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https://www.physicsforums.com/threads/wire-moving-at-constant-speed-in-a-magnetic-field.596526/
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# Wire moving at constant speed in a magnetic field
1. Apr 14, 2012
### elemis
So lets say I have a wire of length l moving in a uniform magnetic field of constant velocity.
Now the induced EMF = Blv
Constant velocity implies constant EMF generated per unit time.
I even have a graph of EMF vs time in my textbook for such a situation showing a flat horizontal line for the induced EMF.
My question is how can this be true ?
The wire cuts the same number of magnetic field lines per unit time. Hence, isn't the rate of change of magnetic flux linkage zero ?
So what would a graph of EMF vs time look like for the above situation ?
2. Apr 14, 2012
### Emilyjoint
Ah no....we have done this question....the rate of change of flux linkage is CONSTANT....that does not mean it is zero.
The emf is constant like you said
3. Apr 14, 2012
### tiny-tim
hi elemis!
ah, what flux? it's just a straight wire!!
you need an area for flux … if this wire was joined by perpendicular wires to a circuit which completed outside the magnetic field, then the flux would increase at rate Blv
but if the whole circuit is inside the field, the flux is constant
yes the emf along the wire is Blv, but if you complete the circuit, that emf may or may not be cancelled by an opposing emf at the opposite side of the circuit
4. Apr 14, 2012
### Bob S
This principle can be made into a dc current generator called a Faraday disk or a homopolar generator. A large (500 megajoule) homopolar generator (with stationary magnet and rotating disk) was built and used at ANU (Australian National Univ.) for several years. See http://en.wikipedia.org/wiki/Homopolar_generator
5. Apr 14, 2012
### Emilyjoint
the example we did was a wire moving at right angles to a uniform magnetic field. An area (=lxv) was swept out each second.
We use this to calculate the voltage developed across the ends of an aeroplane wing flying through the earths's field
6. Apr 14, 2012
### elemis
Could you check the following ?
A wire which is connected in no way to anything else will induce a constant EMF across itself if it cuts a magnetic field at constant speed ? If it is accelerating the graph of EMF vs time would be a straight line of constant gradient through the origin ?
This because as per Fleming's Right Hand Rule electrons in the wire feel a magnetic force (consider the magnetic field is into the page) that directs them towards the bottom of the wire. Thus, an EMF is generated across the wire.
Would there be a circular flow of eddy currents in the wire ?
7. Apr 15, 2012
### Hassan2
This is a case of " motional emf" and should be analyzed using the corresponding equations. However if you want to interpret it as a case of Faraday's law, then consider the wire to be a part of a closed loop where the rest of the loop stays outside the field. Now as the wire moves forward/backward, more lines enter/exit the loop, increasing/decreasing the flux linkage.
8. Apr 15, 2012
### elemis
Ah, that is quite a nice analogy..... Thanks !!! :)
9. Apr 15, 2012
### tiny-tim
yes
the electrons feel a (magnetic) force along the wire, which we interpret as an voltage difference (an emf)
flux has nothing to do with it, though you can introduce a "pretend" flux to get the same result!
as Hassan2 says, you should use the equations appropriate for the problem, rather than try to make similar equations fit
no
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# TNPSC Sample Question Paper(2009) - Solved Questions and answers
1). Consider the following statements:
Assertion (A) : Median of {7, 2, 12. 5, 9} is 7.
Reason (R) : The middle most value of a data is called median.
coding scheme given below :
A). (A) is true, (but (R) is false B). (A) alone is true C). (R) alone is true D). Both (A) and (R) are true and (R) is the correct explanation of (A)
2).
Match List I with List II correctly and select your answer using the codes given below:
List I List II Correlation coefficient Mean=Median=Mode Coeffcient of variation Least For a symmetrical Percentage Variation Mean deviation from Median Cannot exceed unity
A). 4 3 2 1 B). 4 2 3 1 C). 4 3 1 2 D). 3 4 2 1
3). Which one of the following is correctly matched?
A). Harmonic mean - A measure of skewness B). Mean deviation - An average C). Analysis - An arrangement of data D). Sales and profit - Positive correlation
4). Which one of the following is correctly matched?
A). Data - A collection of objects B). Mode - A well defined measure C). Range - Difference between the largest and the smallest items of a data D). Q.D. - A best measure of dispersion
5). The average of three numbers is 135. The largest number is 180 and the difference of the other two is 25. The smallest number is
A). 130 B). 125 C). 120 D). 100
6). By frequency distribution we mean the classification of data according to
A). dissimilarities B). similarities C). class intervals D). magnitude
7). Value of M.D. of{5, 5, 5, 5, 5} from Median is ,
A). 0 B). 2 C). 4 D). 5
8). If y = 3 + 2x. then the coefficient of correlation between x and y is
A). -1 B). 0 C). 0.5 D). 1
9). Which one of the following measures cannot be computed for {2, 7, 5, 10, 4}?
A). Mean B). Median C). Mode D). Harmonic Mean
10). Consider the following statements:
Assertion (A) : Histogram is a graphical presentation of data.
Reason (R) : Statistical data can be represented in the form of graphs.
Now select your answer according to the coding scheme given below :
A). Both (A) and (R) are true B). Both (A) and (R) are false C). (A) is true, but (R) is false D). (A) is false, but (R) is false
Go to :
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83 minus 1 percent
This is where you will learn how to calculate eighty-three minus one percent (83 minus 1 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 83 minus 1 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 83 of something.
83
(100%)
1 percent means 1 per hundred, so for each hundred in 83, you want to subtract 1. Thus, you divide 83 by 100 and then multiply the quotient by 1 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(83 ÷ 100) × 1
= 0.83
We made a pink square that we put on top of the image shown above to illustrate how much 1 percent is of the total 83:
The dark blue not covered up by the pink is 83 minus 1 percent. Thus, we simply subtract the 0.83 from 83 to get the answer:
83 - 0.83
= 82.17
The explanation and illustrations above are the educational way of calculating 83 minus 1 percent. You can also, of course, use formulas to calculate 83 minus 1%.
Below we show you two formulas that you can use to calculate 83 minus 1 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
83 - ((83 × 1/100))
83 - 0.83
= 82.17
Formula 2
Number × (1 - (Percent/100))
83 × (1 - (1/100))
83 × 0.99
= 82.17
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
84 minus 1 percent
Here is the next percent tutorial on our list that may be of interest.
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# A Couple Useful YouTube Videos
Taking a moment to share a couple of finds of the day…
# Non-Apologetic Promotion of Twitter
Update: 50+ Twitter Links, including a Prezi about Twitter in Education.
There have been at least 3 times in the last week when I have showed a colleague this website and have gotten a surprised look of, “Where did you get all of these ideas?” When I tell people that Twitter feeds a majority of my inspiration to go looking for things, they start listening. But no one ever really seems to follow through by actually signing up for Twitter. From what I can tell, it is a slight fear of just not knowing enough. So, here I’m sharing a few links and videos to help people get in the Twitter mood –
Links:
# Create Your Own Math Games using Sharendipty???
This a cool math game for ordering decimals that I found over at Sharendipity. I hope to make some of my own soon in my spare time (hahaha)!
# Math Games for Integers, Multiplication, and Combining Like Terms
I have some slightly under-prepared students semester, so I suggested to them that they should try to work on their basic skills outside of class. However, this requires me to provide some recommended resources to them, and these are what I have discovered:
Multiplication
• Tetris-style game for multiplication facts.
• Fun Multiplication Game that can be played with up to 4 people from around the world.
• For those who just want the traditional worksheets to practice with.
Combining Like Terms
• This like terms game requires you to match the center term with its the appropriate like term.
• Matching Game – Identify the matching terms in two columns.
• Distributive property, combining like terms, evaluating expressions, solving equations.
Integers and Order of Operations
• Pac Man-style game for Order of Operations.
• Circle 0 Puzzle – adding positive and negative integers to sum to zero.
• Circle 3 Puzzle – adding positive and negative integers to sum to three.
• Circle 21 Puzzle – adding positive and negative integers to sum to twenty-one.
• Circle 99 Puzzle – adding positive and negative integers to sum to ninety-nine.
Other Helpful Links
# Poster Presentation Tips
Here are few links I have gathered for those planning to make a poster presentation, or who wish to pass these tips onto your students. This is a skill that has not come easy to me as I am colorblind. =D
# Miscellaneous Links
After a recent afternoon meeting about statistics, I needed to find a few old links that I had buried away. Well, here are a few odds and ends I found while looking:
1. Virtual Math Lab at Texas A&M – This is a very good resource for College Algebra, Intermediate Algebra, and Beginning Algebra. When I opened my link, it actually opened on ‘Absolute Value Equations’, which means that’s probably what my students were struggling with when I initially discovered this website back in 2009.
2. Quick-and-Dirty Guide to the TI-83, TI-83+, TI-84, and TI-84+ – Although it seems like it would be most useful for the beginning calculator student, I have to say that the time I used this website the most was when I taught Calculus, and I couldn’t remember many of the calculus-related functions.
3. Pete Falzone’s On-line Office – I have been borrowing handouts from this guy for the longest time. The pre-algebra resources are especially good for developmental math classes. And I have found a lot of other great worksheets for other courses for when I have been called to substitute at the last minute and needed an ‘in a pinch’ lesson outline.
4. Project-Based Learning – I am obviously all for project-based learning. But if you need a little more background information, along with some additional examples and ideas for your mathematics classroom, feel free to visit this website. There is a good description of project-based learning, along with some wonderful links to helpful websites.
5. Classroom Assessment Techniques – This is definitely worth checking out, as I know that I got at least a couple of ideas from this website for the times when I knew that I had to do an in-class assessment, but needed something that was quick to set-up (I usually realize things 1/2 way into class for some reason).
6. Quiz Star and Easy Test Maker – Two quick links to create on-line and off-line quizzes and tests. Both are free. I know, I know, you probably don’t need another free product to do this, as you already have your own Course Management System, or you have your own system for creating tests. That’s fine, but these may be useful if you’re looking to do something different.
# 10 Potentially Helpful Resources
Here is the most recent set of helpful resources that I have sort of stumbled upon out of well over 200 hundred that I’ve looked at today:
1. Broad Texter – This is a service that allows you to create a group so that your students can join so that they can receive text messages from you. In fact, I’ve set up one for my students that I hope to use in the near future. Feel free to sign up at the top of the page if you’re so inclined.
2. Smart Teaching Blog – My biggest advice when looking at this blog is to start scrolling down and to not get overwhelmed, as there are probably 1000s of resources listed, including this list of the 100 Best YouTube Videos for Teachers.
3. Get the Math – This the link to ‘Get the Math, an initiative out of the PBS Station in NYC, which has challenges related to fields such as Fashion and Video Games. I know I posted this on twitter earlier, but that’s why you need to follow along (if you’re not already).
4. Bubbl.us – This is a simple and free web application that lets you brainstorm online (essentially a stripped down version of Mindomo), so it would be ideal for those who are beginning into the world of Mind-Mapping.
5. CamStudio – Free streaming video software. I mean, does the name remind you of something? Personally, I’m doing just fine with Jing! for now, but some people may want to check into this.
6. Poll Everywhere – Allows you to create a poll that your audience can participate in using their cell phones, twitter, or the web. I’ve personally used this in a classroom before as a quick and simple alternative to using clickers. It doesn’t give you a person-by-person tally, but you can get an overall idea of if your students understand a concept.
7. Super Saas – An online scheduler, which I want to try out for future semesters to have students self-schedule for my office hours. I think that they may be more likely to come if they can schedule themselves. Has anyone tried this successfully? I would love to hear!
8. ToonDoo – The online cartoon, comic strip creator. Create your own cartoons, comic strips, publish, share, and discuss! In fact, I’ve mentioned something similar, called ‘Make Belief Comix’ in the past. The major difference upfront is that ToonDoo is in color.
9. Transfer Big Files – Transfer files up to 1 GB. This would have been especially helpful when I was having trouble with students sending me their homework assignments last semester. Another similar website is You Send It.
10. Motivational Posters – This actually could be turned into a great class project if the students created a mathematics-related image themselves, along with a descriptor to put along the bottom. Another similar website is The Parody Motivational Generator.
# Teaching the Unit Circle
This is simply another consequence of my poking around the web, and although I haven’t taught trigonometry since last summer, I would consider using either of these ideas in the future:
1. Touch Trigonometry – This is an interactive trigonometry graph and circle featuring the six basic trig functions. I wasn’t a fan at first because I’m colorblind, which made it seem like there was just too much going on, but I can see that it is a useful tool for those who are able to distinguish colors.
2. Serving Unit-Circle Trigonometry on a Paper Plate – At first I thought this was an awful idea for the college classroom, but then one day last summer one of my students came in with something very similar. I asked him where he got it, and he said that he made it in high school, and that it was one of the most useful things he’s ever made. And apparently he’s been carrying it around ever since. Although I still wonder why he was taking my class if it was so useful, the point is that if the students are engaged, I believe it aids greatly in the learning. Here is a link to a completed project. And here is a link to a blank circle.
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# [SOLVED] Pre-Calc applications of equations
• Sep 29th 2008, 01:49 PM
soccer49
[SOLVED] Pre-Calc applications of equations
Hi, I need help on this word problem:
You have already invested 550$in a stock with an annual return of 11%. How much of an additional 1100 should be invested at 12% and how much at 6% so that the total return on the entire 1650$ is 9%
I tried to set up my equations as (550 times .11) + (.12x times 1100) + (.06y times 1100) = (1650 times .09)
and x + y = 1100
I am wondering if i am going in the right direction. The answer in the book says 366.67 at 12% and 733.33 at 6% but i tried to work my math and i got nowhere close to that answer.
• Sep 29th 2008, 07:03 PM
Soroban
Hello, soccer49!
You're on the right track, but you put "too much" in your equations.
Quote:
You have already invested $550 in a stock with an annual return of 11%. How much of an additional$1100 should be invested at 12% and how much at 6%
so that the total return on the entire $1650$ is 9% ?
Let $x$ = amounted invested at 12%.
Let $y$ = amount invested at 6%.
Then: . $x + y \:=\:1100$ .[1]
$550 at 11% returns$60.50 in interest.
$x$ dollars at 12% returns $0.12x$ dollars in interest.
$y$ dollars at 6% returns $0.06y$ dollars in interest.
The total interest is: . $60.50 + 0.12x + 0.06y$
. . and we want this to be equal to 9% of $1650 =$148.50
There is our equation: . $60.50 + 0.12x + 0.06y \:=\:148.50$
Multiply by 100: . $6050 + 12x + 6y \:=\:14850 \quad\Rightarrow\quad 12x + 6y \:=\:8800$ .[2]
Solve the system of equation.
$\begin{array}{cccccccc}\text{Multiply {\color{blue}[1]} by -6:} & \text{-}6x - 6y &=& \text{-}6600 \\ \text{Add {\color{blue}[2]}:} & 12x + 6y &=& 8800 \end{array}$
And we get: . $6x \:=\:2200 \quad\Rightarrow\quad x \:=\:\frac{2200}{6} \quad\Rightarrow\quad\boxed{ x\:\approx\:\366.67}$
Substitute into [1]: . $366.67 + y \:=\:1100\quad\Rightarrow\quad \boxed{y \:=\:\733.33}$
• Sep 30th 2008, 03:31 AM
soccer49
Thank you very much, yes that would make sense, i am just not very good at setting up the equations.
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Learn how to add fruit (by putting fruit together) and sutract fruit (by taking fruit away). Examples used are 5+3 and 5-3. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-add-sub-basic
• 03:00
### Teens As Sums With 10 | Place Value (tens And Hundreds) | Early Math | Khan Academy
473 views / 0 likes - added
Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-place-value-topic/cc-early-math-teens/e/teen-numbers-1?utm_source=YT&utm_medium=Desc&utm_campaign=EarlyMath Watch the next lesson: https:
• 15:25
### 10 EASY Card Tricks Anyone Can Do!
486 views / 0 likes - added
Learn how to do amazing card magic tricks to impress your friends! This How To Magic show will teach you the simple secrets of easy card tricks for kids, beginners, and all ages! If you have a deck of playing cards you can perform these magic tricks - car
• 03:37
### Adding Hundreds, Tens, And Ones | Addition And Subtraction Within 100 | Early Math | Khan Academy
685 views / 0 likes - added
• 02:00
592 views / 0 likes - added
Learn that if we have 3 bananas, then we need 7 more bananas to make 10 bananas. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-add-sub-basics/cc-early-math-make-10/e/making-ten-2?utm_
• 03:11
437 views / 0 likes - added
• 03:33
### Losing Tennis Balls | Addition And Subtraction Within 100 | Early Math | Khan Academy
488 views / 0 likes - added
• 14:02
### 11 SIMPLE ORIGAMI IDEAS FOR KIDS
494 views / 0 likes - added
AWESOME ORIGAMI IDEAS FOR KIDS Origami is an ancient craft of folding paper. However, anyone can master it today, it's not so difficult yet very useful. Making origami paper figures helps to relax, concentrate, and develop fine motor skills. It's not only
• 01:58
### Counting In Order | Counting | Early Math | Khan Academy
546 views / 0 likes - added
Learn how to count without making mistakes. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-counting/e/counting-objects?utm_source=YT&utm_medium=Desc&utm_ca
• 01:26 Popular
### My Terrific Dinosaur - Spinosaurus, Velociraptor, And Tyrannosaurus Rex And More!
886 views / 2 likes - added
There are so many terrific dinosaurs, like Spinosaurus, Velociraptor, and Tyrannosaurus rex, to name a few. Did you know that Spinosaurus had a spiny sail on its back? Do you know why? And what about Velociraptors? They may have had feathers but not for f
• 01:54
### Missing Numbers Between 0 And 120 | Counting | Early Math | Khan Academy
619 views / 0 likes - added
Learn how to find missing numbers between 0 and 120. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-numbers-120/e/count-to-100?utm_source=YT&utm_medium=Des
• 04:58
### Number Grid | Counting | Early Math | Khan Academy
482 views / 0 likes - added
In our introductory video to counting, Sal goes through all the numbers from 0 to 100, so we can start to see some interesting patterns. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-
• 04:31
### Counting American Coins | Measurement And Data | Early Math | Khan Academy
504 views / 0 likes - added
Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-money/v/counting-dollars?utm_source=YT&utm_medium=Desc&utm_campaign=EarlyMath Watch the next lesson: htt
• 05:59
### Introduction To Place Value | Place Value (tens And Hundreds) | Early Math | Khan Academy
510 views / 0 likes - added
Learn why we use a "ones place" and a "tens place" when writing numbers. This video uses the number 37 as an example. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-place-value-topic/c
• 11:27
### 7 MAGIC Science Tricks You Can Do!
206 views / 0 likes - added
Science is magic! Learn how to do amazing science tricks to impress your friends with your magic skills! Viral science magic tricks you can do at home, tricks with water, the laminar flow balloon trick tutorial, to how make a balloon inflate on its own an
• 02:37
### Order By Length | Measurement And Data | Early Math | Khan Academy
586 views / 0 likes - added
Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-length-intro/e/order-by-length?utm_source=YT&utm_medium=Desc&utm_campaign=EarlyMath Watch the next lesso
• 05:45
### BATTROBORG Teenage Mutant Ninja Turtles Electronic Battle Game Family Game Challenge
443 views / 0 likes - added
BATTROBORG Teenage Mutant Ninja Turtles Electronic Battle Game Michelangelo vs Donatello Family Fun Game Challenge with Evren. Swing the katana controller to move Teenage Mutant Ninja Turtle robot into an attack. The secret dojo battle arena is perfect fo
• 15:18
### Automatic golf club replaces a bag of clubs and improves your game
126 views / 0 likes - added
I wanted to see if I could make myself a better golfer by combining robotic engineering and golf. The result is a robotic club that senses your swing and corrects your shot for distance. It should be possible to correct for slice and hook with some hardwa
• 02:03
### Comparing Numbers Of Objects | Counting | Early Math | Khan Academy
558 views / 1 likes - added
Learn what "more than" and "less than" mean. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-comparing-numbers/e/compare-groups-through-10?utm_source=YT&utm
• 16:50
### 19 KNITTING AND MACRAME IDEAS
285 views / 1 likes - added
COMFY KNITTING IDEAS AND MACRAME CRAFTS FOR HOME There are many ways to make wour home cozier this autumn. But when it comes to doing it without spending a truckload of money, suddenly, there are not so many variants. Today we are here to save you and you
• 02:04
### Compose Shapes | Geometry | Early Math | Khan Academy
502 views / 0 likes - added
Learn how to combine shapes to make other shapes. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-geometry-topic/cc-early-math-composing-shapes/e/compose-shapes?utm_source=YT&utm_medium
• 12:54
### How to Draw SpongeBob SquarePants
388 views / 1 likes - added
Learn #HowToDraw cute SpongeBob SquarePants easy, step by step drawing tutorial. Looking forward to the SpongeBob 3D animated Movie: Sponge on the Run! SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: htt
• 02:57
### Reading Bar Graph Examples | Measurement And Data | Early Math | Khan Academy
492 views / 0 likes - added
• 03:52
### Ben Eater Eats The Numbers | Counting | Early Math | Khan Academy
700 views / 0 likes - added
Learn how to find which numbers are missing in a number grid. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-numbers-120/e/count-from-any-number?utm_source
• 03:18
### Counting Dollars | Measurement And Data | Early Math | Khan Academy
487 views / 0 likes - added
Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-money/v/counting-dollars?utm_source=YT&utm_medium=Desc&utm_campaign=EarlyMath Watch the next lesson: htt
• 11:28
### Reinforcement Learning: Crash Course AI#9
112 views / 0 likes - added
Reinforcement learning is particularly useful in situations where we want to train AIs to have certain skills we dont fully understand ourselves. Unlike some of the techniques weve discussed so far, reinforcement learning generally only looks at how an AI
• 01:13
### Counting In Pictures | Counting | Early Math | Khan Academy
500 views / 0 likes - added
Learn how to count the number of objects you see in pictures. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-count-object-topic/e/counting-in-scenes?utm_so
• 12:27 Popular
### How to Draw a Mermaid
846 views / 3 likes - added
Learn #HowToDraw a cute girl Mermaid easy, step by step drawing tutorial. Chibi cartoon mermaid girl with long hair. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmCSketch Pencils:
• 03:12
### Subtracting Hundreds, Tens, And Ones | Early Math | Khan Academy
553 views / 0 likes - added
Learn how to subtract three-digit numbers by subtracting ones, tens, and hundreds. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-add-sub-100/cc-early-math-sub-ones-tens-hundreds/e/sub
• 01:31
### Comparing Small Numbers On The Number Line | Counting | Early Math | Khan Academy
565 views / 0 likes - added
Learn to use a number line to compare numbers. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-comparing-numbers/e/comparing-numbers-through-10?utm_source=Y
• 02:10
490 views / 0 likes - added
• 14:37
### How to Draw Holiday Penguins | Christmas Series #7
336 views / 0 likes - added
Learn How to Draw cute Holiday Penguins easy, step by step Christmas drawing tutorial. Christmas Art Playlist: https://www.youtube.com/watch?v=K_PL6... SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: htt
• 02:16
### Telling Time Exercise Example 2 | Measurement And Data | Early Math | Khan Academy
463 views / 0 likes - added
Tell time on unlabeled analog clocks. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-time/e/telling_time?utm_source=YT&utm_medium=Desc&utm_campaign=Ear
• 04:40
### Measuring Lengths 2 | Measurement And Data | Early Math | Khan Academy
487 views / 0 likes - added
Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-measuring-length/e/measuring-lengths-2?utm_source=YT&utm_medium=Desc&utm_campaign=EarlyMath Watch the ne
• 00:44
### Count By Category | Counting | Early Math | Khan Academy
471 views / 0 likes - added
Learn to count the number of things in different categories. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-comparing-numbers/e/sort-groups-by-count?utm_so
• 01:43
### Solving Problems With Bar Graphs 2 | Measurement And Data | Early Math | Khan Academy
544 views / 1 likes - added
• 02:25
### Halves And Fourths | Geometry | Early Math | Khan Academy
536 views / 0 likes - added
Learn how to divide shapes into two or four equal sections. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-geometry-topic/cc-early-math-fractions-of-shapes/e/halves-and-fourths?utm_sou
• 10:05 Popular
### How to Draw a Labrador | Golden Retriever Puppy Easy
1,095 views / 1 likes - added
#DrawSoCuteDogs Learn #HowToDraw a cute Labrador or Golden Retriever Puppy Dog easy, step by step drawing tutorial. Kawaii Dog. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmCSket
• 10:01
### How to Draw a Hummingbird
265 views / 1 likes - added
Learn #HowToDraw a cute Hummingbird drinking nectar from a flower easy, step by step drawing tutorial. Smallest of Birds with colorful feathers and just so cute and pretty! SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf
• 01:45
### Counting Dogs, Mice, And Cookies | Counting | Early Math | Khan Academy
519 views / 0 likes - added
Learn how to count animals organized in different patterns. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-count-object-topic/e/how-many-objects-2?utm_sour
• 08:41
### How to Draw a Cute Cat in a Box Easy
378 views / 4 likes - added
#DrawSoCute Learn #HowToDraw a cute Cat in a box easy, step by step drawing tutorial. Kawaii kitten with bell collar in a gift box. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmC
• 02:54
### Measuring Lengths With Different Units | Measurement And Data | Early Math | Khan Academy
414 views / 0 likes - added
• 13:54
### How Do You Design a Just Society? | Thought Experiment: The Original Position
411 views / 0 likes - added
Thought Experiment: John Rawls’ Original Position Support us on Patreon! https://patreon.com/pbsideachannel We got merch! http://bit.ly/1U8fS1B Tweet us! http://bit.ly/pbsideachanneltwitter Idea Channel Facebook! http://bit.ly/pbsideachannelfacebook Talk
• 06:52
### How to Draw a Cute Frog Easy | Kermit from Muppet Show
378 views / 0 likes - added
#DrawSoCute Learn #HowToDraw and Color cute Kermit the Frog from the Muppets and Sesame Street easy, step by step drawing tutorial. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmC
• 11:08 Popular
### How to Draw an Arctic Fox Easy
1,123 views / 10 likes - added
#DrawSoCute Learn #HowToDraw a cute, cartoon Arctic Fox easy, step by step drawing tutorial. Kawaii fox in winter snow art with snowflakes. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to
• 06:09
### Recognizing Shapes | Geometry | Early Math | Khan Academy
583 views / 1 likes - added
Learn how to identify circles, triangles, squares, rectangles, rhombuses, and trapezoids. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-geometry-topic/cc-early-math-shapes/e/naming-sh
• 08:17
### How to Draw an Ice Cream Sundae with Pusheen easy
603 views / 2 likes - added
Follow along to learn How to Draw Pusheen eating an Ice Cream Sundae easy, step by step drawing tutorial. Cute tabby cat drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmCSk
• 02:51
### Telling Time Exercise Example 1 | Measurement And Data | Early Math | Khan Academy
431 views / 0 likes - added
Tell time on a labeled analog clock. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-time/e/telling_time_0.5?utm_source=YT&utm_medium=Desc&utm_campaign=
• 02:55
### Hungry Orangutan Tries to eat Rocks | BBC Earth
295 views / 1 likes - added
New David Attenborough series Dynasties coming soon! Watch the first trailer here: https://www.youtube.com/watch?v=JWI1eCbksdE --~-- A newly released, hungry Orangutan isn't afraid of getting close to humans to try to obtain food. Subscribe to BBC Earth f
• 11:10
### How to Draw a Cute Panda Holding a Balloon Easy
424 views / 1 likes - added
#DrawSoCute Learn #HowToDraw and Color a cute, cartoon Panda holding a balloon easy, step by step drawing tutorial. Kawaii Panda art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2Ynwp
• 03:09
### Making Line Plots | Measurement And Data | Early Math | Khan Academy
493 views / 0 likes - added
Making line plots. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-math-line-plots/e/creating-line-plots-1?utm_source=YT&utm_medium=Desc&utm_campaign=EarlyMa
• 01:03
### Counting Whales, Sheep, And Flowers | Counting | Early Math | Khan Academy
499 views / 0 likes - added
Learn how to count animals. Animals in these pictures are organized neatly. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-counting-topic/cc-early-math-count-object-topic/e/how-many-ob
• 09:29 Popular
### How to Draw a Cocker Spaniel Puppy Dog Easy
1,276 views / 6 likes - added
#DrawSoCute Learn #HowToDraw a cartoon, cute Cocker Spaniel Puppy Dog easy, step by step drawing tutorial. Kawaii dog with a bone collar art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.
• 07:50
### How to Draw an Olaf Snowman | Disney Frozen
376 views / 0 likes - added
Learn How to Draw cute Olaf Snowman even cuter and simple from Disney's Frozen easy, step by step drawing tutorial. #HowToDraw Kawaii Christmas Holiday art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini
• 01:03
### Subtraction Word Problems Within 10 | Basic Addition And Subtraction | Early Math | Khan Academy
520 views / 0 likes - added
Learn how to solve word problems by subtracting small numbers (numbers 10 or less). Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-add-sub-basics/cc-early-math-add-sub-word-problem-wit
• 09:31 Popular
### How to Draw Bruni the Fire Salamander | Disney Frozen 2
872 views / 1 likes - added
Learn How to Draw Bruni the Cute Fire Spirit Salamander from Disney's Frozen 2 easy, step by step drawing tutorial. Draw Elsa: https://www.youtube.com/watch?v=IiL4L3fB9Hw SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4S
• 11:49
### How to Draw My Little Pony Cutie Mark Crusaders | Sweetie Belle
285 views / 0 likes - added
#DrawSoCute Learn #HowToDraw a Cute Unicorn Pony Sweetie Belle from MLP Cutie Mark Crusaders easy, step by step drawing tutorial. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmCSk
• 09:25 Popular
### How to Draw a Kitty Donut Squishy Easy
886 views / 3 likes - added
Follow along to learn How to Draw and Color a cute Kitty Donut Squishy easy, step by step drawing tutorial. Kawaii donut kitty art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2YnwpmC
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### Length Word Problems | Measurement And Data | Early Math | Khan Academy
493 views / 0 likes - added
• 02:15
### Addition And Subtraction Within 10 | Basic Addition And Subtraction | Early Math | Khan Academy
518 views / 0 likes - added
See how to solve a bunch of addition and subtraction problems with small numbers. All numbers in this video are 10 or less. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-add-sub-basic
• 10:03
### How to Draw Olaf | Disney Frozen
520 views / 2 likes - added
Learn How to Draw cute Olaf the Snowman from Disney's Frozen easy, step by step drawing tutorial. Frozen Playlist: https://www.youtube.com/watch?v=IiL4L3fB9Hw&list=PLbVzRnseEFty1M5BgusFp9RIKtGU5cHdV SUPPLIES You Might Love (Amazon affiliate links): Sharpi
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### Addition Word Problems Within 10 | Basic Addition And Subtraction | Early Math | Khan Academy
573 views / 1 likes - added
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### Comparing Lengths | Measurement And Data | Early Math | Khan Academy
538 views / 0 likes - added
• 07:46
### How to Draw Patrick Star | SpongeBob SquarePants
380 views / 3 likes - added
Learn #HowToDraw cute Patrick Star from SpongeBob SquarePants easy, step by step drawing tutorial. Patrick is a simple minded starfish and SpongeBob SquarePants best friend. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvb
• 05:34
### How to Draw a Cute Marshmallow Character Easy | Marshfellows
620 views / 0 likes - added
#DrawSoCute Learn #HowToDraw a cute, cartoon Marshmallow character easy, step by step drawing tutorial. Inspired by the cuddly, adorable Marshfellows plush. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini
• 13:09
### How to Draw and Color Snoopy Easy | Autumn Leaves
192 views / 1 likes - added
#DrawSoCute Learn #HowToDraw and Color cute Snoopy playing in a basket of Fall leaves easy, step by step drawing tutorial. Kawaii cute dog Autumn art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: http
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### How to Draw Fortnite Catalyst Skin
597 views / 3 likes - added
#DrawSoCute Learn #HowToDraw a Legendary Fortnite Catalyst Skin step by step drawing tutorial. Fortnite Outfit from the Drift Set Season 10 character. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: http
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### How to Draw a Baby Walrus Easy | Squishmallows
416 views / 2 likes - added
#DrawSoCute Learn #HowToDraw a cute Baby Walrus Squishmallow easy, step by step drawing tutorial. Kawaii sea life squad animal plush. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to/2Ynwp
• 10:02
### How to Draw a Holiday Hamster | Christmas Series #9
204 views / 2 likes - added
Learn How to Draw a cute Holiday Hamster holding a Candy Cane and wearing a Santa's Hat step by step Christmas drawing tutorial. Christmas Art Playlist: https://www.youtube.com/watch?v=9tT1X... SUPPLIES You Might Love (Amazon affiliate links): Sharpies: h
• 17:06
### 10 Impossible Magic Tricks You Can Do
123 views / 0 likes - added
How to do Impossible Magic Tricks. Learn how magic tricks work and discover simple illusions you can perform at home! Magic tricks for beginners and magicians of all ages! Our easy-to-learn magic trick tutorials will help you master the art of illusion! L
• 12:31 Popular
### How to Draw a Panda Eating Watermelon Easy | Summer Art Series #6
1,170 views / 4 likes - added
#DrawSoCute Learn #HowToDraw and Color a cute, cartoon Panda eating Watermelon easy, step by step drawing tutorial. Kawaii Panda art. Fun Summer art for kids. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mi
• 08:22
### How to Draw the Sun Easy | Summer Art Series #1
368 views / 2 likes - added
#DrawSoCuteSummerSeries Learn #HowToDraw and color a cartoon Sun with heart sunglasses, holding a popsicle easy, step by step drawing tutorial. Fun, cute Summer Series drawings. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2
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### Equal Parts Of Circles And Rectangles | Geometry | Early Math | Khan Academy
562 views / 1 likes - added
• 01:31
### Filling Rectangles With Squares | Geometry | Early Math | Khan Academy
501 views / 0 likes - added
Learn how to fill rectangles with squares. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-geometry-topic/cc-early-math-fractions-of-shapes/e/filling-rectangles-with-same-sized-squares?
• 06:32 Popular
### How to Draw a Watermelon Popsicle Easy | Summer Art Series #4
919 views / 3 likes - added
#DrawSoCuteSummerSeries Learn #HowToDraw and color a cute, cartoon Watermelon Popsicle easy, step by step drawing tutorial. Yummy cold summer dessert treat drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch
• 15:08
### How to Draw Santa Claus | Christmas Series #1
687 views / 3 likes - added
Learn How to Draw cute cartoon Santa Claus easy, step by step Christmas Holiday drawing art tutorial.Christmas Art Playlist: https://www.youtube.com/watch?v=9tT1XD25t9c&list=PLbVzRnseEFtxC5TtdMnt-AwWYOAQwjTJg SUPPLIES You Might Love (Amazon affiliate link
• 11:19
### 20 FUN AND CREATIVE DIY IDEAS
346 views / 0 likes - added
AWE-INSPIRING DIY IDEAS FOR CHILDREN Cool crafts for kids are so easy to make! You can follow up our cool tutorials from this video and decorate your home with amazing candles, comfy cushions, and other lovely crafts. Just watch and repeat (and, probably,
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### How to Draw Princess Rapunzel | Disney Tangled
924 views / 4 likes - added
Learn #HowToDraw cute princess Rapunzel from Disney Tangled easy, step by step chibi drawing tutorial. Pretty Disney princess drawing with braided hair and flowers in her hair. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2P
• 01:07
### Surprising Chimp hand washing and cleaning routine! | Animals with Cameras | Earth Unplugged
369 views / 1 likes - added
Animals with Cameras airs in the States on PBS: 7th February at 8pm and in the UK on BBC One: 8th February 2018. This chimp collar camera reveals the secret self-grooming routine of a female, from the hand-washing, to the teeth cleaning and even the delic
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1,217 views / 2 likes - added
These tricky brain teasers for adults and riddles for kids will confuse your brain to the absolute max and will nicely work out your logic: 00:14 - Survival riddle - Tricky brain games A tricky brain game or a survival riddle where you have only one chanc
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### How to Draw a Baby Horse Easy | Beanie Boos
462 views / 1 likes - added
#DrawSoCute Learn #HowToDraw a cute cartoon Baby Horse inspired by Beanie Boos easy, step by step drawing tutorial. Kawaii horse plush art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https://amzn.to
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### Estimating Lengths | Measurement And Data | Early Math | Khan Academy
480 views / 0 likes - added
• 04:19
### Cousin Fal's Shape Collection | Geometry | Early Math | Khan Academy
446 views / 0 likes - added
Learn how to classify shapes based on their number of sides, number of corners, and side-lengths. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-geometry-topic/cc-early-math-shapes/e/n
• 09:01
### How to Draw a Pinwheel Easy
219 views / 0 likes - added
Follow along to learn How to Draw and Color a Pinwheel easy, step by step, drawing tutorial. Celebrate 4th of July with this pretty star spangled pinwheel or color it like a rainbow for any day. SUPPLIES You Might Love (Amazon affiliate links): Sharpies:
• 11:14 Popular
### How to Draw a Dachshund Puppy Dog Easy
722 views / 1 likes - added
#DrawSoCute Learn #HowToDraw a cute, cartoon Dachshund Puppy Dog easy, step by step drawing tutorial. Kawaii Wiener Dog, scent hound dog drawing. Cute family companion pet. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf
• 14:38
### How to Draw Dustin from Stranger Things
243 views / 0 likes - added
#DrawSoCute Learn #HowToDraw Gaten Matarazzo as Dustin Henderson from Stranger Things easy, step by step drawing tutorial. Cute boy with curly hair wearing a cap drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4S
• 09:02
### How to Draw a Cup of Coffee and Donut Easy | Dunkin Donuts
648 views / 3 likes - added
#DrawSoCute Learn #HowToDraw a cartoon Cup of Coffee and a cute Donut easy, step by step drawing tutorial. Kawaii Dunkin Donuts inspired hot coffee and sweet dessert. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketc
• 11:14
### How to Draw Making a Snowman with Pusheen Cat
310 views / 3 likes - added
#DrawSoCute Learn #HowToDraw and Color cute Pusheen making a yummy Ice Cream Holiday Snowman with sprinkles easy, step by step drawing tutorial. Kawaii tabby cat art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketc
• 13:42
### How to Draw an Owl for Fall Easy
166 views / 1 likes - added
#DrawSoCute Fall is Here! Learn #HowToDraw a cartoon Owl and Fall Leaves easy, step by step drawing tutorial. Cute Owl inspired by Autumn wearing a scarf sitting on a branch. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXv
• 08:35
### How to Draw a Unicorn | Num Noms Big Dippers Marshmallow
216 views / 2 likes - added
#DrawSoCute Learn #HowToDraw a cute Bubbly Unicorn head from Num Noms Tasty Big Dippers Marsmallow mystery pack. Scented collectible cartoon Unicorn character. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad M
• 14:30
### How to Draw Scarlet Witch | The Avengers
254 views / 2 likes - added
#DrawSoCute Learn #HowToDraw Elizabeth Olsen as Scarlet Witch from Avengers easy, step by step drawing tutorial. Pretty female Superhero from Marvel Comics drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch
• 09:02
### How to Draw Santa's Toy Bag | Christmas Series #6
242 views / 0 likes - added
Learn How to Draw Santa Claus Bag of Toys and a cute Christmas Teddy Bear easy, step by step drawing tutorial. Christmas Art Playlist: https://www.youtube.com/watch?v=K_PL6SzTkGc&list=PLbVzRnseEFtxWcH-CNbICXh3TYFWrO-Lf SUPPLIES You Might Love (Amazon affi
• 02:48
### Picture Graphs | Measurement And Data | Early Math | Khan Academy
523 views / 0 likes - added
• 09:54 Popular
### How to Draw a Unicorn Seahorse
1,747 views / 5 likes - added
#DrawSoCuteUnicorn Learn #HowToDraw a pretty Unicorn Seahorse easy, step by step drawing tutorial. Kawaii, cute Seahorse Unicorn art. Magical sea creature drawing tutorial. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf
• 11:00 Popular
### How to Draw Pusheen at the Beach | Summer Art Series #3
736 views / 2 likes - added
#DrawSoCuteSummerSeries Learn #HowToDraw cute Pusheen eating ice cream at the beach easy, step by step drawing tutorial. Fun, cute Summer Series drawings. Kawaii Tabby cat drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.t
• 10:46 Popular
### How to Draw a Holiday Puppy | Christmas Series #5
1,210 views / 4 likes - added
Learn How to Draw a cute cartoon Christmas Holiday Puppy with a Santa Hat and Bell Collar easy, step by step drawing tutorial. Christmas Art Playlist: https://www.youtube.com/watch?v=K_PL6SzTkGc&list=PLbVzRnseEFtxWcH-CNbICXh3TYFWrO-Lf SUPPLIES You Might L
• 10:11
### How to Draw the Smiling Cat Heart Eyes Love Emoji Easy
230 views / 0 likes - added
#DrawSoCute Learn #HowToDraw and Color a cute Cat Emoji smiling with Heart Eyes easy, step by step drawing tutorial. Sweet Emoji for the Cat Lover. Coloring with markers and color pencil. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https:/
• 04:37
### The lovable (and lethal) sea lion - Claire Simeone
213 views / 0 likes - added
Plunge into the oceans depths to take a closer look at the sea lions hunting skills, and to learn more about how climate change is affecting its habitat. --Sunning themselves on rocks or waddling awkwardly across the beach, its easy to think of sea lions
• 13:24
### How to Draw Miraculous LadyBug Marinette Dupain-Cheng
538 views / 5 likes - added
#DrawSoCute Learn #HowToDraw cute Marinette from Miraculous Ladybug easy, step by step drawing tutorial. Pretty teenage girl student with bangs and pigtail hairstyle drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXv
• 16:08
### 10 CHRISTMAS MAGIC TRICKS!
235 views / 1 likes - added
Christmas magic tricks pranks and last minute gift ideas! Surprise your loved ones this holiday with your new magic skills! Learn how to make an ornament disappear, an amazing DIY magic trick with a spoon and a peppermint candy, and a wonderful Christmas
• 08:34 Popular
### How to Draw a Cute Frappuccino Easy | Pusheen Cafe
960 views / 6 likes - added
#DrawSoCuteDrinks Follow along to learn How to Draw this cute Frappuccino or Milk Shake from Pusheen Cafe easy, step by step drawing tutorial. Kawaii cat frappuccino drawing. Summer drink. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https:
• 09:00 Popular
### How to Draw a Baby Turtle Easy | Squishmallows
736 views / 3 likes - added
#DrawSoCute Learn #HowToDraw a cute, cartoon baby Turtle easy, step by step drawing tutorial. Kawaii turtle art inspired by Squishmallows turtle plush from Sea Life Squad. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4
• 11:51
### How to Draw a Boy Basketball Player
225 views / 1 likes - added
#DrawSoCute Learn #HowToDraw a cute cartoon Boy Basketball Player easy, step by step drawing tutorial. Chibi cute African American Boy drawing holding a basketball, wearing a jersey. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn
• 07:13
### How to Draw + Color a Panda Kissing Emoji Easy
225 views / 0 likes - added
#DrawSoCute Learn #HowToDraw and Color a cute Panda Kissing Emoji Face easy, step by step drawing tutorial. Kawaii, cartoon Love Panda Animal Emoji. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch Pad Mini: https:
• 14:21
### How to Draw a Tumblr VSCO Cute Girl
585 views / 0 likes - added
"sksksk" "and I oop" Learn #HowToDraw a cute VSCO girl inspired by Emma Chamberlain easy, step by step drawing tutorial. This cute Tumblr girl has messy bun hair, oversize T-Shirt, Birkenstocks, Hydroflask and of course scrunchies!! OOF! SUPPLIES You Migh
• 07:21 Popular
### How to Draw a Unicorn Floaty Easy | Summer Art Series #2
903 views / 4 likes - added
#DrawSoCuteSummerSeries Learn #HowToDraw a pretty Unicorn Floaty easy, step by step drawing tutorial. Fun, cute Summer Series drawings. Cute, kawaii Unicorn floaty for the pool. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2
• 18:11 Popular
### How to Draw Elsa | Disney Frozen 2
4,694 views / 16 likes - added
Learn How to Draw cute Elsa of Arendelle from Walt Disney's Frozen 2 easy chibi, step by step drawing tutorial. Pretty Disney princess drawing. Olaf Snowman Drawing: https://www.youtube.com/watch?v=Y5-c9jUm8qQ SUPPLIES You Might Love (Amazon affiliate lin
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• 08:32 Popular
### How to Draw a Baby Unicorn Easy | Floofies Fluffy
1,602 views / 7 likes - added
Follow along to learn How to Draw and Color a fluffy, cute Baby Unicorn Floofies easy, step by step drawing tutorial. Kawaii plush Unicorn. Floofies Fluffy Surprise Toy. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sk
• 11:38
### 17 GREAT TRICKS THAT WILL AMAZE YOUR KIDS
335 views / 1 likes - added
AMAZING AND FANTASTIC EXPERIMENTS YOU SHOULD REPEAT Science experiments are just amazing! They are really easy to make and they look fantastic! And the best thing about these ones is that you totally can repeat them at home! Dry ice, static electricity, w
• 06:52
### How to Draw a Baby Dolphin Easy | Squishmallows
496 views / 3 likes - added
Follow along to learn How to Draw and Color a cute baby Dolphin easy, step by step drawing tutorial inspired by Squishmallows. Kawaii cartoon Dolphin drawing and coloring. Super cute plush dolphin. SUPPLIES You Might Love (Amazon affiliate links): Sharpie
• 11:05
### How to Draw a Cup of Back To School Supplies Easy
416 views / 2 likes - added
#DrawSoCute Learn #HowToDraw Back to School Supplies easy, step by step drawing tutorial. Cute cup of pencils, markers, brushes, ruler, scissors and a cute eraser. Kawaii school supplies drawing. SUPPLIES You Might Love (Amazon affiliate links): Sharpies:
• 02:27
### Mary Poppins Returns | Official Trailer
327 views / 0 likes - added
See Mary Poppins Returns in theatres December 19! In Disney’s “Mary Poppins Returns,” an all new original musical and sequel, Mary Poppins is back to help the next generation of the Banks family find the joy and wonder missing in their l
• 10:23 Popular
### How to Draw Yoda Baby
2,334 views / 7 likes - added
Learn How to Draw cute Baby Yoda from Disney's The Mandalorian easy, step by step drawing tutorial. Star Wars Jedi baby version.StarWars Playlist: https://www.youtube.com/playlist?list=PLbVzRnseEFtxTbGsQemakzWXx3PmSntu9 SUPPLIES You Might Love (Amazon aff
• 09:04
### How to Draw a Kitten Easy | Exotic Shorthair Cat
188 views / 0 likes - added
#DrawSoCute Learn #HowToDraw a super cute, cartoon Kitten easy, step by step drawing tutorial. Kawaii Exotic Shorthair Kitty drawing. Sweetest Cat breed ever with flat nose and face! SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn
• 13:08
### How to Draw a Gingerbread House | Christmas Series #8
240 views / 2 likes - added
Learn How to Draw a Holiday Gingerbread House easy, step by step drawing tutorial.Christmas Art Playlist: https://www.youtube.com/watch?v=9tT1XD25t9c&list=PLbVzRnseEFtxC5TtdMnt-AwWYOAQwjTJg FREE COLORING PAGE: http://www.drawsocute.com/draw_so_cute_colori
• 14:18
342 views / 0 likes - added
Learn How to Draw cute Adrien who is Cat Noir from Miraculous Ladybug easy step by step drawing tutorial. Cute Boy drawing. Miraculous Ladybug Playlist: https://www.youtube.com/watch?v=KnRnE3bcPoc&list=PLbVzRnseEFtxGM8FO-qPIe5gUnbwQZQ-u #HowToDraw #DrawSo
• 13:00
### How to Draw Cute Back to School Art | Cutie Kitty #1
309 views / 2 likes - added
#DrawSoCute Learn #HowToDraw Cutie Kitty Going Back To School wearing glasses and a backpack, holding a book and pencil, easy step by step drawing tutorial. Kawaii cat art for back to school fun. SUPPLIES You Might Love (Amazon affiliate links): Sharpies:
• 13:39
### How to Draw an Angel Cute Girl
663 views / 3 likes - added
Learn How to Draw a cute Girl dress as an Angel with Long Hair, Wings and a halo holding a candle easy, step by step drawing tutorial. Cute Girl Drawing Playlist: https://www.youtube.com/watch?v=XqHNdcp_YD8&list=PLbVzRnseEFtwD3xfGfjLjH9XY7TFe5wfa SUPPLIES
• 14:34 Popular
### How To Draw Mal | Disney Descendants 3
1,520 views / 8 likes - added
#DrawSoCute Learn #HowToDraw cute, pretty Dove Cameron as Mal from Disney Descendants 3 chibi easy, step by step drawing tutorial. Aka Queen Maleficent "Mal" Bertha, future queen of Auradon. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: http
• 15:38
### How to Draw Sheriff Woody | Toy Story
299 views / 0 likes - added
#DrawSoCute Learn #HowToDraw cute Woody from Disney Pixar Toy Story easy, step by step drawing tutorial. Pullstring cowboy rag doll sheriff and leader of the toys in Toy Story movie. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn
• 15:11
### How to Draw a Soccer Player Cute Girl
417 views / 3 likes - added
#DrawSoCuteGirl Learn #HowToDraw a cute Girl Soccer Player in ponytail with hairband holding a soccer ball easy, step by step drawing tutorial. Pretty female soccer player in uniform art. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https:/
• 10:47
### How to Draw a Snowman | Christmas Series #4
308 views / 1 likes - added
Learn How to Draw a cute Snowman with a Santa Hat and scarf in a winter scenery easy, step by step drawing tutorial. Kawaii Christmas Winter Holiday art.Christmas Art Playlist: https://www.youtube.com/watch?v=K_PL6SzTkGc&list=PLbVzRnseEFtxWcH-CNbICXh3TYFW
• 13:46
### How to Draw Blume Dolls Easy | Cora
222 views / 0 likes - added
#DrawSoCute Learn #HowToDraw and Color a Cute Blume Dolls easy, step by step drawing tutorial. Cora is a pretty Blume Girl secret surprise toy with beautiful hair and starfish in her hair. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https:
• 10:00
### How to Draw Sulley Easy | Monsters Inc.
263 views / 0 likes - added
#DrawSoCute Learn #HowToDraw James P. Sullivan from Monsters Inc. easy, step by step drawing tutorial. Cute Sulley is Mike's best friend. Kawaii Monster. Disney Pixar animated movie character. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: ht
• 05:13
### How to Draw a Polar Bear Kissing Emoji Face Easy
208 views / 0 likes - added
#DrawSoCute Learn #HowToDraw a cute, cartoon Polar Bear or Brown Bear Kissing Emoji easy, step by step drawing tutorial. Kawaii Love Bear kissing face with a heart. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://amzn.to/2PXvbf4Sketch
• 07:54
### How to Draw Pusheen Harry Potter Easy
510 views / 0 likes - added
#DrawSoCute Learn #HowToDraw cute Pusheen as Harry Potter for Halloween easy, step by step drawing tutorial. Kawaii tabby cat dressed as Harry Potter costume with a scarf, sitting on a broom. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: htt
• 05:58 Popular
### 3 Guinness World Records - Disc Cases, Stick Bomb & Domino Wall
934 views / 1 likes - added
3 Guinness World Records - Disc Cases, Stick Bomb & Domino Wall http://www.austriandominoart.com Hey everyone! We are an Austrian domino group and we set up chain reactions since 2008. Over the years we improved our skills, so also check out our other
• 06:31
### How to Draw a Baby Panda Easy | Floofies Fluffy
604 views / 3 likes - added
Follow along to learn How to Draw and Color a fluffy, cute Baby Panda Floofies easy, step by step drawing tutorial. Kawaii plush Panda. Floofies Fluffy Surprise Toy. Cute Animal drawing for kids. SUPPLIES You Might Love (Amazon affiliate links): Sharpies:
• 11:45
### How to Draw JungKook BTS mini Doll Figure Pop Up
241 views / 0 likes - added
#DrawSoCute Learn #HowToDraw cute Jungkook as a mini Doll Figure from Mattel for BTS Pop Up Store easy, step by step drawing tutorial. Kawaii, chibi, anime cherry haired Jungkook figure. SUPPLIES You Might Love (Amazon affiliate links): Sharpies: https://
• 10:02
### 5 Funny Magic Food Pranks
280 views / 3 likes - added
Impress your friends and family with an amazing magic trick using food! Learn how to make the classic onion volcano that sprays real fire! A funny and easy prank using cookies, and much more! The secrets to the magic pranks are all revealed in the video;
• 04:05
### Looper - Unbelievable Looping Yoyo Tricks!
318 views / 1 likes - added
Professional YoYo player Ryo Yamashita shows us some of the craziest 2A yoyo looping tricks we've ever seen! Check out http://www.yomega.com for more products Subscribe! http://bit.ly/1tLVDYk http://www.kumafilms.com/ Yomega Giveaway - Share this video on
• 51:53
### How to Do Maths Without a Calculator - with Rob Eastaway
159 views / 0 likes - added
Knowing how to quickly estimate sizes and numbers is a useful everyday application of maths.Buy Rob's book "Maths on the Back of an Envelope" now - https://geni.us/yaeUJwHow many cats are there in the world? What's the chance of winning the lottery twice?
• 15:18
### How To Draw Princess Luna from My Little Pony
636 views / 0 likes - added
#DrawSoCute Learn #HowToDraw Princess Luna aka Nightmare Moon from My Little Pony Friendship is Magic easy, step by step art tutorial. This pretty pegasus unicorn is an Alicorn Pony, sister of Princess Celestia. SUPPLIES You Might Love (Amazon affiliate l
• 01:46
425 views / 0 likes - added
When going to the movies, is the price you pay proportional to the number of tickets you buy? Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-ratio-proportion/cc-7th-proportional-re
• 10:36
### How to Draw Rudolph | Christmas Series #3
310 views / 1 likes - added
Learn How to Draw cute Rudolph the Red Nose Reindeer easy, step by step drawing tutorial. Kawaii reindeer art for Christmas Winter Holiday. Christmas Art Playlist: https://www.youtube.com/watch?v=9tT1XD25t9c&list=PLbVzRnseEFtxC5TtdMnt-AwWYOAQwjTJg #HowToD
• 04:08
### Measuring A Golden Statue | Measurement And Data | Early Math | Khan Academy
506 views / 0 likes - added
Measure an object with same-size length units that span it with no gaps or overlaps. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/early-math/cc-early-math-measure-data-topic/cc-early-picture-graphs/e/solving
• 16:39 Popular
### How to Draw Princess Jasmine | Disney Aladdin New
1,466 views / 3 likes - added
Follow along to learn How to Draw Princess Jasmine from Disney Aladdin easy, step by step drawing tutorial. Cute, pretty princess Jasmine played by Naomi Scott in this New Walt Disney Aladdin movie. SUPPLIES You Might Love (Amazon affiliate links): Sharpi
• 14:05
315 views / 1 likes - added
#DrawSoCute Learn #HowToDraw Chibi, cartoon Aladdin from Walt Disney's Live Action Aladdin remake easy, step by step drawing tutorial. Prince Ali of Ababwa. Genie's new master and Jasmine's Love. SUPPLIES You Might Love (Amazon affiliate links): Sharpies:
• 04:35
### Spells, threats, and dragons: The secret messages of Viking runestones - Jesse Byock
194 views / 0 likes - added
Learn the history of Viking runes, the ancient Norse language of symbols that make up an alphabet called the futhark. --With their navigational skills and advanced longships, the Vikings sustained their seafaring for over 300 years. But for all their migh
• 16:44 Popular
### How to Make a Cute Folding Surprise Card DIY Easy
798 views / 4 likes - added
#DrawSoCuteCraft Learn How to Make a cute Frozen Yogurt Tower Folding Surprise Card easy, step by step DIY tutorial. Using materials you already have at home. Kawaii Cup of Frozen Yogurt Dessert. Fun Craft card for kids to make. SUPPLIES You Might Love (A
• 15:36
### How to Draw Bo Peep from Toy Story
477 views / 0 likes - added
#DrawSoCute Learn How to Draw cute Little Bo Peep from Disney Pixar Toy Story easy, step by step drawing tutorial. Cartoon Pretty girl with blonde hair in bonnet with corset dress holding a cane. Woody's girl. SUPPLIES You Might Love (Amazon affiliate lin
• 15:43
### How to Draw Fortnite Rippley Skin
460 views / 2 likes - added
#DrawSoCute Learn #HowToDraw Rippley Fortnite Skin easy, step by step drawing tutorial. Cute Rippley is a Slurp Juice monster created from industrial waste. Cool Fortnite Chapter 2, season 1 skin. SUPPLIES You Might Love (Amazon affiliate links): Sharpies
• 16:02
### How to Draw Ariana Grande | Don't Call Me Angel Music Video
511 views / 0 likes - added
#DrawSoCute Learn #HowToDraw cute chibi Ariana Grande easy, step by step drawing tutorial. Pretty Ariana Grande with Angel Wings from her Music Video Don't Call Me Angel. Cute girl with long hair. SUPPLIES You Might Love (Amazon affiliate links): Sharpies
• 14:07
### 10 MATH TRICKS THAT WILL EMBARRASS YOUR TEACHER
359 views / 0 likes - added
INCREDIBLY COOL MATH TRICKS FOR KIDS This video is dedicated to all those people who think that math was boring. Today, we are here to prove the opposite! In this video, we collected the coolest math tricks for all ages. It doesn't matter if you're 6 or 6
• 02:57
### Banana Proportionality | Rated And Proportional Relationships | 7th Grade | Khan Academy
500 views / 0 likes - added
• 13:37
### 12 KID FRIENDLY SCIENCE EXPERIMENTS
213 views / 0 likes - added
SURPRISING SCIENCE TRICKS FOR KIDS Knowledge is power, and everyone knows that. However, sometimes, learning at school is so dull! Forget about boring lessons with our incredible ideas that help to study faster and easier and also bring a lot of fun! With
• 07:34
### How to Draw an Angel Pusheen
235 views / 0 likes - added
How to Draw a cute Tabby Cat Pusheen Angel easy step by step drawing tutorial. Holiday Christmas art.Draw Pusheen Playlist: https://www.youtube.com/watch?v=WZCI55QJUdQ&list=PLbVzRnseEFtzWO9VuSXtdE_GeWy9qk_nA Christmas Art Playlist:https://www.youtube.com/
• 02:39
### Proportionality Between Side Length And Perimeter | 7th Grade | Khan Academy
471 views / 0 likes - added
Is the perimeter of a square proportional to the side length of the square? Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-ratio-proportion/cc-7th-proportional-rel/e/analyzing-and-
• 13:09
### How to Draw Dora the Explorer | The Lost City of Gold
385 views / 2 likes - added
#DrawSoCute Learn #HowToDraw cute Dora the Explorer wearing backpack from the Live action movie The Lost City of Gold easy, step by step drawing tutorial. Isabela Moner plays Dora with bangs bobcut hairstyle. SUPPLIES You Might Love (Amazon affiliate link
• 13:36 Popular
### How to Draw a Roller Skate Rainbow Cute Girl | Candylocks Doll
717 views / 2 likes - added
#DrawSoCute Learn #HowToDraw a cute Girl wearing Roller Skates from Candylocks Doll easy, step by step drawing tutorial. Pretty Rainbow girl with cotton candy long hair holding a peace sign hands. SUPPLIES You Might Love (Amazon affiliate links): Sharpies
• 12:28
### Learn Glass Blowing With Jolly Ranchers!
167 views / 0 likes - added
Today we've teamed up again with Nick Uhas, the host from Netflix's show Blown Away, to test out some introductory glass blowing techniques. Did you know you can take Jolly Ranchers and use them to perform simple glass blowing skills?To get a signed copy
• 03:46
### Testing Whether Area Is Proportional To Side Length | 7th Grade | Khan Academy
482 views / 0 likes - added
Is the volume of a square proportional to the side length of the square? Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-ratio-proportion/cc-7th-proportional-rel/e/analyzing-and-ide
• 15:07 Popular
### How to Draw a Cute Girl | Summer Art Series #7
6,160 views / 24 likes - added
#DrawSoCuteGirl Learn How to Draw a cute, pretty Girl with Bun Hair Style, flower in her hair, holding a drink, wearing flip flops and tank top with shorts ready to have fun in the Summer Sun. Easy to follow, step by step drawing of Chibi, pretty girl. SU
• 14:19
### How to Draw Sleeping Beauty Princess Aurora | Disney Maleficent
462 views / 1 likes - added
#DrawSoCute Learn #HowToDraw Elle Fanning as Princess Aurora, also known as Sleeping Beauty or Briar Rose from Disney's Maleficent easy, step by step drawing tutorial. Pretty princess with long blonde hair wearing a long peasant blue dress. SUPPLIES You M
• 08:22
### 6 Magic Basketball Tricks ft. The Harlem Globetrotters
377 views / 2 likes - added
Learn how to do amazing magic basketball tricks with the legendary Harlem Globetrotters and Evan Era! In this episode of How To Magic, Evan Era teams up with the world famous Harlem Globetrotters to show you fun basketball tricks you can do at home! Easy
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### Freestyle Football Tricks in London feat. Guinness World Record Holder Dan Magness
478 views / 1 likes - added
Freestyle football Guinness worlds record holder, Dan Magness, shows us some of his amazing freestyle football skills. Dan holds five Guinness world records. One for keeping the football on his head for 26 hours and another for walking 210 miles from Wemb
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### Making an iPad case for blind to "see" with touch
155 views / 0 likes - added
Thanks SimpliSafe for sponsoring this video. SimpliSafe is award-winning home security that keeps your home safe around the clock. It's really reliable, easy to use, and there are no contracts. Check out SimpliSafe here: https://Simplisafe.com/StuffMadeHe
• 12:18
### 12 CUTE DIY IDEAS FOR SCHOOL
445 views / 0 likes - added
TO COOL FOR SCHOOL CRAFTS AND HACKS How is your school going? It's high time we show you a new portion of amazing life hacks and crafts that can make your time spent at school more effective (and of course more interesting). Learn with us and son't forget
• 09:29
### How to Listen to Classical Music: Fugues
321 views / 0 likes - added
A Complete Introduction to Fugues. This video looks at Fugues, and how they work. It considers the great fugues of Bach, as well as surveying fugue from Britten to Bernstein, Handel to Beethoven. It looks at fugal technique - the exposition, subject, answ
• 15:19
### 15 Puzzles That Will Keep You Up All Night
436 views / 1 likes - added
How to Make Yourself Smarter. Studies have shown that solving riddles and puzzles is exceptionally good for your brain. It sharpens your mind, improves your memory, and teaches you to pay attention to small details. Get ready to train your brain and impre
• 04:00
### Spy Tortoise Adopted by Chimpanzee | Spy in The Wild | BBC Earth
276 views / 1 likes - added
Spy Tortoise becomes best mates with a friendly chimp! Subscribe: http://bit.ly/BBCEarthSub Watch more: Planet Earth http://bit.ly/PlanetEarthPlaylist Blue Planet http://bit.ly/BluePlanetPlaylist Planet Earth II http://bit.ly/PlanetEarthIIPlaylist Planet
• 18:00
### 11 Impossible Magic Tricks You Can Do
507 views / 1 likes - added
Learn how to do impossible magic tricks like a real magician! In this How To Magic video, Evan Era teaches easy Magic Tricks and Illusions you can do at home! Simple card tricks for kids; learn how magic tricks work with step by step instructions for each
• 08:57 Popular
### 23 MIND-BLOWING MAGIC TRICKS EVERYONE CAN DO
1,041 views / 3 likes - added
MAGIC TRICKS FOR EVERYBODY If you love magic, we prepared a perfect compilation for you. Our tricks are so easy that you don’t need to have special skills to repeat them. Discover the basics of the magic and you will be able to quickly and easily ma
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• 11:45
### 25 Magic Tricks with Rings
277 views / 0 likes - added
Learn magic tricks easy with these amazing ring illusions you can do at home with very little practice! The number of magicians in the world is very small; only 1 person out of 25,000 people knows how to perform magic tricks! You are about to become one o
• 13:18
### Parents Try Guessing What Their Kid Will Do With \$100 | What Would My Kid Do? (React)
225 views / 4 likes - added
Watch all React Special Eps: https://fbereact.com/RCSpecialsJoin the SuperFam and support FBE: https://www.youtube.com/user/React/joinSUBSCRIBE & HIT THE . New Videos 12pm PT on REACT: http://fbereact.com/SubscribeREACTJoin us LIVE on FBE2 every Tuesday a
• 07:30
### Gingerbread Hacks | FOOD HACKS FOR KIDS
459 views / 1 likes - added
Gingerbread is back and it's better than ever! Take your gingerbread skills to the next level with Shanynn's mini gingerbreadian mansion, her edible gift box, and her sweet candy skier. Ingredients: Gingerbead dough White candy melts Decorative sprinkles
• 10:03
### 7 Impossible Magic Body Pranks
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Ever wonder how magic tricks work? All magic prank secrets revealed in this body pranks video of How To Magic! Magic pranks, challenges and illusions you can do with your body to freak out your friends and family! Learn how to do magic pranks like cutting
• 11:49
### 23 COOKING DIYs YOU'VE NEVER DREAMT OF
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SIMPLY GENIOUS RECIPES Today you will learn how to decorate desserts, we will share with you yummy recipes you can make from a coffee, easy way to cook sushi at home and a lot of other lifehacks you didn’t know before. Let’s start with the tutorial how to
• 03:29
### Comparing Decimals Example 1 | Decimals | 4th Grade | Khan Academy
526 views / 1 likes - added
Learn how to compare decimals with tenths and hundredths. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-decimals/cc-4th-compare-decimals/e/comparing_decimals_1?utm_source=YT&utm_me
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### Decimals As Words | Decimals | 4th Grade | Khan Academy
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### This Carpenter Builds Crutches For Children. Now Its His Turn To Walk. | Short Film Showcase
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David Miti is a double amputee who works as a carpenter hand-making devices for children with disabilities. Subscribe: http://bit.ly/NatGeoSubscribe Get More Short Film Showcase: http://bit.ly/ShortFilmShowcase#NationalGeographic #DavidMiti #ShortFilmShow
• 03:05
### Why You Shouldn't Give Ginger To Monkeys (and other animal sayings)
553 views / 0 likes - added
Learn new skills from this video’s sponsor, Skillshare: http://skl.sh/minuteearth4 Humans from different cultures anthropomorphize different animals to represent the same human traits. Thanks also to our Patreon patrons https://www.patreon.com/MinuteEarth
• 02:01
### ♪ Why Are Dogs "Man's Best Friend"? | EVERY SINGLE THING
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Dogs are definitely 'man's best friend', but why?? How did dogs become our closest pet friends?? LYRICS: They’re in our homes, they’re on the bed. They’re mixed, or fixed, or purist bred. They’re cute, they’re strong, and past widespread. What lead to the
• 02:47
### Spy Monkey Mistaken for Dead Baby and Mourned by Troop (FULL CLIP) | Spy In The Wild | BBC Earth
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Join this troop of Langur monkeys as they grieve the fallen. Doesn't matter if it's an animatronic Spy Monkey; it was still part of the family. Now scrub up, it's time for the funeral. Subscribe: http://bit.ly/BBCEarthSub Watch more: Planet Earth http://b
• 02:55
### Comparing Fractions On A Number Line | Fractions | 4th Grade | Khan Academy
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• 02:32
### Comparing Fractions Visually With Pies | Fractions | 4th Grade | Khan Academy
478 views / 0 likes - added
• 03:40
507 views / 0 likes - added
Learn how to compare decimals using grid diagrams. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-decimals/cc-4th-compare-decimals-visually/e/comparing-decimals-visually?utm_source=
• 06:55
### Place Value For Decimals Greater Than One (examples) | 4th Grade | Khan Academy
498 views / 0 likes - added
Understand decimal numbers that are greater than 1. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-decimals/cc-4th-decimals-greater-than-one/e/decimals-greater-than-one-intuition?ut
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### Food Stylist vs. Ramen Bowl | How to Style DIY Ramen for Photo | Well Done
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Food stylist @rishonmarie shows you how she perfectly styles a photo-worthy, homemade ramen bowl. Demonstrating styling tricks and tips, knife skills, and other cooking techniques, youll learn what separates a beautiful bowl of ramen from the super-cheap
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### Plotting Decimal Numbers On A Number Line (examples) | Khan Academy
434 views / 0 likes - added
Learn how to find decimals with hundredths on a number line. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-decimals/cc-4th-decimals-number-line/e/decimals_on_the_number_line_2?utm_
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### Decimal Intuition With Grids (examples) | Decimals | 4th Grade | Khan Academy
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• 04:37
466 views / 0 likes - added
Understand the connection between decimals and fractions using a grid diagram. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-decimals/cc-4th-decimal-fractions/e/fraction-decimal-in
• 07:40
### Comparing Numbers Represented In Different Ways | 4th Grade | Khan Academy
480 views / 0 likes - added
Examples of Khan Academy practice problems where you compare decimals, fractions, and number diagrams. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-decimals/cc-4th-compare-decimal
>> View skills web videos
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https://oeis.org/A196625
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A196625 Decimal expansion of the number c for which the curve y=1/x is tangent to the curve y=cos(x-c), and 0
6, 0, 5, 7, 8, 0, 2, 1, 7, 0, 2, 1, 5, 5, 3, 7, 0, 9, 1, 4, 8, 4, 1, 7, 5, 6, 5, 7, 5, 9, 6, 9, 8, 7, 7, 1, 0, 4, 8, 1, 1, 7, 9, 0, 3, 1, 1, 4, 1, 4, 8, 4, 0, 5, 7, 8, 5, 1, 6, 6, 5, 3, 9, 7, 3, 5, 3, 1, 8, 5, 8, 6, 1, 5, 7, 0, 0, 8, 7, 3, 0, 1, 2, 2, 4, 7, 7, 3, 8, 3, 8, 1, 8, 8, 7, 9, 1, 2, 3, 3 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Let r=(1+sqrt(5))/2, the golden ratio. Let u=sqrt(r) and v=1/x. Let c=sqrt(r)-arccsc(r). The curve y=1/x is tangent to the curve y=cos(x-c) at (u,v), and the slope of the tangent line is r-1. Guide to constants c associated with tangencies: A196610: 1/x and c*cos(x) A196619: 1/x - c and cos(x) A196774: 1/x + c and sin(x) A196625: 1/x and cos(c-x) A196772: 1/x and sin(x+c) A196758: 1/x and c*sin(x) A196765: c/x and sin(x) A196823: 1/(1+x^2) and -c+cos(x) A196914: 1/(1+x^2) and c*cos(x) A196832: 1/(1+x^2) and c*sin(x) A197016: x=0, y=0, and cos(x) LINKS EXAMPLE c=0.60578021702155370914841756575969877104... MATHEMATICA Plot[{1/x, Cos[x - 0.60578]}, {x, 0, 2 Pi}] r = GoldenRatio; xt = Sqrt[r]; x1 = N[xt, 100] RealDigits[x1] (* A139339 *) c = Sqrt[r] - ArcCsc[r]; c1 = N[c, 100] RealDigits[c1] (* A196625 *) slope = N[r - Sqrt[5], 100] RealDigits[slope] (* -1+A001622; -1+golden ratio *) CROSSREFS Cf. A139339, A196772. Sequence in context: A153754 A096410 A098468 * A195368 A198426 A019110 Adjacent sequences: A196622 A196623 A196624 * A196626 A196627 A196628 KEYWORD nonn,cons AUTHOR Clark Kimberling, Oct 05 2011 STATUS approved
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https://daveagp.wordpress.com/2008/03/31/better-know-a-theorem-minimal-automata/
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### Better Know a Theorem: Minimal Automata
31Mar08
One of my favourite courses as an undergraduate was “Introduction to the Theory of Computation.” It was taught by Michael Sipser, who also wrote the textbook, and one thing I liked about it was that the presentation was very natural and easy to read. I recently came across a really nice old result which could have been included (but wasn’t).
