.25688 × 10-5 feet per second
• 4 inches per hour = 96 inches per day
• 4 inches per hour = 0.0666664 inches per minute
• 4 inches per hour = 0.001111108 inches per second
• 4 inches per hour = 0.0024384 kilometers per day
• 4 inches per hour = 0.0001016 kilometers per hour
• 4 inches per hour = 1.693332 × 10-6 kilometers per minute
• 4 inches per hour = 2.822224 × 10-8 kilometers per second
• 4 inches per hour = 5.47512 × 10-5 knots
• 4 inches per hour = 9.4138 × 10-14 light speeds
• 4 inches per hour = 8.22804 × 10-8 mach
• 4 inches per hour = 2.4384 meters per day
• 4 inches per hour = 0.1016 meters per hour
• 4 inches per hour = 0.001693332 meters per minute
• 4 inches per hour = 2.822224 × 10-5 meters per second
• 4 inches per hour = 2438400.04 microns per day
• 4 inches per hour = 101600 microns per hour
• 4 inches per hour = 282.24 microns per minute
• 4 inches per hour = 28.24 microns per second
• 4 inches per hour = 0.001515252 miles per day
• 4 inches per hour = 6.32176 × 10-5 miles per hour
• 4 inches per hour = 1.128888 × 10-6 miles per minute
• 4 inches per hour = 1.753648 × 10-8 miles per second
• 4 inches per hour = 2438.4 millimeters per day
• 4 inches per hour = 101.6 millimeters per hour
• 4 inches per hour = 1.693332 millimeters per minute
• 4 inches per hour = 0.02822224 millimeters per second
• 4 inches per hour = 2.666668 yards per day
• 4 inches per hour = 0.1111112 yards per hour
• 4 inches per hour = 0.001851944 yards per minute
• 4 inches per hour = 3.076224 × 10-5 yards per second
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## What is SEC Cosec and cot?
Secant (sec) is the reciprocal of cosine (cos) Cosecant (cosec) is the reciprocal of sin. Cotangent (cot) is the reciprocal of tan.
### How do you find Cosec cot of SEC?
The cotangent of x is defined to be the cosine of x divided by the sine of x: cot x = cos x sin x . The secant of x is 1 divided by the cosine of x: sec x = 1 cos x , and the cosecant of x is defined to be 1 divided by the sine of x: csc x = 1 sin x .
#### How do you graph a Cosec?
To graph y = csc x, follow these steps:
1. Sketch the graph of y = sin x from –4π to 4π, as shown in this figure.
2. Draw the vertical asymptotes through the x-intercepts, as the following figure shows.
3. Draw y = csc x between the asymptotes and down to (and up to) the sine curve, as shown in the following figure.
What does a Cosec graph look like?
The cosecant graph has vertical asymptotes at each value of x where the sine graph crosses the x-axis; we show these in the graph below with dashed vertical lines. Note that, since sine is an odd function, the cosecant function is also an odd function. That is, csc(−x)=−cscx.
How do you write Cosec in Desmos?
cosecant. Click on the circles next to each function, reciprocal function, and it’s asymptote. Note that csc(x) will be in Red and sec(x) will be in Blue. Move the k slider at the bottom to change the position of the asymptote.
## What is SEC a cosec A?
Secant, cosecant and cotangent, almost always written as sec, cosec and cot are trigonometric functions like sin, cos and tan. sec x = 1. cos x. cosec x = 1. sin x.
### How do you find cosec?
The cosecant of an angle in a right triangle is a relationship found by dividing the length of the hypotenuse by the length of the side opposite to the given angle. This is the reciprocal of the sine function.
#### What is secant graph?
As with tangent and cotangent, the graph of secant has asymptotes. This is because secant is defined as. The cosine graph crosses the x-axis on the interval. at two places, so the secant graph has two asymptotes, which divide the period interval into three smaller sections.
What is SEC SEC cosec and cot X?
Sec, Cosec and Cot. Secant, cosecant and cotangent, almost always written as sec, cosec and cot are trigonometric functions like sin, cos and tan. sec x = 1. cos x. cosec x = 1. sin x. cot x = 1 = cos x. tan x sin x. Note, sec x is not the same as cos -1 x (sometimes written as arccos x).
What is the difference between secsecant and cosec?
Secant, cosecant and cotangent, almost always written as sec, cosec and cot are trigonometric functions like sin, cos and tan. sec x = 1. cos x. cosec x = 1. sin x. cot x = 1 = cos x. tan x sin x. Note, sec x is not the same as cos -1 x (sometimes written as arccos x). Remember, you cannot divide by zero and so these definitions are only valid
## What is secsec x = 1 cos x?
sec x = 1 cos x. cosec x = 1 sin x cot x = 1 = cos x tan x sin x. Note, sec x is not the same as cos-1 x (sometimes written as arccos x).
### What are the new trigonometric functions cosec and secant?
This unit looks at three new trigonometric functions cosecant (cosec), secant (sec) and cotan- gent (cot). These are not entirely new because they are derived from the three functions sine, cosine and tangent.
