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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > xpdom3m GIF version
Theorem xpdom3m 6244
Description: A set is dominated by its Cartesian product with an inhabited set. Exercise 6 of [Suppes] p. 98. (Contributed by Jim Kingdon, 15-Apr-2020.)
Assertion
Ref Expression
xpdom3m ((A 𝑉 B 𝑊 x x B) → A ≼ (A × B))
Distinct variable groups: x,A x,B x,𝑉 x,𝑊
Proof of Theorem xpdom3m
StepHypRef Expression
1 xpsneng 6232 . . . . . . 7 ((A 𝑉 x B) → (A × {x}) ≈ A)
213adant2 922 . . . . . 6 ((A 𝑉 B 𝑊 x B) → (A × {x}) ≈ A)
32ensymd 6199 . . . . 5 ((A 𝑉 B 𝑊 x B) → A ≈ (A × {x}))
4 xpexg 4395 . . . . . . 7 ((A 𝑉 B 𝑊) → (A × B) V)
543adant3 923 . . . . . 6 ((A 𝑉 B 𝑊 x B) → (A × B) V)
6 simp3 905 . . . . . . . 8 ((A 𝑉 B 𝑊 x B) → x B)
76snssd 3500 . . . . . . 7 ((A 𝑉 B 𝑊 x B) → {x} ⊆ B)
8 xpss2 4392 . . . . . . 7 ({x} ⊆ B → (A × {x}) ⊆ (A × B))
97, 8syl 14 . . . . . 6 ((A 𝑉 B 𝑊 x B) → (A × {x}) ⊆ (A × B))
10 ssdomg 6194 . . . . . 6 ((A × B) V → ((A × {x}) ⊆ (A × B) → (A × {x}) ≼ (A × B)))
115, 9, 10sylc 56 . . . . 5 ((A 𝑉 B 𝑊 x B) → (A × {x}) ≼ (A × B))
12 endomtr 6206 . . . . 5 ((A ≈ (A × {x}) (A × {x}) ≼ (A × B)) → A ≼ (A × B))
133, 11, 12syl2anc 391 . . . 4 ((A 𝑉 B 𝑊 x B) → A ≼ (A × B))
14133expia 1105 . . 3 ((A 𝑉 B 𝑊) → (x BA ≼ (A × B)))
1514exlimdv 1697 . 2 ((A 𝑉 B 𝑊) → (x x BA ≼ (A × B)))
16153impia 1100 1 ((A 𝑉 B 𝑊 x x B) → A ≼ (A × B))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 97 ∧ w3a 884 ∃wex 1378 ∈ wcel 1390 Vcvv 2551 ⊆ wss 2911 {csn 3367 class class class wbr 3755 × cxp 4286 ≈ cen 6155 ≼ cdom 6156 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 629 ax-5 1333 ax-7 1334 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-8 1392 ax-10 1393 ax-11 1394 ax-i12 1395 ax-bndl 1396 ax-4 1397 ax-13 1401 ax-14 1402 ax-17 1416 ax-i9 1420 ax-ial 1424 ax-i5r 1425 ax-ext 2019 ax-sep 3866 ax-pow 3918 ax-pr 3935 ax-un 4136 This theorem depends on definitions: df-bi 110 df-3an 886 df-tru 1245 df-nf 1347 df-sb 1643 df-eu 1900 df-mo 1901 df-clab 2024 df-cleq 2030 df-clel 2033 df-nfc 2164 df-ral 2305 df-rex 2306 df-v 2553 df-un 2916 df-in 2918 df-ss 2925 df-pw 3353 df-sn 3373 df-pr 3374 df-op 3376 df-uni 3572 df-int 3607 df-br 3756 df-opab 3810 df-mpt 3811 df-id 4021 df-xp 4294 df-rel 4295 df-cnv 4296 df-co 4297 df-dm 4298 df-rn 4299 df-res 4300 df-ima 4301 df-fun 4847 df-fn 4848 df-f 4849 df-f1 4850 df-fo 4851 df-f1o 4852 df-er 6042 df-en 6158 df-dom 6159 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
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# Important Riddle for the World to Ponder: Riddle - The Two Guards
page: 1
6
posted on May, 1 2012 @ 03:07 AM
We have came to the fork in the road, with only two available paths to be taken as a whole for humanity.
One path can lead to paradise, the other to ruins. One would think that the descision should be easy, but that is until, one recognizes that both paths can not be distinguished from one another.
The path we can take that may lead towards paradise will encompass means of governance, morality, and justice... all concluding in the allowance of freedom and prosperity, so that we may no longer be focused on survival but can now thrive as a species.
The other path, also is encompassed by means of governance, morality and justice in the forms of doctrines and dogmas... all concluding in the allowance of the suppression of freedom and prosperity, so that they will be suppressed so much as to bring about moments of revelations and reactions that are only conducive towards ruins.
The face value of these paths are so similar that it's impossible to tell the difference. At this moment you are stuck with having to make a descision, there is no longer time to try and enlighten yourself as to possibly allow you to make the correct descision.
This is then, where the following riddle comes into play:
Riddle: The Two Guards?
You stand at a fork in the road. Next to each of the two forks, there stands a guard. You know the following things:
1. One path leads to Paradise, the other to Death. From where you stand, you cannot distinguish between the two paths. Worse, once you start down a path, you cannot turn back.
2. One of the two guards always tells the truth. The other guard always lies. Unfortunately, it is impossible for you to distinguish between the two guards.
You have permission to ask ONE guard ONE question to ascertain which path leads to Paradise. Remember that you do not know which guard you're asking -- the truth-teller or the liar -- and that this single question determines whether you live or die.
Question:
What one question asked of one guard guarantees that you are led onto the path to Paradise, regardless of which guard you happen to ask?
"If I asked the _other_ guard, which door would he indicate
Regardless of whom you ask, they'll point to the wrong door.
This answer is then to be considered when addressing Einstiens definition of Insanity, in relation towards the answers we are given by Politicians and ourselves.
So, what one question could you answer either of the guards that may give you a hint as to which direction you are to take?
This question has been given the following answer...
Ask a guard, what the other would say is the correct path that will lead towards paradise....
The one always telling the truth, will tell you that the lying guard chooses A... implying that B is the correct choice.
The one always lying, will tell you that the truth telling guard will choose A... because of it being a lie, this would imply that B is the correct choice.
So regardless of who you ask, you will always get the correct answer as to which leads to which, if you are to only ask what the other will say.
This exercise can then be paralleled in the current situation that we are in, and can be played out with in the faculties of ones mind.
Not knowing which path leads to paradise or ruins, we are always asking ourselves questions. We are often incorrect as to these descisions for cognitive bias' have clouded our judgement, so either of these 'guards' in our mind can not be percieved differently, nor are the paths truly that distinguishable.
So now ask yourself, is this the right path for me to take? This is where playing a self appointed 'devils advocate' comes in. If one guard says the other(liar, incorrect descision maker) will choose A and If you ask the other guard(liar, incorrect descision maker) what other(truth, correct) will suggest, he will say A(as to throw you off, and decieve you). Which then means, that B is always the correct descision.
Some have said that insanity is to be defined the way that Albert Einstien put it... The act of repeating the same actions, expecting a different result.
We as a humanity and all the great history that we have, have never made the correct descision. This is because we have always attempted to pretend that we can always distinguish between Guard 1 and 2, and that we can read between the lines as to which path(A & B) is to be taken. One says B, in opposition to what we percieve the other that is saying A as being incorrect, or lying to us. But even to differentiate between A and B is nearly impossible.
But just like the exercise illustrates, the descision to be made that is conducive towards finding paradise is not found in A, so why do we keep choosing A?
Because, we are told that the correct descision is A by the liars. And, that we are told the correct descision is not what the liars have said. Being that we are not all knowing, and many are ignorant to what is conducive towards what, in relation to human nature and allowance of freedom. There is a gap, as to knowing which is which, which is the truth and which is the lie.
This is right where temparance seals the gap. This is where empathy may become the bonding agent to fill this gap. This is where the abandoning of ones dogma and their indoctrinated means of establishing morality, law, and justice.
I'm not implying that there should be no law, that there should be anarchy, or some form of social anomoly. But rather, people should be more willing to say, 'I don't know', it is 'Not another's life I am to dictate the free will of', 'I can only hope that others make the best descisions for themselves.' This is your answer B. The answer that says that I should 'Know thy self', rather than pretend to know which A or B another person is on and that you can distinguish between the two.
This is how you escape the instanity of us dictating one another, expecting different results from the suppression of free will.
There is only dissent, when another's ideology impedes your own ability to act freely.
None of us are all knowing, none of us know the 'best' means... we are always changing, our initial reactions are even changing, our means of percieving one another is changing... paradigm is not set in stone.
I truly shot from the hip with this one, but I feel that there is much to be taken from it. I apologize to those that find this to be just ramblings, and contain no meaning. If that is the case, sit and ponder the mentioned rittle, and see how it applies to your life, government, and your relationship with the world as a whole. In the end, it should be beneficial.
btw, spellcheck is disabled, I apologize for any confusion this may have created when reading this thread.
I would appreciate any and all contributions to this idea, for I may have paradoxially mind fluffed myself LOL!
What does ATS think?
posted on May, 1 2012 @ 03:47 AM
I'd like to slit the throat of the Guard who always lies.
There's no need for that.
posted on May, 1 2012 @ 03:53 AM
It doesn't have to be that the guard always knowingly lies, it's just that what ever comes out of his mouth is incorrect.
This is to represent the duality in all decisions... right or wrong. This is why there's two guards, and exactly why you can not distinguish between the two. Because it would not be a decision then, for everyone would choose right, or the path that leads towards paradise.
That is where admitting that we know nothing, we are all equal in our pursuits, and not to suppress one another comes into play.
Both guards a representation of possibility, to kill one would to kill the other, or collapse dualism that is the construct of everything, which is not an available option at the moment.
So how then do you make a decision... you ask the 1 question and act accordingly.
posted on May, 1 2012 @ 04:02 AM
This is all very true.
So how do we apply this to our everyday life?
How does one go about asking the question? Must we then ask, what would the other person do?
Yikes, you may have just stumbled upon something.
A philosopher I am not.
I go with my first instinct almost always.
posted on May, 1 2012 @ 04:22 AM
Originally posted by Goldcurrent
This is all very true.
So how do we apply this to our everyday life?
How does one go about asking the question? Must we then ask, what would the other person do?
Yikes, you may have just stumbled upon something.
A philosopher I am not.
I go with my first instinct almost always.
How to apply:
It is to be taken in consideration when any answer to such questions arise as to which direction to take. Whether the answer is given by another person, or ourselves... to acknowledge that we simply don't know the correct choice, then gives premise towards going away from the suppressing of anothers freewill.
It's not necessarily about how, but rather, what answers you are willing to accept or the lack of said answer. Sometimes, we should be willing to accept that the answer is not to be known, just as we can not distinguish between the two paths. The answer is not to be found in what has not worked before, which then brings in the definition of insanity provided above.
Our gut reaction, or answer to every question is only based on acknowledging previous answers given to similar questions, which were then all based on a form of doctrine of morality. Yet all these answers have not brought us to a point in which we may all thrive, we are all stuck in survive for this very reason.
What should the other person do:
Know that they are not righteous, infaliable, and/or always correct. When these attributes are considered to be the individuals defining characteristics, that is when the freedoms of others are suppressed or dictated. This is something that empathy will always allow for one to see this is not the correct course of action.
I can't recall exactly what lead me to this questioning, I remember it had something to do with the Mayan elders and egyptian hieroglyphs in relation towards 'end times' and such, as to the specifics... I do not remember.
edit on 1-5-2012 by MESSAGEFROMTHESTARS because: clarification
posted on May, 1 2012 @ 04:22 AM
To try & find the correct path from today's politicians would require that at least one of them would be telling the truth.......
*tumbleweeds
posted on May, 1 2012 @ 05:09 AM
This is a great riddle, one I would never have been able to figure out on my own, but I've heard this one before. Though I kind of wish I hadn't. The way you put it is genius, and I would've considered myself lucky to have first come across it here in your thread! I cannot agree more. I have often pondered this myself, but without tying in the riddle to help it make sense to more people... Even still, I don't know if enough people are open minded enough to give it a try... But we've been trying the same thing for so long and it so obviously isn't working! At this point I'm almost willing to try anything... And that's actually scary... Because most people in my situation will accept whatever feels safe and familiar enough to be "normal", when what we truly need is something vastly different than "the usual"... People tend to think that "if only we can eliminate 'this' such and such enemy, get rid of 'these' kinds of people, then everything will be alright". Well, no... You have to learn to live with them and learn to teach them to live with you. Because there will always be more people who sprout up who you will find some problem with. Eventually, you will find yourself all alone if you could just eliminate everyone who bothers you. Imagine a white supremacist (im white so that's why I use this exampe but you can insert militant black bigot or Muslim extremist or antigay or any other "hater" in place of white supremacist), if he manages to pull of the eradication of all but white people, he will then find faults in other white people. So then he kills off all but his family and friends. Well then, a person like this, will eventually find fault in everyone except maybe his lover and kids, and say, one brother or best friend. Well then, perhaps he comes to realize his best friend has no choice for a mate except his own lover. Now he must die, too. Then after a while, his lover starts to annoy him and he doesn't feel the same way about her anymore. Or one of his kids does something to infuriate him... And eventually he's all alone. What's the point in that? We NEED to stop finding fault in others. We are FARless than perfect ourselves. We NEED to stop fighting our perceived enemies and find ways to become FRIENDS with them. It's NECESSARY for our survival and continuation as a species. How can this be done? Start with a smile... the rest, as they say, is history.
posted on May, 1 2012 @ 09:47 AM
Somebody's been watching too much Labyrinth...
(hoping at least SOMEONE gets that reference)...
True Paradise is a pipedream. We human beings define ourselves by conflicts (whether physical, emotional, etc.). Without these conflicts, life would be extremely BORING. That's right...true paradise would be BORING. Imagine if you could do anything you wanted to, eat anything you wanted to, etc.? What's left? Where's the challenge? You'd be bored out of your skull....
I think this is why I can sign on to the idea of reincarnation. After a bit of this "paradise", the soul simply longs to wipe the slate clean a bit, and give it another go, with all of the conflict that comes with life.
posted on Jun, 8 2012 @ 01:43 AM
The wrong question provides the right answer to the wrong question. The right question has a hidden answer.
Both will tell you the wrong path. In the end, you must choose the narrow path. It's neither one way or the other on the swing of the pendulum. Each side tells you the wrong direction. Choose the other way.
This is the same as esoteric and exoteric. Neither is the correct path. New age Theosophy is one extreme of the esoteric. This is the mystery schools and Freemasonry. They lie. Exoteric is Chruch Dogma and ritual. That amounts to idol worship and control. The path is narrow. Both will tell you the wrong door so you can find the right door: The one that they both deny as correct.
The Rap Star is confused by the contradiction. Choose the narrow way.
Before you blast me for being off: Consider that both guards, when asked, will tell you the wrong door. Just pick the one they don't tell you and ask either one. The one they say is wrong is right. Not a contradiction. The esoteric is a liar. He will say his door is correct. The church will tell you the spirit. They are correct. The spirit is found with Christ and the Church, but not with the Pope or Priest. That's where the Freemasons (Builders who rejected the cornerstone) made their error. The church is Christ. He is the High Priest. This is also where the church made the error on the extreme of exoteric.
Matthew 16:18
18And I say also unto thee, That thou art Peter (Petros), and upon this rock (Petra) I will build my church; and the gates of hell shall not prevail against it.
Petros is the small stone that is rolled away. Petra is the foundation. Christ is where the church is built. Not Peter and not the Pope.
Riddle Solved - In the end, neither one is correct because they both hold a a part. You need both to find the point. You need neither Christ. He is at the crossroads waiting. Christ carries you the rest of the way through faith and repentance. The roads are not needed after you find Christ. This is the excluded middle that you never considered, but was always there.
As above, so below.
A train engineer and a friend argue over the train whistle changing pitch. Which is right? Both. The Doppler Effect is the excluded middle is the answer to both being right and wrong at the same time.
Intellect over emotion. Love over hate. Good over Evil. Choose Christ.
Luke 17:33
Whoever tries to keep his life will lose it, and whoever loses his life will preserve it.
Seeking to avoid death is duplicity. In this case, the true desire would be the preservation of self. Seeking paradise is also self seeking. Christ is the correct desire. Good is its own reward.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 4 * 1/5 = 4/5 = 0.8
Spelled result in words is four fifths.
### How do you solve fractions step by step?
1. Multiple: 4 * 1/5 = 4 · 1/1 · 5 = 4/5
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(4, 5) = 1. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - four multiplied by one fifth = four fifths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Denominator 2
Denominator of a fraction is 5 and numerator is 7. Write the fraction .
• Chocolate to sweets
I have 30 basket sweets and 15 chocolate in a basket. If take 5 sweet out of the basket what is the ratio in simplest form of the chocolate to the numbers of sweet left in the basket?
• Double division
0.25 divided by 1/2 divided by 14
• Which 3
Which is an equivalent ratio for 12:36
• Mass fraction
2 teaspoons (24 g) of sugar are dissolved in a cup with 500 g of tea. Calculate the weight fraction of sugar in the tea.
• Dinner tickets
At the spring festival, 7 out of the first 25 attendees had pre-purchased their dinner tickets. Based on this information, if 600 people attended the spring festival, then how many attendees could be expected to have pre-purchased their dinner tickets?
• Forest nursery
In Forest nursery plant one pine to 1.9 m2. Calculate how many plants are planting in area 362 acres.
• Tractors
6 tractors started plowing the field that they together take 12 days. Agronomist hesitates to A) after 2 days to withdraw 2 tractors, or B) after 3 days to withdraw 3 tractors. Help him which of these two cases will be plowing done sooner.
• Equivalent expressions
A coach took his team out for pizza after their last game. There were 14 players, so they had to sit in smaller groups at different tables. Six players sat at one table and got 4 small pizzas to share equally. The other players sat at the different table
• Strawberries
Father collects strawberries himself in 4 hours, son in 14 hours. How long will it take them along when dad comes to help his son collect strawberries after 3 hours?
• Players - baseball
There are 20 players on each of two baseball teams. If 2/5 of the players on team 1 miss practice and 1/4 of the players on team two miss practice, how many more players from team 1 missed practice then team 2?
• A serving
A serving of rice is 1/2 of a cup. How many servings are there in a 3 1/2-cup bag of rice?
• Penny coins
One 2p coin is 1/8 inch thick. A pile of 2p is 1 1/2 inch high. How many coin are there in the pile
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# Math portfolio | Mathematics homework help
ntroduction: Have you ever ridden or seen a roller coaster in action? Did you know that the algebra that you have learned in this unit is related to the math that engineers use to design roller coasters? Engineers want roller coasters to be fun and scary, but also safe. Directions: For this portfolio, you will use your knowledge of functions to design a roller coaster. You will draw a short roller coaster on graph paper, plot ordered pairs on its path, and determine the slope, or rate of change, along the ride. Roller Coaster Design Instructions Submit responses to the following items, and be sure to show all of your work. 1. Draw the side view of your roller coaster on graph paper. For the sake of simplicity, assume that your roller coaster track never turns left or right. o In order to gain speed, the roller coaster should have an initial climb, at least two hills, and one loop. 2. Label the x- and y-axes. The value of x measures the horizontal distance from the roller coaster cart to the starting point, and the value of y represents the height of the roller coaster cart. 3. Plot ordered pairs on the initial climb and determine the slope. 4. What is the equation of the line that represents your initial climb? 5. What is the domain and range of your roller coaster? 6. Plot ordered pairs at the top and the end of each hill. Find the rate of change to determine which hill is steeper? How do you know that hill is steeper? 7. Is your roller coaster a function? Why or why not?
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https://buildingspeed.org/blog/2012/08/04/pocono-infographics/
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# Pocono: Infographics
If you mouse over the triangle in the upper right-hand side of the Pocono Raceway website, you can see the track dimensions. Those numbers give you a pretty good idea why this track drives crew chiefs crazy.
## Facts and Figures
First – how tight are the turns? The larger the number, the easier it is to make the turn. Second: how much banking is there? The less banking, the harder it is to make a fast turn. The third graph shows you how much force – in pounds – is necessary for the tires to generate when taking each of the three corners.
The factors determining how much turning force you need to take a corner without hitting the wall depends on: the speed, the turn radius, the coefficient of friction, and the banking.
• The turning force is proportional to the square of the speed. If you want to go twice as fast, you need four times as much force.
• The turning force is inversely proportional to the turn radius. The tighter the turn, the more force you need.
• Banking helps you turn. The turning force that has to be generated by the tires is less when there is more banking. (For the sticklers: yes: when the car is on the banking, less of the car’s weight acts perpendicular to the surface, so there isn’t as much of the car’s weight pushing the tires into the track as there would be on a flat track. Even so, having the banking compensates for the reduced normal force.)
If you compare the lengths of the straightaways preceding the turns, the setup makes perfect sense: Although Turn 1 is the tightest turn, it also has the highest banking. You’re coming down a long straightaway, so the steep banking helps you make the turn without slowing down too much.
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http://algebra-test.com/algebra-help/powers/similarity-from-radical.html
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similarity from radical expressions to polynomials
Related topics:
| | simplify radicals calculator | rewrite with positive exponents | radicals explained | 8th grade math taks formula chart | printable number lines | calculator for positive and negative screenings algebra calculator | math variables worksheet | | | | | advanced 10th grade math worksheets
Author Message
Dylen Dartem
Registered: 03.08.2003
From: San Diego, Ca
Posted: Monday 25th of Dec 08:28 I am going to college now. As math has always been my problem area , I purchased the course material in advance. I am plan on learning a handful of topics before the classes begin . Any kind of tools would be highly appreciated that could aid me to start studying similarity from radical expressions to polynomials myself.
Vofj Timidrov
Registered: 06.07.2001
From: Bulgaria
Posted: Tuesday 26th of Dec 09:09 Sounds like your concepts are not clear . Mastering in similarity from radical expressions to polynomials requires that your concepts be concrete. I know students who actually start tutoring juniors in their first year. Why don’t you try Algebrator? I am quite sure, this program will aid you.
Bet
Registered: 13.10.2001
From: kµlt øƒ Ø™
Posted: Wednesday 27th of Dec 19:10 Algebrator indeed is a very good software to help you learn math, sitting at home . You won’t just get the problem solved but the entire solution as well, that’s how concepts are built . And to do well in math, it’s important to have strong concepts. I would highly recommend using this software if you want to finish your assignment on time.
Voumdaim of Obpnis
Registered: 11.06.2004
From: SF Bay Area, CA, USA
Posted: Thursday 28th of Dec 16:14 I remember having problems with adding functions, factoring expressions and leading coefficient. Algebrator is a truly great piece of algebra software. I have used it through several algebra classes - Basic Math, College Algebra and Basic Math. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly recommended.
TCE Hillin
Registered: 24.10.2002
From: UK
Posted: Friday 29th of Dec 08:18 This is most fascinating . Can you suggest from where can I procure the program?
Paubaume
Registered: 18.04.2004
From: In the stars... where you left me, and where I will wait for you... always...
Posted: Saturday 30th of Dec 18:13 Just click here for details : http://www.algebra-test.com/comparison.html. They also claim to provide an unconditional money back guarantee, so you have nothing to lose. Try this and Good Luck!
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Effects of Hematocrit on Blood Flow Through A Stenosed Human Carotid Artery
Authors
• Sefiu Onitilo Department of Mathematical Sciences, Faculty of Science, Olabisi Onabanjo University, Ogun State, Nigeria
• Mustapha Usman Department of Mathematical Sciences, Faculty of Science, Olabisi Onabanjo University, Ogun State, Nigeria
• Deborah Daniel Department of Mathematics and Computer Science, Faculty of Pure & Applied Sciences, Southwestern University, Ogun State, Nigeria
Keywords:
stenosis, carotid artery, wall shear stress, hematocrit, blood viscosity
Abstract
In this paper, the effects of hematocrit of red blood cells on blood flow through a stenosed human carotid artery was considered by taking blood as a Newtonian fluid. The governing equations on blood flow were derived. The mathematical content involved in the equations are the variables of interest such as number of stenosis , percentage of hematocrit of red blood cells in the blood, flow rate, wall shear stress, and viscosity of the blood. Guided by medical data collected on the constraint of blood flow in stenosed human carotid arteries, the governing equations were used to check the effects of pressure gradient, wall shear stress, velocity, and volumetric flow rate of blood in the human carotid arteries. Also, the one-dimensional equation for the steady and axially symmetric flow of blood through an artery was transformed using Einstein’s coefficient of viscosity and hematocrit of red blood cells with the help of the boundary conditions. The effects of hematocrit on the blood flow characteristics are shown graphically and discussed briefly. It was discovered that the resistance increases as the level of hematocrit increases. Also, the wall shear stress decreases with the increase in the hematocrit level of the red blood cells.
2020-08-28
Mathematics
How to Cite
Effects of Hematocrit on Blood Flow Through A Stenosed Human Carotid Artery. (2020). Iraqi Journal of Science, 61(8), 2106-2114. https://doi.org/10.24996/ijs.2020.61.8.25
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A265185 Non-vanishing traces of the powers of the adjacency matrix for the simple Lie algebra B_4: 2 * ((2 + sqrt(2))^n + (2 - sqrt(2))^n). 5
4, 8, 24, 80, 272, 928, 3168, 10816, 36928, 126080, 430464, 1469696, 5017856, 17132032, 58492416, 199705600, 681837568, 2327939072, 7948081152, 27136446464, 92649623552, 316325601280, 1080003158016, 3687361429504, 12589439401984, 42983034748928 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS a(n) is the trace of the 2n-th power of the adjacency matrix M for the simple Lie algebra B_4, given in the Damianou link. M = Matrix[row 1; row 2; row 3; row 4] = Matrix[0,1,0,0; 1,0,1,0; 0,1,0,2; 0,0,1,0]. Equivalently, the trace tr(M^(2k)) is the sum of the 2n-th powers of the eigenvalues of M. The eigenvalues are the zeros of the characteristic polynomial of M, which is det(xI - M) = x^4 - 4x^2 + 2 = A127672(4,x), and are (+-) sqrt((2 + sqrt(2)) and (+-) sqrt((2 - sqrt(2)), or the four unique values generated by 2*cos((2n+1)Pi/8). Compare with A025192 for B_3. The odd power traces vanish. -log(1 - 4 x^2 + 2 x^4) = 8 x^2/2 + 24 x^4/4 + 80 x^6/6 + ... = Sum_{n>0} tr(M^k) x^k / k = Sum_{n>0} a(n) x^(2k) / 2k gives an aerated version of the sequence a(n), excluding a(0), and exp(-log(1 - 4 x + 2 x^2)) = 1 / (1 - 4 x + 2 x^2) is the e.g.f. for A007070. As in A025192, the cycle index partition polynomials P_k(x[1],...,x[k]) of A036039 evaluated with the negated power sums, the aerated a(n), are P_2(0,-a(1)) = P_2(0,-8) = -8, P_4(0,-a(1),0,-a(2)) = P_4(0,-8,0,-24) = 48, and all other P_k(0,-a(1),0,-a(2),0,...) = 0 since 1 - 4 x^2 + 2 x^4 = 1 - 8 x^2/2! + 48 x^4/4! = det(I - x M) = exp(-Sum_{k>0} tr(M^k) x^k / k) = exp[P.(-tr(M),-tr(M^2),...)x] = exp[P.(0,-a(1),0,-a(2),...)x]. Because of the inverse relation between the Faber polynomials F_n(b1,b2,...,bn) of A263916 and the cycle index polynomials, F_n(0,-4,0,2,0,0,0,...) = tr(M^n) gives aerated a(n), excluding a(0). E.g., F_2(0,-4) = -2 * -4 = 8, F_4(0,-4,0,2) = -4 * 2 + 2 * (-4)^2 = 24, and F_6(0,-4,0,2,0,0) = -2(-4)^3 + 6(-4)2 = 80. LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2011-2014. FORMULA a(n) = 2 * ((2 + sqrt(2))^n + (2 - sqrt(2))^n) = Sum_{k=0..3} 2^(2n) (cos((2k+1)Pi/8))^(2n) = 2*2^(2n) (cos(Pi/8)^(2n) + cos(3Pi/8)^(2n)) = 2 Sum_{k=0..1} (exp(i(2k+1)Pi/8) + exp(-i(2k+1)Pi/8))^(2n). E.g.f.: 2 * e^(2x) * (e^(sqrt(2)*x) + e^(-sqrt(2)*x)) = 4 e^(2x) cosh(sqrt(2)*x) = 2 ( exp(4x cos(Pi/8)^2) + exp(4x cos(3 Pi/8)^2) ). a(n) = 4*A006012(n) = 8*A007052(n-1) = 2*A056236(n). G.f.: (4-8*x)/(1-4*x+2*x^2). - Robert Israel, Dec 07 2015 Note the preceding o.g.f. is four times that of A006012 and the denominator is y^4 * A127672(4,1/y) with y = sqrt(x). Compare this with those of A025192 and A189315. - Tom Copeland, Dec 08 2015 MATHEMATICA 4 LinearRecurrence[{4, -2}, {1, 2}, 30] (* Vincenzo Librandi, Dec 06 2015 and slightly modified by Robert G. Wilson v, Feb 13 2018 *) PROG (MAGMA) [Floor(2 * ((2 + Sqrt(2))^n + (2 - Sqrt(2))^n)): n in [0..30]]; // Vincenzo Librandi, Dec 06 2015 (PARI) x='x+O('x^30); Vec((4-8*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Feb 12 2018 CROSSREFS Cf. A006012, A007052, A007070, A025192, A036039, A056236, A127672, A189315, A263916. Sequence in context: A010366 A222356 A214201 * A240530 A303882 A008950 Adjacent sequences: A265182 A265183 A265184 * A265186 A265187 A265188 KEYWORD nonn,easy AUTHOR Tom Copeland, Dec 04 2015 EXTENSIONS More terms from Vincenzo Librandi, Dec 06 2015 STATUS approved
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Last modified August 24 22:49 EDT 2019. Contains 326314 sequences. (Running on oeis4.)
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## Difficulty level
Lecture title:
The simulation of the virtual epileptic patient is presented as an example of advanced brain simulation as a translational approach to deliver improved results in clinics. The fundamentals of epilepsy are explained. On this basis, the concept of epilepsy simulation is developed. By using an iPython notebook, the detailed process of this approach is explained step by step. In the end, you are able to perform simple epilepsy simulations your own.
Difficulty level: Beginner
Duration: 1:28:53
Speaker: : Julie Courtiol
Lecture title:
Explore how to setup an epileptic seizure simulation with the TVB graphical user interface. This lesson will show you how to program the epileptor model in the brain network to simulate a epileptic seizure originating in the hippocampus. It will also show how to upload and view mouse connectivity data, as well as give a short introduction to the python script interface of TVB.
Difficulty level: Intermediate
Duration: 58:06
Speaker: : Paul Triebkorn
Lecture title:
Learn how to simulate seizure events and epilepsy in The Virtual Brain. We will look at the paper: On the Nature of Seizure Dynamics which describes a new local model called the Epileptor, and apply this same model in The Virtual Brain. This is part 1 of 2 in a series explaining how to use the Epileptor. In this part, we focus on setting up the parameters.
Difficulty level: Beginner
Duration: 4:44
Speaker: : Paul Triebkorn
Lecture title:
Introduction to the Mathematics chapter of Datalabcc's "Foundations in Data Science" series.
Difficulty level: Beginner
Duration: 2:53
Speaker: : Barton Poulson
Lecture title:
Primer on elementary algebra
Difficulty level: Beginner
Duration: 3:03
Speaker: : Barton Poulson
Lecture title:
Primer on linear algebra
Difficulty level: Beginner
Duration: 5:38
Speaker: : Barton Poulson
Lecture title:
Primer on systems of linear equations
Difficulty level: Beginner
Duration: 5:24
Speaker: : Barton Poulson
Lecture title:
Primer on calculus
Difficulty level: Beginner
Duration: 4:17
Speaker: : Barton Poulson
Lecture title:
How calculus relates to optimization
Difficulty level: Beginner
Duration: 8:43
Speaker: : Barton Poulson
Lecture title:
Big O notation
Difficulty level: Beginner
Duration: 5:19
Speaker: : Barton Poulson
Lecture title:
Basics of probability.
Difficulty level: Beginner
Duration: 7:33
Speaker: : Barton Poulson
Lecture title:
This lecture covers computational principles that growth cones employ to detect and respond to environmental chemotactic gradients, focusing particularly on growth cone shape dynamics.
Difficulty level: Intermediate
Duration: 26:12
Speaker: : Geoff Goodhill
In this lecture you will learn that in developing mouse somatosensory cortex, endogenous Btbd3 translocate to the cell nucleus in response to neuronal activity and oriented primary dendrites toward active axons in the barrel hollow.
Difficulty level: Intermediate
Duration: 27:32
Speaker: : Tomomi Shimogori
In this presentation, the speaker describes some of their recent efforts to characterize the transcriptome of the developing human brain, and and introduction to the BrainSpan project.
Difficulty level: Intermediate
Duration: 30:45
Lecture title:
How does the brain learn? This lecture discusses the roles of development and adult plasticity in shaping functional connectivity.
Difficulty level: Beginner
Duration: 1:08:45
Speaker: : Clay Reid
Introduction to neurons, synaptic transmission, and ion channels.
Difficulty level: Beginner
Duration: 46:07
2nd part of the lecture. Introduction to cell receptors and signalling cascades
Difficulty level: Beginner
Duration: 41:38
Lecture title:
Introduction to the types of glial cells, homeostasis (influence of cerebral blood flow and influence on neurons), insulation and protection of axons (myelin sheath; nodes of Ranvier), microglia and reactions of the CNS to injury.
Difficulty level: Beginner
Duration: 40:32
Introduction to the origin and differentiation of myelinating cell types, molecular mechanisms defining onset and progression of myelination, demyelination and remyelination after injury.
Difficulty level: Beginner
Duration: 38:52
Lecture title:
This lecture covers: integrating information within a network, modulating and controlling networks, functions and dysfunctions of hippocampal networks, and the integrative network controlling sleep and arousal.
Difficulty level: Beginner
Duration: 47:05
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## Is meter Spelt meter or Metre?
“Metre” is the British spelling of the unit of length equal to 100 cm, and “meter” is the American spelling of the same unit. However, “meter” is also used in British English, but it means something different. A “meter” in British English is an instrument for measuring.
## Why is meter spelled two different ways?
By 1866, when the USA decided to make the metric system legal for internal and international trade, spelling of metre as meter had become so common in the USA that the government had to provide for the two different spellings in all metric laws and regulations.
## How do you spell Metre in Australia?
In UK/Australia/New Zealand English it is metre. These dialects distinguish by spelling between the unit of length (metre) and a measuring instrument (meter). In US English it is meter.
## What’s the difference between center and Centre?
Center and centre have the same meaning. Center is the correct spelling in American English, but British English writers usually prefer centre. Notice that center (and centre) can be a noun, adjective, or a verb. Seeing the two words in real-life examples may help you to visualize how to use them.
100 centimeters
## Does 1m have 100 cm?
Each meter (m) is divided into 100 equal divisions, called centimetre (cm) ie; 1m=100cm. Hence, 1m=100cm .
## Is 1m 100cm?
Convert meter to cm, centimeters to meter (1m = 100cm)
1.71 centimeter
## What is 35g in mL?
Convert 35 grams to ml 67 ml.
## What is bigger mL or CM?
Centimeters cannot be compared to milliliters. A centimeter is a measure of length, while a milliliter is a measure of volume. However, one cubic centimeter is equal to one milliliter.
## How many cm is 50ml?
Convert 50 Milliliters to Cubic Centimeters
mL cc
50.00 50
50.01 50.01
50.02 50.02
50.03 50.03
0.5 meters
## How many cm is a 30ml bottle?
30 ml: nozzle: Total height: 10.8 cm, diameter: 3.3 cm, printed size: length 3 cm height: 4.5 cm.
## Does 1 cm cubed equal 1ml?
A cubic centimetre (or cubic centimeter in US English) (SI unit symbol: cm3; non-SI abbreviations: cc and ccm) is a commonly used unit of volume that corresponds to the volume of a cube that measures 1 cm x 1 cm × 1 cm. One cubic centimetre corresponds to a volume of one millilitre.
## Is 1 cm the same as 1 ml?
These are the same measurement; there is no difference in volume. The primary difference is that milliliters are used for fluid amounts while cubic centimeters are used for solids. No matter what is being measured, 1 cc always equals 1 mL.
## Is 1ML the same as 1g?
If you’re talking about pure water, 1ml is equal to 1g.
## What does 1 ml look like?
In other words 1 milliliter is exactly the same as a little cube that is 1 cm on each side (1 cubic centimeter). If it was uniformly full to exactly 1cm high it would contain 8 of those 1cm × 1cm × 1cm cubes, making 8 cc. But because of its shape it only holds about 5 cc (or 5 ml).
## How can I measure 1 ml at home?
How to Convert Metric Measurements to U.S. Measurements
1. 0.5 ml = ⅛ teaspoon.
2. 1 ml = ¼ teaspoon.
3. 2 ml = ½ teaspoon.
4. 5 ml = 1 teaspoon.
5. 15 ml = 1 tablespoon.
6. 25 ml = 2 tablespoons.
7. 50 ml = 2 fluid ounces = ¼ cup.
8. 75 ml = 3 fluid ounces = ⅓ cup.
## What is 1 ml on a syringe?
In other words, one milliliter (1 ml) is equal to one cubic centimeter (1 cc). This is a three-tenths milliliter syringe. It may be called a “0.3 ml” syringe or “0.3 cc” syringe. It is also known as an insulin syringe.
## How much is 1 ml on a dropper?
So according to the dropper measurements, it is a 1/4ml point on a dropper. Full dropper is 1ml = 7mg of CBD per 200mg 30ml size bottle.
1,000 milligrams
## How much is 5 ml in a dropper?
(1 teaspoon = 5 milliliters.) Find the measurement that matches the dose you need. 3. Squeeze the bulb end and put the tip of the dropper into the medicine bottle.
600 drops
## What is 30ml in MG?
Convert 30 Milliliters to Milligram
30 Milliliters (ml) 30,000 Milligram (mg)
1 ml = 1,000 mg 1 mg = 0.001000 ml
## How many ml is 25 mg?
Convert 25 Milligram to Milliliters
25 Milligram (mg) 0.025000 Milliliters (ml)
1 mg = 0.001000 ml 1 ml = 1,000 mg
## How many ml is 10 drops of essential oil?
5 ml – 100 drops. 7.5 ml – 150 drops. 10 ml – 200 drops. 15 ml – 300 drops.
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# In mathematics, a function defines the relationship between 2 quantities
It is a thing that may be utilised to determine and save data. I will go over some definitions of the function which you could encounter on your work.
Even the job is known as the id function. That is simply called the event f(x) = a. A functionality explained inside this way has a single value that you can employ https://bestresearchpaper.com/ to some factor. However, this definition doesn’t always reflect the full structure of the function. It can not actually show what the role is rather how it can be properly used.
In the event you adjust the value of x ray and x f(x ray ), then it is going to equal a. This could be the first step on how a function can be properly utilized. The second step could be your derivative. Additionally, it may be utilised in order to find the derivative. The third step is to use the quadratic formulation. It’s a system.
Defining a function is not the only approach to make ideas. You are able to modify the worth of help with essay the function that you can predict something. By way of instance, the derivatives of x and x f(x) is used to determine some time show statistics for X.
It can be utilised to gauge the rate of modification of a time series data. This is sometimes utilised in order to draw conclusions.
Statistical functions really are a bit unique. In statistics, you only have an individual price of x. For instance, the Taylor series is really a well-known means to compute some variables. Various other examples include line plots and bubble graphs.
You’ll find a number of ways to construct mathematical designs. A version describes a connection among a pair of worth.
Works that are elaborate have some fine properties. In case there were a function that is elaborate awarded, then you definitely can get a graph. It might seem something like a line that you saw when studying the ground. It’d connect some things.
You will find many different types including the variety that is intricate. Generally in the majority https://www.independence.edu/catalog/school-catalog-iu.pdf of cases, you will need to know the sort of a job, what it does and how to utilize it to compute some details that is handy.
You may specify a role with a complicated kind by choosing a equation that includes the values for x and x f(x ray ). You may make use of the formula to figure out the derivative of this work Whenever you have a function having a type that is complex.
Definition of a function is very important in mathematics. It is utilised to generate roles which produce it possible to calculate some information.
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let n be Nat; :: thesis: for P being Subset of ()
for p1, p2 being Point of () st P is_an_arc_of p1,p2 holds
p1 <> p2
let P be Subset of (); :: thesis: for p1, p2 being Point of () st P is_an_arc_of p1,p2 holds
p1 <> p2
let p1, p2 be Point of (); :: thesis: ( P is_an_arc_of p1,p2 implies p1 <> p2 )
assume P is_an_arc_of p1,p2 ; :: thesis: p1 <> p2
then consider f being Function of I[01],(() | P) such that
A1: f is being_homeomorphism and
A2: f . 0 = p1 and
A3: f . 1 = p2 by TOPREAL1:def 1;
1 in [#] I[01] by ;
then A4: 1 in dom f by ;
A5: f is one-to-one by ;
0 in [#] I[01] by ;
then 0 in dom f by ;
hence p1 <> p2 by ; :: thesis: verum
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YOU START WITH 1000 RIDDLE
Popular Searches
Feel free to use content on this page for your website or blog, we only ask that you reference content back to us. Use the following code to link this page:
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Riddles and Answers © 2019
You Start With 1000 Riddle
You start with 1000 then add 40 add another 1000 then add 30 add another 1000 then add 20 add one more 1000 then add 10 what is your answer?
Hint:
If you said 5000 you're wrong! The answer is 4100, how?
1000 + 40 = 1040
1040 + 1000 + 30 = 2070
2070 + 1000 + 20 = 3090
3090 + 1000 = 4090
4090 + 10 = 4100
Did you answer this riddle correctly?
YES NO
Solved: 64%
Add Your Riddle Here
Have some tricky riddles of your own? Leave them below for our users to try and solve.
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# How to Decimate an Array.
An interesting use case arrived yesterday, I loved the simplicity of the code so I thought I'd share it.
# Code: boom 💣
export function decimateArray(arr, passes = 1, fidelity = 2) {
let tmpArr = arr.filter((_, index) => index % fidelity === 0);
passes--;
if (passes) {
tmpArr = decimateArray(tmpArr, passes, fidelity);
}
return tmpArr;
}
## Use case:
"I have a large array of xy coordinates, I need it to draw freehand on canvas but it's too much data to work with quickly, I just want to draw an approximate polygon. I want the array to be smaller so I can more efficiently loop through the data and I dont care about all of the data just the start and end."
## How?
An array is fed in, if the index of the data is modulus of a passed in fidelity then keep this data, also recursively run this dataset through itself by a given number of passes.
Large array goes in, smaller array with less detail comes out.
Posted on by:
I work at ForgeRock as Staff UI Engineer, I play with all sorts really. Lately WASM is my toy of interest.
### Discussion
I would suggest 4 improvements (3 minor and 1 critical)
# Critical bug
1) Provide the fidelity parameter as argument to the function, otherwise will only have a value other than 2 in the first step of the recursion;
# Minor suggestions
1) Reduce LOC by decreasing by one in the if statement;
2) Remove obsolete else branch;
3) No need to reassign the variable, just return the result right away;
function decimateArray(arr, passes = 1, fidelity = 2) {
const filteredArray = arr.filter((_, index) => index % fidelity === 0);
if (--passes) {
return decimateArray(filteredArray, passes, fidelity);
}
return filteredArray;
}
const parameters = [
Array.from({ length: 100 }, (_, i) => i + 1),
2,
4,
];
const decimatedArray = decimateArray(...parameters);
console.log({ decimatedArray }); // { decimatedArray: [ 1, 17, 33, 49, 65, 81, 97 ] }
BTW sorry if I sound picky...
Array from with a function callback isn't as fast as mapping an empty array from an Array constructor. 🤣
It's not the point that it's buggy, it's important that you can take my crappy code and make something better, success. Still the failure to pass args recursively is something il fix.
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6.2 Exponential Functions
# 6.2 Exponential Functions - Exponential Function Let u be a...
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Exponential Functions Section 6.2
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Defn. of the Natural Exponential Function The inverse of the natural logarithmic function f(x) = ln x is called the natural exponential function : f –1 (x) = ex y = ex iff x = ln y
f –1(x) = ex f(x) = ln x
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l ln( ex ) = x l eln x = x l ln e = 1
Operations with Exponential Functions
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Definition of the Number e
Properties of the Natural Exponential Function
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The Derivative of the Natural
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Unformatted text preview: Exponential Function Let u be a differentiable function of x. 1. 2. [ ] x x e e dx d = [ ] dx du e e dx d u u = Integration Rules for Exponential Functions Let u be a differentiable function of x. 1. 2. c e dx e x x + = ∫ c e du e u u + = ∫ Joke Time How can you stop a skunk from smelling? Hold its nose Why do golfers wear 2 pairs of pants? In case they get a hole in one...