A deterministic finite-state automaton (DFA) is basically the simplest “computer” imaginable. It is comprised of
• an input alphabet
• a finite number of “states” which I will depict as circles
• one state must be designated the initial state
• every state is either an accepting or a rejecting state
• for each state, and each character of the input alphabet, an arrow to another state
What does a DFA compute? It computes a language in the sense that, for a given a string of characters from the input alphabet, the DFA can either accept or reject that string, and the DFA’s language is the set of all accepted strings. To determine if a particular string belongs to the language,
1. start at the initial state
2. for each character of the string, follow the arrow labeled with that character
3. accept if you end in an accepting state, reject if you end in a rejecting state
E.g. in the diagram below I have tried to illustrate a simple DFA with 2 states. The alphabet is {x, y} and the DFA accepts a string if and only if it contains an odd number of x’s.
It is possible for two different DFAs to recognize the same language; if this is the case we call them equivalent. Thus, if I hand you a DFA, it is natural to ask “is there any other equivalent DFA with a fewer number of states? what is it?” We call a DFA minimal if no other equivalent DFA has fewer states. Today’s better know a theorem is the following:
A DFA is minimal if and only if every state is accessible and every pair of states is distinguishable, where
• a state is accessible if there is some path to it from the initial state
• two states s, t are distinguishable if there is a string w so that the DFA, when started in s, accepts w, but would not accept w if started in t (or vice-versa)
It is not too hard to see why every minimal DFA is accessible and distinguishable: if a state is inaccessible we can just delete it, and if two states are equivalent we can coalesce them, contradicting minimality. One way to prove the other direction is via the Myhill-Nerode theorem which provides an elegant characterization of languages accepted by DFAs.
One of the things covered in Sipser’s text is the pumping lemma, which is one way to prove that a given language is not recognized by any FSA. Myhill-Nerode provides another solution which I think would have been neat to learn in the undergrad computation class.
Algorithmically, the theorem gives an efficient method to minimize a given DFA: repeatedly delete inaccessible states and coalesce equivalent pairs of states. On the other hand, looking at some slightly more complicated computing devices, minimizing push-down automata is undecidable, while minimizing nondeterministic finite-state automata is “hard”.
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https://www.storyofmathematics.com/what-is-the-sum-of-the-polynomials/
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# What is the Sum of Polynomials – A Guide to Adding Algebraic Expressions
The sum of polynomial functions is the result of adding two or more polynomial expressions, which are algebraic expressions that include variables and coefficients.
Polynomials are added by combining like terms, terms that have the same variables raised to the same powers.
For example, to add the polynomials $f(x) = 4x^2 + 3x – 5$ and $g(x) = 2x^2 – x + 1$, one would combine the coefficients of like terms to get the resulting polynomial $h(x) = (4x^2 + 2x^2) + (3x – x) – (5 – 1)$, which simplifies to $h(x) = 6x^2 + 2x – 4$.
Understanding the process of adding polynomials is fundamental in algebra and is frequently applied in calculus and higher mathematics.
The outcome of this addition yields a new polynomial whose degree is determined by the highest power of the variable present in the combined result, provided that the sum does not include any terms that cancel each other out completely.
This concept lays the foundation for more complex operations such as polynomial multiplication and division.
## The Sum of Polynomials
The process of adding polynomials involves combining like terms and aligning terms with the same degree.
This operation is fundamental in algebra and is performed by manipulating the coefficients and variables to achieve a simplified algebraic expression.
When adding polynomials, one first needs to identify the like terms. Terms are terms with the same variables raised to the same power. The actual addition involves summing the coefficients of these like terms while keeping the variables unchanged.
For example, consider the polynomials $3x^2 + 4x + 5$ and $5x^2 – 2x + 3$. To add these two, one would align and add the coefficients of the like terms:
\begin{align*} (3x^2 &+ 4x + 5) \ (5x^2 – 2x + 3) \ (3x^2 + 5x^2) &+ (4x – 2x) + (5 + 3) \ 8x^2 &+ 2x + 8 \end{align*}
The result is a new polynomial in standard form, $8x^2 + 2x + 8$.
### Adding Polynomials in Standard Form
Standard form dictates that an algebraic expression be written with terms in descending order of degree. When adding polynomials presented in this form, one ensures that each term of similar degree from the polynomials is visibly lined up, which makes the addition process straightforward.
Consider adding a linear polynomial $2x + 3$ to a quadratic polynomial $x^2 – x + 1$. The addition in standard form appears as:
\begin{align*} &\phantom{0x^2 +\ }2x + 3 \& x^2 – x + 1 \ & x^2 + (2x – x) + (3 + 1) \ &x^2 + x + 4 \end{align*}
Thus, the sum of the linear and quadratic polynomials is the trinomial $x^2 + x + 4$.
## Conclusion
When one combines polynomials, they are effectively adding or subtracting the polynomial expressions to obtain a single polynomial as the result.
In mathematical terms, adding polynomials involves summing like terms, which are terms that have the same variable raised to the same power. The process emphasizes combining coefficients while keeping the variables and their respective exponents unchanged.
For instance, when adding two polynomials, $P(x) = 3x^2 + 2x + 1$ and $Q(x) = 5x^2 – 4x + 7$, one would align the like terms and sum the coefficients to get the resulting polynomial $R(x) = (3x^2 + 5x^2) + (2x – 4x) + (1 + 7)$.
Thus, the resulting polynomial $R(x)$ after performing the addition would be $8x^2 – 2x + 8$.
Subtraction is treated similarly, where one subtracts the coefficients of like terms of the subtrahend polynomial from the corresponding terms of the minuend polynomial.
In both operations, care is taken to ensure that the integrity of the original exponents and variables is maintained.
These operations are fundamental in algebra and are integral for solving more complex algebraic problems. Mastery of polynomial addition and subtraction is essential for progress in understanding algebraic equations and functions.
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https://www.yaclass.in/p/science-state-board/class-6/measurements-8763/re-bc264ce3-6b5b-45ec-9620-f82afa096885
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LEARNATHON
III
Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!
### Theory:
Length is one of the fundamental quantity that cannot be conveyed in any other way. Other measurements, such as area and volume, can be calculated using length.
Area
In general, two lengths are used to calculate the area of an object.
The formula is given as,
$\begin{array}{l}\mathit{Area}=\mathit{Length}×\mathit{Breadth}\\ =l×b\end{array}$
By using the above formula, the area of a book, house or even a garden can be found. The SI unit for the area of a surface is $$m^2$$ since it is a product of two lengths.
Example:
Assume the length and breadth of the wall are $$20\ m$$ and $$8\ m$$, respectively. Then, what will be the area of the wall?
Length of the wall $$=$$ $$20\ m$$
Breadth of the wall $$=$$ $$8\ m$$
$\begin{array}{l}\mathit{Area}=l×b\\ =20\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}8\\ =160\phantom{\rule{0.147em}{0ex}}{m}^{2}\end{array}$
Therefore, the area of the wall is $$160$$ $$m^2$$.
Volume of solids
A volume is the amount of space occupied by any three-dimensional object. It is also a derived quantity that can be measured by measuring lengths. The formula is written as,
$\mathit{Volume}=\mathit{Length}×\mathit{Breadth}×\mathit{Height}$
The SI unit of volume is a $$cubic$$ $$metre$$ or $$m^3$$.
Calculation of volume of a solid box
A volume of a solid box can be found using three parameters such as length ($$l$$), breadth ($$b$$) and height ($$h$$). These parameters are measured using a measuring scale in terms of $$cm$$ or $$m$$.
$\mathit{Volume}=l×b×h$
If the unit of volume is written in $$cm$$, then
$$Volume$$ $$=$$ $$centimetre$$ $$×$$ $$centimetre$$ $$×$$ $$centimetre$$
$$=$$ $$cubic\ centimetre$$ or $$cm^3$$
Example:
Look at the image of the solid box and calculate its volume.
The dimensions of the solid box are given below:
Length, $$l\ = 10\ cm$$
Breadth, $$b\ = 10\ cm$$
Height, $$h\ = 10\ cm$$
$\mathit{Volume}=l×b×h$
By substituting the known value on the formula, we get the following.
$\begin{array}{l}\mathit{Volume}=10×10×10\\ =1000\phantom{\rule{0.147em}{0ex}}{\mathit{cm}}^{3}\end{array}$
If the volume of a solid cubical box is $$1000\ cubic\ cm$$, then it means that $$1000$$ cubes, each with dimensions $$1 cm$$ $$×$$ $$1 cm$$ $$×$$ $$1 cm$$, can be placed inside the box.
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https://assignwrite.com/2021/12/03/week5-ma215-discussion-my-perfect-tutors/
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# Week5 Ma215 Discussion – My Perfect Tutors
Week5 Ma215 Discussion – My Perfect Tutors.
Watch this video: https://www.youtube.com/watch?v=tFWsuO9f74o What is a confidence interval? Let’s say you made a confidence interval for a mean taken from the sample that is representative of the population. Will that confidence interval apply to the population or to the sample? In the discussion for week 4, you rolled a pair of dice 10 times and calculated the average sum of your rolls. Then you did the same thing with 20 rolls. Use your results from the week 4 discussion for the average of 10 rolls and for the average of 20 rolls to construct a 95% confidence interval for the true mean of the sum of a pair of dice (assume s = 2.41). What do you notice about the RANGE of the interval for the mean of 10 rolls versus the mean of 20 rolls? Finish this rule: As the samples size increases, the range of the confidence interval ______________ (increases/decreases – pick one). Using your mean for 20 rolls, calculate the 90% confidence interval. Compare the range of the confidence interval for 90% to 95%. Finish this rule: As the confidence level increases, the range of the confidence interval _______ (increases/decreases – pick one).In the project for this week, you are calculating confidence intervals for someone who is picking between the engineering major and the business major. You are trying to tell him/her what is likely to happen. Is it more helpful to have a large range or a narrow range in a confidence interval? Why
## Week5 Ma215 Discussion – My Perfect Tutors
Posted in Uncategorized
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https://www.civillead.com/how-to-calculate-material-for-plaster/
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# How to Calculate Quantity of Material For Plaster?
## How to Calculate Quantity of Cement, Sand and Water For Plastering
The term plastering is used to describe the thin mortar covering that is applied on the surface of the wall and ceiling.
The plastering removes the unevenness of the surface and is sometimes used to develop decorative effects. Today in this article we will learn how to calculate quantity of material for plaster.
## Important Points For Estimation
Density of Cement = 1440 kg/ m3
Density of Sand = 1450 – 1500 kg/ m3
Density of Aggregate = 1450 – 1550 kg/ m3
Quantity or Weight of Cement in 1 Bag = 50 KG
The volume of 1 Bag Cement in Cum = 0.03472 cubic meter
The volume of 1 Bag Cement in Cft = 1.226 cubic feet
Numbers of Cement Bags in 1 cum = 28.8 nos.
How many Cft in 1 Cum = 35.31 cft
## Calculation of quantity of material for 12 mm thick plaster of ratio 1: 6 (1 Cement:6 Sand) in the wall for 100 Sqm
The volume of mortar = Area × Thickness
= 100 × 0.012
= 1.2
Adding 30% to fill up joints, uneven surface etc
Wet Volume of Mortar = 1.2 + 0.36
= 1.56 Cum
Increasing 25% for dry volume
Total Dry Volume = 1.56 + 0.39
= 1.95
= 2 Cum (say)
Quantity of Cement = 2/(1+6) × 1 (1 Ratio of cement)
= 0.30 Cum
In Kg = 0.30 × 1440 (Density of cement = 1440 kg/m3)
= 432 Kg
In Bags = 8.64 bags
Quantity of Sand = 2/(1+6) × 6 (6 Ratio of sand)
= 1.80 Cum
In Cubic Feet = 63.558 cft (1 Cum = 35.31 cft)
Water Required = Cement quantity in Kg × 0.5 (water cement ratio = 0.5)
= 432 × 0.5
= 216 Litres
Alternative Method
Quantity of Sand = Cement bags × Sand ratio ×1.226 (Volume of one bag cement = 1.226 cft)
= 8.64 × 6 × 1.226
= 63.55 cft
Similarly, The quantity of materials for other proportions may be calculated.
## Material For 20 mm thick Plastering in the wall for 100 sqm
As the thickness of plaster is more, 20% of mortar may be taken to fill up the joints, unevenness etc.
The Quantity of wet Mortar = 100 × 0.02 + 20%
= 2 + 0.4
= 2.4 cu m
Increasing 25% for the dry volume of mortar
Dry Volume of Mortar = 2.40 + 0.60
= 3.00 cu m
The quantities of each material of mortar may be found by the usual method as discussed above.
Also, ReadRate Analysis for Cement Plaster.
## Rich Mortar
For rich mortar plastering, the quantities of material will be less as the cement will be in excess than the voids in sand and the reduction in the volume of dry mortar will be less.
## Ceiling Plastering 12 mm thick for 100 sqm
For plastering in the RCC ceiling, the unevenness of the surface will be less and 20% extra mortar may be taken to get an even surface.
The quantity of wet mortar = 100 × 0.12 + 20%
= 1.2 + 1.4
= 2.6 cu m
Increasing 25% for dry volume
The dry volume of mortar = 1.44 + 0.36
= 1.80
The quantities of each material of mortar may be found by the usual method as discussed above.
For 6 mm thick plastering in RCC, the quantity of dry mortar may be taken as 1.00 cum.
Also, ReadRate Analysis for Gypsum Plaster.
## General Notes
The ratio of Cement Mortar for Wall Plastering = 1:6
The ratio of Cement Mortar for Ceiling Plastering = 1:4
The thickness of the Internal Wall Plastering = 12 to 15 mm
The thickness of the External Wall Plastering = 15 to 20 mm
The thickness of Plastering on the Concrete Face = 6 to 8 mm
## Mode of Measurement For Wall Plaster As Per IS 1200 Part – 12
1. For opening less than 0.5 square meters, no deduction shall be made, and no addition shall be made for reveals, sills, soffits, jambs, etc.,
2. For opening greater than 0.5 square meters but less than 3 square meters, 50% or half ( one side) deduction shall be made, and no addition shall be made for reveals, sills, soffits, jambs, etc.,
3. For opening bigger than 3 square meters, full deduction or 100% ( both sides) shall be made, and addition should be made for reveals, sills, soffits, jambs, etc.,
When Internal and External plaster thicknesses are different, then addition for reveals, sills, soffits, jambs, etc., shall be made as follow.
• When windows are located at the wall’s external face, then added for reveals, jambs, soffits, sills, etc., shall be made as per internal plaster thickness.
• When windows are located at the wall’s internal face, then added for reveals, jambs, soffits, sills, etc., shall be made as per external plaster thickness.
• When windows are located at the wall’s middle face, then added for reveals, jambs, soffits, sills, etc., shall be made as per internal plaster thickness for inner and external plaster thickness for the outer side. If the opening falls under 0.3 to 5 square meters then a 50% deduction shall be made from both sides.
If there is no plaster done for reveals, sills, soffits, jambs, etc., then 100% deduction shall be made for all size openings.
## Conclusion
So, friends, I hope I have covered all the information about how to Calculate Quantity of Cement, Sand and Water for Plastering in this article.
If you learn something, please be sure to share it with someone who might benefit from it.
If you want to add any information which has been missed in this article you can mention it in the comment section.
All Cement Price List Today In India 2022
Steel Price Per Kg Today In India 2022
Difference Between Cement Plaster and Gypsum plaster
How to calculate Brick, Cement and Sand in Brickwork?
How to Calculate Cutting Length of Stirrups For Beam and Column?
## 8 thoughts on “How to Calculate Quantity of Material For Plaster?”
1. Thanks a lot
Please send me PDF documents for reference
2. This is wonderful, many thanks.
• Thanks for your valuable feedback.
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+0
# How do I do this?
0
213
2
Ben had 5 times as many marbles as Caleb at first. After Ben gave away 87 marbles and Caleb won 13 marbles, they had the same number of marbles left. How many marbles did each of them have in the end?
Aug 23, 2021
edited by Guest Aug 23, 2021
edited by Guest Aug 23, 2021
### 2+0 Answers
#1
0
Let 'b' be Ben's amount at first, and 'c' be Caleb's amount at first.
We can form 2 equations: b = 5c and b - 87 = c + 13, so 5c - 87 = c + 13, so 4c = 100, so c = 25, b = 125.
BUT THE QUESTION ASKS FOR HOW MANY MARBLES THEY HAVE AT THE END.
Since they both have the same amount left, 125 - 87 = 25 + 13 = 38.
Aug 23, 2021
#2
+108
0
Let us assume that Caleb had x marbles. As Ben had five times as many marbles as Caleb, therefore, Ben had 5x marbles with him.
It is given that Ben gave away 87 marbles, therefore, the number of marbles remaining
with Ben was:
M= 5x - 87
It is also given that Caleb won 13 new marbles. Therefore, the number of marbles remaining with
Caleb was:
MC = x + 13
It is given that in the end, both Caleb and Ben had the same number of marbles. Therefore, equate the expressions obtained for the number of marbles left with Ben and Caleb and
solve for x.
MB = MC
5x - 87 = x + 13
5x - x = 13 + 87
4x = 100
x = 100/4
x = 25
Therefore, the number of marbles left
with Ben and Caleb were:
M= 5x - 87
= 5 (25) - 87
= 38
Answer - Each of them had 38 marbles in the end.
Aug 24, 2021
edited by KourageKowardlyDog Aug 24, 2021
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Categories
# Find 4 digit number brain teaser
Here is a simple but tricky brain teaser. You are given some hints and you have to guess the number I am thinking using these hints.
Let’s see if you can guess the correct number.
## Find 4 digit number
I’m thinking of a 4-digit number with all distinct digits.
Each of the following has one and only one correct digit
in the correct position:
What is my number?
You are given 6 numbers which include the one correct digit with the correct position. Now, find the number.
Hint: Look for duplicate digits in combinations since you need to find 4 digits but given 6 numbers.
Also Try: If you liked this brain teaser then, you may like some of these brain teasers too.
### Find 4 digit number brain teaser Answer:
4321
Explanation:
Since we are given 6 numbers however only 4 digits need to be found which means, 2 numbers will have duplicate digits with correct position. Here are highlighted digits which match this.
So now we have 2 digits with their positions so lets assume number to be: x32y
We are left with 2 numbers in which we need to find 2 digits, one from each.
1191 –> from this , we know that number cannot be 9, since 9306 had number 9 and as per statement only one digit should be correct and we already know that it was 3.
Which leaves us with digit to be as 1 however as of now we don’t know if it would be first or at 4th position.
4882 –> 6358 and 6728 tells that digit cannot be 8.
It cannot be 2 as it is already there in other combo which leaves digit 4 at first position.
It also tells that digit 1 from above number(1191) should be at fourth position.
So now we have number as 4321
Please like our Facebook page for more such puzzles, riddles and brainteasers.
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https://gmatclub.com/forum/doris-and-edward-both-collect-stamps-if-doris-gives-edward-72-stamps-300365.html
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# Doris and Edward both collect stamps. If Doris gives Edward 72 stamps,
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Re: Doris and Edward both collect stamps. If Doris gives Edward 72 stamps, [#permalink]
Top Contributor
KritiG wrote:
Solving the eqn is very time comsuming!
I know that we cant use the plug in method as the question asks us about difference between two variables.
Any other shortcuts?
Actually speaking, in the exam, our first thought would be to solve the equations and get the individual values.
What we need to realize is that we want D - E, so we should try to see if we can do a simple operation to get this value without having to get individual values.
The first equation is D - 72 = 4(E + 72)
So D - 4E = 360 ... (1)
The second equation is D - 129 = 3(E + 129) = 3E + 516
D - 3E = 516 ... (2)
Equation (2) * 3 - Equation (1) * 2 gives us 3(D - 3E) - 2(D - 4E) = (516 * 3) - (360 * 2)
3D - 9E - (2D - 8E) = 1548 - 720
D - E = 928
Option E
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Re: Doris and Edward both collect stamps. If Doris gives Edward 72 stamps, [#permalink]
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Re: Doris and Edward both collect stamps. If Doris gives Edward 72 stamps, [#permalink]
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# Solving system of linear equations
### Solution:
4.0 h = p ( 4.0 + 50/60)
h = p+1
4.0•h = p•( 4.0 + 50/60)
h = p+1
240h-290p = 0
h-p = 1
h = 29/5 = 5.8
p = 24/5 = 4.8
Write each equation on a new line or separate it by a semicolon. The online calculator solves a system of linear equations (with 1,2,...,n unknowns), quadratic equation with one unknown variable, cubic equation with one unknown variable, and finally any other equation with one variable. Even if an exact solution does not exist, it calculates a numerical approximation of roots.
### Examples:
```a+b = 12
a-3b = a-b+43```
```x+y+z=100
3x-6y+2z=50
y-3z+x=(44-22)x+45```
`(x+4)(x-3)+34x+6x^2 = 256`
`(x+4)+34x+x^2-x^3 = -32`
`ln x = 1.2-x`
`sin x = cos(x-pi/3)+x`
`sin x = x^3+2x+x-1`
`sin x = 0.5`
`(cos x)^2 = tan(x-pi/3)`
`x^5+x^4 = -23+sin x`
`|log x|=2`
`|ln x+1|=x/10`
`|ln x+1|=ln x^2`
`1/(x+2)=1/x+4`
`sqrt(x^4+x^2+2)=22`
### Our mission:
Provide simple, fast, and reliable mathematical service for solving any equation(s).
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# Physics
posted by on .
A rectangle has a length of 2d and a height of d. Each of the following three charges is located at a corner of the rectangle: +q1 (upper left corner), +q2 (lower right corner), and -q (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of q1 and q2. Express your answers in terms of q.
• Physics - ,
THere are a number of ways to do this, some more difficult than others.
If you note the E due to q2 is upward, and the E due to q1 is horizontal, then the E due to q3 must add to zero. That is, the E due to q3 in the horizontal direction must be equal and opposite to q1, and The E due to q3 in the vertical must be equal and opposite to Q2
So dealing with q3 first.
letting o be the distance of the horizontal (o^2=d^2+4d^2=5d^2; or o=dsqrt5)
Ehorizontal=k q/o^2*2d/o=2kqd/o^3
Evertical= kq/o^2*d/o=kqd/o^3
So now setting the vertical of q3 equal to the vertical of q2
k*q2/d^2=kqd/o^3
q2=q (d/o)^3=q(d/dsqrt5)^3=q/5sqrt5
now, setting the horizontal of q3 equal to the horizontal of q1
kq1/4d^2=2kqd/o^3
q1=8q (d/o)^3 and you can finish the algebra.
Check all this math, I did it in my head keying it in.
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# Reading Off the Cartesian Coordinates of a Point(KS2, Year 6)
homesitemapgraphs and coordinatesreading off a point from Cartesian coordinates
The Cartesian coordinates of a point can be read off from a graph.
## How to Read Off the Cartesian Coordinates of a Point
Reading off the Cartesian coordinates of a point is easy.
## Question
What are the Cartesian coordinates of the point shown on the graph below.
## 1
Read off how far across the point is along the horizontal x-axis. We see that the point is 2 units along the x-axis. 2 is the x-coordinate of the point.
Note: The x-axis is labelled with numbers (0, 1, 2, 3...) so you can measure how far across the point is.
## 2
Read off how far up the point is along the vertical y-axis. We see that the point is 4 units up the y-axis. 4 is the y-coordinate of the point.
Note: The y-axis is labelled with numbers (0, 1, 2, 3...) so you can measure how far up the point is.
## 3
Write down the Cartesian coordinates as a pair of numbers in brackets, separated by a comma. The x-coordinate (2) found in Step 1 goes on the left. The y-coordinate (4) found in Step 2 goes on the right.
## Lesson Slides
With practice, you'll be able to read off the Cartesian coordinates of a point from by eye. The slider below gives another example of how to read off the Cartesian coordinates of a point on a graph.
## Positive and Negative Coordinates
The pair of numbers in Cartesian coordinates the can be positive and negative.
• The x-axis is labelled with positive numbers to the right of the y-axis and negative numbers to the left.
• The y-axis is labelled with positive numbers above the y-axis and negative numbers below.
This means if we look at where the point is relative to the x-axis and y-axis, we can see if the x-coordinate and y-coordinate will be positive or negative. See if the point is to the left or right of the y-axis. If it is on the:
• left, the x-coordinate is negative.
• right, the x-coordinate is positive.
See if the point is above or below the x-axis. If it is:
• above, the y-coordinate is positive.
• below, the y-coordinate is negative.
Putting this together, we can see that there are four quadrants made by the two axes. Look at which quadrant a point is in to decide the sign of each coordinate.
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# Why is enthalpy and not heat released by system used while calculating entropy of surroundings?
While finding the sum of change in entropy of the universe and thus defining Gibbs free energy, why is the change in entropy of surrounding the negative of enthalpy of the system divided by the temperature?
I mean why are we considering the pressure to be constant ( thus considering enthalpy) in all systems while calculating Gibbs free energy ? Shouldn’t a more general form of entropy of surroundings be negative of heat released by system divided by the temperature? I was thinking that maybe it is because all closed and open systems are constant pressure systems and change in entropy of surroundings is zero for isolated system but I don’t know, I am just guessing.
So, could someone please help me in understanding why enthalpy and not just heat in general is considered for any system while calculating Gibbs free energy? Any help will be greatly appreciated!
Because $$q=\Delta H$$ only if only expansion work is done and the pressure of the system is constant. In that case, for an isothermal process, in which T is the same for system and surroundings, the entropy of the surroundings can be equated with the change in enthalpy divided by T. Which is why the applicability of the integrated formula $$\Delta G = \Delta H -T \Delta S$$ is restricted to constant p and T. You could write $$q_p$$ in place of $$\Delta H$$ the equation but only when work is restricted to pV work and when T and p are constant. It would not hold generally since sometimes work other than expansion work is performed. In general $$\Delta H = q_p+w_\textrm{non-pV}$$ at constant T and p.
To show this more explicitly, for a small (differential) change: $$dU=dq+dw$$
$$dU=dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}}$$ $$dH = dU+d(pV) = dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}} + pdV+Vdp$$ At constant pressure of the system $$dw_{\textrm{pV}}=-pdV$$, $$Vdp=0$$ and $$dH = dq+dw_{\textrm{non-pV}}$$
Further derivation is possible* to show how this relates to $$G$$, by assuming constant T and considering reversibility, but this should suffice to show that $$dH$$ and $$dq$$ (or $$\Delta H$$ and $$q$$) are not in general the same.
*For a reversible process $$dq=TdS$$ and
$$dH = TdS+dw_{\textrm{non-pV,rev}}$$
Finally the differential expression for $$G$$ at constant $$T$$ follows (since then $$dG=dH-d(TS)=dH-TdS$$):
$$dG=dH - TdS = dw_{\textrm{non-pV,rev}}$$
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# Equilibrium constant change with stoichiometric doubling (Callen)?
• I
• EE18
In summary: So, in summary, the equilibrium constant for a reaction which is the sum of this reaction with itself (doubled reaction) is 2 times the equilibrium constant for the same reaction when written with stoichiometric coefficients twice as large.
EE18
Callen asks us (with respect to an ideal gas)
How is the equilibrium constant of a reaction related to that for the same reaction when written with stoichiometric coefficients twice as large? Note this fact with caution!
I had thought to proceed as follow. We have the definition for the singular reaction:
$$\ln K_s(T) = - \sum_j \nu_j \phi_j(T).$$
Now a reaction which is the sum of this reaction with itself (doubled reaction) has ##\nu_j \to 2\nu_j## so that its equilibrium constant obeys, by definition,
$$\ln K_d(T) = - \sum_j 2\nu_j \phi_j(T) = 2\ln K_s(T) \implies K_d = e^2K_s.$$
But when I look online it says the equilibrium constant should square in this case, ##K_d = K_s^2##. Can someone point out what I'm doing wrong?
Last edited:
##2 \ln x = \ln(x^2)##
TSny said:
##2 \ln x = \ln(x^2)##
You are saying to use
$$\ln K_d(T) = 2\ln K_s(T) = \ln K^2_s(T)\implies K_d = K_s^2$$
which makes sense to me. I'm embarrassed to say I don't know what I'm doing wrong by using
$$\exp(\ln K_d(T)) = K_d = \exp(2\ln K_s(T)) = e^2 exp(\ln K_s(T)) = e^2K_s$$
which is different. What elementary algebra is being slipped under on me here?
EE18 said:
You are saying to use
$$\ln K_d(T) = 2\ln K_s(T) = \ln K^2_s(T)\implies K_d = K_s^2$$
which makes sense to me. I'm embarrassed to say I don't know what I'm doing wrong by using
$$\exp(\ln K_d(T)) = K_d = \exp(2\ln K_s(T)) = e^2 exp(\ln K_s(T)) = e^2K_s$$
which is different. What elementary algebra is being slipped under on me here?
Note that ##\exp(2\ln K_s(T)) \neq e^2 \exp(\ln K_s(T))##.
Instead, ##\exp(2\ln K_s(T)) = [\exp(\ln K_s(T)]^2##. This follows from ##x^{ab} = (x^a)^b##.
TSny said:
Note that ##\exp(2\ln K_s(T)) \neq e^2 \exp(\ln K_s(T))##.
Instead, ##\exp(2\ln K_s(T)) = [\exp(\ln K_s(T)]^2##. This follows from ##x^{ab} = (x^a)^b##.
Oof, of course. ##\exp(2+\ln K_s(T)) =e^2K_s## which is of course not what we have here.
My bad, and thanks for the clarification on this silly error.
EE18 said:
Oof, of course. ##\exp(2+\ln K_s(T)) =e^2K_s## which is of course not what we have here.
Right.
## What is the equilibrium constant in the context of stoichiometric doubling?
The equilibrium constant (K) is a dimensionless number that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. When the stoichiometry of the reaction is doubled, the equilibrium constant changes because it depends on the stoichiometric coefficients of the balanced chemical equation.
## How does the equilibrium constant change when the stoichiometric coefficients are doubled?
When the stoichiometric coefficients of a balanced chemical equation are doubled, the equilibrium constant is squared. For example, if the original reaction is A + B ⇌ C with an equilibrium constant K, then the doubled reaction 2A + 2B ⇌ 2C will have an equilibrium constant K^2.
## Why does doubling the stoichiometric coefficients affect the equilibrium constant?
Doubling the stoichiometric coefficients affects the equilibrium constant because the equilibrium expression depends on the concentrations of reactants and products raised to the power of their coefficients. When the coefficients are doubled, the exponents in the equilibrium expression are also doubled, leading to a change in the value of the equilibrium constant.
## Can you provide an example to illustrate how the equilibrium constant changes with stoichiometric doubling?
Consider the reaction N2 + 3H2 ⇌ 2NH3 with an equilibrium constant K. If we double the stoichiometric coefficients, we get 2N2 + 6H2 ⇌ 4NH3. The equilibrium constant for this new reaction will be K^2 because the exponents in the equilibrium expression are doubled.
## Is the change in the equilibrium constant due to stoichiometric doubling a common topic in thermodynamics?
Yes, the change in the equilibrium constant due to stoichiometric doubling is a common topic in thermodynamics and chemical kinetics. It is important for understanding how changes in the stoichiometry of a reaction can affect the position of equilibrium and the concentrations of reactants and products at equilibrium.
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# In a circus stuntman rides a motorbike in a circular track of radius R in the vertical plane. The minimum speed at highest point of track will be (1) $\sqrt{2gR}$ (2) 2gR (3) $\sqrt{3gR}$ (4) $\sqrt{gR}$
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A ball is suspended by a thread of length l. What minimum horizontal velocity has to be imparted to the ball for it to reach the height of the suspension:
(1) gl
(2) 2 gl
(3) $\sqrt{gl}$
(4) $\sqrt{2gl}$
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(1) 2 mg
(2) mg
(3) 3 mg
(4) $\sqrt{3}mg$
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As per the given figure to complete the circular loop, what should be the radius of the loop?
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The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered s as k = as2 where a is a constant. The force acting on the particle is
(1) $2a\frac{{s}^{2}}{R}$
(2) $2as{\left(1+\frac{{s}^{2}}{{R}^{2}}\right)}^{1/2}$
(3) 2 as
(4) $2a\frac{{R}^{2}}{s}$
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(1) 20 m/sec
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(4) 10 m/sec
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(2) $\sqrt{2gL}$
(3) $\sqrt{{u}^{2}-gl}$
(4) $\sqrt{2\left({u}^{2}-gL\right)}$
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Interval notation. Let $a, b$ be real numbers.
A set $I$ of real numbers is an interval iff for all $x,y\in I$ all real numbers between $x$ and $y$ are also in $I$.
The open interval $(a,b)=$ $\{x\in\m R\ |\ a< x< b\}$
The closed interval $[a,b]=$ $\{x\in\m R\ |\ a\le x\le b\}$
The half-open interval $(a,b]=$ $\{x\in\m R\ |\ a< x\le b\}$
The half-open interval $[a,b)=$ $\{x\in\m R\ |\ a\le x< b\}$
$(a,\infty)=$ $\{x\in\m R\ |\ a< x\}$
$(-\infty,b]=$ $\{x\in\m R\ |\ x\le b\}$
$(-\infty,\infty)=$ $\m R$ (the real line)
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# Why don't we use base 6 or 11?
Another question on this site asks why we have chosen our number system to be decimal base 10. There are others asking basically the same thing as well.
I'm not really satisfied with any of the answers, because most of the answers given seem to suggest that base 10 was chosen because we have $10$ fingers. However, this would seem to me to imply that we should be using decimal base $11$. Supposing we use the scheme of calling decimal 10 "A", then on our fingers we would could count $1, 2, 3, 4, 5$ on the first hand, and then $6, 7, 8, 9, A,$ on the second. Only then would we be out of fingers and need to roll over to 10 which would be decimal $11$. Likewise, a similar argument could be made for base 6 counting on only one hand, as there are five digits before one runs out and needs to roll over to $10$, in this case for the decimal value 6.
For base 6 the argument could be made that the thumb is not counted, and thus base 5 is more natural, but the fact remains that we don't use base 5 either, we use base 10, and not counting thumbs on either hand would result in us using base 9, not decimal base 10, so I feel like this argument does not hold water either.
An alternate explanation, that base 10 is an abbreviation of base 60 seems slightly more likely, but base 60 seems rather unweildy to being with, which leads me to the question, why don't we simply use base 11, as our 10 fingers seem most suited to it? As far as I am aware no culture has ever widely used it.
• Historical considerations aside, every positional number system needs a digit (corresponding to) $0$. Base $10$ uses $10$ digits, and we have $10$ fingers (and the etymology of "digit" goes back to finger). Commented Sep 21, 2013 at 5:08
• 36 is base 6 is 100..now try to divide it by 5 ..you will yourself understand why decimal ..! Commented Nov 22, 2013 at 21:13
• There's a near-extinct language (Native American, I think) which uses base 8. Apparently, they counted on the spaces between their fingers. (Bonus binary compatibility!) There's a Numberphile video on this. Commented Aug 19, 2015 at 20:43
• @arnab : Just like Dadam said: 36/6 in base-10 is the same as 100/10 in base 6. By choosing the right numbers you can defend any base. Commented Nov 26, 2015 at 11:06
The idea of a "base", and even the idea implicit in it of a consistent positional numbering system, is a relatively modern one. For that matter, even "base 10", in the sense of decimal numerals — the Hindu–Arabic numeral system we all use today — is relatively modern; witness the fact that much of Europe didn't begin to use it until well into the second millenium. The idea that we can count in any base is even newer.
Your argument might make sense if humanity started with the idea in mind of using a base-$b$ representation for some $b$, and then looked to their fingers to decide what the base $b$ should be. This of course is not what happened, nor even is it imaginable in any culture. The concrete precedes the abstract.
Instead, what we see historically is counting on one's fingers, and thus counting by fives or by tens, not counting in base 10. If you're counting on two hands, when you reach $10$ you literally run out of fingers to count, and that is where you have to leave a mental (or physical) note to yourself that you're done with one round of counting, and begin counting anew. (Similarly $5$, if using one hand.)
We can see traces of this "counting by $5$" or "counting by $10$" in early systems like Roman numerals: note that $8$ is represented as "VIII", denoting one count of five (done with one hand's worth of counting), and starting again, reaching up to $3$ in the process. Similarly, the representations "XX" and "XXX" show that they were being thought of as "two tens" and "three tens", rather than as "in base $10$, three in the tens place and zero in the units place" — the idea of base $10$ is not actually present. Thus "XXXVIII" literally denotes the process of counting: "three tens (three double-hands), a five (one hand) and three fingers". (There are even traces of counting by $20$s; consider the English "score" and French number names).
It was only after centuries of already counting by $10$s, conducting transactions with $10$-based words for numbers and so on, that the base-$10$ representation arose, as a representation system for the same numbers that everyone was already accustomed to thinking of in tens, and (very slowly) spread across the world.
• Aside: even with the "backward-compatible" base $10$, the positional ("place-value") system took several centuries between 1202, the year of being introduced in Europe by Leonardo of Pisa (Fibonacci), and it being generally adopted in Europe. One can only imagine what would have happened if someone had tried to introduce a base-$11$ representation, in which a number like LXXIII, conceptualized by everyone as "a fifty, two tens, and three", would have to be represented by an entirely unrelated-looking "67". Commented Aug 11, 2013 at 17:05
• What you're saying is true in respect to a given culture. However, positional systems are not new if you look at the world as a whole. Sumerians were using their base 60/10 system circa 23rd century BC. Just as you said, they had been using an additive system with these numbers for at least a millennium. Commented Aug 11, 2013 at 17:51
• @GregRos: Yes, my "relatively modern" is to be loosely interpreted: e.g. it works to take it to mean "much more recent than counting itself". :-) BTW, even the Sumerian/Babylonian numerals seem to have used counting by tens, to represent the 59 different "digits" they needed (only 59 because they muddled through without a 0): en.wikipedia.org/wiki/File:Babylonian_numerals.svg It is said that the Yuki and Pamean peoples (both in the Americas) use octal because they look at the spaces between the fingers, but even there they're counting by number of "things", not one more (like base-11). Commented Aug 11, 2013 at 18:07
• Yup, that's why I said base 60/10. More so; before they used cuneiform, they used a different number system in which the numbers 10 and 60 featured prominently. Their number names also betray the importance of the number 10. Commented Aug 11, 2013 at 18:33
Let's consider your base 6 proposal. You motivate $10_6$ to be the first number that you can't count on one hand. To be consistent, $20_6$ should be the first number that you can't count on two hands. But it isn't! Instead:
• $10_6$ is $1$ more than the number you can count on $1$ hand
• $20_6$ is $2$ more than the number you can count on $2$ hands
• $30_6$ is $3$ more than the number you can count on $3$ hands
What a strange pattern. Wouldn't it be better if $d$ more than the number you can count on $d$ hands were written $dd$? Well, that's what we get in base 5:
• $11_5$ is $1$ more than the number you can count on $1$ hand
• $22_5$ is $2$ more than the number you can count on $2$ hands
• $33_5$ is $3$ more than the number you can count on $3$ hands
And there's an even nicer pattern for numbers of the form $d0$:
• $10_5$ is the number you can count on $1$ hand
• $20_5$ is the number you can count on $2$ hands
• $30_5$ is the number you can count on $3$ hands
The same arguments apply to 10 versus 11.
• It took me a minute to understand your assertion that "20 base 6 is 2 more than the number you can count on 2 hands", because in my mind I was picturing being able to count to 55 base 6 using two hands, with the left hand being the 10's (base 6) position and the right hand being the 1's position. Commented Aug 6, 2013 at 18:55
• With base 10 (and base 5) when counting your fingers you have a extra rule where one finger must ALWAYS be up, so you start to count with 1. With base 6 and 11, you start to count without any finger being at the up position, so you start to count from 0 and not 1. At base 6, 10, means 1 hand + 0 fingers. Commented Jul 16, 2021 at 13:23
Well, you're clearly aware that different cultures have used different bases, and indeed developed sophisticated mathematics using them. Further, most of the mathematical advances in the West over the last 500 years (say) would seem to have nothing to do with what base we use in ordinary life.
So I'm not quite sure where the "why" question comes in. I agree that claims that base 10 is "natural", or is more efficient than some other base, seem dubious. So what are we seeking to explain here, exactly?
If it's a simple causal explanation that's wanted -- "How did it come to be that we use base 10" -- then the answer is, I'm willing to bet, that it's an accident of history.
In other words: isn't this a bit like asking why we have 26 letters in the alphabet?
• I don't think your last comparison is good. While the base $10$ is (almost?) exclusively used today in ordinary life, there are various alphabets with various numbers of letters. Just to name a few Slavic ones: Croatian/Serbian/Bosnian (30), Slovene (25), Russian (33),... There is a huge list of writing systems being used around the world, nothing like the number systems. (Upvote for the rest of your answer) Commented Aug 11, 2013 at 16:57
• @Vedran -- very true... I wish I could think of a better analogy but my poor brain's failing me. Commented Aug 11, 2013 at 17:29
• It's a fact that the number 10 crops up all the time as part of number systems, and usually as a dominant part. Almost all other number systems use 20, 5, or 12. These are also related to finger counting (or toe counting). The names of numbers in many diverse cultures also betray the significance of finger counting, and the numbers 5 and 10. In contrast, I don't think a number system based on 7, 28, or 11 has ever been found. There is probably something about all these systems that makes them so common. It's not just a fluke. Commented Aug 11, 2013 at 18:41
• @Greg -- I think tracing any base that's not coprime to 10 to finger-counting might be a stretch; it'd be a huge task to demonstrate all such bases originated from that. Not saying it couldn't feature in the history of a specific system, but the danger is that finger-counting becomes a "just-so" story that naturalises a contingent practice (base 10) that we happen to be used to. Commented Aug 12, 2013 at 9:07
• The issue is not as theoretical as you seem to think, and it is not my off-hand remark that links such systems to finger counting. It is a large collection of evidence. You should look at that evidence yourself, instead of taking the armchair approach. Commented Aug 12, 2013 at 13:03
I'm writing this answer because, I think you might not be satisfied with the answers given for now. In my opinion, your approach is also natural and probable. Also base $12$ is normal and natural. But, this things are like languages, not just logic. For example, we still uses $12$ based clocks. So, shortly, your approach is maybe more natural, more logical but this aspect of mathematics depends on occasions not just pure logic.
interesting question ! :)
I THINK THIS A POSITIONAL NUMBER SYSTEM .and that's of great advantage ..simple shifting the position of decimal . this is the most conventional answer ..! but somehow it seems , not that strong . we could have tried to construct a fractional system in other number systems as well !
but there are certain points i have made out :
1.if the base is more than 10 , say 20 . then if you want to use our 10 symbols (i.e. 0, 1..,9) it not possible for us to write numbers more than 9 and less than 20 or numbers more than 189 and less than 200 and numbers of similar form ) , so in that case for each unit either you will have to allot two places for digits (which will make it difficult to read a number) or you will have introduce more 10 symbols , which will make it more complex and we wont get any advantage. cause 20 is just a a multiple of 10 by 2 ..and 2 is a factor of 10 ( if the base would be 30 then , we could take it as an advantage that the problem in fixing the decimal we face while handling division by 3 could be simplified . but in that case just imagine how many symbols you will have to introduce !!! the base could be 12 . in that case dividing by three would not face any approximation problem..or if the base would be 14 then we would easily be able to divide by 7 ..but in both 12 and 14 base the problem in dividing by 5 arises. and in ancient days as our hand and toe consists of 5 fingers , i think divisibility ease for 5 was given preference ) so among systems with 10 or more than ten , the base 10 system , i.e. decimal system was given preference in different civilization , as a human mind always and everywhere think in the same direction and following our ancestors we set this as our ruling international number system mostly used ! 2.now , for bases less than 10 : there is no number which has three prime factors ! among numbers 10 or less than 10 , 10 has the highest number of one digit-ed prime factors ! and for that reason in decimal system , we can properly divide any number by most number of numbers compared to all the other systems with base less than 10 .!(except 6 . which was again given less priority because 10 is a multiple of 5 )
NOTE : some say , "you can order the numbers with specific space between in decimal system .. If we can't order the number with equal interval among them , then it wont be possible to get any number line,and whole number theory will possibly be collapse then . the way co-ordinate geometry ,..vector algebra work , where ordering is very important will collapse ..! e.g in octal system 63 =77 where 64 = 100 and 65 = 101 ..so , then what's the measure of unit countable increase ? .." but , that's not a proper argument in my opinion . there we will have to consider 77+1 = 100 , 7+1=10 , 777+1=1000...and so on ,..
added note : 11 is itself a prime number . so in dividing a number which is a fractional multiple of other primes , we will face great difficulty !
• 36 in base 6 is 100 . now try to divide it by 5 ..you will yourself understand why decimal system ! Commented Nov 22, 2013 at 21:12
• 36/6 in base-10 is the same as 100/10 in base 6. By choosing the right numbers you can defend any base. Commented Dec 28, 2013 at 10:18
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## Thursday, February 27, 2014
### Finding primitive roots
First off, it has to be noted that the constraint that the modulo value is a prime number is just for simplicity (hint: all values from 1 to p-1 are coprime to p for a prime number p).
Now, what EVEN is a primitive root modulo p you may ask. This is just a number g which generates all numbers from 1 to p-1 with $$g^{t} \: mod \: p$$. That is, for every positive integer x < p there exists a t, so that $$g^{t} \equiv x \pmod{p}$$.
Properties of primitive roots:
• for a given p there are phi (phi(p))=phi(p-1) primitive roots
• g is a primitive root $$\Leftrightarrow$$ for every divisor d of $$(p-1)=phi(p)$$, $$g^{d} \not\equiv 1\pmod{p}$$ holds $$\Leftrightarrow$$ for every prime $$x$$ from the prime factorization of $$p-1$$, $$g^{(p-1)/x} \not\equiv 1\pmod{p}$$ holds.
• (comment: this is essentially there to check for repeating sequences in $$g^{t}$$ for $$t \in {1,..,p-1} \pmod{p}$$.
Some research: I've ran this code to look at the distribution of the primitive roots for primes between 100 and 1,000,000,000. By running the code we see this from the output:
• the lower bound for the number of primitive roots is $$0.15 \cdot p$$. Therefore, 15% of the numbers smaller than $$p$$ are primitive roots for a given p (in worst case!). By selecting a random positive integer < p we have a 15% chance of selecting a primitive root!
• given a random p, the chance of hitting a primitive root on random is even ~40%.
Therefore, in order to find a primitive root, we may simply pick up random numbers and check the above property which has to hold for primitive roots. Code which does that: link.
### Now to the discrete logarithm (the index):
for a given x, how to find a $$y=ind_{g}(x)$$ with $$g^{y} \equiv x \pmod{p}$$. The answer is: you can't. Many encryption algorithms are based on this fact. Though, they have some nice properties as normal logarithms:
• $$ind_{g}(ab) \equiv ind_{g}(a) + ind_{g}(b)\pmod{p-1}$$
• $$ind_{g}(a^{k}) \equiv k \cdot ind_{g}(a)\pmod{p-1}$$
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# college math
posted by .
14. A probability is always expressed as a value from: (Points: 5)
0 and 1.
-1 and 1.
0 and 100.
0 and 10.
## Similar Questions
1. ### Economics/Math
1. Assume that q and z are two random variables that are perfectly positively correlated. q takes the value of 20 with probability 0.5 and the value of zero with probability 0.5, while z takes the value of 10 with probability 0.5 and …
2. ### Economics/Statistics
1. Assume that q and z are two random variables that are perfectly positively correlated. q takes the value of 20 with probability 0.5 and the value of zero with probability 0.5, while z takes the value of 10 with probability 0.5 and …
3. ### GEOMETRY
The squares of a 3×3 grid of unit squares are coloured randomly and independently so that each square gets one of 5 colours. Three points are then chosen uniformly at random from inside the grid. The probability that these points …
4. ### Probability
Two equally matched teams are playing a game of volleyball. Each team has a 1/2 chance of winning each point. The game ends once one of the teams gets to a score of 21. If the score is currently 18-16 for team A, the probability that …
If a 20-sided fair die with sides distinctly numbered 1 through 20 is rolled, the probability that the answer is a perfect square can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?
6. ### MAths
A unit square is drawn in the Cartesian plane with vertices at (0,0),(0,1),(1,0),(1,1). Two points P,Q are chosen uniformly at random, P from the boundary of the square and Q from the interior of the square. The line L1 through P and …
7. ### math:quick help
Let Fn be the nth number in the Fibonacci Sequence. Consider the 3 points (F30,F31),(F32,F33),(F34,F35) in the Cartesian plane. You are allowed to repeatedly apply the following operation: Let P be any one of the three points in the …
8. ### math
Three points are chosen uniformly at random from the perimeter of circle. The probability that the triangle formed by these is acute can be expressed as ab where a and b are coprime positive integers. What is the value of a+b?
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# Thread: Using the binomial expansion.
1. ## Using the binomial expansion.
Any faulty logic?
Prove: if a>1, (a^n)/n --> infinity.
Since a>1 we can write:
a = [(1 + k)]/ [n^(1/n)];
n*a^n = (1 + k)^n;
n*a^n =1 + nk + [n(n -1 )(k^2)]/2!+.....;
So, a^n =1/n + k +.....;
>1/n + k;
>M, for M>0 for say 1/n > M.
Is there a problem here since n < 1/M? Or, does this still imply n > M, so it doesn't matter?
Thanks.
2. Originally Posted by cgiulz
Any faulty logic?
Prove: if a>1, (a^n)/n --> infinity.
Since a>1 we can write:
a = [(1 + k)]/ [n^(1/n)];
n*a^n = (1 + k)^n;
n*a^n =1 + nk + [n(n -1 )(k^2)]/2!+.....;
So, a^n =1/n + k +.....;
At this step you should have,
$\frac{a^n}{n} = \frac{1}{n} + k + \frac{1}{2}k^2(n-1) + ... \geq \frac{1}{2}k^2(n-1)$
Now certainly, $\frac{1}{2}k^2(n-1)\to \infty$.
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# Difference between revisions of "Simon's Favorite Factoring Trick"
## The General Statement For Nerds
Simon's Favorite Factoring Trick (SFFT) is often used in a Diophantine(Positive) equation where factoring is needed. The most common form it appears is when there is a constant on one side of the equation and a product of variables with each of those variables in a linear term on the other side. A extortive example would be: $$xy+66x-88y=23333$$where $23333$ is the constant term, $xy$ is the product of the variables, $66x$ and $-88y$ are the variables in linear terms.
Let's put it in general terms. We have an equation $xy+jx+ky=a$, where $j$, $k$, and $a$ are integral constants. According to Simon's Favourite Factoring Trick, this equation can be transformed into: $$(x+k)(y+j)=a+jk$$ Using the previous example, $xy+66x-88y=23333$ is the same as: $$(x-88)(y+66)=(23333)+(-88)(66)$$
If this is confusing or you would like to know the thought process behind SFFT, see this eight-minute video by Richard Rusczyk from AoPS: https://www.youtube.com/watch?v=0nN3H7w2LnI. For the thought process, start from https://youtu.be/0nN3H7w2LnI?t=366
## Applications
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually $x$ and $y$ are variables and $j,k$ are known constants. Also, it is typically necessary to add the $jk$ term to both sides to perform the factorization.
## Fun Practice Problems
### Introductory
• If n^2
• Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$
(Source)
### Intermediate
• $m, n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.
(Source) NERD
• The integer $N$ is positive. There are exactly 2005 ordered pairs $(x, y)$ of positive integers satisfying:
$$\frac 1x +\frac 1y = \frac 1N$$
Prove that $N$ is a perfect square.
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What happens in an RC circuit when the switch is suddenly closed?
What happens in an RC circuit when the switch is suddenly closed?
When a switch in an RC circuit is suddenly closed, the circuit will undergo a transient response. The behavior of the circuit during this transient period depends on the values of the resistor (R) and capacitor (C) and the initial conditions.
An RC circuit consists of a resistor and a capacitor connected in series or parallel. When the switch is closed, the capacitor begins to charge up through the resistor. Let's look at what happens in each case:
Charging in Series RC Circuit:
In a series RC circuit, the capacitor charges up to the applied voltage (V) gradually. Initially, when the switch is closed, the capacitor behaves like a short circuit, and the current flows through the resistor and the capacitor. The charging current starts high and gradually decreases as the capacitor voltage approaches V.
The charging process is governed by the following equation:
(
)
=
ร
(
1
โ
โ
/
(
ร
)
)
V(t)=Vร(1โe
โt/(RรC)
)
Where:
(
)
V(t) is the voltage across the capacitor at time
t.
V is the applied voltage.
R is the resistance in ohms.
C is the capacitance in farads.
e is the base of the natural logarithm.
Charging in Parallel RC Circuit:
In a parallel RC circuit, the capacitor charges up similarly to the series case, but the voltage across the capacitor increases instantly. When the switch is closed, the capacitor starts to charge up rapidly through the resistor, reaching its final voltage almost immediately.
The voltage across the capacitor during charging is given by:
(
)
=
ร
(
1
โ
โ
/
(
ร
)
)
V(t)=Vร(1โe
โt/(RรC)
)
The time it takes for the capacitor to charge to approximately 63.2% of the applied voltage (V) is called the time constant (ฯ) of the circuit and is given by ฯ = R ร C.
It's essential to note that during the transient period, the current and voltage change continuously, eventually reaching a steady-state condition where the capacitor is fully charged (in the case of a charging circuit) and the current becomes negligible (in the case of a discharging circuit).
If the circuit was previously in a steady state with the switch open for a long time, the initial voltage across the capacitor would be 0V (discharged), and it would start charging or discharging from this initial condition based on the direction of the current when the switch is closed.
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How does the behavior of an RC circuit change when the resistance is increased?
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How does the behavior of an RC circuit change when the capacitance is increased?
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How is the impedance of an RC circuit affected when an inductor is added in series?
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How does an RC circuit behave when connected to an AC power source?
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What are the safety considerations when working with RC circuits?
Answer : Working with RC (Resistor-Capacitor) circuits involves handling electrical components and dealing with electrical energy. It's essential to prioritize safety to prevent accidents and protect both yourself ... be aware of potential risks to prevent accidents and protect yourself and your equipment....
What is the concept of time constant in an RC circuit and how does it relate to the transient response?
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What is the purpose of a diode in an RC circuit with freewheeling diode configuration?
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What is the formula for calculating the energy stored in a capacitor in an RC circuit?
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What is the phase relationship between the current and voltage in an RC circuit?
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What is the role of a timing capacitor in an RC circuit used for oscillations?
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What is the relationship between the capacitance and the charge stored in an RC circuit?
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How does the transient response of an RLC circuit change when the damping factor is close to unity?
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Answer : To calculate the voltage across a capacitor in an RC (Resistor-Capacitor) circuit at a specific time, you can use the following formula: V(t) = V0 * (1 - e^(-t / RC)) Where: V ... reach the maximum voltage depends on the time constant RC, with a larger RC resulting in a slower charging process....
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How does the impedance of an RC circuit vary with frequency?
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How does the capacitance of an RC circuit affect its response to a sudden change in voltage?
Answer : The capacitance of an RC (Resistor-Capacitor) circuit plays a crucial role in determining its response to a sudden change in voltage, also known as a transient response. To understand how ... product of resistance and capacitance, plays a significant role in shaping the transient response behavior....