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# Find the value of theta so that: $\sin(\theta + 30^\circ ) = \cos 50^\circ$
Can you please explain how to solve this question please, I already have the answer but I do not know the process in achieving it.
Find the value(s) of $\theta$ such that: $\sin(\theta + 30^\circ ) = \cos 50^\circ$.
Method 1: $$\sin(\theta+30^\circ)=\cos 50^\circ$$ $$\implies \cos (90^\circ-(\theta+30^\circ))=\cos 50^\circ$$ $$\implies \cos (60^\circ-\theta)=\cos 50^\circ$$ $$\implies \cos \left(\frac{\pi}{3}-\theta\right)=\cos \frac{5\pi}{18}$$ Writing the general solution as follows $$\frac{\pi}{3}-\theta=2n\pi\pm \frac{5\pi}{18}$$ $$\implies \color{blue}{\theta=\frac{\pi}{3}-\left(2n\pi\pm \frac{5\pi}{18}\right)}$$ Method 2: , $$\sin(\theta+30^\circ)=\cos 50^\circ$$ $$\implies \sin (\theta+30^\circ)=\sin (90^\circ-50^\circ)$$ $$\implies \sin (\theta+30^\circ)=\sin 40^\circ$$ $$\implies \sin \left(\theta+\frac{\pi}{6}\right)=\sin\frac{2\pi}{9}$$ Writing the general solution as follows $$\theta+\frac{\pi}{6}=2n\pi+ \frac{2\pi}{9}$$ $$\implies \theta=2n\pi+\frac{2\pi}{9}-\frac{\pi}{6}$$ $$\implies \color{blue}{\theta=2n\pi+\frac{\pi}{18}}$$ or
$$\theta+\frac{\pi}{6}=(2n+1)\pi- \frac{2\pi}{9}$$ $$\implies \theta=2n\pi+\pi-\frac{2\pi}{9}-\frac{\pi}{6}$$ $$\implies \color{blue}{\theta=2n\pi+\frac{11\pi}{18}}$$
• Why do you convert to radians ? – Yves Daoust Jul 21 '15 at 16:11
• No matter, you may change it in degree as your choice just by substituting $\pi=180^\circ$. Do not worry, it's not a rule – Harish Chandra Rajpoot Jul 21 '15 at 16:44
Hint 1: $\sin(a)=\sin(b)$ iff $a-b=2k\pi$ or $a+b = (2k+1)\pi$ for some $k\in\mathbb{Z}$.
Hint 2: $\cos(40^\circ)=\sin(50^\circ)$.
Hint:
$$\sin \theta = \cos (90^\circ-\theta)$$
$$\cos50^\circ = \sin40^\circ$$
can you solve for $\theta$ using the above?
HINT:
Use $\sin x=\cos(90^\circ-x)$ or $\cos x=\sin(90^\circ\pm x)$
and $\sin y=\sin A\implies y=n180^\circ+(-1)^nA$
or $\cos u=\cos B\implies u= m360^\circ\pm B$ where $m,n$ are integers
Knowing the relation between $\sin(\theta)$ and $\cos(\theta)$ is quite crucial. One of the major relation is that the sine function and cosine function are fairly similar with $90^{\circ}$ difference so, $$Sin(x+90)=cos(x)$$ We are given $x=50$, so $$x+90=30+\theta$$ $$\theta=110$$ or $$180-140=40$$ This is $\theta+30$ so, $$\theta=10^{\circ}$$
• Do you think $\sin(70°+30°)=\cos(50°)$? – Did Jul 21 '15 at 14:01
• @Did sorry for the mistake! Was in a hurry – Socre Jul 21 '15 at 14:38
• And why $180°-140°$ now? – Did Jul 21 '15 at 15:06
• @Did I edited my question – Socre Jul 21 '15 at 15:41
$$\sin(\theta+30°)=\cos(50°)=\sin(40°).$$
Two angles have the same sine when they are equal or supplementary (to a multiple of 360°).
Hence $$\theta=40°-30°+k360°\text{ or }\theta=(180°-40°)-30°+k360°.$$
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2540 Days From February 13, 2024
Want to figure out the date that is exactly two thousand five hundred forty days from Feb 13, 2024 without counting?
Your starting date is February 13, 2024 so that means that 2540 days later would be January 27, 2031.
You can check this by using the date difference calculator to measure the number of days from Feb 13, 2024 to Jan 27, 2031.
January 2031
• Sunday
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January 27, 2031 is a Monday. It is the 27th day of the year, and in the 5th week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2031 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 01/27/2031, and almost everywhere else in the world it's 27/01/2031.
What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 2540 weekdays from Feb 13, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Feb 13, 2024, which falls on a Tuesday. Counting forward, the next day would be a Wednesday.
To get exactly two thousand five hundred forty weekdays from Feb 13, 2024, you actually need to count 3556 total days (including weekend days). That means that 2540 weekdays from Feb 13, 2024 would be November 8, 2033.