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Open In App
# Python | Equal Keys List Summation
Sometimes, while working with dictionaries, we can have a problem in which we have many dictionaries and we are required to sum like keys. This problem seems common, but complex is if the values of keys are list and we need to add elements to list of like keys. Let’s discuss way in which this problem can be solved.
Method : Using list comprehension + items() + sum() This problem can be solved using list comprehension and sum() that can be used to sum the list content and also the items method which can be employed to get the dictionary keys and values.
## Python3
`# Python3 code to demonstrate working of``# Equal Keys List Summation``# Using items() + list comprehension + sum()` `# initializing dictionaries``test_dict1 ``=` `{``'Gfg'` `: [``1``, ``2``, ``3``], ``'for'` `: [``2``, ``4``], ``'CS'` `: [``7``, ``8``]}``test_dict2 ``=` `{``'Gfg'` `: [``10``, ``11``], ``'for'` `: [``5``], ``'CS'` `: [``0``, ``18``]}` `# printing original dictionaries``print``("The original dictionary ``1` `is` `: " ``+` `str``(test_dict1))``print``("The original dictionary ``2` `is` `: " ``+` `str``(test_dict2))` `# Using items() + list comprehension + sum()``# Equal Keys List Summation``res ``=` `{key: ``sum``(value) ``+` `sum``(test_dict2[key]) ``for` `key, value ``in` `test_dict1.items()}` `# printing result``print``("The summation of dictionary values ``is` `: " ``+` `str``(res))`
Output :
```The original dictionary 1 is : {'CS': [7, 8], 'for': [2, 4], 'Gfg': [1, 2, 3]}
The original dictionary 2 is : {'CS': [0, 18], 'for': [5], 'Gfg': [10, 11]}
The summation of dictionary values is : {'CS': 33, 'for': 11, 'Gfg': 27}```
Time Complexity: O(n*n), where n is the length of the list test_dict
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
#### Method#2 : Using for loop
In this method, we first loop through the keys of the first dictionary, and for each key, we check if it exists in the second dictionary. If it does, we sum the values of the two lists using the + operator and store the result in the result_dict. If the key is not present in the second dictionary, we simply sum the values of the list from the first dictionary and store the result in the result_dict. Finally, we loop through the keys of the second dictionary, and for each key that is not already present in the result_dict, we sum the values of the list from the second dictionary and store the result in the result_dict
## Python3
`dict1 ``=` `{``'CS'``: [``7``, ``8``], ``'for'``: [``2``, ``4``], ``'Gfg'``: [``1``, ``2``, ``3``]}``#dictionary input 1``dict2 ``=` `{``'CS'``: [``0``, ``18``], ``'for'``: [``5``], ``'Gfg'``: [``10``, ``11``]}``#dictionary input 2` `result_dict ``=` `{}``#take empty dictionary` `for` `key ``in` `dict1.keys():``#using loop to get keys from dict1`` ``if` `key ``in` `dict2:`` ``result_dict[key] ``=` `sum``(dict1[key] ``+` `dict2[key])``#summation of keys`` ``else``:`` ``result_dict[key] ``=` `sum``(dict1[key])` `for` `key ``in` `dict2.keys():``#using loop to get keys from dict2`` ``if` `key ``not` `in` `result_dict:``#summation of keys`` ``result_dict[key] ``=` `sum``(dict2[key])` `print``(``"The summation of dictionary values is :"``, result_dict)``#print result`
Output
`The summation of dictionary values is : {'CS': 33, 'for': 11, 'Gfg': 27}`
Time complexity: O(n)
Auxiliary Space: O(n)
Method#3 : Using itertools.chain() and itertools.groupby():
Algorithm:
1. Import the necessary modules, itertools.
2.Define two dictionaries.
3.Create an empty dictionary.
4.Merge the keys of both the dictionaries using chain() function and group them using groupby() function.
5.Find the sum of values of common keys in both the dictionaries, using get() function.
6.Store the result in a new dictionary.
7.Print the new dictionary.
## Python3
`from` `itertools ``import` `chain, groupby` `dict1 ``=` `{``'CS'``: [``7``, ``8``], ``'for'``: [``2``, ``4``], ``'Gfg'``: [``1``, ``2``, ``3``]}``dict2 ``=` `{``'CS'``: [``0``, ``18``], ``'for'``: [``5``], ``'Gfg'``: [``10``, ``11``]}``# printing original dictionaries``print``(``"The original dictionary 1 is : "` `+` `str``(dict1))``print``(``"The original dictionary 2 is : "` `+` `str``(dict2))`` ` `merged_dict ``=` `{}``for` `key, group ``in` `groupby(``sorted``(chain(dict1.keys(), dict2.keys()))):`` ``merged_dict[key] ``=` `sum``(chain(dict1.get(key, []), dict2.get(key, [])))` `print``(``"The summation of dictionary values is :"``, merged_dict)``#This code is contributed by Jyothi pinjala.`
Output
```The original dictionary 1 is : {'CS': [7, 8], 'for': [2, 4], 'Gfg': [1, 2, 3]}
The original dictionary 2 is : {'CS': [0, 18], 'for': [5], 'Gfg': [10, 11]}
The summation of dictionary values is : {'CS': 33, 'Gfg': 27, 'for': 11}```
Time Complexity: O(NlogN)
The use of sorted() function makes the time complexity O(NlogN).
The iteration over the merged keys has a time complexity of O(N), as does the iteration over the values of the dictionaries.
The sum() function has a time complexity of O(N).
Space Complexity: O(N)
We are using extra space for the merged dictionary which has the same number of elements as the input dictionaries. Therefore, the space complexity is O(N).
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# One And Two Step Equations Worksheet Pdf
Evolution worksheet middle school. Use inverse operations to isolate the x term.
Free One Step Equations Activity Solving Equations In Algebra 1 One Step Equations Equations Algebra
### That additional step may be something like multiplying the variable by a certain number to get rid of a fraction in front of it.
One and two step equations worksheet pdf. X c21 7 x c15 x c3 x c2 x 6 c15 x 21 7x c21 x 4 7 x 11 27 9x x c6 3x 7 17 x c8 c3x 5 10 x 6 6 4x c 10 x 6 x 18. Click the following links to download one step equations worksheets as pdf documents. Solving one sep equations.
We just have to perform one step in order to solve the equation. For example using the equation 3x 5 11 we will need to perform two steps to find the value of x. K o ya clnlw r icgih7t csf nrse 6s 9e qrvtevdu h d xmcahdrea kw ni mtjhq ivn tf qixnoiwtjeu ip1roeg pa1lxgje xb wrxa b 1 worksheet by kuta software llc kuta software infinite pre algebra name two step equations with integers date period.
6th grade proportions worksheet. Age word problems worksheet pdf. You can customize the worksheets to include one step two step or multi step equations variable on both sides parenthesis and more.
Find here an unlimited supply of printable worksheets for solving linear equations available as both pdf and html files. A one step equation is as straightforward as it sounds. Year 3 english worksheets.
In this case 5 and 11 are our constants. N w2b0 s1o2t 6k yu utya j hslozfatrwpadrdez clulhcs. Reverse pemdas order.
Solving one and two step equations worksheet pdf. The first step would be to get the constant values of the equation by themselves. 3x 1 7 locate the variable term.
The worksheets suit pre algebra and algebra 1 courses grades 6 9. Solving two step equations sol 7 14 example 1 solve 3x 1 7 check. 7th grade geometry problems.
Coloring pages for grade 2. 3x 1 7 3x 1 7 1 1. Two step equation worksheets have a huge collection of printable practice pages to solve and verify the equations involving integers fractions and decimals.
Two step math equations are algebraic problems that require you to make two moves to find the value of the unknown variable. The main difference between one step equationsand two step equations is that one more step you need to do in order to solve a two step equation. Also a number of exercise pdfs on translating two step equations mcqs and word problems based on geometric shapes are given here for additional practice for 7th grade and 8th grade students.
U w2r0g1z2 1 nknudthaw ssodfvtbw8aorle7 ul 3l ic u n p gasl glv 7rviog bh7t8sw ir 8ejs cewrrvke bdm y d tm ra ed se0 cw qiptxhl 1isnbf ti anci ytuev daolwgqembmrkas h1y 4 worksheet by kuta software llc kuta software infinite algebra 1 name two step equations date period solve each equation. Free printable matching worksheets. Coloring page girl 8yr old informal letter worksheets pdf en worksheets interrogative sentences worksheets with answers grade.
Otherwise the rules are the same as before and these equations are just as easy to learn and solve as are the one step ones.
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## Elementary Statistics (12th Edition)
$\alpha=1-0.95=0.05.$ $\sigma$ is not known, hence we use the t-distribution with $df=sample \ size-1=6-1=5$ in the table. $t_{\alpha/2}=t_{0.025}=2.571.$
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# Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of $10$ and the second has a voltage gain of 20. If the input signal is $0.01\;V$ calculate the output a.c signal
$\begin{array}{1 1} 2\;V \\ 4\;V \\ 6\;V \\ 1\;V \end{array}$
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# a02 - CS 135 Fall 2011 Becker Goldberg Kaplan Tompkins...
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Unformatted text preview: CS 135 Fall 2011 Becker, Goldberg, Kaplan, Tompkins, Vasiga Assignment: 2 Due: Tuesday, September 27, 2011 9:00 pm Language level: Beginning Student Files to submit: cond.rkt , grades.rkt , donate-cond.rkt , donate-bool.rkt Warmup exercises: HtDP 4.1.1, 4.1.2, 4.3.1, 4.3.2 Practice exercises: HtDP 4.4.1, 4.4.3, 5.1.4 Policies from Assignment 1 carry forward. For example, your solutions must be entirely your own work, and your solutions will be marked for both correctness and good style. Good style includes qualities such as descriptive names, clear and consistent indentation, appropriate use of helper functions, and documentation (design recipe). For this and all subsequent assignments you are expected to use the design recipe as discussed in class (except for question 1 on this assignment). You must use check-expect for both examples and tests. It is very important that the function names and parameters match ours. You must use the public tests to be sure. The names of the functions will be given exactly. The names of the parameters are up to you, but should be meaningful. The order and meaning of the parameters are carefully specified in each problem. Here are the assignment questions you need to submit. 1. A cond expression can always be rewritten to produce an equivalent expression . That is, the new expression always produces the same answer as the old expression (given the same inputs, of course). For example, the following are all equivalent: ( cond [( > x ) ’ Up ] [( < = x ) ’ Down ]) ( cond [( < = x ) ’ Down ] [( > x ) ’ Up ]) ( cond [( > x ) ’ Up ] [ else ’ Down ]) (There is one more really obvious equivalent expression; think about what it might be.) So far all of the cond examples we’ve seen in class have followed the pattern ( cond [ question1 answer1 ] [ question2 answer2 ] ... [ questionk answerk ]) where questionk might be else ....
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# How can I properly Partition "IntensityData" output from "ComponentMeasurements"? (Recovering the (x, y) dimensions of a rect. morph. component?)
When I ask ComponentMeasurements to return "IntensityData" for some set of rectangular morphological components:
m = MorphologicalComponents[image, Method -> "BoundingBox"];
intensityDataArray = ComponentMeasurements[{m, image}, "IntensityData"][[All, 2]];
The pixel value data is returned as an unpartitioned one-dimensional array. This makes it difficult for me to understand where a particular pixel is in the image. Now, if I know the exact dimensions of each morphological component, I suppose I could fix the problem using Partition. I could also fix the problem with Partition if I could ask for something like a bounding square instead of a bounding rectangular box, though I don't think this is possible (is it?).
Besides manually measuring the dimensions of each rectangular morphological component, is there a workaround to fix this problem?
Letting m consist of strictly rectangular morphological components, it would, for example, be very nice to be able to write something like:
ComponentMeasurements[m, "Dimensions"][[All, 2]]
And have this return something like:
{{5,8}, {87, 40}, {7, 7}, ...}
Meaning that we have morphological components of dimensions $(x, y) = (5 \times 8)$, $(87 \times 40)$, $(7 \times 7)$, and so forth. Here, we're provided ordered values for the extent of the component along the $x$- and $y$-axes, respectively, allowing us to trivially use Partition to restore the cropped section of an image underlying a morphological component (i.e. the image one would have if a morphological component is used as a "cropping mask").
Clarification: Masking approaches are really neat, but not quite what I'm looking for here unless they're part of a larger approach. I need a method of grabbing the pixel data strictly underlying a morphological component (for a set of morphological components) and retaining $(x, y)$ data for each pixel. My downstream procedure then does further image processesing on this rectangular matrix. If I do a matrix multiplication masking procedure, I still need to trim away the dark space to get down to a bounding box (the size of the morphological component mask).
Also, it seems a little wasteful to have to perform multiplication operation on all pixels in the image to accomplish my goal?
• I think "IntensityData" is meant to be used for histogram processing, where the position of the intensities doesn't matter. Couldn't you just measure the "BoundingBox" instead and pass the result to ImageTrim to cut out the right area of the image? You can measure "Mask" to get a sparse array where the pixels of the component are set to 1. (I'm not sure what you're trying to achieve, I that's what I use when I need both the intensity and the location of the pixels in a component.) Commented Nov 2, 2013 at 11:02
• @nikie How do I ask for the dimensions of a rectangular morphological component? If I can do this, I can just use Partition, right? I mean, I can figure this out up to rotation and reflection symmetries using metrics like Elongation, but how do I nail down the orientation as well? Commented Nov 2, 2013 at 11:10
• @nikie I'll also note that I have maybe a few thousand of these morphological components, so I'm kind of hoping for some automated solution. Commented Nov 2, 2013 at 11:16
• Use Partition on the returned IntensityData? I don't think so, because some lines might be longer than others. At least unless the components are all perfectly rectangular. Commented Nov 2, 2013 at 11:34
• @nikie I'm specifying that the morphological components are bounding boxes, and must be rectangular. I'll try to clarify this better in the question. Commented Nov 2, 2013 at 11:35
I'm answering the question you posted in a comment:
but how do I then crop to a bounding box around that component
List I said, the easiest way is probably to measure the "BoundingBox"es and pass them to ImageTrim:
image=ExampleData[{"TestImage","APC"}]
m=MorphologicalComponents[Binarize@ColorNegate[ColorConvert[image,"Grayscale"]]];
Colorize[m]
components=ComponentMeasurements[{m,image},{"Area","BoundingBox"},#1>100&];
ImageTrim[image,#]&/@components[[All,2,2]]
• I don't have the rep. to upvote you, but I follow you now. I didn't know how ImageTrim worked! Commented Nov 2, 2013 at 11:47
• @RVoight Please note that you still have to mask this! Otherwise you work on all the image values and not only on the ones from your component. Commented Nov 2, 2013 at 11:51
• @halirutan: He doesn't if his components are all rectangular (because of Method -> "BoundingBox"). Otherwise, you're right of course. I just wanted to show how BoundingBox and ImageTrim can work together and didn't want to complicate things with masking. Commented Nov 2, 2013 at 11:54
• @nikie Ahh, haven't notice this before. Commented Nov 2, 2013 at 12:06
Why don't you use the "Mask" for a component to separate it in the original image?
image = ExampleData[{"TestImage", "JellyBeans"}];
m = MorphologicalComponents[
Binarize@ColorNegate[ColorConvert[image, "Grayscale"]]];
In this way you can extract from the original image
the 3rd component
ImageTrim as suggested by nikie is better than using ImageTake because you can give it the bounding box directly.
mask = ComponentMeasurements[m, {"Mask", "BoundingBox"}];
• @RVoight You just use "BoundingBox" as additional measure and provide this to ImageTake. Be careful, because the ordering of the min/max box values is different and you have to incorporate the image dimensions in y direction. Commented Nov 2, 2013 at 11:45
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Level 3
# Foreign Tax Credit (Form 1116) on Sale of Foreign Rental Property - Gross Income & Definitely Related Expenses
I'm trying to figure out which figure should go in Line 1a (Gross Income) and which figure should go in Line 2 (Expenses Definitely Related To Income On Line 1a), in setting up Form 1116 for taking a Foreign Tax Credit against the Capital Gain on the sale of a foreign rental property.
Hypothetical example:
Sales Proceeds = 500,000
Original Purchase Price = 200,000
Closing Costs = 20,000
Accumulated Depreciation = 80,000
"Laymen's Gain" (ignoring accounting for depreciation recapture) = 280,000 (i.e. 500,000 less 220,000)
On Form 4797, this is allocated between land and building, the depreciation component only applying to the building. I'm presuming for Form 1116 the totals are all that matter.
So in Form 4797 language, we get:
"Gross Sales Price" = 500,000
"Cost Basis Plus Expense Of Sale" = 220,000 (i.e. 200,000 + 20,000)
"Depreciation Allowed" = 80,000
"Adjusted Basis" = 140,000 (i.e. 220,000 - 80,000)
"Total Gain" = 360,000 (i.e. 500,000 - 140,000)
Which amounts should go to Line 1a and 2 on Form 1116?
Option A: Line 1a = 360,000 and Line 2 = {empty} (i.e. gross income as "all-in" form 4797 total gain)
Option B: Line 1a = 380,000 and Line 2 = 20,000 (i.e. separating closing costs only as a def. related expense)
Or something else entirely where original cost and depreciation are factored into definitely related expenses?
I'm leaning towards Option A simply because no other part of the return seems to treat the cost basis and the expense of sale as two separate amounts. These are combined in Form 4797, so if I was to follow that convention I only have Option A or otherwise I have to put 220,000 into Line 2 and my Line 1a would end up representing sales proceeds rather than gain.
Even if different combinations of Line 1a less Line 2 would give the same answer in principle, you do get slightly different results for Line 6 and Line 7 depending on how you break it down.
Thank you for any input you are able to provide!
x
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# stats: Classical estimates for tables In robCompositions: Compositional Data Analysis
## Description
Some standard/classical (non-compositional) statistics
## Usage
1 2 3 4 5 6 stats( x, margins = NULL, statistics = c("phi", "cramer", "chisq", "yates"), maggr = mean )
## Arguments
x a data.frame, matrix or table margins margins statistics statistics of interest maggr a function for calculating the mean margins of a table, default is the arithmetic mean
## Details
statistics ‘phi’ is the values of the table divided by the product of margins. ‘cramer’ normalize these values according to the dimension of the table. ‘chisq’ are the expected values according to Pearson while ‘yates’ according to Yates.
For the maggr function argument, arithmetic means (mean) should be chosen to obtain the classical results. Any other user-provided functions should be take with care since the classical estimations relies on the arithmetic mean.
## Value
List containing all statistics
Matthias Templ
## References
Egozcue, J.J., Pawlowsky-Glahn, V., Templ, M., Hron, K. (2015) Independence in contingency tables using simplicial geometry. Communications in Statistics - Theory and Methods, 44 (18), 3978–3996.
## Examples
1 2 3 4 5 6 7 8 9 10 data(precipitation) tab1 <- indTab(precipitation) stats(precipitation) stats(precipitation, statistics = "cramer") stats(precipitation, statistics = "chisq") stats(precipitation, statistics = "yates") ## take with care ## (the provided statistics are not designed for that case): stats(precipitation, statistics = "chisq", maggr = gmean)
### Example output
Attaching package: 'pls'
The following object is masked from 'package:stats':
Attaching package: 'robCompositions'
The following object is masked from 'package:robustbase':
alcohol
spring summer autumn winter
30 0.011477143 0.016502131 0.01242841 0.009370268
60 0.016959419 0.007930214 0.01086270 0.012706044
125 0.013801646 0.010077609 0.01173410 0.013242376
250 0.015127419 0.005180031 0.01009488 0.018209008
500 0.010658729 0.005657531 0.01297609 0.018945993
1000 0.009055118 0.002577320 0.01642036 0.018750000
spring summer autumn winter
30 0.006626332 0.009527510 0.007175549 0.005409926
60 0.009791525 0.004578511 0.006271583 0.007335838
125 0.007968384 0.005818310 0.006774686 0.007645489
250 0.008733820 0.002990693 0.005828283 0.010512976
500 0.006153820 0.003266377 0.007491749 0.010938474
1000 0.005227975 0.001488016 0.009480300 0.010825318
spring summer autumn winter
30 153.01576 133.04755 167.37663 144.00181
60 56.48849 49.82127 62.23317 53.27729
125 60.53416 53.10329 66.41816 56.87925
250 64.55863 56.99054 71.04682 60.43666
500 58.29003 51.19979 63.67261 54.28105
1000 49.94205 43.93191 54.27542 46.44123
spring summer autumn winter
30 153.01253 133.04385 167.37368 143.99837
60 56.47979 49.81131 62.22522 53.26803
125 60.52601 53.09396 66.41072 56.87058
250 64.55100 56.98181 71.03985 60.42854
500 58.28154 51.19008 63.66485 54.27201
1000 49.93213 43.92056 54.26636 46.43066
spring summer autumn winter
30 126.17302 78.91394 136.28536 128.97084
60 45.41726 29.33229 49.59548 46.62716
125 49.72864 31.77171 53.95719 50.75981
250 50.25814 32.64018 54.81904 51.07488
500 45.67802 29.40453 49.21792 46.03362
1000 35.67960 23.15341 38.02341 35.79030
robCompositions documentation built on Jan. 13, 2021, 10:07 p.m.
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## Precalculus (6th Edition)
$m=36$
$m$ varies jointly as $z$ and $p$ therefore $m=kzp$ $m=10$ when $z=2$ and $p=7.5$. Substitute these values into $m=kxy$ to obtain: $m=kpz \\10=k(7.5)(2) \\10=15k \\\frac{10}{15}=\frac{15k}{15} \\\frac{2}{3}=k$ Thus, the equation for $m$ is $m=\frac{2}{3}pz$. To find the value of $m$ when $z=6$ and $p=9$, substitute 6 to z and 9 to p in the equation above to obtain: $\require{cancel} \\m=\frac{2}{3}pz \\m=\frac{2}{3}(9)(6) \\m=\frac{2}{\cancel{3}}(\cancel{9}3)(6) \\m=2(3)(6) \\m=36$
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If you're seeing this message, it means we're having trouble loading external resources on our website.
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Lesson 5: Converting units of mass
# US Customary units of weight review (oz & lb)
Review the size of ounces and pounds and how to convert between the two. Then, try some practice problems.
## Customary units of weight
Weight is a quantity of heaviness. For example, you are measuring the weight of your body when you step on to a scale.
In the customary system of measurement, the most common units of weight are the ounce (oz) and pound (lb).
### How much is an ounce?
A slice of bread has a weight of about $1$ ounce:
### How much is a pound?
A soccer ball has a weight of about $1$ pound:
## Practice set 1: Estimating weight
Problem 1A
Identify the most reasonable unit to measure the weight of a refrigerator.
Want to try more problems like this? Check out this exercise.
## Converting pounds to ounces
To convert pounds to ounces we multiply the number of pounds by $16$.
Example:
## Converting ounces to pounds
To convert ounces to pounds we divide the number of ounces by $16$.
Example:
## Practice set 2: Converting units of weight
Problem 2A
Convert.
$9$ pounds $=$
ounces
Want to try more problems like this? Check out these exercises:
Convert larger units to smaller units
Converting between units
## Want to join the conversation?
• how does 8tons = 16000
• bc 8 times 2 is 16 and then you just add the 3 zeros.
• We're no strangers to love
You know the rules and so do I (do I)
A full commitment's what I'm thinking of
You wouldn't get this from any other guy
I just wanna tell you how I'm feeling
Gotta make you understand
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
We've known each other for so long
Your heart's been aching, but you're too shy to say it (say it)
Inside, we both know what's been going on (going on)
We know the game and we're gonna play it
And if you ask me how I'm feeling
Don't tell me you're too blind to see
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
We've known each other for so long
Your heart's been aching, but you're too shy to say it (to say it)
Inside, we both know what's been going on (going on)
We know the game and we're gonna play it
I just wanna tell you how I'm feeling
Gotta make you understand
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
pls upvote
• LOL so funny
• how do I know if I need to multiply or dived?
• To convert "bigger" units to "smaller" ones, you multiply.
To convert "smaller" to bigger, you divide.
Pound is something big, and ounce is something small - smaller parts make one big thing - 16 ounces are in a pound.
To convert pounds to ounces, you multiply by 16.
To convert ounces to pounds, you divide by 16.
• 55+55=?
• 55 + 55 = 110
Explain: we all know 5+5=10 and we add the one from the ten to the other 5+5 and that equals the same number as 5+5=10 so its 110
• how many ounces in a pound? I am braindead:/
• 16 ounces=1 pound
• pi = 3.1413926535897932384626...
• You actually got that wrong. It's 3.141592etc.
• Why is it called U.S Customary Units of Weight?
• Because units change in every country in the world. For example, in Japan, people don't use inches!
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https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2003-July/006313.html
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# [gmx-users] Combined essential dynamics analysis
Bert de Groot bgroot at gwdg.de
Wed Jul 16 13:28:00 CEST 2003
```On Wed, 16 Jul 2003, Ruben Martinez Buey wrote:
> Hi all,
> I´m trying to compare two similar systems (one with a ligand and the other
> without it). I´ve used CONCOORD for both systems. I´ve concatenated the two
> individual trajectories and run g_covar over the concatenated trajectory. Only
> the first 5-7 resulting combined eigenvectors are significant.
> When I project the individual trajectories over these combined eigenvectors, I
> get the next results:
>
> Average projection (mean):
> eigenvec 1: -0.365220 0.457450
> eigenvec 2: 0.137630 -0.169460
> eigenvec 3: 0.137790 -0.319940
> eigenvec 4: 0.068647 -0.078291
> eigenvec 5: 0.087228 0.040858
hmm, I suppose the two trajectories had a different length? Since
otherwise the projections should be symmetrical with respect to zero.
( ie -0.3 0.3 etc)
>
> Mean square fluctations (stddev):
> eigenvec 1: 6.7204 6.7390
> eigenvec 2: 3.9814 4.2909
> eigenvec 3: 3.6093 3.6686
> eigenvec 4: 2.3390 2.2201
> eigenvec 5: 1.8614 1.8681
>
> (Where the first column is the protein with the ligand, and the second column
> is the protein without ligand)
> But the extreme projections over each eigenvector (generated by g_anaeig) are
> exactly the same in both cases. So, what would be the conclusion? There are no
> significant differences in the essential dynamics beetwen both systems? But the
> differences in the average projections are large enough to be significant,
> aren´t they? And the
>
right. although the differences in the averages are small as compared to
the mean square fluctuations, they *may* still be significant.
Look to what kind of motion the first eigenvector corresponds to (ie if eg
a negative projection corresponds to a more closed conformation or
whatever) and then see what that implies for the effect of the ligand
binding to the protein conformation. and repeat this for ev 2 and 3
(the others are certainly not significant)
Bert
____________________________________________________________________________
Dr. Bert de Groot
Max Planck Institute for Biophysical Chemistry
Theoretical molecular biophysics group
Am Fassberg 11
37077 Goettingen, Germany
tel: +49-551-2011306, fax: +49-551-2011089
email: bgroot at gwdg.de
http://www.mpibpc.gwdg.de/abteilungen/071/bgroot
____________________________________________________________________________
```
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https://justaaa.com/statistics-and-probability/88773-assume-that-thermometer-reading-are-normally
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Question
# Assume that thermometer reading are normally distributed with a mean of 0°C and a standard deviation...
Assume that thermometer reading are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A thermometer is randomly selected and tested. (a) Find the probability of a reading less than -2.75. (b) Find the probability of a reading greater than 2.33 (c) Find the probability of a reading between -2.87 and 1.34 (d) Find the 96th percentile
mean = 0
sd = 1
(a) P(x < -2.75)
P(z < -2.75) = 1- P(z < 2.75) = 1- 0.997 = 0.003
P(x < -2.75) = 0.003
(b) P(x > 2.33)
P(z > 2.33) = 1- P(z < 2.33) = 1- 0.9901 = 0.0099
P(x > 2.33) = 0.0099
(c) P(-2.87 < x < 1.34)
P(-2.87 < z < 1.34) = P( z < 1.34) - P( z < - 2.87)
P( z < 1.34) = 0.9099
P( z < - 2.87) = 1- P( z < 2.87) = 1- 0.9979 = 0.0021
P(-2.87 < z < 1.34) = 0.9078
(d) 96th percentile
P(Z < z ) = 0.96
z = 1.751
x = 1.751
96th percentile; x = 1.751
#### Earn Coins
Coins can be redeemed for fabulous gifts.
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https://researchteam101.com/explain-why-a-shewhart-control-chart-is-limited-in-its-ability-to-detect-smaller-process-shifts/
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Explain why a Shewhart control chart is limited in its ability to detect smaller process shifts.
Scenario: Suppose you joined a start-up company to produce paper helicopters which are intended to be an educational toy. The name of the company is California Choppers. The company consists of 5 people (Peter, Meir, Jennifer, Sara, and Selina) who were responsible for the product development, and who are now responsible for production.
Each of the 5 production personnel works for an 8-hour shift with a production output of 20 helicopters. Each of them uses their own tools and measurement equipment, so they are concerned that the performance of helicopters produced on different shifts may vary.
Your role in the company is to provide expertise for quality assurance. As a first step, the team has asked you to institute an SPC process to understand performance variability of flight times when the helicopter is dropped from 5 feet.
The flight duration data was collected for one week (5 shifts) and is shown graphically as a Shewhart control chart. The data from shift 1 (samples 1 to 20) was used to set the control limits.
This data set appears to be under control, but upon close examination, it was found that there may be l process shifts from operator-to-operator. The data is provided on the next page.
Their team leader is convinced that they can control the production with a Shewhart control chart, but the rest of the team is not so sure. So they hired you as an SPC expert! Your first assignment is as follows:
1. Explain why a Shewhart control chart is limited in its ability to detect smaller process shifts. You should use the data from the first five shifts to illustrate the points.
2. Propose an alternate control chart approach, and explain why it’s an improvement.
3. Graph the data using the alternative approach, showing the control limits.
Test-taking advice: This need not be a long document (max of 2 pages), but you need to be careful to articulate your ideas. I’d propose you format with an introductory paragraph, and then for each idea on why the Shewhart chart is limited- state the conjecture, and then explain why. Similarly, for the alternative approach, state your conjecture, and explain why it’s an improvement.
Shift 1 Shift 2 Shift 3 Shift 4 Shift 5 1 Peter 1.644 Meir 1.903 Jennifer 1.510 Sara 1.668 Selina 1.812 2 Peter 1.823 Meir 1.852 Jennifer 1.576 Sara 1.909 Selina 1.729 3 Peter 1.585 Meir 2.060 Jennifer 1.350 Sara 1.807 Selina 1.808 4 Peter 1.464 Meir 2.206 Jennifer 1.561 Sara 1.939 Selina 1.600 5 Peter 1.807 Meir 1.730 Jennifer 1.752 Sara 1.894 Selina 1.638 6 Peter 2.062 Meir 1.715 Jennifer 1.607 Sara 1.232 Selina 1.758 7 Peter 1.213 Meir 1.898 Jennifer 1.208 Sara 1.723 Selina 1.461 8 Peter 1.770 Meir 2.263 Jennifer 1.444 Sara 1.624 Selina 1.756 9 Peter 1.834 Meir 1.960 Jennifer 1.462 Sara 1.433 Selina 1.310 10 Peter 1.581 Meir 1.875 Jennifer 1.895 Sara 1.438 Selina 1.684 11 Peter 1.512 Meir 1.921 Jennifer 1.426 Sara 1.490 Selina 1.640 12 Peter 1.693 Meir 1.688 Jennifer 1.269 Sara 1.588 Selina 1.596 13 Peter 1.751 Meir 1.912 Jennifer 1.463 Sara 1.473 Selina 1.786 14 Peter 1.608 Meir 1.942 Jennifer 1.424 Sara 1.577 Selina 1.891 15 Peter 1.689 Meir 1.421 Jennifer 1.465 Sara 2.046 Selina 1.847 16 Peter 1.970 Meir 2.072 Jennifer 1.101 Sara 1.730 Selina 1.422 17 Peter 1.993 Meir 1.995 Jennifer 1.903 Sara 1.658 Selina 1.984 18 Peter 2.049 Meir 1.852 Jennifer 1.429 Sara 1.565 Selina 1.808 19 Peter 1.551 Meir 1.495 Jennifer 1.570 Sara 2.009 Selina 1.389 20 Peter 1.829 Meir 1.400 Jennifer 1.666 Sara 1.778 Selina 1.847
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# Comoving distance
(Redirected from Comoving coordinates)
In standard cosmology, comoving distance and proper distance are two closely related distance measures used by cosmologists to define distances between objects. Proper distance roughly corresponds to where a distant object would be at a specific moment of cosmological time, which can change over time due to the expansion of the universe. Comoving distance factors out the expansion of the universe, giving a distance that does not change in time due to the expansion of space (though this may change due to other, local factors such as the motion of a galaxy within a cluster). Comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. At other times, the scale factor differs from 1. The Universe's expansion results in the proper distance changing, while the comoving distance is unchanged by this expansion because it is the proper distance divided by that scale factor.
## Comoving coordinates
Although general relativity allows one to formulate the laws of physics using arbitrary coordinates, some coordinate choices are more natural (easier to work with). Comoving coordinates are an example of such a natural coordinate choice. They assign constant spatial coordinate values to observers who perceive the universe as isotropic. Such observers are called "comoving" observers because they move along with the Hubble flow.
A comoving observer is the only observer that will perceive the universe, including the cosmic microwave background radiation, to be isotropic. Non-comoving observers will see regions of the sky systematically blue-shifted or red-shifted. Thus isotropy, particularly isotropy of the cosmic microwave background radiation, defines a special local frame of reference called the comoving frame. The velocity of an observer relative to the local comoving frame is called the peculiar velocity of the observer.
Most large lumps of matter, such as galaxies, are nearly comoving, so that their peculiar velocities (owing to gravitational attraction) are low.
The comoving time coordinate is the elapsed time since the Big Bang according to a clock of a comoving observer and is a measure of cosmological time. The comoving spatial coordinates tell us where an event occurs while cosmological time tells us when an event occurs. Together, they form a complete coordinate system, giving us both the location and time of an event.
Space in comoving coordinates is usually referred to as being "static", as most bodies on the scale of galaxies or larger are approximately comoving, and comoving bodies have static, unchanging comoving coordinates. So for a given pair of comoving galaxies, while the proper distance between them would have been smaller in the past and will become larger in the future due to the expansion of space, the comoving distance between them remains constant at all times.
The expanding Universe has an increasing scale factor which explains how constant comoving distances are reconciled with proper distances that increase with time.
## Comoving distance and proper distance
Comoving distance is the distance between two points measured along a path defined at the present cosmological time. For objects moving with the Hubble flow, it is deemed to remain constant in time. The comoving distance from an observer to a distant object (e.g. galaxy) can be computed by the following formula:
$\chi = \int_{t_e}^t c \; {\mbox{d} t' \over a(t')}$
where a(t′) is the scale factor, te is the time of emission of the photons detected by the observer, t is the present time, and c is the speed of light in vacuum.
Despite being an integral over time, this does give the distance that would be measured by a hypothetical tape measure at fixed time t, i.e. the "proper distance" as defined below, divided by the scale factor a(t) at that time. For a derivation see "standard relativistic definitions" from Davis & Lineweaver 2004.[1]
Definitions
• Many textbooks use the symbol $\! \chi$ for the comoving distance. However, this $\! \chi$ must be distinguished from the coordinate distance r in the commonly used comoving coordinate system for a FLRW universe where the metric takes the form
$\! ds^2 = -c^2 d\tau^2 = - c^2 dt^2 + a(t)^2 \left( \frac{dr^2}{1 - kr^2} + r^2 \left(d\theta^2 + \sin^2 \theta d\phi^2 \right)\right)$.
In this case the comoving coordinate distance $\! r$ is related to $\! \chi$ by $\! \chi = r$ if k=0 (a spatially flat universe), by $\! \chi = \sin^{-1} r$ if k=1 (a positively curved 'spherical' universe), and by $\! \chi = \sinh^{-1} r$ if k=-1 (a negatively curved 'hyperbolic' universe).[2]
• Most textbooks and research papers define the comoving distance between comoving observers to be a fixed unchanging quantity independent of time, while calling the dynamic, changing distance between them proper distance. On this usage, comoving and proper distances are numerically equal at the current age of the universe, but will differ in the past and in the future; if the comoving distance to a galaxy is denoted $\! \chi$, the proper distance $\! d(t)$ at an arbitrary time $\! t$ is simply given by $\! d(t) = a(t) \chi$ where $\! a(t)$ is the scale factor (e.g. Davis & Limeweaver 2004).[1] The proper distance $\! d(t)$ between two galaxies at time t is just the distance that would be measured by rulers between them at that time.[3]
### Uses of the proper distance
Cosmological time is identical to locally measured time for an observer at a fixed comoving spatial position, that is, in the local comoving frame. Proper distance is also equal to the locally measured distance in the comoving frame for nearby objects. To measure the proper distance between two distant objects, one imagines that one has many comoving observers in a straight line between the two objects, so that all of the observers are close to each other, and form a chain between the two distant objects. All of these observers must have the same cosmological time. Each observer measures their distance to the nearest observer in the chain, and the length of the chain, the sum of distances between nearby observers, is the total proper distance.[4]
It is important to the definition of both comoving distance and proper distance in the cosmological sense (as opposed to proper length in special relativity) that all observers have the same cosmological age. For instance, if one measured the distance along a straight line or spacelike geodesic between the two points, observers situated between the two points would have different cosmological ages when the geodesic path crossed their own world lines, so in calculating the distance along this geodesic one would not be correctly measuring comoving distance or cosmological proper distance. Comoving and proper distances are not the same concept of distance as the concept of distance in special relativity. This can be seen by considering the hypothetical case of a universe empty of mass, where both sorts of distance can be measured. When the density of mass in the FLRW metric is set to zero (an empty 'Milne universe'), then the cosmological coordinate system used to write this metric becomes a non-inertial coordinate system in the Minkowski spacetime of special relativity where surfaces of constant Minkowski proper-time τ appear as hyperbolas in the Minkowski diagram from the perspective of an inertial frame of reference.[5] In this case, for two events which are simultaneous according the cosmological time coordinate, the value of the cosmological proper distance is not equal to the value of the proper length between these same events,[6] which would just be the distance along a straight line between the events in a Minkowski diagram (and a straight line is a geodesic in flat Minkowski spacetime), or the coordinate distance between the events in the inertial frame where they are simultaneous.
If one divides a change in proper distance by the interval of cosmological time where the change was measured (or takes the derivative of proper distance with respect to cosmological time) and calls this a "velocity", then the resulting "velocities" of galaxies or quasars can be above the speed of light, c. This apparent superluminal expansion is not in conflict with special or general relativity, and is a consequence of the particular definitions used in cosmology. Even light itself does not have a "velocity" of c in this sense; the total velocity of any object can be expressed as the sum $\! v_{tot} = v_{rec} + v_{pec}$ where $\! v_{rec}$ is the recession velocity due to the expansion of the universe (the velocity given by Hubble's law) and $\! v_{pec}$ is the "peculiar velocity" measured by local observers (with $\! v_{rec} = \dot{a}(t) \chi(t)$ and $\! v_{pec} = a(t) \dot{\chi}(t)$, the dots indicating a first derivative), so for light $\! v_{pec}$ is equal to c (-c if the light is emitted towards our position at the origin and +c if emitted away from us) but the total velocity $\! v_{tot}$ is generally different from c.[1] Even in special relativity the coordinate speed of light is only guaranteed to be c in an inertial frame; in a non-inertial frame the coordinate speed may be different from c.[7] In general relativity no coordinate system on a large region of curved spacetime is "inertial", but in the local neighborhood of any point in curved spacetime we can define a "local inertial frame" in which the local speed of light is c[8] and in which massive objects such as stars and galaxies always have a local speed smaller than c. The cosmological definitions used to define the velocities of distant objects are coordinate-dependent - there is no general coordinate-independent definition of velocity between distant objects in general relativity.[9] The issue of how best to describe and popularize the apparent superluminal expansion of the universe has caused a minor amount of controversy. One viewpoint is presented in Davis and Lineweaver, 2004.[1]
### Short distances vs. long distances
Within small distances and short trips, the expansion of the universe during the trip can be ignored. This is because the travel time between any two points for a non-relativistic moving particle will just be the proper distance (that is, the comoving distance measured using the scale factor of the universe at the time of the trip rather than the scale factor "now") between those points divided by the velocity of the particle. If the particle is moving at a relativistic velocity, the usual relativistic corrections for time dilation must be made.
## References
1. ^ a b c d T.M. Davis, C.H. Lineweaver (2004). "Expanding Confusion: Common Misconceptions of Cosmological Horizons and the Superluminal Expansion of the Universe". Publications of the Astronomical Society of Australia 21 (1): 97–109. arXiv:astro-ph/0310808v2. Bibcode:2004PASA...21...97D. doi:10.1071/AS03040.
2. ^ See pages 9-12 of The Cosmological Background Radiation by Marc Lachièze-Rey and Edgard Gunzig, or p. 263 Measuring the Universe: The Cosmological Distance Ladder by Stephen Webb.
3. ^ see p. 4 of Distance Measures in Cosmology by David W. Hogg.
4. ^ Steven Weinberg, Gravitation and Cosmology (1972), p. 415
5. ^ See the diagram on p. 28 of Physical Foundations of Cosmology by V. F. Mukhanov, along with the accompanying discussion.
6. ^ E.L. Wright (2009). "Homogeneity and Isotropy". E.L. Wright. Retrieved February 2015.
7. ^ see p. 219 of Relativity and the Nature of Spacetime by Vesselin Petkov
8. ^ see p. 94 of An Introduction to the Science of Cosmology by Derek J. Raine, Edwin George Thomas, and E. G. Thomas
9. ^ J. Baez and E. Bunn (2006). "Preliminaries". University of California. Retrieved February 2015.
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Message Font: Serif | Sans-Serif
No. of Recommendations: 2
GNVR = my Graham Number Valuation Range
You can look up the Graham Number in the Fool's wiki. It's a formula derived from Graham's writings. This is not what I think the current stock price should be but if the price were to fall to these levels I would be stupid not to buy shares. Graham (as per my understanding) was always lookng for companies the Market has mispriced and if the price was below his valuations he was picking up shares.
This is also not to say that the current price is to high. Let the trend be your friend. I'm just conducting an exercise and I happen to be using V at this time.
`Fair Value = Square Root of (EPS * Book Value * 22.5) Some say you should use the Tangible Book Value. I do, along with the Normalized EPS thus giving me a range GNVR. I sometimes mix it up some more.Using current Fool CAPS Stats data:EPS = 2.22BV = 34.09First Graham Number = 41.08Using Scottrade Data (they're slow with updated numbers so this data is good as of Sep 30, 2012 and I expect an update soon)EPS = 1.87Normalized EPS = 6.35Book Value = 41.30Tang.Book Value = 6.77Mixing it up to get four different Graham Numbers 41.69 used EPS & BV 76.82 used Normalized EPS & BV 16.88 used EPS & Tang.BV 31.10 used Normalized EPS & Tang.BV `
If I were a shareholder I would want to understand more why the big difference between the Book Value and Tangible Book value. Most companies I look up have a difference but not usually that big a difference.
Mostly this is just for my benefit, my curiosity, and anyone who may care.
I intend to keep this thread running with updated GNVR's over time to watch the trend. So check back on this thread from time to time for an updated GNVR. Normally I only plug into the formula EPS with BV, then Normalized EPS with Tangible Book Value. That's all I'll use for future updates.
As for now with mixing up all that data I have an extreme range of \$16.88 to \$76.82
Foolish best,
blesto
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You are Here: Home >< Maths
# Line integral, vector field Watch
1. Problem is attached.
To do the line integral, I need to find F(r(t)), but I don't understand how to express it. For example I looked at the online notes provided here: http://tutorial.math.lamar.edu/Class...torFields.aspx
I don't understand how F(r(t)) was derived in the first example. I know how to do the rest.
2. (Original post by blah3210)
Problem is attached.
To do the line integral, I need to find F(r(t)), but I don't understand how to express it. For example I looked at the online notes provided here: http://tutorial.math.lamar.edu/Class...torFields.aspx
I don't understand how F(r(t)) was derived in the first example. I know how to do the rest.