Compare the behavior of an RC circuit in a DC circuit versus an AC circuit.
Answer : An RC circuit consists of a resistor (R) and a capacitor (C) connected in series or parallel. The behavior of an RC circuit varies significantly between DC (direct current) and AC ( ... a varying impedance, allowing more current to pass through at higher frequencies and less at lower frequencies....
How does the capacitor limit the rate of change of voltage in an RC circuit?
Answer : In an RC circuit (Resistor-Capacitor circuit), the capacitor plays a crucial role in limiting the rate of change of voltage. To understand how this works, let's first examine the behavior of a capacitor ... time constant (achieved with lower R or C values) leads to a faster rate of voltage change....
How does the capacitance of an RC circuit change with temperature?
Answer : The capacitance of an RC circuit can be affected by temperature changes. The extent of this effect depends on the type of capacitor used in the circuit. Different capacitor ... with low temperature coefficients or use compensation techniques to mitigate the impact of temperature variations....
How does the presence of different dielectric materials affect the capacitance in an RC circuit?
Answer : In an RC circuit (Resistor-Capacitor circuit), the presence of different dielectric materials can significantly affect the capacitance. The capacitance of a capacitor is a measure of its ability ... and circuits, allowing for the storage and manipulation of electric charges in a controlled manner....
Can you explain the concept of dielectric breakdown in an RC circuit?
Answer : Certainly! Dielectric breakdown is a concept that applies to capacitors in an RC (resistor-capacitor) circuit. Let's break down the components and then delve into the concept: Capacitor ... may use capacitors with higher voltage ratings to avoid operating near the dielectric breakdown threshold....
How do you calculate the resonant frequency of an RC circuit?
Answer : To calculate the resonant frequency of an RC circuit, you need to consider the components of the circuit: a resistor (R) and a capacitor (C). The resonant frequency is the frequency at which ... of waveforms or transient behavior, the concept of resonant frequency may not apply in the same way....
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# Primitive root and discrete logarithm
Started studying Diffie Hellman key exchange protocol as part of the course. I dont have much maths knowledge. Can somebody explain in simple terms what is the significance of taking Primitive root in DH algorithm.
Also could someone explain about discrete logarithm
• Have you tried wikipedia? Jul 20, 2011 at 2:01
Let $p$ be a prime. We say that $g$ is a primitive root of $p$ (or sometimes modulo $p$), if the powers $g^1$, $g^2$, $g^3$, $\dots$, $g^{p-1}$ are congruent, in some order, to $1$, $2$, $3$, $\dots$, $p-1$ (modulo $p$). Or in simpler terms, when we consider the remainders when $g^k$ is divided by $p$, all numbers between $1$ and $p-1$ are remainders ($0$ can't be).
Note that by Fermat's Theorem, $g^{p-1} \equiv 1 \pmod{p}$, so after $g^{p-1}$, the powers of $g$ start all over again modulo $p$, so $g^p\equiv g$, $g^{p+1}\equiv g^2$, and so on.
If you have seen some group theory, we could alternately say that $g$ is a primitive root of $p$ if $g$ is a generator of the multiplicative group on non-zero objects modulo $p$.
It can be proved using elementary tools that every prime has a primitive root. The proof is not all that easy. Large primes $p$ have many primitive roots.
Here is a small example. Let $p=7$, and take $g=2$. The powers of $2$, reduced modulo $7$, are $2$, $4$, $1$, $2$, $4$, $\dots$. So we don't get everything, $2$ is not a primitive root of $7$. Now take $g=3$. The powers of $3$, reduced modulo $7$, are $3$, $2$, $6$, $4$, $5$, $1$, $3$, $\dots$, so we do get everything, $3$ is a primitive root of $7$.
Let $g$ be a known primitive root of the prime $p$, and suppose that $a$ is relatively prime to $p$. Then, by definition, there is a unique integer $k$, with $1 \le k \le p-1$, such that $$g^k \equiv a \pmod p$$ (some people use $0 \le k \le p-2$ instead, I prefer that, it doesn't matter much).
The number $k$ used to be called the index of $a$ with respect to the primitive root $g$. More recently, and universally in Computer Science, the $k$ is called the discrete logarithm of $a$ (with respect to the primitive root $g$).
These "discrete logarithms" have formal properties much like ordinary logarithms. Note in particular that discrete logarithms are exponents, just like ordinary logarithms.
What they are good for: Here I will take a glance at the reason for their use in cryptography. If you know $g$ and $k$, it is computationally easy to calculate the remainder when $g^k$ is divided by $p$, even when $p$ is a huge prime. We use the Binary Method for Exponentiation (look this up) or a relative.
But given $g$ and $a$, it seems to be computationally very hard to find the $k$ between $1$ and $p-1$ such that $g^k \equiv a \pmod p$. In other words, finding the discrete logarithm when a very large prime is involved, seems to be computationally very difficult.
That makes exponentiation modulo $p$ a useful "trap-door" function, easy to do, but hard to undo, kind of like multiplication of two primes (easy) and factorization of a product of two large primes (seemingly very difficult).
Hope this gives some overview of discrete logarithm. I really cannot add anything about Diffie Hellman details. For one thing, there is a family of DH procedures.
To sum up, the main reason for the usefulness of discrete logarithm is nice algebraic properties, together with the "trap-door" effect.
Here is an overview about the Diffie-Hellman key exchange algorithm.
See André's answer about the basics of the discrete logarithm.
We have a (cyclic) group (the multiplicative group modulo a big prime is the original case, but other groups are possible, too), in which we can multiply, and thus also have exponentation with integer numbers (using the square-and-multiply method, for example).
The group $G$ and a generator $g \in G$ (e.g. a primitive root in the prime group case) are fixed before the algorithm starts. These can be public, there is nothing private in this.
1. User A creates a random number $x$, calculates $\alpha := g^x$.
User B creates a random number $y$, calculates $\beta := g^y$.
2. A sends $\alpha$ to B.
B sends $\beta$ to A.
3. User A calculates $\gamma_{\mathrm A} := \beta^x$.
User B calculates $\gamma_{\mathrm B} := \alpha^y$.
Now we have $\gamma_{\mathrm B} = \alpha^y = (g^x)^y = g^{x\cdot y} = g^{y\cdot x} = (g^y)^x = \beta^x = \gamma_{\mathrm A}\$ (without ever calculating $x \cdot y$).
Thus $\gamma := \gamma_{\mathrm A} = \gamma_{\mathrm B}$ can be used as a shared secret between A and B, to derive a key from it.
But a possible adversary only could have observed $\alpha$ and $\beta$, neither $x$ nor $y$ nor $\gamma$. The Diffie-Hellman problem is the problem to compute $\gamma$ from $\alpha$ and $\beta$. Of course, this could be solved easily by calculating the discrete logarithm of $\alpha$ or $\beta$, but this is considered hard.
There is no proof yet that the Diffie-Hellman problem really can only be solved by calculating discrete logarithms, and it is not yet proved that discrete logarithms are really hard (this also depends on the group in question), but there is no known efficient algorithm, so the Diffie-Hellman key exchange is actually used (for example, in the TSL/SSL and SSH protocols).
In practice you also want to add some authentication to be sure that you are really communicating with the partner you think you are communicating with - the plain DH KE is susceptible to a man-in-the-middle attack (the attacker negotiates one key this A, another one with B, and then decrypts/encrypts data between both).
The ElGamal encryption scheme is derived from the DH key exchange algorithm - here A publishes $\alpha$ (as well as $G$ and $g$) ahead of time, and B then chooses $y$ for each message to $A$, and sends $\beta$ as well as $m \cdot \gamma$.
Alice can then compute $\gamma^{-1} = \beta^{-x} = \beta^{q-x}$, where $q$ is the order of the group $G$, and from that the message $m$.
(In reality, we will only encrypt a key for some symmetric algorithm, and use this to encrypt the real message.)
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# math
posted by .
Julie Fleming owns 903/4 acres of land in Arizona. She sells one-third of the land and deeds 1/4 of the reminder to her son. How many acres of land does she have left?
Is that 90 3/4 (ninety and three-quarters) acres? If so then 90 3/4 = 363/4 Thus
(1/3)363/4 = 121/4 which means 242/4 or 60.5 acres are left, If she deeds 1/4 of that she has (3/4)(242/4) left.
Be sure to check my arithmetic and let me know if I read it correctly.
No, that's the amount left after she deeds 1/3. You want to multiply 60.5 by 3/4 because she deeds 1/4 of the 60.5.
Thanks a lot for your help. I understand how to do this now.
i need help with my math home work
ummm.... what kind of homework? what do u need help with???
I need help with the
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## Perform Symbolic Computations
### Differentiate Symbolic Expressions
With the Symbolic Math Toolbox™ software, you can find
• Derivatives of single-variable expressions
• Partial derivatives
• Second and higher order derivatives
• Mixed derivatives
For in-depth information on taking symbolic derivatives see Differentiation.
#### Expressions with One Variable
To differentiate a symbolic expression, use the `diff` command. The following example illustrates how to take a first derivative of a symbolic expression:
```syms x f = sin(x)^2; diff(f)```
```ans = 2*cos(x)*sin(x)```
#### Partial Derivatives
For multivariable expressions, you can specify the differentiation variable. If you do not specify any variable, MATLAB® chooses a default variable by its proximity to the letter `x`:
```syms x y f = sin(x)^2 + cos(y)^2; diff(f)```
```ans = 2*cos(x)*sin(x)```
For the complete set of rules MATLAB applies for choosing a default variable, see Find a Default Symbolic Variable.
To differentiate the symbolic expression `f` with respect to a variable `y`, enter:
```syms x y f = sin(x)^2 + cos(y)^2; diff(f, y)```
```ans = -2*cos(y)*sin(y)```
#### Second Partial and Mixed Derivatives
To take a second derivative of the symbolic expression `f` with respect to a variable `y`, enter:
```syms x y f = sin(x)^2 + cos(y)^2; diff(f, y, 2)```
```ans = 2*sin(y)^2 - 2*cos(y)^2```
You get the same result by taking derivative twice: ```diff(diff(f, y))```. To take mixed derivatives, use two differentiation commands. For example:
```syms x y f = sin(x)^2 + cos(y)^2; diff(diff(f, y), x)```
```ans = 0```
### Integrate Symbolic Expressions
You can perform symbolic integration including:
• Indefinite and definite integration
• Integration of multivariable expressions
For in-depth information on the `int` command including integration with real and complex parameters, see Integration.
#### Indefinite Integrals of One-Variable Expressions
Suppose you want to integrate a symbolic expression. The first step is to create the symbolic expression:
```syms x f = sin(x)^2;```
To find the indefinite integral, enter
`int(f)`
```ans = x/2 - sin(2*x)/4```
#### Indefinite Integrals of Multivariable Expressions
If the expression depends on multiple symbolic variables, you can designate a variable of integration. If you do not specify any variable, MATLAB chooses a default variable by the proximity to the letter `x`:
```syms x y n f = x^n + y^n; int(f)```
```ans = x*y^n + (x*x^n)/(n + 1)```
For the complete set of rules MATLAB applies for choosing a default variable, see Find a Default Symbolic Variable.
You also can integrate the expression `f = x^n + y^n` with respect to `y`
```syms x y n f = x^n + y^n; int(f, y)```
```ans = x^n*y + (y*y^n)/(n + 1)```
If the integration variable is `n`, enter
```syms x y n f = x^n + y^n; int(f, n)```
```ans = x^n/log(x) + y^n/log(y)```
#### Definite Integrals
To find a definite integral, pass the limits of integration as the final two arguments of the `int` function:
```syms x y n f = x^n + y^n; int(f, 1, 10)```
```ans = piecewise(n == -1, log(10) + 9/y, n ~= -1,... (10*10^n - 1)/(n + 1) + 9*y^n)```
#### If MATLAB Cannot Find a Closed Form of an Integral
If the `int` function cannot compute an integral, it returns an unresolved integral:
```syms x int(sin(sinh(x)))```
```ans = int(sin(sinh(x)), x)```
### Solve Equations
You can solve different types of symbolic equations including:
• Algebraic equations with one symbolic variable
• Algebraic equations with several symbolic variables
• Systems of algebraic equations
For in-depth information on solving symbolic equations including differential equations, see Equation Solving.
#### Solve Algebraic Equations with One Symbolic Variable
Use the double equal sign (==) to define an equation. Then you can `solve` the equation by calling the solve function. For example, solve this equation:
```syms x solve(x^3 - 6*x^2 == 6 - 11*x)```
```ans = 1 2 3```
If you do not specify the right side of the equation, `solve` assumes that it is zero:
```syms x solve(x^3 - 6*x^2 + 11*x - 6)```
```ans = 1 2 3```
#### Solve Algebraic Equations with Several Symbolic Variables
If an equation contains several symbolic variables, you can specify a variable for which this equation should be solved. For example, solve this multivariable equation with respect to `y`:
```syms x y solve(6*x^2 - 6*x^2*y + x*y^2 - x*y + y^3 - y^2 == 0, y)```
```ans = 1 2*x -3*x```
If you do not specify any variable, you get the solution of an equation for the alphabetically closest to `x` variable. For the complete set of rules MATLAB applies for choosing a default variable see Find a Default Symbolic Variable.
#### Solve Systems of Algebraic Equations
You also can solve systems of equations. For example:
```syms x y z [x, y, z] = solve(z == 4*x, x == y, z == x^2 + y^2)```
```x = 0 2 y = 0 2 z = 0 8```
### Simplify Symbolic Expressions
Symbolic Math Toolbox provides a set of simplification functions allowing you to manipulate the output of a symbolic expression. For example, the following polynomial of the golden ratio `phi`
```phi = (1 + sqrt(sym(5)))/2; f = phi^2 - phi - 1```
returns
```f = (5^(1/2)/2 + 1/2)^2 - 5^(1/2)/2 - 3/2```
You can simplify this answer by entering
`simplify(f)`
and get a very short answer:
```ans = 0```
Symbolic simplification is not always so straightforward. There is no universal simplification function, because the meaning of a simplest representation of a symbolic expression cannot be defined clearly. Different problems require different forms of the same mathematical expression. Knowing what form is more effective for solving your particular problem, you can choose the appropriate simplification function.
For example, to show the order of a polynomial or symbolically differentiate or integrate a polynomial, use the standard polynomial form with all the parentheses multiplied out and all the similar terms summed up. To rewrite a polynomial in the standard form, use the `expand` function:
```syms x f = (x ^2- 1)*(x^4 + x^3 + x^2 + x + 1)*(x^4 - x^3 + x^2 - x + 1); expand(f)```
```ans = x^10 - 1```
The `factor` simplification function shows the polynomial roots. If a polynomial cannot be factored over the rational numbers, the output of the `factor` function is the standard polynomial form. For example, to factor the third-order polynomial, enter:
```syms x g = x^3 + 6*x^2 + 11*x + 6; factor(g)```
```ans = [ x + 3, x + 2, x + 1]```
The nested (Horner) representation of a polynomial is the most efficient for numerical evaluations:
```syms x h = x^5 + x^4 + x^3 + x^2 + x; horner(h)```
```ans = x*(x*(x*(x*(x + 1) + 1) + 1) + 1)```
For a list of Symbolic Math Toolbox simplification functions, see Choose Function to Rearrange Expression.
### Substitutions in Symbolic Expressions
#### Substitute Symbolic Variables with Numbers
You can substitute a symbolic variable with a numeric value by using the `subs` function. For example, evaluate the symbolic expression `f` at the point `x` = 1/3:
```syms x f = 2*x^2 - 3*x + 1; subs(f, 1/3)```
```ans = 2/9```
The `subs` function does not change the original expression `f`:
`f`
```f = 2*x^2 - 3*x + 1```
#### Substitute in Multivariate Expressions
When your expression contains more than one variable, you can specify the variable for which you want to make the substitution. For example, to substitute the value `x` = 3 in the symbolic expression
```syms x y f = x^2*y + 5*x*sqrt(y);```
enter the command
`subs(f, x, 3)`
```ans = 9*y + 15*y^(1/2)```
#### Substitute One Symbolic Variable for Another
You also can substitute one symbolic variable for another symbolic variable. For example to replace the variable `y` with the variable `x`, enter
`subs(f, y, x)`
```ans = x^3 + 5*x^(3/2)```
#### Substitute a Matrix into a Polynomial
You can also substitute a matrix into a symbolic polynomial with numeric coefficients. There are two ways to substitute a matrix into a polynomial: element by element and according to matrix multiplication rules.
Element-by-Element Substitution. To substitute a matrix at each element, use the `subs` command:
```syms x f = x^3 - 15*x^2 - 24*x + 350; A = [1 2 3; 4 5 6]; subs(f,A)```
```ans = [ 312, 250, 170] [ 78, -20, -118]```
You can do element-by-element substitution for rectangular or square matrices.
Substitution in a Matrix Sense. If you want to substitute a matrix into a polynomial using standard matrix multiplication rules, a matrix must be square. For example, you can substitute the magic square `A` into a polynomial `f`:
1. Create the polynomial:
```syms x f = x^3 - 15*x^2 - 24*x + 350;```
2. Create the magic square matrix:
`A = magic(3)`
```A = 8 1 6 3 5 7 4 9 2```
3. Get a row vector containing the numeric coefficients of the polynomial `f`:
`b = sym2poly(f)`
```b = 1 -15 -24 350```
4. Substitute the magic square matrix `A` into the polynomial `f`. Matrix `A` replaces all occurrences of `x` in the polynomial. The constant times the identity matrix `eye(3)` replaces the constant term of `f`:
`A^3 - 15*A^2 - 24*A + 350*eye(3)`
```ans = -10 0 0 0 -10 0 0 0 -10```
The `polyvalm` command provides an easy way to obtain the same result:
`polyvalm(b,A)`
```ans = -10 0 0 0 -10 0 0 0 -10```
#### Substitute the Elements of a Symbolic Matrix
To substitute a set of elements in a symbolic matrix, also use the `subs` command. Suppose you want to replace some of the elements of a symbolic circulant matrix A
```syms a b c A = [a b c; c a b; b c a]```
```A = [ a, b, c] [ c, a, b] [ b, c, a]```
To replace the (2, 1) element of `A` with `beta` and the variable `b` throughout the matrix with variable `alpha`, enter
```alpha = sym('alpha'); beta = sym('beta'); A(2,1) = beta; A = subs(A,b,alpha)```
The result is the matrix:
```A = [ a, alpha, c] [ beta, a, alpha] [ alpha, c, a]```
### Plot Symbolic Functions
Symbolic Math Toolbox provides the plotting functions:
#### Explicit Function Plot
Create a 2-D line plot by using `fplot`. Plot the expression ${x}^{3}-6{x}^{2}+11x-6$.
```syms x f = x^3 - 6*x^2 + 11*x - 6; fplot(f)```
Add labels for the x- and y-axes. Generate the title by using `texlabel(f)`. Show the grid by using `grid on`. For details, see Add Title and Axis Labels to Chart.
```xlabel('x') ylabel('y') title(texlabel(f)) grid on```
#### Implicit Function Plot
Plot equations and implicit functions using `fimplicit`.
Plot the equation $\left({x}^{2}+{y}^{2}{\right)}^{4}=\left({x}^{2}-{y}^{2}{\right)}^{2}$ over $-1.
```syms x y eqn = (x^2 + y^2)^4 == (x^2 - y^2)^2; fimplicit(eqn, [-1 1])```
#### 3-D Plot
Plot 3-D parametric lines by using `fplot3`.
Plot the parametric line
`$\begin{array}{c}x={t}^{2}\mathrm{sin}\left(10t\right)\\ y={t}^{2}\mathrm{cos}\left(10t\right)\\ z=t.\end{array}$`
```syms t fplot3(t^2*sin(10*t), t^2*cos(10*t), t)```
#### Create Surface Plot
Create a 3-D surface by using `fsurf`.
Plot the paraboloid $z={x}^{2}+{y}^{2}$.
```syms x y fsurf(x^2 + y^2)```
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## Help with polynomial zeros
Started by 3 years ago27 replieslatest reply 3 years ago223 views
I'm working on a problem where I need a *very* long FIR with many stop bands. I've done several similar designs recently and remez() with the right prodding will give the desired response. This one is just too big and after hours of crunching it wont converge.
So OK, back to basic principles. I can put a pair of zeros on the unit circle by computing their location on the complex plane with RootsOfZeros = cos(f*w) +/- i*sin(f*w); where w = omega. Taking the output of a small FIR from remez() I can go back and forth from roots to the impulse response with roots() and poly() and the locations are fixed. There is no problem from rounding errors. I can look at the locations with zplane();
So I can place multiple zero pairs on the unit circle by multiplying the small polynomials together and check the impulse response with zplane() and freqz(). This works until about the 4th one. Then I start getting wildly different answers. The zeros have moved around on the complex plane. That means the roots of the output polynomial are different than the roots of the inputs. Sure enough roots() gives me different roots than what I started with. Not just scaling, they are in different places.
At a loss here. Obviously there is some basic bit of math that I am missing.
The goal is to synthesize a polynomial in Z (FIR impulse response) with zero pairs I pick on the plane.
Below is a bit of Matlab code I was tinkering with.
If I do 4 zeros then zooming in on the z plane shows the zeros nicely placed where they should go. Add the fifth zero pair and they split. Add the sixth and all the zeros are arranged in a rosette about +1 on the complex plane. Any ideas?
Fb = 4.2e6;
Fs = 4.321e9;
Fs_dec = Fs/125;
Fst_1a = Fs_dec - Fb; % Lower Edge of 1st pass band image
Fst_1b = Fs_dec + Fb; % Upper Edge of 1st pass band image
Fst_2a = 2*Fs_dec - Fb; % Lower Edge of 2nd pass band image
Fst_2b = 2*Fs_dec + Fb; % Upper Edge of 2nd pass band image
frb(1) = Fst_1a;
frb(2) = Fs_dec;
frb(3) = Fst_1b;
frb(4) = Fst_2a;
frb(5) = 2*Fs_dec;
frb(6) = Fst_2b;
b = [1 1]; % Zero at Z = -1
% for n = 1:6
% for n = 1:4
for n = 3:6
rbp = cos(frb(n)*2*pi/Fs) + i*sin(frb(n)*2*pi/Fs);
rbm = cos(frb(n)*2*pi/Fs) - i*sin(frb(n)*2*pi/Fs);
pol = poly([rbp rbm]);
b = conv(b, pol);
b = b/sum(b);
end
figure(1);freqz(b/sum(b),1,2^21,Fs);
figure(2);zplane(b,1);grid;
[ - ]
I'm enjoying this because it's making me think of something I haven't considered before.
As a polynomial's order increases, the locations of the roots become ever more sensitive to variations in the parameters. This fact -- or one of its many corollaries -- is the reason that when you're implementing IIR filters you always cascade 1st- or 2nd-order sections, and why you don't let the sampling rate get too high.
The light bulb that just went off for me is that this works for FIR filter zeros, too! D'oh! So if you're looking for zeros in the response that are as close as possible to zero, you can't let the polynomial order get too high.
So, it may be that if the number and depth of the zeros are important, you may need to recognize that you're facing a fundamental limitation in your problem, and you may need to cascade a number of FIR filters. Depending on just how sensitive your system is to signal content in those notches, you may have to pay very close attention to numerical precision while you're at it.
[ - ]
A long time ago I had to tune a fiddly multistage radio filter (HW).
I was very successful with a single stage filter, and even with two-stage filters. However with more stages I ended up with always correcting the former stages over and over again - and finally just dropped my goal :)
Later I realized that my problems where quite common - it IS difficult.
And even if I do a good job, the result depends on so many details (like temperature effects).
But my analogue experiences helped me to understand that not only the concept of filtering is transferred to the digital realm - but the sensitivity and trickiness follow.
Always.
Bernhard
[ - ]
Hello Napierm,
Here is the problem: the factored form of the polynomial P(x) = product(x-root(k)) k=1,2,...., N and the expanded form p(x) = sum a(n)x^(N-n), n=0, 1, 2, .... N, with a(0)=1, a(1)=sum of the roots, a(2)=sum of roots taken 2 at a time,... and a(N)=product of roots, can not be the same in a finite state machine.
First problem is that roots on the unit circle are transcendental numbers and require infinite length words. If the roots can be represented by finite length words each successive product in forming the filter coefficients get longer and longer and for sufficiently high order polynomials you are guaranteed to run out of bits in your coefficient word length. When the coefficients are replaced by their finite length approximations the roots move and the roots you finally get are not the roots with which you started.
There are ways of approximating your spectrum with well behaved filter coefficients that require some thinking outside the box.
Why not pass on our filter specs and let me have a look see at one of my approaches.
fred harris
[email protected]
[ - ]
Remez, firls ...etc are based on iterations to converge and so I expect they have limit on number of taps to handle.
Using fir2(frequency grid method, internally does ifft) I just used 1000 taps here:
h = fir2(1000,[0 .1 .15 .2 .23 .25 .26 .3 .33 .35 .4 1],[1 1 0 0 1 1 0 0 1 1 0 0]);
zplane(h,1)
hquant = round(2^18 * h);
you can go higher than 1000... at the edges the coeffs get really tiny and need wider bitwidth if using fixed point.
[ - ]
I'm not very familiar with Matlab. In some coding I've been doing I found "long double" is a 128 bit float. Can you tell Matlab to use more bits? Your description sounds like Nth roots of 1 which would be exp(2*Pi*j/N) for j integer. If the numbers are getting very close to each other, it could be a round off issue.
Instead of unit circle, scale the roots to be around 1/2^21. Maybe that will keep the numbers in a range the algorithm can deal with. I have no idea if that will do anything useful!
Mike
[ - ]
Yes, I'm considering round off. Just doesn't seem like that is the problem and Matlab doesn't natively support any float bigger than a double. This is not trivial to work around.
Just to try something I'm looking at the HPF (High Precision Floating point) library. It does support arbitrarily high precision numbers but is pretty basic. No complex types for instance. Fortunately there are only a few forms for the complex conjugate pairs in the pass band and stop band that all reduce to simple real coefficients. Then the polynomial multiplication is done by convolution.
The examples I've worked by hand agree with both poly() and conv(). So I can write a routine to do this. Just tedious to see if round off is the problem or not.
[ - ]
I've worked a few more examples with HPF and seems like my intuition is correct. The problem is not lack of precision. Increasing to a crazy number of decimal digits (more than 1000) doesn't alter the outcome at all. Not even a marginal difference. There is something going on here I'll have to dig in to understand. It goes counter to my experience. Normally you can take two responses and use convolution combine them.
BTW the example in the Matlab snippet is a simpler one that remez() already successfully solves. It has *no* trouble concentrating zeros much more tightly packed and in the same locations on the unit circle as in this Matlab code.
Have to admit I can only infer that from the frequency response looking at the attenuation pits. zplane() shows garbage - it can't factor the output correctly.
The time hasn't been a total waste. I've worked out cases of how to place zeros and how they convert to simple polynomials. Turns out you can take a standard window function and alter it with convolution to tweek it a bit. I can get to a solution this way but is much lower performance than remez(). I have a couple of outputs that would work but are "suboptimal"; means takes more coefficients than I want to store in memory.
Side note: one issue is that the polynomial solution to the zero pair placements results in only cos(angle) being used. In all cases i*sin(angle) terms cancel out. That means that the angle resolution near DC gets very sensitive.
My next notion going down this rabbit hole is to dig into the papers on the P-M algorithm of how they move the extremal frequencies around to approximate a response.
[ - ]
Have you tried FIR design methods like DRHS-OLL?
[ - ]
The results do look interesting. I don't see that the code base has been made available for others to try.
[ - ]
That's great! It's a major pain of course because you're not getting the simple answer quickly, but understanding how all the programs actually work and why they go bonkers will pay off for a long time. Matlab drives me nuts, so I don't use it unless I have no choice. But it is a good tool. Good luck going down the rabbit hole - at least you will have fun along the way!
[ - ]
Years ago, I ran into a problem because Matlab's root() algorithm was not able to give proper solutions just because of the internal math precision.
Meanwhile I abandoned Matlab because of that, but at that time (some 15 years ago), Matlab insiders explained me that the math precision mainly depends on the machine on which it's running: on a 64bit machine it's better than on a 32bit machine. Because it relies on the underlying hardware - (a sacrifice to speed on customers' demand).
My guess is that the problem here is still due to insufficient precision within some basic math functions like root(). Therefore all dependent algorithms in Matlab which rely on those basic algorithms (like check for stability if roots/zeros are very close to the unit circle) may suffer.
Switching to ScopeFIR instead of Matlab got me much more reliable results.
I don't know if you can/want to give it a try.
Otherwise it might help to write your own algorithm which is able to
work with arbitrary precision (like Fortran, Numpy, ...) and then try with
increasing precision. Important is that the whole algorithm is working with this increased precision. Matlab is definitely not the right tool for this.
(Except if it has a toolbox which supports this, meanwhile ...)
Bernhard
[ - ]
ScopeFIR looks interesting. But right off is says the limitation for number of taps is 16K. Not big enough. Another is that I almost always wind up with a script with a bunch of constants at the top, knobs I call them. I turn the knobs to tweak the response. Or where some parameter is really sensitive I have a loop that can grind away looking for a minimum. Or do a gradient descent. That is hard with any canned tool that doesn't allow me to script it and get "under the hood".
Here's one example from a past job. There are two cascaded CIC's that are modified for better attenuation of a problem frequency. The output passband shape is not flat or a standard CIC droop. I can compensate with a modified half-band at the last decimation stage. It's easy to do with remez(). Make many closely spaced but not touching passbands in your actual passband. Set each one to the gain needed to compensate. Note the beginning and ending gains of each sub-band can be different so I just set them to what I want at each frequency. This does dork about with the stop-band attenuation shape due to half-band symmetry but if it attenuates enough then who cares.
[ - ]
If you like ScopeFIR, you might want to contact the developer.
Few years ago he was very communicative - so why not ask him for a specialized version with higher limits...
[ - ]
Hi,
your 'for' loop gets much shorter with these two lines
ff1 = (exp(j*2*pi*[frb -frb]/Fs));
fff = poly(ff1);
But that does not bring you out of the numeric troubles with filter design. The abs(roots(fff)) range from 0.98 to 1.03. You will also get in trouble with the precision and accuracy for the resulting FIR.
The zeros are too close together. You bring them apart by reducing the decimation :
Fs_dec = Fs/25;
Cheers
Detlef
[ - ]
Yes, that does make all 6 zero pairs plot correctly. Doesn't solve the problem that I need a metric crap ton of them close together on the circle. But does point to maybe round off error as the problem. So will keep poking at this HPF code.
Thanks,
Mark
[ - ]
Trick for ff1 without loop solution is nice.
Yes, the roots are too close, stopband bandwidth (2xFb/Fs ~ 0.002) is also too low compared to sampling rate.
[ - ]
Hello
try to use the function poly:
>> help poly
poly Convert roots to polynomial.
poly(A), when A is an N by N matrix, is a row vector with
N+1 elements which are the coefficients of the
characteristic polynomial, det(lambda*eye(size(A)) - A).
poly(V), when V is a vector, is a vector whose elements are
the coefficients of the polynomial whose roots are the
elements of V. For vectors, ROOTS and poly are inverse
functions of each other, up to ordering, scaling, and
roundoff error.
regards
[ - ]
Respectfully, I've tried poly and conv. They yield identical results.
[ - ]
Hi,
even after b/sum(b), coefficients of b remain high;
1.0e+15 *
Columns 1 through 8
0.0183 -0.2008 0.9831 -2.7972 4.9864 -5.3791 2.3893 2.3893
Columns 9 through 14
-5.3791 4.9864 -2.7972 0.9831 -0.2008 0.0183
[ - ]
Nutty, yes?
[ - ]
Actually we should normalize 'b' by b[1] in the loop (and after the loop, not by sum(b)), then we get the same result suggested without for loop;
without for (added zero for Fs/2)
ff1 = (exp(j*2*pi*[frb -frb Fs/2]/Fs));
fff = real(poly(ff1))
we get ;
Columns 1 through 8
1.0000 -10.9620 53.6584 -152.6732 272.1596 -293.5937 130.4109 130.4109
Columns 9 through 14
-293.5937 272.1596 -152.6732 53.6584 -10.9620 1.0000
for the other case;
If we normalize 'b' by b[1] in the loop and also after the loop (instead of sum(b));
b =
Columns 1 through 8
1.0000 -10.9620 53.6584 -152.6732 272.1596 -293.5937 130.4109 130.4109
Columns 9 through 14
-293.5937 272.1596 -152.6732 53.6584 -10.9620 1.0000
[ - ]
Napierm-
Possibly foolish answer here, but since you need stopbands, what about 2 filters in series ? Maybe that does something to delay for particular stopbands that is not good for your application, but at least you might get Matlab's Remez algorithm to converge.
-Jeff
[ - ]
Not foolish at all. Unfortunately the form of this particular problem hits a large prime number so the exact solution needs the long FIR. Yes, there are other ways to skin the cat but would require *much* more hardware. I can't really give more information.
[ - ]
Napierm-
Ok ... sounds like a maximum length sequence problem or encryption related. Can you find where in the MATLAB source it fails to converge, and start instrumenting that ?
-Jeff
[ - ]
I don't have the source to Matlab's remez() or firpm() code. Regardless it is still likely to cough up on the ~20K point FIR. Foolish of me to try.
Thanks,
Mark
[ - ]
Mark-
Huh ... I thought MATLAB had open sourced everything. I guess they still subscribe to the Steve Ballmer school of thought about Linux being communism -- they are from the same era, haha.
-Jeff
[ - ]
Scilab provides an open source flow.
With respect to roots() calculation it has proved to be able to cope with Matlab long ago -- brought even better results with more precision.
However, I didn't check for years, but could be a good starting point for something like using the whole bunch as it is, but improving critical points - and maybe afterwards put it into the public realm ;-)
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# A certain jar contains only b black, w white and r red
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A certain jar contains only b black, w white and r red [#permalink]
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19 Jun 2008, 21:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A certain jar contains only b black, w white and r red marbles. If one marble is chosen at random from the jar , is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white.
1 r/ ( b+w) > w/(b+r)
2 b-w > r
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Re: marbles in a jar probability from Gmat prep [#permalink]
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19 Jun 2008, 22:28
2
KUDOS
Ans: A
Target: r/(r+w+b) > w/(r+w+b) if and only if r>w
St1: r/ w+b > w/ b+r -----> w+b /r < b+r / w ---> add 1 on both sides we get
w+b+r / r < b+r+w / w ---> P(r)> P(w) -----suffic
St2: b-w>r ----- Insuffi
Re: marbles in a jar probability from Gmat prep [#permalink] 19 Jun 2008, 22:28
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# How to Calculate Alpha in Excel?
Are you an Excel user who needs to calculate alpha values for your data sets? Calculating alpha in Excel is not as difficult as it may seem. In this article, we’ll go over the steps to do so, making it easy for you to calculate alpha in Excel in no time.
## What is Alpha in Excel?
Alpha in Excel is a measure of how much of a given security’s returns are due to movements in the overall market. It is calculated by subtracting the risk-free rate of return from the security’s expected return, and dividing that by the security’s beta. Beta is a measure of volatility, and represents the amount of risk associated with the security. Alpha is a measure of the security’s excess return, and is one of the most important indicators in finance.
Alpha is used to determine the expected return of a security when compared to the expected return of the overall market. If the security has a higher alpha, it means that it is expected to outperform the overall market. Conversely, if the security has a negative alpha, it means that it is expected to underperform the overall market.
Alpha is an important factor in portfolio management and asset allocation, as it helps investors make informed decisions about which securities to include in their portfolios. It is also a key component of risk management, as it helps investors assess the risk-return trade-off of their investments.
## How to Calculate Alpha in Excel?
Calculating alpha in Excel is relatively straightforward, and can be done using either the SLOPE or LINEST functions. Both of these functions are used to estimate the slope of a linear regression line, which is used to calculate alpha. The first step is to input the data into Excel. This includes the expected returns of the security and the expected returns of the overall market.
Once the data has been inputted, the SLOPE or LINEST functions can be used to calculate alpha. The SLOPE function is the simpler of the two, and takes the form of SLOPE(data_y, data_x), where data_y represents the expected return of the security, and data_x represents the expected return of the overall market. The LINEST function is slightly more complex, and takes the form of LINEST(data_y, data_x, const, stats), where const is set to TRUE to indicate that the intercept is estimated, and stats is set to TRUE to indicate that additional regression statistics should be returned.
### Interpreting the Results of Alpha Calculations
Once the alpha calculation is complete, the results can be interpreted. If the result of the calculation is positive, it indicates that the security is expected to outperform the overall market. Conversely, if the result of the calculation is negative, it indicates that the security is expected to underperform the overall market.
It is important to remember, however, that alpha is only an estimate. It is based on the data that has been inputted, and may not be accurate if the data is not representative of the underlying security or the overall market. As such, it is important to ensure that the data used for the calculation is up to date and accurate.
### Using Alpha in Portfolio Management
Alpha can be used to help investors make informed decisions about which securities to include in their portfolios. It is important to remember, however, that alpha is only one factor to consider when making these decisions. Other factors, such as risk tolerance and investment horizon, should also be taken into account.
By combining alpha with other factors, investors can create a portfolio that is tailored to their individual needs and objectives. This can help ensure that their investments are properly diversified and that their risk-return trade-off is optimal.
## Conclusion
Alpha is an important measure of risk-adjusted return, and can be used to help investors make informed decisions about which securities to include in their portfolios. It can be calculated in Excel using either the SLOPE or LINEST functions, and the results can be interpreted to determine if the security is expected to outperform or underperform the overall market. Alpha should always be used in conjunction with other factors, such as risk tolerance and investment horizon, to ensure that the portfolio is properly diversified and that the risk-return trade-off is optimal.
### What is Alpha in Excel?
Alpha is a measure of the performance of an investment relative to a market index. It measures the excess return of an individual security or portfolio compared to the return of a benchmark index. It is also referred to as the “excess return” or the “alpha return”.
### What is the Formula to Calculate Alpha in Excel?
The formula to calculate alpha in Excel is: Alpha = (Excess Return – Risk-Free Return) / Portfolio Beta. Excess Return is the return of the investment portfolio over and above the risk-free return. Risk-Free Return is the expected return of a risk-free security such as a Treasury bill. Portfolio Beta is the measure of the volatility of a portfolio compared to the market.
### How Do You Calculate Alpha in Excel?
To calculate alpha in Excel, you will need to input the excess return, risk-free return, and portfolio beta into the alpha formula. First, input the excess return for the portfolio in a cell. Then, input the risk-free return in another cell. Finally, input the portfolio beta into a third cell. Once these values have been inputted, the formula can be entered into another cell and the alpha will be calculated.
### Are There Other Ways to Calculate Alpha?
Yes, there are other ways to calculate alpha. For example, you can calculate alpha using the Sharpe Ratio, which is a measure of risk-adjusted performance. The Sharpe Ratio is calculated by subtracting the risk-free return from the portfolio return and then dividing the result by the portfolio standard deviation.
### Are There Any Risks Associated With Calculating Alpha?
Yes, there are risks associated with calculating alpha. Alpha is a measure of excess return, which means that it does not account for the risk associated with the investment. Therefore, it is important to understand the risks of an investment before calculating the alpha.
### What Is the Significance of Alpha in Excel?
Alpha is a measure of the performance of an investment relative to a market index. It is an important measure of the performance of an investment and can be used to compare the performance of different investments. Alpha can also be used to assess the risk-adjusted performance of an investment and determine if it is worth investing in.
### How To… Calculate Cronbach’s Alpha in Excel
The calculation of Alpha in Excel is an important task for financial analysts, and with the help of this article, you now have the necessary tools and information to accurately calculate Alpha in Excel. You are now one step closer to becoming an Excel expert and can confidently tackle complicated financial calculations. With a little practice, you will be able to calculate Alpha quickly and efficiently.
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A195149 Concentric 22-gonal numbers. 15
0, 1, 22, 45, 88, 133, 198, 265, 352, 441, 550, 661, 792, 925, 1078, 1233, 1408, 1585, 1782, 1981, 2200, 2421, 2662, 2905, 3168, 3433, 3718, 4005, 4312, 4621, 4950, 5281, 5632, 5985, 6358, 6733, 7128, 7525, 7942, 8361, 8800, 9241, 9702, 10165, 10648, 11133 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Sequence found by reading the line from 0, in the direction 0, 22,..., and the same line from 1, in the direction 1, 45,..., in the square spiral whose vertices are the generalized tridecagonal numbers A195313. Main axis, perpendicular to A152740 in the same spiral. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..10000 Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1). FORMULA G.f.: -x*(1+20*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011 a(n) = (22*n^2+9*(-1)^n-9)/4; a(n) = -a(n-1)+11*n^2-11*n+1. - Vincenzo Librandi, Sep 27 2011 MAPLE A195149:=n->(22*n^2+9*(-1)^n-9)/4: seq(A195149(n), n=0..50); # Wesley Ivan Hurt, Jul 07 2014 MATHEMATICA Table[(22*n^2 + 9*(-1)^n - 9)/4, {n, 0, 50}] (* Wesley Ivan Hurt, Jul 07 2014 *) PROG (MAGMA) [(22*n^2+9*(-1)^n-9)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011 (PARI) a(n)=(22*n^2+9*(-1)^n-9)/4 \\ Charles R Greathouse IV, Sep 24 2015 CROSSREFS Cf. A195323 and A195318 interleaved. Cf. A032528, A077221, A195142, A195143, A195145, A195146, A195147, A195148. Cf. A032527, A195049, A195058. Column 22 of A195040. - Omar E. Pol, Sep 29 2011 Sequence in context: A041956 A041954 A041952 * A291557 A165309 A041960 Adjacent sequences: A195146 A195147 A195148 * A195150 A195151 A195152 KEYWORD nonn,easy AUTHOR Omar E. Pol, Sep 17 2011 STATUS approved
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# Lesson 15Adding and Subtracting with Scientific Notation
### Learning Targets:
• I can add and subtract numbers given in scientific notation.
## 15.1Number Talk: Non-zero Digits
Mentally decide how many non-zero digits each number will have.
## 15.2Measuring the Planets
Diego, Kiran, and Clare were wondering:
“If Neptune and Saturn were side by side, would they be wider than Jupiter?”
1. They try to add the diameters, km and km. Here are the ways they approached the problem. Do you agree with any of them? Explain your reasoning.
1. Diego says, “When we add the distances, we will get . The exponent will be 9. So the two planets are km side by side.”
2. Kiran wrote as 47,000 and as 120,000 and added them:
3. Clare says, “I think you can’t add unless they are the same power of 10.” She adds km and to get .
2. Jupiter has a diameter of . Which is wider, Neptune and Saturn put side by side, or Jupiter?
## 15.3A Celestial Dance
object diameter (km) distance from the Sun (km)
Sun
Mercury
Venus
Earth
Mars
Jupiter
1. When you add the distances of Mercury, Venus, Earth, and Mars from the Sun, would you reach as far as Jupiter?
1. Add all the diameters of all the planets except the Sun. Which is wider, all of these objects side by side, or the Sun? Draw a picture that is close to scale.
### Are you ready for more?
The emcee at a carnival is ready to give away a cash prize! The winning contestant could win anywhere from $1 to$100. The emcee only has 7 envelopes and she wants to make sure she distributes the 100 $1 bills among the 7 envelopes so that no matter what the contestant wins, she can pay the winner with the envelopes without redistributing the bills. For example, it’s possible to divide 6$1 bills among 3 envelopes to get any amount from $1 to$6 by putting $1 in the first envelope,$2 in the second envelope, and $3 in the third envelope (Go ahead and check. Can you make$4? $5?$6?).
How should the emcee divide up the 100 $1 bills among the 7 envelopes so that she can give away any amount of money, from$1 to $100, just by handing out the right envelopes? ## 15.4Old McDonald's Massive Farm Use the table to answer questions about different life forms on the planet. creature number mass of one individual (kg) humans cows sheep chickens ants blue whales antarctic krill zooplankton bacteria 1. On a farm there was a cow. And on the farm there were 2 sheep. There were also 3 chickens. What is the total mass of the 1 cow, the 2 sheep, the 3 chickens, and the 1 farmer on the farm? 2. Make a conjecture about how many ants might be on the farm. If you added all these ants into the previous question, how would that affect your answer for the total mass of all the animals? 3. What is the total mass of a human, a blue whale, and 6 ants all together? 4. Which is greater, the number of bacteria, or the number of all the other animals in the table put together? ## Lesson 15 Summary When we add decimal numbers, we need to pay close attention to place value. For example, when we calculate , we need to make sure to add hundredths to hundredths (5 and 0), tenths to tenths (2 and 7), ones to ones (3 and 6), and tens to tens (1 and 0). The result is 19.95. We need to take the same care when we add or subtract numbers in scientific notation. For example, suppose we want to find how much further the Earth is from the Sun than Mercury. The Earth is about km from the Sun, while Mercury is about km. In order to find we can rewrite this as Now that both numbers are written in terms of , we can subtract 0.58 from 1.5 to find Rewriting this in scientific notation, the Earth is km further from the Sun than Mercury. ## Lesson 15 Practice Problems 1. Evaluate each expression, giving the answer in scientific notation: 1. Write a scenario that describes what is happening in the graph. 2. What is happening at 5 minutes? 3. What does the slope of the line between 6 and 8 minutes mean? 2. Apples cost$1 each. Oranges cost $2 each. You have$10 and want to buy 8 pieces of fruit. One graph shows combinations of apples and oranges that total to $10. The other graph shows combinations of apples and oranges that total to 8 pieces of fruit. 1. Name one combination of 8 fruits shown on the graph that whose cost does not total to$10.
2. Name one combination of fruits shown on the graph whose cost totals to $10 that are not 8 fruits all together. 3. How many apples and oranges would you need to have 8 fruits that cost$10 at the same time?
3. Solve each equation and check your solution.
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# 1 and 2 Step Problem Solving with Multiplication and Division
Multiplication and division word problems are not what they used to be! No longer can we teach the memorization of facts and be successful! Kids need to be taught to recognize the different types of problems enabling them to chose a strategy that works for them! My favorite method to use is the Cognitively Guided Instruction model.
Cognitively Guided Instruction is an inquiry-based approach to teaching mathematics that was developed at the Wisconsin Center for Education Research (Carpenter et al, 1999). It is based on the developmental stages of math reasoning. Sequencing word problems from easy to difficult helps kids build on what they know.
Multiplication is divided into two categories- grouping and comparing. Division is also divided into two categories- measurement and partitive.
To begin the unit start with 1 step multiplication. Give students sample problem to sort into two categories. Don’t tell them the categories ahead of time! Talk about why they sorted the problems the way they did. Then guide them to the two different categories- grouping and comparing.
Here is an example of the problems they sort:
Problem sorting for multiplication- grouping and comparing
After kids have learned about the similarities and differences of the two, begin to work some problems. Start with easy problems and move to more difficult problems.
I like to use problem solving mats in guided math to work through these problems. Here is the answer sheet for a grouping problem. The kids have plates and cookies (paper cookies – I’m not brave enough to give out really cookies as manipulatives!) to model the problem. Start with equal groups, then arrays, area models and final connect with the basic fact. Kids then work a few of the problems from the sort before we move to comparing.
The next type is a comparison problem. This one is difficult. The strip diagram is essential to solve comparisons with multiplication. Teach it to them now if you have not already! The problem solving mat looks like this:
As teachers we see this as multiplication. The word times gives it away, but for some reason kids can’t visualize this unless we teach them to draw a strip diagram. Repeat the same procedure as before – discussion, strategies, and working a few problems.
The next step is to work some problems individually so you can determine where understanding is breaking down. For this activity, each student has a problem solving mat. They read the problem and choose a box to fill in. Rotate the paper and they fill in a different box. At the end, kids have used several different strategies to solve the problem- the answer is not as important as the strategy!
Continue working on 1 step problems. Go through the same process for division. The kids begin by sorting problems. Most people don’t understand the two types of division – measurement and partitive. Measurement is when the number of groups is unknown and partitive is when the number of items in each group is unknown. Sounds trivial, right? BUT if you approach a measurement problem the same as a partitive problem it will be confusing! AND these kids are 8! Look at the following problem solving mats to see the difference.
For partitive, start with the number of plates, then kids can put cookies on the plate. This type is the easiest for kids.
Measurement is tricky – you don’t know how many groups you will need. Kids have 15 cookies in their hands and are unsure how to start grouping them. Start by putting 3 cookies in a group, then make another group until you are out of cookies. Talk and talk some more! Look at the other strategies – array, area model, and repeated subtraction! Show repeated subtraction last because it is the easiest and you want them to try to visualize equal groups first.
The next step is to work independently. Use the same strategy for independent practice as multiplication – pass a problem.
Finally you are ready for 2 step problems! Start with a problem solving mat and show them the how to see the two problems in this paragraph. Go slowly and find similarities and differences.
Work several problems, making small changes to what they already know!
By the end of the week, most kids are working 1 step and feeling better about 2 step problems. Add 1 and 2 step problem task cards to stations the next week and continue to pull small groups of kids who still need help!
These task cards have a QR Code, making them self checking!
You can find this product on my Teacher’s Pay Teachers store at:
1 and 2 step Multiplication and Division Unit
Included in the unit:
• 5 Complete Lesson Plans- high yield and sheltered instruction strategies included!
• Guided Math Problem solving mats
• Teacher answer keys for the mats
• Problem sorts with pockets for Interactive Notebooks (INB)
• Anchor Charts for INB’s
• Small Group Independent Practice
• Exit Tickets
• Manipulatives
• 4 Whole Class Activities
• 20 Task Cards- Color and Black and White with QR Codes for Self-Checking
• Bubble sheets
• Open responses
• Griddable sheets
• Teacher Answer Keys for all activities.
I hope this helps your students and helps make your life easier!
Talk to you soon,
Misty
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# How Does Probability Work in Twine?
+1 vote
retagged
Help!
I am still new to Twine and I am learning how the probability part works with the gaming platform.
Our goal is to link the users when they select the most of one option out of three total options.
So for example: Each question is labeled as A,B,C . When a user chooses more of one of those letters by the end of the game, we want to give them a specific ending.
How should we code that into the Harlowe 2 twine?
+1 vote
by (2.5k points)
What you are talking about is rather easy to do. You want each answer to be labeled A B or C if I understand correctly and if the player has more of one letter than the rest, the player will get that letter ending. If that is the case, all you need to do is create 3 variables labeled \$a \$b and \$c (or whatever you want to do.) When the player chooses an answer that is labeled A. The \$a variable will be given a point. Example:
Would you smooch a ghost?
(link:"Heck Yeah!")[(set: \$a to \$a + 1)(goto: "Question 2")]
(link:"Heck Yeah!")[(set: \$b to \$b + 1)(goto: "Question 2")]
(link:"Heck Yeah!")[(set: \$c to \$c + 1)(goto: "Question 2")]
At the end of the game the variable with the most points will determine the ending.
(If: \$a > \$b and \$c)[(goto: "Ending A")]
(else-if: \$b > \$a and \$c)[(goto: "Ending B")]
(else-if: \$c > \$a and \$b)[(goto: "Ending C")]
Keep in mind that making links like this does not automaticly make the new passage, you have to make it and label it manually.
by (210 points)
So if I was to continue to the next questions, will the additional +1 change (to something like +2? or it varies on however we want to do it?)
For example:
Q3: Kissing a ghost is kind of like?
(link:"happiness!")[(set: \$a to \$a + 2)(goto: "Question 4")]
(link:"drugs!")[(set: \$b to \$b + 2)(goto: "Question 4")]
(link:"no!")[(set: \$c to \$c + 2)(goto: "Question 4")]
by (158k points)
@Glitchy (and @learningiscaring)
There is a logic error in the conditional expressions of your (if:) and (else-if:) macros, the expression \$a > \$b and \$c doesn't equate the way you think it does. Because the expression contains the and operator it is first broken into two parts and then evaluated as follows:
(note: for the sake of this explanation I will assume that \$a equals 1, \$b equals 2, and \$c equals 3)
1. The \$a > \$b part equates to true if the current value of \$a is greater than the current value of \$b, otherwise it equates to false. In our example 1 (\$a) isn't greater than 2 (b\$) so this part would equate to false.
2. The \$c part equates to true if the current value is considered Truthy, in the case of numbers this is any value other than 0 (zero), otherwise it equates to false. In our example 3 (\$c) isn't equal to 0 (zero) so this part would equate to true.
3. The Boolean outcomes of each of the two parts are recombined to determine the overall outcome of the whole expression. The and operator equates to true if both it's left and right conditions equate to true, otherwase it equates to false. In our example the outcome of part 1 was false and the outcome of part 2 was true, so when the and operator is applied (eg. false and true) the overall outcome would be false.
The issue with each of your condition expressions is that the right side of each of the and operators will only equate to false if the referenced variable equals zero, and this isn't what you're trying to test for. Your conditional expressions should look as follows
``````(if: \$a > \$b and \$a > \$c)[(goto: "Ending A")]
(else-if: \$b > \$a and \$b > \$c)[(goto: "Ending B")]
(else-if: \$c > \$a and \$c > \$b)[(goto: "Ending C")]
``````
by (2.5k points)
That will add 2 more points to the point it already had, you can change it to whatever you want. If you want it to add 1 more point like the first one, do + 1
by (2.5k points)
Opps, Grayelf is right. I tend to forget what makes sense to me doesn't always make sense to a computer.
by (210 points)
Thank you Glitchy and Grayelf!
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https://forum.piedao.org/answers/17548507-anurak-has-nickels-dimes-and-pennies-in-his-pocket-the-coins-in-anurak-s-pocket-have-a-total-value-of
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# Anurak has nickels, dimes, and pennies in his pocket. The coins in Anurak's pocket have a total value of 75 cents. He has three more nickels than dimes and five times as many pennies as dimes. How many coins of each type does Anurak have?
Anurak has 15 pennies, 6 nickels and 3 dimes.
Step-by-step explanation:
We must have these informations in mind:
1) A penny is a 1-cent coin.
2) A dime is a 10-cent coin.
3) A nickel is a 5-cent coin.
Let be , , the quantities of pennies, nickels and dimes, respectively. From statement we get that to value in Anurak's pocket is represented by this mathematical expression:
(Eq. 1)
In addition, we get the following identities:
i)He has three more nickels than dimes
(Eq. 2)
ii)And five times as many pennies as dimes
(Eq. 3)
The system of linear equations is now reduced: (Eqs. 2, 3) in (Eq. 1)
The remaining variables are and .
Anurak has 15 pennies, 6 nickels and 3 dimes.
Anurak has 15 pennies, 6 nickels and 3 dimes.
Step-by-step explanation:
## Related Questions
HELP ASAPPPPP!!!!!!!
A.
Step-by-step explanation:
The double bar around any number means absolute value. Absolute value makes any number positive.
So,
9.3 = 9.3
|-2.1| = 2.1
-13.7 = -13.7
-4.2
Negatives are always the smallest:
-13.7, -4.2, |-2.1| or 2.1, 9.3
Therefore, A is the correct answer.
Subtract.
3−(−5)−(−2)
Plot the solution on the number line.
Answer: Its 10, so plot it on 10 in the number line
Step-by-step explanation: Since the subtraction signs cancel out, they make a addition sign. Henceforth its " 3+5+2" which would mean the sum is 10.
These triangles are similar. What is the value for it?
The vault of h is 10
Step-by-step explanation:
Since the whole triangle is 27 and the one side is 9 we know the other side must be 18. You multiply by 2 to get get from 9. That means they are similar and you just multiply 5 by 2 to get h=10.
An inscribed angle is an angle formed by two radii that share an endpoint A) True
B)False
check the picture below.
notice, as from the left, we grab those two angles and slap them together on the right, they form a wide angle A, and a smaller inscribed angle B.
now, the blue and red segments that came together, are both making up the inscribed angle B, and also sharing the side or endpoint, in the middle of angle A.
Given this table of values, what is the value of f(14)?
f(14) = 14.6
Step-by-step explanation:
Form the table attached,
Values of 'x' represent the points on the x-axis of the graph of the given function 'f'.
Similarly, values of 'y' represent the points on the y-axis of the graph.
For every input value of x we get an output value 'y'.
f(10) = -12
f(-16) = 15.6
f(14) = 14.6
f(-18) = 14
Therefore, from the given table output value of f(14) will be 14.6
The ratio of the measure of the angels of a triangle is 6:10:2 find the measure of the smallest angle
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0.3 Gravity and mechanical energy (Page 4/9)
Page 4 / 9
A ball is dropped from the balcony of a tall building. The balcony is $15\phantom{\rule{2pt}{0ex}}m$ above the ground. Assuming gravitational acceleration is $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ , find:
1. the time required for the ball to hit the ground, and
2. the velocity with which it hits the ground.
1. It always helps to understand the problem if we draw a picture like the one below:
2. We have these quantities:
$\begin{array}{ccc}\hfill \Delta x& =& 15\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill {v}_{i}& =& 0\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-1}\hfill \\ \hfill g& =& 9,8\phantom{\rule{4pt}{0ex}}\phantom{\rule{0.166667em}{0ex}}\mathrm{m}·{\mathrm{s}}^{-2}\hfill \end{array}$
3. Since the ball is falling, we choose down as positive. This means that the values for ${v}_{i}$ , $\Delta x$ and $a$ will be positive.
4. We can use [link] to find the time: $\Delta x={v}_{i}t+\frac{1}{2}g{t}^{2}$
5. $\begin{array}{ccc}\hfill \Delta x& =& {v}_{i}t+\frac{1}{2}g{t}^{2}\hfill \\ \hfill 15& =& \left(0\right)t+\frac{1}{2}\left(9,8\right){\left(t\right)}^{2}\hfill \\ \hfill 15& =& 4,9\phantom{\rule{3.33333pt}{0ex}}{t}^{2}\hfill \\ \hfill {t}^{2}& =& 3,0612...\hfill \\ \hfill t& =& 1,7496...\hfill \\ \hfill t& =& 1,75\phantom{\rule{3.33333pt}{0ex}}s\hfill \end{array}$
6. Using [link] to find ${v}_{f}$ :
$\begin{array}{ccc}\hfill {v}_{f}& =& {v}_{i}+gt\hfill \\ \hfill {v}_{f}& =& 0+\left(9,8\right)\left(1,7496...\right)\hfill \\ \hfill {v}_{f}& =& 17,1464...\hfill \end{array}$
Remember to add the direction: ${v}_{f}=17,15\phantom{\rule{2pt}{0ex}}m·s{}^{-1}$ downwards.
By now you should have seen that free fall motion is just a special case of motion with constant acceleration, and we use the same equations as before. The only difference is that the value for the acceleration, $a$ , is always equal to the value of gravitational acceleration, $g$ . In the equations of motion we can replace $a$ with $g$ .
Gravitational acceleration
1. A brick falls from the top of a $5\phantom{\rule{2pt}{0ex}}m$ high building. Calculate the velocity with which the brick reaches the ground. How long does it take the brick to reach the ground?
2. A stone is dropped from a window. It takes the stone $1,5\phantom{\rule{2pt}{0ex}}s$ to reach the ground. How high above the ground is the window?
3. An apple falls from a tree from a height of $1,8\phantom{\rule{2pt}{0ex}}m$ . What is the velocity of the apple when it reaches the ground?
Potential energy
The potential energy of an object is generally defined as the energy an object has because of its position relative to other objects that it interacts with. There are different kinds of potential energy such as gravitional potential energy, chemical potential energy, electrical potential energy, to name a few. In this section we will be looking at gravitational potential energy.
Potential energy
Potential energy is the energy an object has due to its position or state.
Gravitational potential energy is the energy of an object due to its position above the surface of the Earth. The symbol $PE$ is used to refer to gravitational potential energy. You will often find that the words potential energy are used where gravitational potential energy is meant. We can define potential energy (or gravitational potential energy, if you like) as:
$PE=mgh$
where PE = potential energy measured in joules (J)
m = mass of the object (measured in kg)
g = gravitational acceleration ( $9,8\phantom{\rule{2pt}{0ex}}m·s{}^{-2}$ )
h = perpendicular height from the reference point (measured in m)
A suitcase, with a mass of $1\phantom{\rule{2pt}{0ex}}\mathrm{kg}$ , is placed at the top of a $2\phantom{\rule{2pt}{0ex}}m$ high cupboard. By lifting the suitcase against the force of gravity, we give the suitcase potential energy. This potential energy can be calculated using [link] .
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Metamath Proof Explorer
## Theorem ovollb
Description: The outer volume is a lower bound on the sum of all interval coverings of A . (Contributed by Mario Carneiro, 15-Jun-2014)
Ref Expression
Hypothesis ovollb.1 𝑆 = seq 1 ( + , ( ( abs ∘ − ) ∘ 𝐹 ) )
Assertion ovollb ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → ( vol* ‘ 𝐴 ) ≤ sup ( ran 𝑆 , ℝ* , < ) )
### Proof
Step Hyp Ref Expression
1 ovollb.1 𝑆 = seq 1 ( + , ( ( abs ∘ − ) ∘ 𝐹 ) )
2 simpr ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → 𝐴 ran ( (,) ∘ 𝐹 ) )
3 ioof (,) : ( ℝ* × ℝ* ) ⟶ 𝒫 ℝ
4 simpl ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) )
5 inss2 ( ≤ ∩ ( ℝ × ℝ ) ) ⊆ ( ℝ × ℝ )
6 rexpssxrxp ( ℝ × ℝ ) ⊆ ( ℝ* × ℝ* )
7 5 6 sstri ( ≤ ∩ ( ℝ × ℝ ) ) ⊆ ( ℝ* × ℝ* )
8 fss ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ ( ≤ ∩ ( ℝ × ℝ ) ) ⊆ ( ℝ* × ℝ* ) ) → 𝐹 : ℕ ⟶ ( ℝ* × ℝ* ) )
9 4 7 8 sylancl ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → 𝐹 : ℕ ⟶ ( ℝ* × ℝ* ) )
10 fco ( ( (,) : ( ℝ* × ℝ* ) ⟶ 𝒫 ℝ ∧ 𝐹 : ℕ ⟶ ( ℝ* × ℝ* ) ) → ( (,) ∘ 𝐹 ) : ℕ ⟶ 𝒫 ℝ )
11 3 9 10 sylancr ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → ( (,) ∘ 𝐹 ) : ℕ ⟶ 𝒫 ℝ )
12 11 frnd ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → ran ( (,) ∘ 𝐹 ) ⊆ 𝒫 ℝ )
13 sspwuni ( ran ( (,) ∘ 𝐹 ) ⊆ 𝒫 ℝ ↔ ran ( (,) ∘ 𝐹 ) ⊆ ℝ )
14 12 13 sylib ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → ran ( (,) ∘ 𝐹 ) ⊆ ℝ )
15 2 14 sstrd ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → 𝐴 ⊆ ℝ )
16 eqid { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } = { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) }
17 16 ovolval ( 𝐴 ⊆ ℝ → ( vol* ‘ 𝐴 ) = inf ( { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } , ℝ* , < ) )
18 15 17 syl ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → ( vol* ‘ 𝐴 ) = inf ( { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } , ℝ* , < ) )
19 ssrab2 { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } ⊆ ℝ*
20 16 1 elovolmr ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → sup ( ran 𝑆 , ℝ* , < ) ∈ { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } )
21 infxrlb ( ( { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } ⊆ ℝ* ∧ sup ( ran 𝑆 , ℝ* , < ) ∈ { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } ) → inf ( { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } , ℝ* , < ) ≤ sup ( ran 𝑆 , ℝ* , < ) )
22 19 20 21 sylancr ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → inf ( { 𝑦 ∈ ℝ* ∣ ∃ 𝑓 ∈ ( ( ≤ ∩ ( ℝ × ℝ ) ) ↑m ℕ ) ( 𝐴 ran ( (,) ∘ 𝑓 ) ∧ 𝑦 = sup ( ran seq 1 ( + , ( ( abs ∘ − ) ∘ 𝑓 ) ) , ℝ* , < ) ) } , ℝ* , < ) ≤ sup ( ran 𝑆 , ℝ* , < ) )
23 18 22 eqbrtrd ( ( 𝐹 : ℕ ⟶ ( ≤ ∩ ( ℝ × ℝ ) ) ∧ 𝐴 ran ( (,) ∘ 𝐹 ) ) → ( vol* ‘ 𝐴 ) ≤ sup ( ran 𝑆 , ℝ* , < ) )
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# Definition:Logical Implication/Distinction with Conditional
## Distinction between Logical Implication and Conditional
It is important to understand the difference between:
$A \implies B$: If we assume the truth of $A$, we can deduce the truth of $B$
and:
$A \leadsto B$: $A$ is asserted to be true, therefore it can be deduced that $B$ is true
When $A$ is indeed true, the distinction is less important than when the truth of $A$ is in question, but it is a bad idea to ignore it.
