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# physics
posted by .
A 35-kg skier skis directly down a frictionless slope angled at 11° to the horizontal. Assume the skier moves in the negative direction of an x axis along the slope. A wind force with component Fx acts on the skier. What is Fx (in N) if the magnitude of the skier's velocity is.. constant? ..increasing at a rate of 1.0 m/s^2? ..increasing at a rate of 2.0 m/s^2?
Writing the sum of the forces down the slope....
Weight*sinTheta-Windfriction=mass*acceleration.
In the first part, a=0
so..
35*sin(11)-Fx=0
and solvee for Fx?
.. as for the increasing rate.. is it telling me that the 1.0m/s2 is added to the gravity of 9.8m/s2?
• physics -
this is kinda late answer, but still...
35*sin(11)-Fx=0
This works for a = 0
35*sin(11)+ma-Fx=0
So, for increasing rate of a = 1m/s^2
This is going in negative direction, so
35*sin(11)+(35)(1)=Fx
And for a = 22/s^2
35*sin(11)+(35)(2)=Fx
Crunch in calculator for Fx in Newtons
• physics -
Actually, just noticed that was for in -x direction, so change sign on acceleration
35*sin(11)+(35)(-1)=Fx
35*sin(11)+(35)(-2)=Fx
• physics -
ur supposed to multiply the 35sintheta with the acceleration due to gravity (9.8)
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# Help on solving this equation!
• Aug 12th 2012, 09:09 AM
Maxime
Help on solving this equation!
Dear all,
I am having trouble solving this equation, it has been some time since I have done this kind of stuff.
Is there anyone here that can tell me how to solve this equation? It is very basic, but it has been some time since I have done these equations.
Cos(3*pi*x)>ln(x+1)
I just can not get my head around it, I only came this far (Headbang):
ecos(3*pi*x)>x+1
Also, I am having trouble trying to get the integral of sin(x)ecos(x). How does this work with either the reverted product rule or the substitution method?
Thank you!
• Aug 12th 2012, 09:15 AM
Prove It
Re: Help on solving this equation!
Quote:
Originally Posted by Maxime
Dear all,
I am having trouble solving this equation, it has been some time since I have done this kind of stuff.
Is there anyone here that can tell me how to solve this equation? It is very basic, but it has been some time since I have done these equations.
Cos(3*pi*x)>ln(x+1)
I just can not get my head around it, I only came this far (Headbang):
ecos(3*pi*x)>x+1
Also, I am having trouble trying to get the integral of sin(x)ecos(x). How does this work with either the reverted product rule or the substitution method?
Thank you!
For the integral, write \displaystyle \begin{align*} \int{\sin{(x)}e^{\cos{(x)}}\,dx} &= -\int{-\sin{(x)}e^{\cos{(x)}}\,dx} \end{align*} and make the substitution \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*} and the integral becomes \displaystyle \begin{align*} -\int{e^u\,du} \end{align*}. Go from here.
• Aug 12th 2012, 09:28 AM
Maxime
Re: Help on solving this equation!
Thank you! That example really brings out memories of the substitution method.
Anyone any idea of how to solve this for x?:
cos(3*pi*x)>ln(x+1)
• Aug 12th 2012, 09:35 AM
Prove It
Re: Help on solving this equation!
You're not going to be able to get exact solutions. I'd suggest getting wolfram to graph both equations, noting where they cross, and deciding which intervals satisfy your inequality.
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# Thread: Integration by parts using arctan
1. ## (solved) Integration by parts using arctan
Hi, I'm in need of some help as I can't seem to get this problem right.
Im trying to find the integral of 1/(9x^2+36x+81)dx
So, I started by completing the square and got int 1/[9((x+2)^2+5)] and consequently int 1/(9(x+2)^2+45).
I then turned 9(x+2)^2 into (3(x+2))^2 and factored out the 45 leaving me with (1/45) int 1/[1+(3(x+2))^2 I then used u-sub and made u 3(x+2) and eventually got (1/45)*(1/3) int du/1+u^2.
I know the derivative of arctan is dx/1+x^2 and ended up with
(1/135)arctan(3(x+2)). I know this isn't right as the online program I enter my answer in says its wrong.
If someone could help me out it would be much appreciated!
2. Originally Posted by Phaustus
I then turned 9(x+2)^2 into (3(x+2))^2 and factored out the 45 leaving me with (1/45) int 1/[1+(3(x+2))^2
Here's where you went wrong. It should've been:
$\displaystyle \int \frac{1}{(3(x+2))^2+45}~dx = \int \frac{1}{45((\frac{3x+6}{\sqrt{45}})^2 + 1)}$
3. Ok, yea I knew it was some stupid error like that. I got the correct answer now. Thanks for your help!
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http://math.stackexchange.com/questions/67538/can-a-variable-appear-as-a-denominator-in-a-linear-equation-or-is-the-system-non
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# Can a variable appear as a denominator in a linear equation or is the system non-linear?
I am trying to figure out whether the following equation is non-linear or if it's linear, how would I solve it?
$x+\frac{2}{y}=0$
It can be rewritten as $x+y^{-2}=0$ so I guess if this is non-linear.
-
Indeed, the function is not linear. – Arturo Magidin Sep 25 '11 at 21:43
But it would be $xy+2=0$, not $x+y^{-2}=0$. – Ross Millikan Sep 25 '11 at 21:50
Or it would be $x + 2y^{-1}=0$. – Arturo Magidin Sep 25 '11 at 21:54
But if all your $y$'s appear as denominators, you can make a linear equation system by substituting $u=1/y$. – Henning Makholm Sep 25 '11 at 22:31
@DBLim: When you are dealing with a question from a beginner that is trying to puzzle out some of the basics, please don't "correct" what was likely an error rather than a typo as an edit; it's important to point out those errors to prevent them from being committed again. If you simply edit them out, the OP may not realize he had made an algebraic error along the line. – Arturo Magidin Sep 25 '11 at 22:34
It is non-linear, and the criteria of linearity I've put in answer to your previous question. Here $f(x,y) = x+\frac2y$ and for $\alpha = 1,\beta = 1$ we have $$f(x'+x'',y'+y'') = x'+x''+\frac2{y'}+\frac2{y''}\neq x'+x''+\frac2{y'+y''} = f(x',y')+f(x'',y'')$$ thus the equation is non-linear.
Saw the possible misprint in your question: $\frac2y\neq y^{-2} = \frac1{y^2}$.
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Billion is a very big number that is 1 complied with by 9 zeros. That is also known together a thousand millions as 1,000 × 1000,000 = 1000,000,000. The prefix "giga-" in words "gigameter" describes billion, which means 1 gigameter is nothing but a exchange rate meters. Billion has two notations, one is 109 i m sorry is followed in the American system and the various other one is 1012 or 1 followed by 12 zeros i beg your pardon is followed in countless non-American nations like French and also Germany.
You are watching: 1 billion is equal to how many lakhs
1 What is a Billion? 2 Technique come Remember Zeroes in Billion 3 Difference in between Billion, Million, and also Trillion 4 Solved examples on Billion 5 Practice questions on Billion 6 FAQs ~ above Billion
## What is a Billion?
In number systems, once you begin counting numbers from 0 come 9, 9 to 99, 100 come 999, and so on, you will observe that together you relocate forward, you will be presented to 5 digits, 6 digits, 10 digits numbers, and similar higher numbers. A billion is a 10-digit number developed by 1 adhered to by 9 zeros.
### Definition of Billion
A billion is defined together a ten-digit number. That is counted after 100 million and carries forward the chain in the direction of trillions. The is stood for as 109, i m sorry is the the smallest 10-digit number in math.
## Technique come Remember Zeroes in Billion
1 exchange rate is written as 1,000,000,000 and also can be represented in 2 forms.
## Difference Between Billion, Million, and Trillion
As you know in a number system, count is constantly in raising order. So, the boosting order here is million, billion, and also trillion. The means, once we multiply 1000 to million, we acquire billion. Similarly, as soon as we main point 1000 come a billion, we obtain a trillion. The simple difference in between trillion, billion, and also million is clearly shows in the picture below:
### Trillion
A trillion is a million time million which have the right to be created as 1,000,000 × 1,000,000 =1,000,000,000,000. In clinical notation, the is stood for as 1 × 1012. The can likewise be treated as a thousand billion. In simple words, it has actually 12 zeroes.
### Million
A million is a thousand times thousand which deserve to be written as 1,000 × 1,000 = 1,000,000. In clinical notation, that is represented as 1 × 106. In simple words, it has 6 zeroes. The simple flow in between the place value the the digits and how the chain walk on is displayed below:
Think Tank:
1) discover out what these numbers would be called:
a) 1021
b) 1024
c) 1027
d) 1033
2) try to write these number in the increased form.
See more: What Engine Does Napa 1348 Oil Filter Cross Reference, Napa 1348Mp
a) 7,826,657,098
b) 8,013,973,083
### Topics regarded Billion
Check the end these interesting write-ups related come billion.
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SUMMARY / RELATED TOPICS
In axiomatic set theory and the branches of logic and computer science that use it, the axiom of pairing is one of the axioms of Zermelo–Fraenkel set theory. It was introduced by Zermelo as a special case of his axiom of elementary sets. In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: ∀ A ∀ B ∃ C ∀ D In words: Given any set A and any set B, there is a set C such that, given any set D, D is a member of C if and only if D is equal to A or D is equal to B. Or in simpler words: Given two sets, there is a set whose members are the two given sets; as noted, what the axiom is saying is that, given two sets A and B, we can find a set C whose members are A and B. We can use the axiom of extensionality to show. We call the set C the pair of A and B, denote it, thus the essence of the axiom is: Any two sets have a pair. The set is abbreviated, called the singleton containing A. Note that a singleton is a special case of a pair. Being able to construct a singleton is necessary, for example, to show the non-existence of the an infinitely descending chains x = from the Axiom of regularity.
The axiom of pairing allows for the definition of ordered pairs. For any sets a and b, the ordered pair is defined by the following: =. Note that this definition satisfies the condition = ⟺ a = c ∧ b = d. Ordered n-tuples can be defined recursively as follows: =; the axiom of pairing is considered uncontroversial, it or an equivalent appears in just about any axiomatization of set theory. In the standard formulation of the Zermelo–Fraenkel set theory, the axiom of pairing follows from the axiom schema of replacement applied to any given set with two or more elements, thus it is sometimes omitted; the existence of such a set with two elements, such as, can be deduced either from the axiom of empty set and the axiom of power set or from the axiom of infinity. In the absence of some of the stronger ZFC axioms, the axiom of pairing can still, without loss, be introduced in weaker forms. In the presence of standard forms of the axiom schema of separation we can replace the axiom of pairing by its weaker version: ∀ A ∀ B ∃ C ∀ D.
This weak axiom of pairing implies that any given sets A and B are members of some set C. Using the axiom schema of separation we can construct the set whose members are A and B. Another axiom which implies the axiom of pairing in the presence of the axiom of empty set is ∀ A ∀ B ∃ C ∀ D, it differs from the standard one by use of D ∈ A instead of D = A. Using for A and x for B, we get for C. Use for A and y for B, getting for C. One may continue in this fashion to build up any finite set, and this could be used to generate all hereditarily finite sets without using the axiom of union. Together with the axiom of empty set and the axiom of union, the axiom of pairing can be generalised to the following schema: ∀ A 1 … ∀ A n ∃ C ∀ D that is: Given any finite number of sets A1 through An, there is a set C whose members are A1 through An; this set C is again unique by the axiom of extensionality, is denoted. Of course, we can't refer to a finite number of sets rigorously without having in our hands a set t
Gjesing Church Gesing Church, is a modern church in Esbjerg in the southwest of Jutland, Denmark. Designed by architects Niels Munk and Keld Wohlert, the red-brick building with a steeply pitched roof was completed in 1983. In 1979, Gjesing Parish was established in the northern suburbs of Esbjerg; the area had formed part of Bryndum Parish. A temporary building served the new parish until, following a competition between four architectural firms in 1979, a complex designed by Niels Munk and Keld Wohlert of Solrød Strand was inaugurated on 30 January 1983. In 2010, the complex was extended with a handicapped toilet and three offices; the church is adjacent to Esbjerg Storcenter, a shopping centre, in a residential area with apartment buildings and detached houses. Under the same tall roof, the complex consists of the church proper to the east and a parish hall to the west. A corridor connects the two and leads to offices, a classroom for confirmation candidates and atriums; the ground plan, which consists of a porch and waiting room to the west and a sacristy and a chapel to the east, can be described as two large rectangles with two steep half-roofs, set off a little from one another.
Light from glazed rifts in the roofing above comes into the nave from the south and north gables, with a similar approach in the church hall. The entire complex is built of red brick, with zig-zag patterns in the gable brickwork, is roofed with copper; the free-standing bell tower stands to the southwest. Inside, the harmoniously designed nave has an open roof trussed ceiling, finished with white sheeting, whitewashed walls. Three old bricks from the mother church in Bryndum are built into the wall behind the pulpit; the floor is laid with red tiles. A white altar table of pine stands beside the east wall; the font takes the form of an intersected cylinder consisting of four blocks of Bornholm granite. Seen from above, the blocks form a cross; the pulpit, which resembles a normal speaking podium, rises only above the floor. Like the altar, the pulpit and the pews are made of pine. Gjesing Kirke website
Rafael Camacho Guzmán, was a Mexican trade union and political leader, member of the Institutional Revolutionary Party and governor of Queretaro from 1979 to 1985. Rafael Camacho Guzmán was born in the city of Queretaro in 1916, he studied agronomy in Guanajuato. Speaker worked in the XEBZ in Mexico City, it was the official broadcaster of the Presidency of the Republic during the government of Miguel Aleman. He participated in the founding of the Union of the Industry of the Radio and Television and was its director from 1961 to 1979, he was one of the most prominent of the Confederation of Workers of Mexico and close to Fidel Velazquez leaders. He was elected Governor of Queretaro in the 1979 election. During his administration he held various public works such as the "Corregidora of Queretaro" Stadium, the new auditorium Josefa Ortiz de Domínguez and network highways of the Sierra Gorda, he was a Mexican trade union and political leader, member of the Institutional Revolutionary Party and governor of Queretaro from 1979 to 1985.
Rafael Camacho Guzmán began his political career as leader of the Industrial Union of Workers and Television and Radio Artists, which placed as one of the most prominent of the Confederation of Workers of Mexico and close to Fidel Velazquez leaders, this allowed him to occupy various positions of popular election among which are the Senator in 1976 and Governor of Querétaro in 1979. During his administration he held various public works such as the Estadio La Corregidora and highways of the Sierra Gorda, he studied agricultural engineering. Announcer since 1942. Co-founder and secretary general of the Union of Industry Radio and Allied of Mexico.. He was a worker delegate from Mexico to the International Labour Organization, secretary of the Labour Congress and president of the Executive Council of the Inter-American Regional Labor Organization. Of the board of the National Council of Advertising and governor of Querétaro, he died in Mexico City, Federal District, at the age of 82 years. His remains rest in the Pantheon of the Colonia Cimatario
CSS Neuse was a steam-powered ironclad ram of the Confederate States Navy that served in the latter part the American Civil War and was scuttled to avoid capture by advancing Union Army forces. In the early 1960s, she produced 15,000 artifacts from her raised lower hull, the largest number found on a recovered Confederate vessel; the remains of her lower hull and a selection of her artifacts are on exhibit in Kinston, North Carolina at the CSS Neuse Interpretive Center State Historic Site, which belongs to the North Carolina Department of Natural and Cultural Resources. The ironclad is listed on the National Register of Historic Places. A contract for the construction of Neuse was signed on 17 October 1862 between the shipbuilding company of Thomas Howard and Elijah Ellis and the Confederate Navy. Work began in October of that year on the bank across the Neuse River from the small village of Whitehall, North Carolina; the gunboat's design was identical to her sister ironclad CSS Albemarle, but Neuse differed from Albemarle by having four additional gun ports added to her eight-sided armored casemate.
The hull was 158 feet long by 34 feet wide, she was constructed of locally abundant pine, with some 4 inches of oak used as sturdy backing for her 4-inch-thick wrought iron armor. Many delays in construction were incurred by a lack of available materials the iron plate for her armored casemate and deck. Due to continuing iron plate shortages, Neuse became the first of several Southern ironclads built with unarmored decks; this situation was compounded by the Confederate Army exercising priority over the Navy in the use of the South's inadequate railroad system for transporting vital war materiel. Neuse was equipped with two 6.4-inch Brooke rifled cannon. Both cannons were positioned along the ironclad's center-line in the armored casemate, one forward, the other aft; the field of fire for both pivot rifles was 180-degrees, from port to starboard: Each cannon could fire from one of five gun port positions or could deliver a two cannon broadside. Neuse's projectiles consisted of explosive shells, anti-personnel canister shot, grape shot, blunt-nosed, solid wrought iron "bolts" for use against Union armored ships.
Launched in November 1863 while still needing fitting out, Neuse got up steam in April 1864 for duty on the inland waters of North Carolina as part of the force under Commander R. F. Pinkney, CSN. Shortly thereafter, the ironclad grounded off Kinston due to her inexperienced crew, conscripted from the Confederate Army. After that, due to a lack of available Confederate Army shore support, she never left the river area around Kinston, serving instead as a floating ironclad fortification. In March 1865, with Kinston under siege by Union forces, gunpowder trails were laid down which led to a cache of explosives placed in her bow. Neuse burned to just below her waterline and sank into the river mud preventing capture by the advancing Union Army forces, commanded by Major General John M. Schofield. At some point following the war, her sunken hulk, lying in shallow river water and mud, was salvaged of its valuable metals: cannon and their fittings, iron ram, casemate armor, both propellers and their shafts, her steam power plant.
Whatever bits and pieces remained, including her projectiles, lay undisturbed in and around the wreck until Neuse was raised nearly a century later. After nearly a century, the remaining lower hull of the ironclad was discovered and raised in 1963. Neuse's hull was temporarily installed in the Governor Caswell Memorial, beside the river, in Kinston. Since 2013, Neuse and her artifacts have been on display in a new, climate-controlled building in downtown Kinston. There are only four recovered Civil War era ironclad wrecks, CSS Neuse, CSS Muscogee, USS Monitor, USS Cairo. Other Union and Confederate ironclad wreck sites remain untouched; the successful Confederate submarine H. L. Hunley, which sank the Union blockading sloop-of-war USS Housatonic, was recovered and is undergoing extensive restoration and long term conservation at the Warren Lasch Conservation Center in North Charleston, South Carolina. A replica of the CSS Neuse, better known as CSS Neuse II, was the brainchild of Kinston activist and businessman Ted Sampley and built by Alton Stapleford.