Express it as a vector(either in column, row or i, j, k). F is a function taking a 3d vector to a 3d vector. In this case, .
A general point on the line is , so .
3. (Original post by blah3210)
Problem is attached.
To do the line integral, I need to find F(r(t)), but I don't understand how to express it. For example I looked at the online notes provided here: http://tutorial.math.lamar.edu/Class...torFields.aspx
I don't understand how F(r(t)) was derived in the first example. I know how to do the rest.
To find F(r(t)) you need to sub in for each of the components for example if F=(2x,x-y,0) and the path parametrised by r(t)=(x(t),y(t),z(t))=(t,2t,t^2) (for some interval of t) then whenever you see an x you need to sub in for the x component of the path so the 2x of the vector field becomes 2t since x(t)=t. Similarly for the other components and we get F(r(t))=(2t,t-2t,0)=(2t,-t,0).
Hopefully you can see how it works for your example now.
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https://stats.stackexchange.com/questions/397619/distribution-of-maximum-likelihood-estimator
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# Distribution of Maximum Likelihood Estimator
Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)
Let $$L(\theta)$$ be the maximum likelihood function for the distribution $$f(x;\theta)$$
Then after taking sample of size n
$$L(\theta)=f(x_1;\theta)\cdot f(x_2\theta)...f(x_n;\theta)$$
And we want to find $$\theta_{max}$$ such that $$L(\theta)$$ is maximized and $$\theta_{max}$$ is our estimate (once a sample has actually been selected)
Since $$\theta_{max}$$ maximizes $$L(\theta)$$ it also maximizes $$ln(L(\theta))$$
where
$$ln(L(\theta))=ln(f(x_1;\theta))+ln(f(x_2;\theta))...+ln(f(x_n;\theta))$$
Taking the derivative with respect to $$\theta$$
$$\frac{f'(x_1;\theta)}{f(x_1;\theta)}+\frac{f'(x_2;\theta)}{f(x_2;\theta)}...+\frac{f'(x_n;\theta)}{f(x_n;\theta)}$$
$$\theta_{max}$$ would be the solution of the above when set to 0 (after selecting values for all $$x_1,x_2...x_n$$) but why is it normally distributed and how do I show that it's true for large n?
MLE requires $$\frac{\partial \ln L(\theta)}{\partial \theta} = \sum_{i=1}^n \frac{ f'(x_i;\theta)}{f(x_i;\theta)},$$ where $$f'(x_i;\theta)$$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $$g(x;\theta)=\frac{ f'(x;\theta)}{f(x;\theta)}.$$ Then $$\{g(x_i;\theta)\}_{i=1}^n$$ is a new iid sequence of random variables, with $$Eg(x_1;\theta)=0$$. If $$Eg(x_1;\theta)g(x_1;\theta)'<\infty$$, CLT implies, $$\sqrt{n}(\bar{g}_n(\theta)-Eg(x_1;\theta))=\sqrt{n}\bar{g}_n(\theta) \rightarrow_D N(0,E(g(x;\theta)g(x;\theta)'),$$ where $$\bar{g}_n(\theta)=\frac{1}{n} \sum_{i=1}^n g(x_i;\theta).$$ The ML estimator solves the equation $$\bar{g}_n(\theta)=0.$$ It follows that the ML estimator is given by $$\hat{\theta}=\bar{g}_n^{-1}(0).$$ So long as the set of discontinuity points of $$\bar{g}_n^{-1}(z)$$, i.e. the set of all values of $$z$$ such that $$\bar{g}_n^{-1}(z)$$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $$\theta$$.
• $\frac{f'(x)}{f(x)}$ is just another function of x so central limit theorem applies thank you for that Commented Mar 15, 2019 at 1:23
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https://www.caida.org/projects/ark/statistics/monitor/per-au/nonresp_path_length_ccdf.html
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# Archipelago Monitor Statistics
Archipelago (Ark): CAIDA's active measurement infrastructure serving the network research community since 2007.
Statistical information for the topology traces taken by this individual Ark monitor is displayed below. See the main statistics page for the full list of monitors
Prev monitor: pbh2-bt
Prev in AU: avv-au
Prev in Oceania: hlz2-nz
Next monitor: per2-au
Next in AU: per2-au
Next in Oceania: per2-au
# per-au
AARNet
Perth, AU (5)
### CCDF of IP path lengths for non-responding destinations
• percentile 10th 25th 50th 75th 90th Max
IP path length 4 5 8 11 13 34
Use the following link to download the data used to render this graph in ASCII, comma-separated values format here: (CSV output)
### Description
This graph shows the complementary cumulative distribution function (CCDF) of path lengths (number of hops) for probes whose destinations did not respond. The path length is instead the length to the last responding hop.
### Motivation
Examining only traces which have responding destinations gives a more accurate distribution of path lengths, but in typical usage there are several times as many probes which do not have responses. However, the traces without a responding destination still give us a lower bound on path length distribution. For comparison, view the IP path length distribution for responding destinations.
### Background
The complementary cumulative distribution function shows the fraction of collected data points that are greater than a given value. This is backwards from how percentiles are given, as those show the percentage lower than a given value. On this graph, you would find the 80th percentile at the 0.2 Y value. The path length is defined as the number of IP hops required to reach the destination from the Ark monitor. This includes any routing loops that occur during the probing. These values are only used when a response has been received from the destination. In other words, incomplete paths are ignored for the purposes of determining path length.
### Analysis
The shape of the path length CCDF typically doesn't very much from monitor to monitor, as the distribution of path lengths to randomly selected set of destinations is mostly determined by the 'core' of the Internet. Routing loops or large internal networks (on the monitor's side) can shift the path length values higher, but the range doesn't vary dramatically. If a monitor's path length CCDF looks suddenly very different from other monitors', it's quite possible that there's a configuration problem in the local network.
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## College Algebra (11th Edition)
$h=\frac{2A}{B+b}$
Multiply both sides by 2 and divide by $(B+b)$. $2A=h(B+b)$ $\frac{2A}{B+b}=h$ $h=\frac{2A}{B+b}$
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# Question How does the type of surface affect the outcome of the egg’s shell breaking? Hypothesis As we increase the amount of the cushion on the surface,
## Presentation on theme: "Question How does the type of surface affect the outcome of the egg’s shell breaking? Hypothesis As we increase the amount of the cushion on the surface,"— Presentation transcript:
Question How does the type of surface affect the outcome of the egg’s shell breaking? Hypothesis As we increase the amount of the cushion on the surface, then the egg will not break because the surface will not be hard enough of a surface to make the egg break.
Materials 6 eggs (chicken grade AA) 3 different types of surfaces that you will be dropping the eggs onto (fluffy pillow, towel, 26 pages or sheets of newspaper) meter stick or measuring tape
MV-Type of surface that you are dropping the egg onto. RV- Outcome of the egg breaking or not CV- type of egg(chicken, grade AA), height that you are dropping your egg from(4 feet)
Egg Drop Observations Type of SurfaceTrial 1Trial 2 Fluffy PillowThe egg did not break. It bounced, then landed onto the kitchen floor. It also did not break and it also bounced again, but this time landed on the pillow. 5 layer towelIt did not break but it made a loud thump noise. It made a crack in the egg and some liquids oozed out. 26 Pages of NewspaperIt made a big hole and most of the egg insides came out. It completely exploded and everything splattered everywhere.
Procedure 1)Place the pillow on the ground. 2)Use your tape measure and measure 4 ft off the ground. 3) Grab one egg and hold it at 4 feet, then drop the egg onto the pillow. 4) Record your observations. 5) Clean your area and repeat steps 2-4 with a new egg 6) repeat steps 1-5 but with a towel folded into five layers 7) Repeat steps 1-5 but now with 26 layers of newspaper
Conclusion We thought that if we increased the softness of the surface, then the egg will not break. Our results showed the that the egg did not break under the soft surfaces. We can prove this because the egg did not break when on the pillow and it did break under the newspapers. This supports my original hypothesis. Therefore if we increase the softness of the surface, then the egg will not break.
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https://numbersworksheet.com/negative-numbers-and-integers-for-grade-6-worksheets/
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# Negative Numbers And Integers For Grade 6 Worksheets
The Negative Figures Worksheet is the best way to commence training your young ones the idea of bad figures. A negative quantity is any quantity which is lower than absolutely nothing. It can be included or subtracted. The minus signal signifies the negative quantity. You can even create adverse phone numbers in parentheses. Under can be a worksheet to provide you started out. This worksheet has a variety of unfavorable numbers from -10 to 10. Negative Numbers And Integers For Grade 6 Worksheets.
Unfavorable numbers are a lot in whose worth is less than absolutely no
A poor amount includes a worth less than zero. It can be depicted with a quantity line in two ways: together with the positive number created since the initial digit, and with the bad variety published as being the last digit. An optimistic number is created with a in addition sign ( ) prior to it, however it is non-obligatory to write it that way. If the number is not written with a plus sign, it is assumed to be a positive number.
## They are symbolized with a minus indicator
In historic Greece, negative phone numbers were actually not employed. These were ignored, since their math was depending on geometrical concepts. When Western scholars commenced translating ancient Arabic text messages from To the north Africa, they came to acknowledge adverse phone numbers and accepted them. Today, negative figures are represented by way of a minus signal. For additional details on the history and origins of adverse numbers, read through this post. Then, attempt these cases to see how unfavorable figures have advanced as time passes.
## They could be added or subtracted
As you might already know, positive numbers and negative numbers are easy to add and subtract because the sign of the numbers is the same. Negative numbers, on the other hand, have a larger absolute value, but they are closer to than positive numbers are. These numbers have some special rules for arithmetic, but they can still be added and subtracted just like positive ones. Also you can subtract and add negative phone numbers using a number line and utilize the same rules for subtraction and addition as you may do for positive amounts.
## They can be displayed by way of a quantity in parentheses
A negative number is symbolized by a number encased in parentheses. The adverse sign is changed into its binary counterpart, and the two’s enhance is stored in exactly the same place in recollection. Sometimes a negative number is represented by a positive number, though the result is always negative. In these instances, the parentheses should be incorporated. If you have any questions about the meaning of negative numbers, you should consult a book on math.
## They are often divided by a good variety
Negative figures might be multiplied and divided like optimistic numbers. They can be separated by other unfavorable numbers. They are not equal to one another, however. The first time you grow a poor variety from a positive number, you will definately get absolutely nothing as a result. To make the best solution, you need to pick which indicator your answer needs to have. It really is quicker to remember a poor quantity after it is printed in brackets.
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This presentation is the property of its rightful owner.
1 / 11
# The First and Second Derivative Tests PowerPoint PPT Presentation
The First and Second Derivative Tests. Today you will connect concavity with the sign of the second derivative. Silent Board Game. Do NOT Talk!!!!
The First and Second Derivative Tests
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## The First and Second Derivative Tests
Today you will connect concavity with the sign of the second derivative.
### Silent Board Game
Do NOT Talk!!!!
Each of you will label a characteristic on the sketch I am about to draw, one at a time, without talking. It is your job to label the curve as completely as possible. You may use descriptive words, symbolic notation, etc. Each person may only add ONE label to the curve.
### Walk-A-Wave Challenge
• Can you walk a perfect sine wave?
• How can you “walk” so that that you get a nice curvy wave with points of inflection?
• How do you “walk” the concavity of the sine wave?
• When are you walking the fastest?
• When do you slow down?
### The Second Derivative
How does the increasing or decreasing nature of a curve relate to the first derivative? The second derivative?
How does the concavity of the curve relate to the second derivative? The first derivative?
f. Does the direction you walk affect the concavity of the curve?
### First & Second Derivative Tests
• Apply your theories from the last problem to find where is increasing and decreasing. Where is it concave up? Concave down?
### Points of Inflection
A point of inflection is a point where concavity changes. Since the point of inflection deals with concavity, it must be related to the second derivative.
### Partner Exit Slip
Complete the Exit Slip with your partner on the sheet provided.
HW G
See yutmrrw!
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# Math 45 Test 1
### Card Set Information
Author: regishouse ID: 128192 Filename: Math 45 Test 1 Updated: 2012-01-23 04:20:43 Tags: Math 45 Folders: Description: Math 45 test 1 Show Answers:
Home > Flashcards > Print Preview
The flashcards below were created by user regishouse on FreezingBlue Flashcards. What would you like to do?
1. Least Common Factor:
The smallest number that can be divided evenly by each of a group of specified numbers
i.e. Find the LCM of 75, 125
• 75: 75, 150, 225, 300, 375
• 125: 125,250,375
2. Equivilent Fractions:
The Least Common Denominator is the least common multiple of all denominators
i.e. write and as equivilent fractions with the least common denominator
• Find the LCM of 10,12
• LCM = 60
• = (multiply 1 by 6, the number needed to get 10 to 60)
• = ( multiply 1 by 5)
3. Reducing Fractions:
Reduce by finding multiples of both numbers, reduce to the least common factors
i.e
• =
• (divide num & denom by 8)
4. Place Value/ Decimal Numbers:
i.e
Find the place value of 1.694
Round 535.123 to the nearest tenth
• 1.694 = thousandths place
• 535.123 = 535.1
5. Fractions to Decimals:
divide the numerator by the denominator
• 27÷8
• 3.375
6. Decimal to Fraction:
use place value
i.e convert 0.45 to a fraction
• convert 0.45 to a fraction
• (divide by 5)
7. Percents:
Per hundreds (divide by 100)
Decimals to Percents: move decimal place 2 places to the right
Percents to Decimals: move decimal place 2 places to the left
• 14% = 0.14
• 6.57% = .0657
• 0.332 = 33.2%
• 6 = 600%
8. Natural Numbers:
Set of natural / counting numbers
1,2,3,4,5,6,7,8,9....
9. Real Numbers:
sets of real numbers/ all numbers
1, -1, , 0.5
10. Whole Numbers:
All natural numbers and zero
0,1,2,3,4,5,6,7,8,9....
11. Even Numbers:
all numbers divisable by 2 (feminine)
2,4,6,8,10,12...
12. Odd Numbers:
remainder of 1 when divided by 2 (masuline)
1,3,5,7,9,11....
13. Subset:
If all members of one set are also members of a second set, the first set is a subset of the second
i.e B{1,2} A{1,2,3}
B{1,2} A{1,2,3} = BA
the empty set is a subest of A = {} A or A
14. Subsets of Real Numbers:
• Whole Numbers {0,1,2,3,4,5,6}
• Natural Numbers {1,2,3,4,5,6}
• Real Numbers { -1, 0.5, , [-9], 0.5555}
15. Intergers:
All whole numbers and the opposite of every whole number (not including fractions or decimals)
{-4,-3,-2,-1,0,1,2,3,4}
16. Rational vs Irrational Numbers:
Rational Numbers: fractions in which numerator & denominator are intergers
Irrational Numbers: cannot be repeated as fractions of intergers (π, √2)
17. Multiplying with Real Numbers
• (+) (+) = (+)
• (+) (-) = (-)
• (-) (-) = (+)
18. Dividing with Real Numbers
• (+) / (+) = (+)
• (-) / (+) = (-)
• (+) / (-) = (-)
• (-) / (-) = (+)
• (#) / (0) = (undefined)
• multiplicative inverses = reciprocals
must have common denominator
(divide both numbers by 7, find LCM of 98 & 28)
21. Multiplying & Dividing Fractions:
to divide fractions, mulitply by the reciprical
22. Adding & Subtracting Decimal Points:
line up decimals
• -170.826 + 100.7 + 3.02
• -170.826 + 103.72
• -67.106
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https://www.longevitas.co.uk/information-matrix-page/keeping-it-simple
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# Keeping it simple
Which mortality-improvement basis is tougher — a medium-cohort projection with a 2% minimum value, or a long-cohort projection with a 1% minimum? Unless you are an actuary who works with such things, you have little chance of answering this question. This is unsatisfactory in an age of transparency. There are many non-actuaries — regulators, pension-scheme trustees and equity analysts among them — who need to be able to make this kind of judgment. The advent of the CMI's new projection model means that there will be yet more variety in projection bases.
One solution is to quote specimen life expectancies or annuity values. However, different bases have different impacts at varying ages, as shown in Table 1:
Table 1. Specimen factors for level unit annuities to males, valued at a discount rate of 3% per annum. Mortality is according to S1PA from 2011 onwards, with improvements after that year in line with the stated projection.
Mortality projection Age 65 Age 70 Age 75
i) Medium cohort with 2% minimum 14.606 12.042 9.527
ii) Long cohort with 1% minimum 14.381 11.980 9.654
(ii) as percentage of (i) 98.5% 99.5% 101.3%
Table 1 shows that the impact of a projection basis is dependent on age (amongst many other factors), so a quoted life expectancy or annuity factor for age 65 (say) could be quite misleading if the average age in a portfolio is over 70. If we can't quote a specimen annuity value, how else can we eliminate the confounding effects of the portfolio structure and other basis elements?
As it happens, a simple solution is to use the equivalent-annuity calculation. This is documented in Richards and Jones (2004), but its use extends back far earlier than this. The idea is to take a portfolio value and solve for a common basis element to allow simple comparison. When dealing with mortality projections, perhaps the simplest approach is to solve for the equivalent constant annual mortality improvement. This is shown in Table 2:
Table 2. Equivalent annual rates of improvement implied by the projection bases in Table 1.
Mortality projection Age 65 Age 70 Age 75
i) Medium cohort with 2% minimum 2.00% 2.00% 2.00%
ii) Long cohort with 1% minimum 1.66% 1.89% 2.32%
Now things are clearer. We can see that the "medium cohort" bit is a complete misnomer as it has no impact — the medium-cohort improvements are so low that the floor value always applies. We can also see that the "long cohort with 1% minimum" isn't always as tough as it sounds. Using this approach we could compare any two bases we liked by reducing them to the equivalent constant rate of improvement. This would make it much easier for regulators and other users of FSA returns and pension-scheme reports to compare the strengths of mortality-improvement bases.
An equivalent-annuity calculation can be performed over an entire portfolio, thus allowing for the full impact of age distribution and any concentration of liabilities. The end result is a single figure which can be used to compare the strength of a projection basis amongst all portfolios. The method is simple, timeless and does not need to be updated. As such, it can not only be used to compare bases across portfolios, but it can also track the basis strength of a given portfolio over time.
Written by: Stephen Richards
Publication Date:
Last Updated:
## Equivalent annuities in Longevitas and mortalityrating.com
mortalityrating.com uses an equivalent-annuity calculation to turn its bespoke mortality tables into a robust percentage of a standard table for actuarial calculations.
Longevitas uses equivalent annuities to to convert bespoke scheme mortality tables into equivalent percentages of a standard table for communication purposes. Longevitas also has the ability to perform an equivalent-annuity calculation for any given portfolio using the "Equate bases" option.
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https://www.thenational.academy/teachers/programmes/maths-primary-ks2/units/represent-counting-in-nines-as-the-9-times-table/lessons/solve-problems-involving-adjacent-multiples-of-nine
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New
New
Year 4
# Solve problems involving adjacent multiples of nine
I can solve problems involving adjacent multiples of nine.
New
New
Year 4
# Solve problems involving adjacent multiples of nine
I can solve problems involving adjacent multiples of nine.
Share activities with pupils
Share function coming soon...
## Lesson details
### Key learning points
1. Adjacent multiples of nine have a difference of 9 and alternate between odd and even numbers.
2. If you add 9 to a multiple of nine you get the next multiple of nine.
3. If you subtract 9 from a multiple of nine you get the previous multiple of nine.
### Common misconception
Pupils may find it difficult to solve problems or misunderstand questions.
Model clearly how to identify key phrases and words which will identify the operation and equation required for word problems. This could involve drawing bar models or other representations.
### Keywords
• Adjacent multiple - An adjacent multiple is a multiple next to another. It could be before or after.
By this point fluency in 9 times tables will support children in solving these problems. Continue to chorally rehearse the 9 times tables.
Teacher tip
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
Which multiple of 9 is missing from this sequence? 45, , 63, 72
Q2.
What is 12 × 9 =
Q3.
Which symbol goes in the gap to make this correct? 9 × 9 = 10 × 9 ___ 9
+
×
÷
Q4.
Which symbol goes in the gap to make this correct? 2 × 9 = 3 × 9 ___ 9
+
×
÷
Q5.
Complete: I know that 3 × 9 = 27 so 30 × 9 =
Q6.
Complete: I know that 5 × 9 = 45 so 50 × 9 =
## Exit quiz
### 6 Questions
Q1.
Lucas makes shapes using 9 sticks each time. He has made 3 shapes and used 27 sticks. Lucas makes another of the shapes. How many sticks has he used now?
27
28
30
Q2.
There are 7 flowers with 9 petals each. If Jun gives away one flower, how many petals will there be altogether? petals
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https://www.unitconverters.net/length/link-to-decimeter.htm
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Home / Length Conversion / Convert Link to Decimeter
Please provide values below to convert link [li] to decimeter [dm], or vice versa.
### Link to Decimeter Conversion Table
0.01 li0.0201168 dm
0.1 li0.201168 dm
1 li2.01168 dm
2 li4.02336 dm
3 li6.03504 dm
5 li10.0584 dm
10 li20.1168 dm
20 li40.2336 dm
50 li100.584 dm
100 li201.168 dm
1000 li2011.68 dm
### How to Convert Link to Decimeter
1 li = 2.01168 dm
1 dm = 0.4970969538 li
Example: convert 15 li to dm:
15 li = 15 × 2.01168 dm = 30.1752 dm
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CC-MAIN-2024-26
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https://www.farinelliandthekingbroadway.com/2022/05/17/what-is-1mb-in-mb/
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## What is 1Mb in MB?
1Mb = 1/8 MB = 0.125 MB.
### How do you convert MB into MP?
If we want to measure megabytes in a way that they would appear in Windows and other operating systems, then 1 MB/s = 1,048,576 bytes per second. To convert to bits, multiply by 8: 1,048,576 x 8 = 8,388,608 bits per second. This converts to 8.388 mbps (8,388,608 / 1,000,000).
What means 10mb?
10 Mbps delivers internet download speeds at approximately 10 megabits/second and upload speeds up to 1 megabit/second. That means a 10 MB file will take 8 seconds to load. This speed is ideal for small businesses with very few employees, and it functions through a DSL internet connection with a unique IP address.
How many MB is 100mbit?
Mbit to Megabytes Conversion Table
Megabits (Mbit) Megabytes (MB)
97 Mbit 12.125 MB
98 Mbit 12.25 MB
99 Mbit 12.375 MB
100 Mbit 12.5 MB
## Is kB bigger than MB?
KB, MB, GB – A kilobyte (KB) is 1,024 bytes. A megabyte (MB) is 1,024 kilobytes.
### How many KB is 4 MP?
In practical information technology, KB is actually equal to 210 bytes, which makes it equal to 1024 bytes….MB to KB Conversion Table.
Megabytes (MB) Kilobytes (KB) decimal Kilobytes (KB) binary
1 MB 1,000 KB 1,024 KB
2 MB 2,000 KB 2,048 KB
3 MB 3,000 KB 3,072 KB
4 MB 4,000 KB 4,096 KB
How many MB is MP?
Mp or megapixels refers to the size of the image. For example, an image of 4,000 pixels wide by 3,000 pixels high has a total of 12,000,000 pixels (12Mp). MB or megabytes refers to the size of the file used to save the image….Megapixels and Megabytes.
Make OLYMPUS CORPORATION
Capture date Sun, 24 Sep 2017 10:42:32 GMT
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CC-MAIN-2023-40
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https://rdrr.io/cran/gets/man/eqwma.html
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# eqwma: Equally Weighted Moving Average (EqWMA) of the pth.... In gets: General-to-Specific (GETS) Modelling and Indicator Saturation Methods
## Description
The function `eqwma` returns an Equally Weighted Moving Average (EqWMA) of the pth. exponentiated values lagged. Optionally, the absolute values are computed before averaging, and the log of is returned. The function `leqwma` is essentially a wrapper to `eqwma` in which the absolute values are used and the logarithm is applied.
If x is financial return (possibly mean-corrected) and p=2, then this gives the socalled 'historical' model, also known as an integrated ARCH model where the ARCH coefficients all have the same value with sum equal to one. In the log-variance specification the lag of log(EqWMA) is thus a financial volatility proxy. It may be an imperfect proxy compared with high-frequency data (which can also be included as regressors), but - in contrast to high-frequency data - is always available and easy to compute.
## Usage
```1 2 3``` ```eqwma(x, length = 5, lag = 1, start = 1, p = 1, log = FALSE, abs = FALSE, as.vector = TRUE) leqwma(x, length = 5, lag = 1, start = 1, p = 2, as.vector=FALSE) ```
## Arguments
`x` numeric vector, time-series or `zoo` object. Missing values in the beginning and/or at the end of the series is allowed, as they are removed with the `na.trim` command `length` integer or vector of integers each equal to or greater than 1. The length or lengths of the moving window or windows of averages `lag` integer equal to or greater than 0. If 0, then the moving averages are not lagged `start` integer equal to or greater than 1 (default: start=1, i.e. the first observation). Where to start the moving windows of averages `p` numeric value greater than zero. The exponent p in x^p for `eqwma` and in abs(x)^p for `leqwma` `log` logical. If TRUE, then the logarithm of the moving average is returned. If FALSE (default), then the logarithm is not applied `abs` logical. If TRUE, then x is transformed to absolute values before x is exponentiated `as.vector` logical. If TRUE, then a univariate series is returned as a vector. If FALSE, then a univariate series is returnes as a matrix. Note: multivariate series are always returned as a matrix
## Details
The intended primary use of `eqwma` is to construct mixed frequency regressors for the mean specification.
The intended primary use of `leqwma` is to construct volatility proxies for the log-variance specification. The default is the lagged log of an equally weighted moving average of the squared residuals, where each average is made up of m observations. This is equivalent to an integrated ARCH(p) model where the p coefficients are all equal. For further details on the use of log(EqWMA) as a volatility proxy, see Sucarrat and Escribano (2012).
## Value
numeric vector, time series or `zoo` object
## Author(s)
Genaro Sucarrat, http://www.sucarrat.net/
## References
Genaro Sucarrat and Alvaro Escribano (2012): 'Automated Financial Model Selection: General-to-Specific Modelling of the Mean and Volatility Specifications', Oxford Bulletin of Economics and Statistics 74, Issue no. 5 (October), pp. 716-735
`zoo`, `arx`, `getsm`, `getsv`
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16``` ```##generate an iid normal series: set.seed(123) x <- rnorm(100) ##compute lag of EqWMA(20) for x^2: eqwma(x, p=2) ##compute lag of EqWMA(5) and lag of EqWMA(10) for x: eqwma(x, length=c(5,10)) ##compute lag of log(EqWMA(20)) for x^2: leqwma(x) #compute lag of log(EqWMA(5)) and lag of log(EqWMA(8)) #for abs(x)^2: leqwma(x, length=c(4,8)) ```
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http://wiki.multitheftauto.com/wiki/SetVehicleSirens
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# SetVehicleSirens
This function changes the properties of a vehicles siren point.
[[{{{image}}}|link=|]] Note: Although you may be able to add sirens to any vehice, this function may not work. This function fails on certain vehicle models (https://wiki.multitheftauto.com/wiki/Vehicle_IDs#Lua_table_of_vehicles_that_doesn.27t_support_siren_lights).
## Syntax
bool setVehicleSirens ( vehicle theVehicle, int sirenPoint, float posX, float posY, float posZ, float red, float green, float blue, [float alpha = 255, float minAlpha = 0.0] )
OOP Syntax Help! I don't understand this!
Method: vehicle:setSirens(...)
Counterpart: getVehicleSirens
### Required Arguments
• theVehicle: The vehicle to modify
• sirenPoint: The siren point to modify
• posX: The x position of this siren point from the center of the vehicle
• posY: The y position of this siren point from the center of the vehicle
• posZ: The z position of this siren point from the center of the vehicle
• red: The amount of red from 0 to 255
• green: The amount of green from 0 to 255
• blue: The amount of blue from 0 to 255
### Optional Arguments
• alpha: The alpha of the siren from 0 to 255
• minAlpha: The minimum alpha of the light during day time
### Returns
Returns true if the siren point was successfully changed on the vehicle, false otherwise.
## Example
Click to collapse [-]
Server
This example adds a siren for the vehicle, then sets two custom sirens at the top of the vehicle. And then removes the siren when getting out the vehicle.
if (seat == 0) then
setVehicleSirens(vehicle, 1, -0.3, 0, 0.8, 0, 0, 255, 255, 255)
setVehicleSirens(vehicle, 2, 0.3, 0, 0.8, 255, 0, 0, 255, 255)
-- You can also make the sirens flash immediately with 'setVehicleSirensOn(vehicle, true)'
end
end)
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http://oeis.org/A210773
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crawl-data/CC-MAIN-2020-34/segments/1596439738366.27/warc/CC-MAIN-20200808224308-20200809014308-00435.warc.gz
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A210773 Number of partitions of 2^n into powers of 2 less than or equal to 16. 2
1, 2, 4, 10, 36, 201, 1625, 17361, 222241, 3160641, 47594625, 738433281, 11633144321, 184687354881, 2943499290625, 47004182220801, 751333186150401, 12015464030289921, 192200500444954625, 3074832660977745921, 49194319991205396481, 787085099922532597761 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..250 Index entries for linear recurrences with constant coefficients, signature (31,-310,1240,-1984,1024). FORMULA G.f.: (256*x^8-400*x^7-42*x^6-169*x^5-470*x^4+734*x^3-252*x^2+29*x-1) / Product_{j=0..4} (2^j*x-1). a(n) = [x^2^(n-1)] 1/(1-x) * 1/Product_{j=0..3} (1-x^(2^j)) for n>0. MAPLE a:= n-> `if`(n<4, [1, 2, 4, 10][n+1], (Matrix(5, (i, j)-> `if`(i=j-1, 1, `if`(i=5, [1024, -1984, 1240, -310, 31][j], 0)))^(n-4). <<36, 201, 1625, 17361, 222241>>)[1, 1]): seq(a(n), n=0..30); CROSSREFS Column k=4 of A152977. Sequence in context: A243567 A323949 A066278 * A210774 A210775 A210776 Adjacent sequences: A210770 A210771 A210772 * A210774 A210775 A210776 KEYWORD nonn,easy AUTHOR Alois P. Heinz, Mar 26 2012 STATUS approved
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Last modified August 8 19:29 EDT 2020. Contains 336298 sequences. (Running on oeis4.)
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https://altexploit.wordpress.com/tag/temporal/
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# Matter Fields
In classical relativity theory, one generally takes for granted that all there is, and all that happens, can be described in terms of various “matter fields,” each of which is represented by one or more smooth tensor (or spinor) fields on the spacetime manifold M. The latter are assumed to satisfy particular “field equations” involving the spacetime metric gab.
Associated with each matter field F is a symmetric smooth tensor field Tab characterized by the property that, for all points p in M, and all future-directed, unit timelike vectors ξa at p, Tabξb is the four-momentum density of F at p as determined relative to ξa.
Tab is called the energy-momentum field associated with F. The four- momentum density vector Tabξb at a point can be further decomposed into its temporal and spatial components relative to ξa,
Tabξb = (Tmbξmξba + Tmbhmaξb
where the first term on the RHS is the energy density, while the second term is the three-momentum density. A number of assumptions about matter fields can be captured as constraints on the energy-momentum tensor fields with which they are associated.
Weak Energy Condition (WEC): Given any timelike vector ξa at any point in M, Tabξaξb ≥ 0.
Dominant Energy Condition (DEC): Given any timelike vector ξa at any point in M, Tabξaξb ≥ 0 and Tabξb is timelike or null.
Strengthened Dominant Energy Condition (SDEC): Given any timelike vector ξa at any point in M, Tabξaξb ≥ 0 and, if Tab ≠ 0 there, then Tabξb is timelike.
Conservation Condition (CC): ∇aTab = 0 at all points in M.
The WEC asserts that the energy density of F, as determined by any observer at any point, is non-negative. The DEC adds the requirement that the four-momentum density of F, as determined by any observer at any point, is a future-directed causal (i.e., timelike or null) vector. We can understand this second clause to assert that the energy of F does not propagate with superluminal velocity. The strengthened version of the DEC just changes “causal” to “timelike” in the second clause. It avoids reference to “point particles.” Each of the listed energy conditions is strictly stronger than the ones that precede it.
The CC, finally, asserts that the energy-momentum carried by F is locally conserved. If two or more matter fields are present in the same region of space-time, it need not be the case that each one individually satisfies the condition. Interaction may occur. But it is a fundamental assumption that the composite energy-momentum field formed by taking the sum of the individual ones satisfies it. Energy-momentum can be transferred from one matter field to another, but it cannot be created or destroyed. The stated conditions have a number of consequences that support the interpretations.
A subset S of M is said to be achronal if there do not exist points p and q in S such that p ≪ q. Let γ : I → M be a smooth curve. We say that a point p in M is a future-endpoint of γ if, for all open sets O containing p, there exists an s0 in I such that, ∀ s ∈ I, if s ≥ s0, then γ(s) ∈ O; i.e., γ eventually enters and remains in O. Now let S be an achronal subset of M. The domain of dependence D(S) of S is the set of all points p in M with this property: given any smooth causal curve without (past- or future-) endpoint, if its image contains p, then it intersects S. So, in particular, S ⊆ D(S).
Let S be an achronal subset of M. Further, let Tab be a smooth, symmetric field on M that satisfies both the dominant energy and conservation conditions. Finally, assume Tab = 0 on S. Then Tab = 0 on all of D(S).
The intended interpretation of the proposition is clear. If energy-momentum cannot propagate (locally) outside the null-cone, and if it is conserved, and if it vanishes on S, then it must vanish throughout D(S). After all, how could it “get to” any point in D(S)? According to interpretive principle free massive point particles traverse (images of) timelike geodesics. It turns out that if the energy-momentum content of each body in the sequence satisfies appropriate conditions, then the convergence point will necessarily traverse (the image of) a timelike geodesic.
Let γ: I → M be smooth curve. Suppose that, given any open subset O of M containing γ[I], ∃ a smooth symmetric field Tab on M such that the following conditions hold.
(1) Tab satisfies the SDEC.
(2) Tab satisfies the CC.
(3) Tab = 0 outside of O.
(4) Tab ≠ 0 at some point in O.
Then γ is timelike and can be reparametrized so as to be a geodesic. This might be paraphrased another way. Suppose that for some smooth curve γ , arbitrarily small bodies with energy-momentum satisfying conditions (1) and (2) can contain the image of γ in their worldtubes. Then γ must be a timelike geodesic (up to reparametrization). Bodies here are understood to be “free” if their internal energy-momentum is conserved (by itself). If a body is acted on by a field, it is only the composite energy-momentum of the body and field together that is conserved.
But, this formulation for granted that we can keep the background spacetime metric gab fixed while altering the fields Tab that live on M. This is justifiable only to the extent that we are dealing with test bodies whose effect on the background spacetime structure is negligible.
We have here a precise proposition in the language of matter fields that, at least to some degree, captures the interpretive principle. Similarly, it is possible to capture the behavior of light, wherein the behavior of solutions to Maxwell’s equations in a limiting regime (“the optical limit”) where wavelengths are small. It asserts, in effect, that when one passes to this limit, packets of electromagnetic waves are constrained to move along (images of ) null geodesics.
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# Dynamics of Point Particles: Orthogonality and Proportionality
Let γ be a smooth, future-directed, timelike curve with unit tangent field ξa in our background spacetime (M, gab). We suppose that some massive point particle O has (the image of) this curve as its worldline. Further, let p be a point on the image of γ and let λa be a vector at p. Then there is a natural decomposition of λa into components proportional to, and orthogonal to, ξa:
λa = (λbξba + (λa −(λbξba) —– (1)
Here, the first part of the sum is proportional to ξa, whereas the second one is orthogonal to ξa.
These are standardly interpreted, respectively, as the “temporal” and “spatial” components of λa relative to ξa (or relative to O). In particular, the three-dimensional vector space of vectors at p orthogonal to ξa is interpreted as the “infinitesimal” simultaneity slice of O at p. If we introduce the tangent and orthogonal projection operators
kab = ξa ξb —– (2)
hab = gab − ξa ξb —– (3)
then the decomposition can be expressed in the form
λa = kab λb + hab λb —– (4)
We can think of kab and hab as the relative temporal and spatial metrics determined by ξa. They are symmetric and satisfy
kabkbc = kac —– (5)
habhbc = hac —– (6)
Many standard textbook assertions concerning the kinematics and dynamics of point particles can be recovered using these decomposition formulas. For example, suppose that the worldline of a second particle O′ also passes through p and that its four-velocity at p is ξ′a. (Since ξa and ξ′a are both future-directed, they are co-oriented; i.e., ξa ξ′a > 0.) We compute the speed of O′ as determined by O. To do so, we take the spatial magnitude of ξ′a relative to O and divide by its temporal magnitude relative to O:
v = speed of O′ relative to O = ∥hab ξ′b∥ / ∥kab ξ′b∥ —– (7)
For any vector μa, ∥μa∥ is (μaμa)1/2 if μ is causal, and it is (−μaμa)1/2 otherwise.
We have from equations 2, 3, 5 and 6
∥kab ξ′b∥ = (kab ξ′b kac ξ′c)1/2 = (kbc ξ′bξ′c)1/2 = (ξ′bξb)
and
∥hab ξ′b∥ = (−hab ξ′b hac ξ′c)1/2 = (−hbc ξ′bξ′c)1/2 = ((ξ′bξb)2 − 1)1/2
so
v = ((ξ’bξb)2 − 1)1/2 / (ξ′bξb) < 1 —– (8)
Thus, as measured by O, no massive particle can ever attain the maximal speed 1. We note that equation (8) implies that
(ξ′bξb) = 1/√(1 – v2) —– (9)
It is a basic fact of relativistic life that there is associated with every point particle, at every event on its worldline, a four-momentum (or energy-momentum) vector Pa that is tangent to its worldline there. The length ∥Pa∥ of this vector is what we would otherwise call the mass (or inertial mass or rest mass) of the particle. So, in particular, if Pa is timelike, we can write it in the form Pa =mξa, where m = ∥Pa∥ > 0 and ξa is the four-velocity of the particle. No such decomposition is possible when Pa is null and m = ∥Pa∥ = 0.
Suppose a particle O with positive mass has four-velocity ξa at a point, and another particle O′ has four-momentum Pa there. The latter can either be a particle with positive mass or mass 0. We can recover the usual expressions for the energy and three-momentum of the second particle relative to O if we decompose Pa in terms of ξa. By equations (4) and (2), we have
Pa = (Pbξb) ξa + habPb —– (10)
the first part of the sum is the energy component, while the second is the three-momentum. The energy relative to O is the coefficient in the first term: E = Pbξb. If O′ has positive mass and Pa = mξ′a, this yields, by equation (9),
E = m (ξ′bξb) = m/√(1 − v2) —– (11)
(If we had not chosen units in which c = 1, the numerator in the final expression would have been mc2 and the denominator √(1 − (v2/c2)). The three−momentum relative to O is the second term habPb in the decomposition of Pa, i.e., the component of Pa orthogonal to ξa. It follows from equations (8) and (9) that it has magnitude
p = ∥hab mξ′b∥ = m((ξ′bξb)2 − 1)1/2 = mv/√(1 − v2) —– (12)
Interpretive principle asserts that the worldlines of free particles with positive mass are the images of timelike geodesics. It can be thought of as a relativistic version of Newton’s first law of motion. Now we consider acceleration and a relativistic version of the second law. Once again, let γ : I → M be a smooth, future-directed, timelike curve with unit tangent field ξa. Just as we understand ξa to be the four-velocity field of a massive point particle (that has the image of γ as its worldline), so we understand ξnnξa – the directional derivative of ξa in the direction ξa – to be its four-acceleration field (or just acceleration) field). The four-acceleration vector at any point is orthogonal to ξa. (This is, since ξannξa) = 1/2 ξnnaξa) = 1/2 ξnn (1) = 0). The magnitude ∥ξnnξa∥ of the four-acceleration vector at a point is just what we would otherwise describe as the curvature of γ there. It is a measure of the rate at which γ “changes direction.” (And γ is a geodesic precisely if its curvature vanishes everywhere).
The notion of spacetime acceleration requires attention. Consider an example. Suppose you decide to end it all and jump off the tower. What would your acceleration history be like during your final moments? One is accustomed in such cases to think in terms of acceleration relative to the earth. So one would say that you undergo acceleration between the time of your jump and your calamitous arrival. But on the present account, that description has things backwards. Between jump and arrival, you are not accelerating. You are in a state of free fall and moving (approximately) along a spacetime geodesic. But before the jump, and after the arrival, you are accelerating. The floor of the observation deck, and then later the sidewalk, push you away from a geodesic path. The all-important idea here is that we are incorporating the “gravitational field” into the geometric structure of spacetime, and particles traverse geodesics iff they are acted on by no forces “except gravity.”
The acceleration of our massive point particle – i.e., its deviation from a geodesic trajectory – is determined by the forces acting on it (other than “gravity”). If it has mass m, and if the vector field Fa on I represents the vector sum of the various (non-gravitational) forces acting on it, then the particle’s four-acceleration ξnnξa satisfies
Fa = mξnnξa —– (13)
This is Newton’s second law of motion. Consider an example. Electromagnetic fields are represented by smooth, anti-symmetric fields Fab. If a particle with mass m > 0, charge q, and four-velocity field ξa is present, the force exerted by the field on the particle at a point is given by qFabξb. If we use this expression for the left side of equation (13), we arrive at the Lorentz law of motion for charged particles in the presence of an electromagnetic field:
qFabξb = mξbbξa —– (14)
This equation makes geometric sense. The acceleration field on the right is orthogonal to ξa. But so is the force field on the left, since ξa(Fabξb) = ξaξbFab = ξaξbF(ab), and F(ab) = 0 by the anti-symmetry of Fab.
# Quantum Informational Biochemistry. Thought of the Day 71.0
A natural extension of the information-theoretic Darwinian approach for biological systems is obtained taking into account that biological systems are constituted in their fundamental level by physical systems. Therefore it is through the interaction among physical elementary systems that the biological level is reached after increasing several orders of magnitude the size of the system and only for certain associations of molecules – biochemistry.
In particular, this viewpoint lies in the foundation of the “quantum brain” project established by Hameroff and Penrose (Shadows of the Mind). They tried to lift quantum physical processes associated with microsystems composing the brain to the level of consciousness. Microtubulas were considered as the basic quantum information processors. This project as well the general project of reduction of biology to quantum physics has its strong and weak sides. One of the main problems is that decoherence should quickly wash out the quantum features such as superposition and entanglement. (Hameroff and Penrose would disagree with this statement. They try to develop models of hot and macroscopic brain preserving quantum features of its elementary micro-components.)
However, even if we assume that microscopic quantum physical behavior disappears with increasing size and number of atoms due to decoherence, it seems that the basic quantum features of information processing can survive in macroscopic biological systems (operating on temporal and spatial scales which are essentially different from the scales of the quantum micro-world). The associated information processor for the mesoscopic or macroscopic biological system would be a network of increasing complexity formed by the elementary probabilistic classical Turing machines of the constituents. Such composed network of processors can exhibit special behavioral signatures which are similar to quantum ones. We call such biological systems quantum-like. In the series of works Asano and others (Quantum Adaptivity in Biology From Genetics to Cognition), there was developed an advanced formalism for modeling of behavior of quantum-like systems based on theory of open quantum systems and more general theory of adaptive quantum systems. This formalism is known as quantum bioinformatics.
The present quantum-like model of biological behavior is of the operational type (as well as the standard quantum mechanical model endowed with the Copenhagen interpretation). It cannot explain physical and biological processes behind the quantum-like information processing. Clarification of the origin of quantum-like biological behavior is related, in particular, to understanding of the nature of entanglement and its role in the process of interaction and cooperation in physical and biological systems. Qualitatively the information-theoretic Darwinian approach supplies an interesting possibility of explaining the generation of quantum-like information processors in biological systems. Hence, it can serve as the bio-physical background for quantum bioinformatics. There is an intriguing point in the fact that if the information-theoretic Darwinian approach is right, then it would be possible to produce quantum information from optimal flows of past, present and anticipated classical information in any classical information processor endowed with a complex enough program. Thus the unified evolutionary theory would supply a physical basis to Quantum Information Biology.
# Gothic: Once Again Atheistic Materialism and Hedonistic Flirtations. Drunken Risibility.
The machinery of the Gothic, traditionally relegated to both a formulaic and a sensational aesthetic, gradually evolved into a recyclable set of images, motifs and narrative devices that surpass temporal, spatial and generic categories. From the moment of its appearance the Gothic has been obsessed with presenting itself as an imitation.