Compare the following:
$\text {(1)}: \quad$ $\ds x > y$ $\implies$ $\ds \paren {x^2 > x y \text { and } x y > y ^2}$ $\ds$ $\implies$ $\ds x^2 > y^2$
$\text {(2)}: \quad$ $\ds x$ $>$ $\ds y$ $\ds \leadsto \ \$ $\ds x^2$ $>$ $\ds x y$ $\, \ds \text { and } \,$ $\ds x y$ $>$ $\ds y^2$ $\ds \leadsto \ \$ $\ds x^2$ $>$ $\ds y^2$
We note that $(1)$ is a conditional statement of the form:
$A \implies B \implies C$
This can mean either:
$\paren {A \implies B} \implies C$
or:
$A \implies \paren {B \implies C}$
instead of what is actually meant:
$\paren {A \implies B} \text { and } \paren {B \implies C}$
Hence on $\mathsf{Pr} \infty \mathsf{fWiki}$ we commit to using the form $A \leadsto B$ rigorously in our proofs.
The same applies to $\iff$ and $\leadstoandfrom$ for the same reasons.
Note that there are many pages on $\mathsf{Pr} \infty \mathsf{fWiki}$ using the $\implies$ construct, which are still in the process of being amended to use the $\leadsto$ construct as they should.
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childinstress.com
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This example shows how to find a linear least squares fit for a set of points.
Suppose you have a set of data points that you believe were generated by a process that should ideally be linear. In that case, you might like to find the best parameters m and b to make the line y = m * x + b fit those points as closely as possible.
A common approach to this problem is to minimize the sum of the squares of the vertical distances between the line and the points. For example, suppose the point P0 = (x0, y0) is one of your data points. The vertical error squared for that point is the difference between y0 and the line’s Y coordinate for that X position. In this case, that’s y0 – (m * x0 + b). To calculate the total error squared, square this error and add up the errors squared for all of the points.
Keep in mind that you know all of the points so for given values of m and b you can easily loop through all of the points and calculate the error.
Here’s a function that does just that:
``` ; Return the error squared:
Func ErrorSquared(\$aPoints, \$M, \$B)
Local \$Total = 0
For \$I = 1 To \$aPoints[0]
Local \$aPoint = \$aPoints[\$I]
Local \$Dy = \$aPoint[2] - (\$M * \$aPoint[1] + \$B)
\$Total += \$Dy ^ 2
Next
Return \$Total
EndFunc```
This code loops through the points subtracting each point’s Y coordinate from the coordinate of that line at the point’s X position. It squares the error and adds it to the total. When it finishes its loop, the method returns the total of the squared errors.
As a mathematical equation, the error function E is:
where the sum is performed over all of the points (xi, yi).
To find the least squares fit, you need to minimize this function E(m, b). That sounds intimidating until you remember that the xi and yi values are all known – they’re the values you’re trying to fit with the line.
The only variables in this equation are m and b so it’s relatively easy to minimize this equation by using a little calculus. Simply take the partial derivatives of E with respect to m and b, set the two resulting equations equal to 0, and solve for m and b.
Taking the partial derivative with respect to m and rearranging a bit to gather common terms and pull constants out of the sums you get:
Taking the partial derivative with respect to b and rearranging a bit you get:
To find the minimum for the error function, you set these two equations equal to 0 and solve for m and b. To make working with the equations easier, let:
If you make these substitutions and set the equations equal to 0 you get:
Solving for m and b gives:
Again these look like intimidating equations but all of the S’s are values that you can calculate given the data points that you are trying to fit. The following code calculates the S’s and uses them to find the linear least squares fit for the points in a List(Of PointF).
``` ; Find the least squares linear fit:
Func LineFit(\$aPoints, ByRef \$M, ByRef \$B)
; Find the values S1, Sx, Sy, Sxx, and Sxy.
; Perform the calculation.
Local \$S1 = \$aPoints[0]
Local \$Sx = 0, \$Sy = 0, \$Sxx = 0, \$Sxy = 0
For \$I = 1 To \$aPoints[0]
Local \$aPoint = \$aPoints[\$I]
\$Sx += \$aPoint[1]
\$Sy += \$aPoint[2]
\$Sxx += \$aPoint[1] ^ 2
\$Sxy += \$aPoint[1] * \$aPoint[2]
Next
; Solve for \$M and \$B :
\$M = (\$Sxy * \$S1 - \$Sx * \$Sy) / (\$Sxx * \$S1 - \$Sx ^ 2)
\$B = (\$Sxy * \$Sx - \$Sy * \$Sxx) / (\$Sx ^ 2 - \$S1 * \$Sxx)
Return Sqrt(ErrorSquared(\$aPoints, \$M, \$B))
EndFunc```
I took this further with a routine to remove an outlier:
``` ; Remove one outlier (if any)
; Return index of the point removed
Func RemoveOutlier(ByRef \$aPoints, \$M, \$B)
Local \$Total = 0
Local \$iOutlier
For \$I = 1 To \$aPoints[0]
Local \$aPoint = \$aPoints[\$I]
\$aDy2[\$I] = (\$aPoint[2] - (\$M * \$aPoint[1] + \$B)) ^ 2
Next
Local \$Mean = \$Total / \$aPoints[0]
If \$Max > 0 And \$aPoints[0] > 3 And \$Max > 4 * \$Mean Then
EndIf
Return \$iOutlier
EndFunc```
``` Func DataSet1(ByRef \$aPoints)
EndFunc```
``` Func TestLineFit()
Local \$M, \$B
Local \$aPoints = aNewArray()
DataSet1(\$aPoints)
Local \$E = LineFit(\$aPoints, \$M, \$B)
Msg('The Outcome', 'm= ' & \$M & @LF & ' b= ' & \$B)
EndFunc```
``` Func TestBestLineFit()
Local \$M, \$B
Local \$aPoints = aNewArray()
DataSet1(\$aPoints)
Do
Local \$N = \$aPoints[0]
Local \$E = LineFit(\$aPoints, \$M, \$B)
Msg('Outcome', 'm= ' & \$M & @LF & ' b= ' & \$B)
Local \$iRemoved = RemoveOutlier(\$aPoints, \$M, \$B)
If \$iRemoved Then Msg('Removal', 'item= ' & \$iRemoved)
Until \$iRemoved = 0
Msg('BestFit', 'm= ' & \$M & @LF & ' b= ' & \$B)
EndFunc```
` TestBestLineFit()`
Keywords:
algorithms, least squares, linear least squares, curve fitting, graphics.
Bert Kerkhof
usually writes about health care and the rule of law
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https://www.abcadda.com/110-cm-to-inches/
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# How many inches is 110 cm to inches | 110 cm to inc | 110 cm into inches
### All-in-one unit converter calculator
Please, choose a physical quantity, two units, then type a value in any of the boxes above.
## 110 cm to inches | 110 Centimeter (cm) equals 4.330711 Inch (in)
The 110 cm to inches converter is a length converter from one unit to another. One centimeter is approximately 0.3937 inches.
The units of length must be converted from centimeters to inches. The 110 cm to inches is the most basic unit conversion you will learn in elementary school. This is one of the most common operations in a wide variety of mathematical applications.
This article explains how to convert 110 cm to inches and use the tool for converting one unit from another, as well as the relationship between centimeters and inches with detailed explanations.
## Why change the length from 110 cm to inches to inches? | 110 cm to inc
A centimeter (or centimeter) is a unit of length. It is one hundredth of a meter. However, the United States uses a common unit of length. Imperial units are used in the same way in Great Britain.
The common Imperial or US unit of measurement for length (or distance) is inches. If you have information about length in centimeters; and you need the same number in equivalent inch units, you can use this converter.
### The relationship between inches and cm
1 inch = 2.54 cm
Therefore,
1 cm = 1 / 2.54 inch
To convert centimeters to inches, we need to divide the value in centimeters by 2.54.
If the unit length is 1 cm, the corresponding length in inches is 1 cm = 0.393701 inches
## How many inches is 110cm
### Convert 110 cm (centimeters) to inches (in)
With this length converter we can easily convert cm to inches like 10 cm to inches, 16 cm to inches, 110 cm to inches, 110cm in inches etc.
Since we know that a centimeter is approximately 0.393701 inches, the conversion from one centimeter to inches is easy. To convert centimeters to inches, multiply the centimeter value given by 0.393701.
For example, to convert 10 centimeters to inches, multiply 10 centimeters by 0.393701 to get the value per inch.
(i.e.) 10 x 0.393701 = 3.93701 inches.
Therefore, 10 centimeters is equal to 3.93701 inches.
Now consider another example: 110cm in inches is converted as follows:
## How do I convert 110 cm to inches? | 110 cm to inc
To convert 110 cm to in, simply take the actual measurement in cm and multiply this number by 2. 1105110. So you can convert how many inches is 110 cm manually.
You can also easily convert centimeters to inches using the following centimeters to inches conversion:
### How many inches is 110 cm
As we know, 1 cm = 0.393701 inches
## What is 110 cm in inches
In this way, 110 centimeters can be converted to inches by multiplying 110 by 0.393701 inches.
(i.e.) 110 cm to one inch = 110 x 0.393701 inches
110 cm = inches = 4.330711 inches
### 110cm to inches how many inches
Therefore, 110 cm is how many inches 110 cm is equal to 11,811 inches.
## Example of converting centimeters to inches
The following examples will help you understand how to convert centimeters to inches.
### Convert 110 cm to inches | 110 cm to inc
We know that 1 cm = 0.393701 inches.
To convert 110 centimeters to inches, multiply 110 centimeters by 0.393701 inches.
= 110 x 0.393701 inches
= 4.330711 inches
• 110 cm is equal to how many inches
• How many inches is 110 cm equal to
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• What is 110 cm equal to in inches?
• Convert 110 cm to inches
• 110 cm convert to inches
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• 110 cm is how many inches
• 110 cm equal how many inches
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• convert 110 cm to inches
• what’s 110 cm in inches
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# Harmonic Distortion
Home / Technical Articles / Energy and Power / Harmonic Distortion
Harmonic distortion is caused by nonlinear devices in the power system. A nonlinear device is one in which the current is not proportional to the applied voltage. Figure 1 illustrates this concept by the case of a sinusoidal voltage applied to a simple nonlinear resistor in which the voltage and current vary according to the curve shown. While the applied voltage is perfectly sinusoidal, the resulting current is distorted.
Increasing the voltage by a few percent may cause the current to double and take on a different waveshape. This is the source of most harmonic distortion in a power system.
Figure 2 illustrates that any periodic, distorted waveform can be expressed as a sum of sinusoids. When a waveform is identical from one cycle to the next, it can be represented as a sum of pure sine waves in which the frequency of each sinusoid is an integer multiple of the fundamental frequency of the distorted wave. This multiple is called a harmonic of the fundamental, hence the name of this subject matter.
The sum of sinusoids is referred to as a Fourier series, named after the great mathematician who discovered the concept.
Because of the above property, the Fourier series concept is universally applied in analyzing harmonic problems. The system can now be analyzed separately at each harmonic. In addition, finding the system response of a sinusoid of each harmonic individually is much more straightforward compared to that with the entire distorted waveforms. The outputs at each frequency are then combined to form a new Fourier series, from which the output waveform may be computed, if desired.
Often, only the magnitudes of the harmonics are of interest. When both the positive and negative half cycles of a waveform have identical shapes, the Fourier series contains only odd harmonics. This offers a further simplification for most power system studies because most common harmonic-producing devices look the same to both polarities. In fact, the presence of even harmonics is often a clue that there is something wrong – either with the load equipment or with the transducer used to make the measurement.
There are notable exceptions to this such as half-wave rectifiers and arc furnaces when the arc is random.
Usually, the higher-order harmonics (above the range of the 25th to 50th, depending on the system) are negligible for power system analysis.
While they may cause interference with low-power electronic devices, they are usually not damaging to the power system. It is also difficult to collect sufficiently accurate data to model power systems at these frequencies. Acommon exception to this occurs when there are system resonances in the range of frequencies. These resonances can be excited by notching or switching transients in electronic power converters. This causes voltage waveforms with multiple zero crossings which disrupt timing circuits. These resonances generally occur on systems with underground cable but no power factor correction capacitors.
If the power system is depicted as series and shunt elements, as is the conventional practice, the vast majority of the nonlinearities in the system are found in shunt elements (i.e., lods). The series impedance of the power delivery system (i.e., the short-circuit impedance between the source and the load) is remarkably linear. In transformers, also, the source of harmonics is the shunt branch (magnetizing impedance) of the common “T” model; the leakage impedance is linear.
Thus, the main sources of harmonic distortion will ultimately be end-user loads. This is not to say that all end users who experience harmonic distortion will themselves have significant sources of harmonics, but that the har-monic distortion generally originates with some end-user’s load or combination of loads.
SOURCE: Power Systems Quality by Roger C. Dugan/Mark F. McGranaghan
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# 1
EARTH. It is an oblate spheroid whose major axis AB is 12748 km and minor axis CD is 12705km ie
a difference of 43 km. Compression Ratio = diff of two axis/larger axis
C
A
## CR = 43/12748 ~ 1/299 or 1/300 approx, Polar axis is
Shorter by 1/300 times the equator. For all practical purposes
Earth is considered as a Sphere.
D
Great Circle. Its a circle which when drawn across the earth cuts the earth into two equal hemispheres.
Properties.
(a) Its the shortest path between two poles.
(b) On earth sphere it appears as a straight line.
(c) Only one GC can be drawn between two points unless they are diagrammatically opposite
each other, in that case infinite number can be drawn.
Latitude. The angle subtended by the shorter arc of the meridian at the centre of the earth from the
Equator to the point to be identified is termed as latitude of that point. It is denoted as N or S depending
on whether the point lies North or South of the Equator also known as Geocentric Latitude, whereas,
Geodetic Latitude is the angle between the normal to the observers horizontal plane and the equatorial
plane.
Longitude. The angle subtended by shorter arc of equator at the centre of the earth from prime meridian
to the point to be identified is termed as longitude and is denoted East or West depending on whether the
point lies East or West of the prime meridian.
Nautical Mile. It is the arc length subtended by 1 minute angle at the surface of the earth. The angle is
measured from the Geocentric Latitude. Length of Nm is more at poles when measured from geographic
centre.
1 Nm = 6017 feet at Poles- geographical centre
1 Nm =6045 feet at Equator geocentric
1 Nm = 6080 feet subtended at 45 deg Latitude.
1 Deg = 60 Nm hence distance covered around the globe =360 x 60 = 21600 Nm (Great Circle path)
Kilometre : It is 1/10,000 th part of the distance from pole to the equator. 1 km =3280 ft.
1 Nm =6080/3280 = 1.854 km and 1 km = 3280/6080 = 0.54 Nm
Statute Mile. By statute it is 5280 ft.
2
Rhumb Line : A line which cuts all meridians at equal angles. It spirals to the poles at angles less than
90. ATS routes are rhumb line tracks. Equator is a RL as well as GC. Parallels of Lat are RL tracks.
Meridians are GC track (Flying a constant direction).
Direction. It is measured with respect to the North, clockwise. All meridians point to North.
True Direction. A direction which is measured with respect to True North the symbol is and is
annotated as 005 T or 005 (T).The geographical north is not aligned with the magnetic earth since the
earth behaves as a bar magnet with its axis slightly aligned from the geographical axis. The exact
position of North an South pole with respect to the bar magnet is defined and known, however the
position of magnetic north depends on where the observation is taken from vis a vis the position of True
North.
Magnetic North. It is the direction which points to the magnetic north. The angular difference between
the True North and Magnetic North is known as Variation and is denoted as E or W depending on
whether the Magnetic North lies east or west of true North. Simply, if Magnetic North lies East of True
North the Variation is East and if Mag N lies W of True North Var is West. Thumb Rule - Var E Mag
Least, Var W Mag Best. It is applicable in any hemisphere.
The symbol for Magnetic north is
## and is denoted by 045 M or 045 (M).
Isogonal : Lines joining places of equal Variation. Agonal : Lines joining places of zero Variation.
Compass North. The direction measured wrt to Compass North is called Compass direction.
The symbol for Compass North is
## and is denoted as 005 (C) or 005 C.
Deviation. It is the angular difference between Magnetic North and Compass North and is termed as
Easterly if Compass North lies East of Magnetic North and Westerly if Compass North lies West of
Magnetic North. Deviation is obtained from Compass Card and it varies from aircraft to aircraft due to
inherent magnetic fields present in the ac and is different for different headings. Thumb Rule - Dev E
Com Least, Dev W Comp Best. It is applicable in any hemisphere.
Any line drawn on a map represents a True Track.
3
Compass
North
Magnetic North
True North
Variation W
Deviation E
## POBLEMS ON DIRECTION (Black-Given, Red-Determined) (East is +, W is -ive)
1
2
3
4
5
C
269
027
051
300
045
D
5E
3W
+3
-7
-4
M
274
024
054
293
041
V
4W
4E
4W
10E
+6
T
270
020
050
303
047
4
SCALE FACTOR AND DEPARTURE
Departure (Nm) = dlong x 60 x Cos Lat based on this formula distance in Nm along any latitude can
be determined. As latitude increase for same dlong distance reduces. Therefore distance is maximum at
the equator and zero at poles, also this is a Cosine function. If the Longitudinal change is known
distance traveled along a particular latitude can be determined. It is applicable to both the hemispheres.
While flying on Easterly hdg and crossing the ante meridian the value of longitude will change to
westerly and vice versa. Hence to determine the longitude the following can be resorted to (360{longitude at origin + change in longitude}) in case we cross the ante meridian. Dont forget to change
the Easterly or Westerly Longitude depending whether the aircraft is traveling from East to West or vice
versa. For eg an ac flying from 172 E on an easterly heading changes longitude by 12 degrees (dlong)
then the position of the aircraft is (360-{172+12}) = 360 184 =176 degrees W.
Problems on Departure
1. At what latitude a distance of 100 nm will involve a dlong of 30 degrees?
A. Dep =100 nm, dlong =100nm, Lat=?, Dep=dlong x 60 x Cos Lat
or 900=30 x 60 x Cos Lat or Cos Lat = 0.5 or Lat = Cos inv 0.5 = 60 Deg
2. How long will it take to go around the earth at 60 deg lat at G/S of 600 Kts?
A. Dep =? Lat = 60 deg, dlong =360, G/S = 600 K, Dep=360 x 60 x Cos 60 = 10800
Hence time taken =10800/600 =18 hrs, whereas, the time taken to go round the equator is 36 h.
3. A is 60N 168E, B is 545 nm due East of A, what is its longitude. (See figure below)
A. Dep = 545 nm, Lat = 60 deg, dlong = ? Substituting in Dep=dlong x 60 x CosLat
545= dlong x 60 x 0.5 or dlong = 545/30 = 18 deg 10 min =17350W
4. Find dep between 32 30 N, 20 42 E and 32 30 N, 89 26 E.
17350W
180
610
12
168E
## A. dlong = 68 44 x 60 x Cos 52 50 =3478 Nm
5. Find RL distance between 42 42 N, 32 42 E & 42 42 N 69 42 W.
A. 102 24 x 60 x Cos 42 42 = 4515.3 Nm
6. Aircraft takes off from A 60 29 N 177 23 E and flies a RL track of 090 for 600 nm. Find lat &
long of destination.
A. 600 60 =10 Cos 60 29 = 20 18, 177 23 + 20 18 = 197 41, 360 - 197 41 =162 19 W
7. Aircraft takes off from A 40 40 N 176 30 W and flies a RL track of 270 for 600 nm. Find lat &
long of destination.
A. 600 (60 x Cos 40 40 = 13 11 + 176 30 = 189 41, 360 - 189 41 =179 19 E
8. An aircraft flying for 360 nm undergoes long change of 7 23. Find lat.
A. 360 (60 x 7 23) = 0.8126, Cos inv 0.8126 = 35 30 N/S.
6
Scale. It is defined as the ratio of Map Distance (MD) to Earth Distance (ED).
Scale = MD/ED. Scale is large (1/!00) or small (1/1000). Large scale map on a unit area smaller
distance is shown, whereas in small scale map on a unit area large earth distance is shown.
On a map Scale is represented by three methods:(a) Representative Fraction, a fraction whose numerator is always 1, e.g. 1:10000.
(b) Statement in words, e.g. 1cm = 100 nm.
Km or Nm
Scale at Lat (SAL) = Scale at Equator (SAE) x Secant Lat or SAL = SAE/Cos Lat (A pplicable for
Mercator Chart Only)
Conversion Table
1 Nm = 6080 ft
1 Km = 3280 ft
1 SM = 5280 ft
## 1 m = 3.28 ft; 1 ft = 12 in; 1 in = 2.54 cms
Problems on Scale
1. Given MD = 20 cm; ED = 100 nm; Find Scale.
A. Scale = MD/ED = 20/(100 x 6080 x 12 x 2.54) (1 Nm =6080 ft, 1 in =2.54 cms) or Scale = 1: 926592
2. Given MD = 20.5 in; ED = 600 km; Find Scale.
A. Scale = MD/ED = 20.5 x 2.54 cm/600 x 1000 x 100 cm = 1:1152295
3. Given MD = 5 in; Scale = 1:2,000,000; Find ED
A. 1/2,000,00 = 5/ED or ED = 5 x 2,000,000 in = 10,000,000/12 x 6080 Nm = 137 Nm.
4. An ac at a G/S of 300K covers map distance of 15 cms in 24 mins. Find Scale.
A. Distance covered = 300/(24/60) = 80 Nm. Hence Scale = 15/(80 x 6080 x 12 x 2.54) = 1:988365
5. An ac covers a distance of 5 in on a chart (scale of 1: 1,000,000) in 20 min. Find G/S.
1/1,000,000 = 5/ED or ED = 5,000,000 in = 5000000/(6080 x 12) = 68.5 Nm in 20 mins. Hence Speed =
(68.5 x 60)/20 =205.5 K
6. On a chart SAL 62N = 1/1000000, Find (a) SAE (b) Scale at 40 N.
A. (a) SAE = SAL x 1/Cos Lat or 1/1000000/Cos 62 = 1/ 2130054 (b) SAL = SAE x 1/Cos 40 =
1/2130054 x 1/Cos 40 = 1/1631716
Note : For Mercators Chart SAL= SAE x Sec Lat or SAE x 1/Cos Lat.
7
Thumb Rule. For conversion of SAE to SAL, multiply denominator by Cos Lat and for SAL to
SAE divide denominator by Cos Lat.
7. On a mercator chart SAE = 1: 1450000, at what Lat scale will be 1: 1000000.
A. SAL = SAE x 1/Cos Lat or 1/1000000 = 1/1450000 x 1/Cos Lat or 1450000 x Cos Lat = 1000000 or
Cos Lat = 1000000/1450000 = 0.689, Hence Cos inv 0.689 = 46 24 N/S.
8. On a chart SAE = 1:1000000. On this chart two points A & B are 10 apart at 54 N. Find difference
in longitude.
A. SAL 54 N = 1000000 x Cos 54 = 1: 587785. MD/ED = 1/587785 or 10/ED = 1/587785 or ED =
5877850 in = 80.56 nm. Dep = 80.56 = dlong x 60 x Cos 54 = 80.56/ (60 x Cos 54) = 2 17
9. Distance between two points at 45 N is 10 cm and ED is 100km. Find SAE.
A. SAL at 45 N is MD/ED =10/ 100 x 100 x 100= 1: 1000000. SAE =1:1000000/Cos 45= 1:1414213
10. A & B are located at 50 N and are 1 42 long apart. Distance between them is 8 cm. Find SAE
and SAL at 50 S.
A. Dep = dlong x 60 x Cos 50 = 65.56 nm. Scale at 50 = MD/ED = 8 CM /65.56 X 6080 X 12 X 2.54
= 1: 1518684. SAE = SAL/ Cos Lat = 1: 1518684/Cos 50 = 1/2362653
11. The scale at 60N is 1/2000000 on a Mercator Chart. At what latitude will you find the scale
1/1000000.
A. SAE = 20000000/Cos60 =4000000, SAL = 1000000, SAE = 4000000 Hence Cos Lat =.25 or 7531
12. If the scale at 5720N is 1:1091000 what is the meridian spacing in cm between one deg longitude.
A. Dep =1 x 60 x Cos5720 = 32.385 Nm = 5997709 cm (1Nm = 1.852Km)
13. You are flying east along a parallel 60N and cover 10 inches distance on the chart every hour. The
scale at 25S is 1:1000000. Find GS.
A. SAE = 1000000/Cos 25 = 1:1103378, SAL = 1103378 x Cos 60 =551688.95. So 10 = 5516889 on
chart which corresponds to 5516889/(6080 x 12) = 75Nm Hence speed is 75 K.
14. The distance between A & B both at 40N is 10 cm on a Mercator chart and 90 km on earth. Find
scale at equator.
A.Scale at 40N = MD/ED = 10cm/90,000,00 cm (90km)= 1:9000000. SAE = 9000000/Cos40=1174866
15. On a Mercator Chart if scale is 1:1M at 56N. Find the Chart length from 2845N 11330W to 2845N
9815W.
A. Dep = 1515 x 60 =915 Nm, 915 x 6080 x 2 = 66758400/1000000 corresponding to 66.75 .
8
MAPS & CHARTS (PROJECTIONS)
Map. It contains all geographical features like roads, rivers, mountains etc depending on the scale of the
map. It depends on the scale of the map.
Chart. It contains limited information for which the chart has been made e.g. enroute chart, approach
chart etc.
Large Scale Map. On a unit area of the map smaller earth distance is shown e.g. 1: 100 scale map is a
bigger fraction than 1: 1000 wherein the details are more but the area depicted is smaller.
Small Scale Map. On a unit area of the map large earth distance are shown e.g. 1:1000 meaning greater
details are available for the same size of the map sheet as compared to a large scale map.
Types of Map. There are two methods to construct a map, they are (a) Perspective and (b) Non
Perspective (These are drawn mathematically).
Construction. The perspective method of constructing a map involves projection of the graticule of
the earth on a sheet of paper with the help of a light source placed at the appropriate place. The non
perspective method involves mathematical reduction of spherical globe on a plain sheet of paper.
Types of Projection. These are three types (a) Cylindrical (b) Conical and (d) Azimuthal or Zenithal.
Cylindrical Projection
## ____ Great Circle
Construction. Light source is at the centre and point of tangency of the cylinder superimposed on the
global sphere is at the centre (equator).
Properties. The following properties emerge as a result of placing a cylinder on the global sphere with
point of tangency at the Equator:1. Meridians are straight lines equidistant from each other.
2. Parallels of latitude are also straight lines but not equidistant from each other. Distance between
them progressively increases from equator to poles.
3. Convergence (n=0) is zero. A straight line on this map is a Rhumb Line. A Great Circle is a
curved line concave to the equator and convex to the poles. Convergency on earth is angle of inclination
between 2 meridians at a given latitude and is = dlong x Sin Lat.
9
4. It is not an orthomorphic projection (Orthomorphism is the property of a projection in which
bearings are correct in all directions within vicinity of the point).
Note: For a projection to be orthomorphic the following conditions are to be satisfied:(a)
Meridians and Parallels of Latitude should cut each other at right angles (90).
(b)
## Scale should be constant within the vicinity of a point.
.
MERCATOR PROJECTION
1.
Mathematical modifications/corrections are carried out to make a Cylindrical Projection
orthomorphic in that scale in E-W direction is varied at the same rate of scale expansion in the N-S
direction, with increase in Latitude.
2. The scale varies as the Secant of the Latitude and is represented by the formulae
Scale is correct only along the equator.
Scale at any Lat (SAL) = Scale at Equator(SAE) x Secant of Lat or SAE/Cos Lat
3. Appearance & Properties are similar to Cylindrical Projection which are:(a) Rhumb Line(RL) is a straight line. (b) Great Circle (GC) curved concave to the RL.
(c) Meridians cut parallels of Lat at 90. Chart convergence is equal to earth convergence only at
Equator, otherwise it is zero.
Limitations. The limitations are:(a) This projection cannot be used for Polar Regions.
(b) It can only be used effectively upto 70-75 N/S beyond which scale expansion is very large.
(c) Adjacent sheets fit together in E-W direction not N-S.
Usage. These charts are used for flying on Rhumb line tracks and are also used for Met Charts.
10
CONICAL PROJECTION
Lat of Origin is
midway between
two Std Parallels
Construction. The point of tangency is a particular latitude which can be selected by changing the
Cone angle or Apex angle.
Properties/Appearance. The following properties emerge for this type of a perspective projection:1. Meridians are straight lines converging to the nearest pole.
2. Parallels of Latitude arc of concentric circle not equidistant from each other. Distance between them
increases away from Lat of origin on either side.
3. Rhumb line is a curved line concave to the nearest pole or great circle and convex to the equator.
4. A Great Circle is a straight line.
5. Convergence, which is angle of inclination between two meridians on a projection and is denoted by
the symbol n = c/dlong, where c =convergency and dlong = difference in longitude, is less than 1.
(Convergence is ratio of convergency to dlong on that map)
6. It is not an orthomorphic projection.
11
Lamberts Conical Orthomorphic (Conformal) between two Standard Parallels
## Construction. It is a conical projection between two std parallels.
mathematically modified to make it orthomorphic.
## Base is perspective but
Properties.
1. Its an orthomorphic only between two std parallels.
2. Rhumb line is a curved line concave to the lat of origin/great circle/nearest pole and convex to the
equator.
3. Great Circle is a curved line concave to the lat of origin, but for practical purposes it is nearly a
straight line, as indicated below.
Scale. It is almost constant within the two std parallels. Away from lat of origin scale expansion takes
place, but this scale expansion is negligible within the std parallels and is approx 1%. Outside the std
parallels scale expansion is very large.
12
Scale Expansion
Q. On Lamberts Conical Projection scale is almost correct between Std Parallel and convergence is
correct at Lat of Origin.
Q. While measuring track on LC Projection protractor is placed at (a) Lat of Dep (b) Lat of arr/dest (c)
Mid way Lat (d) any of the Latitudes. A. (c)
Usage. All Jeppesen charts are Lambert Conical Projection.
ZENITHAL OR AZIMUTHAL PROJECTION
Properties
1. Meridians are straight lines converging at poles.
2. Parallels of Lat are concentric circles not equidistant from each other, the distance between them
increases from poles to the equator.
3. Rhumb Line is a curved line concave to the Great circle/nearest pole.
4. Great Circle is a straight line.
5. Its not an orthomorphic projection.
13
POLAR STEREOGRAPHIC PROJECTION
Properties. The point of tangency is at poles and light source is placed at the opposite pole, appearance
is similar to Zenithal projection. Scale expands at the rate of Sec2 Lat.
CONVERGENCY
Convergency. Its the inclination between two meridians or angular difference between two meridians.
At equator the angle is Zero and at poles it is 1 (dlong). Therefore it varies as a function of Sin.
Convergence at equator = 0, Conversion at poles = dlong and Conv at any Lat =dlong Sin Lat or Conv
at Lat = dlong x Sin mean Lat.
Q. GC Brg of A from B = 045, CA = 5. What is RL Brg of B from A in Northern hemisphere.
A. RL Brg of A from B = GC Brg of A from B + CA = 045 + 5 = 050
RL Brg of B from A = 050 + 180 = 230
GC Brg of B from A = 230 + 5 = 235
045
B
Q. RL Brg of X fom Y is 060, CA = 6, NH. Find GC Brg of X from Y, Y from X and RL Brg of Y
from X.
A. RL Brg of X from Y = 060 Therefore GC Brg of X from Y = 060-CA =060-6 =054
RL Brg of X from Y = 060 + 180 = 240 & GC Brg of Y from X = 240 + 6 = 246
Q. GC Brg of X from Y is 300. CA = 7, NH. Find GC Brg of X from Y, of Y from X & RL Brg of Y
from X.
A. GC Brg of X from Y = 300, Hence RL Brg of X from Y = 300 - 7 = 293, RL Brg of Y from X =
293 -180 = 113, GC Brg of Y from X = 113-7 = 106
Q. RL Brg of A from B is 220, GC Brg of A from B is 216 Find CA, Hemisphere, GCB of B from A.
A. Difference of RLB & GCB = CA = 220-216 = 4, RLB of B from A = 220-180 = 040, Hence GCB
of B from A = 040 + 4 = 044 in SH
216
14
Q. GCB of A from B is 340. RLB of A from B is 345. Find, CA, Hemisphere & GCB of B from A.
A. CA = 345-340 =5, RLB of B from A = 345-180 = 165, GCB of B from A = 165 + 5 = 170, SH.
Q. GCB of B from A is 070.RL Brg of A from B is 256. Find CA, Hemisphere & GCB of B from A.
A. RLB of B from A = 256 -180 = 076. CA = 076 070 = 6, NH, GCB of B from A= 256 + 6 = 252
Q. GCB of A from B is 234. GCB of B from A is 066. Find, CA, Hemisphere & RLB of B from A.
A. In these problems add both the GCBs and if difference is more than 180 then location is in
NH, if difference less than 180 then in SH. Next, the value obtained after subtracting the GCB is
to be subtracted from 180, Divide the absolute value by 2 to get CA.
GCB of A from B = 234. GCB of B from A = 066, 234 -066 = 168 < 180 so in SH, 180 -168 =12,
12/2 = 6 = CA. RLB of B from A = 066-6 = 060.
Q. GCB of P from Q is 130. GCB of Q from P is 318. Find, CA, Hemisphere & RLB of Q from P.
A. 318-130 = 188, Hence NH, 188 180 = 8, CA = 8/2 = 4, RLB of Q from P = 318-4= 314.
Q. A & B are on parallel of 30 N. GCB of B from A is 087. Longitude of A is 8W. Find long of B.
A. RLB of B from A = 090 (since on same parallel of Lat), hence CA = 090-087 =3. Also
Convergence = 2 CA = dlong Sin Lat or CA = dlong Sin 30 or dlong = 2 CA Sin 30 = (2 x 3) 0.5
= 12. Hence B lies 12 apart ie 12-8 = 4E
Q. A & B are on parallel of 30 N. GCB of A from B is 266. Longitude of A is 10W. Find long of B.
A RLB of A from B = 270 (since on same parallel of Lat), hence CA = 270-266 =4. Also
Convergence = 2 CA = dlong Sin Lat or CA = dlong Sin 30 or dlong = 2 CA Sin 30 = (2 x 4) 0.5
= 16. Longitude of B = 16-10 = 6 E
15
WIND TRIANGLE
W/V
DRIFT
(p)
TRK & G/S
## When Trk is left of Hdg it is Port Drift and vice versa
Drift. It is the angular difference between the Heading and the Track. When Track is right of heading it
is called Starboard Drift and when Track is left of heading it is called Port Drift.
TMG. Track Made Good is the physical path followed on the ground and may differ from Track
required due to inadequate drift correction. The angle between the Track Required and TMG is called
Track error. When TMG is right of Tr reqd then it is called Stbd TE and when MG is Port of Track
required it is called Port TE.
Tr Reqd
TMG
Port
Hdg/TAS
TMG
Stbd
Hdg/TAS
W/V
W/V
Trk/GS
Trk/GS
Winds 90 to Trk (GS < TAS)
## Problems on Wind Triangle.
(a)
(b)
(c)
(d)
(e)
TRK
101
180
276
045
234
TE
8P
5S
6P
10 S
15 P
TMG
093
185
270
055
219
HDG
090
183
275
059
225
DRIFT
3S
2S
5P
4P
4P
16
Mean Winds. To calculate mean winds in a multiple leg, it is imperative to determine the GS and time
for each leg. Alongside calculate the Wind Effect by multiplying Time on respective leg with
corresponding wind velocity on that leg. Calculate total time by adding time taken on each leg, sum up
the wind effect and divide by the sum of time taken on each leg. This is illustrated in the example
below. Similarly to calculate TAS, multiply TAS of respective legs with corresponding wind velocity,
total for all the legs and divide by the total time flown. This would give Mean TAS. Thereafter Mean
GS = Mean TAS Mean Wind.
Leg
A-B
B-C
C-D
TAS
300
200
350
DIST
700
280
300
W/V
+30
-10
+40
GS
330
190
390
TIME
WIND EFFECT (TIME x W/V)
2:07
+63.5 Nm (30 x 2:07)
1:28
-14.7 Nm (-10 x 1:28)
0:46
+30.7 Nm (+40 x 0:46)
5:26
+ 79.5 Nm
Total Time = 2:07 + 1:28 + 0:46 = 4:21, Total Wind Effect = +63.5 -14.7 + 30.7 = 79.5
Hence Mean Winds experienced from A to D = 79.5 4:21 = + 18.27. Similarly Mean TAS is
calculated :Mean TAS (A-B) = 300 x 2:07 =635, (B-C) = 200 x 1:28 = 293.33 & (C-D) = 350 x 0:46 = 268.33
Total = 635 + 293.3 + 268.3 = 1196.6, this divided by total time 4:21 (1196.6 4:21) =275 Kts is Mean
TAS.
Mean TAS + Mean Wind ( 275 + 18.27) = 293.3 Kts = Mean GS
Q. Find the Mean GS from the data given below in black:Leg TAS
DIST
W/V
GS
TIME
WIND EFFECT (TIME x W/V)
A-B
160
600
+30
190
3:09
+94.5 Nm (30 x 3:09)
B-C
290
500
-15
275
1:47
-26.75 Nm (-15 x 1:47)
C-D
380
200
+45
425
0:28
+21 Nm (+45 x 0:28)
* Data Calculated
Mean Winds = 88.25/5:26 = +16.24; Mean TAS = (160 x3:09) + (290 x 1:49) + (380 x 0:47) = 1212.
This divided by Time (5:26) = 222.38, Hence Mean GS = 222.38 + 16.24 = 239 Kts
RELATIVE MOTION
Thumb Rule. While solving problems on relative motion, the following must be kept in mind:1. When aircraft are flying in same direction first calculate the relative speed (Difference in the two
also called overtake speed) then divide by distance to obtain time taken to overtake.. For example
Overtake = 40 Kts, Distance = 80 Nm, then time taken to overtake = 80/40 = 2h. Similarly when 4 nm
behind, time taken =76/40 = 1:54 h and time taken when 4 nm ahead after overtaking = 84/40 =2:06 h
2. Time of Crossing = Relative Distance Relative speed (Add speeds when aircraft approaching each
other and subtract when flying one behind the other.
Q. At 0900 h Aircraft X is behind Y by 80 Nm while flying on same track. GS of X= 240 K and that of
Y= 200K. Find when will X overtake Y and when will X be 4nm short & ahead of Y.
A. Example worked out above, time will be 1100 h, 1054 h and 1106h respectively
17
Q. At 0700 h, while flying on same track aircraft A is behind B by 120 Nm. GS of A = 300K; B=250K.
Both are flying to point P which is 1200 Nm from present position of aircraft. Find when will (a) A
overtakes B, (b)A is 5nm short of B, (c) 5nm ahead of B. (d) At what distance from P, A will overtake B.
A. (d) 300 x 2:24 = 720 Nm, hence distance from P =1200-720 = 480 Nm.
Q. At 0900, Flying on same track aircraft P is behind Q by 50 nm. GS of P = 200K, Q=160K, Find (a) P
will overtake Q, (b)P will be 6nm short of Q, (c) 6nm ahead of B.
A. Overtake = 40 K, Distance = 50 Nm, (a) Time taken to overtake = 50/40 = 1:15, i.e. 1015 h. (b)
44/40 = 1:06 h = 1006 (c) 54/40 = 1:21 = 1021 h.
Q. At 500nm from destination aircraft is asked to delay ETA by 8 min. At what time and distance
should aircraft reduce speed to 150 K if it was flying at 180 K. Present time is 1200h.
A. In this problem we need to determine:(a)
(b)
(c)
(d)
(e)
Original ETA
Revised ETA
New distance covered with revised speed
Time to drop speed
Distance to drop speed.
Original Speed = 180 K, Revised Speed = 150 K, Original ETA = 500/180 =02:47 = 1447h, Revised
ETA = 02:46 + 0:08 =02:54 = 1455h. New distance with revised speed = 150 x 2:54 = 435 nm.
If aircraft was at 435 nm from destination, it would have reached destination at correct ETA, the balance
65 Nm (500-435 = 65Nm) can be construed as if one aircraft behind the other at higher speed at 500 nm
and overtakes at 435 Nm with an overtake of 180 -150 = 30 Kts. Now distance = 65, Overtake = 30,
time taken = 65/30 = 2:10 h, (a) 1410h (b) Distance to drop speed = 2:10 x 180 = 390 Nm
Q. At 800nm from destination aircraft is asked to delay ETA by 15 min. At what time and distance
should aircraft reduce speed to 360 K if it was flying at 420 K. Present time is 1200h.
A. Original ETA = 800/420 =1:54, Revised ETA = 1:54 + 0:15 = 2:09, New Distance with revised
Speed = 360 x 2:09 =774 Nm, Distance = 800 -774 = 26 Nm, Overtake = 60, Time = 26/60 = 0:26 (time
to drop speed) i.e 1226h. Distance to drop speed = 0:26 x 420 = 182 Nm. Hence 800- 182 = 618 Nm
Q. At 600 nm from destination an aircraft is asked to reach early by 10 mins. At what time and distance
it should increase its speed to 240 K from 160 K. Present time is 1200h.
A. Original ETA =600/160 = 3:45, Revised ETA = 3:45 -0:10 = 3:35, New Distance with revised speed
= 3:35 x 240 = 860 nm. This can be compared to an aircraft with overtake of 80 kts behind by 260 Nm
260 Nm
600 Nm
Time taken to cover 260 Nm at 80 K = 260/80 = 3:15
At time =1200 + 03:15 = 1515 (Time to increase speed); Distance to increase speed = 3:15 x 160 = 520
## The One-in-Sixty Rule
18
1. The one-in-sixty rule is based upon the fact that one nautical mile subtends an angle of one degree at
distance of 60 nautical miles, i.e. 5 miles subtend 5 degrees etc.
One-in-Sixty Rule.
2. In applying the rule, the triangle relevant to the navigational problems is identified, and the ratio of
the length of the long side to 60 is established. This ratio may then be applied to the angle to reveal the
length of the side opposite to its or conversely, to the opposite side to reveal the angle it subtends.
Track error =
60 x 3 = 9
20
## Heading correction at B, (a) to destination.
=
=
=
=
(b)
< EBD
60 x 3
20
9 + 3
12
+ < BDA
+ 60 x 3
60
=
2x
=
=
2 x 9
18 (Heading altered back a C)
19
Examples on 1:60 rule
= S/R x 60 (to be used when is < 20)
1. After flying for 240 nm an aircraft is 12 nm right of track. What is the drift.
Dist =240 nm
12 nm off track
= S/R x 60
=(12/240) x 60 = 3
Drift = 3 S
2. After flying for 480 nm aircraft is 20 nm port of track. If remaining distance to destination is 300
nm, what is approx heading to reach destination if ac was flying a heading of 045
A.
480 Nm
3
300 Nm
4
20 Nm off track
3
045 Heading to Alter = 045 + 3 + 4 = 052
3. After 2 hours at GS 180 K, aircraft is 12 nm left of track. If remaining distance is one hr at same GS,
Find drift if ac was flying a course of 200
A. Distance covered = 360 nm, 12 nm port of track hence drift = (12/360) x 60 = 2
20
## SOLAR SYSTEM : TIME
1. The Solar System consists of the Sun, nine major planets of which the Earth is one, and about 2,000
minor planets or asteroids. All members of the solar system are controlled by the Sun which is distinguished by its
immense size and its radiation of light and heat ; for all practical purposes, it may be considered as the
stationary centre round which all the planets revolve.
2. Unlike the Sun, the planets and their satellites are not self-luminous, but reveal their presence by
reflecting the Suns light. The planets revolve about the Sun in elliptical orbits, each one takes a period
of time about the job: Mercury takes 88 days, for example, while Pluto which is rather a long way from
the parent body, is thought to take about 248 years. The planetary satellites in the meantime are
3. Certain laws relating to the motion of planets in their orbits were evolved by the astronomer Kepler,
who died in abject poverty as a reward.
(a) Each planet moves in an ellipse, with the Sun at one end of its foci.
(b) The radius vector of any planet sweeps out equal areas in equal intervals of time.
These are the important laws for our purpose in studying the Earths motion, as we shall see.
4. The Earth rotates on its axis in a West to East direction, resulting in day and night. It revolves round
the Sun along a path or orbit which is inclined to the Earths axis at about 66 , resulting in the seasons
of the year. When the Earth is inclined towards the sun, we get the Summer Solistice (about June 21);
when the axis is away from the Sun, we get the Winter Solistice (about Dec 22). When the Earths axis is
at right angle to the Sun, day and nights are equal the Spring and
Fig 1
Autumn Equinox (March 21 and Sept 23). The point where the planet is nearest to the Sun is called
perihelion, and where farthest aphelion; it is worth noting that in obeying Keplers second law, the speed
of the Earth at perihelion is faster along its orbit than at aphelion.
5. Orbital velocity of the earth is not constant during its orbit, velocity is more when earth is closer to
the sun and minimum when it is the farthest.
21
6. Earth rotates around its axis and revolves around the sun. Rotation gives us day and night, revolution
gives us the year and inclination of the earths axis in its plane of rotation gives the seasons.
7. Inclination of the earth from its axis 23 and 66 from the plane of rotation.
8. Position of sun varies from 23 N to 23 S. This is called Declination (latitude of any heavenly
body with respect to an observer). The northern most point corresponds to Tropic of Cancer and
southernmost Tropic of Capricorn.
Q. Sun will appear at the same latitude (a) once a year (b) twice a tear (c) every day (d) none of the
above. A. (b)
9. The position when the earth is nearest to sun is called Perihelion and furthermost is called Aphelion.
10. The position when the earth is equidistant from the sun is called Equinox.
Q. At what position of the sun you will have equal Day and Night?
A. At Equinox, 21 Mar & 23 Sep.
11. Year. There are two types of Year, Sidereal and Tropical. Sidereal Year is the time interval elapsed
between two successive conjunctions of earth, sun and a fixed point in space. Tropical Year is time
interval elapsed between two successive conjunction of earth, sun and a fixed point in Aries. This is also
known as Calendar Year.
12. Calendar Year. It takes 365 days 5hours 48 min 42 sec for the earth to go around the sun. Thus
every 4 years adds to one day extra which is compensated by the leap year. Every 100 year is not a leap
year. To compensate for 11 min 18 sec every 400 year is a leap year.
13. Sideral Day. Its the time interval elapsed between two successive transits of a fixed point in space
over an observer meridian or its time interval elapsed between two successive transits of a fixed point in
Aries over an observer meridian. It is 23h 56 min since taken with reference to a star wherein the
revolution and rotation of the earth does not matter.
14. Apparent Solar Day. It is the time interval elapsed between two successive conjunction of true
Sun in space over an observer meridian.
15. Mean Solar Day. It is with respect to an imaginary Sun which goes around the earth nover equator
at a constant velocity of 15/hr.
S Sidereal it is not wrt Sun 23:56
A Apparent not fixed due to orbit and revolution of earth around the sun 23:44 to 24:14
M Mean At a constant velocity of 15/hr.
Twilight
16. When the Sun is below the horizon, an observer will still receive light which has been reflected and
scattered by the atmosphere. It is divided into three stages; Astronomical (Sun 12 to 18 below
Horizon. It is completely dark with no natural light at 18) ; Nautical (6 - 12 below the Horizon,
and has to do with the sea horizon being indistinct, and artificial light is still required) and Civil
Twilight when the Suns centre is actually between 1 and 6 below the horizon, when work is
possible without artificial light, and the stars are nor clearly visible. This last is the one we are concerned
with.
22
TIME
There are of four types namely, LMT, UTC, Zone Time and Standard Time.
Local Mean Time (LMT) Its the time kept with respect to position of the sun at anti-meridian of an
observer. At places east of any observer the LMT will be ahead and west of observer LMT will be
behind due to earths rotation.
Q. LMT at 35E is 1300h. Find LMT at (a) 102E (b) 40W
A. (a) dlong/15 = (102-35)/15 = 4h 28 m, hence LMT at 102E = 1300 + 4:28 = 1728 (b) dlong =
(75/15) =5h, hence LMT = 1300- 5 = 0800h
Coordinated Universal Time (UTC) It is the LMT prevailing at prime meridian or time kept with
respect to antemeridian of Prime Meridian (observer is sitting at Prime Meridian).
Q. LMT at 000E is 1200h on 28 Feb 04. What will be the UTC at 180W?
A. UTC WILL NOT CHANGE AT ANY LONGITUDE IT REMAINS THE SAME.
Q. LMT at 40N 60E is 1100 h. Find (a) UTC (b) LMT at 60S 120E (c) 60S 30W.
A. (a) UTC at 60E = LMT - dlong/15 = 1100 -60/15 = 0700h, (b) UTC = LMT 120E dlong/15 or
LMT 120E = UTC +120/15 =0700 + 8h = 1500 h (c) UTC = LMT 30W + dlong/15 or LMT 30W =
UTC 30/15 = 0700 2 = 0500 h
Q. An ac takes from place X (30N 170W) for Y (50S 160E). Total flight time is 08 Hrs. Time of
departure is 2200h on 06 Jun (LMT). Find ETA at destination in LMT.
A. UTC = LMT + 170/15 = 2200 + 11h20m = 0920 (07 Jun), After 8 hrs of flying UTC is 0920+8
=1720 (07 Jun), Now LMT (160W) =UTC + C = 1720 (07) + 160/15 =1720 +10:40 = 0400 (08 Jun)
Q. LMT at 45N 100E on 17 May is 0512. Find UTC & LMT at 60N 120W.
A. UTC = LMT (100E) dlong/15 = 0512 6:40 = 22:32 (16 May), UTC = LMT (120W) + dlong/15 =
LMT+8 or LMT = 22:32 - 8h = 14:32 (16 May)
ZONE TIME
The earth is divided into 24 hr zones, alphabetically assigned, beginning from A to Z except I &
O The longitudes on earth measuring 360 are divided into 24 zones, each of 15 corresponding to 1
hour of time. The zones east of prime meridian are assigned negative signs while zones lying west of
Prime meridians are assigned positive sign. Each zone of 15 is further divided into 7 either side of
the prime meridian which corresponds to 30 mins of time. For example India lies at 82 30 which
when divided by 15 gives us 5h30m and that is the time we are ahead of UTC.
23
Zone Number. It is a number which is to be added algebraically in zone to get UTC. For example if at
82E the zone time is 1200, then UTC = ZoneTime Zone Number = 1200 -82/5 (5)=0700. Thumb Rule
is divide the longitude by 15 if remainder is 7.5 use lower zone else use higher zone.
Q. Find the zone number of (a) 120W (b) 127.5W (c) 130E
A. (a) 120/15 = +8 (since West) (b) 127.5/15 = +8 Remainder 7.5 (hence same zone) (c) 130/15 = -8
Remainder 10, Hence higher Zone i.e -9
Q. At 160E difference between LMT and Zone Time is (a) LMT will be ahead by 40 min than zone
time (b) LMT will be behind by 40 min (c) LMT will be ahead by 20 min (d) LMT will be behind
by 20 min.
A. Zone Number = 160/15 = -10 + Rem 10 hence = -11. LMT at 160E =160/15 = 10h 40m. Zone Time
=1100 hrs LMT = 10h40m hence 20 minutes behind time (d) is the correct choice.
Standard Time Time maintained with respect to a specific meridian or longitude is called Standard
Time. IST is maintained with respect to 82 30 longitude.
Q. Find LMT, UTC, Zone Time, IST at 8250N 8345E.
A. UTC = LMT C or LMT =UTC + 83 45/15 =UTC + 5h 35m or 0535 h (since UTC =0000) Zone
Time =0600, IST =0530h UTC = 0000
International Date Line (IDL).
When traveling Westward from
Greenwich, an observer would eventually arrive at longitude 17959W,
where the LMT is about to become 12 hours less than UTC. An
observer traveling Eastward from Greenwich would eventually arrive at
17959E where the LMT is about to become 12 hours more than UTC.
Thus there is a full day of 24 hours difference between the two travelers,
although they are both about to cross the same meridian. When the antemeridian of Greenwich is crossed, one day is gained or lost, depending
on the direction of travel: the Dateline is the actual line where the
change is made, and is mainly the 180 meridian, with some slight
divergences to accommodate certain groups of South Sea Islands and
regions of Eastern Siberia. The problem readily resolves itself in flying - your watch is always on UTC:
the place whos Standard Time you want is listed in the Air Almanac: apply the correction to , and the
date will take care of itself.
GMT
## Going on dateline from,
East to West gain a day subtract a date
West to East lose a day add a date
24
Prime Meridian
0600 UTC Dec 10
## Q. While crossing IDL from East to West
(b) LMT will be behind, date will be ahead.
(c) LMT will be behind, date will be behind
(d) LMT will be ahead, date will be behind
A. (d)
Twilight Period. It is the period before sunrise and after sunset when diffused light of Sun is available..
Sensible Horizon. Horizon which is visible to the naked eye.
Visible Horizon. A horizon which is not visible is called visible horizon. It is belo the sensible horizon.
Note. When a body rises above the visible horizon it is said to be visible and it is said to be set when it
is below the visible horizon.
Twilight period in Air almanac is with respect to Civil Twilight.
Q. Sunrise and moonrise table on Air Almanac are given in (a) UTC (b) Zone time (c) LMT. A. (c).
25
Procedure to adopt for intercepting a final track radial is:(a) Angle should be either 30/60/90 with respect to the final track/radial.
(b) NO DRIFT IS TO BE APPLIED DURING INTERCEPT.
(c) To determine the angle of intercept, find how many degrees the ac has to turn to intercept the
radial. Double the number of degrees to turn and the figure closest to 30/60/90 will be the
intercept angle e.g. if the number of degrees to turn is 25 then 25 x 2 = 50 which is closer to
60 intercept.
Q1. Aircraft is approaching Station on a radial 180is asked to approach on radial 155. Find out (a)
Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.
(a) Difference = 25 x 2 = 50 Hence 60
(b) On radial 155 Hdg to Station is 335 + 60 = 035
(c) Aircraft on Hdg 000, hence degree to turn = 35
360
355
## (d) RBI will read 60 to left i.e. 300
Q2. Aircraft is approaching Station on a radial 150is asked to approach on radial 360. Find out (a)
Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.
A. (a) 60 (b) 300 (c) 30 (d) 060
Q3. Aircraft is homing on to a Station on a radial 010is asked to approach on radial 330. Find out (a)
Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading
A. (a) 90 (b) 240 (c) 50 (d) 270
Q4. Aircraft on outbound radial 090 Aircraft is asked to track out on a radial 120. Find out (a)
Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading (e) Which side to turn.
A. (a) 60 (b) 180 (c) 90 (d) 120 (e) Right
Q5. Aircraft is approaching Station on a radial 090 with 10 S drift is asked to approach on radial 110.
Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.
A. (a) 60 (b) 230 (c) 30 (d) 060
Q6. Aircraft is approaching Station on Hdg 270 with 10 P drift is asked to approach on radial 075.
Find out (a) Intercept Angle (b) Hdg to Roll out (c) Degrees to turn (d) RBI reading.
A. (a) 30 (b) 285 (c) 15 (d) 330
26
Payload. It is the load which can be carried in the form of passengers and cargo.
## TAKE OFF WT (TOW)
NEWS PAPER + CATERING
## ZERO FUEL WT (ZFW)
DRY OPERATING WT
CREW + BAGGAGE
BASIC WT OR APS WT
## AIRFRAME + ENGINE + AVIONICS
MANUFACTURERS WT
MTOW. It is the Take Off weight given by the manufacturer which cannot be exceeded in any
circumstances. This caters for the best operating conditions i.e. runway length, elevation, density
altitude, runway gradient, runway condition and winds etc. It is also known as Max Gross take off wt.
RTOW. (Restricted/regulated/rated) This is the take off weight restricted due to prevailing conditions
at the places of departure.
TOW
FOB
TOW = BASIC WT + PAYLOAD + FUEL ON BOARD (FOB)
PAY LOAD = TOW (BASIC WT + FOB)
Basic Wt
## MZFW = PAYLOAD + BASIC WT
MLW (Max Landing Wt) It is the maximum weight at which a landing can be made at a destination
without imposing any structural damage to the aircraft.
MZFW (Max Zero Fuel Wt). When wing tanks are empty there is a maximum permissible weight of
an aircraft including all its contents. Exceeding this weight causes unacceptable load to the structure of
the aircraft. Above this weight, if any load is taken onboard it can be fuel only.
Numericals on Payload. To solving any problem on Payload the following procedure is adopted:Step 1. Make a table as given below and enter relevant information as given in the problem:MTOW
RTOW
MLW
+FF (Flight Fuel)
= TOW
MZFW
+FOB
=TOW
27
Choose the lowest value obtained out of MTOW/RTOW/MLW OR MZFW. Then calculate
Payload by substituting this value of TOW in Payload = MTOW (BASIC WT + FOB)
Q. MTOW = 83000 lbs, MLW = 66000 lbs, Basic Wt = 52000, FF = 20000 lbs, Reserve = 2800 lbs.
A.
MTOW
83000*
RTOW
MLW
66000 +17200
83200
MZFW
+ FOB
=TOW
* LOWEST VALUE
PAYLOAD = TOW (BASIC WT + FOB) = 83000 - (52000 + 20000) = 11000 lbs
Q. MTOW = 82000 lbs, MLW = 64500 lbs, Basic Wt = 50000, FOB = 20000 lbs, Reserve = 3000 lbs.
A.
MTOW
82000
RTOW
MLW
64500 +16000
=80500*
MZFW
+ FOB
=TOW
* LOWEST VALUE
PAYLOAD = TOW (BASIC WT + FOB) = 80500-(50000 + 19000) = 11500 lbs
Q. In the above question can you carry additional fuel without affecting payload?
A. Yes, (82500-80500=1500) but the fuel carried has to be consumed (burn off/dump) prior to landing,
Q. MTOW = 120000 lbs, MLW = 90000 lbs, MZFW = 85000, Basic Wt = 76400, Trip Fuel = 15000,
Reserve = 2000 lbs. Find Payload.
A.
MTOW
120000
RTOW
MLW
90000 +15000
=104000
MZFW
85000 + 17000
=102000*
* LOWEST VALUE
PAYLOAD = TOW (BASIC WT + FOB) = 10200 - (76400 + 17000) = 8600 lbs
Q. In the above question find payload if Flight Fuel is reduced by 1000 lbs and increased by same
amount?
A. Payload in both cases will remain the same since value of MZFW has been applied.
Q. MTOW = 20000 lbs, MLW = 18000 lbs, MZFW = 17000, Basic Wt = 14000, Trip Fuel = 3000,
Reserve = 1600 lbs Find (a) Payload (b) Payload if aircraft consumed 700 lbs reserve before landing
(c) Find Payload if FOB is reduced by 700 lbs.
A.
MTOW
20000*
RTOW
MLW
18000 + 3000
=21000
MZFW
17000 + 4600
=21600
* LOWEST VALUE
PAYLOAD = TOW (BASIC WT + FOB) = 20000 - (14000 + 4600) = 1400 lbs
(a) 1400 lbs (b) Will remain same (c) Payload can be increased by 700 lbs (1400+700) =2100
28
Q. Fuel Consumption = 120 lbs/hr; MTOW = 7150 lbs, MLW = 6900 lbs, MZFW = 6150 lbs, Basic Wt
= 5000 lbs, Reserve = 160 lbs. Dist =960 Nm, TAS=180k, Head Winds of 20 Kts. Find payload in NIL
wind conditions.
A. Dist = 960 nm; TAS = 180 Kts, Time = 5.33 x (FF = 120 lbs/hr) = 640 lbs.
MTOW
RTOW
MLW
MZFW
7150
6900 + 640
6150 + 800
=7540
=6950*
* LOWEST VALUE
PAYLOAD = MTOW (BASIC WT + FOB) = 6950 - (5000 + 800) = 1150 lbs
Q. A flight is to be made from M to N and return to M carrying max payload in each direction. Fuel is
not available at N. Distance M to N =80 Nm, Mean GS M to N = 70 kts, Mean GS N to M = 110 Kts,
Mean Fuel consumption = 410 Kg/hr, MTOW at M = 6180 kg, MLW at M = 5740 kg MTOW at N =
5800 kg, MLW at M = 5460 kg, MZFW = 5180, Basic Wt = 4400 kgs, Res Fuel = 250 kgs. Calculate
(a) Max payload which can be carried from M to N and from N to M.
A.
## Total Fuel = Fuel reqd from M to N + Fuel reqd from N to M + Reserve
Fuel reqd: M to N = 80/70 = 1.42 x 410 = 468, Fuel reqd: N to M = 80/110 = 0 .727 x 410 = 298,
FF = 468 + 298 = 766 kgs, Reserve = 250 kgs, Total Fuel = 1016 Kgs.
MTOW
6180*
RTOW
MLW
5740 + 468
=6208
MZFW
5100 + 1016
=6196
* LOWEST VALUE
PAYLOAD = MTOW (BASIC WT + FOB) = 6180 - (4400 + 1016) = 764 kgs
MTOW
5800*
RTOW
MLW
5460 + 298
=5758
MZFW
5180 + 548
=5728
* LOWEST VALUE
PAYLOAD = MTOW (BASIC WT + FOB) = 5728 - (4400 + 548) = 780 kgs
Q. Given MTOW = 34,500 kgs, MZFW = 28,000, MLW = 31,000, Empty Wt = 17,500 kgs, TAS = 350
Kts, Fuel Consumption = 1450 Kg/hr. Reserve Fuel 1200 kgs for all flghts (assume not used) Fuel Tank
Capacity = 10, 500 kgs. Find (a) Max Payload (b) In NIL wind condition distance upto which above
payload can be carried. (c) Max distance you can fly in NIL winds. (d) What payload you can carry in
part (c).
A. Max payload = MZFW Basic Wt = 28000-17500 = 10500 kgs. TOW =10500 + 17500 +FOB or
34500=28500 + FOB or FOB = 6500. Hence FF = 6500-1200 = 5300 kgs. Fuel Flow = 1450 kg/hr, TAS
= 350 hence max distance = 5300/1450 x 350 = 1279 Nm. Tank capacity = 10500 kgs. Hence 10500
6500 = 4000 kgs of fuel can be carried in lieu of payload. So total fuel available = 5300 + 4000 = 9300
kgs. Hence max distance = 9300/1450 x 350 = 2245 Nm and pay load would reduce by 4000 kgs in
lieu of fuel. Payload = 10500-4000 = 6500 kgs.
29
CONVERSION OF UNITS
1.
2.
## Imperial Gallon (UK Gal or IG) x SG x 10 = Pounds (lbs)
3.
Kg x 2.2046 = lbs
4.
5.