If you're counting business days, don't forget to adjust this date for any holidays.
November 2033
• Sunday
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November 8, 2033 is a Tuesday. It is the 312nd day of the year, and in the 312nd week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 2033 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 11/08/2033, and almost everywhere else in the world it's 08/11/2033.
Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date.
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Sunteți pe pagina 1din 3
# EE6340 - Information Theory
## February 21, 2013
1. a) Random variable X= No. of coin tosses till the first head appears. If P(head)=p and P(tail) = q,
P(x = n) = pq n−1
X∞
=⇒ H(X) = − pq n−1 log(pq n−1 )
n=1
p log p pq log q
=− −
1−q p2
= H(p)/p
For p = 21 , H(X) = H(0.5)/0.5 = 2 bits.
b) Best questions are those which have equal chances of being answered as Yes or as No.
P(x = 1) = 12 and P(x=2 or 3 or 4...)= 12 (equally probable).
So first question is, "Is it 1 or not?’".
Similarly,P(x = 2) = 14 and P(x=3 or 4 or 5...)= 14 (equally probable)
So second question is, "Is it 2 or not?’".
The questions proceedP as above.
E(No.of questions) = n=1 n 21n = 2 = H(X).
In general,E(No.of questions) ≥ H(X)
This problem can be interpreted as a source coding problem with 0=no, 1=yes, X=Source,
Y=Encoded sequence.
2. a) H(X, g(X)) = H(X) + H(g(X)|X) (Chain rule)
P P
b) H(g(X)|X) = x p(x)H(g(X)|X = x) = x p(x)0 = 0 (For a given x,g(x) is fixed).
=⇒ H(X, g(X)) = H(X)
c) H(X, g(X)) = H(g(X)) + H(X|g(X)) (Chain rule)
d) H(X|g(X)) ≥ 0 with equality if g(.) is one-to-one.
(a),(b) and (c) =⇒ H(X, g(X)) ≥ H(g(X)
3. Let there be 2 yi ’s y1 and y2 such that for x = x0 , p(x0 , y1 ) > 0 and p(x0 , y2 ) > 0.
=⇒ p(y1 |x0 ) > 0 and p(y2 |x0 ) > 0, neither of them being 0 or 1.
X
H(Y |X) = − p(x, y) log2 p(y|x)
x,y
≥ −p(x0 )p(y1 |x0 ) log2 p(y1 |x0 ) − p(x0 )p(y2 |x0 ) log2 p(y2 |x0 )
>0
since −t log t > 0 for 0 < t < 1. So, H(Y|X)= 0 iff Y is a function of X. Else H(Y|X)>0.
1
4. X=Outcome of world series
Y=No.of games played {4,5,6,7}
1
For Y = 4, there are 2 outcomes {AAAA,BBBB} each with probability 24 .
For Y = 5,there are 2 × 43 = 8 outcomes each with probability 215 .
## for Y = 7,there are 2 × 63 = 40 outcomes each with probability 217 .
Thus,
1
P(Y = 4) =
8
1
P(Y = 5) =
4
5
P(Y = 6) =
16
5
P(Y = 7) =
16
X={AAAA,BBBB,....,BABABAB,....BBBAAAA}
P(BAAAA)= 215 , P(BABABAB)= 217 ,...
There are 2 sequences of length 4, 8 sequences of length 5, 20 sequences of length 6 and 40 sequences
of length 7.
X
H(X) = − p(x) log2 p(x)
x
1 1 1 1
=2 log2 16 + 8 log2 32 + 20 log2 64 + 40 log2 128
16 32 64 128
=5.8125
X
H(Y ) = − p(y) log2 p(y)
y
1 1 5 16 5 16
= log2 8 + log2 4 + log2 + log2
8 4 16 5 16 5
=1.924
## Y=length(X), i.e, Y is a deterministic function of X.
H(Y |X) =0
H(X) + H(Y |X) =H(X, Y ) = H(Y ) + H(X|Y )
=⇒ H(X|Y ) =H(X) − H(Y )
=3.889
2 3 1
5. a) H(X) = 3 log2 2 + 3 log2 3 = 0.918 = H(Y )
1
b) H(X|Y ) = 3 H(X|Y = 0) + 32 H(X|Y = 1) = 0.667 = H(Y |X)
c) H(X, Y ) = H(X) + H(Y |X) = 1.585
d) H(Y ) − H(Y |X) = 0.251
e) I(X; Y ) = H(Y ) − H(Y |X) = 0.251
f)
6. Identically distributed =⇒ H(X1 ) = H(X2 )
H(X2 |X1 ) H(X1 )−H(X2 |X1 ) H(X2 )−H(X2 |X1 ) I(X1 ;X2 )
a) ρ = 1 - H(X1 ) = H(X1 ) = H(X1 ) = H(X1 )
2
H(X) H(Y)
## b) I(X1 ; X2 ) = H(X1 ) − H(X1 |X2 )
But H(X1 |X2 ) ≥ 0 =⇒ I(X1 ; X2 ) ≤ H(X1 ) =⇒ ρ ≤ 1
I(X1 ; X2 ) ≥ 0 =⇒ ρ ≥ 0
∴0≤ρ≤1
c) X1 ,X2 are independent.So H(X1 |X2 ) = H(X1 ) =⇒ I(X1 ; X2 ) = 0
Thus ρ = 0 when X1 and X2 are i.i.d
d) H(X2 |X1 ) = 0 when X2 is a function of X1 =⇒ ρ = 1.