Recent literary theory has extensively probed into the power of the Gothic to evade temporal and generic limits and into the aesthetic, narratological and ideological implications this involves. Officially granting the Gothic the elasticity it has always entailed has resulted in a reconfiguration of its spectrum both synchronically – by acknowledging its influence on numerous postmodern fictions – and diachronically – by rescripting, in hindsight, the history of its canon so as to allow space for ambiguous presences.
Both transgressive and hybrid in form and content, the Gothic has been accepted as a malleable genre, flexible enough to create more freely, in Borgesian fashion, its own precursors. The genre flouted what are considered the basic principles of good prose writing: adherence to verisimilitude and avoidance of both narrative diversions and moralising – all of which are, of course, made to be deliberately upset. Many merely cite the epigrammatic power of the essay’s most renowned phrase, that the rise of the Gothic “was the inevitable result of the revolutionary shocks which all of Europe has suffered”.
The eighteenth-century French materialist philosophy purported the displacement of metaphysical investigations into the meaning of life by materialist explorations. Julien Offray de La Mettrie, a French physician and philosopher, the earliest of materialist writers of the Enlightenment, published the materialist manifesto L’ Homme machine (Man a Machine), that did away with the transcendentalism of the soul, banished all supernatural agencies by claiming that mind is as mechanical as matter and equated humans with machines. In his words: “The human body is a machine that winds up its own springs: it is a living image of the perpetual motion”. French materialist thought resulted in the publication of the great 28-volume Encyclopédie, ou Dictionnaire raisonné des sciences, des arts et des méttrie par une société de gens de lettres by Denis Diderot and Jean Le Rond d’ Alembert, and which was grounded on purely materialist principles, against all kinds of metaphysical thinking. Diderot’s atheist materialism set the tone of the Encyclopédie, which, for both editors, was the ideal vehicle […] for reshaping French high culture and attitudes, as well as the perfect instrument with which to insinuate their radical Weltanschauung surreptitiously, using devious procedures, into the main arteries of French Society, embedding their revolutionary philosophic manifesto in a vast compilation ostensibly designed to provide plain information and basic orientation but in fact subtly challenging and transforming attitudes in every respect. While materialist thinkers ultimately disowned La Mettrie because he ran counter to their systematic moral, political and social naturalism, someone like Sade remained deeply influenced and inspired for his indebtedness to La Mettrie’s atheism and hedonism, particularly to the perception of virtue and vice as relative notions − the result of socialisation and at odds with nature.
# Meillassoux’s Principle of Unreason Towards an Intuition of the Absolute In-itself. Note Quote.
The principle of reason such as it appears in philosophy is a principle of contingent reason: not only how philosophical reason concerns difference instead of identity, we but also why the Principle of Sufficient Reason can no longer be understood in terms of absolute necessity. In other words, Deleuze disconnects the Principle of Sufficient Reason from the ontotheological tradition no less than from its Heideggerian deconstruction. What remains then of Meillassoux’s criticism in After finitude: An Essay on the Necessity of Contigency that Deleuze no less than Hegel hypostatizes or absolutizes the correlation between thinking and being and thus brings back a vitalist version of speculative idealism through the back door?
At stake in Meillassoux’s criticism of the Principle of Sufficient Reason is a double problem: the conditions of possibility of thinking and knowing an absolute and subsequently the conditions of possibility of rational ideology critique. The first problem is primarily epistemological: how can philosophy justify scientific knowledge claims about a reality that is anterior to our relation to it and that is hence not given in the transcendental object of possible experience (the arche-fossil )? This is a problem for all post-Kantian epistemologies that hold that we can only ever know the correlate of being and thought. Instead of confronting this weak correlationist position head on, however, Meillassoux seeks a solution in the even stronger correlationist position that denies not only the knowability of the in itself, but also its very thinkability or imaginability. Simplified: if strong correlationists such as Heidegger or Wittgenstein insist on the historicity or facticity (non-necessity) of the correlation of reason and ground in order to demonstrate the impossibility of thought’s self-absolutization, then the very force of their argument, if it is not to contradict itself, implies more than they are willing to accept: the necessity of the contingency of the transcendental structure of the for itself. As a consequence, correlationism is incapable of demonstrating itself to be necessary. This is what Meillassoux calls the principle of factiality or the principle of unreason. It says that it is possible to think of two things that exist independently of thought’s relation to it: contingency as such and the principle of non-contradiction. The principle of unreason thus enables the intellectual intuition of something that is absolutely in itself, namely the absolute impossibility of a necessary being. And this in turn implies the real possibility of the completely random and unpredictable transformation of all things from one moment to the next. Logically speaking, the absolute is thus a hyperchaos or something akin to Time in which nothing is impossible, except it be necessary beings or necessary temporal experiences such as the laws of physics.
There is, moreover, nothing mysterious about this chaos. Contingency, and Meillassoux consistently refers to this as Hume’s discovery, is a purely logical and rational necessity, since without the principle of non-contradiction not even the principle of factiality would be absolute. It is thus a rational necessity that puts the Principle of Sufficient Reason out of action, since it would be irrational to claim that it is a real necessity as everything that is is devoid of any reason to be as it is. This leads Meillassoux to the surprising conclusion that [t]he Principle of Sufficient Reason is thus another name for the irrational… The refusal of the Principle of Sufficient Reason is not the refusal of reason, but the discovery of the power of chaos harboured by its fundamental principle (non-contradiction). (Meillassoux 2007: 61) The principle of factiality thus legitimates or founds the rationalist requirement that reality be perfectly amenable to conceptual comprehension at the same time that it opens up [a] world emancipated from the Principle of Sufficient Reason (Meillassoux) but founded only on that of non-contradiction.
This emancipation brings us to the practical problem Meillassoux tries to solve, namely the possibility of ideology critique. Correlationism is essentially a discourse on the limits of thought for which the deabsolutization of the Principle of Sufficient Reason marks reason’s discovery of its own essential inability to uncover an absolute. Thus if the Galilean-Copernican revolution of modern science meant the paradoxical unveiling of thought’s capacity to think what there is regardless of whether thought exists or not, then Kant’s correlationist version of the Copernican revolution was in fact a Ptolemaic counterrevolution. Since Kant and even more since Heidegger, philosophy has been adverse precisely to the speculative import of modern science as a formal, mathematical knowledge of nature. Its unintended consequence is therefore that questions of ultimate reasons have been dislocated from the domain of metaphysics into that of non-rational, fideist discourse. Philosophy has thus made the contemporary end of metaphysics complicit with the religious belief in the Principle of Sufficient Reason beyond its very thinkability. Whence Meillassoux’s counter-intuitive conclusion that the refusal of the Principle of Sufficient Reason furnishes the minimal condition for every critique of ideology, insofar as ideology cannot be identified with just any variety of deceptive representation, but is rather any form of pseudo-rationality whose aim is to establish that what exists as a matter of fact exists necessarily. In this way a speculative critique pushes skeptical rationalism’s relinquishment of the Principle of Sufficient Reason to the point where it affirms that there is nothing beneath or beyond the manifest gratuitousness of the given nothing, but the limitless and lawless power of its destruction, emergence, or persistence. Such an absolutizing even though no longer absolutist approach would be the minimal condition for every critique of ideology: to reject dogmatic metaphysics means to reject all real necessity, and a fortiori to reject the Principle of Sufficient Reason, as well as the ontological argument.
On the one hand, Deleuze’s criticism of Heidegger bears many similarities to that of Meillassoux when he redefines the Principle of Sufficient Reason in terms of contingent reason or with Nietzsche and Mallarmé: nothing rather than something such that whatever exists is a fiat in itself. His Principle of Sufficient Reason is the plastic, anarchic and nomadic principle of a superior or transcendental empiricism that teaches us a strange reason, that of the multiple, chaos and difference. On the other hand, however, the fact that Deleuze still speaks of reason should make us wary. For whereas Deleuze seeks to reunite chaotic being with systematic thought, Meillassoux revives the classical opposition between empiricism and rationalism precisely in order to attack the pre-Kantian, absolute validity of the Principle of Sufficient Reason. His argument implies a return to a non-correlationist version of Kantianism insofar as it relies on the gap between being and thought and thus upon a logic of representation that renders Deleuze’s Principle of Sufficient Reason unrecognizable, either through a concept of time, or through materialism.
# Solitude: Thought of the Day 18.0
A reason Nietzsche ponders solitude is that his is largely a philosophy of the future. There is heavy emphasis in Beyond Good and Evil on the temporal nature of the human condition. He posits that “the taste of the time and the virtue of the time weakens and thins down the will.” In order to surpass current modes and fashions in thinking, one must become removed from the present. The new philosopher is necessarily a man of tomorrow and the day after tomorrow and so he is solitary and in contradiction to the ideals of today. Fundamentally, Nietzsche sees current Europe (and especially Germany) as not yet prepared for an overturning of present morality. Although he does predict the time is approaching, there is the overarching sense throughout Beyond Good and Evil that Nietzsche expects (and even embraces) the fact that his philosophy needs a significant passage of time to be understood. His work is lonely. He labors to lay groundwork for the philosophers of the future who will continue on this path someday.
The life of the free spirit is solitary because it requires the recognition of the untruth of life in order to be beyond good and evil. Religion and democratic enlightenment in Europe have forged a herd mentality of mediocrity which has rejected such a possibility. In this society, everyone’s thoughts and morality are given equal merit. Nietzsche despises this because it forces us to reject our nature; both the ugliness and the beauty of it. He tells us that religion is able to teach even the lowliest of people how to place themselves in an illusory higher order of things so they may have the impression that they are content. This herd mentality protects the pack and also makes life palatable. It is also the first enemy of anyone looking to discover their own truths. Nietzsche concludes his book by reflecting on the wonders of solitude. For the free spirit, solitude is life-affirming because the absence of the stifling dogmas of the herd allows for the greatest expansion of one’s sense of self. To be truly beyond good and evil one must be removed from grappling with the order and morality imposed by democratic enlightenment and religion. Only when one stands alone vis-à-vis the herd is greatness and nobleness possible. Upon being removed from the seething torrent of austere and rigid thinking now strangling Europe, the free spirit foments his own morality and thrives.
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# Express 2x^2-28x+53
• May 16th 2011, 02:32 AM
mike789
Express 2x^2-28x+53
Express 2x^2-20x+53 in the form 2(x-p)^2+q , where p and q are integers. not sure how to do this..
i factorised it and got this
2(x^2-10x)+53 plz help thx
• May 16th 2011, 02:45 AM
CaptainBlack
Quote:
Originally Posted by mike789
Express 2x^2-20x+53 in the form 2(x-p)^2+q , where p and q are integers. not sure how to do this..
i factorised it and got this
2(x^2-10x)+53 plz help thx
If:
$\displaystyle 2(x-p)^2+q=2x^2-4px+2p^2+q=2x^2-20x+53$
then $\displaystyle 4p=20$, so $\displaystyle p=5$, and $\displaystyle 2p^2+q=53$ so $\displaystyle q= .... ?$
CB
• May 16th 2011, 04:03 AM
TheCoffeeMachine
Quote:
Originally Posted by mike789
i factorised it and got this
2(x^2-10x)+53 plz help thx
Good start! Now consider (x-5)² = x²-10x+25 which gives x²-10x = (x-5)²-25.
• May 16th 2011, 04:07 AM
jgv115
This question is essentially asking you to express that quadratic in "turning point form". This is done by completing the square.
Alternatively, you can "compare coefficients" described in CaptainBlack's post
• May 16th 2011, 04:19 AM
Romek
$\displaystyle 2x^2 - 20x + 53$
$\displaystyle \equiv 2(x^2 - 10x) + 53$ // Take the 2 outside of the brackets
$\displaystyle \equiv 2((x-5)^2 - 25) + 53$ // Complete the square within the brackets
$\displaystyle \equiv 2(x-5)^2 - 50 + 53$ // Take the -25 outside the brackets
$\displaystyle \equiv 2(x-5)^2 + 3$ // Gather terms
• May 16th 2011, 04:44 AM
CaptainBlack
Quote:
Originally Posted by jgv115
This question is essentially asking you to express that quadratic in "turning point form". This is done by completing the square.
Alternatively, you can "compare coefficients" described in CaptainBlack's post
What I posted is completing the square, but done from first principles rather than remembering the algorithm, which I don't as it is a waste of brain space (same goes for a number of other things taught in common algebra and starting calculus, they are quicker to re-derive than to remember).
CB
• May 16th 2011, 05:08 AM
Ackbeet
Quote:
Originally Posted by CaptainBlack
What I posted is completing the square, but done from first principles rather than remembering the algorithm, which I don't as it is a waste of brain space (same goes for a number of other things taught in common algebra and starting calculus, they are quicker to re-derive than to remember).
CB
Yes, yes. To CB you listen!
• May 16th 2011, 05:27 AM
TheCoffeeMachine
Quote:
Originally Posted by CaptainBlack
What I posted is completing the square, but done from first principles rather than remembering the algorithm, which I don't as it is a waste of brain space (same goes for a number of other things taught in common algebra and starting calculus, they are quicker to re-derive than to remember).
If you are referring to remembering the formula for the coordinates of the turning point, I agree. But what I was hinting at post #3 doesn't really require remembering stuff at all.
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# The Book of Mormon II
June 13, 2020 ☕️ 5 min read
The full version of the problem you can find in part I of the article.
## Recap Challenge
1. We want to read the book in a given number of days: 128.
2. We want to read an integer number of chapters each day (there are more chapters than days), and at least 1 chapter each day.
3. The chapters are very non uniform in length (some very short, a few very long, many in between) so we would like to come up with a reading schedule that minimizes the variance of the length of the days readings (read multiple short chapters on the same day, long chapters are the only one read that day).
4. We want to read through the book in order (no skipping ahead to combine short chapters that are not naturally next to each other)1.
### Objective:
We minimize the variance of the number of verses read per day.
## Shortest Path Approach
Interesting enough this problem can be modelled as a directed network graph:
• chapters form the network nodes.
• edges represent transitions from chapter $i$ to chapter $j$, i.e. they correspond reading all chapters from $i$ to $j-1$.
• the graph is directed, so transitions are only allowed from $i \rightarrow j, \ j\ge i$.
• the edges cost is the proportional to the difference from the actual sum of verses per transition from the average.
• to apply flow modelling techniques we need to add a dummy node as network sink.
• $ch1$ acts as network source.
### Model
Calculate the collection $A$ of edges $(i,j)$ which represent all possible transitions in the network.
$A \ \text{ set of edges}\\ (i,j) \in A | j \ge i\\$
This is analog to calculating the covering sets in part 1. Since the graph is directed the semantic meaning of the edge definition guarantees that every chapter is being read exacty once.
The same optimization as in the set covering approach can be applied in order to reduce the number of allowed network edges. The maximum number of chapters to be read in one day is:
$limit = n-T, \\ i, j: 1,..,limit\\ j \ge i\\$
Any ‘longer’ edges consisting of more chapters would make the number of remaining chapters to be too few to have at least one chapter per remaining day.
#### Variable
Every edge $(i,j)$ has got a flow $x_{i,j}$ and a cost $c_{i,j}$ associated with it. Since we need to put a constraint on the number of nodes in the solution we introduce a helper variable $y_i$:
$x_{i,j}, \ \text{ flow on edge (i,j)}\\ y_i, \ \text{ total flow leaving node i for consumption by other nodes}\\ x_{i,j}, y_i \ge 0\\$
#### Objective
$\min \sum_{(i,j)\in A} c_{i,j} x_{i,j}\\$
#### Parameter:
Node Distance:
The ‘distance’ or ‘cost coefficient’ for flow $x$ is calculated as the variance of the sum of verses of selected chapters from the average number of verses per day $\mu$.
$c_{i,j} = (\sum_{k=i}^{j-1} v_k - \mu)^2\\$
Source and sink flow:
$b_i = \begin{cases} 1, \ \text{ if i=\mathit{ch1}}\\ -1 \ \text{ if i=\mathit{sink}}\\ 0 \ \text{ otherwise} \end{cases}$
#### Constraints
1. classic flow balance constraints
2. enforce the number of nodes being part of the solution, i.e. reading days $T$
Since the total number of reading days $T$ is a parameter, we need to enforce at least $T$ nodes being part of the shortest path’ solution. Therefore we split the classic node balance equation:
$\sum_{(j,i)\in A} x_{j,i} + b_i = y_i \ \forall i\\ y_i = \sum_{(i,j)\in A} x_{i,j} \ \forall i\\ \sum_i y_i = T \\ x_{i,j}, y_i \ge 0$
### Implementation
Since Pyomo allows to use regular Python programming language constructs like if..else, the formulation of boundary conditions which involves trailing or leading time intervals feels very natural for a software developer.
Again, the implementation of the constraints is straight forward:
def flow_balance_in_c(model, ii):
if ii == 1: # source/chap1
b = 1
elif ii == self.N: # sink/dummy chapter
b = -1
else:
b = 0
return sum(model.x[j, i] for (j, i) in model.A if i == ii) + b == model.y[ii]
model.flow_balance_in_c = Constraint(
model.I, rule=flow_balance_in_c
)
model.flow_balance_out_c = Constraint(
model.I,
rule=lambda model, ii: sum(model.x[i, j] for (i, j) in model.A if i == ii) == model.y[ii]
)
model.y_c = Constraint(
rule=lambda model: sum(model.y[i] for i in model.I) == self.T
)
### Result
#### 40 Chapters
• Reading time in days: 8
• Number of chapters: 40, max number of chapters per day: 34
• Number of verses: 1059.0
• Average number of verses per day to be read: 132.38
The CBC solver has no problem solving the model and provides an optimal solution in a split second.
Number of constraints : 83
Number of variables : 833
The sum of deviations from the average number of verses per day is 47. We need to read 122 verses on day 8 (minimum) and 143 verses on day 4 (maximum) in order to have the smoothest reading experience.
#### All Chapters
Reading time in days: 128
Number of chapters: 240, max number of chapters per day: 113
Number of verses: 6603.0
Average number of verses per day to be read: 51.59
Created sets: 20552, verses: 20552
In contrast to the model from the partition set approach model building takes only a few seconds.
Solution [999.828125, 999.828125]
Number of constraints : 481
Number of variables : 20792
Duration: 00:00:01
For more details regarding the result please refer to part 1. The results are identical (as to be expected).
## Summary
Framing the problem as a network model might be not your first choice of thinking. It certainly wasn’t mine.
Choosing the correct abstraction results in dramatic solution performance improvements. The network model seems to be the best approach by far for this problem. Good to have it in our tool belt now.
always be building, by sysid.
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# Book Reviews
Discussion in 'General Electronics Discussion' started by (*steve*), Sep 2, 2012.
1. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator
25,500
2,840
Jan 21, 2010
OK, this is a trial. If you feel there is a really good (or really bad) book that you've read, write a review here.
I will probably incorporate particularly good reviews into this post.
Please also feel free to discuss books in this thread.
I'll start it off.
The Art of Electronics (Horowitz & Hill)
This is my favourite text. It is very readable and requires very limited understanding of higher mathematics.
It is not for the beginner, I'd recommend that you have a pretty good working knowledge of what basic components do before reading this.
The first chapter covers a number of mathematical concepts that are very useful, however they are not required (other than in their most basic form) to understand the rest of the material presented in the book.
2. ### CiaranM
74
1
May 19, 2012
hello! I have the Student Manual version of that book. Its a good book, very informative, and has hand-drawn pictures in it which look good. It also has some humour, e.g. I read a section about op amps, and the authors were talking about dogs being hidden from op amp outputs.
I have several books from Bernard Babani Publishing. You got any? They're inexpensive, and contain great info. Here are a few:
- Practical Electronics Calculations and Formulae: A very useful book which tells you all sorts of equations, stuff about components, S.I. units, how to calculate things, etc.
- Practical Oscillator Circuits: This books shows you a variety of oscillator circuits using ICs such as the 555, 556, TL072 etc. There are also LC circuits. The circuits are explained and they work great. The book is by Andy Flind; anyone know anything about him?
- Practical Electronic Filters: This book explains how filtering works and how to devise filtering circuits. Thanks to this book, I understand that a (passive?) filter is a voltage divider,and I was able to see how placing different components in different parts of a circuit could provide a different response.
There are several books by a guy named R.M. Marston. One of his books I have is Timer/Generator Circuits Manual. There are plenty of interesting circuits to try out. Explanations are provided.
Another book I have is Electronics Calculations Data Handbook. This book features all sorts of tables with values for resistor power ratings, voltage dividers, etc. There are also explanations for circuits and various topics.
3. ### hca
15
1
Aug 28, 2012
I recommend using:
Electrical Engineering: Principles and Applications by Allan R. Hambley (2011)
You can see it here:
http://www.scribd.com/doc/34151864/Electrical-Engineering-Principles-and-Applications
It covers the following topics including many clear worked examples:
Chapters 1-6: Basic Circuit Analysis
ch.1 Introduction
ch. 2 Resistive Circuits
ch. 3 Inductance and Capacitance
ch. 4 Transients
ch. 5 Steady State Sinusoidal Analysis
ch. 6 Frequency Response, Bode Plots and Resonance
Chapters 7-9 Digital Systems
ch. 7 Logic Circuits
ch. 8 Microcomputers
ch. 9 Computer Based Instrumentation Systems
Chapters 10-14 Electronic Devices and Circuits
ch.10 Diodes
ch.11 Amplifiers
ch.12 Field Effect Transistors
ch. 13 Bipolar Junction Transistors
ch.14 Operational Amplifiers
Chapters 15 -17Electromechanics
ch.15 Magnetic Circuits and Transformers
ch. 16 DC Machines
ch.17 AC Machines
The book also has a very good companion website with video solutions.
PDF files of solutions to in chapter questions are available on the companion website (you need to purchase the book to have access to the companion website)
Prerequisites :
The book recommends an understanding of basic physics and single variable calculus.
A very good first text book.
adir figueiredo likes this.
4. ### CiaranM
74
1
May 19, 2012
Yacine Mou likes this.
5. ### Rusty
16
1
Nov 30, 2012
Electric Circuits
by Nilsson, Riedel:
Very nice book for the beginners. Starting out with ohms law, ending with basic op amp circuits, frequency respons and natural/step respons.
Audio Power Amplifiers
by Bob Cordell:
All about power amplifiers. Very easy to read book with lots of practical information about different output stages examples and explanaitions, but not so much theory.
Audio Power Amplifier
by Douglas Self
A little bit more theoretical than Cordell, but still its expected that you know some circuit theory if you want to fully understand it. I prefer this book a bit more than the Cordell. Also easy to read, lots of practical information.
Small Signal Audio Design
by Douglas Self
This is a really really nice book for every audio insterested people. Easy to read and lots of information about different transistor and op amp circuits. All about noise, gain, stability and lots of circuit examples for different use. I strongly recommend this book. Douglas Self knows his stuff.
Design with Operational Amplifiers and Analog Integrated Circuits
by Sergio Franco
This is a good book for instrumentation circuits. All about noise, impedances, stability and frequency response and many different op amp circuits. A good mix of theory and practical information. Not for starters, but if you know basic electronics i strongly recommend this book!
Microelectronic Circuits
by Sedra/Smith
One of my favorite books about basic op amp, diode and transistor circuits. You need to know some basic electronics and frequency respons in advance, but otherwise this book begin with easy circuits and end with a bit more complex designs. Very easy to read, lots of examples, lots of tasks to solve and all in all a really good book.
Digital Design, Applications and VHDL
by Dueck
An easy to read book if you want to learn all about digital circuits. Begins with a beginner level and gets more advanced later in the book. A nice introduction to the digital world.
Hope this was helpfull!
Last edited: Nov 30, 2012
teprojects1 likes this.
6. ### egd_electronic_advisor
1
0
Dec 15, 2012
Good Electronics Books ...
Yes ... there are several of them ...
The BOOK presented here (Horowitz) is an ancient and almost dogmatic book .... Is the BEST one as a reference ...
Even there is several good one ... the most complete (and practical) textbook is
1. Electronic Devices and Circuits (Bogart, Beasley, Rico), almost engineering level
2. Electronic Devices (Conventional Current Version; Floyd) technology and practical level
There is s lot more in the market .... just forget about MAURO (Engineering level) ... one of the worst
EGD
17
0
Jan 27, 2013
8. ### izdane
5
0
Mar 1, 2013
Make: electronics
This is a great book for beginners because it assumes you know nothing
9. ### Proschuno
87
1
Aug 1, 2011
Electronics for Inventors, Paul Scherzy
Not the best for extreme newbies, but will do if you have a decent grasp of algebra, but If you know the higher mathematics he also goes really deep into theory which uses that, so there's a little something for everyone.
http://www.barnesandnoble.com/w/pra...=pla&ean=9780071452816&isbn=9780071452816&r=1
Ch. 1 Introduction
2. Theory
3. Electrical components
4. Semiconductors
5. Optoelectronics
6. Integrated Circuits
7. Op amps
8. Filters
9. Oscillators & timers
10. Power supplies
11. Audio electronics
12. Digital
13. Motors
14. Hands on electronics
Milon likes this.
10. ### NuLED
294
0
Jan 7, 2012
There is a new 3rd edition of the Scherzy text. I will get that later once I am done with the basics, but FYI, in case anyone is about to buy the 2nd edition.
11. ### Rob_K
59
0
Sep 20, 2013
Wow, this fellow was my electronics lecturer 3 years ago during my Mechatronics degree, he is honestly one of the best teachers I have ever had for anything and his book does demonstrate that also.
12. ### duke37
5,364
772
Jan 9, 2011
The Technique of Radio Design by E.E.Zeplar 1943
A classical book for those who wish to know how things were before semiconductors.
13. ### duke37
5,364
772
Jan 9, 2011
Electronic Filter Design by Arthur B. Williams 1981
I think there is a later edition.
Designs of low pass, band pass and high pass filters, passive and active.
Various filter types, Butterworth, Chebyshev, Bessel, Linear Phase,Transitional Gaussian usually up to seven pole.
14. ### ScienceBorn
55
0
Sep 26, 2013
I have to recommend you guys to read "electronics part and its operation", though i authorised the book , it is very good and fascinating for newbies!
15. ### mahone
25
0
Dec 21, 2013
book recommendations
I am always looking for something new to read, so I thought why not start a thread of book recommendations. It may even serve to help somebody out. to start off, the best books I have read are:
electrical circuit theory and technology by john bird.
practical electronics for inventors by paul scherz.
electrical engineering know it all by clive maxfield.
16. ### Supercap2F
550
150
Mar 22, 2014
CMOS Cookbook by Don Lancaster
I really like how Don Lancaster wrote this book! He put very little math in it. It does need some basic understanding of electronics. In the middle of it he put datasheets for CMOS ICs. I have talked with him via email and he seemed nice. In the back it has a TTL to CMOS conversion chart!
OP-AMP Circuits and Principles by Howard M Berlin
The book is from the 90s so it’s kind of dated but is a very good reference book. I have learned a lot on op-amps from reading it. In the back it has datasheets for some op-amps and it also has about 14 experiments. It does need some higher math skills.
Dan
paul taylor likes this.
17. ### Harald KappModeratorModerator
11,826
2,757
Nov 17, 2011
I recently laid my hands on
"The circuit designer's companion"
(Peter Wilson. Newnes, ISBN: 978-0-08-097138-4).
I find this a very comprehensive work for doing actual designs.
It requires that the reader already has a good working knowledge of electronics theory. It supports the conversion from theory to practice.
paul taylor likes this.
18. ### bobbyrae
5
1
May 11, 2014
paul taylor likes this.
11
3
Jun 5, 2014
20. ### dust collector emitter
2
2
Feb 13, 2015
Getting started in electronics was my first book too. I bought an \$80 kit from radioshack about ten years ago that came with the book. It was a solderless breadboard with an ohmeter, phototransistor, and other devices with springs for connections that would hold the wires.
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posted by .
1)A runner's finishing time translated into a z-score of z=2.6
Explain in words whether the runner performed excellently or poorly?
2)Yearly stock returns on India's Sensex Index are well described by a Normal model with a mean of 18.36% and a standard deviation of 14.65%. give the stock return cutoff for the middle 30% of all years.
z-scores mainly range between -3 and 3
z = 2.6 is near the upper end.
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Tries with Compression
This is the task corresponding to exercise 3. Tries with Compression.
Resources
Definitions File
```theory Defs
imports Main
begin
declare [[names_short]]
datatype bt = Lf bool | Nd bt bt
fun \<alpha> :: "bt \<Rightarrow> nat \<Rightarrow> bool list set" where
"\<alpha> (Lf b) d = (if b then {bs. length bs = d} else {})" |
"\<alpha> (Nd l r) d = Cons True ` \<alpha> l (d-1) \<union> Cons False ` \<alpha> r (d-1)"
fun ht :: "bt \<Rightarrow> nat" where
"ht (Lf _) = 0" |
"ht (Nd l r) = max (ht l) (ht r) + 1"
end```
Template File
```theory Submission
imports Defs
begin
fun isin :: "bt \<Rightarrow> bool list \<Rightarrow> bool" where
"isin _ = undefined"
fun inter :: "bt \<Rightarrow> bt \<Rightarrow> bt" where
"inter _ = undefined"
fun compressed :: "bt \<Rightarrow> bool" where
"compressed _ \<longleftrightarrow> undefined"
fun compress :: "bt \<Rightarrow> bt" where
"compress _ = undefined"
end```
Check File
```theory Check
imports Submission
begin
end```
Terms and Conditions
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# Teaspoons to Cubic yards conversion
Result
yd3
Click Result to Copy
tsp=yd3
How did we calculate tsp?
To Calculate, we took the value you submited and divided it by 155,116.054377 to get the result.
(/155,116.054377)
Our Teaspoons to Cubic yards conversion tool is a free converter that enables you to convert from Teaspoons to Cubic yard easily.
## How to Convert Teaspoons to Cubic yard
To convert a Teaspoons volume to a cubic yard volume, divide the volume by the conversion ratio. Since 1 cubic yard is equal to 155,116.054377 tablespoons, you can use this simple formula to convert:
### What is the formula to convert from Teaspoons to Cubic yard?
yd3=tsp / 155,116.054377
### Convert 5tsp to cubic yards
5 tsp = (5 / 155,116.054377) = 0.00003223 yd3
### Convert 10tsp to cubic yards
10 tsp = (10 / 155,116.054377) = 0.00006447 yd3
### Convert 100tsp to cubic yards
100 tsp = (100 / 155,116.054377) = 0.00064468 yd3
## Teaspoon
### What is a Teaspoon?
A teaspoon (symbol: tsp) is a unit of volume based on an item of cutlery. The United States customary teaspoon is equal to exactly 4.928922 mL. The metric teaspoon is equal to 5 mL.
### What is the Teaspoon used for?
The teaspoon is widely used in cooking within certain countries, as well as for measuring pharmaceutic prescriptions. Outside of these applications, the unit is not particularly used, with measurements such as the liter or cubic meter being preferred.
## Cubic yard
### What is a Cubic yard?
A cubic yard (symbol: yd3) is an imperial and United States Customary unit of volume defined as the volume of a cube with measurements 1 yd × 1 yd × 1 yd. It is equal to 27 cubic feet, 0.7645549 cubic meters, and 764.5549 liters.
### What is the Cubic yard used for?
The cubic yard is used to some degree in the United States, the United Kingdom, and Canada. All of these countries also use metric or SI (International System of Units) measurements for volume such as liters, milliliters, and cubic meters.
## How to use our Teaspoons to Cubic yard converter
Follow these 3 simple steps to use our Teaspoons to Cubic yard converter
1. Input the unit of Teaspoons you wish to convert
2. Click on convert and watch this result display in the box below it
3. Click Reset to reset the tablespoon value
## Teaspoons to Cubic yard Conversion Table
teaspoonscubic yards
tsp yd3
| 644
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A286763 Numbers that appear in A195441 at least once for two consecutive indices. 5
1, 30, 210, 330, 2310, 3990, 6090, 14790, 43890, 66990, 82110, 125970, 144210, 181830, 881790, 1009470, 1067430, 1217370, 2284590, 2381190, 17687670, 18888870, 26265030, 35068110, 39544890, 47763870, 115223790, 127652070, 406816410, 497668710, 741110370, 1024748670 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The sequence is infinite; see Cor. 3 in "The denominators of power sums of arithmetic progressions". - Bernd C. Kellner and Jonathan Sondow, May 24 2017 LINKS Table of n, a(n) for n=1..32. Bernd C. Kellner, On a product of certain primes, J. Number Theory 179 (2017), 126-141; arXiv:1705.04303 [math.NT], 2017. Bernd C. Kellner and Jonathan Sondow, Power-Sum Denominators, Amer. Math. Monthly 124 (2017), 695-709; arXiv:1705.03857 [math.NT], 2017. Bernd C. Kellner and Jonathan Sondow, The denominators of power sums of arithmetic progressions, Integers 18 (2018), Article #A95, 17 pp.; arXiv:1705.05331 [math.NT], 2017. EXAMPLE A195441(21) = A195441(22) = 30, so 30 is in the sequence. - Jonathan Sondow, Dec 11 2018 MATHEMATICA Take[#, 32] &@ Union@ SequenceCases[ Table[ Denominator[ Together[ (BernoulliB[n + 1, x] - BernoulliB[n + 1])]], {n, 0, 2000}], w_ /; And[SameQ @@ w, Length@ w >= 2]][[All, 1]] (* Michael De Vlieger, Sep 22 2017, after Jonathan Sondow at A195441 *) PROG (Julia) function A286763_search() A = fmpz[]; a = fmpz(0) for n in 0:10000 u = A195441(n) a == u && push!(A, a) a = u end S = sort([a for a in Set(A)]) S[1:32] end println(A286763_search()) CROSSREFS Cf. A195441, A286516, A286762. Sequence in context: A181629 A346245 A129499 * A347828 A069965 A215627 Adjacent sequences: A286760 A286761 A286762 * A286764 A286765 A286766 KEYWORD nonn,changed AUTHOR Peter Luschny, May 14 2017 STATUS approved
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Last modified September 15 13:51 EDT 2024. Contains 375938 sequences. (Running on oeis4.)
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1
MHT CET 2021 24th September Morning Shift
+1
-0
A beam of light having wavelength $$5400 \mathrm{~A}$$ from a distant source falls on a single slit $$0.96 \mathrm{~mm}$$ wide and the resultant diffraction pattern is observed on a screen $$2 \mathrm{~m}$$ away. What is the distance between the first dark fringe on either side of central bright fringe?
A
4.8 mm
B
1.2 mm
C
2.4 mm
D
3.6 mm
2
MHT CET 2021 24th September Morning Shift
+1
-0
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is $$\pi / 2$$ at point $$\mathrm{A}$$ and $$\pi$$ at point $$\mathrm{B}$$. Then the difference between the resultant intensities at $$\mathrm{A}$$ and $$\mathrm{B}$$ is
A
4I
B
5I
C
2I
D
3I
3
MHT CET 2021 23rd September Evening Shift
+1
-0
In Young's double slit experiment, the intensity at a point where path difference is $$\frac{\lambda}{6}$$ ($$\lambda$$ being the wavelength of light used) is $$I^{\prime}$$. If '$$I_0$$' denotes the maximum intensity, then $$\frac{I}{I_0}$$ is equal to $$\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{\lambda}\right)$$
A
$$\frac{\sqrt{3}}{2}$$
B
$$\frac{4}{3}$$
C
$$\frac{1}{\sqrt{2}}$$
D
$$\frac{1}{2}$$
4
MHT CET 2021 23rd September Evening Shift
+1
-0
In Young's double slit experiment, the distance of $$\mathrm{n}^{\text {th }}$$ dark band from the central bright band in terms of bandwidth '$$\beta$$' is
A
$$\mathrm{n} \beta$$
B
$$(\mathrm{n}-1) \beta$$
C
$$(\mathrm{n}-0.5) \beta$$
D
$$(\mathrm{n}+0.5) \beta$$
EXAM MAP
Medical
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# [Solved] Array-sensor Homework06
Click And Check all our Assignments
10 USD \$
Category: Tag:
1. A stationary signal x[n], that is of zero mean, unity power and temporally uncorrelated,
E[ ,
is passed through a channel with time-discrete impulse response h[k], such that the noiseless received signal u[n] reads as
Lh−1
u[n] = X h[k] · x[n k] ,
k=0
where Lh is the length of the channel’s impulse response. For Lh > 1 the channel will introduce intersymbol interference (ISI), that the receiver wishes to remove by means of a linear minimum mean square error (LMMSE) adaptive equalizer that produces the output y[n]
,
where M is the filter length and w[0],w[1],…,w[M − 1] are the filter coefficients.
• Express the filter output y[n] in terms of the two vectors w = [w[0] w[1] … w[M − 1]]T and u[n] = [u[n] u[n − 1] … u[n M + 1]]T.
• The desired output d[n] of the adaptive filter is chosen to be
d[n] = x[n M + 1] ,
in order to make maximum use of the filter memory. Compute the Wiener solution in terms of a correlation matrix R and a correlation vector p.
• Now compute the correlation coefficients r[m] = E[u[n] · u[n m]] and express the correlation matrix R and correlation vector p from subtask 1b as a function of the channel impulse response h[k].
• For the case M = Lh = 2, with h[0] = a ∈ C, h[1] = b ∈ C and zero elsewhere, compute the Wiener solution for w explicitly in terms of a and b. Now set b = 0 and compute the current error e[n] = d[n] − y[n].
• Now set the filter length to M = 6, and use the channel from subtask 1d to compute the Wiener solution for a = 4 and b = −5 by using matlab. How large is the residual mean square error (MSE)?
• Write a matlab program that will plot the square root of the residual mean squared error as a function of the filter length M in the range 2 ≤ M ≤ 50 in steps of one. Note: Use the semilogy function for the plotting.
2
Figure 1: Frame structure of the transmitted data stream
1. The adaptive equalizer from task 1 shall now be implemented using the LMS algorithm. To this end assume the transmitted data stream is structured in frames as depicted in Figure 1. Each frame starts with a training sequence (pilot) of length LP symbols, which is followed by LD symbols of actual data. At the end of each frame there is a so called guard interval where the transmitter is quiet (transmitting zeros) for LG symbols. In the following assume LG = Lh + M. This will let the programmable receiver FIR-filter start in the zero state for each frame, i.e. the simulation can be carried out on a frame by frame basis. The symbols are chosen out of the set {−1,+1}, i.e. BPSK modulated. The pilot sequence used has length LP = 18, and is defined as:
−1 −1 −1 +1 +1 −1 −1 −1 +1 +1 +1 +1 −1 +1 +1 −1 +1 −1
where the sequence is transmitted going from left to right. In the following assume LD = 150, M = 6, Lh = 2, LG = 8 and the channel coefficients h[0] = 4 and h[1] = −5 as in subtask 1e.
• Write a matlab program that will simulate the advances of LMS adaptive equalization by evaluating thenumber of bit-errors in 30 successive frames. Note, that the weight update by LMS is only performed during the pilot phases. In each frame initialize the LMS with the weight vector obtained from the previous frame. For the very first frame use the all zero vector as the initial weight vector. Set the step size to 0.0015 for the LMS. When computing bit errors, make sure you look at the data portion of the frame only.
• Compare the final weight vector found by the LMS with the theoretical one derived in subtask 1e.
Hints:
• A frame can be generated e.g. like this:
data = sign(randn(Ld,1)); frame = [pilot;data;zeros(Lg,1)];
• The received signal and filter output can be computed e.g. like: u = conv(h,frame); y = conv(conj(w),u);
• Bit detection may be done like this: bits = sign(real(y))
• The desired signal is: d = [zeros(M-1,1);pilot(1:Lp-M+1)];
• Use the stem function for plotting the bit errors
• You can use the LMS program below if you like.
function w = LMS(w0,u,d,s)
%w = LMS(w0,u,d,s)
%LMS alorithm
%INPUT
% w0: [M,1] : Init weight vector, M = Number of taps
% u : [N,1] : Input data vector
% d : [N,1] : Desired output data vector
% s : [1,1] : step-size parameter
%OUTPUT
% w : [M,1] : final tap weight vector
if (nargin<4) error(’Too few parameters’); end;
N = length(u); M = length(w0); w = w0;
x = zeros(M,1); for k=1:N x = [u[k];x(1:M-1)]; e[k] = d[k] – w’*x; dw = s * conj(e[k]) * x; w = w + dw;
end;
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[Solved] Array-sensor Homework06
10 USD \$
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https://www.gamedev.net/forums/topic/295150-how-are-these-called-and-do-they-work-like-this/
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# How are these called, and do they work like this?
This topic is 5050 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hey, This is mostly about collision detection, and I'm sure it already exists: the world is divided into cubes, which are subdivided into more cubes, and so fort. Every cube has a boolean that indicates whether something or someone is in that cube. So, when an object is checked for collision, the game first checks the biggest cube he is in. When that is true, the same for the smaller ones, and so forth. Only when the smallest one where the player is in contains an object, real collision detection is done. When an object enters a smallest cube, that cube sets his flag to true, and tells his parent to do the same, and that parent his parent, and so forth. Now, I have two questions: - How is this called? Are these 'octrees'? (Which is the first name I thought of.. sounds logical too, as every part has 8 corners.) - For leaving the 'cube'. Should the 'cubes' keep a lists of contents, or should the system loop trough all objects to see if that cube is free again? Thanks While I described it in 3D, using the idea in 2D, or even 1D or 4D would be possible too, I guess
yes and yes
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Quote:
Original post by krezyes and yes
To the first question, thanks, but to the second one.. which of them? Keep a list per cube (more memory) or loop trough objects (more cpu)?
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Well you have to keep track of which objects are in the same cube so you can do collision detection between them, but that information will change each frame as the positions of the objects are updated. So in a sense, you do both, however the list of objects per cube is only valid for that single frame.
##### Share on other sites
Each cube/node keeps it's own contents list. You then compare yourself to any other objects inside of your node(s). You would *not* compare yourself to all other objects in the world. As mentioned, this all requires some overhead of updating everyone's positions and the contents list of each node. But for large numbers of evenly-distributed objects, it can be very timesaving.
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# 15706656
## 15,706,656 is an even composite number composed of four prime numbers multiplied together.
What does the number 15706656 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 192 divisors.
15706656 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of one hundred ninety-two divisors.
## Prime factorization of 15706656:
### 25 × 33 × 73 × 53
(2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7 × 53)
See below for interesting mathematical facts about the number 15706656 from the Numbermatics database.
### Names of 15706656
• Cardinal: 15706656 can be written as Fifteen million, seven hundred six thousand, six hundred fifty-six.
### Scientific notation
• Scientific notation: 1.5706656 × 107
### Factors of 15706656
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 12
• Sum of prime factors: 65
### Divisors of 15706656
• Number of divisors d(n): 192
• Complete list of divisors:
• Sum of all divisors σ(n): 54432000
• Sum of proper divisors (its aliquot sum) s(n): 38725344
• 15706656 is an abundant number, because the sum of its proper divisors (38725344) is greater than itself. Its abundance is 23018688
### Bases of 15706656
• Binary: 1110111110101010001000002
• Base-36: 9CNC0
### Squares and roots of 15706656
• 15706656 squared (157066562) is 246699042702336
• 15706656 cubed (157066563) is 3874816999254901948416
• The square root of 15706656 is 3963.1623736607
• The cube root of 15706656 is 250.4347422251
### Scales and comparisons
How big is 15706656?
• 15,706,656 seconds is equal to 25 weeks, 6 days, 18 hours, 57 minutes, 36 seconds.
• To count from 1 to 15,706,656 would take you about thirty-eight weeks!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 15706656 cubic inches would be around 20.9 feet tall.
### Recreational maths with 15706656
• 15706656 backwards is 65660751
• 15706656 is a Harshad number.
• The number of decimal digits it has is: 8
• The sum of 15706656's digits is 36
• More coming soon!
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## Gee, Geoboards are handy!
Sometimes when the kids and I are using specific math manipulatives I get that reminiscent fuzzy feeling from my school days; I remember using some of the manipulatives myself. Still, when we pull out the Geoboards I’m at a total loss. Geoboards were not part of my educational experience – not in elementary school or any time after.
After speaking with a few of my online and off-line friends, it turns out that my lack of Geoboard use is quite the anomaly. And I wonder, how many of you guys used Geoboards in your early education? Surely I’m not alone?
Maybe I had some psychological drama associated with the use of Geoboards in my elementary years. Perhaps a stray Geoband popped-off its peg and nailed me in the eye and I bawled like a baby. And then, maybe the entire class pointed their fingers and laughed at me. And then, when I looked down I realized I was naked – and when I ran out the door I couldn’t find my locker. Obviously I would have blocked such ‘painful’ experience from my memory all this time”¦right?
Or maybe not.
Maybe I dreamed that.
EITHER WAY, my children are not going to have the same (lack of) experience with Geoboards; not now that I’ve learned just how far this simple tool can go.