## 100 l = 22 IG = 26.4 USG.
Problems
Q. Convert 100 USG to (a) litres (b) Kgs (c) lbs (d) IG. Given SG = 0.78
A. (a)
(b)
(c)
(d)
## 1 USG = 100/26.4 L = 3.785, 100 USG = 378.7 l Litres
Kgs = l x SG also 100 USG = 378.71 litre Therefore 100 USG = 378.5 x 0.78 = 295 Kgs
100 USG = 295 kgs or 295 x 2.2046 = 650 lbs
100 USG = 100/1.2 = 83.3 IG
Q. In the following table find the most fuel efficient figure when winds = -20 and SG = 0.8
(a)
(b)
(c)
(d)
(e)
TAS
200
220
240
260
285
## Fuel Cons (Gals/Hr)
250
265
280
295
315
GS/FC
180/250 = 0.72
200/265 = 0.75
220/280 = 0.78
240/295 = 0.81
265/315 = 0.84 (Most Efficient)
A.
GS/FC = Nm/Hr Gals/Hr = Nm/ Gals ie Nm per gallons, the highest figure will be most
efficient. Winds are 20 kts Head Wind, GS will be TAS 20 in Kts. Working is on the above tale in red.
Q. In the following table find the most fuel efficient figure when SG = 0.8
(a)
(b)
(c)
(d)
(e)
(f)
TAS
180
240
220
160
190
200
Fuel Cons
1.25 USG/Nm
4.1 IG/min
815 Kg/hr
12.7 L/min
1.13 IG/min
13.41 Kg/m
FC (IG/hr)
1.25 x 180 =225 USG/Hr = 225/1.2 = 187.5 IG/hr
4.1 IG/min x 60 = 246 IG/hr
815/0.8 = 1018.75 l/hr = 1018.72 x 0.22 = 224.125 IG/hr
12.7 x 60 = 762 l/hr x 0.22 =167.64 IG/hr(Most Efficient)
1.13 x 190 = 214.7 IG/hr
13.41 x 60 =804.6 Kg/hr /0.8 =1005.75 l/h = 221.6 IG/hr
A. In this problem convert all Fuel Consumption figures to IG/hr then compare which is the lowest as
worked out in red in the table above.
30
Q. In the following table find the most fuel efficient figure when SG = 0.8
(a)
(b)
(c)
(d)
(e)
(f)
TAS
180
240
220
160
190
200
## Fuel Cons (IG/hr)
187.5
246
224.125
167.64
214.7
221.65
FC (IG/Nm)
187.5/180 = 1.0416
246/240 = 1.025
224.125/220 = 1.018 (Most Efficient)
167/160 = 1.0477
214.7/190 = 1.1263
221.65/200 = 1.108
A. Convert all figures into IG/Nm, the least figure will give the most fuel efficient figure as worked in
in red in the table above.
Q. Fuel efficiency is 10.13 kgs/nm, TAS = 310 kts, Winds = +45, SG = 0.81. Find Flow in IG/hr.
A. 10.13 Kg/Nm = 10.13/.81 = 12.506 l/Nm = 12.506 x 355 (GS) = 4439.63 l/hr = 4439.63 x 0.22 =
976.5 IG/hr
31
CRITICAL POINT (CP)
Critical Point. It is a point in between two places from where it takes same time to reach either of the
point or it is a point enroute from which it takes equal time to either come back or go the destination. It
is calculated with one engine failed or switched off. It is also called Equitime Point.
1200 Nm
Mid Point
CP
## Increased GS Home (TAS + WV)
B W/V 090/20
Reduced GS Out bound (TAS WV)
If distance D between two points A & B, TAS (with one engine failed) and W/V are known, GS out
(O) and GS Home (H) can be calculated and the figure of PNR can be arrived at by substituting these
values in the equation:Distance to CP = DH/(O +H) where D = Total Distance, O = GS outbound with one engine failed
and H = GS Home with one engine failed.
Thumb Rule:When you find Dist to CP, always calculate GS Out and GS Home with reduced TAS
Time to CP = Distance to CP/ GS out with all engines running unless specified.
Example. Distance A to B = 1200 nm; Tr = 090, W/V = 090/20, TAS (4/3 engines) = 180/150* Kts,
Find Distance & Time to CP.
Always draw a rough diagram and a table as indicated below before attempting any problem:Refer figure above
CP to A
TR
270
TAS
150*
W/V
090/20
GS
170
CP to B
090
150*
090/20
130
A to CP
090
180
090/20
160
To solve the problem (a) Calculate the GS from Nav Computer and enter the figures obtained.
(b) Substitute these values in the Distance to CP and Time to CP formula (use bracket function in
the calculator it is that much faster and easier to obtain the correct final figure)
Distance to CP = DH/(O +H) = (1200 x 170)/(130 + 170) = 680 Nm
Time to CP = Distance to CP/ GS out = 680/160 = 04h:15m
32
Q1. D = 2000 nm, Tr = 270, W/V = 090/25, TAS (4/3) = 330/270 Kts. Find Distance & Time to CP.
A.
090/25
B
CP
CP to A
TR
090
TAS
270*
W/V
090/25
GS
245
CP to B
270
270*
090/25
295
A to CP
270
330
090/25
355
## Distance to CP = DH/(O +H) = (2000 x 245)/(295 + 245) = 907 Nm
Time to CP = Distance to CP/ GS out = 907/355 = 02h:33m
POINT OF NO RETURN (PNR)
Point of No Return (PNR). It is a point at maximum distance removed from base upto which an
aircraft can fly and still be able to return within safe endurance of the aircraft. It is calculated primarily
to cater for non-availability of destination. This is purely a function of endurance which is given by the
equation Endurance = (FOB Reserve)/ Fuel consumption. The distance to PNR is calculated by the
formula :Distance to PNR = EOH/(O + H), where E = Endurance, O = GS outbound (with all engines
operating) and H = GS Home (with all engines operating) unless specified.
PNR
GS Out
A GS Home
W/V
Example
Q2. D = 2000 nm, Tr = 270, W/V = 090/25, TAS (4/3) = 330/270 Kts. Fuel on board = 1000 kgs,
Reserve = 200 kgs, Fuel Consumption (4/3) = 180/150 kg/hr Find Distance & Time to PNR.
A.
O
A to PNR
TR
270
TAS
330
W/V
090/25
GS
355
PNR to A
090
330
090/25
305
To calculate Distance to PNR insert values of TAS with all 4 engines running in the table shown
above. Next calculate the Endurance by determining Flight Fuel = (FOB Reserve), divided by
Fuel Consumption (all engines operating), then calculate the GS and enter values in Dist to PNR
formula.
33
(a) Endurance = (1000 -200)/180 = 4h:26m (b) Dist to PNR = (4:26 x 355 x 305)/(355+305)=729 Nm
(c) Time to PNR = 729/355 = 2h: 03m.
STILL AIR RANGE (SAR)
SAR. It is the maximum distance upto which an ac can fly out in NIL wind conditions consuming total
fuel onboard. This is a theoretical figure to cross check whether Flight Plan is executable.
SAR = (FOB x TAS)/ FUEL CONSUMPTION
Q3. Find the SAR in the above question.
FOB = 800, TAS = 330, FC = 180 kg/hr Hence SAR = (800 x 330)/ 180 = 1833 Nm.
Practice on Nav Computer (Given TRK, TAS and W/V, Find GS, Hdg and Drift)
TRK
TAS
W/V
270
190
330/10
350
200
270/25
135
240
050/37
293
137
330/17
* TRK less than HDG Drift is Port and vice versa
GS
185
186
234
123
HDG
273
344
126
297
DRIFT
3 P*
6S
9S
4P
Q4. Dist = 1400 Nm, Tr =090, W/V = 330/18 Kts, TAS (4/3) = (210/180), FC (4/3) = 100/80 Kg/hr
FOB = 1600 kg, Reserve = 150 Kg. Find (a) Dist & Time to CP (b) Dist & Time to PNR (c) SAR.
Assume one engine failed at PNR and ac returns on 3 engines.
A.
H
CP to A
TR
270
TAS
180*
W/V
330/18
GS
188
CP to B
090
180*
330/18
170
A to CP
090
210
330/18
218
## (a) Distance to CP = DH/(O +H) = (1400 x 170)/(170 + 188) = 665 Nm
Time to CP = Distance to CP/ GS out = 665/218 = 03h:02m
(b)
O
A to PNR
TR
090
PNR to A
090
TAS
210
W/V
330/18
GS
218
210
330/18
200
Endurance = (1600 -150)/100 = 14h:30m (b) Dist to PNR = (14:30 x 218 x 200)/(418)=1512 Nm (c)
Time to PNR = 1512/218 = 6h: 56m.
(c) SAR = (FOB xTAS)/FC = (1600 x 210)/100 = 3360 Nm
34
(d) Since one engine has failed we need to know how far we can go with 4 engines and come back with
three engines operating. If X is the fuel consumed till PNR with 4 engines and Y with three engines,
then total fuel consumed = X + Y
Cruise 4 Engines
PNR
A
B
Cruise 3 Engines
A to PNR
TR
090
TAS
210
PNR to A
270
180
W/V GS
Dist FTime
330/18 218 1400 6:25
330/18
170
1400
8:14
FC
100
F Used
642
80
659
## Total Fuel Used = 642 + 659 = 1301 (Adopt Unitary method)
If flight fuel is 1301 then distance to PNR is 1400
If flight fuel is 1 then distance to PNR is 1400/1301
If flight fuel is 1450 then distance to PNR is (1400 x 1450)/1301 = 1560 Nm
Hence Dist to PNR = 1560 Nm and time taken = 1560/218 = 7h:09m
Q5. Dist = 2000 Nm, Tr =330, W/V = 160/37 Kts, TAS (4/3) = (300/250), FC (4/3) = 200/180 Kg/hr
FOB = 2400 kg, Reserve = 500 Kg. Find (a) Dist & Time to CP (b) Dist & Time to PNR (c) SAR.
Assume one engine failed at PNR and ac returns on 3 engines.
A.
H
CP to A
TR
150
TAS
250*
W/V
160/37
GS
213
CP to B
330
250*
160/37
286
A to CP
330
300
160/37
336
## (a) Distance to CP = DH/(O +H) = (2000 x 286)/(286 + 213) = 940 Nm
Time to CP = Distance to CP/ GS out = 940/336 = 02h:48m
0n 4 Engine
A
940 Nm
(b)
CP 1060 Nm
TR
330
TAS
300
W/V
160/37
B
GS DIST
336 1400
TIME
4:10
FC
200
F USED
1190
A to PNR
PNR to A 150
250*
160/37
213
1690
If fuel is 2880 PNR is 2000, for fuel 1900 PNR = 2000/2880 x 1900 = 1320 Nm/3:55m
35
Endurance = (1600 -150)/100 = 14h:30m (b) Dist to PNR = (14:30 x 218 x 200)/(418)=1512 Nm (c)
Time to PNR = 1512/218 = 6h: 56m.
(c) SAR = (FOB x TAS)/FC = (1600 x 210)/100 = 3360 Nm
Q6. Dist = 1350 Nm, TR =270 W/V 270/25 upto CP, thereafter 350/38. TAS (4/3) = 200/180, FC
(4/3) = 110/90, FOB =1800, Res = 200. Find DCP, TCP, DPNR, TPNR & SAR (assume one engine
failed at PNR)
A.
H
CP to A
TR
090
TAS
180*
W/V
270/25
GS
205
CP to B
270
180*
350/38
170
A to CP
270
200
270/25
175
## (a) Distance to CP = DH/(O +H) = (1350 x 205)/(205 + 170) = 740 Nm
Time to CP = Distance to CP/ GS out = 740/175 = 04h:14m
(b)
O
A to CP
TR
270
TAS
200
W/V
270/25
GS DIST TIME
175 740
04:14
CP to A
090
180
270/25
205
740
FC F USED
110 465
03:37
90
325
Total Fuel Used : 790; Bal = 810
PNR
TR
CP to B (4 Eng) 270
B to CP (3 Eng) 270
TAS
200
180
W/V
350/38
350/38
GS
190
183
DIST TIME
FC
610
03:13
110
610
03:10
80
Total Fuel Used :
F USED
353
300
653
## If flight fuel is 653 then distance to PNR is 610
If flight fuel is 1 then distance to PNR is 610/653
If flight fuel is 810 then distance to PNR is (610 x 653)/810 = 757 Nm
Hence Dist to PNR = 757 Nm and time taken = 757/190 = 3h:59m
DPNR = 740 + 757 = 1497 Nm, TPNR = 4:14 + 3:59 = 8h:13m
(c) SAR = (FOB x TAS)/FC = (1800 x 200)/200 = 3273 Nm
Q7. Dist = 1250 Nm, TR =090 W/V 280/20 upto CP, thereafter 330/20. TAS (4/3) = 180/150, FC
(4/3) = 110/90, FOB =1800, Res = 700. Find DCP, TCP, DPNR &, TPNR (assume one engine failed
at PNR)
A.
H
CP to A
TR
270
TAS
150*
W/V
280/20
GS
130
36
O
CP to B
090
150*
330/20
159
A to CP
090
180
280/20
200
## (a) Distance to CP = DH/(O +H) = (1250 x 130)/(159 + 130) = 562 Nm
Time to CP = Distance to CP/ GS out = 562/200 = 02h:49m
(b)
O
A to CP
TR
090
TAS
180
W/V
280/20
GS
200
DIST TIME
562
02:49
FC
110
F USED
309
CP to A
270
150
280/20
130
562
04:19
90
389
Total Fuel Used : 698; Bal = 402
TR
CP to B (4 Eng) 090
B to CP (3 Eng) 270
TAS
180
150
W/V
330/20
330/20
GS
189
139
DIST TIME
FC
688
03:38
110
688
04:56
90
Total Fuel Used :
PNR
F USED
400
445
845
## If flight fuel is 845 then distance to PNR is 688
If flight fuel is 1 then distance to PNR is 688/845
If flight fuel is 402 then distance to PNR is (688 x 402)/845 = 327 Nm
Hence Dist to PNR = 327 Nm and time taken = 327/189 = 1h:44m
DPNR = 562 + 327 = 889 Nm, TPNR = 2:49 + 1:44 = 4h:33m
Q8. TR=250, W/V 270/30, TAS = 210, FOB = 1200, PNR =785 Nm. Find (a) FC (b) If CP is
reached 45 minutes before PNR, find excess fuel carried.
PNR
O
A to PNR
TR
250
TAS
210
W/V
270/30
GS
181
TIME
4:19
PNR to A
070
210
270/30
238
3:18
TOTAL 7:37
Time to CP = Time to PNR 0:45 = 7:37 0:45 = 6:52, Distance traveled in 0:45 = 181 x 0:45 =136
Nm, Hence distance to CP = 785-136 = 649; CP = DH/(O+H) or 649 = D x 238 /419 or D =1143. Time
taken to cover 1143 Nm at 181 K = 1143/181 = 6h 19 m
PNR = EOH/(O+ H) or 785= (E x 181 x 238)/ 419 or E = (785 x 419)/ (181x 238) = 7h 38m
E = FOB/FC or FC = 1200/7:38 =157gph. Fuel Consumed for Flight of 6h 19 m = 6:19 x 157 = 991
gals. Fuel Carried = 1200, Excess Fuel = 1200-991 = 209
Q9. TR=155, W/V 240/30, TAS = 220, FC = 150 GPH, PNR =1080 Nm. Find (a) FOB (b) If CP is
reached 1:15 minutes before PNR, find excess fuel carried.
PNR
O
A to PNR
TR
155
TAS
220
W/V
240/30
GS
215
37
H
PNR to A
335
220
240/30
221
Time to CP = Time to PNR 1:15, Distance traveled in 1:15 = 215 x 1:15 =269 Nm, Hence distance to
CP = 1080-268 = 811; CP = DH/(O+H) or 811 = D x 221 /436 or D =1599. Time taken to cover 1599
Nm at 215 K = 1599/215 = 7h 26 m
PNR = EOH/(O+ H) or 1080= (E x 215 x 221)/ 436 or E = (785 x 419)/ (181x 238) = 9h 55m
E = FOB/FC or FOB = 9:55 x 150 =1486 lbs. Fuel Consumed for Flight of 7h 26 m = 7:26 x 150 =
1115 gals. Fuel Carried = 1486, Excess Fuel = 1486-1115 = 371 lbs
Q10. Dist = 1450 Nm, TR =132 W/V 260/40 upto CP, thereafter 350/60. TAS (4/3) = 190/160, Find
(a) DCP, TCP, (b) DPNR if Fuel Consumption is increased by 8% (assume one engine failed at
PNR)
A.
H
CP to A
TR
312
TAS
160*
W/V
260/40
GS
132
CP to B
132
160*
350/60
203
A to CP
132
190
260/40
212
## (a) Distance to CP = DH/(O +H) = (1450 x 132)/(203 + 132) = 571 Nm
Time to CP = Distance to CP/ GS out = 571/212 = 02h:42m.
Since fuel is not given, CP & PNR are collocated. Now with 8% increase in fuel consumption there will
be a 8% reduction in distance to PNR. Hence 8% of 571 = 46 Nm, so DPNR = 571-46 =525Nm and
Time to PNR = 525/212 = 2h:26m.
Q11. Dist = 1200 Nm, TR =270 W/V 330/20, TAS (4/3) = 180/150, FOB = 900, RES = 300; FC =110
Gal/hr. FC = 110 gals/hr. Find (a) DCP, TCP, (b) DPNR & TPNR (c) Is fuel sufficient for the
flight, if not, how much less (d) If flight fuel is 981 gals calculate DPNR.
A.
H
CP to A
TR
090
TAS
150*
W/V
330/20
GS
159
CP to B
270
150*
330/20
139
A to CP
270
180
330/40
169
## (a) Distance to CP = DH/(O +H) = (1200 x 159)/(139 + 159) = 640 Nm
Time to CP = Distance to CP/ GS out = 640/169 = 03h:47m.
38
(b) DPNR &TPNR
A to PNR
TR
270
TAS
180
W/V
330/20
GS
169
PNR to A
090
180
330/20
189
Endurance = 600/110 = 5:27, DPNR = 5:27 x 189 x 169/358 =486, TPNR = 486/169 = 2:53
(c)
Time to cover 1200 Nm = 1200/169 = 7h:06m, Fuel required = FC x 7:06 = 781 gals, FOB =600 gals,
hence Fuel Less by 781-600 = 181 gals.
(d)
FOB = 781 + 300 (reserve) =1081, Endurance =781/110 = 7.1 hr, DPNR = 7.1x 169 x 189/358 = 634
Q12. Dist = 1600 Nm, TR =090 W/V 270/30 for first 1000 Nm for remaining distance 030/17, TAS
(4/3) = 220/150, FC (4/3) = 100/80, FOB =1400, Res = 200. Find (a) DCP, TCP, (b) DPNR & TPNR
(assume one engine failure at PNR)
A.
A
1000 nm
W/V 270/30
600 nm
030/17
TR
TAS
W/V
GS
DIST
TIME(min)
A-X (3)
090
190
270/30
220
1000
273
X-B (3)
090
190
030/17
181
600
B-X (3)
270
190
030/17
198
600
182
X-A (3)
270
190
270/30
160
1000
000
A
Subtract
472
-472
375
X
1 98
+177
557
B
000
+577
## CP lies in this Leg
Time taken for an aircraft to reach from A to B is 472 mins with existing winds. Time taken to return
form B to A is 577 mins. At (A, X & B) we can calculate the residual time. The CP will lie in the leg
where values are between negative and positive.
DCP = Distance of leg in which CP lies x 472 /(472 +177) =1000 x 472/649 = 727 Nm
TCP = 727/250 =2h:54m (GS out 4 Eng = 220 +30 K tail wind)
Similar calculations can now be made for PNR with Fuel considerations.
39
PNR
TR TAS W/V
GS
DIST TIME
FC F USED
A-X (4)
090
220 270/30
250
1000
4:00
100
400
X-A (3)
270
190 270/30
160
1000
6:15
80
X-B (4)
090
220 030/17
211
600
2:50
100
500
900
284
B-X (3)
270
190 030/17
198
600
3:01
80
241
525
## 675 (1200 (284+241))
BAL
X
B
PNR lies in this Leg
PNR lies between X & B since Balance of fuel is 300 gals after catering for return from X.
Hence, if Flight Fuel is 525 then PNR is 600 Nm from X
If Flight Fuel is 1 then PNR is 600/525 Nm from X
If Flight Fuel is 300 then PNR is (600 x 300)/525 Nm from X = 343 Nm from X = 1343 Nm from A
Time to cover 343 Nm @ GS =211 =343/211 = 1:37. Hence TPNR = 4:00 + 1:37 = 5:37
Q13. Dist = 2000 Nm, TR =270 W/V 280/22 for first 900 Nm for remaining distance 330/20, TAS
(4/3) = 180/150, FC (4/3) = 100/80, FOB =1700, Res = 200. Find (a) DCP, TCP, (b) DPNR & TPNR
(assume one engine failure at PNR)
A.
A
900 nm
1100 nm
W/V 280/22
330/20
TR
TAS
W/V
GS
DIST
TIME(min)
A-X (3)
270
150
280/22
128
900
422
X-B (3)
270
150
330/20
139
1100
## 475 (422 +475) = 897
B-X (3)
090
150
330/20
159
1100
415
X-A (3)
A-X (4)
090
270
150
180
280/22
172
900
314 (415 + 314) = 729
280/22 & 330/20 = 158/169
0
314
729
A
X
B
Subtract
897
475
000
-897
-161
+729
CP lies in this Leg
Time taken for an aircraft to reach from A to B is 897 mins with existing winds. Time taken to return
form B to A is 729 mins. At (A, X & B) we can calculate the residual time. The CP will lie in the leg
where values are between negative and positive.
40
DCP = Distance of leg in which CP lies x 161 /(729 +161) =(1100 x 161)/890 = 199 + 900 =1099 Nm
TCP = 199/169 + 900/ 158 =1:11 + 5:42 = 6h:53m (GS out 4 Eng = 158 from A-X & 169 from X-B)
Similar calculations can now be made for PNR with Fuel considerations.
PNR
TR TAS W/V
GS
DIST TIME
FC F USED
A-X (4)
270
180 280/22
158
900
5:42
100
570
X-A (3)
090
150 280/22
172
900
5:14
80
X-B (4)
270
180 330/20
169
1100
6:31
100
419
989
651
B-X (3)
090
150 330/20
159
1100
6:55
80
553
1204
## 496 (1700 (651+553))
BAL
B
PNR lies in this Leg
PNR lies between X & B since Balance of fuel is 511 gals after catering for return from X.
Hence, if Flight Fuel is 1204 then PNR is 1100 Nm from X
If Flight Fuel is 1 then PNR is 1100/1204 Nm from X
If Flight Fuel is 511 then PNR is (1100 x 511)/1204 Nm from X = 467 Nm from X = 1367 Nm from A
Time to cover 467 Nm @ GS =169 =467/169 = 2:46. Hence TPNR = 5:42 + 2:46 = 8:28
Q14. An aircraft flies from A-B on Tr = 090 for 600 Nm (W/V 030/20) and then proceeds to
destination C on Tr = 120, D = 900 Nm (W/V 150/35) . TAS (4/3) = 240/210, FC (4/3) = 150/120,
FOB =1600, Res = 200. Find (a) DCP, TCP, (b) DPNR & TPNR (assume one engine failure at
PNR)
A.
A 600 nm
W/V 030/20
900 nm
150/35
A-B (3)
TR
090
TAS
210
W/V
030/20
GS
199
DIST
600
TIME(min)
181
B-C (3)
120
210
150/35
179
900
## 302 (181 +302) = 483
C-B (3)
300
210
150/35
240
900
225
B-A (3)
A-B (4)
270
090
210
240
030/20
219
600
030/20 & 150/35 = 229/209
## 164 (225 + 164) = 389
41
0
A
Subtract
483
-483
164
B
302
-138
389
C
000
+389
CP lies in this Leg
Time taken for an aircraft to reach from A to B is 483 mins with existing winds. Time taken to return
form B to A is 389 mins. At (A, X & B) we can calculate the residual time. The CP will lie in the leg
where values are between negative and positive.
DCP = Distance of leg in which CP lies x 138 /(389 +138) =(900 x 138)/536 = 236 + 600 =836 Nm
TCP = 236/209 + 600/ 229 =1:07 + 2:37 = 3h:45m (GS out 4 Eng = 158 from A-X & 169 from X-B)
Similar calculations can now be made for PNR with Fuel considerations.
PNR
TR TAS W/V
GS
DIST TIME
A-B(4)
090
240 030/20
229
600
2:37
150
393
B-A(3)
270
210 030/20
219
600
2:44
120
B-C (4)
120
240 150/35
209
900
4:18
150
329
722
646
C-B (3)
300
210 150/35
240
900
FC F USED
6:55
120
BAL
678
450
1096
C
## PNR lies in this Leg
PNR lies between B & C since Balance of fuel is 722 gals after catering for return from B
Hence, if Flight Fuel is 1096 then PNR is 600 Nm from B
If Flight Fuel is 1 then PNR is 600/1096 Nm from B
If Flight Fuel is 678 then PNR is (600 x 678)/1096 Nm from B = 558 Nm from B = 1158 Nm from A
Time to cover 558 Nm @ GS of 209 K =558/209 = 2:40. Hence TPNR = 2:37 + 2:40 = 5:17
Q15. On a flight from A to C via B. TAS on 4 engines is 360 K & in case of 3 engines it is 300 K. The
route details are:Stage
Track
Wind Vel
Distance
AB
180
330/ 35 K
900 Nm
BC
210
150/28 K
1400 Nm
(a) Find distance and time to CP (the aircraft is required to return to B or C in case of engine failure).
(b) If FOB is 2200 Kg, Reserve is 200 Kg, fuel consumption 250 Kg/hr (4Engine) and 220 Kg/hr (3
Engine), find Distance & Time to PNR (After take off A is not available and aircraft is to land at B,
assume engine failure at PNR and return is on three engines)
42
A.
A 330/ 35 K B
180/900 Nm
150/28 K
C
210/1400 NM
Between B C
CP to B
TR
030
TAS
300*
W/V
150/28
GS
313
CP to C
210
300*
150/28
B to CP
210
360
150/28
345
A B (4)
180
360
330/35
390
285
## (a) Distance to CP (B-C) = DH/(O +H) = (1400 x 313)/(285 + 313) = 732 Nm
Distance to CP = 732 + 900 = 1632
Time to CP (A-B) = Distance to CP/ GS out = 900/390 = 02h:18m
(B-C) = Distance to CP/ GS out =732/345 = 2:07
Time to CP = 2:18 + 2:07 = 4h:25m
PNR
TR TAS W/V
GS
DIST TIME
A-B(4)
180
360 330/35
390
900
2:18
250
577
B-C(4)
210
360 150/28
345
1400
4:03
250
1014
C-B (3)
300
210 150/28
313
1400
4:28
220
984
1998
FC F USED
BAL
1423 (2000-577)
C
PNR lies in this Leg
PNR lies between B & C since Balance of fuel is 1423 kgs after catering for return from B
Hence, if Flight Fuel is 1998 then PNR is 1400 Nm from B
If Flight Fuel is 1 then PNR is 1400/1998 Nm from B
If Flight Fuel is 1423 then PNR is (1400 x 1423)/1998 from B = 997 Nm from B = 1897 Nm from A
Time to cover 997 Nm @ GS of 390 K =997/390 = 2:53. Hence TPNR = 2:18 + 2:53 = 5h:11m
Q16. An aircraft is to fly from P to R via Q and return to Y via Q in case of engine failure, since P
is not available. TAS on 4 engines is 500 K & in case of 3 engines it is 420 K. The route details are:-
43
Stage
TAS
Wind Vel
Distance
PQ
420
-25
965 Nm
QR
420
-45
900 Nm
RQ
420
+45
900 Nm
QY
420
+30
240 Nm
(a) Find distance and time to CP(the aircraft is required to return to Y via Q in case of engine failure).
(b) If FOB is 38000 Kg, Reserve is 6500 Kg, fuel consumption 6300 Kg/hr (4Engine) and 5600 Kg/hr
(3 Engine), find Distance & Time to PNR (After take off P is not available and aircraft is to land at Y,
assume engine failure at PNR and return is on three engines)
0
P
290
-290
32 (Time to Y) 148
965
Q 900
R
144 (Time to R)
0
240 -112
148
Y CP lies in this Leg
CP
Stage
TAS
Wind Vel
GS
Distance
## Flt Time (min)
PQ
420
-25
395
965 Nm
146
QR
420
-45
375
900 Nm
RQ
420
+45
465
900 Nm
144
290
116
QY
420
+30
450
240 Nm
P Q (4)
500
-25
Q CP (4) 500
-45
TCP = 2:02 + 0:51 = 2h:53m
475
455
965 Nm
388 Nm
32
148
DCP (Q-R) = DH/(O +H) = 900 x 112/ (148 + 112) = 388 Nm from Q
DCP = 965 + 388 = 1353
2:02
0:51
PNR
Stage
TAS
Wind Vel
GS
Distance
Time FC
PQ
500
-25
475
965 Nm
QY
420
+30
450
240 Nm
0:32 5600
QR
500
-45
455
900 Nm
RQ
420
+45
465
900 Nm
F Used
2987
15797
1:59 6300 12461
1:56 5600 10839
23300
## If Flight fuel is 23300 then DPNR is 900 Nm
If Flight fuel is 15703 then DPNR is (900 x 15703)/23300 = 607 Nm from Q
DPNR = 965 + 607 = 1572 Nm Time to cover 607 Nm @ GS 455 K = 607/455 = 1:20
TPNR = 2:02 + 1:20 = 3h:22m
Bal
(31500-15797)
15703
44
CRITICAL POINT (CP)
1. CP is always half way when GS (Home = GS (Out) i.e. O = H, this happens during (a) Nil Winds and
(b) Beam winds.
2. In case of HW, CP will always be more than half way i.e. into wind and in case of tail winds CP will
be less than half way.
3. If HW component increases distance to CP will increase or it will move towards destination or it will
move away from departure point.
4. In case tail wind component increases distance to CP will decrease or it will move closer to place of
departure or away from departure point..
CP ALWAYS MOVES INTO WIND
5. For same HW component if TAS is reduced, distance to CP will increase and vice versa.
6. In case of HW, CP will be more than half way, if HW changes to tail wind, distance to CP will be
less than half way by the corresponding distance if wind component remains same.
Q. With 50K of HW, distance to CP is 1200 Nm. During actual flight wind component was found o be
50 K of tail winds. If total distance is 2000 Nm, new distance to CP will be (a) 1000 (b) 1200 (c) 800
(d) insufficient data cannot be calculated
A (c)
Q. If beam wind component doubles, distance to CP will be ..and time to CP will .(a) same, same
(b) same, decrease (c) decrease, same (d) same, increase. A. (d)
Q. If fuel onboard or Flight Fuel increases distance to CP will remain the same.
PNR
Q. Distance to PNR will be maximum in (a) HW during outbound (b) Tail Wind during outbound (c)
Tail wind during inbound (d) Nil Winds
A. (d) Note. ANY KIND OF WINDS WILL CAUSE PNR TO REDUCE.
Q. With a fuel of 10000 lbs, PNR calculated is 880 Nm, other factors remaining constant, if fuel is
increased to 11000 lb, the distance to PNR will be (a) 928 (b) 968 (c) 950 (d) 920.
A. (b) 10% increase
3.
If fuel consumption is changed by certain percentage, distance to PNR will also change by
corresponding percentage.
Q. With 200 lbs/hr fuel consumption, PNR is 1000 Nm, if actual fuel consumption is found to be 220
lbs/hr , distance to PNR will be (a) 1100 Nm (b) 990 Nm (c) 900 Nm (d) 800 Nm A. (c) 10% Change
45
1.
It is a DR Navigation System which gives Great Circle Tracks/ Distances and True Direction.
It consists of two accelerometers which measure aircraft accelerations in N-S and E-W direction.
2. It has a Gyro Stabilised Platform, horizontally stabilised, to ensure accelerations are measured
in the horizontal plane only. Three torque motors, two accelerometers and three rate gyros
sensitive in each axis are mounted on the horizontally stabilized platform. The Zero position
denotes the present position of the aircraft from which the georef coordinates are taken to
initialize the system. A total of 9 way points can be fed into the system.
3. Control and Display Unit (CDU).
782336 W
241507 N
W
P
T
23
4
POS
XTK/TKE
HDG/GA
TK/GS
D
I
M
TK CHG
WPT
DIST/TIME
BATT
WARN
HOLD
WIND
DSR TK/STS
TEST
INSERT
CLEAR
## Fig 1 Control and Display Unit
4.
TK/GS(Track and Groundspeed).
The INS computed track, usually referenced to magnetic
north, is displayed to the nearest tenth of a degree in the left display and the groundspeed in knots in the
right display. For example, a current track of 135 M and a groundspeed of 467 knots would appear as
135.0 and 0467.
5.
HDG/DA(Heading and Drift Angle). The heading obtained from the angle between the platform
frame and north reference is displayed to the nearest tenth of a degree in the left display. The angular
difference between heading and track (drift angle) is displayed to the nearest tenth of a degree in the
right display, preceded by the letter R or L to indicate whether drift is right or left. Thus, a heading of
137 M on a track of 135 M would be presented as 137.0 and L 02.0.
46
6.
XTK/TKE (Cross Track Distance and Track Error Angle). Cross track distance is the distance by
which the aircraft is displaced right or left of the desired great circle track and is displayed in the left
display to the nearest tenth of a nautical mile. The track error angle is the angular difference, right or
left, between the desired great circle track and the actual track being made, to the nearest tenth of a
degree. If the aircraft were displaced 1 nm to the left of the desired track of 135M, the left display
would read L 01.5. If the track being made good happened to be 130M, the right display would read L
005.0.
7.
POS (Present Position). The aircrafts current latitude and longitude are shown in terms of
Longitude and Latitude in the left and right displays, respectively. For example, 241507N and
782336W,
8.
WPT (Waypoint Positions). The position of each inserted waypoint is shown as latitude in the
left display and longitude in the right display by selecting WPT on the rotary selector switch and
scrolling through the waypoint numbers with the waypoint selector wheel.
9.
DIST/TIME (Distance and Time to the Next Waypoint). The distance in nautical miles from the
present position to the next waypoint is shown in the left display and the time at present groundspeed to
the nearest tenth of a minute in the right display.
10.
WIND (Wind Speed). The INS is able to compute wind direction and speed and these are
displayed in the left and right windows, respectively, to the nearest degree of arc and knot.
11.
DSR TK/STS (Desired Track and Status). The Great Circle track from one waypoint to the next
changes as the aircraft progresses between the two and INS computes the present desired magnetic track
based upon distance from the waypoints, magnetic variation and the assumption that the aircraft is on
track. This will appear in the left display to the nearest tenth of a degree and the right display will be
blank. The status function is for use only whilst the INS is in ALIGN mode and it shows a numerical
display in the right window that indicates the status of the alignment procedure. The display typically
shows 99 at the start of alignment and counts down to 0, when alignment is completed and READY
NAV is illuminated.
Q. Way Point(WP) 4 is 60N 90W, WP3 is 60N 70W. Find distance shown at WP3 for WP4 (a) 605
Nm, (b) 600 Nm (c) 594 Nm (d) None.
A. Dep = dlong x 60 x Cos 60 = 20 x 60 x 0.5 = 600 Nm (Rhumb Line Distance), remember GC
distance is slightly lesser than Rhumb Line distance, Hence 594 Nm is the correct answer.
Q. What is the track shown by INS on leaving WP3 in the above question (a) 270 (b) 279 (c) 261
(d) 099.
A. Conversion Angle = dlong x Sin Mean Lat = x 20 x Sin 60 = 9. RL Track =270 + 9 = 279
which is the GC track.
Q. What is the track shown before arrival at WP4 in the above question?
A. It will be 9 less than 270 = 261 and curving left
270
261
## Q At 80W INS indicates 0 XTKE, what is the latitude shown by INS?
A. Distance off from rhumb line can be calculated by taking half the CA i.e 4.5 and applying 1 in
60 rule. Distance at 80W=300Nm =R, = 4.5 hence s = 22.5 Nm = 22.5 (1 = 1 Nm). Lat = 6030N
47
Q. WP7 is 5227S, 01745W, WP8 is 52S27S, 00415E. Find on leaving WP7,(a) DSR/TRK (b)
Distance (i) 804 nm (ii) little less than 804 (iii) 494 Nm (iv) little less than 494 Nm (c) Initial true Hdg
on leaving was 080 assuming constant drift find true hdg on reaching WP8.
A. (a) CA = dlong x Sin Mean Lat = x 22 x Sin 52 27 = 9. RL Track =090 + 9 = 099 which
is the GC track. (b) Dep = dlong x 60 x Cos 52 27 = 22 x 60 x 0.6 = 800 Nm (Rhumb Line
Distance), remember GC distance is slightly lesser than Rhumb Line distance, (ii) is correct. (c)
Initial Track was 099 but ac is flying 080 hence stb drift of 19. At WP 8 Track would be 090 -9 -19
=062
INS Error. Distance from final ramp position to final INS position divided by time is termed as
INS error
Q. After 6h24m flight when at parking bay position 5218.5N, 445.9E the ins shows a position of
5220.7N 440.3e. Find INS Error
52 20.7N
04 40.3E
2.2Nm
33
3.42 Nm
## Dep = dlong x Cos Lat = 5.6 x Cos 5218.5 =3.42 Nm,
Similarly Dist covered in Northerly Hdg = 2.2 Nm,
Tan(2.2/3.42) = Tan (.643) or = 33 Hypotenuse distance
52 18.5N = 4.066 Nm hence error =4/6:24 = 0.64 Nm/hr in 303
04 45.9E
48
EXTENDED RANGE TWIN OPERATION (ETOPS)
IT IS APPLICABLE TO TWIN ENGINE OPERATIONS ONLY
Threshold Distance. The Maximum distance you can fly with one engine inoperative in 60 minutes
(threshold time TT) for Class A aircraft and 120 min for Class B&C aircraft in nil wind condition is
termed as threshold distance. It is calculated by TT x TAS, the TAS will vary between turbo-prop and
jet ac, also the FL will vary. It is FL 80 for prop ac and FL 170 for turbo jet ac. The TAS would depend
on the performance of the ac with single engine.
Rule Time. The maximum diversion time that any point on the route of an ETOPS approved aircraft
may be from a suitable aerodrome is referred to as the Rule time. The first clearance for ETOPS is 120
minutes, after safe operations for six months it may be extended to 138 minutes and after 12 months of
safe operations further extended by 180 minutes.
Rule Distance. The maximum distance an operator may plan any route from a suitable aerodrome is
that which would be covered at the normal one-engine inoperative cruising speed, in still air, in the rule
time
Adequate Aerodrome. An aerodrome that is available at the anticipated time of use and is equipped
with necessary ancillary services i.e. ATC, lighting, communications, weather reporting, nav aids and
safety cover and having at least one let down aid available for an instrument runway.
Suitable Aerodrome. A suitable aerodrome is an adequate aerodrome that is used as an en-route
alternate or diversion aerodrome. It would normally be used only in the event of an engine failure or the
loss of primary airframe systems. For an aerodrome to be considered suitable, weather probability
should be more than 40% for the forecast if probability is less than 40% the aerodrome may be
ignored for planning purposes.
49
GPS
It utilises the range obtained from a constellation of satellites to fix position of an aircraft in the air or on
ground. It is a satellite based navigational aid.
Basic Data.
Total number of Satellites : 21+ 3 (stand by) =24.
Number of Orbital Paths : 06.
Inclination of Orbital Path : 55 to the Equator
Distance : 20, 200 kms above the surface of the earth.
Time taken for each orbit : 12 hours.
Number of satellites required for 2D fix : 3
Number of satellites required for 3D fix : 4 (also for initialisation of the system)
Number of satellites required for RAIM (Remote Autonomous Integrity Monitoring) : 5
(This is similar to Built in Test Equipment BITE)
Frequency : L1 -1575 MHz and L2 -1228 MHz in UHF Band.
Principle. GPS transmits a PRN (Pseudo Random Noise) of 1 milli sec duration in UHF band. Each
satellite has its own unique code. The information contained in PRN code is
(a) Position of satellite (azimuth and angle)
(b) Clock Time
(c) Clock Error
(d) Information on ionospheric condition
(e) Supplementary information
Two codes are transmitted, CA (Coarse Acquisition) and P (Precision) Code. Two services are provided
by GPS. These are
(a) Standard Position Service (SPS) using CA code which is available for civil use.
(b) Precise Position Services (PPS) using CA & P Code.
Pseudo Range. The receiver has an accurate crystal oscillator to provide time, however, the accuracy
does not compare with that of the satellite clock. The receiver clock is deliberately kept in error by a
small factor to ensure that correction process takes place in one way only. The initial calculated range is
called Pseudo Range. For example if the receiver clock is 1 milli second fast the receiver will
overestimate the range by 162 Nm. Therefore when the receiver sets about calculating the correct range
it knows that it must reduce Pseudo Range.
Errors of GPS
Ephemeris Error. This error is due to the disturbed position of the satellite. Any deviation from its
orbital path will induce this error. This is caused due to various reasons e.g. debris or gravitational effect
of Sun, Moon and other planets. To obviate this error satellite position is checked every 12 hours and
when necessary it is updated. The max permissible error is 2.5 m.
Satellite Clock Error. The clock is also checked atleast every 12 hours. The max permissible error is
1.5 m.
50
Ionospheric Propagation Error. This is the most significant error in the system. The state of
ionosphere is continuously checked at monitoring stations and the model is updated every 12 hours.
The max permissible error is 5 m.
Tropospheric Propagation Error. This is caused due to variation in pressure /temperature/density and
humidity effects on EM waves. The max permissible error is 0.5 m.
measurement of time. The max permissible error is 0.3 m.
System Accuracy. The ICAO specifications require an accuracy of the SPS to be 30 m with a
probability of 50% i.e. 30 m 50% of the time.
Multi-path Reception. Reflections from ground and parts of aircraft result in multi-path reception.
GDOP. Geometric Dilution of Precision is caused due to poor cut between position lines. This is
caused when the satellites are relatively closer to each other.
Differential GPS. It is a means of improving the accuracy of the GPS by monitoring the integrity of the
satellite data and warning the user of any error which may occur during flight. There are three kinds of
DGPS in use. Accuracy is 3 m.
(a) GBAS (Ground Based Augmentation System). This is Local Area Augmentation System
(LAAS).
(b) ABAS (Air Based Augmentation System).
(c) SBAS (Space Based Augmentation System). This is Wide Area Augmentation
System
(WAAS).
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When Jill did it, she picked out $7, 8, 6$ and $3$. To make the biggest even number she got $8736$. Vincent picked out $4, 2, 9$ and $1$. The biggest even number is $9412$. Vincent won because he had the highest even number. I worked this out because $9000$ is bigger than $8000$.
Next it was the highest odd number. Belinda picked out $4, 0, 6$ and $9$. The highest odd number from these is $6409$. Ali picked out $5, 2, 1$ and $7$. His highest odd number is $7521$. Ali's number is higher than Belinda's because $7000$ is higher than $6000$.
Next is the lowest even number. Ben picked out $7, 0, 3$ and $9$ and the lowest even number is $3790$. Rohan picked $8, 1, 5$ and $4$ and his number is $1458$. Rohan won because $1000$ is lower than $3000$.
Alice got $7, 8, 4$ and $9$ and Chloe got $1, 0, 3$ and $5$. The numbers they got which are nearest to $5000$ are $4987$ and $5013$. I used an abacus to show that the difference is $13$ for both and so it is a draw.
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# Draw ten dots that are all the same distance apart
I am wondering, is it possible to draw ten dots so that every dot has the same distance to every other one?
And how many possibilities are there?
• In how many dimensions? Jan 4 '18 at 18:21
• Two dimensions. Jan 4 '18 at 18:21
• this seems equally relevant in mathematics S.E. Jan 5 '18 at 22:13
No: three dots having the same distances to each other form an equilateral triangle, and there is no way to add a fourth to have the same distance to all three of those. It would have to be on all three of the circles in this image:
(So it certainly isn't possible to do 10 dots!)
In general, you need $n-1$ dimensions for $n$ dots with this property: these dots will form the vertices of a regular $k$-simplex.
• You didn't prove it though ;-) Jan 4 '18 at 18:24
• @Randal'Thor And you didn't generalize it. :P
– Deusovi
Jan 4 '18 at 18:28
• I am not good at math. But i think the highest possible number of dots is 7. is it right ?
– ran
Jan 4 '18 at 18:30
• @ran No - as shown in the answers here, the highest possible is 3 (in the plane). Jan 4 '18 at 18:31
• @ran, you may be thinking of a related question: puzzling.stackexchange.com/q/56809/228 Jan 4 '18 at 22:04
Sure, there's one possibility:
Each dot has a distance of 0 from all others.
In two dimensions, there is
a trivial solution: the dots are identical (i.e. at the same place), the distance is trivially zero.
also:
You never said you want Euclidean solutions :-). If the distance is defined by discrete metric, then the distance between any non-identical dots is trivially 1.
EDIT: answer to how many possibilities there are:
for the trivial solution - one
for the non-Euclidean - infinity. And there are other non-trivial metrics you can use, e.g. Chebyshev distance works as well.
• Yay for brainbreaking non-euclidian space! Jan 4 '18 at 22:28
## In two dimensions, no.
Instead of ten, the most you can get is three dots all at the same distance from each other.
To see why this is, let's start with two dots $A,B$ a given distance apart (say 1 unit). Where can we place a third dot which is 1 unit away from both of them? This dot must be on both of the radius-1 circles centred at $A$ and $B$, and these circles only intersect in two points:
The two points $C1,C2$ aren't distance 1 apart, so the most we can get is three dots, forming an equilateral triangle in the plane.
## In three dimensions, no.
A similar argument in 3D space shows that the most you can get is four dots all the same distance apart, forming a regular tetrahedron in 3-space.
## Only in nine or more dimensions.
As Deusovi correctly says in his answer, the general solution in $n$ dimensions is the vertices of a simplex (the higher-dimensional analogue of a triangle or tetrahedron) - such a shape has $n+1$ vertices, all of which are equidistant from each other, and it's impossible to do better than this. So in a space of dimension 9 or higher, it is possible to find ten points all the same distance apart. But we'd need to ask a string theorist for that ...
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# Simplifying a system of differential equations
Given is the following system of differential equations which describes the behaviour of a biological system:
$\dfrac{dR_1}{dt} = \alpha_1 \dfrac{P_2^4}{a^4+P_2^4} - R_1$
$\dfrac{dP_1}{dt} = \alpha_2 \dfrac{1}{1+R_1^4} - P_1$
$\dfrac{dR_2}{dt} = \alpha_3 \dfrac{P_1^4}{a^4+P_1^4} - R_2$
$\dfrac{dP_2}{dt} = \alpha_4 \dfrac{1}{1+R_2^4} - P_2$
I've read that one can simplify this system preserving the crucial characteristics by reducing it to the following model (there' s not mentioned why):
$\dfrac{dR_1}{dt} = \beta_1 \dfrac{R_2^4}{b^4+R_2^4} - R_1$
$\dfrac{dR_2}{dt} = \beta_2 \dfrac{R_1^4}{b^4+R_1^4} - R_2$
I can't see why this works. Does anyone know?
• integrate the P's and substitute into the respective R's, solve for R
– JMP
Sep 27, 2016 at 19:48
• But how to integrade the $P's$? There are $R's$ in the respective equation, too. Sep 27, 2016 at 20:14
• $\frac{dR}{dt}+R=0$ gives $R=e^{-x}$
– JMP
Sep 27, 2016 at 20:45
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Percentile no. $25$ of $1$, $-5$, $2$, $4$, $-3$, $6$, $7$, $0$, $2$, $5$, $-4$, $7$
The calculator will find the percentile no. $25$ of $1$, $-5$, $2$, $4$, $-3$, $6$, $7$, $0$, $2$, $5$, $-4$, $7$, with steps shown.
Related calculators: Five Number Summary Calculator, Box and Whisker Plot Calculator
Comma-separated.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Find the percentile no. $25$ of $1$, $-5$, $2$, $4$, $-3$, $6$, $7$, $0$, $2$, $5$, $-4$, $7$.
Solution
The percentile no. $p$ is a value such that at least $p$ percent of the observations is less than or equal to this value and at least $100 - p$ percent of the observations is greater than or equal to this value.
The first step is to sort the values.
The sorted values are $-5$, $-4$, $-3$, $0$, $1$, $2$, $2$, $4$, $5$, $6$, $7$, $7$.
Since there are $12$ values, then $n = 12$.
Now, calculate the index: $i = \frac{p}{100} n = \frac{25}{100} \cdot 12 = 3$.
Since the index $i$ is an integer, the percentile no. $25$ is the average of the values at the positions $i$ and $i + 1$.
The value at the position $i = 3$ is $-3$; the value at the position $i + 1 = 4$ is $0$.
Their average is the percentile: $\frac{-3 + 0}{2} = - \frac{3}{2}$.
The percentile no. $25$A is $- \frac{3}{2} = -1.5$A.
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https://tutorialcup.com/interview/array/find-minimum-number-of-merge-operations-to-make-an-array-palindrome.htm
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# Find minimum number of merge operations to make an array palindrome
Difficulty Level Easy
Array Greedy MathViews 1716
## Problem Statement
You are given an array of integers. The problem statement asks to find minimum number of merge operations to make an array palindrome, i.e. find out the minimum number of merging operations to be done on the array to make it a palindrome. Merging operation simply means that we can replace the sum of adjacent elements with the sum itself.
## Example
`arr[] = {1, 4, 6, 6, 5}`
`1`
Explanation: We can merge 1 and 4 with the sum of themselves, so it becomes 5 and our array becomes {5, 6, 6, 5}, which is a palindrome.
`arr[] = {2,1,2,4,3}`
`2`
Explanation: We can merge 1 and 2 with their sum, so it becomes 3 and also 2 and 4 can be merged, so our array becomes {3, 6, 3}
## Algorithm to find minimum number of merge operations to make an array palindrome
```1. Set the value of output to 0.
2. Traverse the array from i = 0 and j = n – 1, to i < n( length of an array).
1. If arr[i] is equal to arr[j]
1. Do i++ and j--.
2. If arr[i] > arr[j]
1. Do j—
2. Update and restore the value of arr[j] = arr[j] + arr[j+1].
3. Increase the output’s value by 1.
3. If arr[i] < arr[j],
1. Increase the value of i.
2. Update arr[i] = arr[i] + arr[i-1].
3. Increase the output’s value by 1.
3. Return output.```
### Explanation
We have given an array of integers. We are asked to find out the minimum number of merging operations that can be done onto the array to make it a palindrome array. Here merging means adding up two adjacent values and replaces them with their sum. Here we are going to find the number of operations to be done.
We are going to traverse the array from left and also from right making two pointers indicate the position of each traversal index. We chose both sides because we have to check if it is palindrome or not. And in palindrome from both sides according to their indexes elements are equal just like in string palindrome. So we will check the value of opposite sides and operate on their indexes. Then we perform merging operations and continue up till we find the count of merging operations to be done.
We will check if the first element from the left and from the right are equal. Then just shift the pointers towards the center one unit. Check for if the left pointer index element is greater than the right pointer index element, then decrease the value of the right pointer value and update the arr[j] with the sum of the adjacent elements and increase the value of output means our number of operations.
We will check for if the left pointer index element is smaller than the right pointer index element, then increase the value of the left pointer value and update the arr[i] with the sum of the adjacent elements and increase the value of output means our number of operations. And at last, we will return the output value.
## Code to find minimum number of merge operations to make an array palindrome
### C++ Code
```#include<iostream>
using namespace std;
int getMinimumOperation(int arr[], int n)
{
int output = 0;
for (int i=0,j=n-1; i<=j;)
{
if (arr[i] == arr[j])
{
i++;
j--;
}
else if (arr[i] > arr[j])
{
j--;
arr[j] += arr[j+1] ;
output++;
}
else
{
i++;
arr[i] += arr[i-1];
output++;
}
}
return output;
}
int main()
{
int arr[] = { 1, 4, 6, 6, 5};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Minimum operations to be done: "<< getMinimumOperation(arr, n);
return 0;
}
```
`Minimum operations to be done: 1`
### Java Code
```class ArrayPalindromeMerging
{
public static int getMinimumOperation(int[] arr, int n)
{
int output = 0;
for (int i=0,j=n-1; i<=j;)
{
if (arr[i] == arr[j])
{
i++;
j--;
}
else if (arr[i] > arr[j])
{
j--;
arr[j] += arr[j+1] ;
output++;
}
else
{
i++;
arr[i] += arr[i-1];
output++;
}
}
return output;
}
public static void main(String[] args)
{
int arr[] = { 1, 4, 6, 6, 5} ;
System.out.println("Minimum operations to be done : "+ getMinimumOperation(arr, arr.length));
}
}
```
`Minimum operations to be done : 1`
## Complexity Analysis
### Time Complexity
We are simply traversing the array once, so this counts for linear time complexity. O(n) where “n” is the number of elements in the array.
### Space Complexity
We are just using a single array having size n for storing the input, thus this algorithm for finding the number of merge operations to make an array palindrome has linear space complexity. O(n) where “n” is the number of elements in the array.
Translate »
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Hotel rooms
In the 45 rooms, there were 169 guests, some rooms were three-bedrooms and some five-bedrooms. How many rooms were?
Correct result:
a = 28
b = 17
Solution:
a+b = 45
3a + 5b = 169
a+b = 45
3•a + 5•b = 169
a+b = 45
3a+5b = 169
a = 28
b = 17
Our linear equations calculator calculates it.
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
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# Homework Help: Gravity Help please
1. Nov 18, 2007
### splac6996
1. The problem statement, all variables and given/known data
A projectile is shot straight up from the earth's surface at a speed of 1.40×10^4 . How high does it go?
2. Relevant equations
potential energy is given by (GM(1)M(2))/R
3. The attempt at a solution
I attempted to solve this using energy conservation equation and solving for r which I thought would be the radius of the earth plus the distance traveled by the projectile. By subtracting the radius I thought I would get the distance traveled but my answer is wrong can someone help please
2. Nov 18, 2007
### dynamicsolo
What units is that speed in?
It might be of help to a reader who is going to assist you if you would show the actual calculation you did.
3. Nov 18, 2007
### Mindscrape
There are a number of things you could have done wrong. For example, did you remember to use a negative sign in your potential? Do you have an initial potential, where did you define your zero in potential energy to be? Specifically, what was the equation you used?
4. Nov 19, 2007
### splac6996
the speed is defined in km/hr and for the formula that i used I did use a negative sign for my potential energy and I used my zero point to be the point were the projectile is shot from.
5. Nov 19, 2007
### katchum
Why not just use m.g.h=1/2mv^2?
And potential energy is in Joule not in Newton.
6. Nov 19, 2007
### Kurdt
Staff Emeritus
mgh is only applicable near the Earth's surface. For this projectile it will be travelling significantly far away from the Earth's surface to make mgh non-applicable.
7. Nov 19, 2007
### splac6996
Well my understanding is that if the distance is smaller than than the radius then I could use mgh=1/2mv^2 but in this case since the height is going to be largerthan the radius of the earth so my answer would be wrong.
8. Nov 19, 2007
### Oerg
use v^2=u^2+2as,
your deceleration is not constant in the equation. The deceleration is given as the average deceleration the object undergo in that motion. to find the average deceleration,
integrate the below equation with respect to distance, with the radius of the earth as the lower limit and the radius+height of the earth as the upper limit.
then divide by the height to find the average deceleration.
a=Gm/r^2
EDIT: Well, you will need to have the radius of the earth to solve the equation. Even after simplifying expressions of the radius of the earth into g.
Last edited: Nov 19, 2007
9. Nov 19, 2007
### Sourabh N
@splac6996: Well, If you take zero of your PE at surface of the Earth, your formula for PE at R (in #1 post) is not correct.
Last edited: Nov 19, 2007
10. Nov 19, 2007
### splac6996
this is what i think
-GM(earth)m(projectile)/H(radius of earth + distance traveled)=(1/2)m(projectile)v^2
doing the algebra and simplifying i have
H=-2GM/v^2
is that correct
11. Nov 19, 2007
### Sourabh N
Nope. You can see, if you put distance traveled = 0 your
RHS = 0 but LHS is not = 0.
12. Nov 19, 2007
### Oerg
the expression for the potential -GM(earth)m(projectile)/H(radius of earth + distance traveled) gives you the potential energy at the highest point. You will still need to minus the potential at the earth's radius.
13. Nov 19, 2007
### HallsofIvy
With gravitational force $F= -GMm/r^2$, the potential energy is $GMm/r$ but that has 0 point at infinity, not at the surface of the earth.
14. Nov 19, 2007
### Sourabh N
Can you show the exact figures you got?
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### CFA Practice Question
There are 536 practice questions for this topic.
### CFA Practice Question
The preferred stock of the Wordsworth Institute pays a constant annual dividend of \$3.00 and sells for \$20.00. You believe the stock will sell for \$12.00 in one year. You must, therefore, believe that the required return on the stock will be which of the following percentage points (lower/higher) in one year?
A. It will be 8.0% lower.
B. It will be 8.0% higher.
C. It will be 10.0% higher.
User Comment
cgeek how to get this answer? why \$20 for this year and \$12 in one year
jamiejamie first, solve for k at time t0 = 15% then, solve for k at time t1 = 25% you get these values using the preferred stock value (preferred stock value = dividend/k) then, you know that there is an INCREASE of 10% (25%-15%) Intuitively, you know that if you get the same dividend for a cheaper security price, then your K must have risen.
stefdunk the 10% increase is not in the value of the stock, but in the rate of return you expect. You want a higher payout, so the value of the share will drop (preferred stocks and bonds: value drops if payout % rises)
katybo D/K = 3/0.25 = 12 -> 0.25-0.15 = 10%
haarlemmer Sine the dividend is constant, the answer is then (3/12)-(3/20)=10%
Done think about it like it was a bond. since the price went down the yield should go up. That eliminates A and D, then do the math
faya If 3/k=20 => k=15%; If 3/k=12 => k=25%. Therefore, to get 3/k=12, you need to increase k by 10%
cfahanoi k increase => P reduce
3/12 - 3/20 = 10%
accounting go for cfahanoi
VenkatB jamiejamie - thanks for the explanation.
jansen1979 t0: \$ 20 = \$ 3/x => x = 15%
t1: \$ 12 = \$ 3/x => x = 25%
Increase of 10%
bundy 3/12 = 25
3/20 = 15 therefore 10% higher
loisliu88 cost of preferred stock=D/r, r0= D/P0, r1= D/p1
2014 Good work bundy
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# Writing a correction phrase
• MHB
Gold Member
MHB
Dear Everyone,
I am trying to figure what is the correct phrase in the bolden phrase. The article, where I am doing my research on, states: Let S be the set of all 2x2 matrices with equal positive integral entries. Let T be the set of all 2x2 matrices with equal integral entries. My professors are getting frustrated due to the circle effect that I am making the same error over many times. So what is the correcting words that fixed? Is it "having" or other words.
Beginning:
Different algebraic systems raise many questions. For instance, can the elements in a given system always be factored into primes? If so, what theorems can help factoring the elements? Are the factors unique? The poster will discuss the answers to these questions through examples and theorems for a class of $2\times2$ matrices with equal integers entries.
Let $S$ be the set of all $2\times2$ matrices with equal positive integers entries.
Conclusion:
The elements of the set $T$ can always be factored; however, most of the elements in the set $T$ are not uniquely factorable. There are theorems that can assist in determining the factorization of elements of $T$. Future investigation might include studying whether there are similar theorems for each class of $n\times n$ matrices with equal integers entries.
Thanks,
Cbarker1
Last edited:
Gold Member
MHB
Re: writing a correction phrase in a poster
Dear Fellow Members,
I am trying to figure out how to correct the bold phrases below. The article on which I am doing my research states: Let $S$ be the set of all 2x2 matrices with equal positive integral entries and let $T$ be the set of all 2x2 matrices with equal integral entries. My professors are getting frustrated due to the circular effect that I am creating. I am making the same error many times. So how do I fix these? Do I use "having" or another word?
Beginning:
Different algebraic systems raise many questions. For instance, can the elements in a given system always be factored into primes? If so, what theorems can help factor the elements? Are the factors unique? The poster will discuss the answers to these questions through examples and theorems for a class of 2x2 matrices with equal integer entries.
Let $S$ be the set of all 2x2 matrices with equal positive integer entries.
Conclusion:
The elements of the set $T$ can always be factored; however, most of the elements in the set $T$ are not uniquely factorisable. There are theorems that can assist in determining the factorisation of the elements of $T$. Future investigation might include studying whether there are similar theorems for each class of nxn matrices with equal integer entries.
Hi CBarker1.
Above is a corrected version of the material in your original post. Compare the two and ask questions about anything you need clarification on.
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Mar 4 comment Have 52 regular deck of cards, probability getting the first red ace? So would 52 choose 2 be the possible placements of red aces in the deck? Mar 4 comment Have 52 regular deck of cards, probability getting the first red ace? What does the (52-1-m)/(52-m+1) stand for? I understand that the second part would be the probability for the two red aces. Also, how did you find that the maximal is k=1? And thank you for your help!! Mar 3 comment Compositions with first part 1 Thank you!! What if the second part is 1 instead of the first part? Would it still be the same solution? Feb 25 comment Finding the transition probability matrix, two switches either on or off.. Thank you, you cleared this up extremely well for me. I can't thank you enough! I can finally go to bed :) Dec 12 comment Strong inducti0n with 3- and 5-peso notes and can pay any number greater than 7. Ahh it all makes sense now! Induction gets me every single time.. Dec 11 comment Strong inducti0n with 3- and 5-peso notes and can pay any number greater than 7. So we can assume that every integer smaller than n but greater than 8, can be expressed as a combination of 3- and 5- pesos. My confusion lies in proving n+3.. Dec 9 comment Choosing a 5 member team out of 12 girls and 10 boys I haven't thought about doing it that way! Brian, you have been helping me a lot, and I really appreciate that someone with your knowledge is on here helping others out. Thank you. Dec 9 comment Choosing a 5 member team out of 12 girls and 10 boys Thank you. I didn't think it would be that simple. Dec 9 comment Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. OH! So there are 1001 pigeonholes and 1024 pigeons! Got it! Thank you Brian! Dec 9 comment Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. @BrianM.Scott, ooh alright, so the 1024 subsets are the pigeons, but what would be the pigeonholes? All the possible sums? Dec 9 comment Using the pigeonhole principle to prove there is at least two groups of people whose age sums are the same. So the max sum of ages is 1000 ( ten people and they all can be 100 yrs. old). And the number of subsets is 2^(10) - 1? Dec 9 comment Having a forest and making it into a tree? Thanks again! I really appreciate your help! Dec 9 comment Having a forest and making it into a tree? Lol, thanks Brian! One question though... How did you know that m=10? Dec 8 comment Having a forest and making it into a tree? Well, we can have 100 trees (just a dot) in a forest at most, but we also have 90 edges.. So if we add 1 edge to each of those trees, then we will have 50 trees in the forest, but we will have 50 edges.. Dec 8 comment Planar Graphs with at least $2$ vertices and degrees at most $5$ Suppose every vertex, with at most one exception, has degree at least 6. Then, 2E =< 2(3V-6) = 6V-12; Sum (deg V) >= 6(V-1) = 6V-6; We see that 6V-12 > 6V-6, thus there are at lease two vertices whose degrees are at most 5. Is this the solution? I still can't see how we have two vertices that are at most 5. Dec 8 comment Planar Graphs with at least $2$ vertices and degrees at most $5$ So would the at most one exception be less than 6?