Neuse II is on grounds display at a separate site in Kinston and contains a complete fitted-out interior that shows all shipboard details. Neuse is the only Confederate ironclad that has a full-size replica on display. Since April 2002 Neuse's sister ironclad, CSS Albemarle has had a 3⁄8 scale replica, 63 feet long, at anchor near the Port O' Plymouth Museum in Plymouth, North Carolina; this ironclad replica is capable of sailing on the river. Bisbee, Saxon T.. Engines of Rebellion: Confederate Ir
Regent's Park is one of the Royal Parks of London. It occupies high ground in inner north-west London, administratively split between the City of Westminster and the London Borough of Camden, it contains London Zoo. The Park was designed by James and Decimus Burton, its construction was financed by James after the Crown Estate rescinded its pledge to do so. The park is Grade I listed on the Register of Historic Parks and Gardens; the park has an outer ring road called the Outer Circle and an inner ring road called the Inner Circle, which surrounds the most tended section of the park, Queen Mary's Gardens. Apart from two link roads between these two, the park is reserved for pedestrians; the south and most of the west side of the park are lined with elegant white stucco terraces of houses designed by John Nash and Decimus Burton. Running through the northern end of the park is Regent's Canal, which connects the Grand Union Canal to London's historic docks; the 166-hectare park is open parkland with a wide range of facilities and amenities, including gardens.
The northern side of the park is the home of London Zoo and the headquarters of the Zoological Society of London. There are several public gardens with flowers and specimen plants, including Queen Mary's Gardens in the Inner Circle, in which the Open Air Theatre stands. Winfield House, the official residence of the U. S. Ambassador to the United Kingdom, stands in private grounds in the western section of the park. South of the Inner Circle is dominated by Regent's University London, home of the European Business School London, Regent's American College London and Webster Graduate School among others. Abutting the northern side of Regent's Park is Primrose Hill, another open space which, with a height of 256 feet, has a clear view of central London to the south-east, as well as Belsize Park and Hampstead to the north. Primrose Hill is the name given to the surrounding district; the public areas of Regent's Park are managed by a government agency. The Crown Estate Paving Commission is responsible for managing certain aspects of the built environment of Regent's Park.
The park lies within the boundaries of the City of Westminster and the London Borough of Camden, but those authorities have only peripheral input to the management of the park. The Crown Estate owns the freehold of Regent's Park. In the Middle Ages the land was part of the manor of Tyburn, acquired by Barking Abbey; the 1530s Dissolution of the Monasteries meant Henry VIII appropriated it, under that statutory forfeiture with minor compensation scheme. It has been state property since, it was set aside as a hunting and forestry park, Marylebone Park, from that Dissolution until 1649 after which it was let as small-holdings for hay and dairy produce. Although the park was the idea of the Prince Regent George IV, was named for him, James Burton, the pre-eminent London property developer, was responsible for the social and financial patronage of the majority of John Nash's London designs, for their construction. Architectural scholar Guy Williams has written, "John Nash relied on James Burton for moral and financial support in his great enterprises.
Decimus had showed precocious talent as a draughtsman and as an exponent of the classical style... John Nash needed the son's aid, as well as the father's". Subsequent to the Crown Estate's refusal to finance them, James Burton agreed to finance the construction projects of John Nash at Regent's Park, which he had been commissioned to construct: in 1816, Burton purchased many of the leases of the proposed terraces around, proposed villas within Regent's Park, and, in 1817, Burton purchased the leases of five of the largest blocks on Regent Street; the first property to be constructed in or around Regent's Park by Burton was his own mansion: The Holme, designed by his son, Decimus Burton, completed in 1818. Burton's extensive financial involvement "effectively guaranteed the success of the project". In return, Nash agreed to promote the career of Decimus Burton; such were James Burton’s contributions to the project that the Commissioners of Woods described James, not Nash, as "the architect of Regent's Park".
Contrary to popular belief, the dominant architectural influence in many of the Regent's Park projects – including Cornwall Terrace, York Terrace, Chester Terrace, Clarence Terrace, the villas of the Inner Circle, all of which were constructed by James Burton's company – was Decimus Burton, not John Nash, appointed architectural "overseer" for Decimus's projects. To the chagrin of Nash, Decimus disregarded his advice and developed the Terraces according to his own style, to the extent that Nash sought the demolition and complete rebuilding of Chester Terrace, but in vain. Decimus's terraces were built by his father James; the Regent Park scheme was integrated with other schemes built for the Prince Regent by the triplet of Nash, James Burton, Decimus Burton: these included Regent Street and Carlton House Terrace in a grand sweep of town planning stretching from St. James's Park to Parliament Hill; the scheme is considered one of the first examples of a garden suburb and continues to influence the design of suburbs.
The park was first opened to
The Whole Story is the first compilation album by English singer Kate Bush. Released in November 1986, it was Bush's third UK number one album; the compilation went on to become her best-selling release to date, being certified four times platinum in the United Kingdom. The album includes eleven of Bush's previous singles, with a unreleased track entitled "Experiment IV", released as a single and reached the UK Top 30. A newly recorded version of Bush's debut single; the album mix of "The Man with the Child in His Eyes" features on this album instead of the single version. A home video compilation of the same name was released which includes the promotional videos for each song on the album. In 2014, during Kate Bush's Before the Dawn residency at the Hammersmith Apollo, The Whole Story charted at number 8 in the UK. All tracks are written by Kate Bush. All tracks are written by Kate Bush. Kate Bush – keyboards, producer Ian Cooper – cutting engineer Jon Kelly – producer Andrew Powell – producer
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Q. 2
Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.
Let x = 2 and Δx = 0.01. Then, we get,
f(2.01) = f(x + Δx) = 4(x +Δx)2 + 5 ( x + Δx) + 2
Now, Δy = f (x + Δx) – f(x)
f (x + Δx) = f(x) + Δy
≈ f(x) + f’(x).Δx (as dx = Δx)
f(2.01) ≈ (4x2 + 5x + 2) + (8x + 5) Δx
= [4(2)2 + 5(2) + 2] +[8(2) + 5] (0.01)
= (16 + 10 + 2) + (16 + 5)(0.01)
= 28 + 0.21
= 28.21
Therefore, the approximate value of f (2.01) is 28.21.
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# Derivative of e^x^x^2?
derivative of e^x^x^2??
can someone explain to me how this could be solved?
so far i have:
f(x)=e^x^x^2
lnf(x)= x^2lne^x)
(e and ln cancel?)
f'(x)/f(x)= (x^2)x
f'(x)= f(x) x^3
= (e^x^x^2)(x^3)??
is that right? or do i need to use the power rule or something?
$$y=e^{x^{x^2}}$$
$$lny=x^{x^2}$$
use the fact that if $$Y=u^v$$
then $$\frac{1}{Y}\frac{dY}{dx}=\frac{v}{u}\frac{du}{dx}+\frac{dv}{dx}lnu$$
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https://www.allmath.com/laplace-transform-calculator.php
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Laplace Transform Calculator
To use Laplace Transform calculator, enter the function, and click calculate button
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Laplace Transform Calculator
Laplace transform calculator is used to perform mathematical operations related to Laplace transforms. This Laplace calculator will transform the function in a fraction of a second.
What is Laplace Transform?
Laplace transformation is a technique that allows us to transform a function into a new shape where we can understand and solve that problem easily. It maps a real-valued function into a function of a complex variable. It is very useful to solve differential equations. It is mostly used to transform the analysis of the data values.
Formula of Laplace Transform
The formula of the Laplace transformation is if the domain is given as.
L{f(s)} = ∫0∞ [f(t)e-st ] dt Where f(t) is the function of time. f(s) is the function of frequency. ∫0∞ is the notation of improper integral.
Example of Laplace Transform
Find the Laplace transform of the function f(t)= 3t- 3t+ 3
Solution:
Step 1: Write the function in the Laplace notation
L(3t- 3t+ 3)
Step 2: Apply the transformation separately on each term, we have
= L(3t6) - L(3t2) + L(3)
Step 3: Now take the constant outside form the transformation.
= 3L(t6) - 3L(t2) + 3L(1)
Step 4: Now using the formula (tn)= n!/sn+1
L(t6) =6! /s6+1
L(t6) =720 /s7
L(t2) =2! /s2+1
L(t2) =2 /s3
L(1) = 0! / s0+1
L(1) =1/ s
Step 5: Now put the values in the formula
= 3L(t6) - 3L(t2) + 3L(1)
= 3(720 /s7) - 3(2 /s3) + 3(1/s)
= 2160/s7- 6/s3+ 3/s
Hence the Laplace transformation is 2160/s- 6/s+ 3/s
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# Prove the following lemma(dealing with polynomials)
If $p$ is a prime number and $a_0,a_1,\ldots , a_{p - 1}$ are rational numbers satisfying $$a_0 + a_1 \alpha + a_2 \alpha ^2 + \ldots + a_{ p - 1} \alpha ^{ p - 1} = 0$$ where $$\alpha = \cos(\frac{2 \pi}{p}) + i \cdot \sin(\frac{2 \pi}{p} ) = e^{\frac{2i\pi}{p}}$$ then $a_0 = a_1 = \ldots = a_{p -1}$.
I think that it is helpful to consider:
(i) $g(x) = 1+ X^1 + \ldots + X^{p - 1}$ roots(particularly, $\alpha$ and its conjugate) and the fact that it is irreducible.
(ii) $f(x) = a_0+ a_1X^1 + \ldots + a_{p - 1}X^{p - 1}$.
If $p$ is prime, $1+X+\dots+X^{p-1}$ is the $p$-th cyclotomic polynomial and it is irreducible. Actually it is the minimal polynomial of $\alpha$ and its conjugates.
What can you deduce for $f(X)$?
• Well, we know that $(x-\alpha)(x-\beta) \mid \gcd(f(X),g(X))$ since both $f$ and $g$ are zero for $X=\alpha,\beta$, where $\beta$ is the conjugate of $\alpha$ Jul 13, 2018 at 15:48
• You forget $f(X)$ is the generator of $\{g(X)\in\mathbf Q[X]\mid g(\alpha)=0\}$. Jul 13, 2018 at 15:55
• I am not sure I understand correctly. What do you mean by $f(X)$ generator of $h(X)$, where $h(\alpha)=0$? Jul 13, 2018 at 16:08
• Not of $g(X)$ – a generator of the set of $g(X)$s which vanish at $\alpha$ (this is an ideal in $\mathbf Q[X]$). Jul 13, 2018 at 16:12
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Numerical Examples - Modified Stereographic Conformal Projection
# Numerical Examples for Modified Stereographic Conformal Projection #
## SPHERE #
Given1
Radius of sphere: $R=\;\;$ unit Order of equation: $m=\;\;6$ Center: $\phi_1=\;64.0^\circ$ $\lambda_0=\;-152.0^\circ$ ConstantsA_1-A_m$$B_1-B_m See Table 33, using constants for sphere. Point: \phi=\;° \lambda=\;° Find x, y, k Using equations (26-1) through (26-3) in order,$$ \eqalign{ k' &= 2/\{1+\sin64^\circ\sin60^\circ+\cos64^\circ\cos60^\circ\cos[-150^\circ - (-152^\circ)]\}\cr &= 1.0012864 } \eqalign{ x' &= 1.0012864\cos60^\circ\sin[-150^\circ-(-152^\circ)] \cr &= 0.0174722 } \eqalign{ y' &= 1.0012864\{ \cos64^\circ\sin60^\circ - \sin64^\circ\cos60^\circ\cos[-150^\circ-(-152^\circ)] \} \cr &= -0.0695721 } $$Using equations in (26-6), with j = 2, in order,$$ \eqalign{ r &= 2\times0.0174722 \cr &= 0.0349444 } \eqalign{ s' &= 0.0174722^2 + (-0.0695721)^2 \cr &= 0.0051456 } g_0 = 0 \eqalign{ a_1 &= A_6 + B_6\,i \cr &= 0.3660976-0.2937382\,i } \eqalign{ b_1 &= A_5 + B_5\,i \cr &= 0.0636871-0.1408027\,i } \eqalign{ c_1 &= 6.0\times(0.3660976-0.2937382\,i) \cr &= 2.1965856-1.7624292\,i } \eqalign{ d_1 &= 5.0\times(0.0636871-0.1408027\,i) \cr &= 0.3184355-0.7040135\,i } \eqalign{ a_2 =& b_1 + ra_1 \cr =& (0.0636871-0.1408027\,i)+0.0349444\times(0.3660976-0.2937382\,i) \cr =& 0.0764802-0.1510672\,i } \eqalign{ b_2 =& A_4 +iB_4 - s'a_1\cr =& (-0.0153783-0.1968253\,i)-0.0051456\times(0.3660976-0.2937382\,i) \cr =& -0.0172621-0.1953139\,i } \eqalign{ c_2 =& d_1 + rc_1 \cr =& (0.3184355-0.7040135\,i)+0.0349444\times(2.1965856-1.7624292\,i) \cr =& 0.3951938-0.7656005\,i } \eqalign{ d_2 =& 4\times(A_4+iB_4) - s'c_1 \cr =& 4.0\times(-0.0153783-0.1968253\,i) - 0.0051456\times(2.1965856-1.7624292\,i) \cr =& -0.0728158-0.7782325\,i } $$Incrementing j to 3,4, and 5 for the four variables a_j,b_j,c_j, \text{ and } d_j in the same set of equations,$$ a_3 = b_2+ra_2 = -0.0145895-0.2005928\,i b_3 = A_3 +iB_3 - s'a_2 = 0.0070671+0.0055898\,i c_3 = d_2+rc_2 = -0.0590060-0.8049860\,i d_3 = 3\times(A_3+iB_3) - s'c_2 = 0.0203483+0.0183769\,i a_4 = b_3+ra_3 = 0.0065572-0.0014198\,i b_4 = A_2 +iB_2 - s'a_3 = 0.0053264-0.0030853\,i c_4 = d_3+rc_3 = 0.0182864-0.0097528\,i d_4 = 2\times(A_2+iB_2) - s'c_3 = 0.0108062-0.0040929\,i a_5 = b_4+ra_4 = 0.0055555-0.0031350\,i b_5 = A_1 +iB_1 - s'a_4 = 0.9972186+0.0000073\,i c_5 = d_4+rc_4 = 0.0114452-0.0044337\,i d_5 = 1\times(A_1+iB_1) - s'c_4 = 0.9971582+0.0000502\,i $$Incrementing j to 6 for a_i \text{ and } b_j only,$$ a_6 = b_5+ra_5 = 0.9974127-0.0001022\,i b_6 = g_0 - s'a_5 = -0.0000286+0.0000161\,i $$Using equations (26-7) through (26-9) in order, and with the relationship i^2 = -1,$$ \eqalign{ x+i\,y =& 1.0\times[(0.0174722-0.0695721\,i)(0.9974127-0.0001022\,i)+ \cr & (-0.0000286+0.0000161\,i)] \cr =& 0.0173913-0.0693777\,i } x = 0.01739129\text{ units} y = -0.06937775 \text{ units} \eqalign{ F_2+i\,F_1 =& (0.0174722-0.0695721\,i)(0.0114452-0.0044337\,i) + \cr & (0.9971582+0.0000502\,i) \cr =& 0.9970497-0.0008236\,i } \eqalign{ k &= [0.9970497^2 + (-0.0008236)^2]^{1/2}\times1.0012864 \cr &= 0.9983327 } $$### Inverse Equations # Inversing forward example: Given: R,m, \phi_1, \lambda_0, A_1-A_6,\text{ and }B_1-B_6, for forward example Point: x=\; units y=\; units Find \phi, \lambda Using the Knuth algorithm equations (26-6) with (26 -10), (26-13), and (26-8), but not in that order, the first trial x' = 0.0173913/1.0 and y' = -0.0693777/1.0 . Except for the values of x’ and y’,equations (26-6) are used in the same manner as they were in the forward example, resulting in$$ a_6 = 0.9974119-0.0001021\,i b_6 = -0.0000284+0.0000161\,i c_5 = 0.0114414-0.0044528\,i d_5 = 0.9971586+0.0000493\,i $$Using equations (26-13), (26-8), and (26-10) in order,$$ \eqalign { f(x'+i\,y') =& (0.0173913-0.0693777\,i)(0.9974119-0.0001021\,i) +\cr & (-0.0000284+0.0000161\,i) - (0.0173913-0.0693777\,i)/1.0 \cr =& -0.0000805+0.0001938\,i } \eqalign{ F_2+i\,F_1 =& (0.0173913-0.0693777\,i)(0.0114414-0.0044528\,i) + \cr & (0.9971586+0.0000493\,i) \cr =& 0.9970487-0.0008219\,i } \eqalign{ \Delta(x'+i\,y') &= -(-0.0000805+0.0001938\,i)/(0.9970487-0.0008219\,i) \cr &= 0.0000809-0.0001943\,i } $$The division in equation (26-10) uses the relationship that$$ (a + b\,i)/(c + d\,i) = (ac + bd)/(c^2 + d^2) + [(bc-ad)/(c^2 + d^2)]\,i $$Adding \Delta (x’ + iy’) \text{ to } (x’ + i y’),$$ \eqalign{ (x'+i\,y') &= (0.0173913-0.0693777\,i)+(0.0000809-0.0001943\,i) \cr &= 0.0174722-0.0695720\,i } $$Repeating the above steps with the new values of (x’, y’) until the value of |\Delta(x’+i\,y’)| is 0 to eight decimal places. Finally,$$ x' = 0.017472205 y' = -0.069572049 $$Equations (26-14) through (26-17) may be used in order,$$ \eqalign{ \rho &= [0.0174722^2 + (-0.0695720)^2]^{1/2} \cr &= 0.0717325 } \eqalign{ c &= 2\arctan(0.0717325/2) \cr &= 4.1082071^\circ } \eqalign{ \phi =& \arcsin[\cos4.1082071^\circ\sin64^\circ +(-0.0695720)\times \cr & \sin4.1082071^\circ\cos64^\circ/0.0717325] \cr =& 60.0000026^\circ } \eqalign{ \lambda =& -152^\circ+\arctan[0.0174722\sin4.1082071^\circ/(0.0717325 \cr & \cos64^\circ\cos4.1082071^\circ - (-0.0695720)\sin64^\circ\sin4.1082071^\circ)]\cr =& -149.9999987^\circ }\$
1. Some of parameters for this projection are fixed ↩︎
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# review1(lastsemester) - PHYSICS 231 Review problems for...
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PHY 231 1 PHYSICS 231 Review problems for midterm 1
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PHY 231 2 v(t)=v(0)+at x(t)=x(0)+v(0)t+0.5at 2 Cut problem up in 1s pieces After 1 s: v(1)=0+0x1=0 x(1)=0+0x1+0.5x0X1 2 =0 After 2 s: v(2)=v(1)+at=0+3x1=3 x(2)=x(1)+v(1)t+0.5at 2 =0+0x1+0.5x3x1 2 =1.5 After 3 s: v(3)=v(2)+at=3+2x1=5 x(3)=x(2)+v(2)t+0.5at 2 =1.5+3x1+0.5x2x1 2 =5.5 After 4 s: v(4)=v(3)+at=5-2x1=3 x(4)=x(3)+v(3)t+0.5at 2 =5.5+5x1+0.5x(-2)x1 2 =9.5 What is the displacement at t=4 s. Velocity (m/s) 3 3 1 1 1.5 By drawing: Derive v(t) diagram from a(t) diagram: red line x(t) is area under v(t) diagram:
PHY 231 3 Cross fast 3.30 m/s 6.50 m/s To cross fast: use picture b) V=6.5 m/s so t=x/v=40.3 s (but lands downstream) b) 3.30 m/s 3.30 m/s 6.50 m/s v? Velocity of water Velocity of boat needed to cancel motion of water Total velocity of the boat (I.e. available in still water) Velocity ‘left over’ for crossing river v 2 +3.30 2 =6.50 2 so v= (6.50 2 -3.30 2 )=5.6 m/s Time to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t) Cross straight a) To cross straight: use picture a)
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PHY 231 4 a) Acceleration in vertical direction is ALWAYS g=–9.81 m/s 2 ; TRUE b) See a) FALSE c) It is positive going up, negative going down: FALSE d) It is zero at the start and end and positive everywhere else: TRUE +
PHY 231 5 v x (0)=29.0 m/s 2.19 m x(t)=x 0 +v 0x t = 0+29t=29t v x (t)=v 0x = 29 y(t)=y 0 +v 0y t-0.5gt 2 = 2.19+0t-0.5x9.8t 2 =2.19-4.9t 2 v y (t)=v 0y -gt = 0-9.8t=-9.8t
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This is a fill-in-the-blank practice test with an answer bank.