Geoboards teach basic geometric concepts; problem-solving, exploring shapes, designs, spatial relationships, angles, fractions, area, perimeter, symmetry and so much more (I should write catalog text, no?).
Geoboards can be purchased or hand-made using a square piece of wood and 25 evenly-spaced nails arranged in grid of 5 vertical lines and 5 horizontal lines. The only ‘tool’ you need with Geoboards are Geobands (rubber bands); we use multi-color rubber bands of various sizes.
Though Geoboards are not a good toy for babies, they can be used by Toddlers (with the help of mom and dad) to ‘trace’ shapes with rubber bands already put into place. When Kenny was younger he liked plucking the rubber bands already affixed to the Geoboard to make ‘noise’.
If you are looking for a task to keep your Preschooler busy while you teach your other children, Geoboards are perfect for that task. Make a picture using pattern blocks and have them mimic it on their Geoboard.
If you’re working with them, always name the shapes as they are created and soon they’ll know the names of those shapes and will recognize them in your day-to-day activities in and out of the house.
“Momma, that sign is a hexagon!”
Kenny always enjoyed just being free to make his own designs at that age – in fact, he still likes to make his own designs using Geobands of varying colors and thicknesses.
In Meredith’s work the other day she used her Geoboard to show congruent line segments and map perimeters.
Emelie is 12 and can use the Geoboard for geometric constructions – and in discussing the characteristics of angles, properties of various shapes, similarity and congruency, etc”¦
Either way, if you’re looking for math manipulatives that can go a long way – you’ll definitely get your money’s worth from Geoboards.
About the Author: Heather Sanders writes about faithsizing: the choice to surrender and simplify your life to create peace and balance in all the busyness. Visit her blog, and follow her on Instagram and Pinterest.
## Heather Sanders
Hi. I’m Heather, a freelance writer living in Huntsville, a smallish town on the tail-end of the East Texas Pineywoods. Twenty years ago, I married Jeff, the love of my life, and shortly after, we chose to “go forth and multiply.” We have three kids: Emelie, Meredith and Kenny. We homeschool. It's what we do, and it works for us. Tired of feeling overwhelmed, we recently "faithsized" our family into a 960 square foot lake cabin in need of renovation. I write at HeatherSanders.com about faith and simplifying your desires so you can be content right here and right now
### Heather Sanders' blog:
http://www.heathersanders.com
I never used these but I’m gonna!
• http://www.fromsingletomarried.com Tabitha (FSTM)
Hmm… I’m feeling I must have missed something as a child because I never used them – they look cool though!
tabitha @ http://www.fromsingletomarried.com
• Weldermamma
I’ve never seen them but now I will find some or have my dad make me a few. These are so cool.
• http://wisdomofthemoon.blogspot.com/ Wendy
I vaguely remember these. Not sure if it was from school or just a friend’s house. Cool, though.
• http://www.chocolatechic.wordpress.com chocolatechic
Would I be considered a bad home school mom if I said that I had never even heard of geo boards???
• Susan
Never seen or heard of them. I am going to check these little things out.
• Lisette
I’ve never used one… but they look like a lot of fun.
• http://www.planetmfiles.com Gayle
You’re not alone, Heather. I never used them either. They are cool! I’ll ask my kids if they have ever used them in school.
• http://www.voirvenir.com Muriel
I had never heard of Geoboards, and I graduated High School in 2002. I think they’re not as popular as they would like us to think.
Those sneaky Geoboards…
• Cheryl
Guess what? I’ve never used one or seen one! And I homeschooled my 3 children for 17 years. Now, of course, that means I am really really old! 🙂 but, I have 2 kids who have graduated from college and my 3rd is a senior in college. They survived quite well without these little items but I bet they would have loved math more with them!!!
• Kerry
I have never heard of them either but they look cool. I’m sure my 7 year old would love them.
• Teslaca
I’ve never heard of these things either, but they remind me of those frames that you string stretchy fabric scraps on to make potholders….
• http://myboysmom.blogspot.com dallas
By no means are you alone in your lack of geoboard experience. I had never seen one until got to college and used them in a Math Curriculum and Instruction class! But oh how I love them..they are very versatile and they boys love them!!
• http://butterflygenes.blogspot.com Heathahlee
I have never heard of them! They look like a lot of fun to use, though!
• Autumn
I’d never heard of them before. Sounds difficult.
• Susan
I got one in my manipulatives kit and had no idea what to do with it. It’s sitting in the math box with the yellow clock, scale, number boards…I have the perfect little 3 year old who’d love it…hadn’t thought of that! I’m switching curriculum for my 3rd grader…she doesn’t like Horizon. But I have to wait til March for the level 4 of Teaching Textbooks to come out…maybe she can just play with the geoboard and learn multiplication tables til then!
• Sandra in Phx
I haven’t seen these before either…and starting to think my list of supplies I “need” next year might be as long as my arm…
• http://www.joyfulabode.com/ Joyful Abode: Domesticity by Trial and Error
I LOVED geoboards when I was little… and one day I must’ve talked about that awesome toy a little too much, because my dad helped me make one of my own to use at home.
It was bigger than the school ones, and we painted it red. I got to hammer the nails in myself (which, as a 6 year old, was awesome).
That is seriously one of the best toys ever.
And I’m great at math… coincidence? I think not. That geoboard made me who I am today. (hahaha)
• http://www.ohmystinkinheck.com OMSH
Whoohoo! I WAS NOT ALONE.
And now? Seriously – these things are cool and cheap (Probably cheaper to buy than make – but 6 year olds nailing into boards is an experience worth the cost of wood and nails!)!
• http://www.blackninjakitty.com Jenn H.
We had those, but we made them ourselves for school out of thumbtacks or nails and wood boards that the Dad’s would make for the class. There was never anything “branded”, it was all DIY. This was the 80’s 🙂
• Niki in Baltimore
Never saw them, never heard of them… but they look like they’de be great.
• Niki in Baltimore
Opps… sorry that i spelled “they’d” wrong in the previous comment. I also forget to read what i wrote before i hit submit.
• Mom2Six
I’d never heard of these until one came with my children’s math manipulative set last year. Mostly I just let them play with it–I’ve never really used it. The rubbler bands disappear too quickly around here! I must replace them!
I’ve usually used homemade manipulatives–simple beans and craft sticks may not be as coloful, but they are cheap and work quite well!
• GreatAunt
I’ve never heard of a Geoboard until today. Perhaps I’d have a basic grasp of geometry today if my teachers had used them when I was a kid. 🙂
• http://carsicinkennebec.blogspot.com/ melissa
My girls used them at the Montessorri School when they went. I didn’t know what they were until you told us. I think we will pick up a couple. Thanks.
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# A model rocket with a mass of 2.7kg is launched straight up from the ground at a speed of 110m/s. What is the rocket's mechanical energy when it takes off?
Nov 15, 2015
I found $16 k J$
We can choose the firing position as "zero" for Gravitation Potential Energy (height $h$ zero);
${E}_{\text{mec}} = U + K = m g h + \frac{1}{2} m {v}^{2}$
but $U = 0$ and so:
${E}_{\text{mec}} = 0 + K = K = \frac{1}{2} m {v}^{2} = \frac{1}{2} \left(2.7\right) {\left(110\right)}^{2} = 16 , 335 J \approx 16 k J$
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# Math
posted by .
Hello, I have this question to which it's solution I'm not quite sure of. The question is based on a diagram, and I can't really present it but I'll explain it to my very best, thank you very much for trying to understand my bad presentation of the question.
The diagram is a circle, with center O and points A and B lying on the circle. Radius is 6cm therefore OA=OB=6cm, such that angle AOB = 79 degrees.
Find the area of the shaded segment of the circle contained between the arc AB and the chord [AB].
The solution was a simple sentence, [ (79degrees/360 degrees) x pi x 6^2 ] - [(1/2)(6)(6)sin79]
I understand that [(1/2)(6)(6)sin79] was derived from the rule 1/2absinc and we're finding the area of the triangle.
Therefore, [ (79degrees/360 degrees) x pi x 6^2 ] must be the area of the sector. However, when I looked up the formula for the area of sector, it's (1/2)(delta)(r)^2, where delta is the angle measured in radians.
Primarily, how did [ (79degrees/360 degrees) x pi x 6^2 ] come about then?
• Math -
because the [ (79degrees/360 degrees) x pi x 6^2 ] would be the measure of that section of the circle.
the area of the whole circle would be pi x r^2 but since you're only solving for that section it would be 79/360 times that whole area of the circle to solve for that section.
Think of a circular pizza and you were asked to find the area of one slice and there are 10 slices, to find the area of one slice you would multiply the area of the whole pizza by (1/10) which represents one slice.
Hope this helps.
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# 16.05 Formula Column Syntax
Formula columns are derived from user-defined formulas. Users create these formulas by combining columns (fields) available in the report.
### When might I need to create a custom column?
Custom columns are useful for creating such numeric fields as:
• Values with non-standardized modes of calculation that are therefore not included in UNItekTIME as default columns (for example, overhead cost etc)
• Calculations specific to your organization
• Highly-specialized calculations
### Numeric Formula Examples:
• A formula for 10% Overhead cost in the [Detail Timesheet Report]: (Amount * 10 / 100)
• A formula for calculating profit margin in the [Detail Timesheet Report]. (BillingRate – EmployeeRate) * TotalHours
### Text Formula Examples:
• To identify the location and their departments in a single field, you could combine both location and department by creating a formula like this: (AccountLocation+’-’ DepartmentName)
### Numeric Operators:
OperatorUse this operator to specify…
-Subtraction
Multiplication
/Division
( … )Parentheses
=Two values are equal
<Less than
>Greater than
<=Less than or equal
>=Greater than or equal
!=Not equal
%Modulus
Operators follow standard order of operations rules. For example:
• 2 + 2 * 2 = 6, but (2 + 2) * 2 = 8
• 2 + 2 / 2 = 3, but (2 + 2) / 2 = 2
### Functions:
FunctionsDescriptionSyntaxExample
IIFGets one of two values depending on the result of a logical expressionIif ( expr, truepart, falsepart )Iif (billingrate>500,‘expensive’,‘dear’ )
LENGets the length of a stringLEN ( expression )Len (EmployeeName)
CONVERTConverts particular value to a specified .NET Framework Type.Convert(expression, type)Convert(total, ‘System.Int32’), Convert(id,‘System.String’)
SUBSTRINGGets a sub-string of a specified length, starting at a specified point in the string.SUBSTRING (expression, start, length )SUBSTRING ( phone, 7, 8 )
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# Closed Loop Cooling System
Tags:
1. Aug 9, 2017
### LarrySputnik
Hello! Hopefully someone here can help with this problem:
I have a condenser that holds 35 gallons of water to cool the copper coil on my still. I need to know if I can create a closed loop system of cooling water that allows me to avoid running a chiller. I know that there is probably an equation of Input and Output versus coil temperature and water temperature, but I have unfortunately never come across such an equation throughout my experience.
I can add the following information to help get things started:
-The condenser holds 35 gallons of cooling water
-I will be pumping water from a 2500 gallon tank to the condenser at 5GPM
-The water will be traveling through 3/4" hoses to and from the condenser
-I want to have the water pass through the condenser and return back to the top of the tank without using a chiller.
I simply need to know if 2500 gallons of water stored at, say, 72 degrees F can circulate through a condenser to cool a coil at 5GPM that may reach temperatures of 180 degrees F without heating up too much. If there is a way to calculate that I would appreciate your help.
Thanks!
2. Aug 9, 2017
### CWatters
You would need to know the flow rate of the liquid to be cooled (eg the alcohol). I'm assuming the alcohol it enters the condenser at around 180F but what temperature does it leave at? Alcohol boils at 172F so obviously you want it lower than that but how much lower? This info would allow you to calculate how much energy has to be removed from the alcohol per min.
Then you can calculate how much the cooling water would warm up when pumped through at 5GPM and how fast the 2500Gal tank would warm up.
PS: This info is provided for educational purposes only!
3. Aug 9, 2017
### CWatters
If this is a continuous process (eg like a factory) then the 2500Gal tank will eventually warm up. The temperature it reaches will depend on how much heat the tank can dissipate to it's surroundings, local air temperature etc.
4. Aug 9, 2017
### LarrySputnik
Thank you CWatters!
This is very helpful. So the alcohol will move along into the condenser around 172+F and then be cooled down by the water. If the cooling water is always kept at room temperature then the cooling water is always variable from day to day, season to season. Nevertheless, I would like the cooling water to remove about 100F from the alcohol from the top of the condenser to the output spout. This would give me product coming off the still about 70 to 80 degrees F.
So how do I then calculate how much energy has to be removed per min?
Thank you for getting the ball rolling on this!
5. Aug 9, 2017
### LarrySputnik
The process is only over the course of 4 to 8 hours. I was hoping that a large tank, such as 2500 gal, would allow me at least 8 hours. If I have to go larger though, I just need to know how large I need to go, which is where these calculations will really come in handy.
Thanks!
6. Aug 9, 2017
### CWatters
OK so it goes in at 172-180F and you want it to come out at 70-80F.
We still need the flow rate though.
7. Aug 9, 2017
### LarrySputnik
I'm not sure how to accurately arrive at the flowrate. I am using a copper alembic pot still and all of the calculators I have found are for reflux stills. If I run those calculators it gives me a flow rate of 1.13 g/s, which is highly unlikely.
Perhaps we can suppose that the flow rate of the distillate coming off the condenser is 0.5gpm. It would seem reasonable to collect 1 gallon every 2 minutes for the sake of running this scenario.
Last edited by a moderator: Aug 9, 2017
8. Aug 9, 2017
### JBA
Rather than worry about the flow, etc I think it is easier to work with the alcohol vs water bulk volume heat capacity ratio. Either way, the heat transfer from the tank to atmosphere has to taken into account; and, it will increase the allowable running time in the below analysis.
Ethanol: Heat of vaporization = 364 BTU/lb; Specific heat of liquid = 0.65 BTU/lb °F; Density = 0.01 lb/gal
Water: Heat of vaporization = 970.4 BTU/lb; Specific heat of liquid = 1.00 BTU/lb °F; Density = 8.35 lb/gal
Mass of stored water = 2500 x 8.35 = 20,875 lb
Heat capacity for stored water with 100° temperature change = 20.875 lb x 100 x 1.00 = 20875 BTU
Ethanol condensation and 100° temperature change per gal = (0.01 x 364) + (.01 x 100 x 0.65) = 3.64 + .65 = 68.65 BTU/gal
Ethanol that can be cooled = 20875 BTU / 68.65 BTU/gal = 304 gal
Allowable running time at .5 gal/min = 304/.5 = 608 min / 60 = 10 hrs.
Verification of my calculation will be appreciated.
Edit: While the bulk allowable amount of 304 gal is accurate the actual running time required for this production is less accurate because the increasing temperature of the condenser input water from the heating tank will require increasing rates of water flow to maintain the required 100 degree temperature drop in the condenser.
Last edited: Aug 9, 2017
9. Aug 9, 2017
### CWatters
Edit: Didn't see JBA's reply before writing my effort..
Ok I think we can run some numbers. I'm afraid I work in metric so for my sanity I'll have to do a conversion and then back again..
The temperature reduction required (ΔT) is from 180F (82C) down to 80F (27C) a drop of 82-27 = 55C.
Alcohol (drinking) is Ethyl Alcohol/Ethanol which has the following properties..
Latent Heat LA = 846 kJ/kg
Specific Heat SA = 2.46 kJ/kg·°C
Density = 789 kg/m3
Convert 0.5 gal to kg...
0.5 gpm = 1.89L = 1.89*10-3 m3
mass = volume * density
1.89*10-3 m3 * 789 kg/m = 1.49kg
Mass flow rate (Mfr) = mass/time = 1.49/60 = 0.025kg per second.
So the energy per second (aka power) that has to be removed from the alcohol is....
P (kW) = (LA * Mfr) + (SA * Mfr * ΔT])
= (846 * 0.025) + (2.46 * 0.025 * 55)
= 21.15 + 3.38
= 24.5kW
So now we can look at what happens if we put that much power directly into 1200 gallons of water...
Data for the water..
Mass of Water Mw = 1200 gallons = 4524L = 4524Kg
Specific heat SW = 4.184 kJ/kg·°C
So the temperature rise per second would be
dT/dt = P/(Mw*SW)
= 24.5 * 103/(4524 * 4.184 * 103)
= 1.3 * 10-3 degrees C per second
or
4.6C per hour
So after 8 hours the 1200 gallon tank will have risen about 37C. If it started at say 21C (70F) then it will end up at around 58C (136F).
We need to do one more check because the heat isn't dumped straight into the 1200 gal tank there is a 5gal/min transfer pump between the condenser and the 1200Gal tank. The temperature of water coming out of the condenser will be hotter than that going in...
5gal/min = 0.32L/S = 0.32Kg/S
If you put 24.5kW into 0.32Kg of water the temperature uplift will be..
ΔT = 24.5*103 / (0.32 * 4.184 * 103)
= 18C
So the water coming out of the condenser will be about 18C warmer than it went in.
The issue will be towards the end of an 8 hour day.. The 1200L tank may have risen to around 58C (136F) so that will be the temperature of the water going into the condenser. The water coming out of the condenser will be hotter around 58C + 18C = 76C (168F). The average temperature of the water in the condenser would be (168+136)/2 = 150F. So the alcohol wouldn't be cooled below that.
This all ignores heat loss from the big tank, pipework etc and that might improve the situation. If you need additional cooling perhaps instead of using a proper compressor based chiller you might get away with some car radiators and a fan on the hot side of the condenser (eg to cool the water before it goes into the big tank). You might need something like that anyway to cool the 1200 gal tank back down to 70F overnight for the next run?
Well that's my best stab at the figures. Your mileage may vary!
10. Aug 9, 2017
### LarrySputnik
Thank you CWatters for helping out with this. That is a ton of math to digest and I appreciate the work you put into that. It looks like I'll only barely be able to maintain a temperature that is lower than alcohol boiling point, and for my purposes I need to be much more efficient than that. So a cooler would be necessary.
But, even though this arithmetic is admittedly beyond my understanding, I still have a couple of questions about the numbers that might change the outcome.
First, I see that you are working with 1200 gallons of water in the beginning, but at the end you referenced it as 1200L. I'm sure it is just a typo, but I need to ask anyway just in case it might change the overall answer.
The second thing I wanted to ask, which probably ought to precede the first question, is that the water tank I want to use is 2500 gallons, which is more than twice the number we used for water in this series of equations. If we adjust that number to 2500 gallons of water, does our answer change significantly? I will try to compute this for myself, but your input is paramount here since I am not qualified to make these kinds of computations.
Thank you!
11. Aug 9, 2017
### LarrySputnik
Thank you for chiming in JBA! This assumes that the water flow rate is constant, but you mention that in actuality I would need to increase the water flow beyond 5gpm? If the water flow must remain constant (due to the pump's limitations) would a higher water flow rate of, say 7gpm, make a significant difference?
Thank you for your time and input. It is much appreciated.
12. Aug 10, 2017
### JBA
At this point , I can now see a serious error of my first calculation because I failed to take into account, by oversight, that by reducing the condenser alcohol discharge to the 80°F level with water starting at 72°F would only give a tank heat capacity temperature differential of 8°F, not 100°F; and, as a result, the total heat capacity of the tank is reduced to 20,875 BTU x 8°F / 100°F = 1670 BTU.
As a result, you can only produce your 0.5 gpm of alcohol for: 1670 BTU / 34.325 BTU/min = 48 min = 0.81 Hrs. and after that, as pointed out above, the tank water and heat exchanger alcohol discharge temperatures will continue to rise. For continuous operation at a condenser output temperature of 80°F you will clearly require continuous auxiliary water cooling of 100°F.
As stated above, convection cooling, and evaporation cooling, if you have an open tank, will reduce this requirement to some extent; but, at this point, we do not have enough information to estimate that contribution and its contribution would be variable and dependent upon the ambient weather conditions.
Increasing the water flow rate would be of no value in extending your tank heat capacity limited allowable production time.
I apologize for my ridiculous oversight in my original post calculation.
13. Aug 10, 2017
### CWatters
Yes my mistake. No idea why I used 1200gal. Bit busy at the moment. Will try and comment on JBA's figures and correct mine later today.
14. Aug 11, 2017
### CWatters
Ok correcting this for 2500 Gal...
So now we can look at what happens if we put that much power directly into 2500 gallons of water...
Data for the water..
Mass of Water MW = 2500 gallons = 9463L = 9463Kg
Specific heat SW = 4.184 kJ/kg·°C
So the temperature rise per second would be
dT/dt = P/(Mw*SW)
= 24.5 * 103/(9463 * 4.184 * 103)
= 0.62 * 10-3 degrees C per second
or
2.2C per hour
So if the tank starts at 21C (70F) and must not go above 27C (80F) that gives you only about 3 hours running.
15. Aug 11, 2017
### CWatters
However....
If the flow rate of the pump is 5gal/min then it would take 2500/5 = 500 mins or 8 hours for all the water to be circulated through the condenser. So if you can keep the 2500gal tank stratified (eg not stirred so the hot water stays at the top) then it should work for a lot longer. Ideally you want a tall tank and must feed the returning hot water into the top of the 2500 tank and extract cold water at the bottom. Run the circulating pump as slow as possible while still achieving the required condensate temperature.
So I'm not quite as pessimistic as JBA. If you already have the tank and pump then perhaps worth giving it a go. If you need to buy one then I think you need to weigh up the risks. There is still the issue of cooling down the hot tank over night for the next days run.
16. Aug 11, 2017
### LarrySputnik
This sounds promising. My thoughts were right in line with yours in terms of keeping the warm water at the top. The tank would be pumped from the bottom nozzle over to the bottom of the condenser drum. The warm water at the top of the condenser drum will be fed through a hose that connects to a port at the top of the water tank. While a run will likely last 8 hours, the good news is that the condenser will not be at the temperature of 170+ degrees for more than 4 or 5 of those hours, which means I won't really need to start circulating the water until the condenser starts to get up to temp. This buys me a little more time.
I haven't purchased the water tank yet, but seeing as I'm trying to make moves in order to get my distillery as green as possible you have given me some hope. Thank you! I will post updates to let you know the progress.
17. Aug 12, 2017
### CWatters
Before buying a big tank might be better to just build a big fan cooled radiator. Will need something like that anyway to cool the tank down again. Unless you have another use for hot water?
18. Aug 12, 2017
### NTL2009
I just skimmed most of this, but I think I can add a few practical considerations, based on my related experience chilling 5-6 gallons of wort (raw beer) from near boiling temps to room temperature.
I am not familiar with the design of your condensor, but you will want a "counter-flow" design - the coldest cooling water enters at the exit of the product you are chilling, and the (now warmer) cooling water exits where the hot product enters. That maintains a maximum temperature differential, and allows you (depending on flow rates) to get the product very close to the temperature of your cooling water. Maybe it i already designed this way?
Also, it would be best to not return the warmed water back to the tank. You want to keep that cool, so keep hot/cold separate. And/or, start with tap water and empty that into the tank from the condensor output, your tap water may be cooler than room temp (mine is)?
You could use a two stage condensor - if the limits of the first stage only get you down to say, 85F, and you want 70F~80F, then you could use a chiller, or just a simple ice bath to get those last few degrees out in the second stage. That will take far less ice/energy than applying it to the whole temperature delta. Or three stage - tank water in first, tap water in second, ice/chiller in last/third stage.
A key is to maintain a maximum temperature delta between the two, and good thermal transfer as well of course. And don't underestimate the power of some turbulence in each, without turbulence, a boundary layer builds up, and heat transfer is reduced very significantly.
http://www.engineersedge.com/heat_transfer/parallel_counter_flow_designs.htm
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Parts 8-11
# Proper and Improper Construction of Shapes
by Johannes Kepler
## Part 8: Measuring Shapes
The quantity of a line is knowable if it is either itself immediately measurable.
The quantity of a surface is knowable if its square is measurable by the diameter.
## Part 9: Recreating Measures
The construction of a quantity which is either to be described or to be known is its deduction from the diameter, by permitted means, in Greek [these are called] Kopipa, “practicable.”
So constriction generally yields either description or knowledge. But Description declares mere quantity, whereas knowledge also in addition declares quality or a definite quantity.
A line can be geometrically determined, in Greek raKtri [“fixed”], even though its quality is not yet known intellectually.
On the other hand, a line or lines may be known qualitatively, but that does not yet determine them or make them determinate, that is to say if their quality is common to many other things which are different in quantity. So for such lines description is easy, knowledge very difficult. Einally, many things can be described by some Geometrical means or other; but cannot be knowable by their nature: as knowledge has been defined above.
## Part 10: Proper Construction
This is when the number either of the angles of the shape itself, or of the shape related to it by having either double or half its number of sides, forms the middle term in finding the ratio of the side to the Diameter.
Every regular figure is either:
• itself a triangle or
• can be resolved into triangles by drawing diagonals.
Every such triangle has its 3 angles equal to 2 right angles.
• In the elementary triangle of the Trigon, the angle is 1/3
• In the elementary triangle of the Tetragon, the smallest angle is 1/4
• In the Pentagon, 1/5
• In the Heptagon, 1/7 etc.
Each fraction is that of 2 right angles. It is from the size of the angle that the construction begins.
## Part 11: Improper Construction
This is when the ratio of the side to the diameter cannot immediately be determined Geometrically from the number of the angles, unless the side of another figure is brought in. This extra side is not from the figure with double or half the number of sides [of the original figure].
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# SOLUTION: two cars leave denver at the same time and travel in opposite directions. One car travels 10 mi-h faster than the other car. the cars are 500 mi apart in 5h. how fast is each car g
Algebra -> Algebra -> Customizable Word Problem Solvers -> Travel -> SOLUTION: two cars leave denver at the same time and travel in opposite directions. One car travels 10 mi-h faster than the other car. the cars are 500 mi apart in 5h. how fast is each car g Log On
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Word Problems: Travel and Distance Solvers Lessons Answers archive Quiz In Depth
Question 332226: two cars leave denver at the same time and travel in opposite directions. One car travels 10 mi-h faster than the other car. the cars are 500 mi apart in 5h. how fast is each car goingAnswer by rfer(12660) (Show Source): You can put this solution on YOUR website!x+x+10=100 2x=90 x=45 mph x+10=55 mph
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# CIRCUITO 7408 PDF
O circuito lógico TTL é um dispositivo TTL que possui quatro portas lógicas AND de duas entradas cada porta. Ele é usado, principalmente, em circuitos. jpg ( × pixels, file size: 15 KB, MIME type: image/jpeg). Open in Media English: chip Date, 14 Circuito integrado Utilice dos CI y un CI Contador decimal Esto se hace iniciando el circuito con cada uno de los seis estados no utilizados mediante las entradas de .
Author: Kigor Dougar Country: Serbia Language: English (Spanish) Genre: Love Published (Last): 25 December 2007 Pages: 363 PDF File Size: 9.37 Mb ePub File Size: 13.31 Mb ISBN: 210-4-38776-819-8 Downloads: 39108 Price: Free* [*Free Regsitration Required] Uploader: Kale
As the reverse-bias potential increases in magnitude the input capacitance Cibo decreases Fig. For this particular example, the calculated percent deviation falls well within the permissible range. There will be a change of VB and VC for the two stages if the two voltage divider B configurations are interchanged.
This range includes green, yellow, and orange in Fig. Events repeat themselves after this.
### Construção de um modelo computacional para o circuito de ventilação da Mina Esperança
See circuit diagram above. Such may not be entirely true. Computer Exercises Pspice Simulations 1. For voltage divider-bias-line see Fig. The percent differences are determined with calculated values as the reference. It is essentially the reverse saturation leakage current of the diode, comprised mainly of minority carriers. The voltage-divider configuration is the least sensitive with the fixed-bias configuration very sensitive.
No VPlot data 1. The LCD, however, requires a light source, either internal or external, and the temperature range of the LCD is limited to temperatures above freezing. While in the former case the voltage peaked to a positive 3.
TOP Related CAUSAS DE HEMOTORAX NO TRAUMATICO PDF
For either Q1 or Q2: Common-Emitter DC Bias b. Experimental Determination of Logic States. Y is identical to that of the TTL clock.
The voltage divider bias line is parallel circiuto the self-bias line. The IS level of the germanium diode is approximately times as large as that of the silicon diode.
In addition, the drain current has reversed direction. Majority carriers are those carriers of a material that far exceed the number of any other carriers in the material. Both capacitances are present in both the reverse- and forward-bias directions, but the transition capacitance is the dominant effect for reverse-biased diodes and the diffusion capacitance is the dominant effect for forward-biased conditions.
The levels are higher for hfe but note that VCE is higher also.
### CIRCUITOS INTEGRADOS POR ORDEN NUMERICO
Example of a calculation: V1 12 V The voltage of the TTL pulse was 5 volts. Parallel Clippers continued b. The voltage-divider bias configuration was the least sensitive to variations in Beta.
All the circuit design does is to minimize the effect of a changing Beta in a circuit. To shift the Q point in either direction, it is easiest to adjust the bias voltage VG to bring the circuit parameters within an acceptable range of the circuit design. Logic States versus Voltage Levels a. All the contents of this journal, except where circyito noted, is licensed under a Creative Commons Attribution License.
## File:7408.jpg
Silicon diodes also have a higher current handling capability. Maintain proper bias across Q1 and Q2. This seems not to be the case in actuality. The voltage level of the U2A: Thus in our case, the geometric averages would be: They are the same.
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Determine the final velocity of the first body. m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2. So you could simplify things by assuming an imaginary ball of mass m = 2kg moving upward at 10 m/s instead of the two balls. The can starts at rest, so its initial velocity is 0.0 m/s. We can now use this result to identify elastic collisions in any inertial reference frame. Because the goalie is initially at rest, we know v 2 = 0. A Ball Of Mass 0.4kg Traveling At A Velocity 5m/S Collides With Another Ball Having Mass 0.3kg, Which is At Rest. If the two colliding bodies have equal masses: , then the velocity formulas 13 and 19 simplify. Show that the equal mass particles emerge from a two-dimensional elastic collision at right angles by making explicit use of the fact that momentum is a vector quantity. Ex.2. Apparently for ball to ball collisions the tangential component remains same because no force acts along it. Mass of Moving Object. Elastic collisions can be achieved only with particles like microscopic particles like electrons, protons or neutrons. In several problems, such as the collision between billiard balls, this is a good approximation. Figure 56 shows a 2-dimensional totally inelastic collision. An elastic collision will not occur if kinetic energy is converted into other forms of energy. Let us denote the mass of the body as and insert it into Eqs. 2) A young boy is sledding down a very slippery snow-covered hill. Here is a remarkable fact: Suppose we have two objects with the same mass. The momentum after collision is also found by estimating a change in an object's velocity v after the collision. The formula of elastic collision is - m1u1 + m2u2 = m1v1 + m2v2.
Here's what your final velocity comes out to . Step 5: Switch the colliders' force vectors. = 14.31 m/s. How to calculate final velocity after collision Enter the mass and initial velocity of two different objects undergoing an elastic collision.
In any collision, momentum is conserved.
Step 6: Compose the new vectors into a new velocity: 1) Assumptions: 1) All collisions are elastic. Finally, let the mass and velocity of the wreckage, immediately after the collision, be m1 + m2 and v. Since the momentum of a mass moving with velocity is mass*velocity, and as I said above, Momentum before = Momentum after. It is given as: e = v b f - v a f v b i - v a i; e = 7 - 6 9 - 6; e = 0. If there is some "bounce" but the final kinetic energy is less than the initial kinetic energy then the collision is called inelastic. The 2nd body comes to rest after the collision. After the collision, ball 1 comes to a complete stop.
We did the calculation in the lab frame, i.e., from the point of view of a stationary observer. PseudoCode: RelativeVelocity = ball1.velocity - ball2.velocity; Normal = ball1.position - ball2.position; float dot = relativeVelocity*Normal; dot*= ball1.mass + ball2.mass; Normal*=dot; ball1.velocity += Normal/ball1.mass . m 1 v 1 + m 2 v 2 = ( m 1 + m 2) v , m 1 v 1 + m 2 v 2 = ( m 1 + m 2) v , 8.8. where v is the velocity of both the goalie and the puck after impact. We are all familiar with head-on elastic collisions. And it came out to be negative, that means that this tennis ball got deflected backwards.
the same formula you use in the previous example. . Example 15.6 Two-dimensional elastic collision between particles of equal mass. Conservation of momentum and energy gives you two equations, and you have two unknowns: velocity of A and velocity of the imaginary ball after the . It explains how to solve one dimension elastic collision physics problems. Final velocity of a system in an inelastic collision when masses and initial velocities of the objects involved are given. Final Velocity after a head-on Inelastic collision Calculator. In order for there to be a collision the initial velocity of the club head must be greater than . Elastic collisions equation. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms. If the ball has a mass 5 Kg and moving with the velocity of 12 m/s collides with a stationary ball of mass 7 kg and comes to rest. Final Velocity of body A after elastic collision - (Measured in Meter per Second) - Final Velocity of body A after elastic collision, is the last velocity of a given object after a period of time. Final Velocity after a headon Inelastic collision . As to the rst body, its velocity after a perfectly elastic collision is v0 1 = m .
For an inelastic collision, conservation of momentum is.
A ball sticking to the wall is a perfectly inelastic collision. m/s km/s m/min km/hr yard/s ft/s mile/hr. For head-on elastic collisions where the target is at rest, the derived relationship. The 2nd body comes to rest after the collision.
On the other hand, an elastic collision is one in which the kinetic energy after is the same as the kinetic energy before. m 2 = Mass of 2 nd body. How to Find Momentum After Collision. Inelastic collisions equation. Since momentum is mass times velocity there would be a tendency to say momentum has been conserved. Formulas Used: In an elastic collision both kinetic energy and momentum are conserved. After that, the velocity of the green ball is 5 m/s and the yellow ball was at rest. A molecule of mass m 1 is approaching from infinity with velocity u 1 and collides with mass m 2 moving at velocity u 2. 1.18 m/s. This CalcTown calculator calculates the final velocities of two bodies after a head-on 1-D inelastic collision. The calculator will calculate the final velocities of each object and the total kinetic energy. If we explain in other words, it will be; . Elastic One Dimensional Collision. 1-D Elastic Collisions. g kg ton mg ug ng pg Carat [metric] Stone Ounce (Oz) Grain Pound Dram. Transcript. Equations for post-collision velocity for two objects in one dimension, based on masses and initial velocities: v 1 = u 1 ( m 1 m 2) + 2 m 2 u 2 m 1 + m 2. v 2 = u 2 ( m 2 m 1) + 2 m 1 u 1 m 1 + m 2. Velocity of the second body (after) Velocity of the second body after the head-on elastic collision. The mass, velocity, and initial position of each puck can be modified to create a variety of scenarios The Organic Chemistry Tutor 68,139 views 10:26 The soccer player from the home team (56 kg) approaches the ball with a velocity of 7 Give its equation and unit There are two types of collisionselastic and inelastic There are two types of . You can calculate the new velocities by applying an impulse to each ball. But momentum has changed from +mv to mv. By definition, an elastic collision conserves internal kinetic energy, and so the sum of . In an ideal, perfectly elastic collision, there is no net conversion of . elastic collision: A collision in which all of the momentum is conserved. Hence the velocity after elastic collision for second ball is 14.31 m/s. Solution: We can apply Newton's Third law to do so. What is the velocity of ball 2 after the . The initial velocity of the paintball is 90.0 m/s. An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Initial velocity of body A before the collision .
Solution: For a perfectly elastic collision, kinetic energy is also conserved. U 1 Initial velocity of 1st body. m1v1 + m2v2 = m1v 1 + m2v 2 ( Fnet = 0), where the primes () indicate values after the collision.
In physics, the most basic way to look at elastic collisions is to examine how the. Velocity After Elastic Collision Calculator. Object one is stationary, whereas object two is moving toward object one. They conserve energy and momentum according to the formulas: Conservation of Energy: v 1 2 + v 2 2 = V 1 2 + V 2 2 and Conservation of Momentum: m 1 v 1 + m 2 v 2 = m 1 V 1 + m 2 V 2. A 15 Kg block is moving with an initial velocity of 16 m/s with 10 Kg wooden block moving towards the first block with a velocity of 6 m/s. Figure 15.11 Elastic scattering of identical particles. Solved Examples on Elastic Formula. The value of e is between 0.70 and 0.80. As already discussed in the elastic collisions the internal kinetic energy is conserved so is the momentum. Normal View Full Page View. The following formula is used in the conservation of momentum of two objects undergoing an inelastic collision.
If the collision was perfectly inelastic, e = 0. Many texts expect the student to solve these two formulas simultaneously to find the final . (d)An elastic collision is one in which the objects after impact become stuck together and move with a common velocity. Solution: Given parameters are The two meatballs collide and stick together. How to calculate final velocity after collision Enter the mass and initial velocity of two different objects undergoing an elastic collision. The tennis ball has 3 times the velocity after the collision with the basket . Elastic Collision Formula The following formula is used to calculate the velocities of two objects after an . An elastic collision occurs when both the Kinetic energy (KE) and momentum (p) are conserved. This means. Due to symmetry, balls B will move identically after the collision. An elastic collision is a collision where both the Kinetic Energy, KE, and momentum, p are conserved. Mass of body A - (Measured in Kilogram) - Mass of body A is the measure of the quantity of matter that a body or an object contains. = 204.8. v. 2.
Super elastic collision formula. Answer: (c) Explanation: An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same. objects is the same before and after the collision in this frame. Ex.2.
Ball 1 moves with a velocity of 6 m/s, and ball 2 is at rest. More generally, the expected angle between resulting velocities after an elastic collision is a right angle, but the actual angle observed in Unity is much more acute. The coefficient of restitution can be found after knowing this velocity. Work out the total momentum after the event (after the collision): Work out the total mass after the event (after the collision): Work out the new velocity:. v f is the final velocity. u 2 = Initial Velocity of 2 nd body. Elastic Collision Formula. For example, a ball that bounces back up to its .
If you want to calculate the velocity of the first body . Consider two molecules of mass m 1 and m 2. Perfectly elastic collisions are met when the velocity of both balls after the collision is the same as their . This physics video provides a basic introduction into elastic collisions. A 15 Kg block is moving with an initial velocity of 16 m/s with 10 Kg wooden block moving towards the first block with a velocity of 6 m/s.
Then we get: Velocity of the first body after the collision of two equal masses. For example, the body should not deform or rotate after the collision.
m1 - Mass of object 1; m2 - Mass of object 2; v1i - velocity of object 1 before collision; Show that the equal mass particles emerge from a two-dimensional elastic collision at right angles by making explicit use of the fact that momentum is a vector quantity. The conservation of the total momentum before and after the collision is expressed by: + = +. - The velocity of the ball after the collision is zero. Preview. Two billiard balls collide. b) but actually both went together more or less at the same speed (fig. Created by David SantoPietro. Special case #1: Both collision partners have the same mass. 4 (Elastic and Inelastic Collisions) In-class Practice 6 An elephant on a bike has more momentum than a mouse on a bike moving at the same speed Inelastic collisions Momentum ANSWERS - AP Physics Multiple Choice Practice - Momentum and Impulse Solution Answer 1 The force involved with collision acts only for quite a brief time period The . 391. In the following equations, 1 and 2 indicate the two different objects colliding, unprimed variables indicates those before collision and primed variables indicate those after the collision, p is momentum, KE is kinetic energy, M is mass, and V is velocity . Example 15.6 Two-dimensional elastic collision between particles of equal mass. After the hit, the players tangle up and move with the same final velocity. may be used along with conservation of momentum equation. This formula describes a collision between two bodies.
After the collision, the two objects stick together and move off at an angle to the -axis with speed . Elastic Collision, Massive Projectile In a head-on elastic collision where the projectile is much more massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile and the projectile velocity will be essentially unchanged.. For non-head-on collisions, the angle between projectile and target is always less than 90 degrees. The calculator will calculate the final velocities of each object and the total kinetic energy. After the collision, the velocity of the paintball and can together is 1.18 m/s. Figure 15.11 Elastic scattering of identical particles. Angles in elastic two-body collisions. No headers. In the following equations, 1 and 2 indicate the two different objects colliding, unprimed variables indicates those before collision and primed variables indicate those after the collision, p is momentum, KE is kinetic energy, M is mass, and V is velocity . In any collision, whether it is elastic or inelastic, the total momentum of the system before the collision must be equal to the total momentum of the collision after the collision.
After a collision, both the masses diverts away from each other making an angle with a plane with velocities v 1 and v 2. The Elastic Collision formula of kinetic energy is given by: 1/2 m 1 u 1 2 + 1/2 m 2 u 2 2 = 1/2 m 1 v 1 2 + 1/2 m 2 v 2 2. - The kinetic energy does not decrease. magnitude of its velocity is an elastic collision. A 15 Kg block is moving with an initial velocity of 16 m/s with 10 Kg wooden block moving towards the first block with a velocity of 6 m/s. Login = 204.8. v. 2. P f = mv. In an elastic collision, both momentum and kinetic energy are conserved. An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter.
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## RESEARCH METHODOLOGY WITH ELEMENTS OF STATISTICS METOD. DELLA RIC. CON ELEM. DI STATISTICA
A.Y. Credits
2018/2019 8
Lecturer Email Office hours for students
Marco Bruno Luigi Rocchi
Teaching in foreign languages
Course with optional materials in a foreign language English
This course is entirely taught in Italian. Study materials can be provided in the foreign language and the final exam can be taken in the foreign language.
### Assigned to the Degree Course
Psychology - Sciences and Techniques (L-24)
Curriculum: PERCORSO COMUNE
Date Time Classroom / Location
Date Time Classroom / Location
### Learning Objectives
The course aims to give students a methodological training and basic statistics, not separated from the acquisition of an operational capability in the face of real problems.
### Program
1. Outline of probability (2 hr).
2. Population and sample (2 hr).
3. Measuring scale variables (2 hr).
4. Main indices of position (arithmetic mean, median and mode) and dispersion (variation range, variance, standard deviation, coefficient of variation, quartiles, interquartile range) (4 hr).
5. bivariate statistics: regression and correlation (4 hr).
6. Main distributions: Binomial, Poisson, Gaussian (6 hr).
7. The sampling distributions: distribution of the sample mean and t-distribution (2 hr).
8. Tests: the z test; the Student t test (for one and two samples) (6hr).
9. Confidence intervals for means and proportions (2 hr).
10. Classification of trials (1 hr).
11. The scientific method and unscientific methods (1 hr).
12. The research protocol: the rationale, the literature search, randomization, blindness, experimental design, sample size, assessment tools and their psychometric properties (reliability, validity, responsiveness, least significant difference, sensitivity and specificity), deviations from the protocol and their treatment, critical reading of a scientific paper (16 hr).
None
### Learning Achievements (Dublin Descriptors)
In relation to the methodology of research with elements of statistics,
students should show possession of:
- the mastery of basic skills;
- the understanding of the fundamental concepts of the discipline
- the ability to use knowledge and concepts to reason according to the logic of the discipline.
### Teaching Material
The teaching material prepared by the lecturer in addition to recommended textbooks (such as for instance slides, lecture notes, exercises, bibliography) and communications from the lecturer specific to the course can be found inside the Moodle platform › blended.uniurb.it
### Supporting Activities
Seminars; guided exercises
### Didactics, Attendance, Course Books and Assessment
Didactics
Frontal lessons
Attendance
None; participation is strongly recommended
Course books
Rocchi MBL: Statistica e metodologia della ricerca per le discipline biomediche e psicocomportamentali, Edizioni Goliardiche, Trieste, 2007 (pag. 250)
Assessment
Written test (30 multiple choice quizzes in 30 minutes); the oral exam is optional (after passing the written test )
### Additional Information for Non-Attending Students
Didactics
No differences with respect to attending students
Attendance
No differences with respect to attending students
Course books
No differences with respect to attending students
Assessment
No differences with respect to attending students
### Notes
The student can request to sit the final exam in English with an alternative bibliography. (Course with optional materials in a foreign language)
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# Jennifer came to class wearing her latest purchase and soon everyone was familiar with the scent. What process is being
###### Question:
Jennifer came to class wearing her latest purchase and soon everyone was familiar with the scent. What process is being demonstrated here?
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# Posts by Michelle
Total # Posts: 2,068
1. ### Math
The diameter of a circle is equal to the side of a square. The ratio of the area of the circle to the area of the square is approximately A. pi/2 B. 2/pi C. 2/1 D. 11/14
2. ### math
how many integers are their own multiplicative inverses? a. 0 b. 1 c. 2 d. 3 is it 2 or 3? because it could be 1, -1, and 0
3. ### Chemistry
If the solubility of Ag2CO3 is 1.3x10^-4 mol/L, what is its Ksp? (the answer is said to be 8.8x10^-12) Ag2CO3(s) -----> 2 Ag+ (ag) + CO3 2- Ksp= [Ag+]^2 [CO3^2-] = [1.3x10^-4]^2[1.3x10^-4] =2.197x10^-12 thats what I did, but I found that if you first multiplying the 1.3x10...