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# nLab divided power algebra
Contents
### Context
#### Algebra
higher algebra
universal algebra
# Contents
## Idea
A divided power algebra is a commutative ring $A$ together with an ideal $I$ and a collection of operations $\{\gamma_{n}\colon I\to A\}_{n\in\mathbb{N}}$ which behave like operations of taking divided powers $x\mapsto x^{n}/n!$ in power series.
## Definition
A divided power algebra is a triple $(A,I,\gamma)$ with
• $A$ a commutative ring (with identity);
• $I$ an ideal of $A$;
• $\gamma=\{\gamma_{n}\colon I\to I\}_{n\geq 1}$ an indexed set of functions (of underlying sets);
where we additionally adopt the convention $\gamma_0(x) = 1$ (which is usually not in $I$), and this data is required to satisfy the following conditions:
1. For each $x\in I$, we have $\gamma_{1}(x)=x$.
2. For each $x,y\in I$ and $n\geq 0$, we have
$\gamma_{n}(x+y)=\sum_{k=0}^{n}\gamma_{n-k}(x)\gamma_{k}(y),$
3. For each $\lambda\in A$, each $x\in I$ and $n\geq 0$, we have
$\gamma_{n}(\lambda x)=\lambda^{n}\gamma_{n}(x).$
4. For each $x\in I$ and each $m,n\geq 0$, we have
$\gamma_{n}(x)\gamma_{m}(x)=\frac{(n+m)!}{n!m!}\gamma_{n+m}(x).$
5. For each $x\in I$ and each $m\geq 0$, $n\geq 1$, we have
$\gamma_{m}(\gamma_{n}(x))=\frac{(n m)!}{(n!)^{m}m!}\gamma_{n m}(x).$
For a given $(A,I)$, a divided power structure on $(A,I)$ is a $\gamma$ making $(A, I, \gamma)$ a divided power algebra.
If $A$ is an $R$-algebra for a ring $R$, we call it a divided power $R$-algebra or PD-$R$-algebra.
###### Remark
Some sources include $\gamma_0$ in the data rather than as convention. Some sources give the data as $\gamma_n : I \to A$ typing while including $\gamma_n(x) \in I$ for $n \geq 1$ as an axiom.
## Properties
Genuine powers can be constructed in the expected way from the divided powers, and when $A$ is torsion free, the reverse is true:
###### Proposition
If $(A,I,\gamma)$ is a divided power algebra, then $n! \gamma_n(x) = x^n$ for every $x \in I$ and $n \geq 0$ (taking $x^0:=1$).
###### Proof
It is true for $n=0$ and $n=1$ by definition. For $n \geq 2$, this follows by induction, since $n! \gamma_n(x) = (n-1)! \gamma_{n-1}(x) \cdot 1! \gamma_1(x) = x^{n-1} \cdot x$.
###### Proposition
If $A$ is a commutative, torsion free ring with an ideal $I$ such that $x^n$ is an $(n!)$-th multiple for every $x \in I$ and $n \geq 0$, then $(A,I)$ has a unique divided power structure, and it is given by $\gamma_n(x) = x^n / n!$.
###### Proof
The hypotheses imply the quotients $x^n / n!$ are unique and well-defined, and any divided power structure on $(A,I)$ must be given by that formula. It’s straightforward to check the definition does give a divided power algebra.
So in the torsion free case, the divided power algebras are precisely of the motivating form. In positive characteristic, though, examples can be somewhat more exotic.
## References
Divided power algebras were originally introduced in
Their theory was further developed in Pierre Berthelot‘s PhD thesis (in the context of crystalline cohomology), which was later published as:
• Pierre Berthelot, Cohomologie cristalline des schémas de caractéristique $p \gt 0$, Lecture Notes in Mathematics, Vol. 407, Springer-Verlag, Berlin, 1974. (doi:10.1007/BFb0068636, MR 0384804)
Review:
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Prove by contradiction that if r is irrational, then r^(1/2) is irrational
2. Originally Posted by hebby
Prove by contradiction that if r is irrational, then r^(1/2) is irrational
Where are you stuck? This is straight fwd
Hint: If x is rational, what can you say about x^2?
3. well then r^2 is rational as well, eg)4^2 is rational..then what?...thats it?