Choose the correct answer for each from the list.
Enter only the letter of the correct answer!
A. binomial
B. trinomial
C. 2
D. 3
E. 4
F. 2x2 + 5x + 3
G. -3x2 - x - 12
H. x3 + 3x2 + 3x + 1
I. 4x2 - 2x + 1
J. 2x3 - 8x2 - 8x + 11
K. 3x3 - 3x2 + 2x + 9
L. -3x2 - x + 12
M. 9x2 + 18x + 9
N. -12x3 + 4x2
O. 4x2 - 4x + 1
P. 3x + 4 + [24/(2x - 3)]
Q. x2 + 3x + 2
R. 4x2 - 12
S. -2x3 + 7x2 - 3x + 7
T. x2 - x + 4 + [1 /(x - 1)]
U. x2 - 3x - 10
V. 3x - 5 + [27/(2x -3)]
W. 2x2 + 13x + 15
X. 4x2 - 36
Y. x2 - 2x + 4 - [9/(x + 2)]
Z. 6x3 + 13x2 - 29x + 8
Question 1:
(x + 1)(x + 2) =
Question 2:
(2x3 + 3x2 - 5x + 10) + (x3 - 6x2 + 7x - 1) =
Question 3:
(3x - 1)(2x2 + 5x - 8) =
Question 4:
(x3 - 2x2 + 5x - 3) / (x - 1) =
Question 5:
5x2 - y2 is what kind of polynomial?
Remember to enter a letter from the above list!
Question 6:
(6x3 - 8x2 + 2x) - (4x3 + 10x - 11) =
Question 7:
(2x - 1)2 =
Question 8:
(x - 5)(x + 2) =
Question 9:
15x2 - 13x + 13 has how many terms?
Remember to enter a letter from the above list!
Question 10:
(2x + 6)(2x - 6) =
Question 11:
4x2(-3x + 1) =
Question 12:
(-4x2 + x - 12) + (x2 - 2x) =
Question 13:
(x3 - 1)/(x + 2) =
Question 14:
(x + 1)(x2 + 2x + 1) =
Question 15:
(2x + 3)(x + 5) =
Question 16:
6x2 - 11x - 35x4 + 2 is a polynomial of degree
.
Remember to enter a letter from the above list!
Question 17:
(x3 + 4x2 - 2x + 5) - (3x3 - 3x2 + x - 2) =
Question 18:
(3x + 3)2 =
Question 19:
(6x2 - x + 12) / (2x - 3) =
Question 20:
For a rectangle with length of (x + 1) and width of (3 + 2x), it's area would be
square units.
Relevance
• Anonymous
1Q
2K
3Z
4 not sure
5 not sure
6J
7O
8U
9D (i think)
10X
11N
12G
13 not sure
14H
15W
16 not sure
17S
18M
19 not sure
20F
Source(s): good at math :D
• 5 years ago
It was kinda harsh.... U said no. he said ok u kept goung on about how it would never happen. Probably made him feel bad maybe even kinda embarrassed.... Im sorry but it was kinda mean. U shoulda stopped at no
• Anonymous
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# The product of the units digit, the tens digit, and the
Author Message
Manager
Joined: 10 Oct 2008
Posts: 56
The product of the units digit, the tens digit, and the [#permalink]
### Show Tags
14 Oct 2008, 15:46
The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96. What is the units digit of m?
(1) m is odd.
(2) The hundreds digit of m is 8.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown
### Show Tags
14 Oct 2008, 15:57
C
Lets say xyz is the number
x*y*z = 96
a. xyz is odd
Not Sufficient
b.x = 8
Not Sufficient
Together
8*y*z = 96
y*z = 12
z has to either 1,3,5,7,9
So, 843 is the number
Retired Moderator
Joined: 05 Jul 2006
Posts: 1731
### Show Tags
14 Oct 2008, 16:13
[quote="Jcpenny"]The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96. What is the units digit of m?
(1) m is odd.
(2) The hundreds digit of m is 8.
96 = 3*2^4 ie: = 3*4*8 or 6*4*4 or 6*2*8
if m is odd it has to end in 3 and the only possible combination is (3*4*8)..........suff
2
obviously not suff
A
Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown
### Show Tags
14 Oct 2008, 16:25
how did you guys choose A
m can be 483 or 843 ?????
Retired Moderator
Joined: 05 Jul 2006
Posts: 1731
### Show Tags
14 Oct 2008, 16:31
LiveStronger wrote:
how did you guys choose A
m can be 483 or 843 ?????
EXACTLY THE UNITS DIGIT CAN ONLY = 3...RE READ THE QUESTION
Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown
### Show Tags
14 Oct 2008, 16:43
Oops
Intern
Joined: 23 Sep 2008
Posts: 15
### Show Tags
14 Oct 2008, 23:33
Answer is A. m is odd meaning the units digit will be odd.
And after factoring 96 there is only 3 which is an odd number. Hence A is the answer.
Knowing the hundreds digit gives no clue about the units digit or ay relation.
Manager
Joined: 12 Oct 2008
Posts: 98
### Show Tags
15 Oct 2008, 12:10
could smb explain the question.... Ones like that are just killing me I do not understand what is being asked... "The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96"?????? Pls. help
Retired Moderator
Joined: 05 Jul 2006
Posts: 1731
### Show Tags
15 Oct 2008, 13:56
linau1982 wrote:
could smb explain the question.... Ones like that are just killing me I do not understand what is being asked... "The product of the units digit, the tens digit, and the hundreds digit of the positive integer m is 96"?????? Pls. help
cool down , we all started from somwhere close to where u r right now.
236 * 2 is hundredth digit, 3 is tenth digit , 6 is the units digit
ie: the question is asking the following
the product of 3 intigers < 10 is 96
u need to faactorise 96 into 3 intigers less than 10.
...and so forth
--== Message from GMAT Club Team ==--
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Re: digit &nbs [#permalink] 15 Oct 2008, 13:56
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# The product of the units digit, the tens digit, and the
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# What is the sum of the total voltage drops in a series circuit?
Page Contents
## What is the sum of the total voltage drops in a series circuit?
UNDERSTANDING & CALCULATING SERIES CIRCUITS BASIC RULES The same current flows through each part of a series circuit. The total resistance of a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drops.
## Is power equal in series?
The total power in a series circuit is equal to the SUM of the power dissipated by the individual resistors.
## What are the three rules of series circuits?
In summary, a series circuit is defined as having only one path through which current can flow. From this definition, three rules of series circuits follow: all components share the same current; resistances add to equal a larger, total resistance; and voltage drops add to equal a larger, total voltage.
## How do you find the power of a series?
Power can also be calculated using either P = IV or P=V2R P = V 2 R , where V is the voltage drop across the resistor (not the full voltage of the source).
## Why is current same in a series circuit?
In a series circuit, the current is the same at each resistor. The voltage drop (Iā¢R) will be the same for each resistor since the current at and the resistance of each resistor is the same. Thus the electric potential difference across any one of the bulbs will be the same as that across any one of the other bulbs.
## What are the 3 rules of electricity?
We’ve organized these principles into three basic rules: Rule 1 ā Electricity will always want to flow from a higher voltage to a lower voltage. Rule 2 ā Electricity always has work that needs to be done. Rule 3 ā Electricity always needs a path to travel.
## How is the voltage drop in a series circuit related?
The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor.” This is what we described in the Voltage Drop section above. Voltage drop = Current times Resistor size.
## Is the current the same everywhere in a series circuit?
The current is the same everywhere in a series circuit. The total resistance is equal to the sum of the individual resistance values. The total voltage is equal to the sum of the IR voltage drops across the individual resistances. The total power is equal to the sum of the power dissipated by each resistance.
## How is the total resistance of a series circuit calculated?
The total resistance of a series circuit is equal to the sum of individual resistances.” In a series circuit you will need to calculate the total resistance of the circuit in order to figure out the amperage. This is done by adding up the individual values of each component in series. In this example we have three resistors.
## What are the fundamental laws of Series circuits?
There are three fundamental relationships concerning resistance, current, and voltage for all series circuits. It is important that you learn the three fundamental laws for series circuits. Whenever individual resistances are connected in series, they have the same effect as one large combined resistance.
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# Simple probability question
• Oct 5th 2010, 01:30 AM
Stroodle
Simple probability question
Kim goes to the sports centre each evening and either works out in the gym or has a swim. She never has a swim two evenings in a row. If she has a work-out in the gym one evening, then the next evening she is twice as likely to have a swim as she is to have a work-out in the gym. On a particular Monday evening, she works out in the gym.
What is the probability that she works out in the gym on both the Tuesday and Wednesday evenings of that week?
I can see that Pr(swim|swim)=0, and that Pr(gym|swim)=1. Just can't get my head around how to get Pr(gym|gym) and Pr(swim|gym).
• Oct 5th 2010, 02:12 AM
downthesun01
Since,"if she has a work-out in the gym one evening, then the next evening she is twice as likely to have a swim as she is to have a work-out in the gym."
I would say that:
pr(gym|gym)=1/3
pr(swim|gym)=2/3
Since the sum of the two probabilities must equal 1 because it's stated that she does some sort of work out every evening.
Someone please correct me if I'm wrong.
• Oct 5th 2010, 02:19 AM
Stroodle
Ahh, of course. That's the same answer as in my text :)
Thanks heaps :)
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# Physics - Equilibrium: How would I do this question?A mass m hangs by a cable that is attached to an L shaped rigid bar by a pin at point C. The cable passes over a frictionless pulley as shown....
Physics - Equilibrium: How would I do this question?
A mass m hangs by a cable that is attached to an L shaped rigid bar by a pin at point C. The cable passes over a frictionless pulley as shown. The bar is massless and loaded and supported as shown. Determine the reaction force (magnitude and direction) at A and the mass m such that the bar is in static equilibrium.
Image: http://tinypic.com/r/70anhk/5
valentin68 | College Teacher | (Level 3) Associate Educator
Posted on
The figure is below.
For static equilibrium the sum of force moments with respect to point A, and the sum of forces on the horizontal and vertical axis need to be zero.
Sum of moments with respect to point A:
`F_c*cos(30)*12 +10*2 -100 -5*5 =0`
`F_c =(100+25-20)/(12*cos(30)) =10.1 N`
Sum of forces on x axis:
`F_C*sin(30) +5 -F_(xA) =0`
`F_(xA) =5+10.1*sin(30) =10.05 N`
Sum of forces on y axis:
`F_c*cos(30) +10-F_(yA) =0`
`F_(yA) =10+10.1*cos(30) =18.75 N`
`|F_A| =|10.05*hati +18.75*hatj| =sqrt(10.05^2 +18.75^2) =21.27 N`
To determine the value of the mass m, see the insert figure in the upper left corner.
`G` decomposes on to horizontal and vertical components along the pulley direction (45 degree):
`G_p =G*sin(45)` and `G_n=G*cos(alpha)`
The component `G_n` is canceled by the pulley normal reaction. The component `G_p` further decomposes along the wire at 60 degrees and perpendicular to the wire at 60 degrees (this last component is canceled by the fact that the pulley does not move along the inclined plane).
`F_C =G_p*cos(60-45) =m*g*sin(45)*cos(15) `
Therefore the value of the mass is
`m = F_C/(g*sin(45)*cos(15)) =10.1/(9.81*sin(45)*cos(15)) =1.507 kg`
Answer: the reaction at A is 21.27 N and the mass m is 1.5 kg.
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## Weekend combinatorics work problem (IV)
10 Dec, 2022 at 15:46 | Posted in Statistics & Econometrics | 10 Comments
When my daughter (who studies mathematics) and yours truly solve a combinatorics problem together it takes 12 minutes. If my daughter tries to solve the problem herself it takes her 10 minutes more than it takes when I solve it alone. How long does it take me to solve the problem?
1. Why do you get to cherry-pick away negative solutions? What if the professor starts with an answer and it takes 6 minutes to come up with a problem?
.
Why do you pretend that throwing away half of your math-model predictions is just fine?
12/(t+10) + 12/t = 1 =>
12t + 12(t+10) = t(t+10) =>
24t + 120 = t^2 + 10t =>
14t + 120 = t^2 =>
t^2 – 14t -120 = 0 =>
(t-20)(t+6) = 0
Only applicable root positive, so it takes the professor 20 (minutes) 🙂
• Lars,
.
I think your starting premise is totally conjectural. 🙂
.
I suggest the way to proceed is to speculate about who is the dominant personality, the professor or daughter.
.
Let X = time for professor to complete, therefore time (Y) for daughter to complete is Y = X +10.
If the professor is the dominant personality and totally gets his way then X = 12 and Y = 22.
.
If the daughter is the dominant personality and totally gets her way then Y = 12 and X = 2.
• Interesting that the “fudge” necessary to solve this problem is the nature of the interaction between the father and daughter. You assumed the interaction was additive, I assumed it was multiplicative (efficiency through interaction), and Henry assumed that the interaction was negative (inefficiency through interaction – or interference).
What is so essential illustratively is that the entire problem solution depends on the starting assumption. This represents the object lesson for economic modeling. The result is defined by the starting assumption that only Henry stated explicitly. You and I made our assumptions implicitly, and the assumptions could be slipped by many an unwary reader.
This is illustrative of many economic models: Choose the assumptions that get the desired result.
— John Lounsbury
• John,
.
If Lars’ specification of the problem is based on actual experience, that is, he knows it takes him 20 minutes to solve the problem and his daughter takes 30 minutes and it takes them 12 minutes co-operatively, then his starting premise models the empirical results accurately.
.
If he is speculating about his time to completion, then it is anybody’s guess as to how to proceed.
.
If the results are based on his experience then, in this case, it could be said that two heads are better than one.
.
But as we all know from experience, this is not necessarily always the case. 🙂
• Yes, I agree. My previous statement about your solution was not correct. Your assumption about the effect of a dominant personality led to proof that cooperation produced no improvement, not that the interaction effect was negative. The interactive effect was nil.
The dominant personality effect has been seen in some group case studies I have read over the years. Example: https://pubmed.ncbi.nlm.nih.gov/19159145/
3. Let me address this as a theoretical economist would.
This problem is a simple one involving solving a quadratic equation.
Assume perfect efficiency of interaction between father (Y) and daughter (Z). Then YxZ = 12.
Since Y = Z+10, then
(Z+10)xZ =12.
Z^2 + 10Z -12 = 0
Z = {-10 +/- (10^2 + 4x1x12)^(1/2)}/2
= {-10 +/- (148^(1/2)}/2
= {-10 +/- 12.2}/2
= 1.1 or -11.1
-11.1 is not physically possible.
Therefore, the father alone solves the problem in 1.1 seconds and the daughter in 11.1 seconds.
However, there are few, if any, children who can work perfectly with a parent. Thus perfect efficiency is an imperfect assumption, and frictions must be considered.
Assume frictions amount to 50% for each. Then
(Z-0.5Z)x(Y-0.5Y) = 12
(0.5Z)x(0.5Y) = 12
0.25ZxY = 12
and Y = Z+10 as before.
Substituting and expanding
0.25Zx(Z+10) = 12
0.25Z^2 + 2.5Z -12 = 0
Z^2 + 10Z – 48 = 0
Now
Z = {-10 +/- (10^2 + 4x1x48)^(1/2)}/2
= {-10 +/- (296^(1/2)}/2
= {-10 +/- 17.2}/2
= 3.6 or -13.6, which can be discarded
With frictions, the resulting times for both the daughter and the father increase.
It is left to the econometric technician to determine the observed values of the frictions.
QED
— John Lounsbury
• Correction: The father’s and the daughter’s times have been reversed in the problem statement above. Should be:
“Assume perfect efficiency of interaction between father (Z) and daughter (Y).”
Cross-checking solutions using starting equations show imperfect results due to “rounding errors”.
4. The philosopher’s daughter, a mathematician, deduced that there were numerous theoretical possibilities for x given that z =12 and y = x +10:
If z = x then x = 12
If z = y then x = 2
If z = x + y then x = 1
If z = x + w then x = 12 – w
etc.
.
The philosopher king derided the abstract “axiomatic deductivity” of his mathematician daughter.
However, he himself was utterly incapable of providing any empirical light on this matter or anything else.
Perched in his academic ivory tower high above the real world of practical decisions, he pontificated that:
– The world is dominated by “fundamental uncertainty”.
– No evidence about the past can ever give any guidance regarding the future because “real-world social systems are not governed by stable causal mechanisms or capacities”.
– Science is incapable of providing guarantees that it has found “real causes”. There is always the possibility of other vital and perhaps unobservable causes.
– Science is incapable of providing explanations in terms of “mechanisms, powers, capacities, or causes”. These exist in a “deeper reality” beyond the reach of everyday experience and science. Scientific observation and induction is merely concerned with the measurable aspects of reality.
.
Then a humble econometrician addressed the problem. He carefully observed and studied the philosopher and his daughter doing “combinatorics” both separately and together. He formulated probabilistic hypotheses. Given the paucity and poor quality of the data presented in this post he concluded that there is a 99% probability that ln(x) = 2.5 +/- 2.5, ie
2 hours > x > 1 minute.
• I apologize for leaving your cleverly constructed scenario out of my discussion. It simply did not fit in the narrow construct.
It was hilarious, nonetheless. (Barbed Bayesian)
— John
Sorry, the comment form is closed at this time.
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# Place Value Understanding numerals to hundred thousand.
## Presentation on theme: "Place Value Understanding numerals to hundred thousand."— Presentation transcript:
Place Value Understanding numerals to hundred thousand.