4. ### math
sorry to bother but there's no answer choice that matches 3.0314...
5. ### math
4 to the power of 4/5 A. 48/10 B. 24/10 C. 21/5 D. 4.4 I got B but I'm not sure, any help would be appreciated :-)
6. ### Math
Six teaspoons of extract are mixed with 240 ml of oral suspension what is the final concentration of the suspension?
7. ### Chemistry
269 mL of a 0.2089 M sodium carbonate (Na2CO3) solution is prepared and labelled solution A. Calculate the mass (in grams) of sodium carbonate (molar mass = 105.99 g mol-1) required to prepare Solution.
8. ### he
what are the philosophy of home economics?
9. ### physics
An AC voltage of the form ?v = (65.0 V)sin(340t) is applied to a series RLC circuit. If R = 46.0?, C = 27.0 ?F, and L = 0.300 H, find the following. (a) impedance of the circuit ? (b) rms current in the circuit A (c) average power delivered to the circuit W
10. ### science
a 400 kg block of iron is heated from 295 k to 325 k. how much heat is absorbed by the iron? ( the specific heat is 450 j. /kg. k)
11. ### science
sally and lisa have identical compact cars sally drives on the freeway with a great speed than lisa. which car has more kinetic energy
12. ### Physics
two long, parallel lines each carry 100-A currents are 20 cm apart. 1) calculate the magnetic field at the point C that is 5 cm to the left from the left wire. 2) calculate the magnetic field at the point C that is 5 cm to the right from the left wire. 3) calculate the ...
13. ### physics - magnetism
A positive and a negative particles enter magnetic field at the same angle and with the same velocity. How different would be the path of each particle?
14. ### PHYSICS
by the way i already know the answer, i dont need it. (100 mv) i need to know how to do it or why it is 100 mv
15. ### PHYSICS
I'm having some trouble with a problem on my physics pre-exam, if there is any formula please let me know so i can do a bit of research on it. any help is greatly appreciated in advance. The inner membrane of a cell membrane is negative relative to the outer surface. if 1....
16. ### Algebra 2
Given the exponential function A(x) = P(1 + r)x, what value for r will make the function a decay function? r = ?2.1 r = 2.1 r = 0 r = 0.1
17. ### Chemistry
You have a 35.5 g of a saturated solution of potassium nitrate at 80oC which you cool to 40oC, what mass of potassium nitrate will precipitate out of solution? @80 degrees C solubility is 160g solute/100g H2O @40 degrees C solubility is 60g solute/100g H2O
18. ### calculus
Shouldn't it specify of you are going from the left or the right?
19. ### Personal Fiance
You think that in 8 years it will cost \$247,000 to provide your child a 4-year college education. Will you have enough if you take \$77,000 today and invest it for the next 8 years at 8 percent? Round the answer to the nearest cent. Round FV-factor and FVA-factors to three ...
20. ### Math Circumference to Area???
Help! It say that the circumference is equal to ? over 2 and i have to find the area HOW DO YOU DO THAT???? C=?/2
21. ### Math
Help! It say that the circumference is equal to ? over 2 and i have to find the area HOW DO YOU DO THAT???? C=?/2
22. ### Math
Well huh?? i am right hun
23. ### Math
A B B B B C B A D A A A A B A B C A THIS IS FOR Lesson 9: Equations and Inequalities Unit Test CE 2015 Essential Algebra Readiness (Pre-Algebra) A Unit 7: Equations and Inequalities IM RIGHT TRUST MEE
24. ### Math
WAIT IM SORRY !!! I DID THE WRONG ONE
25. ### Math
1.a 2.a 3.c 4.d 5.c 6.d 7.a 8.d 9.a 10.b 11.a 12.a 13.b 14.b 15.d 16.b 17.b 18.b 19.a 20.a 21.a 22.b 23.b 24.b
26. ### Physics
Oh great! Now I know. Thank you for answering my question.
27. ### Physics
I am curious how the earth work as an omnipresent conductive surface in an electrical distribution system?
28. ### Science
Why is Earth is the omnipresent conductive surface?
29. ### Physics
It saves human life from the danger of electrical shock which can cause death, by blowing a fuse. How can “blowing a fuse” can rescue someone's life? Please answer. Thank you.
30. ### Math
Determine the points at which the function is discontinuous and state the domain using interval notation f(u)=(4-u^2)/(7u+7?-u^2-u?) Please help.
31. ### Math
Lindsey is 5 years less than half her father's age. The sum of their ages is 79. How old is Lindsey?
32. ### math
im sorry kid you got most of them wrong and no one helped you and your mom should not take away your christmas thats so sad.
33. ### Math
You answered this wrong. 3 trays of strawberry plants is \$16.47 not just one strawberry plant. Also the ratio is 3 strawberry plants for every 2 ivy plants. You inverted the ratio. And they want to know how much 10 ivy plants trays + 3 strawberry trays would cost. Shouldn'...
35. ### math
oil consumption is increasing at a rate of 2.2% per year. in how many years will the oil consumption be twice the amount it is today?
36. ### Math
if michael opened an account paying 1% interest for 20 years. will his money double in time? how many years will it take to double?
thank you
38. ### Math
a family of 100 ants invades your home. if its population increases at a rate of 20% per week, in how many weeks will there be 200 ants? 400 ants? and 800 ants?
39. ### math
urban encroachment is causing the area of a forest to decline at a rate of 7% per year. what is the half-life time of the forest?
40. ### math
a daily newspaper had 12,125 subscribers when it began publication. five years later it has 10,100 subscribers. what is the average yearly rate of change in the number of subscribers for the five year period?
im lost
42. ### physics
a golf ball is hit at an angle of 45 degrees with the horizontal. if the initial velocity of the ball is 52m/s , how far will it travel horizontally before striking the ground?
43. ### ss
B C A C D YOUR WELCOME I HOPE YOU GET 5/5
44. ### Social Studies
What are two helpful developments in Georgia and the rest of the country that occurred after the Revolutionary War ?? Please help me I'm comfused
45. ### art
Anynomos and sam your very welcome
46. ### art
A. D. D. there you go hope i helped you
47. ### Geometry
Find m<HIR if m<RIN =80, m<HIR = 8x+2, and m<HIJ = 19x-6
48. ### Math
h(x) = 2-4x A. h(-5) my answer is -62 B. h(4) my answer is -34
49. ### Algebra
No. My classmate and I are working together.
50. ### Algebra
Please can you explain how to solve the problem?
51. ### Physical Science
How will you prepare a laboratory procedure to verify the validity of the hypothesis? Give example.
52. ### Maths age problems
Tom is 6 years older than James. In 5 years time their combined ages willbe36.how old are they now?
53. ### Math-college
Here is a linear inequality to solve: 8x + 8 < 3 (x + 1)
55. ### la
i dont understand compare the victorian period to modern day life like in christmas carol essay
56. ### Physics
A 0.299 kg bead slides on a curved wire, start-ing from rest at point A as described below: A: a crest (hill) 6.15 m high b: a valley between A and C C: a crest (hill) 1.85 m high The acceleration of gravity is 9.8 m/s2 . 1) If the wire is frictionless, find the speed of the ...
59. ### Language Arts
Please help! In an essay discuss the way in which Naomi Shihab Nye treats loneliness in "The Rider" What does loneliness mean to the speaker? How does bicycling help her leave her loneliness behind? Cite passages from the poem to support your points. My answer: ...
60. ### French
Write a paragraph in French of at least 10 sentences talking about a trip you hope to take in the future. What do you hope to do? Write in comp;ete sentences in French.
61. ### Chemistry
If 2.5x10^-3 moles of O2 were produced in the reaction, how many grams of H2O2 had to react?
62. ### social studies
why don' why don't you ask your parents to go back to regular school?
63. ### College Math
What is the change in Michaels total pay for each copy of History is fun he sells? What is Michaels total pay if he doesn't sell any copies of History is fun?
64. ### College Math
Michael is a software salesman. Let y represent his total pay (in dollars). Let x represent the number of copies of History is Fun he sells. Suppose that x and y are related by the equation 1700+90x=y .
65. ### Algebra
30 is the product of Chrissy's height and 2.
66. ### Icd10
A patient presents with hypertrophic scarring to her chin from a burn received one year ago. Scar management has failed. What ICD-10-CM codes are reported? Select one: a. T20.03XA, L90.5 b. T20.03XA, T20.03XS, L91.0 c. L90.5, T20.23XA d. L91.0, T20.03XS D
JKOPK[
68. ### math
From a group of 65 students who failed Business Mathematics in 2010 we found the following reasons: 35 for not attending class (C); 30 for not doing homework (H) and 25 for not visiting tutors (T). Fourteen did not attend class and did not visit tutors; 15 did not visit tutors...
69. ### Technical writing
Identify the sentence that is punctuated correctly. Select one: a. Once inside, my mother's house seemed much larger than it appears on the outside. b. Once inside my mother's house seemed much larger, than it appears on the outside. c. Once inside, my mother's ...
B?
71. ### Technical writing
Identify the sentence that is punctuated correctly. Select one: a. My parents moved to West Virginia when they retired because the price of living is cheaper there. b. My parents moved to West Virginia, when they retired, because the price of living is cheaper there. c. My ...
72. ### Technical writing
Identify the sentence that is punctuated correctly. Select one: a. In December 2013 I visited my aunt for Christmas in Allentown, Pennsylvania, for the first time in 20 years. b. In December 2013, I visited my aunt for Christmas in Allentown, Pennsylvania for the first time in...
73. ### Technical writing
Always use a comma before somebody's name in a sentence. A.true B.false B
74. ### Hcpcs
J codes do not include drug_____. Amount?
75. ### Technical writing
Which is the correct comparative form of easy? Select one: a. easiest b. more easy c. easier d. easyer C
76. ### Technical writing
So is it c because the pad is a subject as well right
77. ### Technical writing
Identify the subject and complete verb in each of the following sentences. A pen and a pad of paper are in the top drawer of my desk. Select one: a. paper/drawer; are b. pen; are c. pen/pad; are d. pen/paper; are D
78. ### Technical writing
Thank you Ms. Sue
79. ### Technical writing
Identify each as fragment, run-on, comma splice, or sentence. Edgar Allen Poe invented the convention of the solution that is overlooked. Because it is so obvious. Select one: a. fragment b. run-on c. comma splice d. correct sentence A
80. ### Finance
Annual credit sales of 7.75 million - divide this by 365 and that gives you the daily credit sales = 26,712 multiplied by the 34 collection period and your total is \$908,219.18
81. ### physical science
A uniform stripe of eureka (resistivity 5-0 × ?10?^(-7) ?m) has a resistance of 0.80 ? per metre and is 0.25 cm wide. What is its thickness? (show work)
82. ### physical science
A block of carbon, 1.0 cm by 2.0 cm by 5.0 cm has a resistance of 0.80 ? between its two smaller faces. What is the resistivity of carbon? (show work)
83. ### physical science
. The maximum allowable resistance for an underwater cable is one hundredth of an ohm per metre. If the resistivity of cooper is 1.54 × ?10?^(-8) ? m, find the least diameter of a copper cable that could be used (show work)
84. ### physical science
What length of German silver wire, diameter 0.050 cm, is needed to make a 28 ?resistor, if the resistivity of German silver is 2.2 × ?10?^(-7) ? m (show work)
85. ### physical science
The resistance of the ohm is very approximately that of a column of mercury 1.06 m long and of uniform cross-section of one hundred of a cm^2. Find the resistivity of mercury.(show work)
86. ### physical science
Taking the resistivity of platinoid as 3.3 × ?10?^(-7) ? m, find the resistance of 7.0 of platinoid wire of average diameter 0.14 cm.
87. ### physical science
The heating elements of an electric toaster is typically made of nichrome wire (an alloy of nickel and chromium). As current passes through the wires, the wires heat up, thus toasting the toast. Estimate the overall resistance of a heating element which is 220 cm long and ...
88. ### pyhsical science
A wire of uniform cross-section has a resistance of R ?. What would be the resistance of a similar wire, made of the same material, but twice as long and of twice the diameter?
89. ### Algebra
can you help me graph the function y=-5(3^x)please? f(x)=-5(3^x) f(x)=-5*3 f(x)=-5(9) =-15 =-45 Is this the correct way to work this problem out and how to I graph?
90. ### Technical writing
Identify the subject and complete verb in each of the following sentences.
91. ### Technical writing
The teacher should have been more patient with her students. Select one: a. students; should have b. teacher: should have c. teacher; have been d. teacher; should have been D.
92. ### Technical writing
In order to have a complete sentence, you need a ________. Select one: a. subject, verb, and a complete thought b. subject and verb c. subject, verb, and a prepositional phrase d. subject, verb, and a dependent clause
93. ### Technical writing
Two independent clauses joined together with no punctuation create a ________. Select one: a. complex sentence b. compound sentence c. run-on d. comma splice
94. ### Chemistry
Given that the battery produces a current of 2.5A that passes through molten zinc chloride for 3 minutes what charge would be dissipated through the electrolyte
95. ### math
find the maximum and/or minimum values of the function f(x) for the indicated value of x f(x)=x^2 -4x+3 2<_ x _>5
96. ### Math 115
Calculate the slope and write an equation for the linear function represented by each of the given table. A W 5 30 7 40 W=_?___ I PUT W=5x+5 and W=5+5x BUT IT SAID IT WAS WRONG.
97. ### Math Velocity
Jeff is standing on a bridge 5M above the river, He picks up a small rock and tosses it into the air, the rock reaches a height of 12M in one second and hits the water in about 2.3 seconds Determine when the rock reaches a height equal to the bridge but on the way down.
98. ### Women should be treated as equals
Obviously we women are the pillars of the world without us the world wouldn't be able to function properly and it is said that behind every successful man is a powerful woman
99. ### Math
MEME_MASTER i did put the CORRECT answers urghhhh
100. ### Math
MC Ren to who??
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https://www.doubtnut.com/question-answer/a-b-c-d-is-a-parallelogram-g-is-the-point-on-a-b-such-that-a-g2-g-b-e-is-a-point-of-d-c-such-that-c--1414851
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# A B C D is a parallelogram, G is the point on A B such that A G=2\ G B ,\ E\ is a point of D C such that C E=2D E\ a n d\ F is the point of B C such that B F=2F Cdot Prove that: a r\ (\ E B G)=a r\ (\ E F C)
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https://www.experts-exchange.com/questions/21875979/Convert-Signed-Big-Endian-short-to-Little-Endian-signed-Short.html
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# Convert Signed Big Endian short to Little Endian signed Short
Posted on 2006-06-06
Medium Priority
1,645 Views
I have a file that contains signed shorts in big-endian format.
How do I convert these to little-endian SIGNED format?
I thought I had this working, but later realized I didn't handle negative values.
Thanks
0
Question by:oxygen_728
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LVL 12
Assisted Solution
rajeev_devin earned 1200 total points
ID: 16841055
Just interchange the bytes
0
LVL 12
Accepted Solution
rajeev_devin earned 1200 total points
ID: 16841119
A technique
union BYTES {
unsigned short number;
unsigned char bytes[2];
};
BYTES b;
b.number contains the number in little-endian.
0
LVL 4
Assisted Solution
ID: 16841424
It is much more efficient if you read them in a batch and use the swab function!
E.g:
#include <cstdlib>
#include <cstdio>
using namespace std;
short inputArray[1024], array[1024];
// now swap the bytes
swab(reinterpret_cast<char*>(inputArray), reinterpret_cast<char*>(array), sizeof inputArray);
Or, if you would like to do it without the extra memory buffer, but with a speed penalty, you can do it 'inplace', like this:
inline inswab(char *array, size_t bytes) {
for (int i = 0; i < bytes; i += 2) {
register char aux = array[i];
array[i] = array[i+1];
array[i+1] = aux;
}
}
See:
http://www.cplusplus.com/ref/cstdlib/swab.html
http://www.codeproject.com/cpp/endianness.asp
0
LVL 17
Assisted Solution
rstaveley earned 200 total points
ID: 16843132
If your application is running on a little-endian system, you can use ntohs and htons to convert to and from the host (little-endian) from and to network (big-endian) byte order. See http://www.google.com/search?q=htons
0
Author Comment
ID: 16845262
raj:
Are you sure it is as simple as interchanging the bytes? There's a sign bit in there somewhere.
I switched over from unsigned shorts to signed shorts and my results are screwy with the numbers that should be negative
0
Author Comment
ID: 16845284
well, that doesnt seem far from what I'm doing:
ifstream in;
in.open("N36W117.hgt",ios::in | ios::binary);
int x;
short Val = 0;
int counter = 0;
Vertices = new SimpleVertex[1201*1201];
while (counter < 1201 * 1201){
Val = (Val << 8) | (Val >> 8); //<----------------------------------------------------
Vertices[counter].x = counter % 1201;
Vertices[counter].y = Val;
Vertices[counter].z = counter / 1201;
counter++;
}
0
Author Comment
ID: 16845642
I think I found part of the problem;
short Val;
Val = 0xa300
Val = Val >> 8;
Val will equal ffa3 instead of 00a3
Must be a sign issue.
0
Author Comment
ID: 16845927
Ya, apparently I was using a short too early - there was some funky stuff going on with the bit shifting.
Val >> 8 SOMETIMES would fill in the 'new' bits (on the left) with f's instead of 0's (hex)
It's all fixed now, I used bytes - and I learned about this strange union structure thanks to rajeev.
Thanks guys
0
LVL 53
Expert Comment
ID: 16846546
>> Val >> 8 SOMETIMES would fill in the 'new' bits (on the left) with f's instead of 0's (hex)
that "SOMETIMES" happens when the left most bit is set to 1 ... by shifting, you shift in that same bit on the left side. That doesn't happen with unsigned integer types though.
The rule with converting between little and big endian is to either use unsigned types, or to avoid shifting when using signed types.
Any way you inter-change bytes would be ok ... and I see you already have some examples given by the others :)
0
LVL 4
Expert Comment
ID: 16848096
Note: if you use "unsigned short", the problem of filling 1's in the left should not happen. It only happens with signed shift, because of negative numbers in 2's-complement!
0
Author Comment
ID: 16848288
Perhaps I could have used unsigned shorts, then performed a cast to a signed short? I have a feeling that conversion does something goofy to overcome the negativity though =\
0
LVL 53
Expert Comment
ID: 16849669
Can I ask why you need to extract certain bits from a signed short ? It sounds like it's something that could be done easier differently.
0
LVL 17
Expert Comment
ID: 16850253
> Val >> 8 SOMETIMES would fill in the 'new' bits (on the left) with f's instead of 0's (hex)
The problem occurred when you made the original assignment, which set the signed bit.
--------8<--------
#include <stdio.h>
int main()
{
short Val;
Val = 0xa300; // Same as assigning -23808 to the short
printf("%hd %hx %x\n", Val, Val, Val);
Val = Val >> 8;
printf("%hd %hx %x\n", Val, Val, Val);
}
--------8<--------
The shift right operator drags the signed bit into the most significant bit, when you operate on signed integers.
0
Author Comment
ID: 16850388
Well, I'm just reading in data in big-endian signed-short format... and I need the data in little-endian signed-short format.
0
Author Comment
ID: 16850394
rstavlely - I thought I had read somewhere that 0's are always 'dragged in' ... I don't remember reading anything about the sign bit being dragged in - thus my confusion =)
Thanks for clearing things up!
0
LVL 53
Expert Comment
ID: 16850441
>> Well, I'm just reading in data in big-endian signed-short format... and I need the data in little-endian signed-short format.
My apologies - I mixed up two questions ... ignore my question :)
>> I thought I had read somewhere that 0's are always 'dragged in'
That's true for unsigned types
0
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# Square Room Tiles
The square floor of a square room was covered with colored 1 square unit tiles.The covering tiles were designed to have different (solid) colored tiles that forms rectangular areas but none of these areas have the same dimension with another area. How many tiles are needed to cover the smallest floor?
• do they have to be rectangular areas or can the tiles form square areas of the same colour? Jun 24, 2017 at 18:55
• I would advise disallowing square areas because then you can just have '1' as the answer. See my answer Jun 24, 2017 at 19:05
• Yes,squares are rectangles with the same dimensions. theyre not allowed
– TSLF
Jun 24, 2017 at 19:11
• Let me check whether I've understood everything correctly, with the help of comments on BG's now-deleted answer. We have a room of size n by n units, for some currently unknown n. It's tiled by rectangles; let's say k of them. So there are 2k rectangle side-lengths, and no two of these are equal. We want the smallest n for which this can be done, and the corresponding value of k. Is that right? Are the tiles' side-lengths required to be integers? Jun 24, 2017 at 20:19
• yes ,whole tiles. no cuts
– TSLF
Jun 25, 2017 at 3:40
As others have already said, the arrangement with the fewest rectangles has 5 rectangles. This is because:
• No rectangle can use up the whole side of the floor. If one did, the rectangular arrangement used in the remainder of the floor could be scaled to fill the whole floor with one fewer rectangle.
• You therefore need a different rectangle in each corner of the floor.
• Each rectangle has at least 3 neighbours (otherwise on its two internal sides one of the neighbours has a matching dimension)
• Four rectangles (one in each corner) cannot work as both diagonally opposite pairs would need to be neighbours.
So we need at least 5 rectangles. If the fifth rectangle were in the middle of a side, the two adjacent corner rectangles would still need to be adjacent to their diagonally opposite corners to get three neighbours, so that is not possible. The fifth rectangle must therefore be an interior rectangle like this:
With 5 rectangles, the smallest dimensions they could have are the numbers 1 to 10, so the square floor has dimensions of at least 11x11. I think there are only two solutions for 11x11.
Using the letters in the diagram, we clearly have:
b+c=11
d+e=11
f+g=11
h+a=11
a+e+i=11
c+g+j=11
It turns out that all further equations you can set up (e.g. for total area) are linearly dependent on these.
{b,c},{d,e},{f,g},{h,a} are four of the five pairs {1,10},{2,9},{3,8},{4,7},{5,6}. This means that {i,j} is the fifth unused pair, and we can deduce that i+j=11.
So a+e=j and c+g=i. Obviously i,j>=3. This rules out {i,j}={1,10} or {2,9}.
If j=5, then the only choices for {a,e} are {2,3} and {1,4}. So {a,h,d,e} is either {2,9,3,8} or {1,10,4,7}, and {b,c,f,g} is the other. In neither case can we get c+g=6. The same goes for i=5, so {i,j}={5,6} is not possible.
If j=3, then the only choice for {a,e} is {1,2}. So {a,h,d,e}={1,10,2,8}, and {b,c,f,g}={4,7,5,6}. From this we cannot get c+g=8. The same goes for i=3, so {i,j}={3,8} is not possible.
This leaves only {i,j}={4,7}. We can assume wlog that i=4, j=7 because we can rotate everything by 90 degrees to swap i and j. Then {c,g}={1,3}. We can rotate everything by 180 degrees to swap c and g, so we can assume wlog that c=1, g=3. We have {a,e}={2,5}, and a=2 leads to the solution above, and a=5 gives the following solution:
In text form, the two solutions are:
2x10,1x6,5x8,3x9,4x7
5x10,1x9,2x8,3x6,4x7
For completeness, suppose we allow some of the areas to be squares, but disallow the trivial solution consisting of a single area. Much of the above argument works, until the point where it was deduced that {i,j} is the remaining pair. It is now possible for i=j to equal only one of the numbers in the remaining pair. One that choice is made, similar arguments to before then either prove there is no solution, or produce 2 solutions. I won't go through the cases in detail, but just list the solutions:
1x8,3x7,4x9,2x10,6x6
1x9,2x7,4x8,3x10,6x6
1x5,6x4,7x9,2x10,3x3
1x9,2x4,7x5,6x10,3x3
1x6,5x3,8x7,4x10,2x2
1x7,4x3,8x6,5x10,2x2
2x7,4x3,8x5,6x9,1x1
2x5,6x3,8x7,4x9,1x1
I found no solution when i=j are equal to 4, 5 or greater than 6.
• -are there also 2 solutions using nxn rectangle if allowed?
– TSLF
Jun 25, 2017 at 19:20
• @TSLF: I've added the 8 solutions I found if squares are allowed. Jun 26, 2017 at 4:50
The smallest floor can be covered in
5 rectangles.
I know this because
You cannot have a square area covered in 4 rectangles because there is no way to put 4 rectangles in a square grid without at least 2 of them sharing an edge. I have found an example for 5 rectangles for an 11 by 11 unit room. In that example you have (in clockwise order) a 6*3 rectangle, a 2*8 rectangle, a 1*9 rectangle and a 5*10 rectangle on the edges. This leaves a 4*7 rectangle in the center. I know that an 11 by 11 unit room is the smallest room made up of 5 rectangles because for 5 rectangles you need 10 dimensions and they cannot be 0 or the width of the square.
Floor Tiles layout:
• Wrong, actually - check the central rectangle, it ix in fact 5x7. That duplicates the 7 measurement of the 1x7. Fitting 5 rectangles into 11x11 may be possible but this layout is not a valid solution. (@TSLF you may want to unAccept this until it is fixed)
– Rubio
Jun 25, 2017 at 4:32
• I have amended my answer now. Hopefully I didn't make another stupid mistake. Jun 25, 2017 at 7:23
• @ZadokStorkey-Looks like Jaap's edit was 14min earlier
– TSLF
Jun 25, 2017 at 19:21
The first key observation is that
where two rectangles meet, they can never do so along a complete edge of both -- for then they would have to have a dimension in common.
This means that
the number of rectangles (what I called k in a clarification-seeking comment on the original question, and will continue to call k now) must be at least 5; and I think there is "combinatorially/topologically" exactly one pattern with only 5 rectangles.
Here is a pretty small way to do it with that layout. I don't have a proof that this makes the room size (what I called n in that comment) as small as possible, and indeed it may very well not be. I doubt it's possible to do much better, though.
A.A.A.A.A.B.B.B.B.B.B.B.B. A.A.A.A.A.B.B.B.B.B.B.B.B. A.A.A.A.A.B.B.B.B.B.B.B.B. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. E.E.E.E.E.E.E.E.E.E.E.D.D.
The dimensions are:
13x13 (for the room); 12x5, 8x3, 9x6, 10x2, 11x1 for the rectangles.
(Assuming I read this correctly; dimensions equivalent means both dimensions are equivalent) If square sub tiles are allowed, then $3*1$ can be tiled with $2*1$ and $1*1$. Otherwise, tile an $5*1$ with a $3*1$ and a $2*1$
• It needs to be a square room though Jun 24, 2017 at 18:58
• Oh sure no doubt something looked wrong. Jun 24, 2017 at 18:58
• Hi and welcome to the site! Have you taken the tour yet? Jun 24, 2017 at 18:59
• OK deleting this as I don't even understand the question properly. Jun 24, 2017 at 19:02
• (But you didn't actually delete it. You might want to before it gets more downvotes.)
– Rubio
Jun 25, 2017 at 1:55
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Combinatorial Optimization Easy or Hard
COP's are interesting because they include both the easiest and hardest problems to solve. The difference is described by the partition of all optimization problems into two sets, P and NP-Complete. The classification of problems into these two sets and by other distinctions is the subject of complexity theory and is far beyond this simple introduction. The subject is important to OR practice, however, because a particular problem will probably be tractable if it is identified as a member of P, while the problem will probably be difficult or impossible to solve if it is identified as a member of NP-Complete. A problem in the set P can be solved by an algorithm whose solution time grows as a polynomial function of the size of the problem. For example the shortest path problem for a network with having n nodes and with non-negative arc lengths can be solved with Dijkstra's algorithm. If implemented correctly, the computation time is bounded from above by a constant times the number of nodes squared. We say the solution time is order or . Since the function in the parentheses is a polynomial, this is a polynomial algorithm and the shortest path problem with positive arc lengths is in the set P. We say that this is an easy problem because large problems can be solved in reasonable times with the proper algorithm and on a fast computer. A variety of special case COP's fall in the set P including versions of the assignment, minimal spanning tree, shortest path tree, pure network flow programming and linear programming problems. For the problems in the set NP-Complete, no polynomial algorithms have been found. As an example, the time for enumerating all the solutions of a binary IP with n variables is . The function in the parentheses is exponential in n. This is bad news. Although it may be possible to solve problems by enumeration for some small n, it is clear that for some larger n the enumeration will be impossible. One might ask, "why not solve the problem with a more efficient method such as branch and bound or with cutting planes?" The answer is that these methods also have exponential bounds. In fact, no polynomial algorithm has been found for the binary IP problem. If one could find one, the problem would be in class P. In fact, most researchers would say that it is unlikely that such an algorithm will ever be discovered. Again, one might ask, why not wait for a faster computer to be invented? Surely there will be one. The answer is that no matter how fast computers become, exponential growth will eventually make problems of some size unsolvable. On the following pages we call problems in the NP-Complete class hard problems. What does this say to someone who must solve a COP? If the problem can be reduced to a member of the class P, just apply the associated the algorithm. Even if a problem is in P it may not be easy to solve because appropriate algorithms are not always easy to find, easy to program or inexpensive to purchase. For a given COP it is often not easy to tell if it can or cannot be reduced to P. Even if the COP is identified as a member of P, a slight variation of the problem, perhaps a change in form for the objective function or the addition of a constraint, may cause the COP to be no longer reducible to P. It is easy to find the shortest path from node a to node b in network, but if the path is required to have at least a specified number of arcs, the problem becomes hard. If the problem cannot be reduced to a problem in P, does that mean it is in the class NP-Complete? We can prove that a problem is in NP-Complete if we can take some problem that is known to be NP-Complete and reduce it to the problem at hand. Again, this is not always easy to determine. If the analyst is not able to classify a problem, if it cannot be reduced to a member of P or one does not have a convenient algorithm, is all lost? The answer is that a problem can be solved to optimality if it is not too big. If it is too big, then there are heuristic methods that may yield acceptable solutions.
Operations Research Models and Methods
Internet
by Paul A. Jensen
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# How to back transform a folded power transformation?
I work with a dependent variable with values from 0 to 1 and because I have a lot of 0 and 1 in this variable, i'm doing a folded power transformation with the following formular: y = x^0.5 - (1-x)^0.5. In order to be able to describe the effect strength of the DV in a regression later on, I want to back transform the results of the folded power transformation.
I saw here: How to back transform a folded root?, that @Izy had a similar question. But the formula he gets to back transform the folded power transformation, which is this formula:c=((1−(f(p))2)/2)^2 doesn't work for me. Does anyone know what the problem is or what I am doing wrong? I would be grateful for any advice.
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# A source of monochromatic radiation
Question:
A source of monochromatic radiation of wavelength $400 \mathrm{~nm}$ provides $1000 \mathrm{~J}$ of energy in 10 seconds. When this radiation falls on the surface of sodium, $x \times 10^{20}$ electrons are ejected per second. Assume that wavelength $400 \mathrm{~nm}$ is sufficient for ejection of electron from the surface of sodium metal. The value of $x$ is________
(Nearest integer)
$\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}\right)$
Solution:
Total energy provided by
Source per second $=\frac{1000}{10}=100 \mathrm{~J}$
Energy required to eject electron $=\frac{h c}{\lambda}$
$=\frac{6.626 \times 10^{-34}}{400 \times 10^{-9}} \times 3 \times 10^{8}$
Number of electrons ejected
$=\frac{100}{\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9}}}$
$=\frac{400 \times 10^{-7} \times 10^{26}}{6.626 \times 3}$
$=\frac{40 \times 10^{-20}}{6.626 \times 3}$
$=2.01 \times 10^{20}$
#### Leave a comment
None
Free Study Material
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Saturday, February 23, 2008
What's the percentage of "adders-across" in Numeracy?
In the past, I've given diagnostics before a unit so as to be able to compare pre- and post-instruction scores. Now, in the spirit of differentiation, I'm going to go one step further.
The next unit is about adding and subtracting fractions and mixed numbers. On my diagnostic, I wanted to see what percent of the students are still "adders-across" (#25 down: snakes that are bad at math). That would be 68/80, or 85%. The remaining 12 students could all do the basic algorithms, but most stumbled on the more complicated mixed number subtraction problem.
So here's the plan. In each class, I will assign one of the non-adders-across (NAA) to an adder-across (AA), tasking the NAA to help the AA learn over the coming lessons. If I see that they remain on task during practice time, the NAA will not have to take the quizzes, earning an automatic 100% on them. This seems reasonable, since they have already shown me they know the skill. Additionally, if the AA passes the quizzes (i.e. becomes an NAA!) then the NAA helper will earn some oh-so-coveted extra credit points. This way, the NAA has strong incentive to help, but there is no penalty if the AA doesn't make enough improvement.
Since almost no students showed mastery of the mixed number subtraction problems, every one will need to take that quiz when we get to it.
Now, the only thing that remains is to pair up the NAAs with the AAs effectively. I need to factor in personality, motivation, and so forth. Also, this experiment really highlights the imbalance between classes, even though we try to avoid any tracking (a constant difficulty in a small school). Here are the numbers of NAAs by period... Period 1: 5, Period 2: 4, Period 4: 2, Period 6: 1.
jd2718 said...
So adding across, in its own way, is pretty cool. It doesn't give a sum, but it is a quick way of getting a number in between two numbers.
1/2 and 2/3 ? 3/5 is in between them.
2 and 3? That's 2/1 and 3/1. 5/2 is between them.
I don't know if that helps your kids at all (I kind of doubt it), but it's still pretty cool.
Good luck
continuities said...
Looking at the number of NAA's in each section, pairing will be interesting. Unless of course you have some really small sections. Good Luck!
Dan Greene said...
jd - That's cool - I never thought about that before. But you're right, I think it would just confuse the issue for the students more than anything right now. Maybe I can go back to this toward the end of the unit and see if they can explain why it works.
Continuities - my Numeracy sections are at 20 students each (we keep the remedial classes as small as possible). So not all students can be paired up with someone who already has the adding skill. It will be interesting, because I don't want to make simple high-low pairs.. I want students who can effectively collaborate, which can be hard to predict sometimes. My 6th period has my weakest students in it - only 1 already can add fractions. But, since the start of the new semester, there have been a few kids added to that class, and they have been a breath of fresh air. They don't have the skills, but they have the motivation, and they are helping bring some of the others up. I love that feeling when you have a class that you used to dread going to, and now becomes one you look forward to. It's amazing what an impact switching a few students out of/into a room can have on the culture of a class.
Mr. K said...
This is one thing my students learned well this year. (The fact that i'd scheduled 3 months just for fractions might have something to do with it.)
We used fraction strip manipulatives to start. They had a lot of experience with these, so not a lot of prep work was needed. We started off just adding fractions with like denominators - they worked in pairs, each one finding their part of the sum and putting them together.
They quickly realized (during the recording part - making them write what they see is important) that they just added the numerators, and the denominators stayed the same. I asked them often enough that they soon had a mantra of "during adding, the denominator stays the same". Note that I didn't give them this as a mnemonic - they figured it out and I just asked them enough to cement it for the next step.
That was asking them to try the fraction strips with different denominators. Confusion! They couldn't figure out which denominator to use! Some tried adding the two numbers together, but quickly realized on their own that meant a new strip which obviously wasn't the sum of the two.
Fortunately, they had equivalent fractions down cold at this point, so when i rewrote one of them with a new denominator, they quickly got it, and were able to rewrite the second one. Everyone got the having to rewrite part after only a couple of tries - as long as I was providing them with the common denominator. They referred to this as "making the pieces the same size", since that's how it looked on their fraction strips.
It was at *this* point that I let them teach each other. Some were stuck with just multiplying the two denominators together, and having to simplify later. Others got the idea of using the least common multiple (why introduce a new term for something they already knew how to do?) and happily used that as their new denominator.
I had numbers like your 6th period for adders across, and now have 90% of my kids not only doing it right, but understanding why it has to work that way. None of them will use the phrase "lowest common denominator", but they can all do it.
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# Voltage Divider
The Swiss Army Knife of Circuits
The questions below are due on Monday February 18, 2019; 11:59:00 PM.
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Note that this link will take you to an external site (https://oidc.mit.edu) to authenticate, and then you will be redirected back to this page.
Back to Exercise 02
Joy Divider
We'll now study/derive/analyze/play with one primary type of circuit: The voltage divider. In order to figure this circuit out, however we need to run through some more basic circuit theory first.
In last week's exercises, we studied circuit network theory, the idea that components when connected together are subject to two major laws:
• Kirchoff's Voltage Law: The voltage around any closed loop in a circuit must sum to zero
• Kirchoff's Current Law: The currents into any junction in a circuit must sum to zero
Given a network topology and sufficient starting information, it should always be possible to define the voltages across and the currents through all components in any circuit.
## 1) Resistors
Many components constrain/dictate how the current through them and the voltage across them are related, and this can further dictate what a circuit settles at when built. One part we'll consider today is the resistor.
In schematic form,1 resistors take on the form of a zig-zag:
A resistor asserts that the voltage across it, v, is proportional to the current through it, i, times its Resistance, R (measured in Ohms). Many people learn Ohm's Law, and believe that as long as you violently shout "V = IR" then you'll be good, but there's more to it than that. The direction/orientation really, really matter. As we've drawn it in the figure above, Ohm's Law states:
v = i\cdot R
What Ohm's Law tells us is that current (i) will flow from areas of high potential to lower potential. If v is positive that means that i will also be positive. If v is negative, that means i will also be negative.
We can also draw out a resistor and state Ohm's Law in terms of node voltages (again a concept introduced last week):
Here Ohm's Law would be written as:
v_a - v_b = i\cdot R
The two v variables represent electrical potential (which we often call voltage) at the top "node" and bottom "node". If the voltage at node A, v_a is larger than the voltage at node B, v_b, that means current is going to be positive. If v_a is lower than v_b, then the current i will be negative.
If v_a =11\text{V}, v_b=2\text{V}, and R=190\Omega, what would current i be (in Amps)?
Now, what if v_a =-11\text{V}, v_b=-2\text{V}, and R=190\Omega, what would current i be (in Amps)?
This should emphasize a really important point that i can be either positive or negative. What does it mean for i to be positive? You just need to think of i as a frame of reference2
## 2) Ground
When working with node voltages, we need to define one node as being 0.0 Volts. We'll often call this node the ground of our circuit.
When we define one node as ground, we then define all other voltages with respect to that ground node. So if you had a circuit where v_1 = 10\text{V} and v_2 = 5\text{V} and we decided to define the v_2 node as Ground, what would we now say the voltage at v_1 is (in volts)?
And what would the current through that resistor connected to nodes v_1 and v_2 be if R = 50\Omega and current is defined as going from node 1 to node 2(in Amps)?
Now what if we had a circuit where v_1 = 1\text{V} and v_2 = 9\text{V} and we decided to define the v_2 node as Ground, what would we now say the voltage at v_1 is (in volts)?
And what would the current through that resistor be if R = 50\Omega (in Amps)?
A lot of times when you have a fixed positive voltage we call it V_{cc}. In many of our ESP32-based circuits our V_{cc}=3.3\text{V}, for reference. A resistor going from V_{cc} to ground could therefore be drawn like so:
What would the current i be in the circuit above if V_{cc}=3.3\text{V} and the resistance R is 787\Omega? Just to make life interesting provide your answer in milliAmps.
## 3) Voltage Divider
OK, now what happens when you connect two resistors in series like what is shown below? Let's define everything for the top resistor with underscore a and everything with the bottom resistor with underscore b. Drawing it out and labeling gives the following
Let's redo our math from earlier for each resistor.
v_{a1}-v_{a2} = i_aR_a
and
v_{b1}-v_{b2} = i_bR_b
Current flowing through R_a (downward) must equal the current flowing through R_b (downward). This should sort of be intuitive from looking at this figure and this is known as Kirchoff's Current Law. We'll call this common current i_{total}.
i_a = i_b = i_{total}
We should also be able to recognize that the voltage described by v_{a2} and v_{b1} are referring to the same node and therefore (they refer to teh same electrical spot):
v_{a2}=v_{b1}= v_{mid}
When we substitute these new insights into our we get:
v_{a1} - v_{mid} = i_{total}R_a
and
v_{mid} - v_{b2} = i _{total}R_b
Let's now assume that we're in a more specific circuit situation where our bottom voltage (v_{b2}) is actually at GND, which means we can call it 0. Also let's assume that our top voltage v_{a1} is now just V_{cc}. Further, the voltage v_{mid} can be called v_{out}.
The two equations above become:
V_{cc} - v_{out} = i_{total}R_a
and
v_{out} = i_{total}R_b
Shuffling the second equation around so that it expresses:
i_{total} = \frac{v_{out}}{R_b}
and then substitutin that into the first equation:
V_{cc}-v_{out} = \frac{v_{out}R_a}{R_b}
which then becomes:
V_{cc} = v_{out} +\frac{v_{out}R_a}{R_b}
which can be rewritten as:
V_{cc} = v_{out}\left(\frac{R_a+R_b}{R_b}\right)
Then rearranging so our equation is more like
v_{out}
as an output we have:
V_{out} = V_{cc}\left(\frac{R_b}{R_a+R_b}\right)
This is known as the voltage divider equation and is extremely versatile since it fully describes the voltage divider circuit above.
Yes, it looks very simple and its math is very simple, but we can do many things with it, and we'll use it in coming labs and homeworks.
## 4) Practice
Use the voltage divider equation to solve specific cases of the voltage divider circuit below.
What is the output voltage v_{out} (in Volts) if V_{cc}=5V, R_a = 8\Omega, and R_b=9\Omega?
What is the output voltage v_{out} (in Volts) if V_{cc}=15V, R_a = 14\Omega, and R_b=5\Omega?
What is the output voltage v_{out} (in Volts) if V_{cc}=11V, R_a = 9\Omega, and R_b=10\Omega?
## 5) The Voltage Divider and Programming
Doing the same math again and again repeatedly gets boring. We've been doing lots of C++ programming. Just to keep the Python muscles limber, let's use some Python to write a function that will do the math for us. Define a function voltage_divider that takes in three inputs vcc, ra, and rb and returns the output voltage v_{out} in Volts.
def voltage_divider(vcc,ra,rb): pass
## 6) Next Steps
We can now use the voltage divider circuit to generate many types of sensor circuits. By replacing one of the "fixed" resistors with a device who's resistance varies as a function of some external signal (heat, light, touch, whatever), you can have a circuit that with input. This forms what we call a transducer. For example in the circuit below, the resistance of the lower element is a function of light. As a result, the voltage out v_{out} is a function of the light! The ESP32 can measure this voltage using a call to analogRead(which we'll talk about in future weeks).
In circuits like these it'll actually is productive to use the voltage divider in a backwards form (and this is where our understanding of this circuit becomes really useful. We'll be getting readings of the voltage V_{out} and we'll need to use those to extract what value R_{sensor} (R_b) is in order to use that to figure out the temperature, amount of light etc...
So give this a try real quick: What must the value of R_b be (in Ohms), if V_{cc}=11V, R_a = 9.00\Omega, and V_{out}=5.79V?
Now create a function resistanceExtractor. It should take in three values, vcc, ra, and vout, correspondign to V_{cc}, R_a, and v_{out}, respectively, .and return back the calculated value of R_b.
def resistanceExtractor(vcc,ra,vout): pass #your code here
Back to Exercise 02
Footnotes
1In the US, anyways, in many other countries a box is used...I'm looking at you UK. (click to return to text)
2Do not dismiss the fact that sign is important. If someone told you that your bank account changed by either positive or negative \$1,000,000 the sign would really matter to you. Same goes for voltages and currents! Moreso even. Money can't buy happiness, but voltage and current can. (click to return to text)
This page was last updated on Monday February 11, 2019 at 02:15:42 PM (revision 3b75fe6).
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# Famous Mathematicians
### Task 135 ... Years 4 - 10
#### Summary
Using the popular context of a word find puzzle students become familiar with the names and life span of 18 historic mathematicians. They discover that some are male and some are female and that there have been famous mathematicians in many centuries. From here students are encouraged to research more deeply into the life of one from the list. The bigger challenge is to find names and details of three mathematicians alive today.
#### Materials
• Board, marking pen and cloth
• List of names of mathematicians
#### Content
• history of mathematics
• data representation and analysis
#### Iceberg
A task is the tip of a learning iceberg. There is always more to a task than is recorded on the card.
We would not expect students to become authors without knowing something of the life and work of famous writers. Similarly if students are learning to work like a mathematician, then it is especially appropriate to come into contact with the lives and work of professionals in the field. This task starts that process.