4. Originally Posted by hebby
well then r^2 is rational as well, eg)4^2 is rational..then what?...thats it?
yep !!
so if sqrt(r) is rational => r is rational (Contradiction !!)
5. Thanks, but we r going the other way around....ie if r is rational then sqrt r is rational.
6. Originally Posted by hebby
Thanks, but we r going the other way around....ie if r is rational then sqrt r is rational.
Absolutely not.
We are saying "if sqrt r is rational then r is rational"
Plz note the difference carefully
7. but the questions says if r is irrational then the sqrt r is irrational so we have to work this way right?
8. Originally Posted by hebby
but the questions says if r is irrational then the sqrt r is irrational so we have to work this way right?
dnt confuse yourself. both the statements are equivalent. one implies the other and vice-versa.
if sqrt r is rational then r is rational (we proved it)
but r is irrational (as per the question)
that mean sqrt r can't be rational because otherwise r will be rational (a contradiction)
hence sqrt r must be irrational !!
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# Speed Control of D.C. Motors
The relationship given below gives the speed of a D.C. motor
The above equation shows that the speed depends upon the supply voltage V, the armature circuit resistance Ra, and the field flux Ф, which is produced by the field current. In practice, the variation of these three factors is used for speed control. Thus, there are three general methods of speed control of D.C. Motors.
1. Resistance variation in the armature circuit: This method is called armature resistance control or Rheostat control.
2. Variation of field flux Ф
This method is called field flux control.
3. Variation of the applied voltage.
This method is also called armature voltage control.
## 1. Armature resistance control (Rheostat Control):
Figure: (a) Speed control of a d.c. Shunt motor by armature resistance control.
(b) Speed control of a D.C. Series motor by armature resistance control.
In this method, a variable series resistor Re is put in the armature circuit. The figure (a) above shows the process of connection for a shunt motor. In this case, the field is directly connected across the supply and therefore the flux Ф is not affected by variation of Re.
Figure (b) shows the method of connection of external resistance Re in the armature circuit of a D.C. series motor. In this case, the current and hence the flux is affected by the variation of the armature circuit resistance.
The voltage drop in Re reduces the voltage applied to the armature, and therefore the speed is reduced.
This method has the following drawbacks:
1. In the external resistance Re a large amount of power is wasted.
2. Control is limited to give speed below normal and increase of speed cannot is obtained by this method.
3. For a given value of Re, the speed reduction is not constant but varies with the motor load.
This method is only used for small motors.
## 2. Variation of field flux Ф (Field flux control):
Since the field current produces the flux, and if we control the field current then the speed can be controlled. In the shunt motor, speed can be controlled by connecting a variable resistor Rc in series with the shunt field winding. In the diagram below resistor, Rc is called the shunt field regulator.
Figure: (a) Speed control of a D.C. shunt motor by variation of field flux.
(b) The diverter in parallel with the series of D.C. Motor.
gives the shunt field current
Any of the one methods can vary the field current of the series motor:
• A variable resistance Rd is connected in parallel with the series field winding. The resistor connected in parallel is called the diverter. A portion of the main current is diverted through Rd.
• The second method uses a tapped field control.
Here the ampere-turns are varied by varying the number of field turns. This arrangement is used in electric traction.
Figure: Tapped series field on D.C. motor
The advantages of field control are as follows:
• This is an easy and convenient method.
• The power loss in the shunt field is small because shunt field current Ish is very small.
## 3. Armature Voltage control:
We can control the speed of the D.C. motors by varying the applied voltage to the armature. Ward-Leonard system of speed control works on this principle of armature voltage control. In this system, M is the main dc motor whose speed is to be controlled, and G is a separately excited dc generator. The generator G is driven by a 3- phase driving motor which may be an induction motor or asynchronous motor. The combination of ac driving motor and the dc generator is called the motor-generator (M-G) set.
Figure: Ward-Leonard drive
### Advantages of Ward-Leonard Drives:
1. This drive has a smooth speed control of dc motors over a wide range in both directions.
2. It has inherent regenerative braking capacity.
3. By using an overexcited synchronous motor as the drive for the dc generator, the lagging reactive volt-amperes of the plant are compensated. Therefore the overall power factor of the plant improves.
### Drawbacks of classical Ward-Leonard system:
1. Its initial cost is high because of the use of two additional machines (M-G set) of the same rating as the main dc motor.
2. It has a large size and weight.
3. It requires more floor area and costly foundation.
4. Very frequent maintenance is required.
5. The losses are higher because of lower efficiency.
6. Its drive produces more noise.
### Feedback
• Send your Feedback to [email protected]
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# Thread: cos(x)/x amplitude vs squeeze
1. ## cos(x)/x amplitude vs squeeze
I was presented a question on
$\displaystyle \frac{cos(x)}{x}$ as x approaches infinity.
My approach was to separate the amplitude
$\frac{1}{x} cos(x)$
My explination was since the amplitude was going to zero as x approached infinity, the limit was 0.