Thousands period, Ones period __ __ __,__ __ __ Each period has 3 places separated by a comma 457, 326 Ones period Thousand period,
Circle the ONES PERIOD 456,987 302,450 584,023 20,584 117,503 2,066 Circle the THOUSANDS PERIOD 102,456 985,639 357,560 74,006 56,220 962,475
EACH PERIOD HAS 3 PLACES Thousands Period, Ones Period HUNDREDS TENS ONES, HUNDRED THOUSDAND TEN THOUSANDS ONE THOUSANDS 1 4 7, 3 6 9
Identify the underlined place ONES TENS HUNDREDS ONE THOUSANDS TEN THOUSANDS HUNDRED THOUSANDS 125,022 358,258 908,300 562,004 652,858 525,001 45,280 400,563 63,820 905,307 450,258 360,859
Value depends on which digit is in which place. 123,456 2 in the ten thousands place is 20,000 2 groups of 10,000 is 20,000 2 X 10,000 = 20,000 Digit times place equals the value
Write the value of the underlined digit 567,930 765,087 312,076 56,008 960,520 165,089 7,810 852,063 97,250 235,411 9,005 650,759 45,326 682, 258 1,357
Tell the value of each digit 457,235 =400,00 + 50,000 + 7,000 + 200 + 30 + 5 5,637 = _______ + _______ + _______ + ______ 56,891 = ________ + ________ + ______ + _____ + _____ 250,693 = ________ + ________ + ______ + ______ + ____ + ____ 924,784 = ________ + ________ + ______ + ______ + ____ + ____ 506,982 = ________ + ________ + _______ + ____ + _____ + _____
628, 340 Read the number in the largest period and tell the name of the period. Six hundred twenty-eight thousand, Then read the next period Three hundred forty Six hundred twenty-eight thousand, three hundred forty
Read each numeral 5,224 45,369 10,589 12,584 8,692 78,264 20,589 4,026 16,008 145,236 236,574 568,026 863,025 405,668 804,358 150,870 800,638 801,500 869,400 500,000 900,600 78,025 40,650 709,206 Let’s do some more!
WRITING LARGER NUMBERS IN STANDARD FORM Two hundred twelve thousand, one hundred sixteen What is in the thousand period? 212, Two hundred twelve What is in the ones period?One hundred sixteen 116 Sixteen thousand, three hundred fifty -seven ____________16,357 Thirty-one thousand, nine hundred twenty one __________ 31,921 NOTICE THE COMMA
Can you write each in standard form? 1. Sixty-four thousand, five hundred seventy-one __________64,571 2. One hundred nineteen thousand, six hundred eleven _______ 119,611 3. Eight hundred thirty-two thousand, nine hundred sixty- three _____________________ 832,963 4. Two hundred ten thousand, three hundred forty-five _______210,345 5. Four hundred twenty-seven thousand, seven hundred eighty -eight _____________________ 427,788
1. Six hundred nine thousand, ninety-four _______________ 609,094 2. Five hundred thousand, two ____________500,002 3. Fifty-six thousand, three hundred _____________56,300 4. Seven hundred thousand ____________700,000 5. Twenty- four thousand, one hundred eight ___________ 24,108
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### Problem 1: Multiples of 3 and 5
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
### Problem 2: Even Fibonacci numbers
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
### Problem 3: Largest prime factor
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
### Problem 4: Largest palindrome product
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
### Problem 5: Smallest multiple
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
### Problem 6: Sum square difference
The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
### Problem 7: 10001st prime
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
### Problem 8: Largest product in a series
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
### Problem 9: Special Pythagorean triplet
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
### Problem 10: Summation of primes
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
### Problem 11: Largest product in a grid
In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
### Problem 12: Highly divisible triangular number
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
### Problem 13: Large sum
Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690
### Problem 14: Longest Collatz sequence
The following iterative sequence is defined for the set of positive integers:
nn/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
### Problem 15: Lattice paths
Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.
How many such routes are there through a 20×20 grid?
### Problem 16: Power digit sum
215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 21000?
### Problem 17: Number letter counts
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
### Problem 18: Maximum path sum I
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
### Problem 19: Counting Sundays
You are given the following information, but you may prefer to do some research for yourself.
• 1 Jan 1900 was a Monday.
• Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
### Problem 20: Factorial digit sum
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
### Problem 21: Amicable numbers
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where ab, then a and b are an amicable pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
### Problem 22: Names scores
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.
What is the total of all the name scores in the file?
### Problem 23: Non-abundant sums
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
### Problem 24: Lexicographic permutations
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
### Problem 25: 1000-digit Fibonacci number
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the first term in the Fibonacci sequence to contain 1000 digits?
### Problem 26: Reciprocal cycles
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5 1/3 = 0.(3) 1/4 = 0.25 1/5 = 0.2 1/6 = 0.1(6) 1/7 = 0.(142857) 1/8 = 0.125 1/9 = 0.(1) 1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
Euler discovered the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² − 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, −79 and 1601, is −126479.
n² + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |−4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
### Problem 28: Number spiral diagonals
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
### Problem 29: Distinct powers
Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
### Problem 30: Digit fifth powers
Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
1634 = 14 + 64 + 34 + 44
8208 = 84 + 24 + 04 + 84
9474 = 94 + 44 + 74 + 44
As 1 = 14 is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
### Problem 31: Coin sums
In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
It is possible to make £2 in the following way:
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
How many different ways can £2 be made using any number of coins?
### Problem 32: Pandigital products
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
### Problem 33: Digit canceling fractions
The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.
If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
### Problem 34: Digit factorials
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
### Problem 35: Circular primes
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
### Problem 36: Double-base palindromes
The decimal number, 585 = 10010010012 (binary), is palindromic in both bases.
Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
(Please note that the palindromic number, in either base, may not include leading zeros.)
### Problem 37: Truncatable primes
The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
### Problem 38: Pandigital multiples
Take the number 192 and multiply it by each of 1, 2, and 3:
192 × 1 = 192
192 × 2 = 384
192 × 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
### Problem 39: Integer right triangles
If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p ≤ 1000, is the number of solutions maximised?
### Problem 40: Champernowne's constant
An irrational decimal fraction is created by concatenating the positive integers:
0.123456789101112131415161718192021...
It can be seen that the 12th digit of the fractional part is 1.
If dn represents the nth digit of the fractional part, find the value of the following expression.
d1 × d10 × d100 × d1000 × d10000 × d100000 × d1000000
### Problem 41: Pandigital prime
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
What is the largest n-digit pandigital prime that exists?
### Problem 42: Coded triangle numbers
The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?
### Problem 43: Sub-string divisibility
The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
• d2d3d4=406 is divisible by 2
• d3d4d5=063 is divisible by 3
• d4d5d6=635 is divisible by 5
• d5d6d7=357 is divisible by 7
• d6d7d8=572 is divisible by 11
• d7d8d9=728 is divisible by 13
• d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
### Problem 44: Pentagon numbers
Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.
Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
### Problem 45: Triangular, pentagonal, and hexagonal
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal Hn=n(2n−1) 1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
### Problem 46: Goldbach's other conjecture
It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
9 = 7 + 2×12
15 = 7 + 2×22
21 = 3 + 2×32
25 = 7 + 2×32
27 = 19 + 2×22
33 = 31 + 2×12
It turns out that the conjecture was false.
What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
### Problem 47: Distinct primes factors
The first two consecutive numbers to have two distinct prime factors are:
14 = 2 × 7
15 = 3 × 5
The first three consecutive numbers to have three distinct prime factors are:
644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19.
Find the first four consecutive integers to have four distinct prime factors. What is the first of these numbers?
### Problem 48: Self powers
The series, 11 + 22 + 33 + ... + 1010 = 10405071317.
Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.
### Problem 49: Prime permutations
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
What 12-digit number do you form by concatenating the three terms in this sequence?
### Problem 50: Consecutive prime sum
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
### Problem 51: Prime digit replacements
By replacing the 1st digit of the 2-digit number *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.
Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.
### Problem 52: Permuted multiples
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
### Problem 53: Combinatoric selections
There are exactly ten ways of selecting three from five, 12345:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
In combinatorics, we use the notation, 5C3 = 10.
In general,
nCr = n!r!(n−r)! ,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.
It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
How many, not necessarily distinct, values of nCr, for 1 ≤ n ≤ 100, are greater than one-million?
### Problem 54: Poker hands
In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:
• High Card: Highest value card.
• One Pair: Two cards of the same value.
• Two Pairs: Two different pairs.
• Three of a Kind: Three cards of the same value.
• Straight: All cards are consecutive values.
• Flush: All cards of the same suit.
• Full House: Three of a kind and a pair.
• Four of a Kind: Four cards of the same value.
• Straight Flush: All cards are consecutive values of same suit.
• Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.
Consider the following five hands dealt to two players:
Hand Player 1 Player 2 Winner 1 5H 5C 6S 7S KDPair of Fives 2C 3S 8S 8D TDPair of Eights Player 2 2 5D 8C 9S JS ACHighest card Ace 2C 5C 7D 8S QHHighest card Queen Player 1 3 2D 9C AS AH ACThree Aces 3D 6D 7D TD QDFlush with Diamonds Player 2 4 4D 6S 9H QH QCPair of QueensHighest card Nine 3D 6D 7H QD QSPair of QueensHighest card Seven Player 1 5 2H 2D 4C 4D 4SFull HouseWith Three Fours 3C 3D 3S 9S 9DFull Housewith Three Threes Player 1
The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.
How many hands does Player 1 win?
### Problem 55: Lychrel numbers
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
Not all numbers produce palindromes so quickly. For example,
349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337
That is, 349 took three iterations to arrive at a palindrome.
Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).
Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.
How many Lychrel numbers are there below ten-thousand?
NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.
### Problem 56: Powerful digit sum
A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
Considering natural numbers of the form, ab, where a, b < 100, what is the maximum digital sum?
### Problem 57: Square root convergents
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
By expanding this for the first four iterations, we get:
1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
### Problem 58: Spiral primes
Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
### Problem 59: XOR decryption
Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.
For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both "halves", it is impossible to decrypt the message.
Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.
Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher.txt (right click and 'Save Link/Target As...'), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.
### Problem 60: Prime pair sets
The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.
Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.
### Problem 61: Cyclical figurate numbers
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:
Triangle P3,n=n(n+1)/2 1, 3, 6, 10, 15, ... Square P4,n=n2 1, 4, 9, 16, 25, ... Pentagonal P5,n=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal P6,n=n(2n−1) 1, 6, 15, 28, 45, ... Heptagonal P7,n=n(5n−3)/2 1, 7, 18, 34, 55, ... Octagonal P8,n=n(3n−2) 1, 8, 21, 40, 65, ...
The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.
1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.
Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.
### Problem 62: Cubic permutations
The cube, 41063625 (3453), can be permuted to produce two other cubes: 56623104 (3843) and 66430125 (4053). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.
Find the smallest cube for which exactly five permutations of its digits are cube.
### Problem 63: Powerful digit counts
The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.
How many n-digit positive integers exist which are also an nth power?
### Problem 64: Odd period square roots
All square roots are periodic when written as continued fractions and can be written in the form:
√N = a0 + 1 a1 + 1 a2 + 1 a3 + ...
For example, let us consider √23:
√23 = 4 + √23 — 4 = 4 + 1 = 4 + 1 1√23—4 1 + √23 – 37
If we continue we would get the following expansion:
√23 = 4 + 1 1 + 1 3 + 1 1 + 1 8 + ...
The process can be summarised as follows:
a0 = 4, 1√23—4 = √23+47 = 1 + √23—37 a1 = 1, 7√23—3 = 7(√23+3)14 = 3 + √23—32 a2 = 3, 2√23—3 = 2(√23+3)14 = 1 + √23—47 a3 = 1, 7√23—4 = 7(√23+4)7 = 8 + √23—4 a4 = 8, 1√23—4 = √23+47 = 1 + √23—37 a5 = 1, 7√23—3 = 7(√23+3)14 = 3 + √23—32 a6 = 3, 2√23—3 = 2(√23+3)14 = 1 + √23—47 a7 = 1, 7√23—4 = 7(√23+4)7 = 8 + √23—4
It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
The first ten continued fraction representations of (irrational) square roots are:
√2=[1;(2)], period=1
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√13=[3;(1,1,1,1,6)], period=5
Exactly four continued fractions, for N ≤ 13, have an odd period.
How many continued fractions for N ≤ 10000 have an odd period?
### Problem 65: Convergents of e
The square root of 2 can be written as an infinite continued fraction.
√2 = 1 + 1 2 + 1 2 + 1 2 + 1 2 + ...
The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].
It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.
1 + 1 = 3/2 2
1 + 1 = 7/5 2 + 1 2
1 + 1 = 17/12 2 + 1 2 + 1 2
1 + 1 = 41/29 2 + 1 2 + 1 2 + 1 2
Hence the sequence of the first ten convergents for √2 are:
1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
The first ten terms in the sequence of convergents for e are:
2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.
Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.
### Problem 66: Diophantine equation
Consider quadratic Diophantine equations of the form:
x2 – Dy2 = 1
For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.
It can be assumed that there are no solutions in positive integers when D is square.
By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
32 – 2×22 = 1
22 – 3×12 = 1
92 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1
Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
### Problem 67: Maximum path sum II
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.
NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether! If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
### Problem 68: Magic 5-gon ring
Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.
Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.
It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.
Total Solution Set 9 4,2,3; 5,3,1; 6,1,2 9 4,3,2; 6,2,1; 5,1,3 10 2,3,5; 4,5,1; 6,1,3 10 2,5,3; 6,3,1; 4,1,5 11 1,4,6; 3,6,2; 5,2,4 11 1,6,4; 5,4,2; 3,2,6 12 1,5,6; 2,6,4; 3,4,5 12 1,6,5; 3,5,4; 2,4,6
By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.
Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a "magic" 5-gon ring?
### Problem 69: Totient maximum
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
n Relatively Prime φ(n) n/φ(n) 2 1 1 2 3 1,2 2 1.5 4 1,3 2 2 5 1,2,3,4 4 1.25 6 1,5 2 3 7 1,2,3,4,5,6 6 1.1666... 8 1,3,5,7 4 2 9 1,2,4,5,7,8 6 1.5 10 1,3,7,9 4 2.5
It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
### Problem 70: Totient permutation
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.
Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
Find the value of n, 1 < n < 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.
### Problem 71: Ordered fractions
Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
It can be seen that 2/5 is the fraction immediately to the left of 3/7.
By listing the set of reduced proper fractions for d ≤ 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.
### Problem 72: Counting fractions
Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
It can be seen that there are 21 elements in this set.
How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?
### Problem 73: Counting fractions in a range
Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
It can be seen that there are 3 fractions between 1/3 and 1/2.
How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?
### Problem 74: Digit factorial chains
The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
1! + 4! + 5! = 1 + 24 + 120 = 145
Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:
169 → 363601 → 1454 → 169
871 → 45361 → 871
872 → 45362 → 872
It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,
69 → 363600 → 1454 → 169 → 363601 (→ 1454)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)
Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
### Problem 75: Singular integer right triangles
It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.
12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)
In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.
120 cm: (30,40,50), (20,48,52), (24,45,51)
Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
### Problem 76: Counting summations
It is possible to write five as a sum in exactly six different ways:
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
How many different ways can one hundred be written as a sum of at least two positive integers?
### Problem 77: Prime summations
It is possible to write ten as the sum of primes in exactly five different ways:
7 + 3
5 + 5
5 + 3 + 2
3 + 3 + 2 + 2
2 + 2 + 2 + 2 + 2
What is the first value which can be written as the sum of primes in over five thousand different ways?
### Problem 78: Coin partitions
Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can separated into piles in exactly seven different ways, so p(5)=7.
OOOOO OOOO O OOO OO OOO O O OO OO O OO O O O O O O O O
Find the least value of n for which p(n) is divisible by one million.
### Problem 79: Passcode derivation
A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.
The text file, keylog.txt, contains fifty successful login attempts.
Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible secret passcode of unknown length.
### Problem 80: Square root digital expansion
It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.
The square root of two is 1.41421356237309504880..., and the digital sum of the first one hundred decimal digits is 475.
For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.
### Problem 81: Path sum: two ways
In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down, is indicated in bold red and is equal to 2427.
$$\begin{pmatrix} \color{red}{131} & 673 & 234 & 103 & 18\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\ 630 & 803 & \color{red}{746} & \color{red}{422} & 111\\ 537 & 699 & 497 & \color{red}{121} & 956\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$
Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by only moving right and down.
### Problem 82: Path sum: three ways
NOTE: This problem is a more challenging version of Problem 81.
The minimal path sum in the 5 by 5 matrix below, by starting in any cell in the left column and finishing in any cell in the right column, and only moving up, down, and right, is indicated in red and bold; the sum is equal to 994.
$$\begin{pmatrix} 131 & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\ 630 & 803 & 746 & 422 & 111\\ 537 & 699 & 497 & 121 & 956\\ 805 & 732 & 524 & 37 & 331 \end{pmatrix}$$
Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the left column to the right column.
### Problem 83: Path sum: four ways
NOTE: This problem is a significantly more challenging version of Problem 81.
In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down, is indicated in bold red and is equal to 2297.
$$\begin{pmatrix} \color{red}{131} & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & \color{red}{150}\\ 630 & 803 & 746 & \color{red}{422} & \color{red}{111}\\ 537 & 699 & 497 & \color{red}{121} & 956\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$
Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by moving left, right, up, and down.
### Problem 84: Monopoly odds
In the game, Monopoly, the standard board is set up in the following way:
GO A1 CC1 A2 T1 R1 B1 CH1 B2 B3 JAIL H2 C1 T2 U1 H1 C2 CH3 C3 R4 R2 G3 D1 CC3 CC2 G2 D2 G1 D3 G2J F3 U2 F2 F1 R3 E3 E2 CH2 E1 FP
A player starts on the GO square and adds the scores on two 6-sided dice to determine the number of squares they advance in a clockwise direction. Without any further rules we would expect to visit each square with equal probability: 2.5%. However, landing on G2J (Go To Jail), CC (community chest), and CH (chance) changes this distribution.
In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail.
At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square.
• Community Chest (2/16 cards):
2. Go to JAIL
• Chance (10/16 cards):
2. Go to JAIL
3. Go to C1
4. Go to E3
5. Go to H2
6. Go to R1
7. Go to next R (railway company)
8. Go to next R
9. Go to next U (utility company)
10. Go back 3 squares.
The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square, and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction between "Just Visiting" and being sent to JAIL, and we shall also ignore the rule about requiring a double to "get out of jail", assuming that they pay to get out on their next turn.
By starting at GO and numbering the squares sequentially from 00 to 39 we can concatenate these two-digit numbers to produce strings that correspond with sets of squares.
Statistically it can be shown that the three most popular squares, in order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO (3.09%) = Square 00. So these three most popular squares can be listed with the six-digit modal string: 102400.
If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.
### Problem 85: Counting rectangles
By counting carefully it can be seen that a rectangular grid measuring 3 by 2 contains eighteen rectangles:
Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.
### Problem 86: Cuboid route
A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the opposite corner. By travelling on the surfaces of the room the shortest "straight line" distance from S to F is 10 and the path is shown on the diagram.
However, there are up to three "shortest" path candidates for any given cuboid and the shortest route doesn't always have integer length.
By considering all cuboid rooms with integer dimensions, up to a maximum size of M by M by M, there are exactly 2060 cuboids for which the shortest route has integer length when M=100, and this is the least value of M for which the number of solutions first exceeds two thousand; the number of solutions is 1975 when M=99.
Find the least value of M such that the number of solutions first exceeds one million.
### Problem 87: Prime power triples
The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:
28 = 22 + 23 + 24
33 = 32 + 23 + 24
49 = 52 + 23 + 24
47 = 22 + 33 + 24
How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?