The solution to the word find is:
Answers to the remainder of the card depend on individual research results. The web offers plenty of information in response to any of the names on the list. A significant site for information about women mathematicians is: http://www.agnesscott.edu/lriddle/women/women.htm
One way of extending the task is to challenge students to calculate the age of each listed mathematician.
• Which mathematician lived for the shortest time?
• Which mathematician lived for the longest time?
• Make a graph showing the names and ages of these mathematicians in order.
• Using the data as a guide, is it possible to decide whether male or female mathematicians live longer?
#### Whole Class Investigation
Tasks are an invitation for two students to work like a mathematician. Tasks can also be modified to become whole class investigations which model how a mathematician works.
The word puzzle can easily become a class activity by first recreating the word find board using the table making tool of a word processor. Include the names on the same page. Print a copy for each pair of students and ask them to try the puzzle. Display the same word finder on the Interactive White Board. As names are found students 'cross them out' on the whiteboard. There are 18 names to find so try to give each pair a chance to cross off at least one on the board.
Use the exercise to begin a discussion of current knowledge of these people - emphasise people - and to encourage thought about what their lives may have been like.
• How do you think the life of Euclid might have been the same as, or different from, the life of Fermat?
Assign one of the name to each pair and suggest that each pair is now expected to find out information about the life of their mathematician. As a class develop a list of research questions that everyone could use. Try to raise the intellectual level of a few questions above that of just collecting facts; perhaps by including questions such as:
• In what ways do you think your mathematician's life was successful?
• In what ways do you think your mathematician's life could have been better?
• In what ways do you think your mathematician's life made the world a better place for others?
• Is there any part of your mathematician's life that you think they would change if they could? Explain.
At least one teacher has continued the use of mathematician's names as a group name for a pair of students doing other investigations. See Becky & Lydia's report in the PowerPoint at Settlebeck High School section of the Recording & Publishing section in Mathematics Task Centre.
Teachers could also include this investigation in a regular Mathematician's Friday, an idea illustrated by the work of Catherine La Franchi and staff of St. Vincent de Paul Primary School.
Maths300 Lesson 124, Famous Mathematicians uses the faces and names of six famous mathematicians as collectable cards distributed as a promotional give away in Pythagoras Popcorn. In this context, the investigation shifts to the time it is likely to take to collect a complete set. An additional underlying purpose of this lesson is to explore how a mathematician applies the strategy of make a model to a real-life situation. The lesson includes information that can readily be turned into posters of each mathematician. Of course, the information from the students research as described above could also be the source of the posters and cards that stimulate this lesson.
#### Is it in Maths With Attitude?
Maths With Attitude is a set of hands-on learning kits available from Years 3-10 which structure the use of tasks and whole class investigations into a week by week planner.
The Famous Mathematicians task is an integral part of:
• MWA Space & Logic Years 7 & 8
The Famous Mathematicians lesson is an integral part of:
• MWA Chance & Measurement Years 9 & 10
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"
">
# From the adjoining figure find the values of x and y:"
Given :
The triangle with an internal angle one internal angle 50° and an exterior angle 120° is given.
To do :
We have to find the value of x and y.
Solution :
We know that,
The sum of the angles on a straight line is 180°.
The sum of the angles in a triangle is 180°.
From the figure,
$y+120°=180°$
$y = 180°-120°$
$y = 60°$
$x+50°+y = 180°$
$x+50°+60° = 180°$
$x+110° = 180°$
$x = 180°-110°$
$x = 70°$.
The values of x and y are 70° and 60° respectively.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
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# chemistry
posted by on .
thank you for any help possible.
Given a Mass Percent 13.3% and a Mass Solvent of 325.2 g, find Mass of Solute and Mass of Solution.
Mass%=13.3%
Mass solvent= 325.2g
solve for the Mass solute and Mass solution. This is all the information I am given.
• chemistry - ,
mass% = [(g solute/g solute + g solvent)]*100
Let x = grams solute
13.3 = 100*(x/x + 325.2)
Solve for x = g solute.
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# CHAPTER 8 Stocks and Their Valuation
Document Sample
``` CHAPTER 9
Stocks and Their Valuation
Features of common stock
Determining common stock values
Preferred stock
9-1
Facts about common stock
Represents ownership
Ownership implies control
Stockholders elect directors
Directors elect management
Management’s goal: Maximize the
stock price
9-2
Intrinsic Value and Stock Price
Outside investors, corporate insiders, and
analysts use a variety of approaches to
estimate a stock’s intrinsic value (P0).
In equilibrium we assume that a stock’s price
equals its intrinsic value.
Outsiders estimate intrinsic value to help
determine which stocks are attractive to
Stocks with a price below (above) its
intrinsic value are undervalued
(overvalued).
9-3
Determinants of Intrinsic Value
and Stock Prices (Figure 1-1)
9-4
Different approaches for estimating the
intrinsic value of a common stock
Dividend growth model
Corporate value model
Using the multiples of comparable
firms
9-5
Dividend growth model
Value of a stock is the present value of the
future dividends expected to be generated by
the stock.
^ D1 D2 D3 D
P0 ...
(1 rs )1
(1 rs ) 2
(1 rs ) 3
(1 rs )
9-6
Constant growth stock
A stock whose dividends are expected to
grow forever at a constant rate, g.
D1 = D0 (1+g)1
D2 = D0 (1+g)2
Dt = D0 (1+g)t
If g is constant, the dividend growth formula
converges to:
^ D 0 (1 g) D1
P0
rs - g rs - g
9-7
Future dividends and their
present values
t
\$ D t D0 ( 1 g )
Dt
0.25 PVD t
( 1 r )t
P0 PVD t
0 Years (t)
9-8
What happens if g > rs?
If g > rs, the constant growth formula
leads to a negative stock price, which
does not make sense.
The constant growth model can only be
used if:
rs > g
g is expected to be constant forever
9-9
If rRF = 7%, rM = 12%, and b = 1.2,
what is the required rate of return on
the firm’s stock?
Use the SML to calculate the required
rate of return (rs):
rs = rRF + (rM – rRF)b
= 7% + (12% - 7%)1.2
= 13%
9-10
If D0 = \$2 and g is a constant 6%,
find the expected dividend stream for
the next 3 years, and their PVs.
0 1 2 3
g = 6%
D0 = 2.00 2.12 2.247 2.382
1.8761
rs = 13%
1.7599
1.6509
9-11
What is the stock’s intrinsic value?
Using the constant growth model:
ˆ D1 \$2.12
P0
rs - g 0.13 - 0.06
\$2.12
0.07
\$30.29
9-12
What is the expected market price
of the stock, one year from now?
D1 will have been paid out already. So,
P1 is the present value (as of year 1) of
D2, D3, D4, etc.
^ D2 \$2.247
P1
rs - g 0.13 - 0.06
\$32.10
Could also find expected P1 as:
^
P1 P0 (1.06) \$32.10
9-13
What are the expected dividend yield,
capital gains yield, and total return
during the first year?
Dividend yield
= D1 / P0 = \$2.12 / \$30.29 = 7.0%
Capital gains yield
= (P1 – P0) / P0
= (\$32.10 - \$30.29) / \$30.29 = 6.0%
Total return (rs)
= Dividend Yield + Capital Gains Yield
= 7.0% + 6.0% = 13.0%
9-14
What would the expected price
today be, if g = 0?
The dividend stream would be a
perpetuity.
0 1 2 3
rs = 13%
...
2.00 2.00 2.00
^ PMT \$2.00
P0 \$15.38
r 0.13
9-15
Supernormal growth:
What if g = 30% for 3 years before
achieving long-run growth of 6%?
Can no longer use just the constant growth
model to find stock value.
However, the growth does become
constant after 3 years.
9-16
Valuing common stock with
nonconstant growth
0 r = 13% 1 2 3 4
s
...
g = 30% g = 30% g = 30% g = 6%
D0 = 2.00 2.600 3.380 4.394 4.658
2.301
2.647
3.045
4.658
46.114 \$
P3 \$66.54
^ 0.13 - 0.06
54.107 = P0
9-17
Find expected dividend and capital gains
yields during the first and fourth years.
Dividend yield (first year)
= \$2.60 / \$54.11 = 4.81%
Capital gains yield (first year)
= 13.00% - 4.81% = 8.19%
During nonconstant growth, dividend yield
and capital gains yield are not constant,
and capital gains yield ≠ g.
After t = 3, the stock has constant growth
and dividend yield = 7%, while capital
gains yield = 6%.
9-18
Nonconstant growth:
What if g = 0% for 3 years before long-
run growth of 6%?
0 r = 13% 1 2 3 4
s
...
g = 0% g = 0% g = 0% g = 6%
D0 = 2.00 2.00 2.00 2.00 2.12
1.77
1.57
1.39
2.12
20.99 \$
P3 \$30.29
^ 0.13 - 0.06
25.72 = P0
9-19
Find expected dividend and capital gains
yields during the first and fourth years.
Dividend yield (first year)
= \$2.00 / \$25.72 = 7.78%
Capital gains yield (first year)
= 13.00% - 7.78% = 5.22%
After t = 3, the stock has constant
growth and dividend yield = 7%,
while capital gains yield = 6%.
9-20
If the stock was expected to have
negative growth (g = -6%), would anyone
buy the stock, and what is its value?
The firm still has earnings and pays
dividends, even though they may be
declining, they still have value.
^ D1 D0 ( 1 g )
P0
rs - g rs - g
\$2.00 (0.94) \$1.88
\$9.89
0.13 - (-0.06) 0.19
9-21
Find expected annual dividend and
capital gains yields.
Capital gains yield
= g = -6.00%
Dividend yield
= 13.00% - (-6.00%) = 19.00%
Since the stock is experiencing constant
growth, dividend yield and capital gains
yield are constant. Dividend yield is
sufficiently large (19%) to offset a negative
capital gains.
9-22
Corporate value model
Also called the free cash flow method.
Suggests the value of the entire firm
equals the present value of the firm’s
free cash flows.
Remember, free cash flow is the firm’s
after-tax operating income less the net
capital investment
FCF = NOPAT – Net capital investment
9-23
Applying the corporate value model
Find the market value (MV) of the firm,
by finding the PV of the firm’s future
FCFs.
Subtract MV of firm’s debt and preferred
stock to get MV of common stock.
Divide MV of common stock by the
number of shares outstanding to get
intrinsic stock price (value).
9-24
Issues regarding the
corporate value model
Often preferred to the dividend growth
model, especially when considering number
of firms that don’t pay dividends or when
dividends are hard to forecast.
Similar to dividend growth model, assumes at
some point free cash flow will grow at a
constant rate.
Terminal value (TVN) represents value of firm
at the point that growth becomes constant.
9-25
Given the long-run gFCF = 6%, and
WACC of 10%, use the corporate value
model to find the firm’s intrinsic value.
0 r = 10% 1 2 3 4
...
g = 6%
-5 10 20 21.20
-4.545
8.264
15.026 21.20
398.197 530 = = TV3
0.10 - 0.06
416.942
9-26
If the firm has \$40 million in debt and
has 10 million shares of stock, what is
the firm’s intrinsic value per share?
MV of equity = MV of firm – MV of debt
= \$416.94 - \$40
= \$376.94 million
Value per share = MV of equity / # of shares
= \$376.94 / 10
= \$37.69
9-27
Firm multiples method
Analysts often use the following multiples
to value stocks.
P/E
P / CF
P / Sales
EXAMPLE: Based on comparable firms,
estimate the appropriate P/E. Multiply this
by expected earnings to back out an
estimate of the stock price.
9-28
What is market equilibrium?
In equilibrium, stock prices are stable and
there is no general tendency for people to
buy versus to sell.
In equilibrium, two conditions hold:
The current market stock price equals its
^
intrinsic value (P0 = P0).
Expected returns must equal required returns.
^ D1
rs g rs rRF (rM - rRF )b
P0
9-29
Market equilibrium
Expected returns are determined by
estimating dividends and expected
capital gains.
Required returns are determined by
estimating risk and applying the CAPM.
9-30
How is market equilibrium
established?
If price is below intrinsic value …
The current price (P0) is “too low” and
offers a bargain.
Buy orders will be greater than sell
orders.
P0 will be bid up until expected return
equals required return.
9-31
How are the equilibrium
values determined?
Are the equilibrium intrinsic value and
expected return estimated by
managers or are they determined by
something else?
Equilibrium levels are based on the
market’s estimate of intrinsic value and
the market’s required rate of return, which
are both dependent upon the attitudes of
the marginal investor.
9-32
Preferred stock
Hybrid security.
Like bonds, preferred stockholders
receive a fixed dividend that must be
paid before dividends are paid to
common stockholders.
However, companies can omit
preferred dividend payments without
fear of pushing the firm into
bankruptcy.
9-33
If preferred stock with an annual
dividend of \$5 sells for \$50, what is the
preferred stock’s expected return?
Vp = D / rp
\$50 = \$5 / rp
^ = \$5 / \$50
rp
= 0.10 = 10%
9-34
```
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https://byjus.com/question-answer/karan-spends-rs-5000-on-food-which-is-left-frac-5-8-right-th-of-1/
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Question
# Karan spends Rs.5000 on food which is (58)th of his income. Karan's income is ___ (in rupees).
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Solution
## Let Karan's income be expressed as Rs. x. Given, 58 of Karan's income =x×58= Rs.5000 So, x= Rs.5000×85 ⇒ Rs.1000×8= Rs.8000
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pancake sorting
I'm having a bit of trouble with my assignment; I was given a task of coming up with my own solution to the pancake problem.
I've gotten most of my code down, except for this one part (following is in pseudocode):
``````//assuming input is an array of [0...n-1] size
int maxValue = -infinity
for int i <- 0 to n-1 do
{
for int j <-i to n-1 do
{
if A[j] > maxValue
{
maxValue <- A[j]
maxPos <- j
if ((maxPos == n-1) && (maxPos > i))
{
flip(i) //flipping starting from index i
}
/*the following is the bit i'm stuck on
i know that should be able to flip the max value IN the array
(but not the end) to the n-1 term.
On the next iteration of the loop, i flip the maxValue (now held in the last
element) into the slot that is either at the beginning of the array, or at the
element closest to the elements already sorted */
maxValue <- -infinity
``````
And sorry, for the random short code, i pressed sumbit too early on when i was typing =(.
-
what is maxValue? and what is the value of n? – Arun P Johny Jan 26 '13 at 7:47
N is the number of items in the array. MaxValue is the max value in the array for that particular iteration – docaholic Jan 26 '13 at 18:03
I think you are stuck with `infinity`. You can just take it as a `large integer or long number`.
-
Pancake sort:
Pancake sorting is the colloquial term for the mathematical problem of sorting a disordered stack of pancakes in order of size when a spatula can be inserted at any point in the stack and used to flip all pancakes above it.
A pancake number is the maximum number of flips required for a given number of pancakes.
flip(array, i): Reverse array from 0 to i.
Pseudo code Pancake sort:
1) iMax = Index of Largest element in unsorted array.
Find index of the maximum element index in arr[0..unsorted_array_size -1].
2) Call flip(array, iMax)
It will flip all element of array from 0 to iMax index. The largest element will be the first element in the array.
3) Call flip(array, unsorted_array_size -1)
Flip complete unsorted array which result to put the largest element at the end of the unsorted array.
Complexity : Total O(n) flip operations are performed in where each flip will take O(n) time. Therefore complexity is O(n^2).
http://javaexplorer03.blogspot.in/2015/11/pancake-sort-in-java.html
-
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https://www.teachoo.com/8269/2464/Ex-5.8--1/category/Ex-5.8/
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Ex 5.8
Chapter 5 Class 6 Understanding Elementary Shapes
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 5.8, 1 Examine whether the following are polygons. If any one among them is not, say why?A polygon is Simple - Does not cross itself Closed - Has no open ends Made up of line segments (a) Since the figure is not closed, It is not a polygon (b) It is a polygon as it is Simple Closed Made up of line segment It is a polygon of 6 sides Since the figure is not made up of line segments, It is not a polygon Since the figure is not made up of line segments, It is not a polygon
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https://www.programmingmag.com/product/homework-1-amath-563-solved/
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Sale!
Homework 1 AMATH 563 solved
Original price was: \$30.00.Current price is: \$25.00.
Category:
5/5 - (1 vote)
Problems
Hint: read chapters 2 and 3 of Kreyszig.
1. Prove that πΆ([π, π]) equipped with the πΏ
2
([π, π]) norm is not a Banach space.
2. If (π1, β Β· β1) and (π2, β Β· β2) are normed spaces, show that the (Cartesian) product space π = π1 Γ π2
becomes a normed space with the norm βπ₯β = max(βπ₯1β1, βπ₯2β2) where π₯ β π is defined as the tuple
π₯ = (π₯1, π₯2) with addition and scalar multiplication operations: (π₯1, π₯2) + (π¦1, π¦2) = (π₯1 + π₯2, π¦1 + π¦2)
and πΌ(π₯1, π₯2) = (πΌπ₯1, πΌπ₯2).
3. Show that the product (composition) of two linear operators, if it exists, is a linear operator.
4. Let π : π β π be a linear operator and dimπ = dimπ = π < +β. Show that Range(π) = π if and
only if π
β1
exists.
5. Let π be a bounded linear operator from a normed space π onto a normed space π . Show that if there
is a positive constant π such that βπ π₯β β₯ πβπ₯β for all π₯ β π then π
β1
exists and is bounded.
6. Consider the functional π(π₯) = maxπ‘β[π,π] π₯(π‘) on πΆ([π, π]) equipped with the sup norm. Is this functional
linear? is it bounded?
7. Let π be a Banach space and denote its dual as π*
. Show that βπβ : π β¦β supβπ₯β=1 |π(π₯)| is a norm
on π*
.
8. Prove the Schwartz inequality on inner product spaces: |β¨π₯, π¦β©| β€ βπ₯β Β· βπ¦β for all π₯, π¦ β π, where
equality holds if and only if π₯, π¦ are linearly dependent.
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https://www.studymode.com/essays/Acid-Base-Titration-Lab-1256187.html
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Top-Rated Free Essay
# Acid-Base Titration Lab
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Acid-Base Titration Lab
Acid-Base Titration
Objectives:
1. To titrate a hydrochloric acid solution of unknown concentration with standardized 0.10M sodium hydroxide.
2. To utilize the titration data to calculate the molarity of the hydrochloric acid.
Materials:
Procedure:
Data and Calculations:
Table 1: Volume of NaOH Required to Neutralize 10.00mL of Unknown HCl Molarity of NaOh | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Initial Volume of NaOH(mL) | 0.00 | 11.00 | 20.85 | 30.45 | Final Volume of NaOH(mL) | 11.00 | 20.85 | 30.45 | 39.98 | Volume of NaOH used(mL) | 11.00 (Cancel out) | 9.85 | 9.60 | 9.53 |
Average Volume of NaOH = (9.85+9.60+9.53)/3 = 9.66mL
Sample Calculations:
(9.85+9.60+9.53)/3 = 9.66mL The average volume of NaOH used.
Calculations:
1. Moles NaOH = M x V = (0.1M) (0.00966L) = 0.000966 moles
2. Moles HCl = moles NaOH 0.000966 moles -> 9.66x10-4
3. NaOH + HCl = NaCl + H2O Moles NaOH = M x V = (0.1M)(0.00966L) = 0.000966 moles Moles HCl = moles NaOH [HCl] = moles/volumes = (0.000966)/ (0.0096L) [HCl] = 0.1M
Follow-up Questions:
1. It will have no effect because the phenolphthalein only changes color depending on the pH level. Adding substances that will not change the pH level will have no effect.
2. We rinsed out the buret with NaOH, it is to neutralize any leftover acids that may have existed from previous experiments that the buret may have been used in.
4. When we added the NaOH, it instantly neutralized the HCl but because of HCl having more moles inside the beaker, the excess HCl instantly reverting the system back into a base.
5. [HCl] = 0.1M pH = -log[HCl] pH = -log(0.1M) pH = 1
Conclusion:
By using the titration data, we found out that the molarity of HCl is equal to the concentration of NaOH. This happens because the system is in a one-to-one relationship between the two compounds. If we use the same amount of volume of HCl and NaOH, and mix the two, it will reach to a neutral pH value.
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Gaussian integers are complex numbers of the form a+bi where a and b are both integers. In base -1+i, all Gaussian integers can be uniquely represented using the digits 0 and 1, without the need for a symbol to denote sign.
For instance, 1100 in base -1+i represents the decimal number 2, since
1*(-1+i)^3 + 1*(-1+i)^2 + 0*(-1+i)^1 + 0*(-1+i)^0
= (2+2i) + (-2i) + 0 + 0
= 2
Input will be two Gaussian integers in base -1+i represented using the digits 01. This can take one of the following forms:
• Two separate digit strings,
• Two decimal integers consisting of 01 representing the base -1+i numbers (e.g. 1100 for 2 in base -1+i),
• Two binary integers representing the base -1+i numbers (e.g. decimal 12 or 0b1100 for 2 in base -1+i)
• A single string separating two digit strings/binary integers by a single non-alphanumeric separator (e.g. 1100 1100 or 12,12 for 2+2)
Output the sum of the two Gaussian integers, also in base -1+i and represented using the digits 01 (in one of the formats allowed as input, not necessarily the same choice). The output is allowed to contain a finite number of leading zeroes.
Your function or program must terminate within 2 seconds for inputs of at most 30 digits each.
• You may assume that the input contains no extraneous leading zeroes. For the special case of 0, you may choose either 0 or the empty string as the representation.
## Test cases
0, 0 => 0 # 0 + 0 = 0
0, 1 => 1 # 0 + 1 = 1
1, 1 => 1100 # 1 + 1 = 2
1100, 1100 => 111010000 # 2 + 2 = 4
1101, 1101 => 111011100 # 3 + 3 = 6
110111001100, 1110011011100 => 0 # 42 + (-42) = 0
11, 111 => 0 # i + (-i) = 0
11, 110 => 11101 # i + (-1-i) = -1
10101, 11011 => 10010 # (-3-2i) + (-2+3i) = (-5+i)
1010100101, 111101 => 1110100000100 # (-19+2i) + (3-4i) = (-16-2i)
Longer test cases:
11011011010110101110010001001, 111100010100101001001010010101 => 0
111111111111111111111111111111, 111111111111111111111111111111 => 100100100100100100100100100100
101101110111011101110111011101, 101101110111011101110111011101 => 11101001010001000100010001000100011100
100100010101001101010110101010, 100010011101001011111110101000 => 110000110010101100001100111100010
• No digit lists? Mar 23, 2016 at 14:48
• @CatsAreFluffy No digit lists, sorry. Mar 23, 2016 at 15:09
• You can save one byte by changing -1+i to i-1 in the title. Mar 23, 2016 at 18:11
• Now we need a conversion the other way around. :P Mar 23, 2016 at 18:35
• There are 1100 types of people in the world. Those who understand binary, those who don't, those who confuse it with ternary, those who confuse it with base 4, those who confuse it with base 5, those who confuse it with base -1+i, those who confuse it with base 6, those who confuse it with base 7, those who confuse it with base 8, those who confuse it with base 9... Mar 27, 2016 at 9:51
# Python 2, 989791 84 bytes
s=input();L=1
for _ ins*8:s+=1098*int(str(s).translate('0011'*64));L*=10
print s%L
This does I/O in decimal. The integers have to be separated by the non-alphanumeric character +.
Thanks to @xnor for golfing off 2 bytes!
Try it on Ideone.
### How it works
In Arithmetic in Complex Bases, the author shows how to add and multiply complex numbers in bases of the form -n + i.
For base -1 + i, addition is done similarly to regular, binary addition with carry, with two differences:
• Instead of carrying 1 to the next higher position, we carry 110 to the next three.
• Carry digits can propagate indefinitely. However, without leading zeroes, the sum a + b has at most eight digits more than the maximum of a and b.
We proceed as follows.
1. First, we add a and b as if their digits were decimal digits.
For a = 10101 and b = 11011, this gives 21112.
2. Next, we form a new number by replacing the digits larger than 1 with a 1, others with a 0.1
For the sum 21112, this gives 10001.
3. For each digit larger than 1, we have to subtract 2 from that digit and carry 110 to the next three higher positions. Since 1098 = 10 * 110 - 2, we can achieve this by multiplying the result from step 2 by 1098, then adding that product to the sum.2
For the sum 21112, this gives 21112 + 1098 * 10001 = 21112 + 10981098 = 11002210.
4. We repeat steps 2 and 3 a total of d * 8 times, where d is the number of digits of a + b. 3
For the initial sum 21112, the results are
11002210
12210010
1220010010
122000010010
12200000010010
1220000000010010
122000000000010010
12200000000000010010
1220000000000000010010
122000000000000000010010
12200000000000000000010010
1220000000000000000000010010
122000000000000000000000010010
.
.
.
5. We take the final sum modulo 10d + 8, discarding all but the last d + 8 digits.
For the initial sum 21112, the final result is 10010.
1 This is achieved with translate. Repeating the string 0011 64 times makes one repetition line up with the sequence of ASCII characters 0123, achieving the desired replacement.
2 Note that the digits of this sum cannot exceed 3 (initial value 1 plus two 1's from carries).
3 This happens to work for d = 1, and d * 8 > d + 8 otherwise. The code may repeat the steps (d + 1) * 8 times, since s has a trailing L if s is a long integer.
• This is deep magic. What format is input() expecting? (I get 21112 when I input 10101, 11011.) Mar 23, 2016 at 15:20
• Never mind; that was running a version translated (unsuccessfully, it seems) to Python 3. It works fine under Python 2. Mar 23, 2016 at 15:28
– anon
Mar 23, 2016 at 18:50
• @QPaysTaxes I've edited my answer. Mar 24, 2016 at 2:12
• @Dennis Now could you explain why that works? For example, why d+8 and not, say, d+9? How????
– anon
Mar 24, 2016 at 2:15
# Pyth, 34 bytes
_shM.u,%J/eMeN\12-+PMeNm.B6/J2k,kQ
Try it online: Demonstration or Test Suite (takes quite a while). It should satisfy the time restriction though easily, since the online compiler is quite slow in comparison with the normal (offline) compiler.
### Explanation:
My algorithm is basically an implementation of addition with carrying. But instead of carrying 1, I have to carry 110 (1100 in base -1+i is the same as 2 in base -1+i). This works mostly fine, but you can get stuck in an infinite loop printing zeros. For instance if you are adding 1 with 11 and currently have the carry 110. So I basically add until I get stuck in a loop and then stop. I think that a loop that a loop will always print zeros and therefore this should be fine.
_shM.u,%J/eMeN\12-+PMeNm.B6/J2k,kQ implicit: Q = input list of strings
,kQ create the pair ["", Q]
.u modify the pair N (^) until loop:
, replace N with a new pair containing:
eN N[1] (the remaining summand)
eM take the last digits of each summand
/ \1 count the ones
J store the count in J
%J 2 J % 2 (this is the first element of the new pair)
PMeN remove the last digit of each summand
+ m /J2 and add J / 2 new summand:
.B6 with the value "110" (binary of 6)
- k remove empty summand
.u returns all intermediate results
hM extract the digits
s sum them up to a long string
_ reverse
# Python 2, 69 67 bytes
f=lambda a,b:a*a+b*b^58and 2*f(a*b%2*6,f(a/2,b/2))|a+b&1if a else b
I/O is done with base 2 integers.
-2 thanks @Dennis.
• I take it a*a+b*b^58==0 when a and b are inverses? How does that work?
– xnor
Mar 23, 2016 at 23:59
• @xnor No, a*a+b*b==58 when one of them is 3 and the other is 7. Mar 24, 2016 at 0:06
• It's not obvious me to that (3,7) is the only pair that gives a cycle and needs special-casing. If it is true, then you only actually need to check (a,b)==(3,7) in that order, since (7,3) recurses to (3,7), and maybe there's a shorter expression for that.
– xnor
Mar 24, 2016 at 0:09
• Now this is guaranteed to confuse anyone who doesn't know (or forgets) that (a) ^ is XOR, not exponentiation, or (b) XOR has lower precedence than +. Mar 24, 2016 at 8:03
# Jelly, 2928262421 20 bytes
DBḅ1100ḌµDL+8µ¡Dṣ2ṪḌ
This does I/O in decimal. The integers have to be separated by the non-alphanumeric character +.
### Background
In Arithmetic in Complex Bases, the author shows how to add and multiply complex numbers in bases of the form -n + i.
For base -1 + i, addition is done similarly to regular, binary addition with carry, with two differences:
• Instead of carrying 1 to the next higher position, we carry 110 to the next three.
• Carry digits can propagate indefinitely. However, without leading zeroes, the sum a + b has at most eight digits more than the maximum of a and b.
We proceed as follows.
1. First, we add a and b as if their digits were decimal digits.
For a = 10101 and b = 11011, this gives 21112.
2. For each digit larger than 1, we have to subtract 2 from that digit and carry 110 to the next three higher positions. We can achieve this by converting each decimal digit to binary, the resulting binary arrays from base 1100 to integer, and interpret the resulting list of 0's, 1's, 1100's and 1101's as a non-canonical base 10 number.1
For the sum 21112, this gives 21112 + 1098 * 10001 = 21112 + 10981098 = 11002210.
3. We repeat steps 2 a total of d + 8 times, where d is the number of digits of a + b.
For the initial sum 21112, the results are
11002210
12210010
1220010010
122000010010
12200000010010
1220000000010010
122000000000010010
12200000000000010010
1220000000000000010010
122000000000000000010010
12200000000000000000010010
1220000000000000000000010010
122000000000000000000000010010
4. We discard all but the last d + 8 digits from the final result. This is achieved by discarding everything after the last 2.2
For the initial sum 21112, the final result is 10010.
### How it works
DBḅ1100ḌµDL+8µ¡Dṣ2ṪḌ Main link. Argument: a + b (implicit sum)
µ µ¡ Execute the chain before the first µ n times, where n is
the result of executing the chain before the second µ.
D Convert a + b to base 10.
L Length; count the decimal digits.
+8 Add 8 to the number of digits.
D Convert the initial/previous sum to base 10.
B Convert each digit (0 - 3) to binary.
ḅ1100 Convert each binary array from base 1100 to integer.
Ḍ Interpret the resulting list as a base 10 number.
D Convert the final sum to base 10.
ṣ2 Split at occurrences of 2.
Ṫ Select the last chunk.
Ḍ Convert from base 10 to integer.
1 Note that the digits of this sum cannot exceed 3 (initial value 1 plus two 1's from carries).
2 This works because the last digit that will cancel out cannot be a 3.
## Retina, 100 bytes
r+(.*)(\d|(?!\4))( .*)(.?)
$2$4:$1$3
T 0
+1:11(1*:1*)11
:$1 ^:* ::: }:(1*:1*:)11 1:1$1
(1)*:
$#1 This takes the input separated with a comma. The output always starts with three leading zeroes. Try it online! I really wonder if there's a shorter solution for the first stage... • No, no, the score is perfect as it is ;) Mar 23, 2016 at 18:29 • Nice score of -2i! – anon Mar 23, 2016 at 18:51 • Wow. I did not see this solution when I posted mine... Much more superior than my solution. Apr 29, 2016 at 14:07 • @KennyLau I was just looking at it and thinking "hm, I guess I should've added an explanation at some point..." Apr 29, 2016 at 14:09 • ...-2i? This is decimal but the program uses a base that not. Dec 28, 2017 at 15:42 ## Python 3, 289 bytes This performs digitwise addition from least to most significant digit (in other words, the exact same algorithm you were taught in primary school). The differences are that (a) it's in binary, not decimal, so you carry whenever a digit is 2 or more, and (b) 1 + 1 = 1100, not 10. Actually, it's also necessary to note that 11 + 111 = 0, or else sums that should become zero will never terminate. from collections import* def a(*s,p=0): r=defaultdict(int,{0:0}) for S in s: n=0 for d in S[::-1]:r[n]+=d=='1';n+=1 while p<=max(r): while r[p]>1: r[p]-=2 if r[p+1]>1<=r[p+2]:r[p+1]-=2;r[p+2]-=1 else:r[p+2]+=1;r[p+3]+=1 p+=1 return str([*map(r.get,sorted(r))])[-2::-3] More golfing is surely possible. • How certain are you that your "zero detector" is sufficient? – Yakk Mar 23, 2016 at 19:55 • @Yakk: On a scale of one to peer-reviewed-journal, maybe give it a no-counterexamples-yet? Mar 24, 2016 at 7:43 ## JavaScript (ES6), 146 126 bytes r=n=>n&&n%2-r(n>>=1)-i(n) i=n=>n&&r(n>>=1)-i(n) g=(x,y,b=(x^y)&1)=>x|y&&b+2*g(b-x+y>>1,b-x-y>>1) (x,y)=>g(r(x)+r(y),i(x)+i(y)) g converts a Gaussian integer (real and imaginary parts) into base i-1, while r and i converts a base i-1 integer into a Gaussian integer (real and imaginary parts respectively). Once the conversions are in place, I just have to do the arithmetic. Edit: Saved 20 bytes by calculating the real and imaginary parts separately. C++ 416 bytes, plus #include <vector>\n#include <algorithm>\n (another 40) using I=int;using v=std::vector<I>;void r(v&x){v r{rbegin(x),rend(x)};x=r;}v a(v L,v R){r(L);r(R);L.resize(std::max(L.size(),R.size()));for(int&r:R)L[&r-R.data()]+=r;while(1){L.resize(L.size()+3);auto it=find(rbegin(L),rend(L),2);if(it==rend(L))break;I i=-1+it.base()-begin(L);i&&L[i+1]&&L[i-1]/2?L[i+1]=L[i]=L[i-1]=0:(++L[i+2],++L[i+3],L[i]=0);}L.erase( std::find(rbegin(L),rend(L),1).base(),end(L));r(L);return L;} or, with more whitespace: using I=int; using v=std::vector<I>; void r(v&x){v r{rbegin(x),rend(x)};x=r;} v a(v L,v R) { r(L);r(R); L.resize(std::max(L.size(),R.size())); for(int&r:R) L[&r-R.data()]+=r; while(1) { L.resize(L.size()+3); auto it=find(rbegin(L), rend(L), 2); if(it==rend(L)) break; I i=-1+it.base()-begin(L); i&&L[i+1]&&L[i-1]/2? L[i+1]=L[i]=L[i-1]=0 : (++L[i+2],++L[i+3],L[i]=0); } L.erase( std::find(rbegin(L),rend(L),1).base(), end(L)); r(L); return L; } Barely golfed. It takes input as a vector of ints, and returns a vector of ints. # Retina, 157151134133124 123 bytes 1 byte off thanks to Martin Büttner. (.+),(.+)$.1$*0$2,$.2$*0$1, 1 0x +(0x*)(,.*)0(x*),$2,$1$3
{,
(^|0x0xx0xx)
000
(0x*)(0x*)(0x*0)xx
$1x$2x\$3
)^0+
0
0x
1
Try it online!
Converts to unary, and then repeat the following replacements (shown here in decimal):
122 -> 000
0002 -> 1100 (this can also be 0012 -> 1110 and 1112 -> 2210 or even 2222 -> 3320 or even 3333 -> 4431)
Basically, when larger than two: take away two, add nothing in the previous digit, add one to the previous digit, then add one to the previous digit.
In pseudocode:
if(a[n]>2):
a[n] -= 2;
a[n-2] += 1;
a[n-3] += 1;
### Unary implementation:
Each digit (e.g. 3) is shown as the number of xs (e.g. xxx) and then prefixed with 0.
For example, 1234 would be expressed as 0x0xx0xxx0xxxx.
This leaves 0 unchanged, as 101 would be expressed by 0x00x.
Since initially and finally, there is only 0 and 1, the conversion could be easily done by 1->0x and 0x->1`.
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http://mathlair.allfunandgames.ca/tictactoe.php
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New: Read my new book, How to Ace the Multiple-Choice Mathematics Test, available as an e-book for Kindle.
# Tic-tac-toe
Math Lair Home > Topics > Tic-tac-toe
[The Traditional Game | Variations | Generalized Tic-Tac-Toe | Other Stuff]
Tic-tac-toe (also spelled tictactoe, ticktacktoe, or tick-tack-toe, and also referred to as noughts and crosses or X's and O's) is one of the world's simplest games. It's also one of the most ubiquitous, with game boards found on toys, in graffiti, and even children's playground equipment (right). One could wonder how there could be any mathematical interest in it whatsoever. After all, everyone who is old enough knows that, with best play, neither player can force a win in standard tic-tac-toe (i.e. on a 3×3 grid). Still, investigating tic-tac-toe can be a useful introduction to combinatorial game theory. Furthermore, there are several variations that could be played, such as altering the board or the winning conditions, that could make the game interesting. For example, the first player has an easy win with three-in-a-row tic-tac-toe on a 4×4 (or larger) board.
Since everyone already knows that, with best play, Tic-Tac-Toe is a draw, I won't perform a detailed analysis of the best strategy for the game here. I will ask this question, though: If you are the second player, what first move should you make to help ensure a draw?
If the first player chooses the centre square to start, the second player, in order to not lose, must select one of the corner squares. If the first player plays first in the corner square, the second player must select the centre square. If the first player selects a side square, there are several possibilities. See the images on the left for more information.
If you're the first player, you might think that your probability of winning is better if you take the corner square, but that's not necessarily true unless your opponent is playing at random, because the only move that leads to a draw (take the centre square) is easier to find.
## Variations
Here are some interesting variations of tic-tac-toe that you may want to try out. Think about whether one player or another has a winning strategy, a plan they can follow that guarantees that they will win every time:
Each player may put down either an X or an O on each of their turns, and may change their mind from turn to turn. The winner is the one who finishes any row, column, or diagonal of all X's or all O's.
Magic Square Tic-Tac-Toe
Instead of X's and O's, the numbers 1 through 9 are used. Each number may be used only once. The winner is the one to get the numbers in any row, column, or diagonal to add up to 15. Question: Can you determine a winning strategy for this variant?
Last One Wins
On a player's turn, that player marks as many non-empty spaces as they like, as long as they are in the same row or column (not diagonal). Whoever fills in the last space wins.
Avoidance Tic-Tac-Toe (also known as Misère Tic-Tac-Toe or Toe-Tac-Tic)
Same order of play as the standard game, except that a player who completes three in a row loses.
Drawbridge
Same order of play as the standard game, only one player wins if the game is a draw, while the other wins if either completes three in a row.
Movable Markers
Players have three counters each and take turns placing them on the grid. If neither has won after all six counters are down, players may move one of their counters somewhere else as their move.
Hot
Each of the words HOT, HEAR, TIED, FORM, WASP, BRIM, TANK, SHIP, and WOES is printed on a card. Players take turns withdrawing cards from the pile. The first to hold three cards containing the same letter wins. An alternate version of this game uses the cards FISH, SOUP, HORN, KNIT, VOTE, ARMY, CHAT, SWAN, and GIRL. Question: Why is this listed under tic-tac-toe variants? Think about it.
The Numbers Game (or Fifteen)
Players take turns selecting a number between 1 and 9. Each number may be selected only once. The first person to accumulate three numbers whose sum is 15 is the winner. Question: Why is this listed under tic-tac-toe variants? The magic squares page may provide a hint.
Multi-Player Tic-Tac-Toe
For three to six people, try a giant board, one that is at least 10×10, perhaps infinite. Four in a row wins this variant.
Ultimate Tic-Tac-Toe
Ultimate tic-tac-toe is played on a 3×3 board, but each square contains a 3×3 board of its own. Players alternate turns marking individual squares on one of the small boards. If you win one of the little boards, you can mark the big square as X or O. What makes this game interesting is that you can't simply choose any of the nine boards to play on; you must choose the board located in the large square corresponding to the small square that your opponent just marked. So, for example, if your opponent just marked the upper left corner in the middle board, you must play somewhere in the upper left board.
3-D Tic-Tac-Toe
Play on a 4×4×4 board; the goal is to get 4 in a row. There are a lot of computer games of this version of tic-tac-toe. One of the earliest was a video game for the Atari 2600.
## Generalized Tic-Tac-Toe
Martin Gardner once wrote a Scientific American column (found in his book Fractal Music, Hypercards and More...; see the bibliography for more information on the book) on "Generalized Ticktacktoe". The generalization is as follows: Choose a polyomino (such polyominoes are called "animals" by Frank Harary, who devised this generalization) and declare its formation to be the objective of the tic-tac-toe game. Each player tries to fill in cells that will form the desired animal. Rotations and reflections are okay.
An interesting idea is to look at each of the polyominoes and find two properties of that polyomino: The length of the side of the smallest square on which the first player can force a win (b), and the number of moves required on this board (m). Here are some "animals" of 1 to 4 cells (aside: polyominoes of four cells are called tetrominoes, and should be familiar to anyone who as played Tetris), their "names", and their b and m values.
Note that the 2×2 square, nicknamed "Fatty", is a "loser". That is, the first player can never force Fatty on a board of any size. Note that any polyomino containing any smaller loser is also a loser. This makes sense, since if you can never force a 2×2 square, for example, you can't force (say) a 3×3 square either, since to construct a 3×3 square you have to create a 2×2 square. As a matter of fact, most polyominoes of size 5 or greater are losers. There are only three pentominoes and (probably) one hexomino that are winners:
Since almost every heptomino (size 7) or larger will contain at least two different hexominoes, and there is no more than one winning hexomino, it is not too hard to show that all 107 order-7 animals contain a smaller loser and thus are losers themselves.
Another question: Is there ever a winning strategy for the second player? This is a pretty easy question to answer. Assume that, for some shape, the second player does have a strategy. Then the first player could win by starting with some irrelevant move and thereafter following the second player's winning strategy. Making an extra move is never a liability in tic-tac-toe, even generalized tic-tac-toe. We have reached a contradiction, so our assumption that the second player can ever have a winning strategy is false. The moral of the story is: go first!
## Other Thoughts
One last tic-tac-toe item: If you play a normal game of tic-tac-toe with someone else and let him/her go first, what is the probability that s/he will start with an X? It is certainly greater than 50%. As a matter of fact, it's probably close to 100%. Interesting, eh?
However, I have seen a tic-tac-toe set in the toy section of a department store that contained five O's and four X's, which would require O to go first. Maybe at least one toy designer thinks differently than most people...
Sources used (see bibliography page for titles corresponding to numbers): 44.
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×
Contest
EASY
# PREREQUISITES
Greedy Algorithms
# PROBLEM
You need to find the representation of a given integer N as a sum of powers of two up to 211 with the smallest number of summands (repetitions in representation are allowed).
# QUICK EXPLANATION
The problem can be solved by greedy algorithm: at each step we should take the largest possible menu that we can. To prove this noticed that if we order two equal menus, say 4 and 4, then we can order instead one menu of cost 8. In this problem you can implement this approach in any way you like but the easiest and fastest way gives the following pseudo-code:
res = 0
for i = 11 to 0
res = res + N div 2i
N = N mod 2i
# EXPLANATION
At first we reveal the answer and then will prove it in detail. Write N as Q * 2048 + R, where Q and R are non-negative integers and R < 2048. Then the answer is Q + bitCount(R), where bitCount(X) is the number of ones in binary representation of X. If R = 2h[1] + 2h[2] + ... + 2h[K] is a binary representation of R then the optimal representation of N is N = 2048 + ... + 2048 + 2h[1] + 2h[2] + ... + 2h[K] where we have Q copies of 2048. Let’s call this approach formula approach.
Another way to come up with this answer is to use Greedy Algorithm. That is, at each step you should take the largest possible summand among {1, 2, 4, 8, ..., 2048} that is not greater than the current value of N and then subtract it from N. In fact, this problem is a special case of Change-making problem and in general it should be solved using Dynamic Programming or Integer Linear Programming but this set of coin values is special and admit using of Greedy Algorithm as we will see below.
Now we discuss why both of these approaches are correct.
# 1. Formula Approach.
Let’s prove that formula approach is correct. Consider some representation of N as a sum of allowed powers of two. Let there is exactly C[K] occurrences of 2K in this representation. Then we have
N = C[0] * 1 + C[1] * 2 + C[2] * 4 + ... + C[11] * 2048
Note that the total number of summands here is C[0] + C[1] + ... + C[10] + C[11]. Assume that for some K < 11 we have C[K] >=2, that is, we have at least two copies of 2K in the representation of N. Since K < 11 then the price 2K+1 is allowed. Hence we can replace two copies of 2K by one copy of 2K+1 not changing the value of sum but decreasing total number of summands. Thus, for optimal representation we should have
C[K] <= 1 for all K < 11....................(1)
We will show that under the constraints (1) representation of N is uniquely determined. Of course this unique representation will be the optimal one.
At first note that
R = C[0] * 1 + C[1] * 2 + C[2] * 4 + ... + C[10] * 1024 <= 1 + 2 + 4 + ... + 1024 = 2047 < 2048.