I was told this is not correct at all, which i fail to understand how my approach is not correct at all. In fact i can not imagine a single situation with cos(x) where if the amplitude is zero, the over all cos function would do anything but go to zero.
The problem, I was told, was that i did not use the squeeze theorem. i was expected to show
$\frac{-1}{x} < \frac{cos(x)}{x} < \frac{1}{x}$
I do not understand how my approach was not correct at all vs using the squeeze theorem. Anyone explain where my logic if flawed?
2. ## Re: cos(x)/x amplitude vs squeeze
technically, your method uses the product of limits ...
$\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x} = \lim_{x \to \infty} \frac{1}{x} \cdot \lim_{x \to \infty} \cos{x}$
since $\displaystyle \lim_{x \to \infty} \cos{x}$ does not exist, the product of limits does not show that the overall limit is zero ... you would have to use the fact that $\displaystyle -1 \le \cos{x} \le 1$, hence the need for the squeeze theorem.
3. ## Re: cos(x)/x amplitude vs squeeze
Originally Posted by friesr
I was presented a question on
$\displaystyle \frac{cos(x)}{x}$ as x approaches infinity.
My approach was to separate the amplitude
$\frac{1}{x} cos(x)$
My explination was since the amplitude was going to zero as x approached infinity, the limit was 0.
I was told this is not correct at all, which i fail to understand how my approach is not correct at all. In fact i can not imagine a single situation with cos(x) where if the amplitude is zero, the over all cos function would do anything but go to zero.
The problem, I was told, was that i did not use the squeeze theorem. i was expected to show
$\frac{-1}{x} < \frac{cos(x)}{x} < \frac{1}{x}$
I do not understand how my approach was not correct at all vs using the squeeze theorem. Anyone explain where my logic if flawed?
Actually your method IS correct, it's just worded in an incorrect way. You need to say that the product of a function that goes to 0 with a BOUNDED function has a limit of 0.
4. ## Re: cos(x)/x amplitude vs squeeze
$\displaystyle \lim_{x\to\infty} \frac{1}{x}$ is zero and the $\displaystyle \lim_{x\to\infty} Cos(x)$ does not exist. Then $\displaystyle 0 * DNE = DNE$ is the problem with my approach. This is why I would have needed to state that Cos(x) was bounded so that it would have existed or used squeeze on the whole function?
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# Get the Percentage Compared to a Total
Options
✭✭✭✭
I know I'm overthinking this but I haven't worked much with %'s quite yet in SmartSheet.
I'm basically just needed to get a percentage of how much Soft Costs and Hard Costs take from the total budget
So I have the [Construction Loan Amount] column which is my total overall. I need to take the [Soft Cost \$\$] and [Hard Cost \$\$] column totals and calculate how much those two totals take up from the main total in terms of percentage.
Ex:
## Best Answer
• ✭✭✭✭✭✭
edited 12/11/23 Answer ✓
Options
What do you mean by "main total" or "total budget"?
Is it the "construction loan amount"/"total overall"?
• So in the example, \$10,000 of the \$40,000 is 25% and \$20,000 is 50%
• If so, the formula is =[Soft Cost \$\$]/[Construction Loan Amount \$]
• Then set the column as a %
• Repeat for Hard Cost just changing the column name
Or do you mean how the soft/hard costs are related to just each other?
• So in the example, \$10,000 of the \$30,000 is 33% and \$20,000 is 66%.
• If so, the formula is =[Soft Cost \$\$]/([Soft Cost \$\$]+[Hard Cost \$\$])
• Then set the column as a %
• Repeat for Hard Cost just changing the first column name
## Answers
• ✭✭✭✭✭✭
edited 12/11/23 Answer ✓
Options
What do you mean by "main total" or "total budget"?
Is it the "construction loan amount"/"total overall"?
• So in the example, \$10,000 of the \$40,000 is 25% and \$20,000 is 50%
• If so, the formula is =[Soft Cost \$\$]/[Construction Loan Amount \$]
• Then set the column as a %
• Repeat for Hard Cost just changing the column name
Or do you mean how the soft/hard costs are related to just each other?
• So in the example, \$10,000 of the \$30,000 is 33% and \$20,000 is 66%.
• If so, the formula is =[Soft Cost \$\$]/([Soft Cost \$\$]+[Hard Cost \$\$])
• Then set the column as a %
• Repeat for Hard Cost just changing the first column name
• ✭✭✭✭
Options
Your first example was the correct one. I slapped that formula in and that seemed to do the trick so much appreciated! :)
• ✭✭✭✭✭✭
Options
Happy to help.
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template!
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# Generalized Mean Value Theorem
• May 28th 2006, 07:58 AM
TexasGirl
Generalized Mean Value Theorem
Let a<b be elements of R, and let f,g:[a,b]-->R be continuous on [a,b] and differentiable on ]a,b[. Assume that f(a) is less than or equal to g(a) and that f'(x)<g'(x) for x element of ]a,b[. Show that f(x)<g(x) for x element of ]a,b].