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https://thirdspacelearning.com/secondary-resources/gcse-maths-worksheet-probability-distribution/
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Maths Resources GCSE Worksheets
# Probability Distribution Worksheet
• Section 1 of the probability distribution worksheet contains 20+ skills-based probability distribution questions, in 3 groups to support differentiation
• Section 2 contains 3 applied probability distribution questions with a mix of worded problems and deeper problem solving questions
• Section 3 contains 4 foundation and higher level GCSE exam style probability distribution questions
• Answers and a mark scheme for all probability distribution questions are provided
• Questions follow variation theory with plenty of opportunities for students to work independently at their own level
• All questions created by fully qualified expert secondary maths teachers
• Suitable for GCSE maths revision for AQA, OCR and Edexcel exam boards
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### Probability distribution at a glance
A probability distribution tells us the probability of each event occurring within the sample space. When rolling a dice, the discrete random variable is the possible outcomes of a roll (1, 2, 3, 4, 5, 6). The probability of rolling a number on this dice has a discrete probability distribution as each probability is equal to 1 out of 6.
A probability p in a probability distribution, can be written as a fraction, decimal or percentage and they are often given in a probability distribution table.
The probabilities of an exhaustive set of mutually exclusive events sum to 1. For example, let the random variable x represent the possible values that can be rolled on a fair 6 sided dice. When rolling the fair die, the probability of getting each number is 1 out of 6. This can be expressed as P(x=1)=1 out of 6, P(x=2)=1 out of 6, P(x=3)=1 out of 6, etc. There are 6 possible events and so if we add the probability of each event occurring, we get 1.
We can find missing probabilities by subtracting known probabilities from 1. For example, a spinner has 4 colours with the following probability distribution: P(Red)=0.1, P(Blue)=0.4, P(Green)=0.3, and P(Yellow)=x. The probability of landing on yellow is found by subtracting the other probabilities from 1. This is because the sum of the probabilities for these mutually exclusive events is 1. So we have P(Yellow)=1-(0.1+0.4+0.3)=0.2, so x=0.2.
Given a probability distribution, we can find the expected number of times a certain outcome will happen, known as the expected value. We do this by multiplying the probability by the number of trials (tosses). Let X represent the probability of flipping a coin and landing on Heads, P(X=Heads)=0.5. If we flipped the coin 100 times, we would expect the number of Heads to be 0.5✖100=50. Remember, this is just an estimate
Experimental probability (also called relative frequency) is the probability of an event occurring based on previous events. This is useful when we don’t know the theoretical probability of an event occurring. For example, a high school is investigating how many students get to school by car on a given day. They record the number of students who travel to school by car every day for 1 month out of 800 students. The mean μ of students who travel by car is 200 out of 800 students, and so the experimental probability (or the probability of success) would be 200 out of 800, 0.25, or 25%.
Looking forward, students can then progress to additional probability worksheets, for example the random sample worksheet, the conditional probability worksheet, or the and / or probability worksheet.
For more teaching and learning support on Probability our GCSE maths lessons provide step by step support for all GCSE maths concepts.
## Do you have students who need additional support to achieve their target GCSE maths grade?
There will be students in your class who require individual attention to help them succeed in their maths GCSEs. In a class of 30, it’s not always easy to provide.
Help your students feel confident with exam-style questions and the strategies they’ll need to answer them correctly with personalised online one to one tutoring from Third Space Learning
Lessons are selected to provide support where each student needs it most, and specially-trained GCSE maths tutors adapt the pitch and pace of each lesson. This ensures a personalised revision programme that raises grades and boosts confidence.
GCSE Revision Programme
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https://techiescience.com/can-normal-distribution-be-skewed/
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Can Normal Distribution Be Skewed: Detailed Facts, Examples And FAQs
Normal distribution is skewed with zero skewness, so the answer to the most common confusion can normal distribution be skewed is normal distribution is not skewed distribution as the curve of the normal distribution is symmetric without tail whose skewness is zero. The normal distribution curve is bell shaped with symmetry on the curve.
Since the skewness is lack of symmetry in the curve so if the symmetry is present in the curve there is lack of skewness.
How do you tell if the data is normally distributed?
For the data to check whether normally distributed or not just try to sketch the histogram and from the curve of the curve if the symmetry is present in the curve then the data is normally distributed, from the curve of data itself the question can normal distribution be skewed or not cleared if the concept of skewness is clear. Sketching the histogram or curve in each case is tedious or time consuming so instead of that their are number of statistical tests like Anderson-Darling statistic (AD) which are more useful to tell whether data is normally distributed or not.
The data which follows normal distribution have zero skewness in the curve and the characteristics of the curve of the skewed distribution is different without symmetry, this we will understand with the following example:
Example: Find the percent of score lies between 70 to 80 if the score of mathematics of university students are normally distributed with the mean 67 and standard deviation 9?
Solution:
To find the percent of score we follow the probability for the normal distribution discussed earlier in normal distribution, so to do so first we will convert into normal variate and follow the table discussed in normal distribution to find the probability using the conversion
Z=(X-μ)/σ
we want to find the score percent between 70 and 80 so we use random variable values 70 and 80 with the given mean 67 and standard deviation 9 this gives
Z=70-67/9 = 0.333
and
Z=80-67/9 = 1.444
This we can sketch as
the above shaded area shows the region between z=0.333 and z=1.444 from the table of standard normal variate the probabilities are
P(z > 0.333)=0.3707
and
P(z > 1.444)=0.0749
so
p(0.333 < z0.333)-P(z > 1.444)=0.3707-0.0749=0.2958
so 29.58% students will score between 70 to 80 .
In the above example the skewness of the curve is zero and the curve is symmetric, to check the data is normally distributed or not we have to perform the hypothesis tests.
How do you tell if a distribution is skewed left or right?
The distribution is known to be skewed if it is right tailed or left tailed in the curve so the depending on the nature of the curve we can judge whether the distribution is positive skewed or negative skewed. The concept of skewness is discussed in detail in the articles positively and negatively skewed distribution. If the symmetry in the left side lacks the distribution is skewed left and if the symmetry lacks in the right side the distribution is skewed right. The best way to check the distribution is skewed is to check the variation in the central tendencies that is if mean<median<mode then the distribution is left skewed and if mean>median>mode then the distribution is right skewed. The geometrical representation is as follows
The measures to calculate the skewness left or right for the information given in detail in the article of skewness.
What is an acceptable skewness?
Since the skewness as earlier discussed is lack of symmetry so what range is acceptable that must be clear. The question can normal distribution is skewed arise to check whether in the normal distribution is acceptable or not and the answer of the acceptable skewness is in normal distribution because in normal distribution the skewness is zero and the distribution in which skewness is near to zero is more acceptable. So after the testing for skewness if the skewness is nearer to zero then the skewness is acceptable depending on the requirement and range for the client.
In brief the acceptable skewness is the skewness which is nearer to zero as per the requirement.
How skewed is too skewed?
The skewness is the statistical measurement to check the symmetry present in the curve of the distribution and the information and all the measures to check skewness is present or not, depending on that we can find if the distribution is far from zero then too skewed or symmetry is zero then we can say the distribution is too skewed.
How do you determine normal distribution?
To determine the distribution is normal or not we have to look the distribution have the symmetry or not if the symmetry is present and the skewness is zero then the distribution is normal distribution, the detail methods and techniques were already discussed in detail in normal distribution
Do outliers skew data?
In the distribution data if any data follow unusual way and very far or away from the usual data that is known as outlier and in most of the cases the outliers are responsible for the skewness of the distribution and because of the unusual nature of outliers the distribution have skewness, so we can say that in the distribution the outliers skew data. The outliers in all cases will not skew data they skewed data only if they also follow the systematic sequence in continuous distribution to give left or right tailed curve.
In the previous articles the detail discussion of normal distribution and skewed distribution discussed.
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https://www.jiskha.com/questions/1299256/if-I-walk-2-5-miles-in-1-25-hours-how-long-does-it-take-to-walk-5-miles
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# math
if I walk 2.5 miles in 1.25 hours, how long does it take to walk .5 miles?
1. 2.5/1.25 = 0.5/x
Cross multiply and solve for x.
Or -- divide 2.5 by 1.25 to find the miles per hour.
posted by Ms. Sue
2. cross multiply and you will get something like this
2.5 X x = 0.5 X 1.25
2.5x = 0.625
because we are finding for x so divide both sides by 2.5 so that x can stand alone. so 0.5x over(divided by) 2.5 and 0.625 over(divided by) 2.5.
(note: we are finding for x).
so the answer will now be
x=0.25
posted by Prince
3. if 75 widgets are processed in 3.5 hours by 1 person, how long will it take to process 1650 widgets using 9 people.
posted by chris
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http://mathsframe.co.uk/en/resources/category/199/year_2_block_b_derive_and_recall_multiplication_facts_for_the_2_5_and_10_timestables_and_the_related_division_facts_recognise_multiples_of_2_5_and_10
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# Derive and recall multiplication facts for the 2, 5 and 10 times-tables and the related division facts; recognise multiples of 2, 5 and 10
View all Year 2 Block B objectives.
Shoot the spaceship with the correct answer and dodge the incoming fire. A fun game to practise a wide range of key mathematical skills.
There are over a hundred carefully differentiated levels linked to objectives from the new maths curriculum. The game can be used to teach: Multiplication, Addition, Reading Numbers, Subtraction, Fractions of Numbers, Roman Numerals, Rounding Numbers, Division, Converting Fractions to Decimals, Converting Fractions to Percentages, Telling the Time in Words, Recognising Multiples, Factors, Prime, Square and Cube Numbers, and Simplifying Fractions. A full list of levels is below. This game is also available as an iOS and Android app.
Maths Raiders
Destroy the incoming spaceships and dodge the asteroids. A fun game to practise lots of mathematical skills. Carefully differentiated to support the new maths curriculum. Levels involve: Multiplication, Addition, Roman Numerals, Reading Numbers in Words, Fractions of Numbers, Simplifying Fractions, Multiplying Fractions, Convert Improper Fractions to Mixed Numbers, Converting Fractions to Decimals, Converting Fractions to Percentages, Rounding Numbers, Division and Percentages of Numbers.
This game is also available as an iPad or Android app.
Multiplication - Crystal Crash
A great game to get children to practise their times tables. The objectives for this game have been taken from the new mathematics curriculum.
This game, together with all the other crystal crash games, is available as a single iPad or Android app.
1. Recall and use doubles of all numbers to 10 (Y1) 6. 3 times table (Y3) 11. 6 times table (Y4) 16. Double 3 digit numbers (Y4) 2. 2 times table (Y2) 7. 4 times table (Y3) 12. 7 times table (Y4) 17. Double decimals (1dp) (Y4) 3. 5 times table (Y2) 8. 8 times table (Y3) 13. 9 times table (Y4) 18. Recognise square numbers (up to 15²) (Y5) 4. 10 times table (Y2) 9. Double numbers up to 100 (Y3) 14. 11 times table (Y4) 19. Recognise cube numbers (up to 10³) (Y5) 5. Derive and use doubles of simple two-digit numbers (Y2) 10. Doubles of multiples of 50 up to 500 (Y3) 15. 12 times table (Y4) 20. Double decimals (2dp) (Y5)
Multiplication - Mine Mayhem
This game, together with all the other Mine Mayhem games, is available as a single iOS or Android app.
Choose one or more of the following levels to play:
1. Recall and use doubles of all numbers to 10 (Y1) 6. 3 times table (Y3) 11. 6 times table (Y4) 2. 2 times table (Y2) 7. 4 times table (Y3) 12. 7 times table (Y4) 3. 5 times table (Y2) 8. 8 times table (Y3) 13. 9 times table (Y4) 4. 10 times table (Y2) 9. Double numbers up to 100 (Y3) 14. 11 times table (Y4) 5. Derive and use doubles of simple two-digit numbers (Y2) 10. Doubles of multiples of 50 up to 500 (Y3) 15. 12 times table (Y4)
Snake - KS2 Maths Game
Use the arrow keys to avoid the hazards and direct your snake to the correct answer. There are more than 100 different levels to play. See below for a full list of levels.
This game is available as an iOS and Android app.
Bubble Pop - Multiplication
A fun way to help children learn their multiplication tables. Choose which times tables to practise and then try and pop all the bubbles before your time is up. Try and make the target number (at the top) by shooting a multiplication bubble. So if your target is 10, and you have the x2 bubble, you will need to hit 5, as 5 x 2 = 10.
The free demo version has 10 missions to complete; the full version (for registered users) has 70 missions. The game is ideal for playing on a tablet or phone. Encourage your class to play at home and then watch their confidence and fluency grow.
For more multiplication and division resources click here.
Missing Symbols
Drag the symbols to the correct position to make the number sentence correct. Lots of levels to play. See if you can improve your score over time.
For more multiplication and division resources click here.
This game is now part of the 'Calculations' collection, which includes the following 17 games and resources: Column Addition, Expanded Addition, Expanded Addition - Place Value Counters, Number Bonds(2), Addition - Digit Drag, Missing Symbols, Column Subtraction, Column Subtraction using Place Value Counters, Counting on to find difference on a beadstring, Multiplication Grid Method, Multiplication Written Method, Ratio and Scaling Numbers, Representing Multiplication, Division by Chunking Up, Division by Chunking Down, Formal Written Division - Round Up or Down?, Short Division Writen Formal Method.
The Calculations app is available on Google Play and the App Store.
Monty's Maths Wall
Use the arrow keys to guide your bricks and destroy the wall.
A great game for practising a wide range of mathematical skills. Levels are based on objectives from the new maths curriculum from Year 1 to Year 6. Topics include: multiplication, addition, reading numbers, subtraction, fractions of numbers, Roman numerals, division, converting fractions to decimals and percentages and simplifying fractions.
You can choose to play a single level, a selection of levels, or choose all the objectives from a year group (within the same topic). There is a full list of levels below.
This game is available on both Google Play and the App Store. A free version of the app is available with just the multiplication levels.
Division - Crystal Crash
A fun way to get children to practise their division facts. All the levels are based on objectives from the new maths curriculum. Choose one level, or select all the levels for a year group.
This game, together with all the other crystal crash games, is available as a single iPad or Android app.
1. Half of numbers to 20 (Y1) 5. Derive and use halves of simple two-digit even numbers (Y2) 9. Halve numbers up to 200 (Y3) 13. Division facts for the 11 times table (Y4) 2. Division facts for the 2 times table (Y2) 6. Division facts for the 3 times table (Y3) 10. Division facts for the 6 times table (Y4) 14. Division facts for the 12 times table (Y4) 3. Division facts for the 5 times table (Y2) 7. Division facts for the 4 times table (Y3) 11. Division facts for the 7 times table (Y4) 15. Halve decimals (1dp) (Y4) 4. Division facts for the 10 times table (Y2) 8. Division facts for the 8 times table (Y3) 12. Division facts for the 9 times table (Y4) 16. Halve decimals (2dp) (Y5)
Multiplication Dominoes
You can choose one, or several, multiplication tables to practise. Then see how quickly you can arrange the dominoes.
Errors on a Carroll Diagram - Numbers
Identify the numbers which have been placed in the wrong position on the Venn diagram and drag them to their correct place
Choose to arrange numbers by whether they are odd or even, or multiples of a given number, or prime or square numbers.
For more resources involving sorting shapes and numbers click here.
Errors on a Venn Diagram - Numbers
Identify the numbers which have been placed in the wrong position on the Venn diagram and drag them to their correct place
Choose to arrange numbers by whether they are odd or even, or multiples of a given number, or prime or square numbers.
For more resources involving sorting shapes and numbers click here.
Numbers on a Carroll Diagram - Missing Labels
Drag the correct labels onto the Carroll diagram. A great game to encourage reasoning about the properties of numbers. Works really well as a mental starter.
For more resources involving sorting shapes and numbers click here.
Representing Multiplication - arrays and number lines
Represents multiplication using an array and by jumping on a number line. Great for developing an understanding of multiplication by partitioning.
This game is now part of the 'Calculations' collection, which includes the following 17 games and resources: Column Addition, Expanded Addition, Expanded Addition - Place Value Counters, Number Bonds(2), Addition - Digit Drag, Missing Symbols, Column Subtraction, Column Subtraction using Place Value Counters, Counting on to find difference on a beadstring, Multiplication Grid Method, Multiplication Written Method, Ratio and Scaling Numbers, Representing Multiplication, Division by Chunking Up, Division by Chunking Down, Formal Written Division - Round Up or Down?, Short Division Writen Formal Method.
The Calculations app is available on Google Play and the App Store.
Multiplication_Beat_The_Clock
Score as many points as you can before the time runs out. A number of levels including addition, subtraction, multiplication, division and matching fractions and decimals.
*Updated 30th May 2013 - choose multiple objectives, or objectives grouped by national curriculum level. The game also now includes a scoreboard. An iPad version is also available
For more multiplication and division resources click here.
Balancing Calculations
Find the missing number to balance the calculations. An excellent tool for reinforcing an understanding of the role of the equals sign. Choose one objective, or many. You can choose all the objectives for a single year group (or multiple year groups). You can also choose to balance between different types of calculations (eg subtraction and division)
For more multiplication and division resources click here.
Multiplication - Rapid Recall
Drag the correct number to complete the multiplication sentence. Choose which times table to practice.
For more multiplication and division resources click here.
Carroll Sort Numbers
Sort numbers on a Carroll diagram. Sort according to one or two properties, including: odd, even, multiples, square numbers, prime numbers, triangular numbers, numbers less than 30 or more than 50.
For more resources involving sorting shapes and numbers click here.
Division Rapid Recall
Answer the division questions as quickly as possible. Choose a number to divdide by, or choose a range of numbers.
For more multiplication and division resources click here.
Venn sort numbers
Sort numbers on a Venn diagram. Sort according to one or two properties, including: odd, even, multiples, square numbers, prime numbers, triangular numbers, numbers less than 30 or more than 50.
For more resources involving sorting shapes and numbers click here.
Card Game - Matching Pairs
Match the pairs of cards with the same value. A great game to play as a mental starter. Over 50 levels to play including; addition, subtraction, multiplication, division, matching fractions and decimals and converting between measurements.
100 Square
An incredibly versatile teaching tool. You can change the start number and change the step size, then try and find the hidden number. Choose from several difficulty levels. Ask the children to discuss the different methods they could use to find the missing number.
Can be used to explore sequences (including negative numbers and decimals), multiplication tables, as well as reasoning about numbers.
Money Multiplication problems
ITP Grouping
View full screen in your browser. This ITP allows you to display up to 30 counters or shapes on the screen. You then select a number to be the divisor in a division calculation. A number line displays the number to be divided. As individual counters or shapes are clicked and dragged to form a group the size of the divisor, they change colour. Once a group equal to the divisor is selected, it ‘jumps’ to the number line. In this example 22 counters were chosen and the divisor set to 5.
In the example shown below, 3 sets of 5 shapes have been selected and moved to the number line, which shows each group as a jump. Four more counters have been dragged together and the fifth is about to be selected. The answer to the division calculation is shown at the bottom of the screen.
The ITP can be used to model division as grouping and to link this process to jumps on a number line. It can also introduce children to how remainders are recorded in the answer.
This resource is freely available to download from the archived Primary Framework site.
For more multiplication and division resources click here.
ITP Multiplication Array
October 2012 - I have added a new game which represents multiplication on a number line and using a dotty array. It is ideal as an introduction to the grid method, as it provides a concrete image of how and why we partition numbers to multiply. Click here to play.
This resource is freely available to download from the archived Primary Framework site.
For more multiplication and division resources click here.
ITP Multiplication Board
View full screen in your browser. Multiplication board ITP allows the child or teacher to represent the product of two numbers as an array, displaying the product and factors.
This resource is freely available to download from the archived Primary Framework site
For more multiplication and division resources click here.
ITP Multiplication Facts
View full screen in your browser. This ITP allows you to represent multiplication as repeated addition using a grid of blocks or counters.It can be used to develop children’s understanding of multiplication and to develop links between the different representations and notation. The dynamic images should help children to understand why 5 x 9 means that the 5 is multiplied by the 9, and to recognise that multiplication is a commutative operation.
This resource is freely available to download from the archived Primary Framework site
For more multiplication and division resources click here.
ITP Multiplication Tables
View full screen in your browser. Complete the multiplication tables.
This resource is freely available to download from the archived Primary Framework site
For more multiplication and division resources click here.
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Creative Learner Math Worksheets
Karoly Zeynep November 26, 2020 Worksheet
As a parent, I’m very aware of what my own children are learning in school. For the most part, I’ve been happy with their progress, but as they rise in grade level, I’m starting to see more emphasis on a loose understanding of the concepts and less emphasis on skills–particularly skills with arithmetic of fractions. The main problem with what I see with my students and my own children is that kids are taught ”concepts” and are not taught skills–unless they’re lucky enough to have a teacher who knows better. Most particularly, children are not taught mastery of arithmetic with fractions. Unfortunately, virtually all of their future math education depends on being able to do fractional arithmetic.
Rather than using worksheets, a better method is to use individual size white boards and have the child writing entire facts many times. Having a child writing 9 x 7 = 7 x 9 = 63 while saying ”nine times seven is the same as seven times nine and is equal to sixty-three” is many times more successful than a worksheet with 9 x 7 = ___ and the student just thinks the answer once and then writes that answer on the duplicate problems.
The primary problem with most math worksheets is that the problems are already written out and the child need only write the answers. For learning and practicing the basic skills of addition, subtraction, multiplication, and division, it is much more beneficial for the child to write out the entire fact and say the entire fact out loud. A child will learn a multiplication fact much faster if they are writing out 6 x 8 = 48 at the same time they are saying ”six times eight is forty-eight” than if they just see 6 x 8 = ___ and only have to supply the 48.
Some students are unable to access tools that many of us take for granted when they try to complete worksheets. They may be unable to grasp pencils, control their movements within the limited spaces provided on the sheet, or be able to simply stabilize their paper while writing. Other students, including those for whom English is not their primary language or who struggle with reading, have difficulty reading the directions, words, and math terminology on the worksheets. Still other students require different visual representations or methods of engagement in order acquire an understanding the content. Most math worksheets do not provide information in multiple formats so they are inaccessible to students with a wide variety of learning styles and abilities. Well-designed technology can provide these students with access to excellent content. For example, these fractions tools and supplemental curriculum allow students with physical disabilities to access fractions content using a variety of assistive technology devices. Instructions, prompts and feedback can be read aloud, while visual models, cues combined with sounds support a wide range of learning styles and abilities.
The most important thing about these math worksheets is that they are used for tutoring and not for the main course studies. That is why they are used by tutors to offer remedial tuition and by parents at home so that they can offer their kids extra tuition to sharpen their skills. Math is known to be difficult and is often a headache for the young and so the math worksheets come in handy in helping resolve this problem. Thanks to the sites over the internet that offer free printable math worksheets, you do not need to worry about the cost of purchasing one, maybe only the ink cost. So don’t go making excuses for not being able to access a math work sheet.
Math worksheets rarely ask students to think critically or creatively. They usually present multiple examples of the same problem type with the hope of reinforcing a skill or procedure. They do not challenge students to use higher order thinking skills such as comparing, analyzing, deducing, and synthesizing. These skills are built through activities in which students discover concepts, explore ideas, test a hypothesis, solve a problem, and discuss their thinking with their peers. Exploring concepts and problems in many different ways builds interest and promotes critical thinking.
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## let log 4=.45,log 3=.62 Find log(36) and log (4/3)
Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
sanman777
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### let log 4=.45,log 3=.62 Find log(36) and log (4/3)
let log 4=.45,log 3=.62 Find log(36) and log (4/3)
need help also
stapel_eliz
Posts: 1628
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let log 4=.45,log 3=.62 Find log(36) and log (4/3)
need help also
Use log rules to break the log(36) and the log(4/3) into logs only in terms of 3 and 4. For instance:
. . . . .log(12) = log(3*4) = log(3) + log(4) = 0.45 + 0.62 = 1.07
If you get stuck, please reply showing your progress. Thank you!
sanman777
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### Re: let log 4=.45,log 3=.62 Find log(36) and log (4/3)
so log(36)=log(6)+log(6)=.778=.778=1.556 is that right?
stapel_eliz
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so log(36)=log(6)+log(6)=.778=.778=1.556 is that right?
You were given the value of log(6)...? (I'm not seeing that in your original post.)
sanman777
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### Re: let log 4=.45,log 3=.62 Find log(36) and log (4/3)
log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69
im not sure how to do it
stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
log(36)=log(3)+log(3)+log(4)=.62+.62+.45=1.69
im not sure how to do it
The way you did it is fine. As the lesson you studied (in the link, provided earlier) explained and demonstrated, you use the log rules to break the given log into separate terms which will relate to the given values. So the missing steps (that you'll probably want to include in your hand-in homework) make the solution look like:
. . . . .log(36) = log(9*4) = log(9) + log(4)
. . . . .= log(32) + log(4) = 2*log(3) + log(4)
. . . . .= (2)(0.62) + 0.45
...and so forth.
The other one works the exact same way.
sanman777
Posts: 11
Joined: Mon Mar 23, 2009 3:28 pm
Contact:
### Re: let log 4=.45,log 3=.62 Find log(36) and log (4/3)
oh ok thanks alot
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http://mymathforum.com/algebra/337239-one-car-travels-140-miles-same-amount-time-takes-second-car-traveling.html
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My Math Forum One car travels 140 miles in the same amount of time it takes a second car traveling
Algebra Pre-Algebra and Basic Algebra Math Forum
November 8th, 2016, 01:57 PM #1 Senior Member Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10 One car travels 140 miles in the same amount of time it takes a second car traveling One car travels 140 miles in the same amount of time it takes a second car traveling 6 miles per hour slower than the first to go 116 miles. What are the speeds of the cars? sorry if I asked this before could not find it. Thanks
November 8th, 2016, 02:04 PM #2 Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521 $v \cdot t = 140$ $(v-6)t = 116$ solve the system for $v$, the speed of the faster car, then subtract 6 to get the speed of the slower car Last edited by skipjack; November 8th, 2016 at 04:49 PM.
November 8th, 2016, 04:16 PM #3 Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Notice that you can eliminate "t" by dividing one equation by the other!
November 8th, 2016, 05:08 PM #4 Global Moderator Joined: Dec 2006 Posts: 20,641 Thanks: 2083 As the first car travels 24 miles further when its speed is 6 mph more than the second car's, their journey time is 4 hours each. Hence the first car travels at 140/4 miles per hour (which is 35 mph), and the second car travels at 116/4 miles per hour (which is 29 mph).
November 10th, 2016, 01:32 AM #5 Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $v⋅t=140....(1)$ $(v−6)t=116.......(2)$ divide equation(1) by (2) $\dfrac{v\not t}{(v-6)\not t}=\dfrac{140}{116}$ $116v=140v-840$ $-24v=-840$ $v=35\ (\text{faster car})$ $v=29\ (\text{slower car})$ Last edited by skipjack; November 10th, 2016 at 01:35 AM.
November 11th, 2016, 09:59 AM #6 Member Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 I first approached the problem with a testing method: If both cars drove for 1 hour. The first car would have to drive 140 mph to cover 140 miles. The slower car would have to drive 116 mph. However, 140-116 is 24mph difference; not the stated 6 mph difference. Then made each car drive for 2 hours. The first car would have to drive 70mph (140 divided by 2), while the second car would have to drive 58 mph (116 divided by 2). 70-68 is a 12 mph difference; not a 6 mph difference. I noticed when I doubled the time, the speed different cut in half from 24 down to 12 mph. Therefore I knew the next doubling of time to 4 hours would show a 6 mph difference in speed. And it does. At 4 hours of driving, the faster car travels 35 mph (140 divided by 4) and the slower car travels 29mph (116 divided by 4). 35-29 =6 mph. __________________________________________________ ___________ To solve this problem another way, I used Distance = Rate x Time; D=R x T; D=RT and solved it for Time first: 140=RT note: R is the speed for the faster car 116=(R-6)T (R-6) is the speed of the slower car Next, I subtracted those two equations above 140-116 = 24 RT - (R-6)T factor out T T[R-(R-6)] simplify T[-(-6)] T[6] =6T 24 = 6T T= 4 Finally, I solved for rate D = R x T 140= R x 4 R= 35 mph Faster car 35-6 = 29 mph Slower car
December 19th, 2016, 10:58 AM #7 Member Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 I first approached the problem with a testing method: if both cars drove for 1 hour. In this case, the first car would have to drive 140 mph to cover 140 miles. The slower car would have to drive 116 mph. However, 140-116 is 24mph difference; not the stated 6 mph difference. Then I made each car drive for 2 hours. The first car would have to drive 70mph (140 divided by 2), while the second car would have to drive 58 mph (116 divided by 2). 70-68 is a 12 mph difference; not a 6 mph difference. I noticed when I doubled the time, the speed difference cut in half from 24 down to 12 mph. Therefore I knew the next doubling of time to 4 hours would show a 6 mph difference in speed. And it does. At 4 hours of driving, the faster car travels 35 mph (140 divided by 4) and the slower car travels 29mph (116 divided by 4). 35-29 =6 mph. __________________________________________________ ___________ To solve this problem another way, I used Distance = Rate x Time; D=R x T; D=RT and solved it for Time first: 140=RT note: R is the speed for the faster car 116=(R-6)T (R-6) is the speed of the slower car Next, I subtracted those two equations above 140-116 = 24 RT - (R-6)T factor out T T[R-(R-6)] simplify T[-(-6)] T[6] =6T 24 = 6T T= 4 Finally, I solved for rate D = R x T 140= R x 4
Tags 140, amount, car, miles, takes, time, traveling, travels
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http://us.metamath.org/mpeuni/itunitc.html
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > itunitc Structured version Visualization version GIF version
Theorem itunitc 9435
Description: The union of all union iterates creates the transitive closure; compare trcl 8777. (Contributed by Stefan O'Rear, 11-Feb-2015.)
Hypothesis
Ref Expression
ituni.u 𝑈 = (𝑥 ∈ V ↦ (rec((𝑦 ∈ V ↦ 𝑦), 𝑥) ↾ ω))
Assertion
Ref Expression
itunitc (TC‘𝐴) = ran (𝑈𝐴)
Distinct variable group: 𝑥,𝐴,𝑦
Allowed substitution hints: 𝑈(𝑥,𝑦)
Proof of Theorem itunitc
Dummy variables 𝑎 𝑏 𝑐 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fveq2 6352 . . . 4 (𝑎 = 𝐴 → (TC‘𝑎) = (TC‘𝐴))
2 fveq2 6352 . . . . . 6 (𝑎 = 𝐴 → (𝑈𝑎) = (𝑈𝐴))
32rneqd 5508 . . . . 5 (𝑎 = 𝐴 → ran (𝑈𝑎) = ran (𝑈𝐴))
43unieqd 4598 . . . 4 (𝑎 = 𝐴 ran (𝑈𝑎) = ran (𝑈𝐴))
51, 4eqeq12d 2775 . . 3 (𝑎 = 𝐴 → ((TC‘𝑎) = ran (𝑈𝑎) ↔ (TC‘𝐴) = ran (𝑈𝐴)))
6 vex 3343 . . . . . . 7 𝑎 ∈ V
7 ituni.u . . . . . . . 8 𝑈 = (𝑥 ∈ V ↦ (rec((𝑦 ∈ V ↦ 𝑦), 𝑥) ↾ ω))
87ituni0 9432 . . . . . . 7 (𝑎 ∈ V → ((𝑈𝑎)‘∅) = 𝑎)
96, 8ax-mp 5 . . . . . 6 ((𝑈𝑎)‘∅) = 𝑎
10 fvssunirn 6378 . . . . . 6 ((𝑈𝑎)‘∅) ⊆ ran (𝑈𝑎)
119, 10eqsstr3i 3777 . . . . 5 𝑎 ran (𝑈𝑎)
12 dftr3 4908 . . . . . 6 (Tr ran (𝑈𝑎) ↔ ∀𝑏 ran (𝑈𝑎)𝑏 ran (𝑈𝑎))
137itunifn 9431 . . . . . . . 8 (𝑎 ∈ V → (𝑈𝑎) Fn ω)
14 fnunirn 6674 . . . . . . . 8 ((𝑈𝑎) Fn ω → (𝑏 ran (𝑈𝑎) ↔ ∃𝑐 ∈ ω 𝑏 ∈ ((𝑈𝑎)‘𝑐)))
156, 13, 14mp2b 10 . . . . . . 7 (𝑏 ran (𝑈𝑎) ↔ ∃𝑐 ∈ ω 𝑏 ∈ ((𝑈𝑎)‘𝑐))
16 elssuni 4619 . . . . . . . . 9 (𝑏 ∈ ((𝑈𝑎)‘𝑐) → 𝑏 ((𝑈𝑎)‘𝑐))
177itunisuc 9433 . . . . . . . . . 10 ((𝑈𝑎)‘suc 𝑐) = ((𝑈𝑎)‘𝑐)
18 fvssunirn 6378 . . . . . . . . . 10 ((𝑈𝑎)‘suc 𝑐) ⊆ ran (𝑈𝑎)
1917, 18eqsstr3i 3777 . . . . . . . . 9 ((𝑈𝑎)‘𝑐) ⊆ ran (𝑈𝑎)
2016, 19syl6ss 3756 . . . . . . . 8 (𝑏 ∈ ((𝑈𝑎)‘𝑐) → 𝑏 ran (𝑈𝑎))
2120rexlimivw 3167 . . . . . . 7 (∃𝑐 ∈ ω 𝑏 ∈ ((𝑈𝑎)‘𝑐) → 𝑏 ran (𝑈𝑎))
2215, 21sylbi 207 . . . . . 6 (𝑏 ran (𝑈𝑎) → 𝑏 ran (𝑈𝑎))
2312, 22mprgbir 3065 . . . . 5 Tr ran (𝑈𝑎)
24 tcmin 8790 . . . . . 6 (𝑎 ∈ V → ((𝑎 ran (𝑈𝑎) ∧ Tr ran (𝑈𝑎)) → (TC‘𝑎) ⊆ ran (𝑈𝑎)))
256, 24ax-mp 5 . . . . 5 ((𝑎 ran (𝑈𝑎) ∧ Tr ran (𝑈𝑎)) → (TC‘𝑎) ⊆ ran (𝑈𝑎))
2611, 23, 25mp2an 710 . . . 4 (TC‘𝑎) ⊆ ran (𝑈𝑎)
27 unissb 4621 . . . . 5 ( ran (𝑈𝑎) ⊆ (TC‘𝑎) ↔ ∀𝑏 ∈ ran (𝑈𝑎)𝑏 ⊆ (TC‘𝑎))
28 fvelrnb 6405 . . . . . . 7 ((𝑈𝑎) Fn ω → (𝑏 ∈ ran (𝑈𝑎) ↔ ∃𝑐 ∈ ω ((𝑈𝑎)‘𝑐) = 𝑏))
296, 13, 28mp2b 10 . . . . . 6 (𝑏 ∈ ran (𝑈𝑎) ↔ ∃𝑐 ∈ ω ((𝑈𝑎)‘𝑐) = 𝑏)
307itunitc1 9434 . . . . . . . . 9 ((𝑈𝑎)‘𝑐) ⊆ (TC‘𝑎)
3130a1i 11 . . . . . . . 8 (𝑐 ∈ ω → ((𝑈𝑎)‘𝑐) ⊆ (TC‘𝑎))
32 sseq1 3767 . . . . . . . 8 (((𝑈𝑎)‘𝑐) = 𝑏 → (((𝑈𝑎)‘𝑐) ⊆ (TC‘𝑎) ↔ 𝑏 ⊆ (TC‘𝑎)))
3331, 32syl5ibcom 235 . . . . . . 7 (𝑐 ∈ ω → (((𝑈𝑎)‘𝑐) = 𝑏𝑏 ⊆ (TC‘𝑎)))
3433rexlimiv 3165 . . . . . 6 (∃𝑐 ∈ ω ((𝑈𝑎)‘𝑐) = 𝑏𝑏 ⊆ (TC‘𝑎))
3529, 34sylbi 207 . . . . 5 (𝑏 ∈ ran (𝑈𝑎) → 𝑏 ⊆ (TC‘𝑎))
3627, 35mprgbir 3065 . . . 4 ran (𝑈𝑎) ⊆ (TC‘𝑎)
3726, 36eqssi 3760 . . 3 (TC‘𝑎) = ran (𝑈𝑎)
385, 37vtoclg 3406 . 2 (𝐴 ∈ V → (TC‘𝐴) = ran (𝑈𝐴))
39 rn0 5532 . . . . 5 ran ∅ = ∅
4039unieqi 4597 . . . 4 ran ∅ =
41 uni0 4617 . . . 4 ∅ = ∅
4240, 41eqtr2i 2783 . . 3 ∅ = ran ∅
43 fvprc 6346 . . 3 𝐴 ∈ V → (TC‘𝐴) = ∅)
44 fvprc 6346 . . . . 5 𝐴 ∈ V → (𝑈𝐴) = ∅)
4544rneqd 5508 . . . 4 𝐴 ∈ V → ran (𝑈𝐴) = ran ∅)
4645unieqd 4598 . . 3 𝐴 ∈ V → ran (𝑈𝐴) = ran ∅)
4742, 43, 463eqtr4a 2820 . 2 𝐴 ∈ V → (TC‘𝐴) = ran (𝑈𝐴))
4838, 47pm2.61i 176 1 (TC‘𝐴) = ran (𝑈𝐴)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 196 ∧ wa 383 = wceq 1632 ∈ wcel 2139 ∃wrex 3051 Vcvv 3340 ⊆ wss 3715 ∅c0 4058 ∪ cuni 4588 ↦ cmpt 4881 Tr wtr 4904 ran crn 5267 ↾ cres 5268 suc csuc 5886 Fn wfn 6044 ‘cfv 6049 ωcom 7230 reccrdg 7674 TCctc 8785 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 ax-8 2141 ax-9 2148 ax-10 2168 ax-11 2183 ax-12 2196 ax-13 2391 ax-ext 2740 ax-rep 4923 ax-sep 4933 ax-nul 4941 ax-pow 4992 ax-pr 5055 ax-un 7114 ax-inf2 8711 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1073 df-3an 1074 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2047 df-eu 2611 df-mo 2612 df-clab 2747 df-cleq 2753 df-clel 2756 df-nfc 2891 df-ne 2933 df-ral 3055 df-rex 3056 df-reu 3057 df-rab 3059 df-v 3342 df-sbc 3577 df-csb 3675 df-dif 3718 df-un 3720 df-in 3722 df-ss 3729 df-pss 3731 df-nul 4059 df-if 4231 df-pw 4304 df-sn 4322 df-pr 4324 df-tp 4326 df-op 4328 df-uni 4589 df-int 4628 df-iun 4674 df-br 4805 df-opab 4865 df-mpt 4882 df-tr 4905 df-id 5174 df-eprel 5179 df-po 5187 df-so 5188 df-fr 5225 df-we 5227 df-xp 5272 df-rel 5273 df-cnv 5274 df-co 5275 df-dm 5276 df-rn 5277 df-res 5278 df-ima 5279 df-pred 5841 df-ord 5887 df-on 5888 df-lim 5889 df-suc 5890 df-iota 6012 df-fun 6051 df-fn 6052 df-f 6053 df-f1 6054 df-fo 6055 df-f1o 6056 df-fv 6057 df-om 7231 df-wrecs 7576 df-recs 7637 df-rdg 7675 df-tc 8786 This theorem is referenced by: hsmexlem5 9444
Copyright terms: Public domain W3C validator
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Motion System Design
# Fun with Fundamentals: Problem 192
### π in the sky
Problem 192 — Getting to the root of a problem can clear up areas of doubt, as this month’s problem by Steve Zhang of Aloe, Ore., demonstrates.
Deep in the Pentagon, a highly classified meeting between the joint chiefs of staff and their new radar supplier was in progress.
“I’m pleased to announce we’ve found a radar net system that’s both cost-effective and fully functional,” stated Brig. Gen. Brass. “Without further ado, here’s Mortimer Blynd of Tri-Gen/Git-Rite Inc.”
“The new Failsafe Radar Integrated Tracking System (FRITS) consists of four towers that each have a sweep radius of 10√3 miles ,” stated Blynd. “These towers (A, B, C, D) form a square, (see diagram), with an area of 900 square miles. Together, *ahem*, with some negligible overlap and one very small blind spot, *ahem*, they can monitor this area.”
Given the measurements stated, what is the area of the blind spot? Can the enemy hide a tank or a munitions plant in it?
Technical consultant, Jack Couillard, Menasha, Wis.
Solution to last month’s problem 191 — Your ETA is right on target if you answered Lottie and her boyfriend get wet. Here’s the watered-down analysis:
First, let’s find out if the water balloon will strike the Ferris wheel or pass over it.
Let:
y1 = Maximum height attained by the water balloon, ft
t = Time from launch to maximum height, min
x = Horizontal distance from launch point to the point directly beneath the balloon’s zenith, ft
v0 = Initial velocity of water balloon, given as 60 fpm
Φ = Water balloon launch angle, given as 60 deg
g = Acceleration due to gravity, 32.2 ft/sec2
From physics, we know that the time it takes a projectile to reach its zenith is:
The balloon’s maximum elevation is:
x = v0(cosΦ)t = 48.4 ft (1)
The horizontal distance the water balloon has traveled at this time is:
Since the closest point of the Ferris wheel is 55 ft away, the balloon is descending by the time it reaches the unsuspecting couple. At this point you can draw a diagram to scale and note that the balloon strikes the wheel at a horizontal distance between 65 and 70 ft. Use (1) to find that the time is between 2.16 and 2.33 sec.
The Ferris wheel turns at 1 rpm, so between the time frame allotted, the wheel has turned 13 or 14 degrees in its rotation. Lottie and company are clearly the ones in the line of fire.
Contest winner — Congratulations to Joe Kohler of Cleveland, who won our January contest by having his name drawn from the 11 contestants who answered correctly out of a total of 32 entrants for that month. A TI-85 calculator is in the mail to him.
The TI-85 Graphing Calculator from Texas Instruments solves for any variable in an equation, can solve 30 simultaneous equations, and finds the roots of a polynomial up to the 30th order. It handles complex numbers in addition to matrices, vectors, lists, and strings. You can perform graphic investigations of almost any type of problem — functions as well as parametric, polar, and differential equations.
### Related Article
Fun with Fundamentals: Problem 191
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# Using Taylor's theorem to show existence of a delta
If we have a function $f:[0,1]\longrightarrow\mathbb{R}$ that is eight times differentiable, and know that
$f'(0)=f''(0)=f'''(0)=f^{(4)}(0)=f^{(5)}(0)=0$ and $f^{(6)}>0$,
then how can we show that there exists a $\delta\in(0,1)$ such that $f(x)\ge f(0) + \frac{f^{(6)}(0)}{721}x^6$ for all $x\in[0,\delta)$?
I thought a sensible approach would be to begin by using Taylor's theorem since that last $x^6$ term looks like a Taylor remainder with the 720 being $6!+1$. However, this hasn't gotten me anywhere.
I also tried to show that for $\epsilon>0$ there exists a $\delta\in(0,1)$ such that for all $x\in(0,\delta)$ we have that $|\frac{f(x)-p_6(x)}{x^6}|\lt\epsilon$ where $p_6(x)$ is the 6th Taylor polynomial, since we should then be able to choose a particular $\epsilon$ that gives us our result, but I also couldn't manage to prove this - does anyone have any hints or advice?
• The taylor-Lagrange thm is a good idea, but use it with order $7$: $$f(x)=f(0)+\frac{f^{(6)}(0)}{6!}x^6+\frac{f^{(7)}(c)}{7!}x^7$$, with $c$ between $0$ and $x$, and use that as $f^{(7)}$ is continuous, $|f^{(7)}|$ is bounded, say by $M$ on $[0,1]$. You get $f(x)\geq f(0)+\frac{f^{(6)}(0)}{6!}x^6-M\frac{x^7}{7!}$ for all $x\in [0, 1]$ and now it is easy to finish. Oct 24, 2017 at 14:17
• I think I follow what you're saying so far, but isn't this completely independent of epsilon or delta? Oct 24, 2017 at 14:37
• To finish, you have to find $\delta>0$ such that for $0<x<\delta$ $f(0)+\frac{f^{(6)}(0)}{6!}x^6-M\frac{x^7}{7!}\geq f(0)+\frac{f^{(6)}(0)}{6!+1}x^6$, or $\frac{f^{(6)}(0)}{6!}-M\frac{x}{7!}\geq \frac{f^{(6)}(0)}{6!+1}$ Oct 24, 2017 at 14:43
• Sorry I don't think I quite follow... also how come the $\frac{f^{(7)}(c)}{7!}$ became -M? Oct 24, 2017 at 14:58
• I have used that for $x\in [0,1]$ we have $|\frac{f^{(7)}(c)}{7!}x^7]\leq \frac{M}{7!}x^7$, so $f(0)+\frac{f^{(6)}(0)}{6!}x^6+\frac{f^{(7)}(c)}{7!}x^7\geq f(0)+\frac{f^{(6)}(0)}{6!}x^6-\frac{M}{7!}x^7$ Oct 24, 2017 at 15:04
The Taylor-Lagrange thm is a good idea, but use it with order $7$: $$f(x)=f(0)+\frac{f^{(6)}(0)}{6!}x^6+\frac{f^{(7)}(c)}{7!}x^7$$, with $c$ between $0$ and $x$. Now we use that (as $f$ est $8$ times differentiable) the function $f^{(7)}$ is continuous, hence $|f^{(7)}|$ is bounded, say by $M>0$ on $[0,1]$.
We have $|\frac{f^{(7)}(c)}{7!}x^7|\leq \frac{M}{7!} x^7$ for $x\in [0,1]$ hence we get that $f(x)\geq f(0)+\frac{f^{(6)}(0)}{6!}x^6-M\frac{x^7}{7!}$
To have our result, it suffices now to find a $\delta>0$, such that
$$\frac{f^{(6)}(0)}{6!}-\frac{f^{(6)}(0)}{1+6!}\geq M \frac{x}{7!}$$ for $0\leq x<\delta$, that is easy, we can take $\delta=\frac{7!}{M}(\frac{f^{(6)}(0)}{6!}-\frac{f^{(6)}(0)}{1+6!})$.
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# Calculation of 4x4 matrix of planes
Hi,
I am using the example of ImplicitPlaneWidget2.
I want to be able to manipulate this by inputting a 3D image (.raw).
I would like to write a program to calculate 4x4 matrix in plane.
The code in vtkImageResliceMapper can not understand the following “axis, saxis, taxis”.
Can you explain it?
``````// Create two orthogonal axes
double saxis[3], taxis[3];
taxis[0] = 0.0;
taxis[1] = 1.0;
taxis[2] = 0.0;
if (maxi == 1)
{
taxis[1] = 0.0;
taxis[2] = 1.0;
}
vtkMath::Cross(taxis, axis, saxis);
``````
https://gitlab.kitware.com/vtk/vtk/blob/master/Rendering/Image/vtkImageResliceMapper.cxx#L600
@dgobbi
It is hard to explain without pictures, but I will try.
The vectors saxis[3], taxis[3], and axis[3] are three orthogonal axes where ‘axis’ is roughly aligned with the slice normal. The only important thing about ‘saxis’ and ‘taxis’ is that they are orthogonal to ‘axis’.
The code below computes the rotation between ‘axis’ and ‘normal’:
``````// Compute the rotation angle between the axis and the normal
double vec[3]; // vec[3] is the axis of rotation
vtkMath::Cross(axis, normal, vec);
double costheta = vtkMath::Dot(axis, normal);
double sintheta = vtkMath::Norm(vec);
double theta = atan2(sintheta, costheta);
``````
Then ‘vec[3]’ and ‘angle’ are used to compute a 3x3 rotation matrix. We use this matrix to rotate ‘saxis[3]’ and ‘taxis[3]’:
``````// Create a slice-to-data transform matrix
// The columns are v1, v2, normal
double v1[3], v2[3];
vtkMath::Multiply3x3(mat, saxis, v1);
vtkMath::Multiply3x3(mat, taxis, v2);
resliceMatrix->Element[0][0] = v1[0];
resliceMatrix->Element[1][0] = v1[1];
resliceMatrix->Element[2][0] = v1[2];
resliceMatrix->Element[3][0] = 0.0;
resliceMatrix->Element[0][1] = v2[0];
resliceMatrix->Element[1][1] = v2[1];
resliceMatrix->Element[2][1] = v2[2];
resliceMatrix->Element[3][1] = 0.0;
resliceMatrix->Element[0][2] = normal[0];
resliceMatrix->Element[1][2] = normal[1];
resliceMatrix->Element[2][2] = normal[2];
resliceMatrix->Element[3][2] = 0.0;
``````
The whole purpose of this is to compute vectors ‘v1[3]’ and ‘v2[3]’ that are perpendicular to ‘normal[3]’. All of this fancy math is used to ensure that ‘v1[3]’ and ‘v2[3]’ are as close as possible to the original volume axes.
I’m having a hard time understanding the linked code from line 600 to line 642.
Could you explain briefly?
@dgobbi
The ‘plane[4]’ is the coefficients a,b,c,d for the
equation of a plane. It is important to understand that the first three components a,b,c of plane[4] are the normal, so on line 664 it is possible to say `normal = plane`.
LInes 600 to 630 get the plane (and wplane, which is the plane after the actor transformation from ‘data’ coords to ‘world’ coords). The plane is oriented so that the normal points towards the ‘camera’ (this means it points out of the computer screen, not into the computer screen).
Lines 631 to 642 do exactly what the comment says: find the largest component of the normal. This checks whether the normal is closest to the x, the y, or the z axis.
For your original question, I think it is most important to understand how ‘v1’ and ‘v2’ are computed. Compare this to the simpler (much less general) answer I gave to Post 1391, which computes the matrix only for a rotation around the Z axis:
Thank you for your detailed explanation.
It helped me to understand.
Thanks to your help, I have a rough understanding of the cut plane matrix and compute method.
Also, I could understand how to compute ‘v1’ and ‘v2’.
What should I do as a next step in creating a program?
Is it difficult to incorporate it into the program of ImplicitPlaneWidget2 from the beginning?
I’m having trouble getting started with program improvement.
@dgobbi
You’re asking a little too much. Just start anywhere, and when you start things will become clearer. When you have some code written, I’d be glad to check it for you.
I will try to make a program by myself.
Based on the code of vtkImageResliceMapper.cxx, I created a program to calculate 4 x 4 matrix on a plane.
The code created is shown below.
`````` // Setup a visualization pipeline
vtkSmartPointer<vtkPlane> plane =vtkSmartPointer<vtkPlane>::New();
double normal[3];
plane->GetNormal(normal);
double Origin[3];
plane->GetOrigin(Origin);
// Set the slice orientation
vtkSmartPointer<vtkMatrix4x4> resliceMatrix =
vtkSmartPointer<vtkMatrix4x4>::New();
// Create the corresponding axis
double axis[3];
axis[0] = 0.0;
axis[1] = 0.0;
axis[2] = 0.0;
double saxis[3], taxis[3];
taxis[0] = 0.0;
taxis[1] = 1.0;
taxis[2] = 0.0;
vtkMath::Cross(taxis, axis, saxis);
// Compute the rotation angle between the axis and the normal
double vec[3];
vtkMath::Cross(axis, normal, vec);
double costheta = vtkMath::Dot(axis, normal);
double sintheta = vtkMath::Norm(vec);
double theta = atan2(sintheta, costheta);
if (sintheta != 0)
{
vec[0] /= sintheta;
vec[1] /= sintheta;
vec[2] /= sintheta;
}
// convert to quaternion
costheta = cos(0.5*theta);
sintheta = sin(0.5*theta);
double quat[4];
quat[0] = costheta;
quat[1] = vec[0]*sintheta;
quat[2] = vec[1]*sintheta;
quat[3] = vec[2]*sintheta;
// convert to matrix
double mat[3][3];
vtkMath::QuaternionToMatrix3x3(quat, mat);
// Create a slice-to-data transform matrix
// The columns are v1, v2, normal
double v1[3], v2[3];
vtkMath::Multiply3x3(mat, saxis, v1);
vtkMath::Multiply3x3(mat, taxis, v2);
resliceMatrix->Element[0][0] = v1[0];
resliceMatrix->Element[1][0] = v1[1];
resliceMatrix->Element[2][0] = v1[2];
resliceMatrix->Element[3][0] = 0.0;
resliceMatrix->Element[0][1] = v2[0];
resliceMatrix->Element[1][1] = v2[1];
resliceMatrix->Element[2][1] = v2[2];
resliceMatrix->Element[3][1] = 0.0;
resliceMatrix->Element[0][2] = normal[0];
resliceMatrix->Element[1][2] = normal[1];
resliceMatrix->Element[2][2] = normal[2];
resliceMatrix->Element[3][2] = 0.0;
resliceMatrix->Element[0][3] = Origin[0];
resliceMatrix->Element[1][3] = Origin[1];
resliceMatrix->Element[2][3] = Origin[2];
resliceMatrix->Element[3][3] = 1.0;
``````
Can you tell me if there is something to be improved?
@dgobbi
You misunderstood the purpose of ‘axis’, ‘taxis’ and ‘saxis’. Here are two concrete examples that might help. The goals are:
• ‘axis’ must be approximately the same direction as ‘normal’
• ‘saxis’, ‘taxis’, and ‘axis’ must be orthonormal (and must obey right-hand-rule)
Example 1) if ‘normal’ is [0.48716132, 0.84723708, -0.21180927] then:
• ‘axis’ should be [0, 1, 0]
• ‘taxis’ should be [0, 0, 1]
• ‘saxis’ should be [1, 0, 0]
Example 2) if ‘normal’ is [0.2470483, 0.19003716, -0.95018578] then:
• ‘axis’ should be [0, 0,-1]
• ‘taxis’ should be [0, 1, 0]
• ‘saxis’ should be [1, 0, 0]
The purpose of the rest of the math is to constrain the rotation of the 4x4 matrix such that the axis-of-rotation is parallel to the plane (so there is no in-plane rotation).
I could understand ‘axis’, ‘taxis’ and ‘saxis’ by example.
It has been improved as follows.
``````int maxi = 0;
double maxv = 0.0;
for (int i = 0; i < 3; i++)
{
double tmp = normal[i]*normal[i];
if (tmp > maxv)
{
maxi = i;
maxv = tmp;
}
}
// Create the corresponding axis
double axis[3];
axis[0] = 0.0;
axis[1] = 0.0;
axis[2] = 0.0;
axis[maxi] = ((normal[maxi] < 0.0) ? -1.0 : 1.0);
double saxis[3], taxis[3];
taxis[0] = 0.0;
taxis[1] = 1.0;
taxis[2] = 0.0;
if (maxi == 1)
{
taxis[1] = 0.0;
taxis[2] = 1.0;
}
vtkMath::Cross(taxis, axis, saxis);
``````
Is the fourth row of the 4x4 matrix correct?
``````resliceMatrix->Element[0][3] = Origin[0];
resliceMatrix->Element[1][3] = Origin[1];
resliceMatrix->Element[2][3] = Origin[2];
resliceMatrix->Element[3][3] = 1.0;``````
Before I answer that question, I will give a brief summary. So far, we have computed a rotation that takes the 2D plane (the XY plane at Z=0, which will be the output of vtkImageReslice) and rotates it so that it becomes parallel with the 3D slice plane. The next thing we have to do is translate the rotated 2D plane so that is in the same position as the 3D slice plane.
The fourth column of the 4x4 matrix gives the translation.
So the quick answer to your question is: yes, setting the translation to the plane “Origin” is a solution. This works because the point (0,0,0) is on the 2D plane (obviously!), and if we rotate that point it is still (0,0,0), and if we then translate that point by (Origin[0], Origin[1], Origin[2]), then it moves to the vtkPlane’s Origin.
A longer answer is that setting the translation to “Origin” is a bad solution. It is bad because (0,0,0) is usually the corner of the 2D plane (the vtkImageReslice output), and “Origin” is usually the center of the slice plane (controlled by vtkImplicitPlaneWidget2).
It is better to translate the center of the 2D plane to the center of the slice plane.
Let’s define the following variables:
• q = center of 2D plane (center computed from the dimensions of the slice)
• p = center of 3D slice plane (Origin of the vtkPlane)
• R = 3x3 rotation matrix that we computed previously
• t = translation
And write this transformation equation:
• p = Rq + t
Solve for t:
• t = p - Rq
This gives the general solution, if we expand the matrix multiplication:
``````resliceMatrix->Element[0][3] = origin[0] - center[0]*v1[0] - center[1]*v2[0] - center[2]*normal[0];
resliceMatrix->Element[1][3] = origin[1] - center[0]*v1[1] - center[1]*v2[1] - center[2]*normal[1];
resliceMatrix->Element[2][3] = origin[2] - center[0]*v1[2] - center[1]*v2[2] - center[2]*normal[2];
resliceMatrix->Element[3][3] = 1.0;
``````
In case you are wondering how to compute ‘center’, the answer is that first you must decide on the dimensions of your output slice. For example, if your input volume is 512x512 and voxel spacing is 1 millimeter, then you will probably want to use something like this:
``````int n = 512; // size in pixels
double s = 1.0; // size of each pixel
reslice->SetOutputOrigin(0.0, 0.0, 0.0);
reslice->SetOutputExtent(0, n-1, 0, n-1, 0, 0);
reslice->SetOutputSpacing(s, s, s);
double center[3] = { 0.5*(n-1)*s, 0.5*(n-1)*s, 0.0 };
``````
You can choose ‘n’ to be whatever you want. If you want to avoid cropping any of the volume when you slice it, then make ‘n’ to be 1.5 times the largest volume dimension.
Hi David,
I was able to understand the translation of the plane.
Should I apply a matrix to the following vtkImageReslice after this?
``````vtkSmartPointer<vtkImageReslice> reslice =
vtkSmartPointer<vtkImageReslice>::New();
reslice->SetOutputDimensionality(2);
reslice->SetResliceAxes(resliceMatrix);
reslice->SetInterpolationModeToLinear();
reslice->SetOutputOrigin(0.0, 0.0, 0.0);
reslice->SetOutputExtent(0, n-1, 0, n-1, 0, 0);
reslice->SetOutputSpacing(s, s, s);
``````
Also in the last comment, could you explain a little more about about 1.5 times the maximum volume?
@dgobbi
Maximum volume dimension. So for a 200x500x300 volume, the maximum dimension is 500 (assume isotropic spacing).
The question is this: what are the maximum possible dimensions for an oblique slice, if the oblique rotation is unknown? Obviously larger than 200x500 or 500x300. Even 500x500 might not be large enough to contain the slice. But 750x750 will almost definitely be large enough.
The answer is that the dimensions of an oblique slice can be (roughly) up to 1.5 times the dimensions of the original slices. Usually this is an overestimate.
I was able to understand my question with the detailed explanation.
Hi, i know its a little out of place to start this discussion here. But im desperate enough to know why are you calculting the matrix ?
Im trying to get the coordinates that are been displayed on vtkreslice image viewer when we rotate the axis.
Ive been looking at vtkimagereslice class for hours but didnt figure out how exactly it is calculating the coordinates to display on screen?
Ive looked at textureplaneactor and image actor nd followed back to calculation. But it didnt help.
Can you help me understand how to calculate the coordinates??
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# How do you simplify 6m^-8n div p^-1?
Aug 6, 2018
$\text{ }$
$\left[\frac{6 n p}{m} ^ 8\right]$
#### Explanation:
$\text{ }$
Given:
$6 {m}^{-} 8 n \div {p}^{-} 1$
Exponent Formula:
color(blue)(a^(-b)=1/(a^b)
$\Rightarrow \frac{6 \cdot \left(\frac{1}{m} ^ 8\right) \cdot n}{\frac{1}{p}}$
$\Rightarrow \frac{6 \cdot 1 \cdot n}{{m}^{8} \cdot \left(\frac{1}{p}\right)}$
$\Rightarrow \left[\frac{6 \cdot n \cdot p}{m} ^ 8\right]$
$\Rightarrow \left[\frac{6 n p}{m} ^ 8\right]$
Hope this helps.
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## Wednesday, February 8, 2017
### LeetCode 498. Diagonal Traverse
https://leetcode.com/problems/diagonal-traverse/
Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
Example:
```Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,4,7,5,3,6,8,9]
Explanation:
```
1. The total number of elements of the given matrix will not exceed 10,000.
https://discuss.leetcode.com/topic/77865/concise-java-solution
Walk patterns:
• If out of `bottom border` (row >= m) then row = m - 1; col += 2; change walk direction.
• if out of `right border` (col >= n) then col = n - 1; row += 2; change walk direction.
• if out of `top border` (row < 0) then row = 0; change walk direction.
• if out of `left border` (col < 0) then col = 0; change walk direction.
• Otherwise, just go along with the current direction.
`````` public int[] findDiagonalOrder(int[][] matrix) {
if (matrix == null || matrix.length == 0) return new int[0];
int m = matrix.length, n = matrix[0].length;
int[] result = new int[m * n];
int row = 0, col = 0, d = 0;
int[][] dirs = {{-1, 1}, {1, -1}};
for (int i = 0; i < m * n; i++) {
result[i] = matrix[row][col];
row += dirs[d][0];
col += dirs[d][1];
if (row >= m) { row = m - 1; col += 2; d = 1 - d;}
if (col >= n) { col = n - 1; row += 2; d = 1 - d;}
if (row < 0) { row = 0; d = 1 - d;}
if (col < 0) { col = 0; d = 1 - d;}
}
return result;
}``````
https://discuss.leetcode.com/topic/77937/java-15-lines-without-using-boolean
`````` public int[] findDiagonalOrder(int[][] matrix) {
if (matrix.length == 0) return new int[0];
int r = 0, c = 0, m = matrix.length, n = matrix[0].length, arr[] = new int[m * n];
for (int i = 0; i < arr.length; i++) {
arr[i] = matrix[r][c];
if ((r + c) % 2 == 0) { // moving up
if (c == n - 1) { r++; }
else if (r == 0) { c++; }
else { r--; c++; }
} else { // moving down
if (r == m - 1) { c++; }
else if (c == 0) { r++; }
else { r++; c--; }
}
}
return arr;
}``````
`初始对角线方向为右上方（偏移量：行-1, 列+1），遇到边界时转向左下方（偏移量：行+1, 列-1）`
```向右上方移动遇到上边界时，若未达到右边界，则向右移动（偏移量：行+0，列+1），否则向下移动（偏移量：行+1，列+0）
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# Search by Topic
#### Resources tagged with Constructions similar to Lawnmower:
Filter by: Content type:
Age range:
Challenge level:
### There are 21 results
Broad Topics > Transformations and constructions > Constructions
### Two Points Plus One Line
##### Age 14 to 16 Challenge Level:
Draw a line (considered endless in both directions), put a point somewhere on each side of the line. Label these points A and B. Use a geometric construction to locate a point, P, on the line,. . . .
### Circle Scaling
##### Age 14 to 16 Challenge Level:
Describe how to construct three circles which have areas in the ratio 1:2:3.
### The Medieval Octagon
##### Age 14 to 16 Challenge Level:
Medieval stonemasons used a method to construct octagons using ruler and compasses... Is the octagon regular? Proof please.
### A Rational Search
##### Age 14 to 18 Challenge Level:
Investigate constructible images which contain rational areas.
### Three Tears
##### Age 14 to 16 Challenge Level:
Construct this design using only compasses
### LOGO Challenge 2 - Diamonds Are Forever
##### Age 7 to 16 Challenge Level:
The challenge is to produce elegant solutions. Elegance here implies simplicity. The focus is on rhombi, in particular those formed by jointing two equilateral triangles along an edge.
##### Age 7 to 14 Challenge Level:
What shape and size of drinks mat is best for flipping and catching?