Hence
2048 * C[11] <= N < 2048 * (C[11] + 1)
or
C[11] <= N / 2048 < C[11] + 1
which means by one of the definition of floor function that C[11] = floor(N / 2048) = Q. So C[11] is uniquely determined under the constraints (1) and equals to Q.
Further note that due to (1) C[10]C[9]...C[1]C[0] is a binary representation of R and hence C[0], C[1], ..., C[10] are also uniquely determined under the constraints (1).
Thus we have found this unique representation.
# 2. Greedy Approach.
Now let’s see why greedy approach produces the same solution. Clearly at first several steps we will take the 2048 until we get a number strictly less than 2048. Then we will consider powers of two in descending order starting from 1024 and take the current power of two if it is not greater than the current value of N. It is quite clear that this process generates exactly the binary representation of N. Thus the whole representation coincides with the constructed above.
There are several ways how to implement this algorithm in this problem. First one is simple but slower in general. Namely, we have an outer loop of the form while (N > 0). Then at each iteration of this loop we simply check in one loop all allowed powers of two in descending order until we find the power of two, say 2X, that is not greater than the current value of N. After that we decrease N by 2X and increase answer by 1. The complexity of this approach is O(N / 2K-1 + K2) in the worst test case. This is because at first N / 2K-1 steps we have only one iteration of inner loop (we break at 2048) and then we have at most K steps for each of which we have at most K steps in the inner loop.
In second method we swap outer and inner loop of the first method. So we iterate over allowed powers of two in descending order and for each power of two we have an inner loop of the form while (N >= 2X) in the body of which we do the same as for the first method, that is, decrease N by 2X and increase answer by 1. The complexity of this method is O(N / 2K-1 + K). Strictly speaking the number of basic operations in this method is O(answer + K). N / 2K-1 + K is an upper bound for the answer.
Analyzing second implementation of the greedy approach it is easy to see how to make it faster. For each power of two we have the following inner loop
while N >= 2X do
N = N - 2X
res = res + 1
Clearly this equivalent to
res = res + N div 2X
N = N mod 2X
Thus we obtain third implementation of the greedy algorithm with complexity O(K). Next we have bitCount(R) = C[0] + C[1] + ... + C[10]. Hence the total number of summands in this representation is Q + bitCount(R) as was announced earlier. The complexity of this method is O(K), where K = 12 is the total number of powers of two that we are allowed to use.
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question asked: 13 Oct '16, 22:54
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last updated: 07 Jun '17, 17:15
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# Mathematics / Foundation / Measurement and Geometry / Using units of measurement
Version 8
Curriculum content descriptions
Use direct and indirect comparisons to decide which is longer, heavier or holds more, and explain reasoning in everyday language (ACMMG006)
Elaborations
• comparing objects directly, by placing one object against another to determine which is longer or by pouring from one container into the other to see which one holds more
• using suitable language associated with measurement attributes, such as ‘tall’ and ‘taller’, ‘heavy’ and ‘heavier’, ‘holds more’ and ‘holds less’
General capabilities
• Literacy Literacy
• Numeracy Numeracy
• Critical and creative thinking Critical and creative thinking
ScOT terms
Volume (Dimensions), Length, Mass
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CC-MAIN-2024-10
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en
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https://plainmath.net/integral-calculus/6432-use-areas-to-evaluate-the-integral-int-a-11b-2s-ds-0-less-then-a-less-then-b
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crawl-data/CC-MAIN-2023-23/segments/1685224656869.87/warc/CC-MAIN-20230609233952-20230610023952-00018.warc.gz
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DofotheroU
2021-03-07
Use areas to evaluate the integral.
falhiblesw
Step 1
Consider the integrals,
Step 2
To solve the given integrals,
${\int }_{a}^{11b}2sds=2{\int }_{a}^{11b}sds$
$=2{\left[\frac{{s}^{2}}{2}\right]}_{a}^{11b}$
$=2\cdot \frac{1}{2}\cdot {\left[{s}^{2}\right]}_{a}^{11b}$
$={\left(11b\right)}^{2}-{a}^{2}$
$=121{b}^{2}-{a}^{2}$
Jeffrey Jordon
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http://metamath.tirix.org/mpests/prdstotbnd.html
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# Metamath Proof Explorer
## Theorem prdstotbnd
Description: The product metric over finite index set is totally bounded if all the factors are totally bounded. (Contributed by Mario Carneiro, 20-Sep-2015)
Ref Expression
Hypotheses prdsbnd.y ${⊢}{Y}={S}{⨉}_{𝑠}{R}$
prdsbnd.b ${⊢}{B}={\mathrm{Base}}_{{Y}}$
prdsbnd.v ${⊢}{V}={\mathrm{Base}}_{{R}\left({x}\right)}$
prdsbnd.e ${⊢}{E}={\mathrm{dist}\left({R}\left({x}\right)\right)↾}_{\left({V}×{V}\right)}$
prdsbnd.d ${⊢}{D}=\mathrm{dist}\left({Y}\right)$
prdsbnd.s ${⊢}{\phi }\to {S}\in {W}$
prdsbnd.i ${⊢}{\phi }\to {I}\in \mathrm{Fin}$
prdsbnd.r ${⊢}{\phi }\to {R}Fn{I}$
prdstotbnd.m ${⊢}\left({\phi }\wedge {x}\in {I}\right)\to {E}\in \mathrm{TotBnd}\left({V}\right)$
Assertion prdstotbnd ${⊢}{\phi }\to {D}\in \mathrm{TotBnd}\left({B}\right)$
### Proof
Step Hyp Ref Expression
1 prdsbnd.y ${⊢}{Y}={S}{⨉}_{𝑠}{R}$
2 prdsbnd.b ${⊢}{B}={\mathrm{Base}}_{{Y}}$
3 prdsbnd.v ${⊢}{V}={\mathrm{Base}}_{{R}\left({x}\right)}$
4 prdsbnd.e ${⊢}{E}={\mathrm{dist}\left({R}\left({x}\right)\right)↾}_{\left({V}×{V}\right)}$
5 prdsbnd.d ${⊢}{D}=\mathrm{dist}\left({Y}\right)$
6 prdsbnd.s ${⊢}{\phi }\to {S}\in {W}$
7 prdsbnd.i ${⊢}{\phi }\to {I}\in \mathrm{Fin}$
8 prdsbnd.r ${⊢}{\phi }\to {R}Fn{I}$
9 prdstotbnd.m ${⊢}\left({\phi }\wedge {x}\in {I}\right)\to {E}\in \mathrm{TotBnd}\left({V}\right)$
10 eqid ${⊢}{S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)={S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)$
11 eqid ${⊢}{\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)}$
12 eqid ${⊢}\mathrm{dist}\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)=\mathrm{dist}\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)$
13 fvexd ${⊢}\left({\phi }\wedge {x}\in {I}\right)\to {R}\left({x}\right)\in \mathrm{V}$
14 totbndmet ${⊢}{E}\in \mathrm{TotBnd}\left({V}\right)\to {E}\in \mathrm{Met}\left({V}\right)$
15 9 14 syl ${⊢}\left({\phi }\wedge {x}\in {I}\right)\to {E}\in \mathrm{Met}\left({V}\right)$
16 10 11 3 4 12 6 7 13 15 prdsmet ${⊢}{\phi }\to \mathrm{dist}\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)\in \mathrm{Met}\left({\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)}\right)$
17 dffn5 ${⊢}{R}Fn{I}↔{R}=\left({x}\in {I}⟼{R}\left({x}\right)\right)$
18 8 17 sylib ${⊢}{\phi }\to {R}=\left({x}\in {I}⟼{R}\left({x}\right)\right)$
19 18 oveq2d ${⊢}{\phi }\to {S}{⨉}_{𝑠}{R}={S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)$
20 1 19 syl5eq ${⊢}{\phi }\to {Y}={S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)$
21 20 fveq2d ${⊢}{\phi }\to \mathrm{dist}\left({Y}\right)=\mathrm{dist}\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)$
22 5 21 syl5eq ${⊢}{\phi }\to {D}=\mathrm{dist}\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)$
23 20 fveq2d ${⊢}{\phi }\to {\mathrm{Base}}_{{Y}}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)}$
24 2 23 syl5eq ${⊢}{\phi }\to {B}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)}$
25 24 fveq2d ${⊢}{\phi }\to \mathrm{Met}\left({B}\right)=\mathrm{Met}\left({\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)\right)}\right)$
26 16 22 25 3eltr4d ${⊢}{\phi }\to {D}\in \mathrm{Met}\left({B}\right)$
27 7 adantr ${⊢}\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\to {I}\in \mathrm{Fin}$
28 istotbnd3 ${⊢}{E}\in \mathrm{TotBnd}\left({V}\right)↔\left({E}\in \mathrm{Met}\left({V}\right)\wedge \forall {r}\in {ℝ}^{+}\phantom{\rule{.4em}{0ex}}\exists {w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
29 28 simprbi ${⊢}{E}\in \mathrm{TotBnd}\left({V}\right)\to \forall {r}\in {ℝ}^{+}\phantom{\rule{.4em}{0ex}}\exists {w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}$
30 9 29 syl ${⊢}\left({\phi }\wedge {x}\in {I}\right)\to \forall {r}\in {ℝ}^{+}\phantom{\rule{.4em}{0ex}}\exists {w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}$
31 30 r19.21bi ${⊢}\left(\left({\phi }\wedge {x}\in {I}\right)\wedge {r}\in {ℝ}^{+}\right)\to \exists {w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}$
32 df-rex ${⊢}\exists {w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}↔\exists {w}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
33 rexv ${⊢}\exists {w}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)↔\exists {w}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
34 32 33 bitr4i ${⊢}\exists {w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}↔\exists {w}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
35 31 34 sylib ${⊢}\left(\left({\phi }\wedge {x}\in {I}\right)\wedge {r}\in {ℝ}^{+}\right)\to \exists {w}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
36 35 an32s ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge {x}\in {I}\right)\to \exists {w}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
37 36 ralrimiva ${⊢}\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {w}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
38 eleq1 ${⊢}{w}={f}\left({x}\right)\to \left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)↔{f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\right)$
39 iuneq1 ${⊢}{w}={f}\left({x}\right)\to \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)=\bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)$
40 39 eqeq1d ${⊢}{w}={f}\left({x}\right)\to \left(\bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}↔\bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)$
41 38 40 anbi12d ${⊢}{w}={f}\left({x}\right)\to \left(\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)↔\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)$
42 41 ac6sfi ${⊢}\left({I}\in \mathrm{Fin}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {w}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({w}\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {w}}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\to \exists {f}\phantom{\rule{.4em}{0ex}}\left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)$
43 27 37 42 syl2anc ${⊢}\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\to \exists {f}\phantom{\rule{.4em}{0ex}}\left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)$
44 elfpw ${⊢}{f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)↔\left({f}\left({x}\right)\subseteq {V}\wedge {f}\left({x}\right)\in \mathrm{Fin}\right)$
45 44 simplbi ${⊢}{f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\to {f}\left({x}\right)\subseteq {V}$
46 45 adantr ${⊢}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\to {f}\left({x}\right)\subseteq {V}$
47 46 ralimi ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{f}\left({x}\right)\subseteq {V}$
48 47 ad2antll ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{f}\left({x}\right)\subseteq {V}$
49 ss2ixp ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{f}\left({x}\right)\subseteq {V}\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\subseteq \underset{{x}\in {I}}{⨉}{V}$
50 48 49 syl ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\subseteq \underset{{x}\in {I}}{⨉}{V}$
51 fnfi ${⊢}\left({R}Fn{I}\wedge {I}\in \mathrm{Fin}\right)\to {R}\in \mathrm{Fin}$
52 8 7 51 syl2anc ${⊢}{\phi }\to {R}\in \mathrm{Fin}$
53 fndm ${⊢}{R}Fn{I}\to \mathrm{dom}{R}={I}$
54 8 53 syl ${⊢}{\phi }\to \mathrm{dom}{R}={I}$
55 1 6 52 2 54 prdsbas ${⊢}{\phi }\to {B}=\underset{{x}\in {I}}{⨉}{\mathrm{Base}}_{{R}\left({x}\right)}$
56 3 rgenw ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{V}={\mathrm{Base}}_{{R}\left({x}\right)}$
57 ixpeq2 ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{V}={\mathrm{Base}}_{{R}\left({x}\right)}\to \underset{{x}\in {I}}{⨉}{V}=\underset{{x}\in {I}}{⨉}{\mathrm{Base}}_{{R}\left({x}\right)}$
58 56 57 ax-mp ${⊢}\underset{{x}\in {I}}{⨉}{V}=\underset{{x}\in {I}}{⨉}{\mathrm{Base}}_{{R}\left({x}\right)}$
59 55 58 syl6eqr ${⊢}{\phi }\to {B}=\underset{{x}\in {I}}{⨉}{V}$
60 59 ad2antrr ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to {B}=\underset{{x}\in {I}}{⨉}{V}$
61 50 60 sseqtrrd ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\subseteq {B}$
62 27 adantr ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to {I}\in \mathrm{Fin}$
63 44 simprbi ${⊢}{f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\to {f}\left({x}\right)\in \mathrm{Fin}$
64 63 adantr ${⊢}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\to {f}\left({x}\right)\in \mathrm{Fin}$
65 64 ralimi ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{f}\left({x}\right)\in \mathrm{Fin}$
66 65 ad2antll ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{f}\left({x}\right)\in \mathrm{Fin}$
67 ixpfi ${⊢}\left({I}\in \mathrm{Fin}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{f}\left({x}\right)\in \mathrm{Fin}\right)\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\in \mathrm{Fin}$
68 62 66 67 syl2anc ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\in \mathrm{Fin}$
69 elfpw ${⊢}\underset{{x}\in {I}}{⨉}{f}\left({x}\right)\in \left(𝒫{B}\cap \mathrm{Fin}\right)↔\left(\underset{{x}\in {I}}{⨉}{f}\left({x}\right)\subseteq {B}\wedge \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\in \mathrm{Fin}\right)$
70 61 68 69 sylanbrc ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\in \left(𝒫{B}\cap \mathrm{Fin}\right)$
71 metxmet ${⊢}{D}\in \mathrm{Met}\left({B}\right)\to {D}\in \mathrm{\infty Met}\left({B}\right)$
72 26 71 syl ${⊢}{\phi }\to {D}\in \mathrm{\infty Met}\left({B}\right)$
73 rpxr ${⊢}{r}\in {ℝ}^{+}\to {r}\in {ℝ}^{*}$
74 blssm ${⊢}\left({D}\in \mathrm{\infty Met}\left({B}\right)\wedge {y}\in {B}\wedge {r}\in {ℝ}^{*}\right)\to {y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
75 74 3expa ${⊢}\left(\left({D}\in \mathrm{\infty Met}\left({B}\right)\wedge {y}\in {B}\right)\wedge {r}\in {ℝ}^{*}\right)\to {y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
76 75 an32s ${⊢}\left(\left({D}\in \mathrm{\infty Met}\left({B}\right)\wedge {r}\in {ℝ}^{*}\right)\wedge {y}\in {B}\right)\to {y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
77 76 ralrimiva ${⊢}\left({D}\in \mathrm{\infty Met}\left({B}\right)\wedge {r}\in {ℝ}^{*}\right)\to \forall {y}\in {B}\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
78 72 73 77 syl2an ${⊢}\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\to \forall {y}\in {B}\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
79 78 adantr ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \forall {y}\in {B}\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
80 ssralv ${⊢}\underset{{x}\in {I}}{⨉}{f}\left({x}\right)\subseteq {B}\to \left(\forall {y}\in {B}\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}\to \forall {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}\right)$
81 61 79 80 sylc ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \forall {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
82 iunss ${⊢}\bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)\subseteq {B}↔\forall {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{y}\mathrm{ball}\left({D}\right){r}\subseteq {B}$
83 81 82 sylibr ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)\subseteq {B}$
84 62 adantr ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to {I}\in \mathrm{Fin}$
85 60 eleq2d ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \left({g}\in {B}↔{g}\in \underset{{x}\in {I}}{⨉}{V}\right)$
86 vex ${⊢}{g}\in \mathrm{V}$
87 86 elixp ${⊢}{g}\in \underset{{x}\in {I}}{⨉}{V}↔\left({g}Fn{I}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in {V}\right)$
88 87 simprbi ${⊢}{g}\in \underset{{x}\in {I}}{⨉}{V}\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in {V}$
89 df-rex ${⊢}\exists {z}\in {f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)↔\exists {z}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)$
90 eliun ${⊢}{g}\left({x}\right)\in \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)↔\exists {z}\in {f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)$
91 rexv ${⊢}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)↔\exists {z}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)$
92 89 90 91 3bitr4i ${⊢}{g}\left({x}\right)\in \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)↔\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)$
93 eleq2 ${⊢}\bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\to \left({g}\left({x}\right)\in \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)↔{g}\left({x}\right)\in {V}\right)$
94 92 93 syl5bbr ${⊢}\bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\to \left(\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)↔{g}\left({x}\right)\in {V}\right)$
95 94 biimprd ${⊢}\bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\to \left({g}\left({x}\right)\in {V}\to \exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
96 95 adantl ${⊢}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\to \left({g}\left({x}\right)\in {V}\to \exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
97 96 ral2imi ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\to \left(\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in {V}\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
98 97 ad2antll ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \left(\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in {V}\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
99 88 98 syl5 ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \left({g}\in \underset{{x}\in {I}}{⨉}{V}\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
100 85 99 sylbid ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \left({g}\in {B}\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
101 100 imp ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)$
102 eleq1 ${⊢}{z}={y}\left({x}\right)\to \left({z}\in {f}\left({x}\right)↔{y}\left({x}\right)\in {f}\left({x}\right)\right)$
103 oveq1 ${⊢}{z}={y}\left({x}\right)\to {z}\mathrm{ball}\left({E}\right){r}={y}\left({x}\right)\mathrm{ball}\left({E}\right){r}$
104 103 eleq2d ${⊢}{z}={y}\left({x}\right)\to \left({g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)↔{g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)$
105 102 104 anbi12d ${⊢}{z}={y}\left({x}\right)\to \left(\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)↔\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
106 105 ac6sfi ${⊢}\left({I}\in \mathrm{Fin}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\exists {z}\in \mathrm{V}\phantom{\rule{.4em}{0ex}}\left({z}\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({z}\mathrm{ball}\left({E}\right){r}\right)\right)\right)\to \exists {y}\phantom{\rule{.4em}{0ex}}\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
107 84 101 106 syl2anc ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to \exists {y}\phantom{\rule{.4em}{0ex}}\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)$
108 ffn ${⊢}{y}:{I}⟶\mathrm{V}\to {y}Fn{I}$
109 simpl ${⊢}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\to {y}\left({x}\right)\in {f}\left({x}\right)$
110 109 ralimi ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{y}\left({x}\right)\in {f}\left({x}\right)$
111 108 110 anim12i ${⊢}\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\to \left({y}Fn{I}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{y}\left({x}\right)\in {f}\left({x}\right)\right)$
112 vex ${⊢}{y}\in \mathrm{V}$
113 112 elixp ${⊢}{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)↔\left({y}Fn{I}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{y}\left({x}\right)\in {f}\left({x}\right)\right)$
114 111 113 sylibr ${⊢}\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\to {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)$
115 114 adantl ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)$
116 85 biimpa ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to {g}\in \underset{{x}\in {I}}{⨉}{V}$
117 ixpfn ${⊢}{g}\in \underset{{x}\in {I}}{⨉}{V}\to {g}Fn{I}$
118 116 117 syl ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to {g}Fn{I}$
119 118 adantr ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {g}Fn{I}$
120 simpr ${⊢}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\to {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
121 120 ralimi ${⊢}\forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
122 121 ad2antll ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
123 86 elixp ${⊢}{g}\in \underset{{x}\in {I}}{⨉}\left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)↔\left({g}Fn{I}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}{g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)$
124 119 122 123 sylanbrc ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {g}\in \underset{{x}\in {I}}{⨉}\left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
125 simp-4l ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {\phi }$
126 50 ad2antrr ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\subseteq \underset{{x}\in {I}}{⨉}{V}$
127 126 115 sseldd ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {y}\in \underset{{x}\in {I}}{⨉}{V}$
128 125 59 syl ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {B}=\underset{{x}\in {I}}{⨉}{V}$
129 127 128 eleqtrrd ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {y}\in {B}$
130 simp-4r ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {r}\in {ℝ}^{+}$
131 fveq2 ${⊢}{y}={x}\to {R}\left({y}\right)={R}\left({x}\right)$
132 131 cbvmptv ${⊢}\left({y}\in {I}⟼{R}\left({y}\right)\right)=\left({x}\in {I}⟼{R}\left({x}\right)\right)$
133 132 oveq2i ${⊢}{S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)={S}{⨉}_{𝑠}\left({x}\in {I}⟼{R}\left({x}\right)\right)$
134 20 133 syl6eqr ${⊢}{\phi }\to {Y}={S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)$
135 134 fveq2d ${⊢}{\phi }\to \mathrm{dist}\left({Y}\right)=\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)$
136 5 135 syl5eq ${⊢}{\phi }\to {D}=\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)$
137 136 fveq2d ${⊢}{\phi }\to \mathrm{ball}\left({D}\right)=\mathrm{ball}\left(\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)\right)$
138 137 oveqdr ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {y}\mathrm{ball}\left({D}\right){r}={y}\mathrm{ball}\left(\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)\right){r}$
139 eqid ${⊢}{\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)}$
140 eqid ${⊢}\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)=\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)$
141 6 adantr ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {S}\in {W}$
142 7 adantr ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {I}\in \mathrm{Fin}$
143 fvexd ${⊢}\left(\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\wedge {x}\in {I}\right)\to {R}\left({x}\right)\in \mathrm{V}$
144 metxmet ${⊢}{E}\in \mathrm{Met}\left({V}\right)\to {E}\in \mathrm{\infty Met}\left({V}\right)$
145 15 144 syl ${⊢}\left({\phi }\wedge {x}\in {I}\right)\to {E}\in \mathrm{\infty Met}\left({V}\right)$
146 145 adantlr ${⊢}\left(\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\wedge {x}\in {I}\right)\to {E}\in \mathrm{\infty Met}\left({V}\right)$
147 simprl ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {y}\in {B}$
148 134 fveq2d ${⊢}{\phi }\to {\mathrm{Base}}_{{Y}}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)}$
149 2 148 syl5eq ${⊢}{\phi }\to {B}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)}$
150 149 adantr ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {B}={\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)}$
151 147 150 eleqtrd ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {y}\in {\mathrm{Base}}_{\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)}$
152 73 ad2antll ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {r}\in {ℝ}^{*}$
153 rpgt0 ${⊢}{r}\in {ℝ}^{+}\to 0<{r}$
154 153 ad2antll ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to 0<{r}$
155 133 139 3 4 140 141 142 143 146 151 152 154 prdsbl ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {y}\mathrm{ball}\left(\mathrm{dist}\left({S}{⨉}_{𝑠}\left({y}\in {I}⟼{R}\left({y}\right)\right)\right)\right){r}=\underset{{x}\in {I}}{⨉}\left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
156 138 155 eqtrd ${⊢}\left({\phi }\wedge \left({y}\in {B}\wedge {r}\in {ℝ}^{+}\right)\right)\to {y}\mathrm{ball}\left({D}\right){r}=\underset{{x}\in {I}}{⨉}\left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
157 125 129 130 156 syl12anc ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {y}\mathrm{ball}\left({D}\right){r}=\underset{{x}\in {I}}{⨉}\left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)$
158 124 157 eleqtrrd ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to {g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)$
159 115 158 jca ${⊢}\left(\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\wedge \left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\right)\to \left({y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\wedge {g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)\right)$
160 159 ex ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to \left(\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\to \left({y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\wedge {g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)\right)\right)$
161 160 eximdv ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to \left(\exists {y}\phantom{\rule{.4em}{0ex}}\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\to \exists {y}\phantom{\rule{.4em}{0ex}}\left({y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\wedge {g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)\right)\right)$
162 df-rex ${⊢}\exists {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)↔\exists {y}\phantom{\rule{.4em}{0ex}}\left({y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\wedge {g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)\right)$
163 161 162 syl6ibr ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to \left(\exists {y}\phantom{\rule{.4em}{0ex}}\left({y}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({y}\left({x}\right)\in {f}\left({x}\right)\wedge {g}\left({x}\right)\in \left({y}\left({x}\right)\mathrm{ball}\left({E}\right){r}\right)\right)\right)\to \exists {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)\right)$
164 107 163 mpd ${⊢}\left(\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\wedge {g}\in {B}\right)\to \exists {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)$
165 164 ex ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \left({g}\in {B}\to \exists {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)\right)$
166 eliun ${⊢}{g}\in \bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)↔\exists {y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)\phantom{\rule{.4em}{0ex}}{g}\in \left({y}\mathrm{ball}\left({D}\right){r}\right)$
167 165 166 syl6ibr ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \left({g}\in {B}\to {g}\in \bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)\right)$
168 167 ssrdv ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to {B}\subseteq \bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)$
169 83 168 eqssd ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}$
170 iuneq1 ${⊢}{v}=\underset{{x}\in {I}}{⨉}{f}\left({x}\right)\to \bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)=\bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)$
171 170 eqeq1d ${⊢}{v}=\underset{{x}\in {I}}{⨉}{f}\left({x}\right)\to \left(\bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}↔\bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}\right)$
172 171 rspcev ${⊢}\left(\underset{{x}\in {I}}{⨉}{f}\left({x}\right)\in \left(𝒫{B}\cap \mathrm{Fin}\right)\wedge \bigcup _{{y}\in \underset{{x}\in {I}}{⨉}{f}\left({x}\right)}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}\right)\to \exists {v}\in \left(𝒫{B}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}$
173 70 169 172 syl2anc ${⊢}\left(\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\wedge \left({f}:{I}⟶\mathrm{V}\wedge \forall {x}\in {I}\phantom{\rule{.4em}{0ex}}\left({f}\left({x}\right)\in \left(𝒫{V}\cap \mathrm{Fin}\right)\wedge \bigcup _{{z}\in {f}\left({x}\right)}\left({z}\mathrm{ball}\left({E}\right){r}\right)={V}\right)\right)\right)\to \exists {v}\in \left(𝒫{B}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}$
174 43 173 exlimddv ${⊢}\left({\phi }\wedge {r}\in {ℝ}^{+}\right)\to \exists {v}\in \left(𝒫{B}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}$
175 174 ralrimiva ${⊢}{\phi }\to \forall {r}\in {ℝ}^{+}\phantom{\rule{.4em}{0ex}}\exists {v}\in \left(𝒫{B}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}$
176 istotbnd3 ${⊢}{D}\in \mathrm{TotBnd}\left({B}\right)↔\left({D}\in \mathrm{Met}\left({B}\right)\wedge \forall {r}\in {ℝ}^{+}\phantom{\rule{.4em}{0ex}}\exists {v}\in \left(𝒫{B}\cap \mathrm{Fin}\right)\phantom{\rule{.4em}{0ex}}\bigcup _{{y}\in {v}}\left({y}\mathrm{ball}\left({D}\right){r}\right)={B}\right)$
177 26 175 176 sylanbrc ${⊢}{\phi }\to {D}\in \mathrm{TotBnd}\left({B}\right)$
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Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up? New to KöMaL?
# Problem S. 31. (December 2007)
S. 31. Someone made up a 5×5 table containing equations. Each cell of the table contains either a basic arithmetic operation (any of the 4 operations), a variable (a single letter) or the equality sign.
See the example:
When reading rows (from left to right), columns and the two diagonals (downwards) of the table, we get a system of 8 equations. During the solution of this system, one can always find an equation from which a suitable unknown can be expressed (independently of the other equations), or a suitable simple relation between two unknowns can be derived.
In the example above, one sees from the last line that d=0 or e=1. The first column implies a=0, so b=c due to the first row. Proceeding in a similar way, the solution is a=d=0 with b=c=e=arbitrary; or a=0, e=1 and b=c=1+d with d arbitrary.
Write your program to solve an arbitrary (syntactically correct) table. The first input parameter of your program is the name of the text file containing the table. The input file consists of 5 rows. Each row contains the successive entries of a row in the table (that is, mathematical symbols or unknowns) separated by a space. The output of your program on the standard output should consist of a list of values of the variables in alphabetical order (e.g. a=3.25'' or the value of a is arbitrary''). If the system can not be solved, a warning should be issued.
The source code of your program (s31.pas, s31.cpp, ...) together with a short documentation (s31.txt, s31.pdf, ...) should be submitted, also containing a short description of your code and the name of the developer environment to use.
Proposed by András Tálas, 12th grade student, Budapest
(10 pont)
Deadline expired on May 15, 2008.
### Statistics:
2 students sent a solution. 4 points: 2 students.
Problems in Information Technology of KöMaL, December 2007
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Examples
Chapter 7 Class 12 Integrals
Serial order wise
### Transcript
Example 36 Evaluate ∫1▒(𝑥^4 −𝑥)^(1/4)/𝑥^5 𝑑𝑥 ∫1▒(𝑥^4 − 𝑥)^(1/4)/𝑥^5 𝑑𝑥 Taking 𝑥^4 common from numerator =∫1▒〖(𝑥^4 )^(1/4) (1 − 𝑥/𝑥^4 )〗^(1/4)/𝑥^5 𝑑𝑥 = ∫1▒(𝑥 (1 − 1/𝑥^3 )^(1/4))/𝑥^5 𝑑𝑥 = ∫1▒(1 − 1/𝑥^3 )^(1/4)/𝑥^4 𝑑𝑥 Let t = 1 − 1/𝑥^3 𝑑𝑡/𝑑𝑥=3/𝑥^4 Substituting values, =∫1▒〖𝑡 〗^(1/4)/3 𝑑𝑡 = 1/3 (𝑡^( 1/(4 ) + 1)/(1/4 +1))+ C = 1/3 ((𝑡^( 5/4) )/(5/4))+ C = 4/15 〖𝑡 〗^(5/4)+ C = 𝟒/𝟏𝟓 (𝟏−𝟏/𝒙^𝟑 )^( 𝟓/𝟒)+ C
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## 255202900000000
255,202,900,000,000 (two hundred fifty-five trillion two hundred two billion nine hundred million) is an even fifteen-digits composite number following 255202899999999 and preceding 255202900000001. In scientific notation, it is written as 2.552029 × 1014. The sum of its digits is 25. It has a total of 18 prime factors and 324 positive divisors. There are 98,560,000,000,000 positive integers (up to 255202900000000) that are relatively prime to 255202900000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 15
• Sum of Digits 25
• Digital Root 7
## Name
Short name 255 trillion 202 billion 900 million two hundred fifty-five trillion two hundred two billion nine hundred million
## Notation
Scientific notation 2.552029 × 1014 255.2029 × 1012
## Prime Factorization of 255202900000000
Prime Factorization 28 × 58 × 29 × 88001
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 18 Total number of prime factors rad(n) 25520290 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,202,900,000,000 is 28 × 58 × 29 × 88001. Since it has a total of 18 prime factors, 255,202,900,000,000 is a composite number.
## Divisors of 255202900000000
324 divisors
Even divisors 288 36 36 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 324 Total number of the positive divisors of n σ(n) 6.58726e+14 Sum of all the positive divisors of n s(n) 4.03523e+14 Sum of the proper positive divisors of n A(n) 2.0331e+12 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59751e+07 Returns the nth root of the product of n divisors H(n) 125.524 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,202,900,000,000 can be divided by 324 positive divisors (out of which 288 are even, and 36 are odd). The sum of these divisors (counting 255,202,900,000,000) is 658,725,570,935,460, the average is 20,331,036,139,98.,333.
## Other Arithmetic Functions (n = 255202900000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 98560000000000 Total number of positive integers not greater than n that are coprime to n λ(n) 770000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7938579266860 Total number of primes less than or equal to n r2(n) 144 The number of ways n can be represented as the sum of 2 squares
There are 98,560,000,000,000 positive integers (less than 255,202,900,000,000) that are coprime with 255,202,900,000,000. And there are approximately 7,938,579,266,860 prime numbers less than or equal to 255,202,900,000,000.
## Divisibility of 255202900000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 1 0 7
The number 255,202,900,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (255202900000000)
Base System Value
2 Binary 111010000001101100001110001100110011110100000000
3 Ternary 1020110121011021220001101202021
4 Quaternary 322001230032030303310000
5 Quinary 231422221014400000000
6 Senary 2302434332130331224
8 Octal 7201541614636400
10 Decimal 255202900000000
12 Duodecimal 24758046a29b14
20 Vigesimal 14i8h5650000
36 Base36 2igmldil8g
## Basic calculations (n = 255202900000000)
### Multiplication
n×i
n×2 510405800000000 765608700000000 1020811600000000 1276014500000000
### Division
ni
n⁄2 1.27601e+14 8.50676e+13 6.38007e+13 5.10406e+13
### Exponentiation
ni
n2 65128520168410000000000000000 16620987219686720389000000000000000000000000 4241724139326988134761928100000000000000000000000000000000 1082500301356251420256834860711490000000000000000000000000000000000000000
### Nth Root
i√n
2√n 1.59751e+07 63430.1 3996.88 760.987
## 255202900000000 as geometric shapes
### Circle
Diameter 5.10406e+14 1.60349e+15 2.04607e+29
### Sphere
Volume 6.96218e+43 8.18429e+29 1.60349e+15
### Square
Length = n
Perimeter 1.02081e+15 6.51285e+28 3.60911e+14
### Cube
Length = n
Surface area 3.90771e+29 1.6621e+43 4.42024e+14
### Equilateral Triangle
Length = n
Perimeter 7.65609e+14 2.82015e+28 2.21012e+14
### Triangular Pyramid
Length = n
Surface area 1.12806e+29 1.9588e+42 2.08372e+14
## Cryptographic Hash Functions
md5 9dfbe5d9005c233eafa3c1acfb337e10 5aaf3bbe0320f7b1ff8bcca325ae5a1a34fbdba1 547cacc3bf5ed924f510cffd2b62f0236182d438702a8601cccebe9ee45b1cb3 f0b329abddfcb21d61df01fc413cf0fd25c1cf348d178f7e13bfc18b79071088f55318e215772e758e17fad64b96a626022c902903384642b1fb52e9980533c7 1126dd19560acff2e757b38367f93e05d94bde26
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# isposdef inconsistent results, numerically unstable?
#1
I am checking here before opening an issue.
Using the above jld file holding a symmetric matrix A, the following results are very weird:
julia> d = load("posdef_symm_sparse_matrix.jld");
julia> A = d["A"];
julia> typeof(A)
Symmetric{Float64,SparseMatrixCSC{Float64,Int64}}
julia> isposdef(A.data)
false
julia> isposdef(Array(A.data))
false
julia> isposdef(Array(A))
true
julia> Array(A) ≈ Array(A.data)
true
julia> Array(A) == Array(A.data)
false
julia> min(eigvals(Array(A.data))...)
0.00017220258699615022
julia> min(eigvals(Array(A))...)
0.0001722025869961005
Is there a more reliable way to check if a sparse matrix is positive definite or not in v0.6.2? The eigmin function fails on sparse matrices in v0.6.2. And finding the whole eigenvalue decomposition is overkill.
#2
I thought I just mentioned it here:
But I guess not. Order of operations matters when taking into account floating point error. Multithreading can change the order of operations, making the multithreaded results stochastic in the last few values due to changing floating point error. I’m not sure what eigs is calling but I would assume that’s what’s happening here.
I would’ve suggested iterative solvers, but they find the eigenvalues in the other order…
#3
True but I thought with Float64, I can get more than just 3 decimal places! The smallest eigenvalue is not even that small. If the error from machine precision overshoots, then that’s a numerical instability of the algorithm, so perhaps something can be done about this.
Well if they work on negative definite matrices, I could run them on -A instead! I didn’t check that path yet, but I thought there must be a simpler and reliable way of doing that very simple operation.
#4
The check for A.data fails because the underlying data is not exactly symmetric. The Symmetric{Float64,SparseMatrixCSC{Float64,Int64}} type enforces symmetry but unfortunately the isposdef method doesn’t work for Symmetric{Float64,SparseMatrixCSC{Float64,Int64}} on 0.6.x. This is fixed on 0.7 so things should be better soon. Meanwhile, you could convert to a SparseMatrixCSC, i.e.
julia> isposdef(convert(SparseMatrixCSC{Float64,Int}, A))
true
#5
Is this really the reason? The discrepancy is very very small.
julia> norm(Array(A.data) - Array(A.data)', 1)
3.4701245783114176e-17
This works thanks.
#6
3 decimal places? Your post shows it consistently getting 0.0001722025869961 and just having errors/changes in the last 3 decimal places if I read it correctly?
Iterative solvers will order them by norm. I didn’t know about eigmin but now that I heard of it, I’m curious what it’s doing.
#7
Oh I meant that the smallest eigenvalue is only 0 if you take the first 3 decimal places. So the isposdef function is treating it as less than or equal to 0 when it is clearly not if only you take the fourth decimal place into account. So even a discrepancy as small as 3.4701245783114176e-17 in the matrix can make the isposdef miss the real minimum eigenvalue by a good 0.0001722 or so thinking it is actually 0, which seems too unstable for my taste.
#8
Are you sure it’s even checking the minimum eigenvalue? I’m not seeing that. It checks if it’s Hermitian and if so then it uses LAPACK’s dpotrf to do a Cholesky factorization and then grabs something from it (I don’t know what the second return is there). From the result it seems like an algorithm bug (that was fixed on v0.7 according to @andreasnoack) and not something due to numerical instability in minimum eigenvalue calculations.
#9
I have no idea what it’s doing, but mathematically if the algorithm is led to conclude that the matrix is not positive definite, that’s equivalent to using an eigenvalue algorithm that misses the minimum eigenvalue by a good margin. I was just using mathematical intuition to reason about what the algorithm should return, perhaps not the most direct/relevant discussion though!
Another simple way to view the weirdness of isposdef above is as a classifier that will mis-classify a data point if you perturb its features by 3.4701245783114176e-17! I wouldn’t use such a classifier in any critical application.
#10
Our definition of positive definiteness in isposdef is symmetric (Hermitian) and “can be Cholesky factorized” which is pretty close to having positive eigenvalues but not as robust. We have had some discussions about using a tolerance for the symmetry check but we concluded that it was better to make the users enforce symmetry via the Symmetric wrapper instead of coming up with a possibly arbitrary tolerance for symmetry. Hence, as long as you keep your matrix wrapped in the Symmetric type, a small permutation shouldn’t change the result of isposdef unless the matrix is almost singular.
#11
I see so it’s failing the first condition, makes sense now. I was under the wrong impression that it was using an unstable algorithm and making a wrong conclusion. isposdef didn’t work on Symmetric in v0.6.2, but it works in v0.7 so I won’t face this issue again in the future I guess as long as I use a Symmetric wrapper before checking for isposdef. Thanks again!
#12
But shouldn’t the definition be that the symmetric component can be Cholesky factorized? Being positive definite has nothing to do with being symmetric. But (A+A')/2 being positive definite implies and is implied by A being positive definite.
#13
There are two different definitions. The one you mention here and the one I mentioned. We are using the latter. To cite Wikipedia
More generally, an n \times n Hermitian matrix M is said to be positive definite if …
Some authors use more general definitions of “positive definite” that include some non-symmetric real matrices, or non-Hermitian complex ones.
It has come up a couple of times previously but not many times so it seems that in most applications matrices are in fact symmetric (or supposed to be) when people are checking for positive definiteness. Using the stricter definition also makes the implementation slightly cheaper and simpler.
#14
This statements says that if M is nxn and Hermitian -> it is positive definite when… . It doesn’t say that every positive definite matrix is Hermitian!
May you point me to one established, preferably old, linear algebra book/reference which uses the “less general” definition of positive definiteness?
Perhaps an option would make everyone happy? Also the user has the option of using a Symmetric wrapper, and if not, the function will check for symmetry first thing, so computing the symmetric component will only be relevant in the following cases:
1. The user has entered a non-symmetric matrix and is expecting isposdef to use the the general definition. This is like me now because I expect isposdef to return true even if the matrix is not perfectly symmetric.
2. The user has wrongly entered a non-symmetric matrix where he/should should have entered a symmetric one, then probably they should have used a Symmetric wrapper if the matrix is supposed to be symmetric or they should have checked for symmetry separately. I think even those people would agree that a non-symmetric matrix can be positive definite at least according to some authors as you mention, and if they don’t want their matrix to be treated as a non-symmetric one, they should go through the extra “burden” of using a Symmetric wrapper.
An option which defaults to the cheaper behavior would make me happy, but if it defaults to the more general/correct behavior I would be even happier. I am genuinely interested what @stevengj thinks about this.
#15
Although I’m on your side here, the opposition is formidable.
Limitation by association with quadratic forms goes back at least to Gantmacher (1950’s). Springer’s Encyclopedia of Mathematics is in this camp. Gil Strang has promulgated the “restricted” version in his textbooks, which may have poisoned influenced the Julia developers.
Most others (e.g. Golub & van Loan, LAPACK, Mathematica) are careful to write “symmetric (or Hermitian) positive definite” where that applies.
Since it seems obvious to many of us that definiteness is a feature of bilinear forms without regard to symmetry, the LinAlg documentation should be clarified if your suggestions are not followed. Or maybe there should just be separate functions: ishpd, isspd, isposdef?
#16
I see, that explains the division. I don’t see a reason to lean to one side though. I am a fan of Golub & van Loan’s masterpiece and there is a very clear distinction between the concept of symmetry and the concept of definiteness in their book.
I second these suggestions.
#17
I would be very interested in learning how non-symmetric positive (semi) definite matrices are useful in practice.
#18
If you are solving a system of equations Ax = b and A is unsymmetric and positive definite or indefinite, this will dictate the choice of decomposition or iterative solver to be used, so it is important to know the class of A. Just because many problems have A as symmetric doesn’t mean that there isn’t a whole theory for unsymmetric matrices! A quick Google search brought up Markov chains as an application for solving unsymmetric systems of equations.
#19
By your argument, one might conclude that there is no need for the isposdef function altogether, but I need it! And maybe someone will need to check the positive definiteness of some unsymmetric matrix to make sure it is invertible for example before solving the system, there is no reason to assume that that won’t happen! Yes it’s expensive, and yes there are probably other methods to do this which one might use as an optimization or to analytically prove the class of a matrix, but the argument here is to ensure that isposdef is a convenient tool for checking positive definiteness of any matrix, symmetric or not, because that’s what a group of people will justifiably think it is supposed to do.
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# physics
posted by .
A jogger accelerates from rest to 5.64 m/s in 3.36 s. A car accelerates from 24.6 to 39.8 m/s also in 3.36 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.36 s?
• physics -
a. a = (V-Vo)/t = (5.64-0)/3.36=1.68 m/s^2.
d1 = 0.5a*t^2 = 0.5*1.68*(3.36)^2=9.48 m.
b. a = (39.8-24.6)/3.36 = 4.52 m/s^2.
d2 = 0.5*4.52*(3.36)^2 = 25.5 m.
c. d2 d1 = 25.5 9.48 = 16.0 m.
• physics -
c. d2-d1 = 25.5-9.48 = 16.0 m.
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A jogger accelerates from rest to 2.7 m/s in 2.2 s. A car accelerates from 39.0 to 44.0 m/s also in 2.2 s. (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does …
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# Are all line integrals zero by divergence theorem?
Suppose there is a vector field A. Now the line integral over some curve c equals double integration of curl A over surface S enclosed by C. Now if we put the divergence theorem in this then it becomes triple integral of divergence of curl Adv over the volume enclosed by the surface. Now divergence of curl of vector A becomes zero. Now so the volume integral should be zero. So the surface and line integral is also zero. So we can do this for any line integral (where the vector is defined in the surface). So does it imply all line integrals are zero. Where am I doing the mistake? Sorry for mathjax.
• What are the hypotheses on using the divergence theorem? Aug 4, 2018 at 0:54
• @JasonDeVito the vector should be valid throughout the region. The surface should enclose a volume. The region has to be simply connected. Aug 4, 2018 at 0:56
• If the surface encloses a volume, does it have a boundary curve? Aug 4, 2018 at 0:59
• @JasonDeVito it should have a plane not a curve I guess. Except sphere or paraboloid etc. Aug 4, 2018 at 1:00
• Nope, no exceptions. If a surface encloses a volume, it has no boundary curve. So there is no way to apply the divergence theorem after applying Stokes's theorem Aug 4, 2018 at 1:20
In order to apply the divergence theorem, your surface $S$ needs to enclose a solid. This, in particular, implies that $S$ itself has no boundary. But using Stokes's theorem relates the integral over a surface to an integral over its boundary. Thus, there is no surface $S$ for which both theorems apply.
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