Can this be done using Generalized Mean Value Theorem? And either way, could somebody give me some pointers?
Thanks a bunch!
• May 28th 2006, 08:19 AM
CaptainBlack
Quote:
Originally Posted by TexasGirl
Let a<b be elements of R, and let f,g:[a,b]-->R be continuous on [a,b] and differentiable on ]a,b[. Assume that f(a) is less than or equal to g(a) and that f'(x)<g'(x) for x element of ]a,b[. Show that f(x)<g(x) for x element of ]a,b].
Can this be done using Generalized Mean Value Theorem? And either way, could somebody give me some pointers?
Thanks a bunch!
Consider $h:[a,b] \rightarrow \manthbb{R}$ such that
$\forall x \in [a,b]\ h(x)=f(x)-g(x)$.
Then $h$ is strictly decreasing in $[a,b]$, and $h(a)\le 0$.
And the rest should be trivial.
RonL
• May 28th 2006, 08:51 AM
TexasGirl
thanks
thanks a bunch =)
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Open in App
Not now
Magnetism Formula
• Last Updated : 08 Mar, 2022
Magnetism is a scientific process that is induced by an electromagnetic field. Magnetic fields are generated by the magnetic moments and electric currents of elementary particles. Magnetic fields have a strong attraction for ferromagnetic materials. As a result, they can be easily attracted towards permanent magnets. Cobalt, nickel, iron, and their alloys are examples of ferromagnetic compounds. Temperature, as well as other factors such as magnetic field and pressure, influence a material’s magnetic state.
What is Magnetic Field?
A magnetic field is formed around a wire when an electric current is passed through it. This magnetic field also creates concentric rings around the wire. Furthermore, the magnetic field’s direction is determined by the current’s direction.
Furthermore, we may use the ‘right-hand rule,’ which involves pointing the right-hand thumb towards the direction of the current flow. Furthermore, the magnetic field aligns up with the orientation of your folded fingers. The magnitude of the field is determined by the quantity of current and distance from the wire.
Formula
The magnetic effect of an electric field is the property that produces an electric field around the conductor. Prof. H.C Oersted established this phenomenon for the first time in 1820. When he maintained the magnetic needle adjacent to a current-carrying wire, he saw deflection. Ampere’s law indicates the direction in which the needle deflects. The magnetic field’s strength is given by the following formula:
B =
where,
• μ0 depicts the allowance of free space (μ0 ​= 4π×10−7T.m/A).
• I refers to the magnitude of current in amperes.
• d and l refer to distance and wire’s length respectively.
Sample Questions
Question 1. Does the emf in a wire in a magnetic field depend upon the resistance of the loop? If not, why?
Solution:
Since, emf =
Here, Ï• = B x A
As a result, the resistance of the loop does not concern the magnitude of the emf produced. Only the area or size of the loop influences it.
Question 2. The strength of a uniform magnetic field through a 2 m long wire with 7 A current is 3.5 T. Find the force on the wire if it makes an angle of 30 degrees.
Solution:
We know, F = ILBsinθ
Here, I = 7 A, L = 2 m, B = 3.5 T, θ = 30°
⇒ F = (7 A)(2 m)(3.5 T)(sin30°)
⇒ F = 24.5 kg m/s2
Question 3. Consider a loop of wire whose surface is tangential to your vantage point. A constant current flows through the loop in a clockwise direction. What would happen if a magnetic field pointing in your direction were activated?
Solution:
This can be answered using the concept of induction. The current flowing through the loop will adjust to counteract a change in magnetic field. Initially, the loop’s magnetic field is pointing in a direction away from the viewer. As a result, if the external magnetic field is unexpectedly enabled in the direction opposite (towards the spectator), the current in the circuit will reverse this change and boost while staying right to left.
Question 4. While a consistent magnetic flux points into the page, an electron moves at a constant speed to the right somewhere along the plane of the page. Find the direction of the force on it.
Solution:
This question necessitates the use of the right-hand thumb rule. Point your right hand’s fingers in the position of the electron’s velocity (to the right). Turn your thumb in the magnetic field’s direction (into the page). Your palm ought to be approaching the force acting on the electron. However, because electrons are negative, this direction must be overturned, implying that the force is directed upwards through the plane of the page.
Question 5. The strength of a uniform magnetic field through a 0.5 m long wire with 3 A current is 8 T. Find the force on the wire if it makes an angle of 30 degrees.