### Problem 88: Product-sum numbers
A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1 × a2 × ... × ak.
For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.
For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.
k=2: 4 = 2 × 2 = 2 + 2
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.
In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.
What is the sum of all the minimal product-sum numbers for 2≤k≤12000?
### Problem 89: Roman numerals
The rules for writing Roman numerals allow for many ways of writing each number (see About Roman Numerals...). However, there is always a "best" way of writing a particular number.
For example, the following represent all of the legitimate ways of writing the number sixteen:
IIIIIIIIIIIIIIII
VIIIIIIIIIII
VVIIIIII
XIIIIII
VVVI
XVI
The last example being considered the most efficient, as it uses the least number of numerals.
The 11K text file, roman.txt (right click and 'Save Link/Target As...'), contains one thousand numbers written in valid, but not necessarily minimal, Roman numerals; that is, they are arranged in descending units and obey the subtractive pair rule (see About Roman Numerals... for the definitive rules for this problem).
Find the number of characters saved by writing each of these in their minimal form.
Note: You can assume that all the Roman numerals in the file contain no more than four consecutive identical units.
### Problem 90: Cube digit pairs
Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.
For example, the square number 64 could be formed:
In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49, 64, and 81.
For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.
However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.
In determining a distinct arrangement we are interested in the digits on each cube, not the order.
{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers.
How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
### Problem 91: Right triangles with integer coordinates
The points P (x1, y1) and Q (x2, y2) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ.
There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive; that is,
0 ≤ x1, y1, x2, y2 ≤ 2.
Given that 0 ≤ x1, y1, x2, y2 ≤ 50, how many right triangles can be formed?
### Problem 92: Square digit chains
A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
For example,
44 → 32 → 13 → 10 → 11
85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89
Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
How many starting numbers below ten million will arrive at 89?
### Problem 93: Arithmetic expressions
By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four arithmetic operations (+, −, *, /) and brackets/parentheses, it is possible to form different positive integer targets.
For example,
8 = (4 * (1 + 3)) / 2
14 = 4 * (3 + 1 / 2)
19 = 4 * (2 + 3) − 1
36 = 3 * 4 * (2 + 1)
Note that concatenations of the digits, like 12 + 34, are not allowed.
Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different target numbers of which 36 is the maximum, and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number.
Find the set of four distinct digits, a < b < c < d, for which the longest set of consecutive positive integers, 1 to n, can be obtained, giving your answer as a string: abcd.
### Problem 94: Almost equilateral triangles
It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the almost equilateral triangle 5-5-6 has an area of 12 square units.
We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.
Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion (1,000,000,000).
### Problem 95: Amicable chains
The proper divisors of a number are all the divisors excluding the number itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As the sum of these divisors is equal to 28, we call it a perfect number.
Interestingly the sum of the proper divisors of 220 is 284 and the sum of the proper divisors of 284 is 220, forming a chain of two numbers. For this reason, 220 and 284 are called an amicable pair.
Perhaps less well known are longer chains. For example, starting with 12496, we form a chain of five numbers:
12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)
Since this chain returns to its starting point, it is called an amicable chain.
Find the smallest member of the longest amicable chain with no element exceeding one million.
### Problem 96: Su Doku
Su Doku (Japanese meaning number place) is the name given to a popular puzzle concept. Its origin is unclear, but credit must be attributed to Leonhard Euler who invented a similar, and much more difficult, puzzle idea called Latin Squares. The objective of Su Doku puzzles, however, is to replace the blanks (or zeros) in a 9 by 9 grid in such that each row, column, and 3 by 3 box contains each of the digits 1 to 9. Below is an example of a typical starting puzzle grid and its solution grid.
0 0 39 0 00 0 1 0 2 03 0 58 0 6 6 0 00 0 14 0 0 0 0 87 0 00 0 6 1 0 20 0 07 0 8 9 0 00 0 82 0 0 0 0 28 0 00 0 5 6 0 92 0 30 1 0 5 0 00 0 93 0 0
4 8 39 6 72 5 1 9 2 13 4 58 7 6 6 5 78 2 14 9 3 5 4 87 2 91 3 6 1 3 25 6 47 9 8 9 7 61 3 82 4 5 3 7 28 1 46 9 5 6 8 92 5 34 1 7 5 1 47 6 93 8 2
A well constructed Su Doku puzzle has a unique solution and can be solved by logic, although it may be necessary to employ "guess and test" methods in order to eliminate options (there is much contested opinion over this). The complexity of the search determines the difficulty of the puzzle; the example above is considered easy because it can be solved by straight forward direct deduction.
The 6K text file, sudoku.txt (right click and 'Save Link/Target As...'), contains fifty different Su Doku puzzles ranging in difficulty, but all with unique solutions (the first puzzle in the file is the example above).
By solving all fifty puzzles find the sum of the 3-digit numbers found in the top left corner of each solution grid; for example, 483 is the 3-digit number found in the top left corner of the solution grid above.
### Problem 97: Large non-Mersenne prime
The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 26972593−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2p−1, have been found which contain more digits.
However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×27830457+1.
Find the last ten digits of this prime number.
### Problem 98: Anagramic squares
By replacing each of the letters in the word CARE with 1, 2, 9, and 6 respectively, we form a square number: 1296 = 362. What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: 9216 = 962. We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).
What is the largest square number formed by any member of such a pair?
NOTE: All anagrams formed must be contained in the given text file.
### Problem 99: Largest exponential
Comparing two numbers written in index form like 211 and 37 is not difficult, as any calculator would confirm that 211 = 2048 < 37 = 2187.
However, confirming that 632382518061 > 519432525806 would be much more difficult, as both numbers contain over three million digits.
Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.
NOTE: The first two lines in the file represent the numbers in the example given above.
### Problem 100: Arranged probability
If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2.
The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs.
By finding the first arrangement to contain over 1012 = 1,000,000,000,000 discs in total, determine the number of blue discs that the box would contain.
### Problem 101: Optimum polynomial
If we are presented with the first k terms of a sequence it is impossible to say with certainty the value of the next term, as there are infinitely many polynomial functions that can model the sequence.
As an example, let us consider the sequence of cube numbers. This is defined by the generating function,
un = n3: 1, 8, 27, 64, 125, 216, ...
Suppose we were only given the first two terms of this sequence. Working on the principle that "simple is best" we should assume a linear relationship and predict the next term to be 15 (common difference 7). Even if we were presented with the first three terms, by the same principle of simplicity, a quadratic relationship should be assumed.
We shall define OP(k, n) to be the nth term of the optimum polynomial generating function for the first k terms of a sequence. It should be clear that OP(k, n) will accurately generate the terms of the sequence for nk, and potentially the first incorrect term (FIT) will be OP(k, k+1); in which case we shall call it a bad OP (BOP).
As a basis, if we were only given the first term of sequence, it would be most sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u1.
Hence we obtain the following OPs for the cubic sequence:
OP(1, n) = 1 1, 1, 1, 1, ... OP(2, n) = 7n−6 1, 8, 15, ... OP(3, n) = 6n2−11n+6 1, 8, 27, 58, ... OP(4, n) = n3 1, 8, 27, 64, 125, ...
Clearly no BOPs exist for k ≥ 4.
By considering the sum of FITs generated by the BOPs (indicated in red above), we obtain 1 + 15 + 58 = 74.
Consider the following tenth degree polynomial generating function:
un = 1 − n + n2n3 + n4n5 + n6n7 + n8n9 + n10
Find the sum of FITs for the BOPs.
### Problem 102: Triangle containment
Three distinct points are plotted at random on a Cartesian plane, for which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.
Consider the following two triangles:
A(-340,495), B(-153,-910), C(835,-947)
X(-175,41), Y(-421,-714), Z(574,-645)
It can be verified that triangle ABC contains the origin, whereas triangle XYZ does not.
Using triangles.txt (right click and 'Save Link/Target As...'), a 27K text file containing the co-ordinates of one thousand "random" triangles, find the number of triangles for which the interior contains the origin.
NOTE: The first two examples in the file represent the triangles in the example given above.
### Problem 103: Special subset sums: optimum
Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).
If S(A) is minimised for a given n, we shall call it an optimum special sum set. The first five optimum special sum sets are given below.
n = 1: {1}
n = 2: {1, 2}
n = 3: {2, 3, 4}
n = 4: {3, 5, 6, 7}
n = 5: {6, 9, 11, 12, 13}
It seems that for a given optimum set, A = {a1, a2, ... , an}, the next optimum set is of the form B = {b, a1+b, a2+b, ... ,an+b}, where b is the "middle" element on the previous row.
By applying this "rule" we would expect the optimum set for n = 6 to be A = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25}, with S(A) = 115 and corresponding set string: 111819202225.
Given that A is an optimum special sum set for n = 7, find its set string.
NOTE: This problem is related to Problem 105 and Problem 106.
### Problem 104: Pandigital Fibonacci ends
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order). And F2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.
Given that Fk is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
### Problem 105: Special subset sums: testing
Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).
For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159, 161, 139, 158} satisfies both rules for all possible subset pair combinations and S(A) = 1286.
Using sets.txt (right click and "Save Link/Target As..."), a 4K text file with one-hundred sets containing seven to twelve elements (the two examples given above are the first two sets in the file), identify all the special sum sets, A1, A2, ..., Ak, and find the value of S(A1) + S(A2) + ... + S(Ak).
NOTE: This problem is related to Problem 103 and Problem 106.
### Problem 106: Special subset sums: meta-testing
Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).
For this problem we shall assume that a given set contains n strictly increasing elements and it already satisfies the second rule.
Surprisingly, out of the 25 possible subset pairs that can be obtained from a set for which n = 4, only 1 of these pairs need to be tested for equality (first rule). Similarly, when n = 7, only 70 out of the 966 subset pairs need to be tested.
For n = 12, how many of the 261625 subset pairs that can be obtained need to be tested for equality?
NOTE: This problem is related to Problem 103 and Problem 105.
### Problem 107: Minimal network
The following undirected network consists of seven vertices and twelve edges with a total weight of 243.
The same network can be represented by the matrix below.
A B C D E F G A - 16 12 21 - - - B 16 - - 17 20 - - C 12 - - 28 - 31 - D 21 17 28 - 18 19 23 E - 20 - 18 - - 11 F - - 31 19 - - 27 G - - - 23 11 27 -
However, it is possible to optimise the network by removing some edges and still ensure that all points on the network remain connected. The network which achieves the maximum saving is shown below. It has a weight of 93, representing a saving of 243 − 93 = 150 from the original network.
Using network.txt (right click and 'Save Link/Target As...'), a 6K text file containing a network with forty vertices, and given in matrix form, find the maximum saving which can be achieved by removing redundant edges whilst ensuring that the network remains connected.
### Problem 108: Diophantine reciprocals I
In the following equation x, y, and n are positive integers.
1x + 1y = 1n
For n = 4 there are exactly three distinct solutions:
15 + 120 = 14 16 + 112 = 14 18 + 18 = 14
What is the least value of n for which the number of distinct solutions exceeds one-thousand?
NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.
### Problem 109: Darts
In the game of darts a player throws three darts at a target board which is split into twenty equal sized sections numbered one to twenty.
The score of a dart is determined by the number of the region that the dart lands in. A dart landing outside the red/green outer ring scores zero. The black and cream regions inside this ring represent single scores. However, the red/green outer ring and middle ring score double and treble scores respectively.
At the centre of the board are two concentric circles called the bull region, or bulls-eye. The outer bull is worth 25 points and the inner bull is a double, worth 50 points.
There are many variations of rules but in the most popular game the players will begin with a score 301 or 501 and the first player to reduce their running total to zero is a winner. However, it is normal to play a "doubles out" system, which means that the player must land a double (including the double bulls-eye at the centre of the board) on their final dart to win; any other dart that would reduce their running total to one or lower means the score for that set of three darts is "bust".
When a player is able to finish on their current score it is called a "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s and double bull).
There are exactly eleven distinct ways to checkout on a score of 6:
D3 D1 D2 S2 D2 D2 D1 S4 D1 S1 S1 D2 S1 T1 D1 S1 S3 D1 D1 D1 D1 D1 S2 D1 S2 S2 D1
Note that D1 D2 is considered different to D2 D1 as they finish on different doubles. However, the combination S1 T1 D1 is considered the same as T1 S1 D1.
In addition we shall not include misses in considering combinations; for example, D3 is the same as 0 D3 and 0 0 D3.
Incredibly there are 42336 distinct ways of checking out in total.
How many distinct ways can a player checkout with a score less than 100?
### Problem 110: Diophantine reciprocals II
In the following equation x, y, and n are positive integers.
1x + 1y = 1n
It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n for which the total number of distinct solutions exceeds one hundred.
What is the least value of n for which the number of distinct solutions exceeds four million?
NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.
### Problem 111: Primes with runs
Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by 22, and so on. But there are nine 4-digit primes containing three ones:
1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111
We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit, N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.
So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit, there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275. It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.
In the same way we obtain the following results for 4-digit primes.
Digit, d M(4, d) N(4, d) S(4, d) 0 2 13 67061 1 3 9 22275 2 3 1 2221 3 3 12 46214 4 3 2 8888 5 3 1 5557 6 3 1 6661 7 3 9 57863 8 3 1 8887 9 3 7 48073
For d = 0 to 9, the sum of all S(4, d) is 273700.
Find the sum of all S(10, d).
### Problem 112: Bouncy numbers
Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.
Find the least number for which the proportion of bouncy numbers is exactly 99%.
### Problem 113: Non-bouncy numbers
Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million that are not bouncy and only 277032 non-bouncy numbers below 1010.
How many numbers below a googol (10100) are not bouncy?
### Problem 114: Counting block combinations I
A row measuring seven units in length has red blocks with a minimum length of three units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square. There are exactly seventeen ways of doing this.
How many ways can a row measuring fifty units in length be filled?
NOTE: Although the example above does not lend itself to the possibility, in general it is permitted to mix block sizes. For example, on a row measuring eight units in length you could use red (3), black (1), and red (4).
### Problem 115: Counting block combinations II
NOTE: This is a more difficult version of Problem 114.
A row measuring n units in length has red blocks with a minimum length of m units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.
Let the fill-count function, F(m, n), represent the number of ways that a row can be filled.
For example, F(3, 29) = 673135 and F(3, 30) = 1089155.
That is, for m = 3, it can be seen that n = 30 is the smallest value for which the fill-count function first exceeds one million.
In the same way, for m = 10, it can be verified that F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value for which the fill-count function first exceeds one million.
For m = 50, find the least value of n for which the fill-count function first exceeds one million.
### Problem 116: Red, green or blue tiles
A row of five black square tiles is to have a number of its tiles replaced with coloured oblong tiles chosen from red (length two), green (length three), or blue (length four).
If red tiles are chosen there are exactly seven ways this can be done.
If green tiles are chosen there are three ways.
And if blue tiles are chosen there are two ways.
Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of replacing the black tiles in a row measuring five units in length.
How many different ways can the black tiles in a row measuring fifty units in length be replaced if colours cannot be mixed and at least one coloured tile must be used?
NOTE: This is related to Problem 117.
### Problem 117: Red, green, and blue tiles
Using a combination of black square tiles and oblong tiles chosen from: red tiles measuring two units, green tiles measuring three units, and blue tiles measuring four units, it is possible to tile a row measuring five units in length in exactly fifteen different ways.
How many ways can a row measuring fifty units in length be tiled?
NOTE: This is related to Problem 116.
### Problem 118: Pandigital prime sets
Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different sets can be formed. Interestingly with the set {2,5,47,89,631}, all of the elements belonging to it are prime.
How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?
### Problem 119: Digit power sum
The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.
We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.
You are given that a2 = 512 and a10 = 614656.
Find a30.
### Problem 120: Square remainders
Let r be the remainder when (a−1)n + (a+1)n is divided by a2.
For example, if a = 7 and n = 3, then r = 42: 63 + 83 = 728 ≡ 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.
For 3 ≤ a ≤ 1000, find rmax.
### Problem 121: Disc game prize fund
A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.
If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.
Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.
### Problem 122: Efficient exponentiation
The most naive way of computing n15 requires fourteen multiplications:
n × n × ... × n = n15
But using a "binary" method you can compute it in six multiplications:
n × n = n2
n2 × n2 = n4
n4 × n4 = n8
n8 × n4 = n12
n12 × n2 = n14
n14 × n = n15
However it is yet possible to compute it in only five multiplications:
n × n = n2
n2 × n = n3
n3 × n3 = n6
n6 × n6 = n12
n12 × n3 = n15
We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
For 1 ≤ k ≤ 200, find m(k).
### Problem 123: Prime square remainders
Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)n + (pn+1)n is divided by pn2.
For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
The least value of n for which the remainder first exceeds 109 is 7037.
Find the least value of n for which the remainder first exceeds 1010.
The radical of n, rad(n), is the product of the distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.
If we calculate rad(n) for 1n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:
Unsorted Sorted n rad(n) n rad(n) k 1 1 1 1 1 2 2 2 2 2 3 3 4 2 3 4 2 8 2 4 5 5 3 3 5 6 6 9 3 6 7 7 5 5 7 8 2 6 6 8 9 3 7 7 9 10 10 10 10 10
Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 and E(6) = 9.
If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).
### Problem 125: Palindromic sums
The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 62 + 72 + 82 + 92 + 102 + 112 + 122.
There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 02 + 12 has not been included as this problem is concerned with the squares of positive integers.
Find the sum of all the numbers less than 108 that are both palindromic and can be written as the sum of consecutive squares.
### Problem 126: Cuboid layers
The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.
If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.
However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.
We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8.
It turns out that 154 is the least value of n for which C(n) = 10.
Find the least value of n for which C(n) = 1000.
### Problem 127: abc-hits
The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.
We shall define the triplet of positive integers (a, b, c) to be an abc-hit if:
1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
2. a < b
3. a + b = c
For example, (5, 27, 32) is an abc-hit, because:
1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
2. 5 < 27
3. 5 + 27 = 32
4. rad(4320) = 30 < 32
It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for c < 1000, with ∑c = 12523.
Find ∑c for c < 120000.
### Problem 128: Hexagonal tile differences
A hexagonal tile with number 1 is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an anti-clockwise direction.
New rings are added in the same fashion, with the next rings being numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows the first three rings.
By finding the difference between tile n and each its six neighbours we shall define PD(n) to be the number of those differences which are prime.
For example, working clockwise around tile 8 the differences are 12, 29, 11, 6, 1, and 13. So PD(8) = 3.
In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and 10, hence PD(17) = 2.
It can be shown that the maximum value of PD(n) is 3.
If all of the tiles for which PD(n) = 3 are listed in ascending order to form a sequence, the 10th tile would be 271.
Find the 2000th tile in this sequence.
### Problem 129: Repunit divisibility
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
The least value of n for which A(n) first exceeds ten is 17.
Find the least value of n for which A(n) first exceeds one-million.
### Problem 130: Composites with prime repunit property
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
Find the sum of the first twenty-five composite values of n for which
GCD(n, 10) = 1 and n − 1 is divisible by A(n).