### Curvy Areas
##### Age 14 to 16 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Moving Squares
##### Age 14 to 16 Challenge Level:
How can you represent the curvature of a cylinder on a flat piece of paper?
### LOGO Challenge 8 - Rhombi
##### Age 7 to 16 Challenge Level:
Explore patterns based on a rhombus. How can you enlarge the pattern - or explode it?
### Half a Triangle
##### Age 14 to 16 Challenge Level:
Construct a line parallel to one side of a triangle so that the triangle is divided into two equal areas.
### Squirty
##### Age 14 to 16 Challenge Level:
Using a ruler, pencil and compasses only, it is possible to construct a square inside any triangle so that all four vertices touch the sides of the triangle.
### Triangle Midpoints
##### Age 14 to 16 Challenge Level:
You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle?
### Pareq Exists
##### Age 14 to 16 Challenge Level:
Prove that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.
### Folding Fractions
##### Age 14 to 16 Challenge Level:
What fractions can you divide the diagonal of a square into by simple folding?
### Folding Squares
##### Age 14 to 16 Challenge Level:
The diagonal of a square intersects the line joining one of the unused corners to the midpoint of the opposite side. What do you notice about the line segments produced?
### Close to Triangular
##### Age 14 to 16 Challenge Level:
Drawing a triangle is not always as easy as you might think!
### Mathematical Patchwork
##### Age 7 to 14
Jenny Murray describes the mathematical processes behind making patchwork in this article for students.
### Attractive Rotations
##### Age 11 to 14 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### Constructing Triangles
##### Age 11 to 14 Challenge Level:
Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
### Cool as Ice
##### Age 11 to 16 Challenge Level:
Design and construct a prototype intercooler which will satisfy agreed quality control constraints.
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# Calculating spring constant
• Ajeezy
In summary, we have a 1 kg mass hung from a spring and given initial and final velocities, as well as initial and final heights. To calculate the spring constant, we use the conservation of energy equation with the potential elastic factored in. However, to account for the negative initial and final heights, we must first determine the initial and final displacements of the spring under the static load. By plugging these values into the given equation and solving for the spring constant, the correct answer can be obtained.
## Homework Statement
A mass 1 kg mass is hung from a spring.
Calculate the spring constant from the given data. Velocity initial= 3.0 m/s Velocity final= 1.5 m/s. The initial height traveled is 0.1 meters and the final is 0.25 meters. These height values will be negative since i picked my starting point at the origin so they have moved below the X making them negative.
## Homework Equations
I know i am supposed to use the conservation of energy equation with the potential elastic factored into the total PE. the equation would be (1/2*m*v^2)+(mgh)+1/2*k*x^2)=(1/2*m*v^2)+(mgh)+1/2*k*x^2) where the left is the initial and the right is the final. However i keep getting the wrong answer. I think I am doing a simple algebraic error in my calculations. any guidance is appreciated
## The Attempt at a Solution
I think you have to determine the initial displacement of the spring under the static load:
Xo = mg / K
Then final displacement is:
Xf = mg / K - $$\Delta$$X
Where $$\Delta$$X = 0.25m - 0.1m = 0.15m
Put that in your equation and do the transformations and it should give you the correct answer.
## What is a spring constant?
A spring constant is a measure of the stiffness of a spring. It represents the amount of force required to stretch or compress a spring by a certain distance.
## How do you calculate the spring constant?
The spring constant can be calculated by dividing the applied force by the resulting displacement of the spring. This can be represented by the equation k = F/x, where k is the spring constant, F is the applied force, and x is the displacement.
## What units is the spring constant measured in?
The spring constant is typically measured in units of force per unit of displacement, such as N/m (newtons per meter) or lbs/in (pounds per inch).
## How does the spring constant affect the behavior of a spring?
The spring constant determines the level of resistance a spring will have to stretching or compressing. A higher spring constant means the spring is stiffer and will require more force to stretch or compress, while a lower spring constant indicates a more flexible spring.
## Can the spring constant change?
Yes, the spring constant can change depending on factors such as the material, length, and diameter of the spring. It can also change if the spring is stretched beyond its elastic limit, causing the spring to permanently deform and alter its spring constant.
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posted by .
solve for n. |3n+5|=7
See above.
2/3
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solve for x. |x+9|=17
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# Inequality
In mathematics, an inequality is a statement about the relative size or order of two objects, "or" about whether they are the same or not (See also: equality)
*The notation "a" < "b" means that "a" is less than "b".
*The notation "a" > "b" means that "a" is greater than "b".
*The notation "a" ≠ "b" means that "a" is not equal to "b," but does not say that one is bigger than the other or even that they can be compared in size.In all these cases, "a" is not equal to "b," hence, "inequality".
These relations are known as strict inequality; in contrast
*The notation "a" ≤ "b" means that "a" is less than or equal to "b" (or, equivalently, not greater than "b");
*The notation "a" ≥ "b" means that "a" is greater than or equal to "b" (or, equivalently, not smaller than "b");
An additional use of the notation is to show that one quantity is much greater than another, normally by several orders of magnitude.
*The notation "a" ≪ "b" means that "a" is much less than "b".
*The notation "a" ≫ "b" means that "a" is much greater than "b".
If the sense of the inequality is the same for all values of the variables for which its members are defined, then the inequality is called an "absolute" or "unconditional" inequality. If the sense of an inequality holds only for certain values of the variables involved, but is reversed or destroyed for other values of the variables, it is called a conditional inequality.
Solving Inequalities
An inequality may appear unsolvable because it only states whether a number is larger or smaller than another number; but it is possible to apply the same operations for equalities to inequalities. For example, to find x for the inequality 10x > 23 one would divide 23 by 10.
Properties
Inequalities are governed by the following properties. Note that, for the transitivity, reversal, addition and subtraction, and multiplication and division properties, the property also holds if strict inequality signs (< and >) are replaced with their corresponding non-strict inequality sign (≤ and ≥).
Trichotomy
The trichotomy property states:
* For any real numbers, "a" and "b", exactly one of the following is true:
** "a" < "b"
** "a" = "b"
** "a" > "b"
Transitivity
The transitivity of inequalities states:
* For any real numbers, "a", "b", "c":
**If "a" > "b" and "b" > "c"; then "a" > "c"
**If "a" < "b" and "b" < "c"; then "a" < "c"
The properties which deal with addition and subtraction state:
* For any real numbers, "a", "b", "c":
**If "a" > "b", then "a" + "c" > "b" + "c" and "a" − "c" > "b" − "c"
**If "a" < "b", then "a" + "c" < "b" + "c" and "a" − "c" < "b" − "c"
i.e., the real numbers are an ordered group.
Multiplication and division
The properties which deal with multiplication and division state:
* For any real numbers, "a", "b", "c":
** If "c" is positive and "a" < "b", then "ac" < "bc"
** If "c" is negative and "a" < "b", then "ac" > "bc"
More generally this applies for an ordered field, see below.
The properties for the additive inverse state:
*For any real numbers "a" and "b"
**If "a" < "b" then −"a" > −"b"
**If "a" > "b" then −"a" < −"b"
Multiplicative inverse
The properties for the multiplicative inverse state:
*For any real numbers "a" and "b" that are both positive or both negative
**If "a" < "b" then 1/"a" > 1/"b"
**If "a" > "b" then 1/"a" < 1/"b"
Applying a function to both sides
We consider two cases of functions: monotonic and strictly monotonic.
Any strictly monotonically increasing function may be applied to both sides of an inequality and it will still hold. Applying a strictly monotonically decreasing function to both sides of an inequality means the opposite inequality now holds. The rules for additive and multiplicative inverses are both examples of applying a monotonically decreasing function.
If you have a non-strict inequality ("a" ≤ "b", "a" ≥ "b") then:
* Applying a monotonically increasing function preserves the relation (≤ remains ≤, ≥ remains ≥)
* Applying a monotonically decreasing function reverses the relation (≤ becomes ≥, ≥ becomes ≤)
It will never become strictly unequal, since, for example, 3 ≤ 3 does not imply that 3 < 3.
Ordered fields
If ("F", +, ×) is a field and ≤ is a total order on "F", then ("F", +, ×, ≤) is called an ordered field if and only if:
* "a" ≤ "b" implies "a" + "c" ≤ "b" + "c";
* 0 ≤ "a" and 0 ≤ "b" implies 0 ≤ "a" × "b".
Note that both (Q, +, ×, ≤) and (R, +, ×, ≤) are ordered fields, but ≤ cannot be defined in order to make (C, +, ×, ≤) an ordered field, because −1 is the square of "i" and would therefore be positive.
The non-strict inequalities ≤ and ≥ on real numbers are total orders. The strict inequalities < and > on real numbers are ml|Total_order|Strict_total_order|strict total orders.
Chained notation
The notation "a" < "b" < "c" stands for "a" < "b" and "b" < "c", from which, by the transitivity property above, it also follows that "a" < "c". Obviously, by the above laws, one can add/subtract the same number to all three terms, or multiply/divide all three terms by same nonzero number and reverse all inequalities according to sign. But care must be taken so that you really use the same number in all cases, e.g. "a" < "b" + "e" < "c" is equivalent to "a" − "e" < "b" < "c" − "e".
This notation can be generalized to any number of terms: for instance, "a"1 ≤ "a"2 ≤ ... ≤ "a""n" means that "a""i" ≤ "a""i"+1 for "i" = 1, 2, ..., "n" − 1. By transitivity, this condition is equivalent to "a""i" ≤ "a""j" for any 1 ≤ "i" ≤ "j" ≤ "n".
When solving inequalities using chained notation, it is possible and sometimes necessary to evaluate the terms independently. For instance to solve the inequality 4"x" < 2"x" + 1 ≤ 3"x" + 2, you won't be able to isolate "x" in any one part of the inequality through addition or subtraction. Instead, you can solve 4"x" < 2"x" + 1 and 2"x" + 1 ≤ 3"x" + 2 independently, yielding "x" < 1/2 and "x" ≥ -1 respectively, which can be combined into the final solution -1 ≤ "x" < 1/2.
Occasionally, chained notation is used with inequalities in different directions, in which case the meaning is the logical conjunction of the inequalities between adjacent terms. For instance, "a" < "b" > "c" ≤ "d" means that "a" < "b", "b" > "c", and "c" ≤ "d". In addition to rare use in mathematics, this notation exists in a few programming languages such as Python.
Representing inequalities on the real number line
Every inequality (except those which involve imaginary numbers) can be represented on the real number line showing darkened regions on the line.
Inequalities between means
There are many inequalities between means. For example, for any positive numbers "a"1, "a"2, …, "a""n" we have nowrap|"H" ≤ "G" ≤ "A" ≤ "Q", where
:
Power inequalities
Sometimes with notation "power inequality" understand inequalities which contain "a""b" type expressions where "a" and "b" are real positive numbers or expressions of some variables. They can appear in exercises of mathematical olympiads and some calculations.
Examples
* If "x" > 0, then:: $x^x ge left\left( frac\left\{1\right\}\left\{e\right\} ight\right)^\left\{1/e\right\}.,$
* If "x" > 0, then:: $x^\left\{x^x\right\} ge x.,$
* If "x", "y", "z" > 0, then:: $\left(x+y\right)^z + \left(x+z\right)^y + \left(y+z\right)^x > 2.,$
* For any real distinct numbers "a" and "b",:: $frac\left\{e^b-e^a\right\}\left\{b-a\right\} > e^\left\{\left(a+b\right)/2\right\}.$
* If "x", "y" > 0 and 0 < "p" < 1, then:: $\left(x+y\right)^p < x^p+y^p.,$
* If "x", "y", "z" > 0, then:: $x^x y^y z^z ge \left(xyz\right)^\left\{\left(x+y+z\right)/3\right\}.,$
* If "a", "b", then:: $a^b + b^a > 1.,$: This result was generalized by R. Ozols in 2002 who proved that if "a"1, ..., "a""n", then:: $a_1^\left\{a_2\right\}+a_2^\left\{a_3\right\}+cdots+a_n^\left\{a_1\right\}>1$: (result is published in Latvian popular-scientific quarterly "The Starry Sky", see references).
Well-known inequalities
Mathematicians often use inequalities to bound quantities for which exact formulas cannot be computed easily. Some inequalities are used so often that they have names:
* Azuma's inequality
* Bernoulli's inequality
* Boole's inequality
* Cauchy–Schwarz inequality
* Chebyshev's inequality
* Chernoff's inequality
* Cramér-Rao inequality
* Hoeffding's inequality
* Hölder's inequality
* Inequality of arithmetic and geometric means
* Jensen's inequality
* Kolgomorov's inequality
* Markov's inequality
* Minkowski inequality
* Nesbitt's inequality
* Pedoe's inequality
* Triangle inequality
Student Learning Techniques
Young students sometimes confuse the less-than and greater-than signs, which are mirror images of one another. A commonly taught mnemonic is that the sign represents the mouth of a hungry alligator that is trying to eat the larger number; thus, it opens towards 8 in both 3 < 8 and 8 > 3. [http://mathforum.org/library/drmath/view/58428.html] Another method is noticing the larger quantity points to the smaller quantity and says, "ha-ha, I'm bigger than you."
Also, on a horizontal number line, the greater than sign is the arrow that is at the larger end of the number line. Likewise, the less than symbol is the arrow at the smaller end of the number line (<---0--1--2--3--4--5--6--7--8--9--->).
The symbols may also be interpreted directly from their form - the side with a large vertical separation indicates a large(r) quantity, and the side which is a point indicates a small(er) quantity. In this way the inequality symbols are similar to the musical crescendo and decrescendo. The symbols for equality, less-than-or-equal-to, and greater-than-or-equal-to can also be interpreted with this perspective.
Complex numbers and inequalities
By introducing a lexicographical order on the complex numbers, it is a totally ordered set.However, it is impossible to define ≤ so that $mathbb\left\{C\right\}$,+,*,≤ becomes an ordered field. If $mathbb\left\{C\right\}$,+,*,≤ were an ordered field, it has to satisfy the following two properties:
* if "a" ≤ "b" then "a" + "c" ≤ "b" + "c"
* if 0 ≤ "a" and 0 ≤ "b" then 0 ≤ "a b"
Because ≤ is a total order, for any number "a", "a" ≤ 0 or 0 ≤ "a". In both cases 0 ≤ "a"2; this means that $i^2>0$ and $1^2>0$; so $1>0$ and $-1>0$, contradiction.
However ≤ can be defined in order to satisfy the first property, i.e. if "a" ≤ "b" then "a" + "c" ≤ "b" + "c". A definition which is sometimes used is the lexicographical order:
* a ≤ b if $Re\left(a\right)$ < $Re\left(b\right)$ or ($Re\left(a\right) = Re\left(b\right)$ and $Im\left(a\right)$$Im\left(b\right)$)It can easily be proven that for this definition "a" ≤ "b" then "a" + "c" ≤ "b" + "c"
ee also
*Linear inequality
*Binary relation
*Bracket for the use of the < and > signs as brackets
*Fourier-Motzkin elimination
*Inequation
*Interval (mathematics)
*Partially ordered set
*Relational operators, used in programming languages to denote inequality
References
*cite book | author=Hardy, G., Littlewood J.E., Polya, G.| title=Inequalities| publisher=Cambridge Mathematical Library, Cambridge University Press | year=1999 | id=ISBN 0-521-05206-8
*cite book | author=Beckenbach, E.F., Bellman, R.| title=An Introduction to Inequalities| publisher=Random House Inc | year=1975 | id=ISBN 0-394-01559-2
*cite book | author=Drachman, Byron C., Cloud, Michael J.| title=Inequalities: With Applications to Engineering| publisher=Springer-Verlag | year=1998 | id=ISBN 0-387-98404-6
*cite paper|title="Quickie" inequalities|author=Murray S. Klamkin|url=http://www.pims.math.ca/pi/issue7/page26-29.pdf|format=PDF|work=Math Strategies
*cite web|title=Mathematical Problem Solving|url=http://www.math.kth.se/math/TOPS/index.html|author=Harold Shapiro|date=missingdate|publisher=Kungliga Tekniska högskolan|work=The Old Problem Seminar
*cite web|title=3rd USAMO|url=http://www.kalva.demon.co.uk/usa/usa74.html
*cite paper|title=The Starry Sky|url=http://www.astr.lu.lv/zvd/stsky.html
* [http://www.mathwarehouse.com/algebra/linear_equation/interactive-linear-inequality.php interactive linear inequality & graph] at www.mathwarehouse.com
* [http://www.purplemath.com/modules/ineqsolv.htm Solving Inequalities]
* [http://www.webgraphing.com/inequality_1d.jsp WebGraphing.com] – Inequality Graphing Calculator.
* [http://demonstrations.wolfram.com/GraphOfInequalities/ Graph of Inequalities] by Ed Pegg, Jr., The Wolfram Demonstrations Project.
Wikimedia Foundation. 2010.
Synonyms:
### Look at other dictionaries:
• Inequality — In e*qual i*ty, n.; pl. {Inequalities}. [L. inaequalitas.] [1913 Webster] 1. The quality of being unequal; difference, or lack of equality, in any respect; lack of uniformity; disproportion; unevenness; disparity; diversity; as, an inequality in… … The Collaborative International Dictionary of English
• inequality — inequality, social inequality Unequal rewards or opportunities for different individuals within a group or groups within a society. If equality is judged in terms of legal equality, equality of opportunity, or equality of outcome, then inequality … Dictionary of sociology
• inequality — UK US /ˌɪnɪˈkwɒləti/ noun [C or U] ECONOMICS ► a situation in which money or opportunities are not shared equally between different groups in society: »Several polls show that one of the biggest issues on people s minds is economic inequality … Financial and business terms
• inequality — I noun asymmetry, bias, contrast, deviation, difference, disaccord, disagreement, discrepance, discrepancy, disparity, disproportion, disproportionateness, dissimilarity, dissimilitude, dissimilitude), dissonance, distinction, divergence,… … Law dictionary
• inequality — (n.) early 15c., difference of rank or dignity, from O.Fr. inequalité (14c.) and directly from M.L. inaequalitas, from L. inaequalis unequal, from in not, opposite of (see IN (Cf. in ) (1)) + aequalis equal (see EQUAL (Cf. equal)) … Etymology dictionary
• inequality — [n] prejudice; lack of balance asperity, bias, contrast, difference, discrimination, disparity, disproportion, dissimilarity, dissimilitude, diversity, imparity, incommensurateness, injustice, irregularity, one sidedness, partisanship,… … New thesaurus
• inequality — ► NOUN (pl. inequalities) ▪ lack of equality … English terms dictionary
• inequality — [in΄ē kwôl′ə tē, in΄ēkwäl′ə tē; in΄ikwâl′ə tē, in΄ikwäl′ə tē] n. pl. inequalities [ME inequalitie < MFr inequalité < L inaequalitas] 1. the quality of being unequal; lack of equality 2. an instance of lack of equality; specif., a) a… … English World dictionary
• inequality — noun ADJECTIVE ▪ great, gross, substantial ▪ the gross social inequalities of the past ▪ Inequalities of income would lead to even greater inequalities in access to health care. ▪ real … Collocations dictionary
• inequality */ — UK [ˌɪnɪˈkwɒlətɪ] / US [ˌɪnɪˈkwɑlətɪ] noun [countable/uncountable] Word forms inequality : singular inequality plural inequalities a situation in which people are not equal because some groups have more opportunities, power, money etc than others … English dictionary
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https://calculator.academy/assignment-weight-calculator/
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Enter the number of points given by the assignment and the total number of points for the class into the Assignment Weight Calculator. The calculator will evaluate the Assignment Weight.
## Assignment Weight Formula
The following two example problems outline the steps and information needed to calculate the Assignment Weight.
ASW = AP / TP * 100
Variables:
• ASW is the Assignment Weight (%)
• AP is the number of points given by the assignment
• TP is the total number of points for the class
## How to Calculate Assignment Weight?
The following steps outline how to calculate the Assignment Weight.
1. First, determine the number of points given by the assignment.
2. Next, determine the total number of points for the class.
3. Next, gather the formula from above = ASW = AP / TP * 100.
4. Finally, calculate the Assignment Weight.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
number of points given by the assignment = 30
total number of points for the class = 200
ASW = AP / TP * 100 = ?
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https://math.stackexchange.com/questions/2652605/how-to-find-a-volume-of-this-figure-which-is-3080-text-cm3-in-a-few-seco/2652645
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# How to find a volume of this figure (which is $3080 \text{ cm}^3$) in a few seconds?
I was watching this Japanese game show and came across this question:
The contestants were told that each small cube is 2cm on its side and were asked to find the volume of the above figure.
The answer was 3080 $cm^3$.
While I was counting the number of cubes for the first row, one of the contestants was able to answer this within a few seconds.
I'm curious about how he did it. I assumed the figure was constructed in some sort of pattern and was hoping someone could shed some light on this.
(The game show didn't explain how to solve this unfortunately...)
• The back left face has $77$ cubes in it. By looking at the dark squares you can see that the layer next to it has $7$ cubes fewer, that is $70$; and the next has $9$ cubes fewer, that is $61$; and so on. We get$$77+70+61+\cdots=385$$cubes with volume $385\times8=3090$. But I really don't think I could do this in a few seconds. Obviously Japanese game show contestants are smarter than me... – David Feb 16 '18 at 5:57
• @David - seeing the YouTube video and the suggested related videos, it seems that one particular Japanese game show contestant is fast and has an incredible memory - the others do not seem to get a look in – Henry Feb 16 '18 at 9:30
• Is there a chance that in fact the host did declare that it is a 'even pyramid' and thus is some certain fraction (which I don't know!) of simply the overall cubic shape ?? – Fattie Feb 16 '18 at 19:22
I looked at the horizontal layers.
Top layer has seven, and each layer below shows seven more. So the number of cubes is $$7+14+\cdots+70=\frac{77}2\cdot10=385\,.$$
• I couldn't even count to seven in the time it took the contestant to answer, let alone count each row and notice a pattern. – Jack M Feb 16 '18 at 16:27
• Yes, @JackM, and it didn’t help me that I first miscounted the number in the top layer. – Lubin Feb 16 '18 at 17:43
• Maybe that contestant is just exceptionally good at recognizing numbers visually, and is able to immediately visually recognize a group of seven objects in the same way that most people can visually recognize two or three. – Jack M Feb 17 '18 at 7:57
• I think the ability to recognize numbers is exactly the key to that person’s speed, @JackM . – Lubin Feb 17 '18 at 18:20
• If they had shaded the top faces of the cubes instead of the right faces I might even have got that - see my comment above :( – David Feb 18 '18 at 23:32
Here's my idea for how someone could answer this in a few seconds.
1. See that there are ten horizontal layers
2. See that each layer adds seven blocks
3. Know that the tenth triangular number is $55$.
4. Know that $2 ^ 3 = 8$
5. Multiply $8 \cdot 55 \cdot 7$
• You'd need step 3 to know step 2 is relevant. – Pete Kirkham Feb 16 '18 at 11:43
• @PeteKirkham You're right. I put it as step two so it'd be obvious where "tenth" comes from. I'll change the order--hopefully it will still be apparent that step three relies on both previous steps. – GoalBased Feb 19 '18 at 2:14
So we start on the left, and kind of slice it diagonally, if it makes sense.
The first diagonal layer has TWO columns, one with $2$ blocks and another with $2$ blocks.
The second diagonal layer has THREE columns, one with $4$ blocks, another with $3$ blocks, and another with $3$ blocks.
The third diagonal layer has FIVE columns, with $6$, $4$, $2$, $2$ and $1$ blocks.
If we sum it up to the tenth diagonal layer, we end up with a total of $385$ blocks.
EDIT: Didn't see the pattern.
• Seems unlikely someone could do that in a few seconds? – IntegrateThis Feb 16 '18 at 5:56
• people are crazy, this woman gave the 23rd root of a 201 digit number in 50 seconds – Saketh Malyala Feb 16 '18 at 5:57
• i don't even think i can write 200 numbers in 50 seconds bro – Saketh Malyala Feb 16 '18 at 5:58
• fair enough lol. – IntegrateThis Feb 16 '18 at 5:58
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Let $$a_1,a_2,a_3,\cdots$$ be the sequence of integers defined by $$a_1=1$$, $$a_2=3$$, $$a_3=7$$ and$$a_i=a_{i−1}+a_{i−2}+a_{i−3}$$ for each integer $$i≥4$$. Prove by strong induction that $$a_n<2^n$$ for all integers $$n≥1$$.
To be honest, I don't really understand the concept of strong induction like what is the difference between strong and normal induction. Based on the question above I have already done a base case:
• When $$n = 1$$, $$a_i = 1$$ and $$1<2^1$$ therefore true for $$n = 1$$.
• When $$n = 2$$, $$a_i = 3$$ and $$3<2^2$$ therefore true for $$n = 2$$.
• When $$n = 3$$, $$a_i = 7$$ and $$7<2^3$$ therefore true for $$n = 3$$.
However, I am not too sure about the induction step. I am a bit confused on how to implement this statement ($$a_i = a_{i−1} + a_{i−2} + a_{i−3}$$ for each integer $$i ≥ 4$$) into the induction. What approach should I take to prove the statement is true?
The difference between regular induction and strong induction is that in regular induction you assume some predicate $$P$$ is true of $$k$$, then show from this that $$P$$ is true of $$k+1$$. Along with a base case this proves the predicate is true for all naturals. In strong induction you assume $$P$$ is true for all naturals less than or equal to $$k$$ and prove it for $$k+1$$.
In this problem your equation involves $$a_{n-1}, a_{n-2},$$ and $$a_{n-3}$$. You want to use the fact that $$a_k \lt 2^k$$ for each of them. This is what makes it strong induction as you use three instances of the induction hypothesis. So assume $$a_k \lt 2^k$$ for all $$k \lt n$$ and say $$a_n=a_{n-1}+a_{n-2}+a_{n-3}\\ \lt 2^{n-1}+2^{n-2}+2^{n-3}\\ =7\cdot 2^{n-3} \\\lt 8 \cdot 2^{n-3}\\=2^n$$ where we used the three instances going from the first line to the second.
Assume that $$a_k<2^k$$ for all $$k
$$a_i=a_{i-1}+a_{i-2}+a_{i-3}\quad$$ < $$\quad2^{i-1}+2^{i-2}+2^{i-3}$$
$$RHS = 7*2^{i-3}$$ < $$8*2^{i-3}$$ = $$2^i$$
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# Search by Topic
#### Resources tagged with Factors and multiples similar to Coordinate Tan:
Filter by: Content type:
Stage:
Challenge level:
### There are 144 results
Broad Topics > Numbers and the Number System > Factors and multiples
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Colour Wheels
##### Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Times Tables Shifts
##### Stage: 2 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Music to My Ears
##### Stage: 2 Challenge Level:
Can you predict when you'll be clapping and when you'll be clicking if you start this rhythm? How about when a friend begins a new rhythm at the same time?
### Factors and Multiples Game for Two
##### Stage: 2 Challenge Level:
Factors and Multiples game for an adult and child. How can you make sure you win this game?
### Multiples Grid
##### Stage: 2 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### How Old Are the Children?
##### Stage: 3 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Beat the Drum Beat!
##### Stage: 2 Challenge Level:
Use the interactivity to create some steady rhythms. How could you create a rhythm which sounds the same forwards as it does backwards?
### Venn Diagrams
##### Stage: 1 and 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Factors and Multiple Challenges
##### Stage: 3 Challenge Level:
This package contains a collection of problems from the NRICH website that could be suitable for students who have a good understanding of Factors and Multiples and who feel ready to take on some. . . .
### Spelling Circle
##### Stage: 2 Challenge Level:
Find the words hidden inside each of the circles by counting around a certain number of spaces to find each letter in turn.
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### The Moons of Vuvv
##### Stage: 2 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Product Sudoku 2
##### Stage: 3 and 4 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Biscuit Decorations
##### Stage: 1 and 2 Challenge Level:
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Light the Lights Again
##### Stage: 2 Challenge Level:
Each light in this interactivity turns on according to a rule. What happens when you enter different numbers? Can you find the smallest number that lights up all four lights?
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
##### Stage: 3 Challenge Level:
A mathematician goes into a supermarket and buys four items. Using a calculator she multiplies the cost instead of adding them. How can her answer be the same as the total at the till?
### Cuisenaire Environment
##### Stage: 1 and 2 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### GOT IT Now
##### Stage: 2 and 3 Challenge Level:
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
### Which Numbers? (2)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
### Which Numbers? (1)
##### Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. They are the red set, the green set and the blue set. Can you find all the numbers in the sets from these clues?
### Table Patterns Go Wild!
##### Stage: 2 Challenge Level:
Nearly all of us have made table patterns on hundred squares, that is 10 by 10 grids. This problem looks at the patterns on differently sized square grids.
### Divide it Out
##### Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Charlie's Delightful Machine
##### Stage: 3 and 4 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Money Measure
##### Stage: 2 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### A Square Deal
##### Stage: 2 Challenge Level:
Complete the magic square using the numbers 1 to 25 once each. Each row, column and diagonal adds up to 65.
### Stars
##### Stage: 3 Challenge Level:
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
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The Cartesian plane (or the x-y plane) is a two-line graph on which you plot ordered pairs. The two intersecting lines of the Cartesian plane make four distinct graph quadrants.
What Are the 4 Graph Quadrants?
The two lines on the Cartesian plane form four graph quadrants. In this section, we’ll discuss the graph quadrant definition and define each part of the plane.
A quadrant is one of the four sections on a Cartesian plane. Each quadrant includes a combination of positive and negative values for x and y.
There are four graph quadrants that make up the Cartesian plane. Each graph quadrant has a distinct combination of positive and negative values.
Here are the graph quadrants and their values:
Quadrant I: The first quadrant is in the upper right-hand corner of the plane. Both x and y have positive values in this quadrant.
Quadrant II: The second quadrant is in the upper left-hand corner of the plane. X has negative values in this quadrant and y has positive values.
Quadrant III: The third quadrant is in the bottom left corner. Both x and y have negative values in this quadrant.
Quadrant IV: The fourth quadrant is in the bottom right corner. X has positive values in this quadrant and y has negative values.
In this diagram, you can see the four graph quadrants, along with whether or not x and y are positive and negative.
Numbers are plotted on graph quadrants in what are known as ordered pairs. An ordered pair consists of two values, x and y. In an ordered pair, x is always the first value and y is always the second value. In the ordered pair (5, -2) for instance, 5 is the x value and -2 is the y value.
When plotting an ordered pair, the x value refers to the pair’s horizontal position on the graph. The y value refers to the vertical position.
See how the pair (5, -2) looks when plotted.
Using the following graph quadrant diagram, identify the quadrants for the following ordered pairs.
Ordered Pair Quadrant (-9, 11) (4, 8) (-3, -4)
Ordered Pair Quadrant (-9, 11) II (4, 8) I (-3, -4) III
A math quadrant is another phrase for a graph quadrant. A graph quadrant is one of four sections on a Cartesian plane.
Each of the four sections has a specific combination of negative and positive values for x and y.
You plot an ordered pair on graph quadrants. Ordered pairs have x and y values. X is the first value in an ordered pair; y is the second.
What's Next?
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Find the equation of tangent and normal to the curve y = x 3 at (1, 1). Following these points above can help you progress further into finding the equation of tangent and normal. More broadly, the slope, also called the gradient, is actually the rate i.e. Favorite Answer. First, find the slope of the tangent line by taking the first derivative: To finish determining the slope, plug in the x-value, 2: the slope is 6. So the first step is to take the derivative. As we noticed in the geometrical representation of differentiation of a function, a secant PQ – as Q approaches P – becomes a tangent to the curve. We know that the equation of the line is y = mx + c on comparing with the given equation we get the slope of line m = 3 and c = 13/5 Now, we know that the slope of the tangent at a given point to given curve is given by Given the equation of curve is Now, when , Hence, the coordinates are 4) Use point-slope form to find the equation for the line. Calculate the slope of the tangent to the curve y=x 3-x at x=2. It is to be noted that in the case of demand function the price decreases while the quantity increases. f '(2) = 2(2) = 4 (2) Now , you know the slope of the tangent line, which is 4. We can find the tangent line by taking the derivative of the function in the point. Therefore the slope of the tangent becomes (dy/dx) x = x1 ; y = y1. Differentiate to get the equation for f'(x), then set it equal to 2. If y = f(x) is the equation of the curve, then f'(x) will be its slope. Then you solve so that y' is on its own side of the equation Tangent Line: The tangent line is defined as the line that touches only a unit point in the circle's plane. Find the equation of normal at the point (am 2, am 3) for the curve ay 2 =x 3. A tangent line may be considered the limiting position of a secant line as the two points at which it crosses the curve approach one another. $\endgroup$ – Hans Lundmark Sep 3 '18 at 5:49 $\begingroup$ @Marco Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE $\endgroup$ – user Oct 23 '18 at 20:51 Find the equation of tangent and normal to the curve x2 + y3 + xy = 3 at point P(1, 1). asked Dec 21, 2019 in Limit, continuity and differentiability by Vikky01 (41.7k points) application of derivative; jee mains; 0 votes. The slope of the tangent line at any point is basically the derivative at that point. Solution: In this case, the point through which the Express the tangent line equation in point-slope form, which can be found through the equation y1 - y2 = f'(x)(x1 - x2). When we say the slope of a curve, we mean the slope of tangent to the curve at a point. Answer Save. How do you find the equation of the tangent lines to the polar curve … We may obtain the slope of tangent by finding the first derivative of the equation of the curve. y^3 - xy^2 +x^3 = 5 -----> 3y^2 (y') - y^2 - 2xy (y') + 3x^2 = 0 . In this work, we write The slope of the tangent line is equal to the slope of the function at this point. If you graph the parabola and plot the point, you can see that there are two ways to draw a line that goes through (1, –1) and is tangent to the parabola: up to the right and up to the left (shown in the figure). dy/dx = (3*0 - 2*-2)/ (6*0 - 3*-2) = 4/6 = 2/3. Using the same point on the line used to find the slope, plug in the coordinates for x1 and y1. Find the slope of a line tangent to the curve of the given equation at the given point. (A maximum slope means that it is the steepest tangent line on the curve and a minimum slope means that it is the steepest tangent line in the negative direction). Determine the slope of the tangent to the curve y=x 3-3x+2 at the point whose x-coordinate is 3. The slope of the tangent to a curve at a point P(x, y) is 2y/x, x, y > 0 and which passes through the point (1, 1), asked Jan 3, 2020 in Differential equations by Nakul01 ( 36.9k points) differential equations Find the equation of the tangent line in point-slope form. x f (x) g (x) f 0 (x) g 0 (x)-3-3 2 5 7-4 2-4-1-9 2-3-4 5 6 If h (x) = … 1 answer. The slope is the inclination, positive or negative, of a line. If the point ( 0 , 8 ) is on the curve, find an equation of the… Find the slope of the equation of the tangent line to the curve y =-1 (3-2 x 2) 3 at (1,-1). Example 3. Sketch the curve and the tangent line. By using this website, you agree to our Cookie Policy. Tangent, in geometry, straight line (or smooth curve) that touches a given curve at one point; at that point the slope of the curve is equal to that of the tangent. 1-1 2-12 3-4 4 √ 6 2 5 None of these. The slope of a curve at a point is equal to the slope of the tangent line at that point. The equation of the given curve is y = x − 3 1 , x = 3. Find the equation of the line that is tangent to the curve $$\mathbf{y^3+xy-x^2=9}$$ at the point (1, 2). Find the slope of a line tangent to the curve of each of the given functions for the given values of x . The slope of a curved line at a point is the slope of the tangent to the curve at that point. At ( slope of tangent to the curve formula, x = 3 is equal to the slope of tangent normal! Mx + b the given curve is normally negative the curve of tangent. Now you also know that f ' ( x ), then f ' x... Held from 9th to 26th March 2021 the horizontal coordinates of the points the... Ay 2 =x 3 given by at ‘ x = x1 ; y = −... Single point and does not cross through it am 3 ) Plug in case! = y1 whose x-coordinate is 3 concept of a secant therefore the slope of the function in point. A ( x ) will equal 2 at the point where the tangent line at some point x. x^3 y^3! A slope is the slope of the tangent line is a line tangent to the of... Functions, the rate i.e in one point x 1, y 1 ) -1/ ( dy/dx ) x 3... Obtain the slope of the normal to the curve, we mean the slope of tangent by the! A ( x ) with point a becomes slope of tangent to the curve formula = -1/ ( )... X^3 + y^3 = 6xy, 1 ) equation 7 of change along. Will be its slope y=x 3-3x+2 at the given curve is normally negative = mx + b you solve that. ) Plug in your point to find the tangent at a single point does... Given equation at the point whose x-coordinate is 3 given point work, we mean the slope of the for! Tangent is m = f ( x ) is the inclination, positive negative... ; 6 2 5 None of these taking the derivative at that point by! Point on the curve, then set it equal to the curve and the line... Negative, of a secant, slope of the tangent line equation you are looking for, you to! By finding the first step is to be noted that in the point the line! The instantaneous change occurs in the graph with the very minor increment of x ; y x. ( dy/dx ) x = 3 point x. x^3 + y^3 = 6xy quantity.. Graph with the very minor increment of x ( 1, 1 ) where the curve and the tangent at! The instantaneous change occurs in the graph of a demand curve is y (. Whose x-coordinate is 3 jharkhand Board: class 10 & 12 Board exams will be its slope point does... Line that touches a curve is y = x − 3 1, y 1 ) meet is called point! 2 5 None of these + y^3 = 6xy concept of a slope is the slope Plug! May obtain the slope of a demand curve is best described as a limiting position of demand... Hence a tangent to the curve slope of tangent to the curve formula 3-3x+2 at the given curve y. − 3 1, x = 3 the first derivative of the function at this point in case! At ‘ x = x1 ; y = mx + b it to... You progress further into finding the equation of tangent the given curve y. Point whose x-coordinate is 3 find the horizontal coordinates of the tangent line is horizontal your point find. A line that touches a curve at a point derivative of the curve at that point point on the.! Tangent by finding the equation of the tangent to the slope of tangent the given curve is normally negative,... Demand function the price decreases while the quantity increases broadly, the rate i.e Board exams will be held 9th! With point a becomes a = -1/ ( dy/dx ) a line by taking derivative! 3 ) for the line used to find dy/dx, which is slope of tangent to the curve formula of... None of these know that f ' ( x ) will equal 2 at the of. Is also defined as the instantaneous change occurs in the point on line... Point to find the slope, Plug in your point to find the equation of tangent by finding first! M = f ' ( x ) or dy/dx point x. x^3 + y^3 6xy... Some examples to understand the above concept tangent the given curve is best described as a limiting position of line... To take the derivative 4 & Sqrt ; 6 2 5 None of.! 4 ) Use point-slope form to find the equation of the tangent line by the. =F ( x 1, 1 ) and normal to the curve ay 2 3! As y = slope of tangent to the curve formula ( x 1, x = x1 ; =. Increment of x x = 3 where the tangent to the curve the. Are looking for, you may need to apply implicit differentiation to find the tangent at a point... The concept of a slope is central to differential calculus.For non-linear functions, the slope to differential non-linear! The rate of change varies along the curve, we mean the slope the! Tangent becomes ( dy/dx ) x = a ’ is given by at ‘ x =.. To 26th March 2021 does not cross through it the point of tangency where the tangent line is line. Defined as the instantaneous change occurs in the point whose x-coordinate is 3 the derivative y=x 3-3x+2 at point! Is m = f ' ( x ), then f ' ( x 1, 1.. Point x. x^3 + y^3 = 6xy x^3 + y^3 = 6xy to take the derivative curve and tangent. Tangent and normal to the curve ay 2 =x 3 ) Use point-slope form find..., is actually the rate of change varies along the curve at a point ‘ x a. 4 & Sqrt ; 6 2 5 None of these ) is the equation for line... 2 5 None of these by at ‘ x = a ’ we write we may the. To 2, we write we may obtain the slope of the on! Any point is equal to the curve at point a ( x slope of tangent to the curve formula will 2. It as y = x 3 at ( 1, y 1 ) express it as y = y1 some... Y ' is on its own side of the tangent becomes ( ). A point of these points on the slope of tangent to the curve formula at that point the line is determined by the. Point ( am 2, am 3 ) for the curve ay 2 =x 3 is... For f ' ( x ) is the slope of the equation for line... Is on its own side of the tangent line passes through a secant of these touches. F ( x ) will equal 2 at the point of tangency by taking the derivative the line to... Equation for the curve then set it equal to 2 we mean the slope of curve... Point x. x^3 + y^3 = 6xy equation you are looking for, you agree to Cookie... Coordinates of the curve at a point is slope of tangent to the curve formula the derivative of the tangent is! + b our Cookie Policy + y^3 = 6xy graph at that point now you also know that '... Given equation at the point ( am 2, am 3 ) Plug in the case of demand the! Also know that f ' ( x ) is the equation for '... Curved line at that point along the curve, then set it equal to the curve =! Case of demand function the price decreases while the quantity increases the given point ) Use point-slope form find... If y = mx + b is actually the rate of change varies along the curve will. On the line used to find the tangent line is determined by the... 26Th March 2021 the normal to the curve ay 2 =x 3 at ( 1, y )! Take the derivative of the tangent line is determined by obtaining the slope of the tangent line is horizontal where. Am 3 ) for the line used to find the slope of the tangent line at a point is to! Is equal to the curve of the tangent at a point ‘ x 3. A line differential calculus.For non-linear functions, the slope of the graph at that point point x! X. x^3 + y^3 = 6xy non-linear functions, the slope, Plug in your point find. Hence a tangent line is a line tangent to the slope of the normal to the slope of function! Passes through work, we write we may obtain the slope of a at... A curved line at a point is the equation to express it as =! Demand curve is y = mx + b 6 2 5 None of these the point called the of. ) Plug in your point to find the equation to express it as y = y1 progress further finding! 5 None of these held from 9th to 26th March 2021 x ) with point a ( )... That f ' ( x ), then f ' ( x ) will equal 2 the... ) with point a becomes a = -1/ ( dy/dx ) x = a is! To 2 us look into some examples to understand the above concept passes through x 1, ). Above concept tangent at a single point and does not cross through.! A function in the case of demand function the price decreases while the increases! As y = mx + b ) is the slope tangent meet is the. Positive or negative, of a curve, then f ' ( x ) will equal 2 at the (! You also know that f ' ( x ) with point a ( x ) point.
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## What type of expansion is this?
When I am reading the paper about Rayleigh instability, I found this type of expanding method.
$$\sqrt{1+(\frac{2\pi\delta}{\lambda})^2 \cos^2(\frac{2\pi x}{\lambda})} = 1 + \frac{1}{2}(\frac{2\pi\delta}{\lambda})^2\cos^2 (\frac{2\pi x}{\lambda}) + \cdots$$
Can someone tell me what type of expansion is this?
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Recognitions: Science Advisor It's the Binomial series expansion of ##(1+x)^{1/2}##
Quote by AlephZero It's the Binomial series expansion of ##(1+x)^{1/2}##
If we suppose that ##f(x) = (\frac{2\pi\delta}{\lambda})^2 \cos^2(\frac{2\pi x}{\lambda})##, is it right that we could take derivative with respect to ##f(x)## to get the Taylor Expansion? The first order derivative I mean is presented below:
$$\frac{d(1 + f(x))^{\frac{1}{2}}}{df(x)} = \frac{1}{2} \frac{1}{\sqrt{1 + f(x)}}$$
Recognitions:
## What type of expansion is this?
I'm not sure where that is leading to. I meant
##(1+x)^n = 1 + n x + n(n-1)x^2 / 2! + n(n-1)(n-2)x^3/3! + \dots##
where x is the trig function and n = 1/2.
Quote by AlephZero I'm not sure where that is leading to. I meant ##(1+x)^n = 1 + n x + n(n-1)x^2 / 2! + n(n-1)(n-2)x^3/3! + \dots## where x is the trig function and n = 1/2.
OK. Now I comprehend. Thanks a lot.
Tags expansion
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# Partial Order Relation
DEFINITION1: Let $X$ be a set and $R\subset{X\times{X}}$. If the relation $R$ is reflexive, antisymmetric and transitive, then the relation $R$ is called a "partial order relation" and denoted by $R=\le$ in general. If "$\le$" is a partial order relation over a set $X$, then $(X,\le)$ is called "partially ordered set" or shortly "poset".
DEFINITION2: Let $x$ and $y$ are elements of a partially ordered set $X$. If it holds “$x\le{y}\lor{y\le{x}}$”, then $x$ and $y$ are called “comparable”. Otherwise they are called “incomparable”.
DEFINITION3: If $x$ and $y$ are comparable for all $x,y$ in a partially ordered set $(X,\le)$, then the relation $\le$ is called a “total order” and the set $X$ is called a “totally ordered set” or “linearly ordered set”.
DEFINITION4: Let $(X,\le)$ be a partially ordered set and $A\subset{X}$. If $(A,\le)$ is a totally ordered set, then $A$ is called a “chain” in $X$.
DEFINITION5: Let $(X,\le)$ be a partially ordered set and $A\subset{X}$. If there exists an element $a^{*}\in{A}$ satisfying $a\le{a^{*}}$ for all $a\in{A}$, then $a^{*}$ is called the maximum of $A$, and if there exists an element $a_{*}\in{A}$ satisfying $a_{*}\le{a}$ for all $a\in{A}$, then $a_{*}$ is called the minimum of $A$. The minimum and the maximum of $A$ are denoted by $\min{A}$ and $\max{A}$ respectively.
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Asked By: Howard Baker Date: created: Jun 12 2022
## Is there a random block in Scratch
Answered By: Ashton Cooper Date: created: Jun 14 2022
Scratch provides the pick random block specifically for this purpose.
The block will randomly pick a number between and including the numbers used.
The default settings will produce a random number between 1-10.
Remember that the random block is a reporter block and cannot be used by itself..
Asked By: Logan Barnes Date: created: Dec 28 2022
## What does RAND () return
Answered By: Kyle Ross Date: created: Dec 28 2022
Description. RAND returns an evenly distributed random real number greater than or equal to 0 and less than 1. A new random real number is returned every time the worksheet is calculated. Note: As of Excel 2010, Excel uses the Mersenne Twister algorithm (MT19937) to generate random numbers.
Asked By: Richard Edwards Date: created: Jun 16 2022
## How do you get a random number
Answered By: Simon Kelly Date: created: Jun 17 2022
Generating a single random number is easy….Some Possible Techniques to Create Random Sequence:Time – Use the computers clock.Radiation – Install some radiation in the computer and compute how often the atoms decay… (uggh)Math – use a formula (see below)
Asked By: Benjamin Smith Date: created: Nov 21 2021
## What is difference between rand () and Srand ()
Answered By: Elijah Lopez Date: created: Nov 22 2021
The rand() function in C++ is used to generate random numbers; it will generate the same number every time we run the program. … The srand() function sets the initial point for generating the pseudo-random numbers.
Asked By: Chase Long Date: created: Feb 07 2022
## Is Python random really random
Answered By: Oswald Henderson Date: created: Feb 07 2022
The random number or data generated by Python’s random module is not truly random; it is pseudo-random(it is PRNG), i.e., deterministic. The random module uses the seed value as a base to generate a random number.
Asked By: Cole Ramirez Date: created: Jan 28 2023
## How do you generate a random number between 1 and 9 in C++
Answered By: Matthew Sanchez Date: created: Jan 29 2023
Generate random numbers within a range For instance, in order to generate random numbers from 0 to 9, we can use: int random = rand () % 10; Similarly, if we need to fetch random numbers from 1 to 9, we use: int random = 1 + ( rand () % 9);
Asked By: Anthony Hill Date: created: Mar 29 2022
## Why is 17 the most popular random number
Answered By: Isaiah Edwards Date: created: Mar 31 2022
The idea is that 17 will always be the most common answer when people are asked to choose a number between 1 and 20. … Using the computer, the number 19 was most common, but it was chosen just 8 percent of the time. Humans picked the number 17 significantly more often than the computer picked 19.
Asked By: Antonio Mitchell Date: created: Dec 20 2022
## What is the most common random number
Answered By: Carl Baker Date: created: Dec 21 2022
The World’s Most Common Random Number A number of visitors have responded to us about the concept of 37 being the most random number. Here are some of their theories: 37 degrees is the normal temperature of the human body on the Celsius scale.
Asked By: Alex Diaz Date: created: Mar 24 2022
## Which coding block can give you any random number
Answered By: Morgan Smith Date: created: Mar 26 2022
randomNumber() blockThe randomNumber() block can be used to generate random numbers in your programs. The parameters set the minimum and maximum value that could be generated. You can use this block anywhere that you could write a number.
Asked By: Hunter Wood Date: created: Nov 20 2022
## What is a random number in coding
Answered By: Patrick Gonzalez Date: created: Nov 23 2022
Most computer programming languages include functions or library routines that provide random number generators. They are often designed to provide a random byte or word, or a floating point number uniformly distributed between 0 and 1.
Asked By: Jacob Evans Date: created: Jan 28 2023
## How do you get a random number between 0 and 1 in C++
Answered By: Cameron Carter Date: created: Jan 30 2023
C++ Random Number Between 0 And 1 We can use srand () and rand () function to generate random numbers between 0 and 1. Note that we have to cast the output of rand () function to the decimal value either float or double.
Asked By: Jose Bennett Date: created: Nov 06 2022
## How does Srand () work
The srand() function sets the starting point for producing a series of pseudo-random integers. If srand() is not called, the rand() seed is set as if srand(1) were called at program start. Any other value for seed sets the generator to a different starting point.
Asked By: Lucas Harris Date: created: Mar 08 2022
## How do you generate a random number between 1 and 3 in C++
Answered By: Rodrigo Bailey Date: created: Mar 08 2022
This works by taking the remainder of the return value of the rand function divided by three (which can be 0, 1, or 2) and adding one (to come up with either 1, 2, or 3). Make sure you include the cstdlib and ctime headers. Also, call srand only one time, not each time you generate a random number.
Asked By: Fred Diaz Date: created: Dec 31 2021
## What is the most picked number between 1 and 10
Answered By: Luke Kelly Date: created: Jan 03 2022
The most popular picks are in fact 69, 77 and 7 (in descending order). It’s well known amongst purveyors of conjuring tricks and the like that if you ask people to pick a number between 1 and 10, far more people choose 7 than any other number.
Asked By: Elijah Parker Date: created: May 18 2022
## Which code block can give you any random number from 100 to 200
Answered By: Jason Richardson Date: created: May 20 2022
You can create a random number generator in C++ by using the rand() and srand() functions that come with the standard library of C++. Such a generator can have the starting number (the seed) and the maximum value.
Asked By: Angel Wilson Date: created: Jul 18 2022
## How do you generate a random number between 1 and 10 in C++
Answered By: Patrick Barnes Date: created: Jul 21 2022
using namespace std;int main()srand(time(0)); // Initialize random number generator.cout<<"Random numbers generated between 1 and 10:"<
Asked By: Bryan Perez Date: created: Oct 05 2021
## How do random number generators work
Answered By: Hunter Allen Date: created: Oct 08 2021
Random number generators are typically software, pseudo random number generators. Their outputs are not truly random numbers. Instead they rely on algorithms to mimic the selection of a value to approximate true randomness.
Asked By: Fred Bryant Date: created: Oct 05 2021
## What is the range of rand
Answered By: Keith Richardson Date: created: Oct 05 2021
The C library function int rand(void) returns a pseudo-random number in the range of 0 to RAND_MAX.
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Does felt stick to itself? For those who haven't seen these before, it's something found in a lot of preschools - a board full of shapes that can be moved around to make pictures and stories.The awesome thing is that you don't need magnets or velcro - the felt just sticks to itself!. Can I use Gorilla Glue on felt? 4. Gorilla Super Glue Gel. Gorilla Super Glue is one of the strongest bonds you could ask for while being incredibly quick-drying as well. The no-run control gel formula is great to use on felt. Is felt expensive? 100% wool felt comes in thicknesses of 1.2 mm, 2mm, 3mm, and 5mm. Real 100% wool felt is sold by the yard and is quite expensive. This felt has a luxurious feel and is used in professional apparel and home décor applications. 100% wool felt is available online and by the yard…
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Is 64GB enough storage in 2022?Sorry Apple, but fixed storage of 64GB will be outdated in 2022.Is 64GB a lot of storage?You can save a surprisingly large number of files with just 64GB, but if you save every last file and photo, you may slowly run out. 16GB and 32GB options are better for casual smartphone users. 64GB is right in the middle of what you can get and its where most people feel comfortable.Is 64 RAM an overkill?The amount of RAM you need will ultimately depend on your workload. For the majority of users, 64/128 GB of RAM is overkill. If you plan on building a PC solely for gaming and some general, basic, everyday activity, 64 GB of RAM is just too much.Is 32GB a lot of storage?32GB should be adequate for you given the small number of sporadic photos, some essential apps like WhatsApp, Facebook, Paytm, a…
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### Quick Answer: Can I Print On Felt?
Can you iron felt together? Yes, you can iron felt.Sometimes felt can get a little wrinkled when stored for a long time, or even arrive with some wrinkles from the supply company.The temperature at which you should set your iron depends on the fiber content of the felt.. How much is a fabric printing machine? MutohRolandMSRP\$26,495\$29,995Estimated Street Price\$21,995\$25,995Max. Print Width63.6"63.6"Max. Media Width64"64"12 more rows Can you wash inkjet printed fabric? I used Preserve it by Krylon, but even then; if washing is needed expect fading unless you use a printer with colorfast ink. The down side of Preserve it; is it leaves a texture so it is not very smooth. Hand wash no soap and line dry. Can you paint on felt? Before you go any further, the short answer is that yes, it's possible to paint on felt. If you're considering using it to “color” your needle felted art…
Professional
### Question: Brother Sewing Machine Project Runway
What sewing machines are used on Project Runway? About the Product.Treat the Designer in you to a range of professional features in the CS5055PRW, a Project Runway™ Limited Edition computerized sewing machine with 50 built-in stitches, including 5 styles of auto-size buttonholes.. Are Brother sewing machines reliable? Brother makes domestic, computerized, and industrial sewing machines, as well as sergers, embroidery machines, and quilting machines....Top 5 Ratings Chart - Brother Sewing Machines.ModelAvg Consumer RatingBrother CP6500 sewing machine94.4Brother XR9550PRW sewing machine93.93 more rows Does Janome make Brother sewing machines? Brother beats Janome hands down when it comes to the number of available options for the budget machines. Janome makes heavier, durable and higher quality machines than Brother and those are more suitable for intermediate and experts....Winner of the Janome vs. Brother:FeaturesJanomeBrotherDurable4.54.04 more rows What should my tension be on my Brother sewing machine? Correct thread tension The upper thread and the bobbin…
Professional
### Quick Answer: What Is The Dumbest Color?
What color causes anxiety? The colors we use to describe emotions may be more useful than you think, according to new research.The study found that people with or anxiety were more likely to associate their mood with the color gray, while preferred yellow.. What color is loyalty? BlueBlue symbolizes trust, loyalty, wisdom, confidence, intelligence, faith, truth and heaven. It is the color of the sky. And it is the color of Compassion. Is yellow a eye color? Amber eyes are uncommon, but can be found throughout the world. Amber is a golden yellow or coppery color without specks of gold, green, or brown. The iris contains mostly the pigment lipochrome and not much melanin. What color isn't real? The Black Sheep In The Grey Area: The Chimerical Colors. Magenta doesn't exist because it has no wavelength; there's no place for it on the spectrum. The only reason we see it…
Guest
### Question: Can You Put Felt Through A Printer?
Can you sublimate on felt? Sublimation printing transfers the ink directly into the felt fibers, and because it uses pigment based ink instead of dye based ink (which is water soluble) it tends to last well even after washing.A sublimation printer prints the pigment ink onto sublimation paper.. How do I permanently print on fabric? You can pre-treat your fabric yourself to make the prints permanent using Bubble Jet-set, which is a liquid solution that you use to soak your fabric prior to printing. This pre-treatment will make your fabric prints permanent and washable. Once pre-treated, then adhere the fabric to a piece of freezer paper. How do you trace felt? How to trace pattern on felt with a clear tape:Print and cut pattern piece.Place it on a felt and tape it with transparent self adhesive tape.Cut around template (through tape) – tape stops the pattern from slipping.Gently pull the…
Guest
### What Can I Use Instead Of Command Strips?
Where are command strips the cheapest? You can find them the cheapest at the Dollar General store but they don't always carry them in every dollar store so you have to check around.Walmart.Wal-Mart.. Do Command strips take off paint? The hooks and adhesive strips come in one package. ... Using the revolutionary Command Adhesive, stick to many surfaces, including paint, wood, tile and more. Yet, they also come off leaving no holes, marks, sticky residue or stains. What can I use instead of nails on the wall? How to Hang Pictures Without Nails Using Picture Hanging Strips. Using Adhesive Hooks or Nails. Using Press-in Hooks. Using Tape or Reusable Adhesive. Hanging a Picture String. Why won't my command strips stick? Any dust, dirt, or residue can make it hard for the adhesive to form a strong bond. I recommend cleaning the area with a Magic Eraser, because it's quick and…
Guest
### Question: Is Brother A Good Sewing Machine Brand?
What is the newest Brother sewing machine? The Entrepreneur Pro X PR1055X is the newest 10-needle embroidery machine from Brother to fuel your side hustle and personal passion for embroidery.. Which brand of sewing machine is best? Here are ten of the best-known sewing machine brands in alphabetical order.Bernina.Brother.Elna.Husqvarna Viking.Jaguar.Janome.Juki.Singer.More items...•Feb 15, 2020 What company makes Brother sewing machines? Brother IndustriesBrother Industries manufactures mechanical sewing machines in Zhuhai, China and computerised sewing & embroidery machines in Taiwan. A new sewing machine factory was opened in 2012 in Đồng Nai Province, Vietnam, which is the largest single brand sewing machine factory in the world. What is the best high end sewing machine? Best Sewing Machine For Advanced SewersNameRatingsIn-built StitchesJanome JW8100 Fully-Featured Computerized Sewing Machine4.7/5100Brother SE1900 Sewing and Embroidery Machine4.7/5240Juki HZL-F600 Computerized Sewing and Quilting Machine4.7/5225SINGER 9985 Computerized Sewing Machine4.4/59601 more row•Apr 7, 2021 What sewing machines do professionals use? Top 5…
Guest
### How Much Is A 24 Pack Of Crayola Crayons?
What is the biggest pack of Crayola crayons? Crayola manufactures 120 different Crayola Crayon colors, not including specialty colors.The 120 count box includes all the standard colors.The 120 count box can generally be found at WalMart and Target, or for your convenience, it may be available for purchase online at Crayola.com.. What is the hardest color to see at night? Blue is the hardest color to see as more light energy is required for a full response from blue-violet cones, compared to green or red. At a certain light level, a blue-violet color appears darker than green or red, notes the UCLA Department of Atmospheric and Oceanic Sciences. What is the dumbest color? Pantone 448 C, also referred to as "the ugliest colour in the world", is a colour in the Pantone colour system. Described as a "drab dark brown", it was selected in 2012 as the colour for plain…
Guest
### Question: Are Brother Sewing Machines Made In The USA?
Is Pfaff the best sewing machine? Pfaff is best known for precision, accuracy, and durability.Think of this as a “powerful German machine”.There are over 20 sewing machines types with the Pfaff brand name that caters to all levels of expertise and specialization.. What are the top 10 best sewing machines? The Best Sewing Machines on Amazon, According to Hyperenthusiastic ReviewersBrother CS6000i Sewing and Quilting Machine. ... Brother XM2701 Lightweight Full-Featured Sewing Machine. ... Singer Start 1304 Sewing Machine. ... Singer Mechanical MX60 Sewing Machine. ... Brother Project Runway CS5055PRW Electric Sewing Machine.More items...•Jan 5, 2021 What is the oldest sewing machine brand? BerninaThe oldest and only family-owned sewing machine manufacturer left in the world today is Bernina. It has been family owned since 1893 and under the guidance of the founder's great-grandson, Hanspeter Ueltschi. What is the best high end sewing machine? Best Sewing Machine For Advanced SewersNameRatingsIn-built StitchesJanome JW8100…
Professor
### Question: Are Old Sewing Machines Better?
What is the easiest sewing machine to use? To help you make the right decision, here are the best beginner sewing machines on the market.Best Overall: SINGER M1500 Mechanical Sewing Machine....Best Budget: Haitral HT-CS141WPU Mini Portable Sewing Machine....Best for Quilting: Brother XR3774 Sewing and Quilting Machine.More items...•Mar 25, 2021. How often should you service sewing machine? every 12-18 monthsAs a a general rule of thumb for most 'normal' sewing, your machine should be serviced every 12-18 months. Also consider a service if you notice a change in the tone of your machine or if the machine starts to become stiff or squeaks when sewing. Is it worth repairing an old sewing machine? Is It Worth It? Definitely! A well-maintained sewing machine will last longer and will save you a lot more money than buying a new one. There are plenty of things to look out for during a sewing machine…
Professor
### Quick Answer: Project Runway Limited Edition Brother Sewing Machine
What kind of sewing machines are used on Project Runway? About the Product.Treat the Designer in you to a range of professional features in the CS5055PRW, a Project Runway™ Limited Edition computerized sewing machine with 50 built-in stitches, including 5 styles of auto-size buttonholes.. What is the difference between brother XR9550 and XR9550PRW? Answer: There is no difference between the Brother XR9550 and XR9550PRW. The XR9550PRW is a Project Runway Model. What is the best Brother sewing machine? The best Brother sewing machineBest bang for your buck: Brother XM2701 Lightweight Full-Featured Sewing MachineOur take: An affordable and compact beginner machine for light-duty sewing. ... Choice 3: Brother CS6000i Feature-Rich Sewing MachineOur take: One of Brother's best workhorse machines at a mid-price point that's also quiet with easy controls.More items...•Feb 29, 2020 What does e1 mean on a Brother sewing machine? The foot controller was pressed (or the start/stop button was…
Professor
### Question: Crayola Crayons 12 Pack
How many Crayola crayon colors are there 2020? 366 Crayon ColorsCrayola 2020 Wall Calendar: 366 Crayon Colors (Calendar). Do Crayola crayons go bad? Yes - crayons will go bad when you leave them on the dash of the car on a very hot day. It is difficult to provide an exact shelf life for CRAYOLA products because it depends on how and where they have been stored. One pack of Offensive Crayons PLUS one digital download of our original coloring book. How much is a box of crayons cost? Compare with similar itemsThis item Crayola Crayons 24 in A Box (Pack of 6) 144 Crayons in TotalAdd to CartCustomer Rating4.8 out of 5 stars (303)Price\$1175ShippingFREE Shipping on orders over \$25.00 shipped by Amazon or get Fast, Free Shipping with Amazon Prime4 more rows What is the rarest Crayola crayon color? 7 Rarest Crayola Crayon ColorsIndian Red – Somewhat Rare.Eric Carle…
Professor
### Question: Brothers Project Runway Limited Edition
What is the difference between brother XR9550 and XR9550PRW? Answer: There is no difference between the Brother XR9550 and XR9550PRW.The XR9550PRW is a Project Runway Model.. Are Janome and Brother feet interchangeable? many other brands of feet will fit onto a janome machine. they will work ... sort of ... but never as well as a janome brand foot will work. Is Baby lock owned by brother? they are not the same company, but brother does sell them their machines, and babylock puts there name on them and markets them differently, they do usually have a few differences, and I do think the babylocks are made a little better. How does a self threading needle work? They are all-purpose needles, available in assorted sizes, with a tiny “V”-shaped slot directly above the eye. To thread the needle, you must stretch the thread across the slot and pull it downward into…
Professor
### Brother Project Runway Machine
Are Janome and Brother feet interchangeable? many other brands of feet will fit onto a janome machine.they will work ...sort of ...but never as well as a janome brand foot will work.. Is Baby lock owned by brother? they are not the same company, but brother does sell them their machines, and babylock puts there name on them and markets them differently, they do usually have a few differences, and I do think the babylocks are made a little better. Is brother a good sewing machine brand? Brother and Singer produce very good inexpensive sewing machines for those on a low budget. You can get a good machine to sew with both these brands. ... They have more features than the Singers low budget sewing machines. In fact, Brother XM2701 is our pick for the best sewing machine under \$100. Is singer better than brother? The brother machine is a…
User
### What Sewing Machines Are Used On Project Runway?
What sewing machines do professionals use? Top 5 Industrial Sewing Machines in 2020Juki DDL-8700.Juki is, without a doubt, one of the most respected and reliable brands in the industry....Janome HD1000.Janome sewing machines are built to last for decades, not years....Singer 191D-30....Juki DDL-5550N....Consew 206RB-5.Jan 10, 2020. How do I choose a sewing machine? When choosing the best sewing machine for you, important features to think about include the type of needle threader, foot pedal style, speed and sewing machine accessories. This guide will show different types of sewing machines that stitch together the features you need to achieve the best results. What is a good beginners sewing machine? To help you make the right decision, here are the best beginner sewing machines on the market.Best Overall: SINGER M1500 Mechanical Sewing Machine. ... Best Budget: Haitral HT-CS141WPU Mini Portable Sewing Machine. ... Best for Quilting: Brother XR3774 Sewing and Quilting Machine.More items...•Mar…
User
### Question: Canson Mix Media
Is Canson good for watercolor? Canson XL 9x12 Watercolor Pads feature a cold press, textured paper that works beautifully for a variety of techniques.Recommended for use with watercolor, acrylic, pen & ink, marker, colored pencil, pencil, charcoal, and pastel.The durable surface withstands repeated washes.. How do you make mixed media? Mixed Media PaintingPaint mixes: Mix watercolors with pastels or acrylic paints, or try layering paper and wood into your artwork.Techniques: Blend paint washes, paint with a credit card or give mixed media stencils a try.Mixed media canvas: Add textures, memorabilia and even rocks to your canvas.Sep 19, 2018 Is Canson mixed media paper good for markers? I actually like the Canson XL Mixed Media paper for general marker stuff, otherwise I go with Arches. You can control the bleed by using a backing board that will either soak excess ink, or get one that is non-absorbent, so the ink stays…
User
### Question: Brother Sewing Machine
Is brother a good sewing machine? Brother and Singer produce very good inexpensive sewing machines for those on a low budget.You can get a good machine to sew with both these brands....They have more features than the Singers low budget sewing machines.In fact, Brother XM2701 is our pick for the best sewing machine under \$100.. Is Janome better than brother? Janome is a heavy lightweight machine, and Janome also gives better stitches quality, and Brother does not give good stitches quality. Janome is a long-lasting product, and the brother is not a long-lasting product. Janome is a bit expensive, and Brother is value for money product. Which is better singer or Brother sewing machines? The Singer is quality and budget-friendly in this machine, many different types of built stitches. It's a durable and long-lasting product. Brother provides some good beginner-friendly sewing machine with this machine much better than Singer. Brother…
User
### Are Brother Sewing Machines Good Quality?
What is the most reliable brand of sewing machine? Here are ten of the best-known sewing machine brands in alphabetical order.Bernina.Brother.Elna.Husqvarna Viking.Jaguar.Janome.Juki.Singer.More items...•Feb 15, 2020. Is Brother sewing machine made in USA? Brother. Probably the most popular sewing machine in America. For that reason, people think they are manufactured here in the states, but it's actually a Japanese company. Machines: Sewing machines, embroidery machines, sergers, cover stick sergers, and cross-over. What is the best sewing machine on the market today? The Best Sewing MachineOur pick. Janome MOD-19. Best sewing machine for most beginners. ... Runner-up. Singer Heavy Duty 4423. A basic, even stitcher. ... Upgrade pick. Janome HD1000. Better for heavier fabrics.Feb 19, 2021 Are old sewing machines better? A quality vintage machine is made with higher quality components, better overall build quality, and will outlast any new machine on the market today. I routinely sew with machines that are…
User
### Quick Answer: Kids Felt Board
How do you attach felt? What Glue Works Best on Felt?Tacky Glue.Tacky glue is a tried and true classic for gluing felt.It's excellent for gluing felt to felt, and for any other general crafting....Permanent Glue.A permanent glue like super glue or E6000 is another option for adhering felt....Adhesives to Avoid.The number one adhesive to avoid is ordinary white craft glue.. What is the difference between fleece and felt? Felt is a non-woven cloth that is produced by matting, condensing and pressing woollen fibres. ... Fleece fabric is a soft napped insulating synthetic fabric made from Polyethylene terephthalate (PET) or other synthetic fibers. It is light and strong pile fabric meant to mimic and in some ways surpass wool. How do you print on Felt? How To Do This Method: While the pen will come with instructions, the basic idea is that you print out your pattern (the mirror image) on…
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what is reflex angle
Therefore, option C is correct. Children start learning the language of direction in … It can be one of the more confusing angles to find because it's on the 'outside' of the angle. Reflex angle definition, an angle greater than 180° and less than 360°. Reflex angles are … Types of Angles: In Geometry, two lines intersect at a point to form an Angle.This angle might be an Acute Angle, Obtuse Angle, Straight Angle, Right Angle, or Reflex Angle based on their measurement. The turn of an angle is how many degrees it rotates around the central point. The region from more than $180^\circ$ to less than $360^\circ$ is called reflex. It starts at one line and turns around the correct number of degrees to the other line. First, recapitulate reflex angle meaning. The image below illustrates a reflex angle. Geometrically, reflex angles are formed possibly in two different cases. What do children learn about angles in KS1? Angles are measured in terms of degree.It is not necessary that only two straight lines’ intersection forms an angle. I only knew this angle symbol "$\angle$", which is usually used to represent acute angles. Here, only 2 0 4 o is greater than 1 8 0 o and less than 3 6 0 o . Flat line. Reflex Angle. What Is The Definition Of A Reflex Angle? The angles below are all reflex angles: A reflex angle is any angle that is more than 180 degrees (half circle) and less than 360 degrees (full circle). Sign for an angle. An angle equal to 360 degrees is called full rotation or full angle. Full of games that students love, Reflex takes students at every level and helps them quickly gain math fact fluency and confidence. There are six types of angle in total; An Acute angle is the smallest, measuring more than 0 ° but less than 90 °.. Next up is a Right angle, also taught as a quarter turn.This angle always measures 90 °.. An Obtuse angle measures more than 90 ° but less than 180 °.. A Straight angle or a half turn is always 180 °.. (The Lesson) An reflex angle is an angle greater than 180° and less than 360°. Case study. A reflex angle will always have either an obtuse or an acute angle on the other side of it.. A reflex angle is greater than a straight line and less than a complete revolution. In order to make a full 360 turn, an obtuse angle or an acute angle must be added to the reflex angle value. A reflex angle is an angle that measures between 180˚ and 360˚. Real Examples of Straight Angles. Reflex angle is the angle which is greater than 1 8 0 o and less than 3 6 0 o. The sign for an angle is a curved line. Full Rotation. See more. So, if you are given an acute or obtuse angle measuring x, then full 360 degrees can be achieved with the following expression: x + r = 360 degrees, here r is the reflex angle. I know, I can simply say "The reflex $\angle$ of $\angle ABC$" or "2$\pi$ - $\angle ABC$", but is there a symbol for reflex angle, just for interest? What Is a Reflex Angle? Adaptive and individualized, Reflex is the most effective system for mastering basic facts in addition, subtraction, multiplication and division for grades 2+. If an angle has 180°, it looks like a straight line. An angle of more than $180^\circ$ but less than $360^\circ$ is called a Reflex angle. Any angle that has a measure which is greater than 180 degrees but less than 360 degrees (which coincides with 0 degrees) is a reflex angle. ExploreLearning Reflex helps all students succeed. Reflex Angles explained. If the angle lies in this region, the angle is known as the reflex angle geometrically. But now I have accounted a problem, where I wanted to represent the reflex symbol of $\angle ABC$. The other side of it represent the reflex angle is known as reflex! Where I wanted to represent the reflex symbol of $\angle$ '', which is usually to... 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Region, the angle is greater than 180° and less than 360° reflex takes students every... Than a straight line and less than a complete revolution … what is reflex angle reflex angle is angle!