Solution:
We know, F = ILBsinθ
Here, I = 4 A, L = 0.5 m, B = 8 T, θ = 30°
⇒ F = (4 A)(0.5 m)(8 T)(sin30°)
⇒ F = 8 kg m/s2
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https://oeis.org/A255551
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A255551 Lucky / Unlucky array, shifted version, read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ... 19
2, 4, 3, 6, 5, 7, 8, 11, 19, 9, 10, 17, 39, 27, 13, 12, 23, 61, 57, 45, 15, 14, 29, 81, 91, 97, 55, 21, 16, 35, 103, 121, 147, 117, 85, 25, 18, 41, 123, 153, 199, 181, 177, 109, 31, 20, 47, 145, 183, 253, 243, 277, 225, 139, 33, 22, 53, 165, 217, 301, 315, 369, 345, 295, 157, 37, 24, 59, 187, 247, 351, 379, 471, 465, 447, 325, 175, 43 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 2,1 COMMENTS Note how in comparison to A255545, the even numbers on the first row have been shifted one step left, "pushing" term 1 out of the array proper. This was done to obtain a better alignment with arrays like A083221 and A255127 associated with other sieves, from which one may then induce permutations like A255553 by cross-referencing. The starting offset of the sequence giving the terms in square array is 2. However, we can tacitly assume that a(1) = 1 when the sequence is used one-dimensionally as a permutation of natural numbers. LINKS FORMULA For row = 1, A(row,col) = 2*col; For row > 1 and col = 1, A(row,col) = A000959(row); otherwise, A(row,col) = A255543(row,col-1). EXAMPLE The top left corner of the array: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 3, 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83 7, 19, 39, 61, 81, 103, 123, 145, 165, 187, 207, 229, 249, 271, 291 9, 27, 57, 91, 121, 153, 183, 217, 247, 279, 309, 343, 373, 405, 435 13, 45, 97, 147, 199, 253, 301, 351, 403, 453, 507, 555, 609, 661, 709 15, 55, 117, 181, 243, 315, 379, 441, 505, 571, 633, 697, 759, 825, 889 21, 85, 177, 277, 369, 471, 567, 663, 757, 853, 949, 1045, 1141, 1239, 1333 25, 109, 225, 345, 465, 589, 705, 829, 945, 1063, 1185, 1305, 1423, 1549, 1669 31, 139, 295, 447, 603, 765, 913, 1075, 1227, 1377, 1537, 1689, 1843, 1999, 2155 33, 157, 325, 493, 667, 835, 999, 1177, 1347, 1513, 1687, 1861, 2029, 2205, 2367 ... PROG (Scheme) (define (A255551 n) (if (<= n 1) n (A255551bi (A002260 (- n 1)) (A004736 (- n 1))))) (define (A255551bi row col) (cond ((= 1 row) (+ col col)) ((= 1 col) (A000959 row)) (else (A255543bi row (- col 1))))) ;; Other code as in A255543. CROSSREFS Inverse: A255552. Variant of array A255545. (See also A255543). Row 1: A005843 (even numbers). Column 1: 2 followed by A000959(2..) (Lucky numbers from their second term onward). Main diagonal: A255550. Similar arrays: A083221, A255127. Associated permutations: A255553, A255554. Sequence in context: A278512 A278511 A255552 * A332072 A330340 A328696 Adjacent sequences: A255548 A255549 A255550 * A255552 A255553 A255554 KEYWORD nonn,tabl AUTHOR Antti Karttunen, Feb 26 2015 STATUS approved
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Discover a lot of information on the number 48014: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 48014
Is 48014 a prime number? No
Is 48014 a perfect number? No
Number of divisors 4
List of dividers 1, 2, 24007, 48014
Sum of divisors 72024
Prime factorization 2 x 24007
Prime factors 2, 24007
## How to write / spell 48014 in letters?
In letters, the number 48014 is written as: Forty-eight thousand fourteen. And in other languages? how does it spell?
48014 in other languages
Write 48014 in english Forty-eight thousand fourteen
Write 48014 in french Quarante-huit mille quatorze
Write 48014 in spanish Cuarenta y ocho mil catorce
Write 48014 in portuguese Quarenta e oito mil quatorze
## Decomposition of the number 48014
The number 48014 is composed of:
2 iterations of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
Other ways to write 48014
In letter Forty-eight thousand fourteen
In roman numeral
In binary 1011101110001110
In octal 135616
In US dollars USD 48,014.00 (\$)
In euros 48 014,00 EUR (€)
Some related numbers
Previous number 48013
Next number 48015
Next prime number 48017
## Mathematical operations
Operations and solutions
48014*2 = 96028 The double of 48014 is 96028
48014*3 = 144042 The triple of 48014 is 144042
48014/2 = 24007 The half of 48014 is 24007.000000
48014/3 = 16004.666666667 The third of 48014 is 16004.666667
480142 = 2305344196 The square of 48014 is 2305344196.000000
480143 = 110688796226744 The cube of 48014 is 110688796226744.000000
√48014 = 219.1209711552 The square root of 48014 is 219.120971
log(48014) = 10.77924791403 The natural (Neperian) logarithm of 48014 is 10.779248
log10(48014) = 4.6813678881305 The decimal logarithm (base 10) of 48014 is 4.681368
sin(48014) = -0.86213843761448 The sine of 48014 is -0.862138
cos(48014) = -0.50667278828417 The cosine of 48014 is -0.506673
tan(48014) = 1.7015684630195 The tangent of 48014 is 1.701568
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