### Problem 131: Prime cube partnership
There are some prime values, p, for which there exists a positive integer, n, such that the expression n3 + n2p is a perfect cube.
For example, when p = 19, 83 + 82×19 = 123.
What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.
How many primes below one million have this remarkable property?
### Problem 132: Large repunit factors
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k.
For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these prime factors is 9414.
Find the sum of the first forty prime factors of R(109).
### Problem 133: Repunit nonfactors
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
Let us consider repunits of the form R(10n).
Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).
Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).
### Problem 134: Prime pair connection
Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p1 whilst also being divisible by p2.
In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive primes, p2 > p1, there exist values of n for which the last digits are formed by p1 and n is divisible by p2. Let S be the smallest of these values of n.
Find ∑ S for every pair of consecutive primes with 5 ≤ p1 ≤ 1000000.
### Problem 135: Same differences
Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n, for which the equation, x2y2z2 = n, has exactly two solutions is n = 27:
342 − 272 − 202 = 122 − 92 − 62 = 27
It turns out that n = 1155 is the least value which has exactly ten solutions.
How many values of n less than one million have exactly ten distinct solutions?
### Problem 136: Singleton difference
The positive integers, x, y, and z, are consecutive terms of an arithmetic progression. Given that n is a positive integer, the equation, x2y2z2 = n, has exactly one solution when n = 20:
132 − 102 − 72 = 20
In fact there are twenty-five values of n below one hundred for which the equation has a unique solution.
How many values of n less than fifty million have exactly one solution?
### Problem 137: Fibonacci golden nuggets
Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + ..., where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ; that is, Fk = Fk−1 + Fk−2, F1 = 1 and F2 = 1.
For this problem we shall be interested in values of x for which AF(x) is a positive integer.
Surprisingly AF(1/2) = (1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + ... = 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ... = 2
The corresponding values of x for the first five natural numbers are shown below.
x AF(x) √2−1 1 1/2 2 (√13−2)/3 3 (√89−5)/8 4 (√34−3)/5 5
We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.
Find the 15th golden nugget.
### Problem 138: Special isosceles triangles
Consider the isosceles triangle with base length, b = 16, and legs, L = 17.
By using the Pythagorean theorem it can be seen that the height of the triangle, h = √(172 − 82) = 15, which is one less than the base length.
With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b ± 1.
Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 and b, L are positive integers.
### Problem 139: Pythagorean tiles
Let (a, b, c) represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length c.
For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 square can be tiled with twenty-five 1 by 1 squares.
However, if (5, 12, 13) triangles were used then the hole would measure 7 by 7 and these could not be used to tile the 13 by 13 square.
Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?
### Problem 140: Modified Fibonacci golden nuggets
Consider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk−1 + Gk−2, G1 = 1 and G2 = 4; that is, 1, 4, 5, 9, 14, 23, ... .
For this problem we shall be concerned with values of x for which AG(x) is a positive integer.
The corresponding values of x for the first five natural numbers are shown below.
x AG(x) (√5−1)/4 1 2/5 2 (√22−2)/6 3 (√137−5)/14 4 1/2 5
We shall call AG(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365.
Find the sum of the first thirty golden nuggets.
### Problem 141: Investigating progressive numbers, n, which are also square
A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
For example, 58 divided by 6 has quotient 9 and remainder 4. It can also be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common ratio 3/2).
We will call such numbers, n, progressive.
Some progressive numbers, such as 9 and 10404 = 1022, happen to also be perfect squares.
The sum of all progressive perfect squares below one hundred thousand is 124657.
Find the sum of all progressive perfect squares below one trillion (1012).
### Problem 142: Perfect Square Collection
Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x − y, x + z, x − z, y + z, y − z are all perfect squares.
### Problem 143: Investigating the Torricelli point of a triangle
Let ABC be a triangle with all interior angles being less than 120 degrees. Let X be any point inside the triangle and let XA = p, XC = q, and XB = r.
Fermat challenged Torricelli to find the position of X such that p + q + r was minimised.
Torricelli was able to prove that if equilateral triangles AOB, BNC and AMC are constructed on each side of triangle ABC, the circumscribed circles of AOB, BNC, and AMC will intersect at a single point, T, inside the triangle. Moreover he proved that T, called the Torricelli/Fermat point, minimises p + q + r. Even more remarkable, it can be shown that when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO also intersect at T.
If the sum is minimised and a, b, c, p, q and r are all positive integers we shall call triangle ABC a Torricelli triangle. For example, a = 399, b = 455, c = 511 is an example of a Torricelli triangle, with p + q + r = 784.
Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli triangles.
### Problem 144: Investigating multiple reflections of a laser beam
In laser physics, a "white cell" is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out.
The specific white cell we will be considering is an ellipse with the equation 4x2 + y2 = 100
The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing, allowing the light to enter and exit through the hole.
The light beam in this problem starts at the point (0.0,10.1) just outside the white cell, and the beam first impacts the mirror at (1.4,-9.6).
Each time the laser beam hits the surface of the ellipse, it follows the usual law of reflection "angle of incidence equals angle of reflection." That is, both the incident and reflected beams make the same angle with the normal line at the point of incidence.
In the figure on the left, the red line shows the first two points of contact between the laser beam and the wall of the white cell; the blue line shows the line tangent to the ellipse at the point of incidence of the first bounce.
The slope m of the tangent line at any point (x,y) of the given ellipse is: m = −4x/y
The normal line is perpendicular to this tangent line at the point of incidence.
The animation on the right shows the first 10 reflections of the beam.
How many times does the beam hit the internal surface of the white cell before exiting?
### Problem 145: How many reversible numbers are there below one-billion?
Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not allowed in either n or reverse(n).
There are 120 reversible numbers below one-thousand.
How many reversible numbers are there below one-billion (109)?
### Problem 146: Investigating a Prime Pattern
The smallest positive integer n for which the numbers n2+1, n2+3, n2+7, n2+9, n2+13, and n2+27 are consecutive primes is 10. The sum of all such integers n below one-million is 1242490.
What is the sum of all such integers n below 150 million?
### Problem 147: Rectangles in cross-hatched grids
In a 3x2 cross-hatched grid, a total of 37 different rectangles could be situated within that grid as indicated in the sketch.
There are 5 grids smaller than 3x2, vertical and horizontal dimensions being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is cross-hatched, the following number of different rectangles could be situated within those smaller grids:
1x1: 1
2x1: 4
3x1: 8
1x2: 4
2x2: 18
Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles could be situated within 3x2 and smaller grids.
How many different rectangles could be situated within 47x43 and smaller grids?
### Problem 148: Exploring Pascal's triangle
We can easily verify that none of the entries in the first seven rows of Pascal's triangle are divisible by 7:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
However, if we check the first one hundred rows, we will find that only 2361 of the 5050 entries are not divisible by 7.
Find the number of entries which are not divisible by 7 in the first one billion (109) rows of Pascal's triangle.
### Problem 149: Searching for a maximum-sum subsequence
Looking at the table below, it is easy to verify that the maximum possible sum of adjacent numbers in any direction (horizontal, vertical, diagonal or anti-diagonal) is 16 (= 8 + 7 + 1).
−2 5 3 2 9 −6 5 1 3 2 7 3 −1 8 −4 8
Now, let us repeat the search, but on a much larger scale:
First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator":
For 1 ≤ k ≤ 55, sk = [100003 − 200003k + 300007k3] (modulo 1000000) − 500000.
For 56 ≤ k ≤ 4000000, sk = [sk−24 + sk−55 + 1000000] (modulo 1000000) − 500000.
Thus, s10 = −393027 and s100 = 86613.
The terms of s are then arranged in a 2000×2000 table, using the first 2000 numbers to fill the first row (sequentially), the next 2000 numbers to fill the second row, and so on.
Finally, find the greatest sum of (any number of) adjacent entries in any direction (horizontal, vertical, diagonal or anti-diagonal).
### Problem 150: Searching a triangular array for a sub-triangle having minimum-sum
In a triangular array of positive and negative integers, we wish to find a sub-triangle such that the sum of the numbers it contains is the smallest possible.
In the example below, it can be easily verified that the marked triangle satisfies this condition having a sum of −42.
We wish to make such a triangular array with one thousand rows, so we generate 500500 pseudo-random numbers sk in the range ±219, using a type of random number generator (known as a Linear Congruential Generator) as follows:
t := 0
for k = 1 up to k = 500500:
t := (615949*t + 797807) modulo 220
sk := t−219
Thus: s1 = 273519, s2 = −153582, s3 = 450905 etc
Our triangular array is then formed using the pseudo-random numbers thus:
s1
s2 s3
s4 s5 s6
s7 s8 s9 s10
...
Sub-triangles can start at any element of the array and extend down as far as we like (taking-in the two elements directly below it from the next row, the three elements directly below from the row after that, and so on).
The "sum of a sub-triangle" is defined as the sum of all the elements it contains.
Find the smallest possible sub-triangle sum.
### Problem 151: Paper sheets of standard sizes: an expected-value problem
A printing shop runs 16 batches (jobs) every week and each batch requires a sheet of special colour-proofing paper of size A5.
Every Monday morning, the foreman opens a new envelope, containing a large sheet of the special paper with size A1.
He proceeds to cut it in half, thus getting two sheets of size A2. Then he cuts one of them in half to get two sheets of size A3 and so on until he obtains the A5-size sheet needed for the first batch of the week.
All the unused sheets are placed back in the envelope.
At the beginning of each subsequent batch, he takes from the envelope one sheet of paper at random. If it is of size A5, he uses it. If it is larger, he repeats the 'cut-in-half' procedure until he has what he needs and any remaining sheets are always placed back in the envelope.
Excluding the first and last batch of the week, find the expected number of times (during each week) that the foreman finds a single sheet of paper in the envelope.
Give your answer rounded to six decimal places using the format x.xxxxxx .
### Problem 152: Writing 1/2 as a sum of inverse squares
There are several ways to write the number 1/2 as a sum of inverse squares using distinct integers.
For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:
In fact, only using integers between 2 and 45 inclusive, there are exactly three ways to do it, the remaining two being: {2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}.
How many ways are there to write the number 1/2 as a sum of inverse squares using distinct integers between 2 and 80 inclusive?
### Problem 153: Investigating Gaussian Integers
As we all know the equation x2=-1 has no solutions for real x.
If we however introduce the imaginary number i this equation has two solutions: x=i and x=-i.
If we go a step further the equation (x-3)2=-4 has two complex solutions: x=3+2i and x=3-2i.
x=3+2i and x=3-2i are called each others' complex conjugate.
Numbers of the form a+bi are called complex numbers.
In general a+bi and abi are each other's complex conjugate.
A Gaussian Integer is a complex number a+bi such that both a and b are integers.
The regular integers are also Gaussian integers (with b=0).
To distinguish them from Gaussian integers with b ≠ 0 we call such integers "rational integers."
A Gaussian integer is called a divisor of a rational integer n if the result is also a Gaussian integer.
If for example we divide 5 by 1+2i we can simplify in the following manner:
Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i.
The result is .
So 1+2i is a divisor of 5.
Note that 1+i is not a divisor of 5 because .
Note also that if the Gaussian Integer (a+bi) is a divisor of a rational integer n, then its complex conjugate (abi) is also a divisor of n.
In fact, 5 has six divisors such that the real part is positive: {1, 1 + 2i, 1 − 2i, 2 + i, 2 − i, 5}.
The following is a table of all of the divisors for the first five positive rational integers:
n Gaussian integer divisors with positive real part Sum s(n) of these divisors 1 1 1 2 1, 1+i, 1-i, 2 5 3 1, 3 4 4 1, 1+i, 1-i, 2, 2+2i, 2-2i,4 13 5 1, 1+2i, 1-2i, 2+i, 2-i, 5 12
For divisors with positive real parts, then, we have: .
For 1 ≤ n ≤ 105, ∑ s(n)=17924657155.
What is ∑ s(n) for 1 ≤ n ≤ 108?
### Problem 154: Exploring Pascal's pyramid
A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.
Then, we calculate the number of paths leading from the apex to each position:
A path starts at the apex and progresses downwards to any of the three spheres directly below the current position.
Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).
The result is Pascal's pyramid and the numbers at each level n are the coefficients of the trinomial expansion (x + y + z)n.
How many coefficients in the expansion of (x + y + z)200000 are multiples of 1012?
### Problem 155: Counting Capacitor Circuits
An electric circuit uses exclusively identical capacitors of the same value C.
The capacitors can be connected in series or in parallel to form sub-units, which can then be connected in series or in parallel with other capacitors or other sub-units to form larger sub-units, and so on up to a final circuit.
Using this simple procedure and up to n identical capacitors, we can make circuits having a range of different total capacitances. For example, using up to n=3 capacitors of 60 F each, we can obtain the following 7 distinct total capacitance values:
If we denote by D(n) the number of distinct total capacitance values we can obtain when using up to n equal-valued capacitors and the simple procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ...
Find D(18).
Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +...,
whereas when connecting them in series, the overall capacitance is given by:
### Problem 156: Counting Digits
Starting from zero the natural numbers are written down in base 10 like this:
0 1 2 3 4 5 6 7 8 9 10 11 12....
Consider the digit d=1. After we write down each number n, we will update the number of ones that have occurred and call this number f(n,1). The first values for f(n,1), then, are as follows:
n f(n,1) 0 0 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 2 11 4 12 5
Note that f(n,1) never equals 3.
So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The next solution is n=199981.
In the same manner the function f(n,d) gives the total number of digits d that have been written down after the number n has been written.
In fact, for every digit d ≠ 0, 0 is the first solution of the equation f(n,d)=n.
Let s(d) be the sum of all the solutions for which f(n,d)=n.
You are given that s(1)=22786974071.
Find ∑ s(d) for 1 ≤ d ≤ 9.
Note: if, for some n, f(n,d)=n for more than one value of d this value of n is counted again for every value of d for which f(n,d)=n.
### Problem 157: Solving the diophantine equation 1/a+1/b= p/10n
Consider the diophantine equation 1/a+1/b= p/10n with a, b, p, n positive integers and ab.
For n=1 this equation has 20 solutions that are listed below:
1/1+1/1=20/10 1/1+1/2=15/10 1/1+1/5=12/10 1/1+1/10=11/10 1/2+1/2=10/10 1/2+1/5=7/10 1/2+1/10=6/10 1/3+1/6=5/10 1/3+1/15=4/10 1/4+1/4=5/10 1/4+1/20=3/10 1/5+1/5=4/10 1/5+1/10=3/10 1/6+1/30=2/10 1/10+1/10=2/10 1/11+1/110=1/10 1/12+1/60=1/10 1/14+1/35=1/10 1/15+1/30=1/10 1/20+1/20=1/10
How many solutions has this equation for 1 ≤ n ≤ 9?
### Problem 158: Exploring strings for which only one character comes lexicographically after its neighbour to the left
Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed.
Examples are 'abc', 'hat' and 'zyx'.
When we study these three examples we see that for 'abc' two characters come lexicographically after its neighbour to the left.
For 'hat' there is exactly one character that comes lexicographically after its neighbour to the left. For 'zyx' there are zero characters that come lexicographically after its neighbour to the left.
In all there are 10400 strings of length 3 for which exactly one character comes lexicographically after its neighbour to the left.
We now consider strings of n ≤ 26 different characters from the alphabet.
For every n, p(n) is the number of strings of length n for which exactly one character comes lexicographically after its neighbour to the left.
What is the maximum value of p(n)?
### Problem 159: Digital root sums of factorisations
A composite number can be factored many different ways. For instance, not including multiplication by one, 24 can be factored in 7 distinct ways:
24 = 2x2x2x3
24 = 2x3x4
24 = 2x2x6
24 = 4x6
24 = 3x8
24 = 2x12
24 = 24
Recall that the digital root of a number, in base 10, is found by adding together the digits of that number, and repeating that process until a number is arrived at that is less than 10. Thus the digital root of 467 is 8.
We shall call a Digital Root Sum (DRS) the sum of the digital roots of the individual factors of our number.
The chart below demonstrates all of the DRS values for 24.
FactorisationDigital Root Sum
2x2x2x3
9
2x3x4
9
2x2x6
10
4x6
10
3x8
11
2x12
5
24
6
The maximum Digital Root Sum of 24 is 11.
The function mdrs(n) gives the maximum Digital Root Sum of n. So mdrs(24)=11.
Find ∑mdrs(n) for 1 < n < 1,000,000.
### Problem 160: Factorial trailing digits
For any N, let f(N) be the last five digits before the trailing zeroes in N!.
For example,
9! = 362880 so f(9)=36288
10! = 3628800 so f(10)=36288
20! = 2432902008176640000 so f(20)=17664
Find f(1,000,000,000,000)
### Problem 161: Triominoes
A triomino is a shape consisting of three squares joined via the edges. There are two basic forms:
If all possible orientations are taken into account there are six:
Any n by m grid for which nxm is divisible by 3 can be tiled with triominoes.
If we consider tilings that can be obtained by reflection or rotation from another tiling as different there are 41 ways a 2 by 9 grid can be tiled with triominoes:
In how many ways can a 9 by 12 grid be tiled in this way by triominoes?
In the hexadecimal number system numbers are represented using 16 different digits:
0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
The hexadecimal number AF when written in the decimal number system equals 10x16+15=175.
In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 and A are all present.
Like numbers written in base ten we write hexadecimal numbers without leading zeroes.
How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits 0,1, and A present at least once?
### Problem 474: Last digits of divisors
For a positive integer n and digits d, we define F(n, d) as the number of the divisors of n whose last digits equal d.
For example, F(84, 4) = 3. Among the divisors of 84 (1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84), three of them (4, 14, 84) have the last digit 4.
We can also verify that F(12!, 12) = 11 and F(50!, 123) = 17888.
Find F(106!, 65432) modulo (1016 + 61).
### Problem 475: Music festival
12n musicians participate at a music festival. On the first day, they form 3n quartets and practice all day.
It is a disaster. At the end of the day, all musicians decide they will never again agree to play with any member of their quartet.
On the second day, they form 4n trios, each musician avoiding his previous quartet partners.
Let f(12n) be the number of ways to organize the trios amongst the 12n musicians.
You are given f(12) = 576 and f(24) mod 1 000 000 007 = 509089824.
Find f(600) mod 1 000 000 007.
### Problem 476: Circle Packing II
Let R(a, b, c) be the maximum area covered by three non-overlapping circles inside a triangle with edge lengths a, b and c.
Let S(n) be the average value of R(a, b, c) over all integer triplets (a, b, c) such that 1 ≤ a ≤ b ≤ c < a + b ≤ n
You are given S(2) = R(1, 1, 1) ≈ 0.31998, S(5) ≈ 1.25899.
Find S(1803) rounded to 5 decimal places behind the decimal point.
### Problem 477: Number Sequence Game
The number sequence game starts with a sequence S of N numbers written on a line.
Two players alternate turns. At his turn, a player must select and remove either the first or the last number remaining in the sequence.
The player score is the sum of all the numbers he has taken. Each player attempts to maximize his own sum.