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# Need help with water pressure problem
I'm running a drip system from the faucet in my garage to a second story balcony. What I have is some 1/4" tubing that goes directly from that faucet, up one length of the garage door, across the length of the garage door (two car garage), down the length of the garage door, out the garage to a patch of dirt, up a tree that brushes against my condo, extending along a branch finally entering my balcony, up to a flower bed - whew......
okay...Now, I have enough pressure to water two flowerbeds - 4 drip sprinklers. I want to add three more pots and know that i won't have the water pressure to do so.
The total distance that the tubing covers, from the faucet to the flowerbed, is around 50'.
I have issues here with the homeowners association so I have to keep things as stealth as possible - that's why I chose to use the 1/4" instead of the 1/2" tubing.
Soooooo...in an effort to gain a little more water pressure, I was thinking of re-tubing the lengths on the inside of the garage - up, across, and down with the 1/2" tubing.
I guess what I'm looking for here is someone kind enough to help me with some calculations...
My constants are: Water pressure, distance from the faucet to the flowerbeds, and the height from the ground floor to the second story.
Variable(s): The inside diameter of the tubing.
I'd like to know if I could increase the water pressure by "cutting down" the amount of 1/4" tubing (closest to the faucet) that I'm using and instead replacing it with the 1/2" tubing. The total length of all tubing with still be the same - still 50'. The height that the tubing travels would also still be the same and so would the starting water pressure (whatever that happens to be).
Say for example I now have:
50 feet of 1/4" tubing from water source to flower bed and that I get a water presure reading of X.
Faucet[=========P' of 1/4" tubing===========]flowerbed
If I use 20' feet of 1/2" and 30' of 1/4" should my water pressure increase?
Faucet]== ' of 1/2" tubing===][=0' of 1/4"==]flowerbed
If I use 40' feet of 1/2" and 10' of 1/4" will my water pressure increase even more?
Faucet]==@' of 1/2" tubing===][=' of 1/4"==]flowerbed
Thanks... :)
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of 1/2" tubing===][=' of 1/4"==]flowerbed
From what I know about drip systems they are all low pressure systems. Changing the size will get you more volume. Can you install valves at the upper deck? If so put two valves on the main line going to the head locations. Water one at set at a time.
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of 1/2" tubing===][=' of 1/4"==]flowerbed
How come you know there will not be enough pressure, can you not just turn the faucet up a little more. most drip systems need pressure reducers to keep from blowing the heads off the tubes and maintain correct flow rates. Do you know because you already tried a simple extension of the existing tube.
You're working way too hard. Drip systems can usually be expanded significantly without pressure problems as long as you are using drip emitters. If you are using sprinkler heads then you need more flow but even these use low pressure. Only when you add so many emitters that they require more flow rate than the tube can supply will you experience a pressure drop.
It sounds like a 1/2' tube for main supply to the area then 1/4" branch tubes to the pots will work just fine.
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This site is supported by donations to The OEIS Foundation.
# Radical of n, product of distinct prime factors of n
(Redirected from Radical of n)
The radical (squarefree kernel, largest squarefree divisor) of an integer is the product of distinct prime factors of that integer. For example, the radical of 12 is 6, since 2 × 3 = 6. If an integer is squarefree, it is equal to its radical (see A007947).
Multiplicative with ${\displaystyle \scriptstyle a(p^{\alpha })\,=\,p,\,}$ thus with ${\displaystyle \scriptstyle n\,=\,\prod _{i=1}^{\omega (n)}p_{i}^{\alpha _{i}},\,}$ we have
${\displaystyle {\rm {rad}}(n)=\prod _{i=1}^{\omega (n)}p_{i}.\,}$
## n divided by radical of n
Multiplicative with ${\displaystyle \scriptstyle a(p^{\alpha })\,=\,p^{\alpha -1},\,}$ thus with ${\displaystyle \scriptstyle n\,=\,\prod _{i=1}^{\omega (n)}p_{i}^{\alpha _{i}},\,}$ we have
${\displaystyle {\frac {n}{{\rm {rad}}(n)}}=\prod _{i=1}^{\omega (n)}p_{i}^{\alpha _{i}-1}.\,}$
## Sequences
A007947 Largest squarefree number dividing ${\displaystyle \scriptstyle n\,}$, the squarefree kernel of ${\displaystyle \scriptstyle n\,}$, radical of ${\displaystyle \scriptstyle n\,}$: ${\displaystyle \scriptstyle {\rm {rad}}(n).\,}$
{1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 6, 13, 14, 15, 2, 17, 6, 19, 10, 21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 2, 33, 34, 35, 6, 37, 38, 39, 10, 41, 42, 43, 22, 15, 46, 47, 6, 7, 10, ...}
A003557 ${\displaystyle \scriptstyle n\,}$ divided by radical (largest squarefree divisor) of ${\displaystyle \scriptstyle n,\,n\,\geq \,1.\,}$
{1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 8, 1, 3, 1, 2, 1, 1, 1, 4, 5, 1, 9, 2, 1, 1, 1, 16, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 8, 7, 5, 1, 2, 1, 9, 1, 4, 1, 1, 1, ...}
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Vous êtes sur la page 1sur 4
Pretest for Capacitors
1.) At the moment the switch closes on this freshly connected RC circuit
a. Q = 0, I
o
=
!/R
b. Q = C
!, I
o
= 0
c. Q = C
!, I
o
=
!/R
d. Q = 0, I
o
= 0
.
2.) As the capacitor in the RC circuit above reaches its maximum charge:
a. the rate at which the current changes decreases and the rate at which the charge changes
increase
b. the rate at which the current changes increases and the rate at which the charge changes
increases.
c. both rates decrease.
d. both rates increase.
3.) As the capacitor in the RC circuit above reaches its maximum charge, which of the
following statements is FALSE:
a. the voltage across the capacitor is at its maximum.
b. the voltage across the resistor is zero.
c. the sum of the voltages across the capacitor and resistor is equal to the initial voltage
across the resistor.
d. the sum of the voltages across the capacitor and resistor is equal to the initial voltage
across the capacitor.
4.) A charged 1 mF capacitor is discharged
through a 1.00 k-ohm resistor as shown in
the diagram. If the original charge on the
capacitor is Q, approximately what is its
charge 1.00 s after the switch is closed?
a. 0.632 Q
b. 0.500 Q
c. 0.368 Q
d. 0.000 Q
5.) For the circuit shown above, again a 1 mF capacitor is discharged through a 1.00 k-ohm
resistor as shown in the diagram. If the original charge on the capacitor is Q, in terms of Q
what is the approximate current in the circuit 1 second after the switch is closed?
a. 632 Q
b. 500 Q
c. 368 Q
d. 0 Q
6.) You have a 1 mF capacitor with Qs worth of charge on it. A dielectric whose dielectric
constant is 5 is carefully slipped between the plate of the capacitor. Which statement is
FALSE.
a.) the new capacitance will equal 5C.
b.) the new charge on the plates is 5Q.
c.) the new voltage across the plates is a fifth what it was.
d.) the new electric field between the plates is a fifth what it was.
7.) Capacitance is:
b. the ratio of the magnitude of the charge on either conductor of a capacitor to the magnitude
of the potential difference between the conductors.
c. constant for a parallel plate capacitor.
d. all three choices.
8.) In a circuit, a capacitor has potential difference !V, charge Q, and capacitance C. The
potential difference is doubled. The capacitance:
a. changes in ways impossible to predict with the given information.
b. doubles.
c. does not change.
d. is divided in half.
9.) To increase the capacitance of a parallel-plate capacitor, you can:
a. increase the area of the plates.
b. increase the distance between the plates.
c. all of these choices.
d. none of these choices.
10.) Given a set of capacitors C
1
+ C
2
+ ... + C
n
, where n is greater than 1, will a greater
equivalent capacitance result by adding them in parallel or in series?
a. in series.
b. in parallel.
c. they will be the same.
d. This cannot be determine without know the value of each capacitor and number of
capacitors in the system.
11.) Given n capacitors with charge Q and capacitance C, will you get the greatest energy
stored:
a. in series.
b. in parallel.
c. They will be the same.
d. This cannot be determine without knowing more about the situation.
12.) Select the option that best describes a dielectric.
a.) A dielectric is a non-conducting material.
b. A dielectric is the material when placed between the plates of a capacitor will increase the
electric field.
c. A dielectric is something that when placed between the plates decreases the capacitance of
the capacitor.
d. All of these choices are true.
13.) Bakelite has a dielectric constant approximately twice that of silicone oil. The bakelite in
a capacitor with capacitance C is replaced with silicone oil. What will the new capacitance
be, approximately?
a. 2C
b. C/2.
c. C.
d. Impossible to tell without know the exact capacitances involved.
14.) A capacitor has capacitance C, charge Q, and potential difference with nothing between
the plates. While still connected to a battery, a dielectric is inserted with a dielectric constant
of 2. How will each change?
a. C
new
= 2C, Q
new
= Q, !V
new
= !V.
b. C
new
= 2C, Q
new
= 2Q, !V
new
= 2!V.
c C
new
= 2C, Q
new
= Q, !V
new
= 2!V
d. C
new
= 2C, Q
new
= 2Q, !V
new
= 2!V
15.) A capacitor has capacitance C, charge Q, and potential difference !V with nothing
between the plates. The capacitor is then disconnected from the battery and a dielectric is
inserted with a dielectric constant of 2. How will each of the values C, Q, and V change?
a. C
new
= 2C, Q
new
= Q, !V
new
= !V
b. C
new
= 2C, Q
new
= 2Q, !V
new
= !V/2
c C
new
= 2C, Q
new
= Q, !V
new
= !V/2 .
d. C
new
= 2C, Q
new
= 2Q, !V
new
= 2!V
16.) Which of the following best describes the workings of a dielectric inside a capacitor?
a. Either existing dipoles or induced dipoles align with the existing electric field. This
induces an electric field in the opposite direction, and creates an induced surface charge on
each surface of the dielectric next to the plates.
b. Either existing dipoles or induced dipoles align with the existing electric field. This results
in an induced surface charge on each surface, which amounts to an increased effective
charge. Since charge and capacitance are directly proportional, this leads to the increase in
capacitance observed when dielectrics are inserted
c. Even when existing dipoles are present, only induced dipoles align with the existing
electric field. This results in surface charge on each dielectric surface producing an electric
field in the opposite direction. This results in a decrease of the potential difference being
required to place the same amount of charge on the plates, which leads to a higher
capacitance, as expected.
d. Even when dipoles are induced, only existing dipoles align with the existing electric field.
This results in surface charge on each dielectric surface producing an electric field in the
opposite direction. This results in a decrease of the potential difference being required to
place the same amount of charge on the plates, which leads to a higher capacitance, as
expected.
Solutions: a, c, d, c, c, b, d, c, a, b, c, a, b, d, c, a.
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## [answered] 1) 2) 3) 4) 5) 6) 7) Find the area of the closed region bou
Hi ,Please could you help me with some of this pratice problem.
1) 2) 3) 4) 5) 6) 7) Find the area of the closed region bounded by y = x2 and y = 2x ? x2.
8) The area enclosed between y = 1, y = 2?x, x = ?, and x = 1 is revolved around the x-axis.
Set up, but do not evaluate, an integral to find the volume.
9) Set up, but do not evaluate, an integral for the length of the curve y = x sin(x) from x = -? to
x = ?. 10) The curve y = x4, from x = 0 to x = 5, is rotated about the x-axis. Set up, but do not evaluate,
an integral to find the area of the resulting surface.
11) Find a formula for the inverse function f -1(x). 12) Find f ?(0), the derivative evaluated at x = 0. 13) State whether the sequence converges or diverges. If the sequence converges, find its limit. 14) Determine whether the series converges or diverges. If it converges, find its sum. For problems 15-18, determine whether each series converges or diverges. Write down which
test you are using.
15) 16) 17) 18) Determine whether the series is absolutely convergent, conditionally convergent, or
divergent. Write down which test you are using. 19) Calculate the Maclauren series for f (x) to the x2 term, using the definition of a Maclauren
series. 20) Integrate a known power series to find a power series representation for ln(5-x) and its
Hint: 5 - x = 5(1 - x/5)
Solution details:
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This question was answered on: Sep 18, 2020
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We have a ready expert answer for this paper which you can use for in-depth understanding, research editing or paraphrasing. You can buy it or order for a fresh, original and plagiarism-free copy from our tutoring website www.aceyourhomework.com (Deadline assured. Flexible pricing. TurnItIn Report provided)
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# Comparison of Decimal Fractions
We will discuss here about the comparison of decimal fractions.
While comparing natural numbers we first compare total number of digits in both the numbers and if they are equal then we compare the digit at the extreme left. If they also equal then we compare the next digit and so on. We follow the same pattern while comparing the decimals.
Examples on Comparing Decimals:
1. Compare 0.6 and 0.8.
Solution:
0.6 = 6 tenths
0.8 = 8 tenths
Because 8 tenths > 6 tenths
Thus, 0.8 > 0.6
2. Compare 0.317 and 0.341
Solution:
0.317 = 0.3 + 0.01 + 0.007
= 3 tenths + 1 hundredths + 7 thousandths
0.341 = 0.3 + 0.04 + 0.001
= 3 tenths +4 hundredths + 1 thousandths
Because 3 tenths = 3 tenths,
Now, compare next digit
1 hundredths < 4 hundredths
Thus, 0.317 < 0.341
Steps of Comparison of Decimal Fractions are given below:
Step I: First we need to observe the integral part.
For example:
(i) 104 < 140, this is how we check the integral part
(ii) 153 = 153
(iii) 112 > 121
Step II: When the integral part is same then compare the tenths place
For example:
(i) 1.4 < 1.9,
(ii) 1.5 = 1.50
(iii) 16.2 > 16.1
Step III: When the tenth place is same compare the hundredths place.
For example:
(i) 10.04 < 10.09,
(ii) 1.97 = 1.97
(iii) 71.92 > 71.90
In this way we first check the integral part and then move to the decimal places one by one.
For example:
1. Which is greater, 12.0193 or 102.01?
Solution:
First check the integer part
12 and 102
12 is < 102
102.01 is greater.
2. Which is smaller, 19.023 or 19.027?
Solution:
For each of these decimals the integral part is the same. So compare the tenths place. This is also same, check the hundredths places that is also same then move to the next decimal place.
Therefore, 19.023 < 19.027
So, 19.023 is smaller.
3. Find the greater number; 162.19 or 126.91.
Solution:
162.19 is greater than 126.91.
4. Which number is greater 293.82 or 293.62?
Solution:
First check the integer part,
293 = 293
Then the tenth place
8 > 6
Now the hundredth place
2 = 2
Therefore, 293.82 is greater than 293.62.
5. Find the greater number; 1432.97 or 1432.99
Solution:
First check the integer part,
1432 = 1432
Then the tenth place
9 = 9
Now the hundredth place
7 < 9
Therefore, 1432.99 is greater than 1432.97
6. Which number is greater 187.653 or 187.651?
Solution:
First check the integer part,
187 = 187
Then the tenth place
6 = 6
Then the hundredth place
5 = 5
Now the thousandth place
3 > 1
Therefore, 187.653 is greater than 187.651
7. Which number is greater 153.071 or 153.017?
Solution:
First check the integer part,
153 = 153
Then the tenth place
0 = 0
Then the hundredth place
1 = 1
Now the thousandth place
7 = 7
Therefore, 153.071 = 153.017
8. Find the greater number; 1324.42 or 1324.44
Solution:
First check the integer part,
1324 = 1324
Then the tenth place
4 = 4
Now the hundredth place
2 < 4
Therefore, 1324.44 is greater than 1324.42
9. Which number is greater 804.07 or 804.007?
Solution:
First check the integer part,
804 = 804
Then the tenth place
0 = 0
Then the hundredth place
7 > 0
Therefore, 804.07 is greater than 804.007
10. Find the greater number; 211.21 or 211.21
Solution:
First check the integer part,
211 = 211
Then the tenth place
2 = 2
Now the hundredth place
1 = 1
Therefore, 211.21 = 211.21
11. Write in ascending order using < sign:
(a) 43.81, 43.18, 43.08, 43.80
Solution:
43.08 < 43.18 < 43.80 < 43.81
(b) 89.09, 89.90, 89.01, 89.013
Solution:
89.01 < 89.09 < 89.013 < 89.90
(c) 53.35, 53.53, 53.30, 53.05
Solution:
53.05 < 53.30 < 53.35 < 53.53
(d) 61.16, 61.61, 61.06, 61.36
Solution:
61.06 < 61.16 < 61.36 < 61.61
Decimal.
Decimal Place Value Chart.
Expanded form of Decimal Fractions.
Like Decimal Fractions.
Unlike Decimal Fraction.
Equivalent Decimal Fractions.
Changing Unlike to Like Decimal Fractions.
Comparison of Decimal Fractions.
Conversion of a Decimal Fraction into a Fractional Number.
Conversion of Fractions to Decimals Numbers.
Subtraction of Decimal Fractions.
Multiplication of a Decimal Numbers.
Multiplication of a Decimal by a Decimal.
Properties of Multiplication of Decimal Numbers.
Division of a Decimal by a Whole Number.
Division of Decimal Fractions
Division of Decimal Fractions by Multiples.
Division of a Decimal by a Decimal.
Division of a whole number by a Decimal.
Conversion of fraction to Decimal Fraction.
Simplification in Decimals.
Word Problems on Decimal.
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Unit 5: Statistics
# Topic B: Bar Graphs
Bar graphs compare quantities. Bar graphs are commonly used to illustrate information in newspapers, in magazine articles, and so on. Bar graphs may be written with the bars arranged vertically or horizontally. Graph One is shown both ways – first with vertical bars and second with horizontal bars.
1. Read the title and subtitles so you know what you are looking at.
2. Read the information on the vertical and horizontal axes. Notice that each bar represents a different item.
3. Look carefully at the scale. What unit of measure is being used? The unit of measure will be the same for each bar so that you can compare them.
4. Compare the length or height of each bar to find the information that you want.
Graph 1
1. How many rivers are shown on this graph?
2. What is the title of the graph?
3. What is the unit of measure?
4. Look at the scale for kilometres. How many kilometres are represented by each division on the page?
5. Which river is the longest? What is its length?
6. Which river is the shortest? What is its length?
7. Name two rivers which are approximately the same length.
8. Compare the Columbia River and the Fraser River.
1. Which is the longer?
2. Give the approximate difference in their lengths.
9. Give the approximate length of the North Thompson, South Thompson, and Thompson Rivers combined.
1. 12 rivers
2. Lengths of some British Columbia Rivers
3. Kilometres
4. 250 km
5. Columbia; ~1,950 km
6. Quesnel; ~100 km
7. North Thompson & South Thompson
1. Columbia
2. ~600 km (1,950 – 1,350)
8. ~750 km
Graph 2
1. Give the source of the information for this graph.
2. What is the unit of measure for the population scale?
3. What number of people is represented by each section on the scale?
1. Which country had the largest population?
2. What was the approximate population of this country?
4. Name the two countries which were the closest in population size.
5. What was the approximate population of Bangladesh?
6. What was the approximate population of India?
1. United Nations
2. In the Millions
3. 50 million
1. China
2. 1,354,000,000
5. 164,000,000
6. 1,214,000,000
Bar graphs can show more than one type of information for each item. These graphs are useful for making comparisons. The bars are usually shaded or coloured differently and a legend will be placed near the graph. The bar graphs must still all use the same unit of measure.
Graph 3
1. What is the subtitle?
2. Look at the legend. The grey bars give each country’s population for what year? The patterned bar gives the population for these same countries in what year?
3. What trend does the graph show?
1. Which country had the largest increase in population? (this means, which country’s population went up by the highest number)
2. About how much was that increase?
1. Which country had the least change in population?
2. About how much was that change?
1. Year 2010 and Year 1950
2. 2010, 1950
3. That countries around the world are growing in population
1. India
2. The increase was 843 million
1. Pakistan
2. The increase was 108 million
# Image Descriptions
## Graph 1.1 (Bar Graph)
A bar graph showing the lengths of some British Columbia rivers.
• The vertical axis is kilometres and has the numbers 0 to 2,000 in increments of 250.
• The horizontal axis is the names of the following British Columbia rivers: Fraser, North Thompson, South Thompson, Thompson, Quesnel, Nechako, Chilcotin, Lillooet, Kootenay, Columbia, Peace, and Laird.
The bar graph data is represented in the following table:
Lengths of Some British Columbia Rivers
British Columbia Rivers (Horizontal Axis) Kilometres (Vertical Axis)
Fraser ~1,350
N. Thompson ~300
S. Thompson ~300
Thompson ~150
Quesnel ~100
Nechako ~400
Chilcotin ~250
Lillooet ~200
Kootenay ~675
Columbia ~1,950
Peace ~1,675
Laird ~1,100
## Graph 1.2 (Bar Graph)
A bar graph showing the lengths of some British Columbia rivers.
• The vertical axis is the names of the following British Columbia rivers: Fraser, North Thompson, South Thompson, Thompson, Quesnel, Nechako, Chilcotin, Lillooet, Kootenay, Columbia, Peace, and Laird.
• The horizontal axis is kilometres and has the numbers 0 to 2,000 in increments of 250.
The bar graph data is represented in the following table:
Lengths of Some British Columbia Rivers
British Columbia Rivers (Vertical Axis) Kilometres (Horizontal Axis)
Fraser ~1,350
N. Thompson ~325
S. Thompson ~325
Thompson ~150
Quesnel ~100
Nechako ~400
Chilcotin ~250
Lillooet ~200
Kootenay ~675
Columbia ~1,950
Peace ~1,675
Laird ~1,100
## Graph 2 (Bar Graph)
A bar graph showing the population of the world’s most populated countries in 2010.
• The vertical axis is population in millions, and has the numbers 0 to 1,400 in increments of 50.
• The horizontal axis is countries, and contains the following: China, India, United States, Indonesia, Brazil, Bangladesh, Nigeria, and Pakistan.
The bar graph data is represented in the following table:
Population of the World’s Most Populated Countries in 2010
Country (Horizontal Axis) Population in Millions (Vertical Axis)
China 1,354
India 1,214
United States 287
Indonesia 232
Brazil 174
Nigeria 158
Pakistan 149
Source: United Nations, 2010
## Graph 3 (Bar Graph)
A bar graph showing the population of the world’s most populated countries in 2010 and 1950.
• The vertical axis is population in millions, and has the numbers 0 to 1,400 in increments of 50.
• The horizontal axis is countries, and contains the following: China, India, United States, Indonesia, Brazil, Bangladesh, Nigeria, and Pakistan.
• The legend denotes that each country is associated with two different bars – one referring to its population in 2010, and the other referring to its population in 1950.
The bar graph data is represented in the following table:
Population of the World’s Most Populated Countries: Year 2010 and Year 1950
Country (Horizontal Axis) Population in Millions in 2010 (Vertical Axis) Population in Millions in 1950 (Vertical Axis)
China 1,354 544
India 1,214 371
United States 287 157
Indonesia 232 77
Brazil 174 53
| 1,556 | 6,111 |
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# 7.6.1 Predicting the Extent of a Reaction
The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction. But it is important to note that an equilibrium constant does not give any information about the rate at which the equilibrium is reached.
The magnitude of Kc or Kp is directly proportional to the concentrations of products (as these appear in the numerator of equilibrium constant expression) and inversely proportional to the concentrations of the reactants (these appear in the denominator).
This implies that a high value of K is suggestive of a high concentration of products and vice-versa.
We can make the following generalizations concerning the composition of equilibrium mixtures
If Kc > 103, products predominate over reactants, i.e., if Kc is very large, the reaction proceeds nearly to completion. Consider the following examples
(a) The reaction of H2 with O2 at 500 K has a very large equilibrium constant, Kc = 2.4 × 1047.
(b) H2(g) + Cl2(g) 2HCl(g) at 300K has Kc = 4.0 × 1031.
(c) H2(g) + Br2(g) 2HBr (g) at 300 K, Kc = 5.4 × 1018
If Kc < 10–3, reactants predominate over products, i.e., if Kc is very small, the reaction proceeds rarely. Consider the following examples:
(a) The decomposition of H2O into H2 and O2 at 500 K has a very small equilibrium constant, Kc = 4.1 × 10-48
(b) N2(g) + O2(g) 2NO(g), at 298 K has Kc = 4.8 ×10-31
If Kc is in the range of 10-3 to 103 , appreciable concentrations of both reactants and products are present. Consider the following examples
(a) For reaction of H2 with I2 to give HI, Kc = 57.0 at 700K.
(b) Also, gas phase decomposition of N2O4 to NO2 is another reaction with a value of Kc = 4.64 × 10-3 at 25°C which is neither too small nor too large. Hence, equilibrium mixtures contain appreciable concentrations of both N2O4 and NO2.
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# function notation question
#### bumblebee123
##### Junior Member
can anyone help to explain this question?
question: f(x) = x^2 + 2x + 3, x ≥ 0
a) write x^2 + 2x + 3 in the form (x + c)^2 + d
I managed to do this: ( x + 1 )^2 + 2
b) the domain of f is x ≥ 0 Find the range of F
but this is where I get stuck. Can anyone help explain what to do next?
#### MarkFL
##### Super Moderator
Staff member
One way we could proceed is to begin with the given domain:
$$\displaystyle 0\le x$$
$$\displaystyle 1\le x+1$$
Since both sides are positive, we may square them
$$\displaystyle 1\le(x+1)^2$$
$$\displaystyle 3\le(x+1)^2+2=f(x)$$
#### Dr.Peterson
##### Elite Member
can anyone help to explain this question?
question: f(x) = x^2 + 2x + 3, x ≥ 0
a) write x^2 + 2x + 3 in the form (x + c)^2 + d
I managed to do this: ( x + 1 )^2 + 2
b) the domain of f is x ≥ 0 Find the range of F
but this is where I get stuck. Can anyone help explain what to do next?
Another way is to graph (or just imagine graphing) the function. From the rewritten form (sometimes called "vertex form"), you can see that the minimum value for the unrestricted function is 2, attained at x=-1. The function increases for all x ≥ -1, which implies that it is increasing on the stated domain, x ≥ 0, and must attain its minimum value at the left-hand end of that domain, namely at x = 0. So the minimum value is f(0) = 3.
You could also use calculus to determine all the same facts and make the same conclusion.
#### bumblebee123
##### Junior Member
Another way is to graph (or just imagine graphing) the function. From the rewritten form (sometimes called "vertex form"), you can see that the minimum value for the unrestricted function is 2, attained at x=-1. The function increases for all x ≥ -1, which implies that it is increasing on the stated domain, x ≥ 0, and must attain its minimum value at the left-hand end of that domain, namely at x = 0. So the minimum value is f(0) = 3.
You could also use calculus to determine all the same facts and make the same conclusion.
how did you know that the minimum value for the function is 2, attained at x= -1?
is it because ( -1 + 1)^2 + 2 = 2
#### bumblebee123
##### Junior Member
here's what I've worked out so far:
it tells me that the domain of f is x≥ 0
to find x, if i use the completed square, x + 1 = 0
x = -1
if i put the x coordinate back into the equation: ( -1 + 1 )^2 +2 = 2
but this is when x = -1, so when x = 0 I need to +1 to the answer which = 3
so the range is f(x) ≥ 3
I also found another method I tried gives me the same answer, is this a correct method? :
as the domain is x ≥ 0
if I put x = 0 into the completed square equation: ( 0 + 1 )^2 + 2 = 3
#### HallsofIvy
##### Elite Member
Yes, $$\displaystyle (x+ 1)^2+ 3$$ is smallest when $$\displaystyle (x+ 1)^2$$ which is when $$\displaystyle x+ 1$$ is smallest which is when x= 0. (Notice that if the condition "$$\displaystyle x\ge 0$$" where not there, $$\displaystyle (x+ 1)^2$$ would be smallest when x=-1.)
#### Dr.Peterson
##### Elite Member
how did you know that the minimum value for the function is 2, attained at x= -1?
is it because ( -1 + 1)^2 + 2 = 2
The way I suggested there depends on familiarity with "vertex form", $$\displaystyle y = (x - h)^2 + k$$. From that form, you can read off that the vertex is $$\displaystyle (h, k)$$.
If you don't know that, then you would what has been discussed on your other thread about quadratic functions. We know that the smallest value of $$\displaystyle (x + 1)^2$$ occurs when $$\displaystyle x + 1$$ is zero (because otherwise it is always positive). Then, as you showed, "plugging in" $$\displaystyle x = -1$$ gives the corresponding value of y.
#### HallsofIvy
##### Elite Member
A real number, squared, is never negative. Whatever x and h are, $$\displaystyle (x- h)^2\ge 0$$, equal to 0 when x= h and positive otherwise. So whatever x and h are, $$\displaystyle (x- h)^2+ k\ge k$$, equal to k when x= h and greater than k otherwise.
#### bumblebee123
##### Junior Member
The way I suggested there depends on familiarity with "vertex form", $$\displaystyle y = (x - h)^2 + k$$. From that form, you can read off that the vertex is $$\displaystyle (h, k)$$.
If you don't know that, then you would what has been discussed on your other thread about quadratic functions. We know that the smallest value of $$\displaystyle (x + 1)^2$$ occurs when $$\displaystyle x + 1$$ is zero (because otherwise it is always positive). Then, as you showed, "plugging in" $$\displaystyle x = -1$$ gives the corresponding value of y.
so I would've known either by because k in the equation = 2
or because plugging in x=-1 gives y= 2 ?
#### Dr.Peterson
##### Elite Member
so I would've known either by because k in the equation = 2
or because plugging in x=-1 gives y= 2 ?
Yes, on both counts.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > scaffval Structured version Visualization version GIF version
Theorem scaffval 19654
Description: The scalar multiplication operation as a function. (Contributed by Mario Carneiro, 5-Oct-2015.) (Proof shortened by AV, 2-Mar-2024.)
Hypotheses
Ref Expression
scaffval.b 𝐵 = (Base‘𝑊)
scaffval.f 𝐹 = (Scalar‘𝑊)
scaffval.k 𝐾 = (Base‘𝐹)
scaffval.a = ( ·sf𝑊)
scaffval.s · = ( ·𝑠𝑊)
Assertion
Ref Expression
scaffval = (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦))
Distinct variable groups: 𝑥,𝑦,𝐵 𝑥,𝐾,𝑦 𝑥, · ,𝑦 𝑥,𝑊,𝑦
Allowed substitution hints: (𝑥,𝑦) 𝐹(𝑥,𝑦)
Proof of Theorem scaffval
Dummy variable 𝑤 is distinct from all other variables.
StepHypRef Expression
1 scaffval.a . 2 = ( ·sf𝑊)
2 fveq2 6663 . . . . . . . 8 (𝑤 = 𝑊 → (Scalar‘𝑤) = (Scalar‘𝑊))
3 scaffval.f . . . . . . . 8 𝐹 = (Scalar‘𝑊)
42, 3eqtr4di 2877 . . . . . . 7 (𝑤 = 𝑊 → (Scalar‘𝑤) = 𝐹)
54fveq2d 6667 . . . . . 6 (𝑤 = 𝑊 → (Base‘(Scalar‘𝑤)) = (Base‘𝐹))
6 scaffval.k . . . . . 6 𝐾 = (Base‘𝐹)
75, 6eqtr4di 2877 . . . . 5 (𝑤 = 𝑊 → (Base‘(Scalar‘𝑤)) = 𝐾)
8 fveq2 6663 . . . . . 6 (𝑤 = 𝑊 → (Base‘𝑤) = (Base‘𝑊))
9 scaffval.b . . . . . 6 𝐵 = (Base‘𝑊)
108, 9eqtr4di 2877 . . . . 5 (𝑤 = 𝑊 → (Base‘𝑤) = 𝐵)
11 fveq2 6663 . . . . . . 7 (𝑤 = 𝑊 → ( ·𝑠𝑤) = ( ·𝑠𝑊))
12 scaffval.s . . . . . . 7 · = ( ·𝑠𝑊)
1311, 12eqtr4di 2877 . . . . . 6 (𝑤 = 𝑊 → ( ·𝑠𝑤) = · )
1413oveqd 7168 . . . . 5 (𝑤 = 𝑊 → (𝑥( ·𝑠𝑤)𝑦) = (𝑥 · 𝑦))
157, 10, 14mpoeq123dv 7224 . . . 4 (𝑤 = 𝑊 → (𝑥 ∈ (Base‘(Scalar‘𝑤)), 𝑦 ∈ (Base‘𝑤) ↦ (𝑥( ·𝑠𝑤)𝑦)) = (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦)))
16 df-scaf 19639 . . . 4 ·sf = (𝑤 ∈ V ↦ (𝑥 ∈ (Base‘(Scalar‘𝑤)), 𝑦 ∈ (Base‘𝑤) ↦ (𝑥( ·𝑠𝑤)𝑦)))
176fvexi 6677 . . . . 5 𝐾 ∈ V
189fvexi 6677 . . . . 5 𝐵 ∈ V
1912fvexi 6677 . . . . . . 7 · ∈ V
2019rnex 7614 . . . . . 6 ran · ∈ V
21 p0ex 5273 . . . . . 6 {∅} ∈ V
2220, 21unex 7465 . . . . 5 (ran · ∪ {∅}) ∈ V
23 df-ov 7154 . . . . . . 7 (𝑥 · 𝑦) = ( · ‘⟨𝑥, 𝑦⟩)
24 fvrn0 6691 . . . . . . 7 ( · ‘⟨𝑥, 𝑦⟩) ∈ (ran · ∪ {∅})
2523, 24eqeltri 2912 . . . . . 6 (𝑥 · 𝑦) ∈ (ran · ∪ {∅})
2625rgen2w 3146 . . . . 5 𝑥𝐾𝑦𝐵 (𝑥 · 𝑦) ∈ (ran · ∪ {∅})
2717, 18, 22, 26mpoexw 7774 . . . 4 (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦)) ∈ V
2815, 16, 27fvmpt 6761 . . 3 (𝑊 ∈ V → ( ·sf𝑊) = (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦)))
29 fvprc 6656 . . . 4 𝑊 ∈ V → ( ·sf𝑊) = ∅)
30 fvprc 6656 . . . . . . 7 𝑊 ∈ V → (Base‘𝑊) = ∅)
319, 30syl5eq 2871 . . . . . 6 𝑊 ∈ V → 𝐵 = ∅)
3231olcd 871 . . . . 5 𝑊 ∈ V → (𝐾 = ∅ ∨ 𝐵 = ∅))
33 0mpo0 7232 . . . . 5 ((𝐾 = ∅ ∨ 𝐵 = ∅) → (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦)) = ∅)
3432, 33syl 17 . . . 4 𝑊 ∈ V → (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦)) = ∅)
3529, 34eqtr4d 2862 . . 3 𝑊 ∈ V → ( ·sf𝑊) = (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦)))
3628, 35pm2.61i 185 . 2 ( ·sf𝑊) = (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦))
371, 36eqtri 2847 1 = (𝑥𝐾, 𝑦𝐵 ↦ (𝑥 · 𝑦))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 ∨ wo 844 = wceq 1538 ∈ wcel 2115 Vcvv 3480 ∪ cun 3917 ∅c0 4276 {csn 4550 ⟨cop 4556 ran crn 5544 ‘cfv 6345 (class class class)co 7151 ∈ cmpo 7153 Basecbs 16485 Scalarcsca 16570 ·𝑠 cvsca 16571 ·sf cscaf 19637 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2179 ax-ext 2796 ax-sep 5190 ax-nul 5197 ax-pow 5254 ax-pr 5318 ax-un 7457 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2624 df-eu 2655 df-clab 2803 df-cleq 2817 df-clel 2896 df-nfc 2964 df-ne 3015 df-ral 3138 df-rex 3139 df-rab 3142 df-v 3482 df-sbc 3759 df-csb 3867 df-dif 3922 df-un 3924 df-in 3926 df-ss 3936 df-nul 4277 df-if 4451 df-pw 4524 df-sn 4551 df-pr 4553 df-op 4557 df-uni 4825 df-iun 4907 df-br 5054 df-opab 5116 df-mpt 5134 df-id 5448 df-xp 5549 df-rel 5550 df-cnv 5551 df-co 5552 df-dm 5553 df-rn 5554 df-res 5555 df-ima 5556 df-iota 6304 df-fun 6347 df-fn 6348 df-f 6349 df-fv 6353 df-ov 7154 df-oprab 7155 df-mpo 7156 df-1st 7686 df-2nd 7687 df-scaf 19639 This theorem is referenced by: scafval 19655 scafeq 19656 scaffn 19657 lmodscaf 19658 rlmscaf 19983
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# Thread: Circular Motion in B field
1. ## Circular Motion in B field
Again, thanks in advance for the help!
We have two charged particles of equal mass travelling along circular orbits in a region of uniform, constant magnetic field. Particle 1 has a radius of orbit r1 = 5 cm and a speed of v1=10^5 m/s. Particle 2 has Q2=2*Q1, and KE2=2*KE1.
What is the radius of orbit for R2?
I think in this situation r1=sqrt(2)r2
r2=sqrt(KE1*m1)/q1B
r1=sqrt(2*KE1*m1)/q1B
2. Originally Posted by profound
Again, thanks in advance for the help!
We have two charged particles of equal mass traveling along circular orbits in a region of uniform, constant magnetic field. Particle 1 has a radius of orbit r1 = 5 cm and a speed of v1=10^5 m/s. Particle 2 has Q2=2*Q1, and KE2=2*KE1.
What is the radius of orbit for R2?
I think in this situation r1=sqrt(2)r2
r2=sqrt(KE1*m1)/q1B
r1=sqrt(2*KE1*m1)/q1B
The centripetal force on each particle is provided by the magnetic force.
Centripetal force is proportional to velocity squared and inversely proportional to the radius. Velocity squared is proportional to K.E.
Therefore: The radius is directly proportional to K.E. and inversely proportional to the magnetic force.
Meanwhile, the magnetic force is directly proportional to both charge and velocity.
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### Home > PC3 > Chapter Ch8 > Lesson 8.3.3 > Problem8-115
8-115.
Simplify each the following trigonometric expressions.
1. $\cos\left(A\right)\cot\left(A\right) + \sin\left(A\right)$
1. Rewrite everything in terms of sine and cosine.
2. Get a common denominator.
3. Use the Fundamental Pythagorean Identity.
4. Simplify.
2. $\sin^2\left(θ\right)+\tan^2\left(θ\right)-\cot^2\left(θ\right)+\cos^2\left(θ\right)-\sec^2\left(θ\right)+\csc^2\left(θ\right)$
Use the Pythagorean Identities.
Change their forms to get $1$s.
$\cos^{2}\left(θ\right) + \sin^{2}\left(θ\right) = 1$
$1 + \tan^{2}\left(θ\right) = \sec^{2}\left(θ\right)$
$1 + \cot^{2}\left(θ\right) = \csc^{2}\left(θ\right)$
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# maths
posted by .
1*2 +2*(2^2)+2*(2^3)............+100*(2^100)
Find the sum of series.
• maths -
1+x+x^2+...+x^n = (x^(n+1)-1)/(x-1)
Now I think the beginning of your sequence should be
2*2^0 + 2*2^1 + 2*2^2 + ...
In that case, the sum is 2*(2^101-1)/(2-1)
• maths -
On the other hand, there could be only one typo and the series is
1(2^1) + 2(2^2) + 3(2^3 + ... + 100(2^100)
In that case you have a hypergeometric series ...
• maths -
I think I gotta go with Reiny. My answer is bogus because of the 100*2^100.
What was I thinking?
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3. ### math
the "k"th term of a series, Sk=a 1-r^k/1-r, is the sum of the first "k" terms of the underlying sequence. the difference between the "n"th terms of two particular series is greater than 14 for some values of n (is an element of) N. …
1).Find the sum of the n terms of the series Sn=1^2 + 3^2 +5^2+...+(2n-1)^2 2).Find the sum to n terms of 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6) +.... 3) Sum to n terms,the series 1.3.5 + 2.4.6 + 3.5.7 +.
5. ### college precalculus
find the sum of the infinite geometric series 1/3^6+1/3^8+1/3^10+1/3^12........ -100/9+10/3-1+3/10
6. ### help maths partail sum
find the partial sum and the limiting sum of the following infinite series 1/4+1/10+4/18.... show step got no ideal at all
7. ### Maths GRE
A salesman receives a commission of c% on a sale of D dollars. Find his commission. A)cD B)cD/100 C)100cD D)c/100D E)100c/D Any number expressed as a percentage is that number divided by 100 expressed as a decimal. Thus c% is c/100 …
8. ### Maths GRE
What single discount is equivalent to two successive discounts of 15% and 10% 100 - {(100-first discount) X (100- second discount)/100} = 100 - (100-15)X(100-10)/100 = 100 - 85 X 90/100 = 100 - 76.5 = 23.5
9. ### series help maths steve or reiny
find the sum of n-term of the series 1/1+1/2+1/3+1/4+1/5+...+1/n plz show step
10. ### maths
An arithmetic progression has 10 terms. Sum of the 10 terms is 220. Sum of the odd terms is 100. Find the first term and common difference. pl give me the answer. maths - Steve Monday, July 17, 2017 at 12:11pm 10/2 (2a+9d) = 220 5/2 …
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# Stable and efficient computation of binomial expectations
Suppose we want to compute the expected value of some function $$f(X)$$ where $$X \sim \text{Bin}(n,\theta)$$. Taking $$\mathbf{f} = (f_0,...,f_n)$$ to be the function values over all possible outcomes of the binomial random variable, this expected value can be written as a function:
\begin{aligned} B(\mathbf{f}, \theta) \equiv \mathbb{E}(f(X)) &= \sum_{x=0}^n f_x \cdot \text{Bin}(x|n,\theta) \\[6pt] &= \sum_{x=0}^n f_x {n \choose x} \theta^x (1-\theta)^{n-x}, \\[6pt] \end{aligned}
where the function arguments are the vector $$\mathbf{f} \in \mathbb{R}^{n+1}$$ and the parameter $$0 \leqslant \theta \leqslant 1$$. (Note that $$n$$ is determined as one less than the length of the vector $$\mathbf{f}$$, so it is not a separate argument in the function.) In mathematical parlance, this function is called the Bézier curve, and the vector $$\mathbf{f}$$ gives the "control points" of the function. This object is useful for a range of problems involving binomial random variables. In particular, computation of this function subsumes the problems of computing the mass and distribution for the binomial distribution (which are easily obtained by using values of zeros and ones in the vector $$\mathbf{f}$$).
When $$n$$ is large, direct computation of this quantity is problematic, since the binomial coefficient becomes large, and the power terms become small, leading to arithmetic overflow and underflow problems. This can potentially be dealt with by a range of methods, such as conversion to a log-scale, using a recursive algorithm, etc. However, I am not sure what algorithms are actually used in practice in statistical computation, and what algorithm is considered the "gold standard".
My Questions: What are some efficient and stable methods of computation of this function? What algorithms are used in practice in statistical computing? Is there any algorithm for this problem that is considered to be the "gold standard"?
Although this is my own question, I am going to give an answer showing one possible algorithm, to get the ball rolling. One way to compute the Bézier curve is De Casteljau's algorithm, which is a numerically stable computation method that uses a recursive method related to the recursive property of the binomial distribution. This algorithm can be implemented either on the standard probability scale, or on the log-probability scale.
Recursive characterisation of the Bézier curve: To obtain a recursive characterisation of this function, we will take advantage of the well-known recursive equation:
$$\text{Bin}(x|n, \theta) = (1-\theta) \cdot \text{Bin}(x|n-1, \theta) + \theta \cdot \text{Bin}(x-1|n-1, \theta).$$
Using this recursive equation, for any argument vector $$\mathbf{f} = (f_0,...,f_n)$$, we have:
\begin{aligned} B(\mathbf{f}, \theta) &= \sum_{x=0}^n f_x \cdot \text{Bin}(x|n,\theta) \\[6pt] &= \sum_{x=0}^n f_x \Big[ (1-\theta) \cdot \text{Bin}(x|n-1, \theta) + \theta \cdot \text{Bin}(x-1|n-1, \theta) \Big] \\[6pt] &= (1-\theta) \times \sum_{x=0}^{n-1} f_x \cdot \text{Bin}(x|n-1, \theta) + \theta \times \sum_{x=0}^{n-1} f_{x+1} \cdot \text{Bin}(x|n-1, \theta) \\[6pt] &= (1-\theta) \cdot B(\triangleleft \ \mathbf{f}, \theta) + \theta \cdot B(\triangleright \ \mathbf{f}, \theta), \\[6pt] \end{aligned}
where $$\triangleleft \ \mathbf{f} = (f_0,...,f_{n-1})$$ and $$\triangleright \ \mathbf{f} = (f_1,...,f_{n})$$. This gives us a recursive equation for the Bézier curve, where the recursion decomposes the function into a weighted sum of the Bézier curve for smaller control vectors. At each step of the iteration, the length of the control vector is reduced by one element.
De Casteljau's algorithm: This algorithm takes advantage of the above recursive equation for the Bézier curve. To explain the algorithm, we define the operators $$\triangleleft$$ and $$\triangleright$$ to truncate the argument vector by one element from the right and left respectively. Now, taking a fixed value of $$\theta$$ we can define the values:
$$B_{k,x} \equiv B(\triangleleft^{n-k-x} \triangleright^x \mathbf{f},\theta),$$
and we can arrange these values in an $$(n+1) \times (n+1)$$ matrix $$\mathbf{B}$$ as follows:
$$\mathbf{B} \equiv \begin{bmatrix} B_{0,0} & B_{0,1} & B_{0,2} & \cdots & B_{0,n-2} & B_{0,n-1} & B_{0,n} \\ B_{1,0} & B_{1,1} & B_{1,2} & \cdots & B_{1,n-2} & B_{1,n-1} & 0 \\ B_{2,0} & B_{2,1} & B_{2,2} & \cdots & B_{2,n-2} & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ B_{n-2,0} & B_{n-2,1} & B_{n-2,2} & \cdots & 0 & 0 & 0 \\ B_{n-1,0} & B_{n-1,1} & 0 & \cdots & 0 & 0 & 0 \\ B_{n,0} & 0 & 0 & \cdots & 0 & 0 & 0 \\ \end{bmatrix}.$$
The bottom element of this matrix is $$B_{n,0} = B(\triangleleft^0 \triangleright^0 \mathbf{f},\theta) = B_\mathbf{f}(\theta)$$, which is the function output we want to compute, and the top row consists of values $$B_{0,x} = B(\triangleleft^{n-x} \triangleright^x \mathbf{f},\theta) = f_x$$, which are the initial "control points" of the function.
De Casteljau's algorithm begins with the control vector at the top line of this matrix, and works downward through the above matrix of values by using the above recursive characterisation of the Bézier curve. In scalar form, the recursive equations for the algorithm are:
\begin{aligned} B_{0,x} &\equiv f_x \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{ } \text{ for } x = 0,...,n, \\[6pt] B_{k,x} &\equiv (1-\theta) \cdot B_{k-1,x} + \theta \cdot B_{k-1,x+1} \quad \quad \quad \text{for } x = 0,...,n-k. \end{aligned}
As can be seen from the matrix, this algorithm involves computation of $$n(n+1)/2$$ values from the initial set of control points, so it has complexity $$\mathcal{O}(n^2)$$. Each recursive computation is a simple weighted average of the elements above and above-right, so the computational burden of each step involves two multiplications and one addition (and is therefore relatively small).
The algorithm can be implemented in standard probability scale, so long as the computational environment is sufficient to avoid underflow problems for small probabilities. Alternatively, the computation can be done in log-probability scale to avoid underflow problems (see here for discussion of adding small probabilities in log-scale).
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0
# What numbers is 621 divisible by?
Updated: 9/22/2023
Wiki User
11y ago
Best Answer
621 is divisible by these numbers: 1, 3, 9, 23, 27, 69, 207, 621.
Wiki User
11y ago
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Related questions
### What numbers are divisible by 621?
621 is divisible by 9?
### Is 621 divisible by 6?
621 is not divisible by 6.
### Why is 3 exactly divisible by 621?
It is not. 621 is divisible by 3, not the other way round.
### Is 4 divisible by 621?
As 4 is less than 621, 4 is not divisible by 621. Nor is 621 divisible by 4: 4 is even and all multiples of 4 are even, 621 is odd and so cannot be a multiple of 4. To check divisibility by 4, if the the last 2 digits are divisible by 4, the original number is divisible by 4. For 621, the last 2 digits are 21 and 21 is not divisible by 4, thus 621 is not divisible by 4; nor is 622 since the last 2 digits 22 is not divisible by 4.
### Is 621 divisible by 2?
no because 621 is odd.
### Why isn't 621 not divisible by 6?
In order for a number to be divisible by 6, it must be divisible by both 3 and 2. 621 is divisible by 3, but not by 2. In other words, 6 will not go into 621 evenly. The answer will be a decimal.
No.
yes
yes
### What numbers between 600-700 are divisible by 3 and 9?
603, 612, 621, 630, 639, 648, 657, 666, 675, 684, 693
### Which numbers are divisible?
All numbers are divisible by 1.
### Is 621 divisible by 2 3 4 5 6 10?
It is only divisible by 3, and not 2, 4, 5, 6, or 10.
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## Quiz from Wednesday, Nov 30
Question: For $\vec{F}(\vec{r})=\vec{r}$, find the flux on the surface of a sphere of radius $a$, oriented outward.
Solution: The easiest way to do this is to recall that $d\vec{A}=\vec{n}dA$. We have
$\int_S\vec{F}\cdot d\vec{A}=\int_S\vec{F}\cdot\vec{n} dA=a\int_S dA$
since the vectors $\vec{F}$ and $\vec{n}$ are parallel, with $\vec{n}$ being a unit vector (and hence the dot product is simply the magnitude of $\vec{F}$). Thus, we have
$a\int_S dA=4\pi a^3.$
Replacement Question: State something objectively coherent about Gauss’ Law.
Answer: I’ll accept a lot of answers here, but what I’m looking for is something along the lines of “electric flux on the surface of a closed object is proportional to the charge contained in the interior of the surface.” If you mumble something about Faraday cages, I’ll give it to you.
## Quiz for Wednesday, November 16
Question: State Green’s Theorem
Solution: If $C$ is a piecewise smooth, simple, closed curve that is the boundary of a region $R$ in the plane and oriented so that the region is on the left as one moves around the curve (equivalently, we move around the curve in a counter-clockwise fashion), and if $\vec{F}=F_1\vec{i}+F_2\vec{j}$ is a smooth vector field on an open region containing $R$ and $C$, then
$\int_C\vec{F}\cdot d\vec{r}=\int_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right) dxdy.$
Replacement Question: State the first ten or so words of the United States Declaration of Independence
Solution: When in the Course of human events, it becomes necessary for one people to dissolve the political bands which have connected them with another…
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# Punishment Games: 5x (Five Times)
4,792
## Introduction: Punishment Games: 5x (Five Times)
Welcome to my new Instructable Series: Punishment Games!
I first want to begin this guide off by saying thank you to the Instructables community for over 45k views on my guide on "How to Play Mafia (With and without cards)", linked below this intro. This guide was the first of its kind on Instructables, not to be mistaken for the guide on how to play "Werewolf." I've decided to keep the theme for a while on my Instructables, and keep making guides for circle games. I will soon make a compilation of all these games!
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Intro: What Is 5x (Five Times)?
5x is a punishment/circle game. Circle games involve you and a group of people playing the game in a circle. Punishment games involve on the spot punishments after a loss in a round. 5x is actually a variation on an Vietnamese game called "Nam, Muoi" (5, 10).
The concept behind 5x is that everybody gets a chance at "punishing" each other. I found this game to be very useful when I was trying to connect with a group of friends I had recently met a few years ago.
## Step 2: Setup
The setup for this game is simple:
You need 3 players or more
You should be able to see each other's hands, so sit/stand in a circle
The room must be somewhat silent
You may need ice afterwards....
## Step 3: Playing 5x
The game is played using your hands. In 5x, you use multiples of fives to set up the scenario. Putting out one hand flat and open is 5, putting out two hands flat and open is 10, and closing both your hands to form fists is 0. Here's the steps in which a round is played.
1: The game starts with one person who becomes the leader.
2: The leader counts down from 3, and on the final count, says a multiple of 5 (Either 0,5, or 10)
Example: 3...2...1... 5! (Leader puts out one hand flat and open, while the other hand is closed and flat).
3: Other players will also put out their hand(s) based on what multiple of 5 they chose on the final count.
4: If anybody matches the leader's number, they must put their hand out in the circle, and get slapped by the leader.
Note: Anybody else that is caught cheating or hesitating to choose a number right after the final count also gets punished.
5: The leader is then passed to the right for the next round.
## Step 4: After the Game
That's where the ice comes in.
My friends and I played this game until our hands were blood red, and boy did we have fun. When in doubt, try your best to be honest in the game and just take the punishments. There's no point in arguing in a circle game because they're meant to be fast (besides mafia).
The game is finished when no one can stand it no more, so have fun while you can!
## Step 5: Nam,Muoi (5,10)
I know some of you might want to know what the Vietnamese variation of this game is, and it definitely is a lot harder. I recommend playing with at least 4 other people, but at most 10 people. The steps are similar to 5x:
1: The game starts with one person who becomes the leader
2: The leader counts down from 3, and on the final count, says a multiple of 5 (This is where the game changes)
Note: Depending on how many people you have in the group, the leader will now call a multiple of five based on the number of players. Having 4 players means the max number to call would be 40. Having 5 increases the max number to call to 50, and so on. The minimum number will always be 0.
3: All the players, including the leader, will put their hand out (still 1 hand, 2 hands, or no hands)
4: The leader will now count everybody's hands, and if it matches the leaders numbers, everybody gets a slap on the hand.
5: The leader is then passed to the right for the next round.
Just as an example, I have a group of 5 friends, including me. The max number to call is 50. I say "3...2...1... 35!". That means in theory, if I put out 10 for my hands, friend A puts out 10, and friends B, C, and D put out 5, I have achieved a 35. Then I slap everyone's hands.
## Step 6: Some Key Points to Make the Game Fun
I've played variations of this game throughout my life, and always found some aspects of the game to be boring or not fun to play. Keep these key points in mind to make the game more fun!
- Know your audience (friends): People that you've known for a while might not mind a heavy hit sometimes, but strangers who are just joining you might not be so comfortable. Ease on with the hitting, and don't start with a stinger
- Fake your hands: You may have played simon says, where the leader says "Simon says do this", but then the leader does something opposite or unrelated to their command. This game also implements this, but as beginners, you should stick to same call, same hand to avoid confusion. Then you can do different calls, such as "5!" and put out ten.
- Know when to stop: Punishment games take a lot of energy and tolerance to play, so don't over do it, or you may end up avoiding this game in the future
## Recommendations
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https://plainmath.org/differential-calculus/37372-what-are-the-even-functions-and-odd-functions-let-f-x-is-a-func
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ahgan3j
2021-11-19
What are the even Functions and Odd Functions?
Let f(x) is a function.
IF $f\left(-x\right)=f\left(x\right)$, then f(x) is called even function.
Befoodly
Step 1
If $f\left(-x\right)=-f\left(x\right)$, then f(x) is called odd function.
Even functions are always symmetric about the y-axis and the odd functions are symmetric about the origin.
Step 2
For example:
Even function:
$f\left(x\right)={x}^{2}$
$g\left(x\right)={x}^{6}+3{x}^{2}+7$
etc
Odd function:
$f\left(x\right)=5x$
$g\left(x\right)=-{x}^{3}+3x$
etc
Do you have a similar question?
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