If N = 4 and S = {1, 2, 10, 3}, then each player maximizes his score as follows:
• Player 1: removes the first number (1)
• Player 2: removes the last number from the remaining sequence (3)
• Player 1: removes the last number from the remaining sequence (10)
• Player 2: removes the remaining number (2)
Player 1 score is 1 + 10 = 11.
Let F(N) be the score of player 1 if both players follow the optimal strategy for the sequence S = {s1, s2, ..., sN} defined as:
• s1 = 0
• si+1 = (si2 + 45) modulo 1 000 000 007
The sequence begins with S = {0, 45, 2070, 4284945, 753524550, 478107844, 894218625, ...}.
You are given F(2) = 45, F(4) = 4284990, F(100) = 26365463243, F(104) = 2495838522951.
Find F(108).
### Problem 478: Mixtures
Let us consider mixtures of three substances: A, B and C. A mixture can be described by a ratio of the amounts of A, B, and C in it, i.e., (a : b : c). For example, a mixture described by the ratio (2 : 3 : 5) contains 20% A, 30% B and 50% C.
For the purposes of this problem, we cannot separate the individual components from a mixture. However, we can combine different amounts of different mixtures to form mixtures with new ratios.
For example, say we have three mixtures with ratios (3 : 0 : 2), (3 : 6 : 11) and (3 : 3 : 4). By mixing 10 units of the first, 20 units of the second and 30 units of the third, we get a new mixture with ratio (6 : 5 : 9), since:
(10·3/5 + 20·3/20 + 30·3/10 : 10·0/5 + 20·6/20 + 30·3/10 : 10·2/5 + 20·11/20 + 30·4/10) = (18 : 15 : 27) = (6 : 5 : 9)
However, with the same three mixtures, it is impossible to form the ratio (3 : 2 : 1), since the amount of B is always less than the amount of C.
Let n be a positive integer. Suppose that for every triple of integers (a, b, c) with 0 ≤ a, b, cn and gcd(a, b, c) = 1, we have a mixture with ratio (a : b : c). Let M(n) be the set of all such mixtures.
For example, M(2) contains the 19 mixtures with the following ratios:
{(0 : 0 : 1), (0 : 1 : 0), (0 : 1 : 1), (0 : 1 : 2), (0 : 2 : 1),
(1 : 0 : 0), (1 : 0 : 1), (1 : 0 : 2), (1 : 1 : 0), (1 : 1 : 1),
(1 : 1 : 2), (1 : 2 : 0), (1 : 2 : 1), (1 : 2 : 2), (2 : 0 : 1),
(2 : 1 : 0), (2 : 1 : 1), (2 : 1 : 2), (2 : 2 : 1)}.
Let E(n) be the number of subsets of M(n) which can produce the mixture with ratio (1 : 1 : 1), i.e., the mixture with equal parts A, B and C.
We can verify that E(1) = 103, E(2) = 520447, E(10) mod 118 = 82608406 and E(500) mod 118 = 13801403.
Find E(10 000 000) mod 118.
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When a satellite travels in a geosynchronous orbit around the Earth, it needs to travel at a certain orbiting radius and period to maintain this orbit. Because the radius and period are related, you can use physics to calculate one if you know the other.
The period of a satellite is the time it takes it to make one full orbit around an object. The period of the Earth as it travels around the sun is one year. If you know the satellite’s speed and the radius at which it orbits, you can figure out its period.
You can calculate the speed of a satellite around an object using the equation
The satellite travels around the entire circumference of the circle — which is
if r is the radius of the orbit — in the period, T. This means the orbital speed must be
giving you
If you solve this for the period of the satellite, you get
You, the intuitive physicist, may be wondering: What if you want to examine a satellite that simply stays stationary over the same place on the Earth at all times? In other words, a satellite whose period is the same as the Earth’s 24-hour period? Can you do it?
Such satellites do exist. They’re very popular for communications because they’re always orbiting in the same spot relative to the Earth; they don’t disappear over the horizon and then reappear later. They also allow for the satellite-based global positioning system, or GPS, to work.
In cases of stationary satellites, the period, T, is 24 hours, or about 86,400 seconds. Can you find the distance a stationary satellite needs be from the center of the Earth (that is, the radius) to stay stationary? Using the equation for periods, you see that
Plugging in the numbers, you get
If you take the cube root of this, you get a radius of
This is the distance the satellite needs to be from the center of the Earth. Subtracting the Earth’s radius of
you get
which converts to about 22,300 miles. This is the distance from the surface of the Earth geosynchronous satellites need to orbit. At this distance, they orbit the Earth at the same rate the Earth is turning, which means that they stay put over the same piece of real estate.
In practice, it’s very hard to get the speed just right, which is why geosynchronous satellites have either gas boosters that can be used for fine-tuning or magnetic coils that allow them to move by pushing against the Earth’s magnetic field.
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## Rolling without slip
andres13_x
Posts: 1
Joined: Fri Jan 26, 2018 2:23 pm
### Rolling without slip
Hello, I have a question about this problem, I am not sure how to approach it. This is the problem:
When a disk rolls in a surface whitout slip, the velocity of the disk's edge (where it contacts the surface) is zero with respect the surface and the friction force is less than the maximum allowable of Us.N, where Us is the coefficient of static friction and N is the normal force exerted by the surface against the disk. Determine:
A)the maximum value of the force P such that the disk rolls without slip
B) angular acceleration of the disk for this maximum value of P
DATA:
m=1.8 kg
R=20cm
Us=0.25
The force P is placed in the center of disk and it goes to the right.
My attempt:
I know that when a disk rolls without slip, the condition Vcm= w.r must be satisfied, and also I read that in that condition, the friction force must be zero (But I am not sure about that last thing).
So setting my second newton's law we have the next (Ignoring the last thing I wrote):
P-fr= m.Acm (Where Acm is the acceleration in the mass center)
Now, I use the next equation:
Torque= I.α
Where Torque= Fr. R
So I would have: Fr. R = I.α
the moment of inertia for a disk is I=1/2 MR^2
So replacing I have:
Fr. R = 1/2MR^2.α
Canceling one R in each side I have:
Fr = 1/2MR.α
I also know α= Acm/R (Due the condition of rolling without slip)
So I have
Fr =1/2 MR.Acm /R (Where again I can cancel out both R's)
Fr =1/2 M.Acm
Solving for Acm: Acm= 2Fr/M
Now replacing this thing in the second newton's law I have this:
P- Fr = M. (2Fr/M)
Canceling both M and solving por P i have this:
P=3FR
Obiously I can find Fr with Fr= N.Us, But I am not sure if all that process is okay.
And also, I don't why angular acceleration depends of P, because I think I could find it with
Fr = 1/2MR.α
I would be happy if someone could help me with that
Vimal
Posts: 2
Joined: Mon Jan 29, 2018 6:05 am
### Re: Rolling without slip
The first one you did is correct .
For the 2nd one we have to just place the value of fr. in the eqn. : P = 3fr.
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Q. 93.7( 6 Votes )
If the equation (
If the roots are equal then d = b2 – 4ac = 0
Here a = (a2 + b2), b = 2 (ac + bd), c = (c2 + d2)
D = b2 – 4ac = 0
b2 = 4ac
{– 2(ac + bd)}2 = 4{(a2 + b2) (c2 + d2)}
4(a2c2 + b2d2 + 2acbd) = 4(a2c2 + a2d2 + b2c2 + b2d2)
2acbd = a2d2 + b2c2
a2d2 + b2c2 – 2abcd = 0
(ad – bc)2 = 0
ad – bc = 0
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#### Confidence Intervals for Independent Samples t-Test (Jump to: Lecture | Video )
We use the independent samples t-test to test if two sample means are different from one another. After the t-test, confidence intervals can be constructed to estimate how large that mean difference is.
Imagine we already have this data from a previous t-test:
Construct a 95% confidence interval for the difference of these two means.
Above are the equations for the lower and upper bounds of the confidence interval.
We already know most of the variables in the equation, but what should we put for t? First, we need to calculate the degrees of freedom for both samples:
Now, we'll use the smaller degree of freedom value to look up the t value. Go to the t-table and look up the critical value for a two-tailed test, alpha = 0.05, and 29 degrees of freedom. You should find a value of 2.0452. Now, we can finish calculating the lower and upper bounds:
We are 95% confident that the mean difference between sample 1 and sample 2 is between 6.65 and 9.35.
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# In the given figure, a || gm ABCD and a rectangle ABEF are of equal area. Then,
Question:
In the given figure, a || gm ABCD and a rectangle ABEF are of equal area. Then,
(a) perimeter of ABCD = perimeter of ABEF
(b) perimeter of ABCD < perimeter of ABEF
(c) perimeter of ABCD > perimeter of ABEF
(d) perimeter of $A B C D=\frac{1}{2}$ (perimeter of $A B E F$ )
Solution:
(c) perimeter of ABCD > perimeter of ABEF
Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures.
In ||gm ABCD, we have:
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Why are residuals Normally distributed?
When they assume the $$\varepsilon_i$$ have a normal distribution, with mean $$0$$ and variance $$\sigma^2,$$ then the estimated errors $$\hat\varepsilon$$ have a normal distribution with mean $$0$$ and variance $$\operatorname{Var}(\hat\varepsilon).$$
But what is the connection? Why, when errors are normally distributed, the estimated residuals also are normally distributed?
• Because this is a statistics site and we have been around a fairly long time, you can find a great deal written about this with a site search. The key phrase "hat matrix" is especially helpful.
– whuber
Commented Jul 13, 2022 at 13:36
The connection is made through a specific procedure: namely, ordinary least squares. When you suppose a response $$Y$$ is a random variable related to other variables $$X$$ in the form
$$Y = X\beta + \varepsilon$$
where $$\varepsilon$$ is assumed to have a joint Normal distribution (plus some other assumptions that don't matter here), then the least squares solution is
$$\hat\beta = (X^\prime X)^{-}X^\prime Y = (X^\prime X)^{-}X^\prime (X\beta + \epsilon) = \beta + J\varepsilon$$
for a matrix $$J = (X^\prime X)^{-}X^\prime$$ that is computed only from $$X.$$ The residuals, or estimated errors, are the differences
$$\hat\varepsilon = Y - \hat Y = (X\beta + \varepsilon) - X\hat\beta = \varepsilon - XJ\varepsilon = (\mathbb I - H)\varepsilon$$
where $$H = XJ = X(X^\prime X)^{-}X^\prime$$ is the "hat matrix." This exhibits the residuals as linear combinations of $$\varepsilon$$ (with coefficients given by $$\mathbb I - H$$). Linear combinations of jointly Normal variables are Normal, QED.
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# Conditional Expectation Help
• Dec 8th 2010, 04:59 PM
theprestige
Conditional Expectation Help
Suppose you flip one fair coin and roll one fair six-sided die. Let X be the number showing on the die, and let Y = 1 if the coin is heads with Y = 0 if the coin is tails. Let Z = XY.
Compute E(Z):
Am I correct when I do E(Z) = E(X)E(Y) = (3.5)(1/2) = 7/4
Also, how do i do this?
Compute E(Z|X = 4)
Thanks
• Dec 9th 2010, 04:52 AM
Focus
Quote:
Originally Posted by theprestige
Suppose you flip one fair coin and roll one fair six-sided die. Let X be the number showing on the die, and let Y = 1 if the coin is heads with Y = 0 if the coin is tails. Let Z = XY.
Compute E(Z):
Am I correct when I do E(Z) = E(X)E(Y) = (3.5)(1/2) = 7/4
I am guessing X and Y are independent, in which case you are correct.
Quote:
Also, how do i do this?
Compute E(Z|X = 4)
Thanks
Given that you rolled a 4, what is the expectation of Z? There are only 2 values Y could take with equal probability, so look at what happens at each. You could look at this as $\mathbb{E}[4Y]$.
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# Calculate the average acceleration of 2 or more objects?
1. Jan 9, 2009
### mrmazw
can i take calc. the average acceleration (m/s2) of 3 objects by adding them and dividing the answer by 3 if the measuretime is more or less equal for the 3 objects? or what should i do different?
do you not have to take into account that you are adding and dividing logarithms? and can you do it with any number of objects?
2. Jan 9, 2009
### Staff: Mentor
What do you mean by logarithms? Accelerations can be averaged as you say. You will want to do a 3-dimensional average, in general, so calculate the 3 averages (one in each of the x, y and z dimensions, for example).
3. Jan 9, 2009
### mrmazw
Thank you Berkeman, I made a mistake and meant exponentials. I took several vibration measurements of persons driving cars. Due to the amount of unavoidable variables I wanted to average the x,y,z directions (separately for each of the axis of course) for each of the multiple measurements taken per person. Because acceleration means exponential growth, I thought I could not add them and divide them by the number of measurements, can I?
4. Jan 9, 2009
### Staff: Mentor
A constant acceleration results in a linearly increasing velocity, and an exponentially growing displacement. The accelerations are not exponential in themselves.
I'm not understanding what your experiment was, so it's hard for me to give you a firm answer about how to do the math. You also have to be careful about acceleratsions of the drivers and accelerations of the vehicle, and be consistent about what coordinate system you are using. You probably need to use an inertial frame of reference, which would not be the car if it is accelerating or going in a circle.
I guess if you could post a diagram or more clearly state where the sensors were and what the vehicle and driver were doing, we would have a better chance at guiding your calculations.
5. Jan 9, 2009
### mrmazw
many thanks for your assistance, it is already very helpfull. if i run into problems with the calc. I will add them to this threat later. thank you
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# Quick Answer: What Does Linear Regression Tell You?
## How do you tell if a linear model is a good fit?
In general, a model fits the data well if the differences between the observed values and the model’s predicted values are small and unbiased.
Before you look at the statistical measures for goodness-of-fit, you should check the residual plots..
## When should you use linear regression?
Linear regression is the next step up after correlation. It is used when we want to predict the value of a variable based on the value of another variable. The variable we want to predict is called the dependent variable (or sometimes, the outcome variable).
## How do you calculate simple linear regression?
The Linear Regression Equation The equation has the form Y= a + bX, where Y is the dependent variable (that’s the variable that goes on the Y axis), X is the independent variable (i.e. it is plotted on the X axis), b is the slope of the line and a is the y-intercept.
## What does linear regression predict?
Statistical researchers often use a linear relationship to predict the (average) numerical value of Y for a given value of X using a straight line (called the regression line). If you know the slope and the y-intercept of that regression line, then you can plug in a value for X and predict the average value for Y.
## How do you interpret linear regression?
What is Linear Regression? Linear regression, at it’s core, is a way of calculating the relationship between two variables. It assumes that there’s a direct correlation between the two variables, and that this relationship can be represented with a straight line.
## What is best fit line in linear regression?
Line of best fit refers to a line through a scatter plot of data points that best expresses the relationship between those points. Statisticians typically use the least squares method to arrive at the geometric equation for the line, either though manual calculations or regression analysis software.
## What is predicted value in regression?
We can use the regression line to predict values of Y given values of X. For any given value of X, we go straight up to the line, and then move horizontally to the left to find the value of Y. The predicted value of Y is called the predicted value of Y, and is denoted Y’.
## What are two major advantages for using a regression?
The two primary uses for regression in business are forecasting and optimization. In addition to helping managers predict such things as future demand for their products, regression analysis helps fine-tune manufacturing and delivery processes.
## What is a good R squared value?
Any study that attempts to predict human behavior will tend to have R-squared values less than 50%. However, if you analyze a physical process and have very good measurements, you might expect R-squared values over 90%.
## Which regression model is best?
Statistical Methods for Finding the Best Regression ModelAdjusted R-squared and Predicted R-squared: Generally, you choose the models that have higher adjusted and predicted R-squared values. … P-values for the predictors: In regression, low p-values indicate terms that are statistically significant.More items…•
## How do you know if a linear relationship is statistically significant?
If the p-value is less than the significance level (α = 0.05),Decision: Reject the null hypothesis.Conclusion: There is sufficient evidence to conclude there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero.
## How do you interpret a regression graph?
Interpreting the slope of a regression line The slope is interpreted in algebra as rise over run. If, for example, the slope is 2, you can write this as 2/1 and say that as you move along the line, as the value of the X variable increases by 1, the value of the Y variable increases by 2.
## Can linear regression be used for prediction?
You can use regression equations to make predictions. Regression equations are a crucial part of the statistical output after you fit a model. … However, you can also enter values for the independent variables into the equation to predict the mean value of the dependent variable.
## How do you determine the best multiple regression model?
When choosing a linear model, these are factors to keep in mind:Only compare linear models for the same dataset.Find a model with a high adjusted R2.Make sure this model has equally distributed residuals around zero.Make sure the errors of this model are within a small bandwidth.
## What does a regression analysis tell you?
Use regression analysis to describe the relationships between a set of independent variables and the dependent variable. Regression analysis produces a regression equation where the coefficients represent the relationship between each independent variable and the dependent variable.
## How do you know if a regression model is good?
If your regression model contains independent variables that are statistically significant, a reasonably high R-squared value makes sense. The statistical significance indicates that changes in the independent variables correlate with shifts in the dependent variable.
## How do you interpret B in linear regression?
If the beta coefficient is significant, examine the sign of the beta. If the beta coefficient is positive, the interpretation is that for every 1-unit increase in the predictor variable, the outcome variable will increase by the beta coefficient value.
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# Quick Answer: What Is The Difference Between Class Size And Class Width?
## What is a class size in statistics?
Class size is the average number of students per class, calculated by dividing the number of students enrolled by the number of classes..
## What is a class width?
Class width refers to the difference between the upper and lower boundaries of any class (category). Depending on the author, it’s also sometimes used more specifically to mean: The difference between the upper limits of two consecutive (neighboring) classes, or.
## How do you determine your class size?
In inclusive form, class limits are obtained by subtracting 0.5 from lower limitand adding 0.5 to the upper limit. Thus, class limits of 10 – 20 class interval in the inclusive form are 9.5 – 20.5. Class size: Difference between the true upper limit and true lower limit of a class interval is called the class size.
## What is the class size of 0 4?
Hence, the class size of 0 – 4 = 4, 5 – 9 = 4 and 10 – 14 = 4.
## What is a class limit?
Class limits, class boundaries, class marks. The lower class limit of a class is the smallest data value that can go into the class. … Class limits have the same accuracy as the data values; the same number of decimal places as the data values. Class boundaries. They are halfway points that separate the classes.
## What is a class interval?
Definition. The class intervals are the subsets into which the data is grouped. The width of the class intervals will be a compromise between having intervals short enough so that not all of the observations fall in the same interval, but long enough so that you do not end up with only one observation per interval.
## How do you find the mean of a class interval?
To calculate the mean of grouped data, the first step is to determine the midpoint of each interval, or class. These midpoints must then be multiplied by the frequencies of the corresponding classes. The sum of the products divided by the total number of values will be the value of the mean.
## What is the formula of class size?
Hence, ClassMark=actualupperlimit+actual lowerlimit2 and classsize=actualupperlimit – actuallowerlimit .
## How do you find the minimum class size?
The class size of any class interval is the difference between the lower and upper limit of the given class interval and the minimum class size is the smallest class size. In the given data. It can be observed that the minimum class size is 5 for the interval 45 – 50.
## What is the class size of 25 35?
Answer: Class Mark of 25 – 35 is 30. Class mark is the average of the limits of the class interval.
## Which class has the highest frequency?
(ii) The class 200-225 is having the highest frequency (which is 140).
## How do you determine a class in statistics?
Calculating Class Interval The first step is to determine how many classes you want to have. Next, you subtract the lowest value in the data set from the highest value in the data set and then you divide by the number of classes that you want to have: Example 1: Group the following raw data into ten classes.
## What is the size of the class interval?
The frequency of a class interval is the number of data values that fall in the range specified by the interval. The size of the class interval is often selected as 5, 10, 15 or 20 etc. Each class interval starts at a value that is a multiple of the size.
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# How do you solve 2/(x-1) + (x-2)/3=4/(x-1) ?
Oct 13, 2015
$x = - 1 \text{ }$ or $\text{ } x = 4$
#### Explanation:
$\frac{2}{x - 1} + \frac{x - 2}{3} = \frac{4}{x - 1}$
The first thing to do here is find the common denominator of the three fractions. This will allow you to get rid of the denominators altogether,
The least common multiple for the three expressions that serve as denominators will be $3 \cdot \left(x - 1\right)$.
This means that the first fraction must be multiplied by $1 = \frac{3}{3}$, the second $1 = \frac{x - 1}{x - 1}$, and the third by $1 = \frac{3}{3}$.
$\frac{2}{x - 1} \cdot \frac{3}{3} + \frac{x - 2}{3} \cdot \frac{x - 1}{x - 1} = \frac{4}{x - 1} \cdot \frac{3}{3}$
$\frac{6}{3 \left(x - 1\right)} + \frac{\left(x - 2\right) \left(x - 1\right)}{3 \left(x - 1\right)} = \frac{12}{3 \left(x - 1\right)}$
This is equivalent to
$6 + \left(x - 2\right) \left(x - 1\right) = 12$
Expand the parantheses, group like terms, and move all the terms on one side of the equation to get
$6 + {x}^{2} - 2 x - x + 2 - 12 = 0$
${x}^{2} - 3 x - 4 = 0$
You can factor this quadratic by writing
${x}^{2} + x - 4 x - 4 = 0$
$x \cdot \left(x + 1\right) - 4 \left(x + 1\right) = 0$
$\left(x + 1\right) \left(x - 4\right) = 0$
This means that you must have
$x + 1 = 0 \implies x = \textcolor{g r e e n}{- 1}$
or
$x - 4 = 0 \implies x = \textcolor{g r e e n}{4}$
Do a quick check to make sure that the calculations are correct
$x = - 1 \implies \frac{2}{- 1 - 1} + \frac{- 1 - 2}{3} = \frac{4}{- 1 - 1}$
$- 1 - 1 = - 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$
and
$x = 4 \implies \frac{2}{4 - 1} + \frac{4 - 2}{3} = \frac{4}{4 - 1}$
$\frac{2}{3} + \frac{2}{3} = \frac{4}{3} \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$
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0
# What is 0.6 as a fraction?
Wiki User
2011-08-04 02:00:56
Expressed as a proper fraction in its simplest form, 0.6 is equal to 3/5 or three fifths.
Wiki User
2011-08-04 02:00:56
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Wiki User
2014-12-18 20:39:56
- .06 as a fraction is - 6/100. In simplest terms it is - 3/50.
Wiki User
2013-07-18 20:14:00
0.06 would be 6/100 or simplified to 3/50
Wiki User
2013-07-30 03:02:20
-0.6 as a fraction in lowest terms is: -3/5
Micheal Daugherty
Lvl 2
2020-10-30 20:05:10
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Q: What is 0.6 as a fraction?
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# Ch005A - Appendix 5A The Term Structure of Interest Rates...
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Appendix 5A: The Term Structure of Interest Rates, Spot Rates, and Yields to Maturity 5A.1 a. The present value of any coupon bond is the present value of its coupon payments and face value. Match each cash flow with the appropriate spot rate. For the cash flow that occurs at the end of the first year, use the one-year spot rate. For the cash flow that occurs at the end of the second year, use the two-year spot rate. P = C 1 / (1+r 1 ) + (C 2 +F) / (1+r 2 ) 2 = \$60 / (1.1) + (\$60 + \$1,000) / (1.11) 2 = \$54.55 + \$860.32 = \$914.87 The price of the bond is \$914.87. b. The yield to the maturity is the discount rate, y , which sets the cash flows equal to the price of the bond. P = C 1 / (1+ y ) + (C 2 +F) / (1+ y ) 2 \$914.87 = \$60 / (1+ y ) + (\$60 + \$1,000) / (1+ y ) 2 y = .1097 = 10.97% The yield to maturity is 10.97%. 5A.2 The present value of any coupon bond is the present value of its coupon payments and face value. Match each cash flow with the appropriate spot rate. P = C 1 / (1+r 1 ) + (C 2 +F) / (1+r 2 ) 2 = \$50 / (1.10) + (\$50 + \$1,000) / (1.08) 2 = \$45.45 + \$900.21 = \$945.66 The price of the bond is \$945.66. 5A.3 Apply the forward rate formula to calculate the one-year rate over the second year. (1+r 1 ) × (1+ f 2 ) = (1+r 2 ) 2 (1.09) ×
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Probability and Statistics K-6
Craig Loewen, Faculty of Education, Lethbridge University
In January 1998 I was asked to give a workshop at a local school on the Probability and Statistics strand in the Western Canadian Protocol at the kindergarten to grade 6 levels. To do so I created the following pages, one for each objective for each grade level (there are seventeen). On each page is the objective, followed by a problem, a manipulative-type activity, and a game. Some activities require special cards, spinners etc. which were appended.
Objective 1 Predict the chance of an event happening, using the terms never, sometimes, always. Objective 2 Describe the likelihood of an outcome, using such terms as likely, unlikely, expect, probability. Objective 3 Make a prediction based on a simple probability experiment. Objective 4 Describe the likelihood of an outcome, using such terms as likely, less likely, chance. Objective 5 Conduct a probability experiment, choose an appropriate recording method, and draw conclusions from the results. Objective 6 Identify an outcome as possible, impossible, certain, uncertain.F Objective 7 Compare outcomes as equally likely, more likely, less likely. Objective 8 Design and conduct experiments to answer one's own questions. Objective 9 List all possible outcomes of an experiment involving a single event. Objective 10 Describe events, using the vocabulary of probability: best, worst; probable, improbable; always, more likely, less likely, equally likely, never. Objective 11 Conduct probability experiments, and explain the results, using the vocabulary of probability. Objective 12 Conduct probability experiments to demonstrate that results are not influenced by such factors as the age, experiences or skills of the participant. Objective 13 Distinguish between experimental and theoretical probability for single events. Objective 14 Make the connection between the number of faces for various dice, and the probability of a single event. Objective 15 Calculate theoretical probability, using numbers between 0 and 1. Objective 16 Demonstrate that different outcomes may occur when repeating the same experiment. Objective 17 Compare experimental results with theoretical results.
Go to Math Central
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## How do you find the factors of 32?
See explanation.
To look for factors of a specified number you start from the smallest possible factors.
##32## is an evan number, so ##2## is the smallest factor:
##32-:2=16##
Again ##16## is even, so ##2## is also the factor of ##16##:
##16-:2=8##
##8-:2=4##
##4-:2=2##
##2## is prime, so it only is divisible by ##1## and itself.
Finally we can write that: ##32=2xx2xx2xx2xx2=2^5##
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# You WANT me to make a paper airplane??? A lesson in calculating the speed of an object.
## Presentation on theme: "You WANT me to make a paper airplane??? A lesson in calculating the speed of an object."— Presentation transcript:
You WANT me to make a paper airplane??? A lesson in calculating the speed of an object.
What is speed? Speed is a measure of how fast an object moves.
How is speed measured? Speed is measured by a unit of distance divided by a unit of time.
How do I calculate the speed of an object? In order to calculate the speed of an object, do the following: –Measure the distance the object went. –Measure the time that the object went. –DIVIDE THE DISTANCE BY THE TIME!
Example: Distance: 20 miles Time:4 hours Speed = 20 miles( 20 / 4 = 5 ) 4 hours The speed of this object is 5 miles per hour (mph).
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2014-08-06T14:38:21-04:00
Equation of a circles is given as
(x - xc )² + ( y - yc )² = r²
center is (xc, yc) radius is r origin is (0,0)
since center is at the origin xc = 0 yc = 0
circle is x² + y² = r²
radius of the circle is the distance between the point (4, 1) and (0,0) origin.
distance is √ [ (4-0)² + (1-0)² ] = √17
equation is x² + y² = 17
2014-08-06T14:41:26-04:00
Center of the circle is at the point (0, 0).
Circle passes through (4,1), so radius is r:
a=4, b = 1
Equation of circle:
(canonical form)
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Does $E^2=(mc^2)^2+(pc)^2$ hold for light travelling in an optically dense medium?
The rest mass of photon $$m_0=0$$ and photon travels at the speed of light in vacuum. So the energy of photon in vacuum is given by $$E_{vacuum}^2=(m_0c^2)^2+(pc)^2=(pc)^2$$ $$E_{vacuum}=pc=\gamma m_0c^2$$ As photon travels at the speed of light $$\gamma= \frac{1}{\sqrt{1-c^2/c^2}}=∞$$ and $$m_0=0$$, so energy takes the form $$E_{vacuum}=∞\times0$$ which can have a finite value.
But if light travels in glass ($$μ=1.5$$) the speed of light becomes $$\frac{2c}{3}$$ so $$\gamma= \frac{1}{\sqrt{1-\frac{4c^2}{9c^2}}}=1.34$$ a finite value. In this case, energy of photon in glass becomes $$E_{glass}=pc=\gamma m_0\frac{c}{μ}c=1.34\times0\times\frac{c}{μ}\times c=0$$
Why am I getting the energy of photon in glass as zero when I apply the equation $$E^2=(m_0c^2)^2+(pc)^2$$?
• Apr 28, 2021 at 17:17
When we say, the speed of a light in a medium, we are actually referring to the average velocity of light, I'll explain myself.
Think about only one photon reaching the "dense medium", as we know from quantum electrodynamics, light interacts with charged particles, and this dense medium, is nothing but a lot of charged particles: protons, neutrons(even they interact with light because they are "made of" charged quarks), electrons...
So if this photon enters the medium, it will interact, or not, with the rest of the particles, resulting that the time it takes the photon to escape from the medium, may, or may not fulfill the relation c=L/t with L the lenght of the medium, I mean, it may interact with no particle, it may interact "a bit" or it may interact "a lot". So the velocity of light in a medium is more like the average velocity it takes a photon to escape from the medium.
Now think about neutrinos, they are massive particles, so they do not travel at speed of light, but their interaction with matter is so weak, that in some mediums, we see that they go faster than light. https://physics.stackexchange.com/questions/94138/do-neutrinos-travel-faster-than-light-in-air#:~:text=Neutrinos%20will%20travel%20faster%20than,%2Fs%20and%20n%3E1.
Also I want to remark that your calculation is wrong. For massless particles, the relation $$E = \gamma m$$ does not hold, instead you should use $$E =\frac{p}{\beta}$$ with $$\beta = v$$ (all this with units c=1, I'm sorry).
What you should finally understand is that you are using microscopic arguments, to directly solve a macroscopic problem. Understand macroscopic behaviour from microscopic behaviour is possible but complicate, and needs to use average of certain properties between a lot of complicated stuff, for example the famous $$T\propto mv^2$$ from the kinetic theory of gases that relates the temperature of a gas with the squared average velocity of its particles.
I hope I made my answer clear, if not whe can still comment on this!
• Photons do not interact "a bit" with a medium, they either get absorbed or not. If a photon interacts with a medium then it cannot "escape it" because it's been absorbed. What escapes are re emitted photons. The only way a photon could interact without getting absorbed would be by changing frequency, but that's not the case here. Apr 29, 2021 at 2:30
• @ManudeHanoi When I said "a bit" I was meaning that they interact with "a bit" of particles in the medium. Yes, they can be absorbed and reemitted but this process may change photon frequency, changing the photon, so I think this is a different problem.There are other proceses with may change photon energy, in fact when photons interacts with matter, they may lose energy transferring it tho the matter, but the process we are interested here is that in which the energy of the photons does not change, just its direction, and this is elastic scattering of photons with charged particles. Apr 29, 2021 at 7:01
• Please correct me if im wrong but even elastic scattering is done though absobtion and emission. The direction of a photon doesnt change, it gets absorbed and another one is re emitted. From raylegh scat. wiki: "The particle[of matter], therefore, becomes a small radiating dipole whose radiation we see as scattered light" Apr 29, 2021 at 15:55
• @ManudeHanoi I think we are talking about different phenomena, I'm talking about scattering between to particles, for example photon + electron --> photon + electron in which photon momentum only changes its direction, and maybe you are talking about the photon being absorbed by an atom and then reemitted? In the case I'm talking, I'm not sure if I'm correct 100% but yes, you could say that photon and electron are absorbed, but, by the vacuum and then emitted also by the vacuum. Apr 29, 2021 at 22:35
• @ManudeHanoi but when emitted by vacuum is this photon actually different? Well, quantum mechanically not really(open to discussion), but classically, we just see light entering the medium, and getting out with a different direction. Apr 29, 2021 at 23:08
from refractive index wiki
In optics, the refractive index (also known as refraction index or index of refraction) of a material is a dimensionless number that describes how fast light travels through the material. It is defined as
n = c/v ,
where c is the speed of light in vacuum and v is the phase velocity of light in the medium
Matter is mostly made of vacuum and light always goes at the speed of light so the speed of light doesnt change in an "optically dense medium", it's the phase velocity that changes. How you could experimentally distinguish the speed of light from the phase velocity in an optically dense medium, that I wish to know.
• I don't quite follow you. Does photon travel in a optically dense medium with reduced speed or does it still travel at the speed of light ? Apr 28, 2021 at 16:41
• physics.stackexchange.com/q/466 Apr 28, 2021 at 16:46
• physics.stackexchange.com/questions/1307/… Apr 28, 2021 at 16:47
• Group velocity, $v_{gr}=d\omega/d k$, also changes along with the phase velocity. Moreover, if your refractive index does not depend on light frequency, group velocity and phase velocities are the same Apr 28, 2021 at 21:02
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## Conversion formula
The conversion factor from kilometers to yards is 1093.6132983377, which means that 1 kilometer is equal to 1093.6132983377 yards:
1 km = 1093.6132983377 yd
To convert 179.6 kilometers into yards we have to multiply 179.6 by the conversion factor in order to get the length amount from kilometers to yards. We can also form a simple proportion to calculate the result:
1 km → 1093.6132983377 yd
179.6 km → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 179.6 km × 1093.6132983377 yd
L(yd) = 196412.94838145 yd
The final result is:
179.6 km → 196412.94838145 yd
We conclude that 179.6 kilometers is equivalent to 196412.94838145 yards:
179.6 kilometers = 196412.94838145 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 5.0913140311804E-6 × 179.6 kilometers.
Another way is saying that 179.6 kilometers is equal to 1 ÷ 5.0913140311804E-6 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred seventy-nine point six kilometers is approximately one hundred ninety-six thousand four hundred twelve point nine four eight yards:
179.6 km ≅ 196412.948 yd
An alternative is also that one yard is approximately zero times one hundred seventy-nine point six kilometers.
## Conversion table
### kilometers to yards chart
For quick reference purposes, below is the conversion table you can use to convert from kilometers to yards
kilometers (km) yards (yd)
180.6 kilometers 197506.562 yards
181.6 kilometers 198600.175 yards
182.6 kilometers 199693.788 yards
183.6 kilometers 200787.402 yards
184.6 kilometers 201881.015 yards
185.6 kilometers 202974.628 yards
186.6 kilometers 204068.241 yards
187.6 kilometers 205161.855 yards
188.6 kilometers 206255.468 yards
189.6 kilometers 207349.081 yards
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# oldhomework 03-solutions - howard(cah3459 – oldhomework...
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Unformatted text preview: howard (cah3459) – oldhomework 03 – Turner – (56705) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A circular arc has a uniform linear charge density of- 6 nC / m. The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 2 2 ◦ 2 . 4 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? Correct answer: 42 . 2277 N / C. Explanation: Let : λ =- 6 nC / m =- 6 × 10- 9 C / m , Δ θ = 220 ◦ , and r = 2 . 4 m . θ is defined as the angle in the counter- clockwise direction from the positive x axis as shown in the figure below. 1 1 ◦ 1 1 ◦ r # E θ First, position the arc symmetrically around the y axis, centered at the origin. By symmetry (in this rotated configuration) the field in the x direction cancels due to charge from opposites sides of the y-axis, so E x = 0 . For a continuous linear charge distribution, # E = k e dq r 2 ˆ r In polar coordinates dq = λ ( r dθ ) , where λ is the linear charge density. The positive y axis is θ = 90 ◦ , so the y component of the electric field is given by dE y = dE sin θ . Note: By symmetry, each half of the arc about the y axis contributes equally to the electric field at the origin. Hence, we may just consider the right-half of the arc (beginning on the positive y axis and extending towards the positive x axis) and multiply the answer by 2. Note: The upper angular limit θ = 90 ◦ . The lower angular limit θ = 90 ◦- 110 ◦ =- 20 ◦ , is the angle from the positive x axis to the right-hand end of the arc. E =- 2 k e λ r 90 ◦- 20 ◦ sin θ dθ ˆ =- 2 k e λ r [cos (- 20 ◦ )- cos (90 ◦ )] ˆ . Since k e λ r = (8 . 98755 × 10 9 N · m 2 / C 2 ) × (- 6 × 10- 9 C / m) (2 . 4 m) =- 22 . 4689 N / C , E =- 2 (- 22 . 4689 N / C) × [(0 . 939693)- (0)] ˆ = 42 . 2277 N / C ˆ # E = 42 . 2277 N / C . howard (cah3459) – oldhomework 03 – Turner – (56705) 2 Alternate Solution: Just solve for # E in a straight forward manner, positioning the beginning of the arc on the positive x axis (as shown in the original figure in the question). θ is still defined as the angle in the counter- clockwise direction from the positive x axis. E x =- k e λ r 220 ◦ ◦ cos θ dθ ˆ ı =- k e λ r [sin (220 ◦ )- sin(0 ◦ )] ˆ ı =- (- 22 . 4689 N / C) × [(- . 642789)- . 0] ˆ ı = [- 14 . 4427 N / C] ˆ ı, E y =- k e λ r 220 ◦ ◦ sin θ dθ ˆ =- k e λ r [cos (0 ◦ )- cos (220 ◦ )] ˆ =- (- 22 . 4689 N / C) × [1 .- (- . 766043)] ˆ = [39 . 681 N / C] ˆ , # E = E 2 x + E 2 y = (- 14 . 4427 N / C) 2 + (39 . 681 N / C) 2 1 / 2 = 42 . 2277 N / C ....
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# Ratio Formula
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## Different Types of Ratio Formulas
While studying mathematics, we come across new concepts of comparing two more quantities with the same unit. In general terms, we consider the numerical values of the quantities and compare them. By performing a ratio between the terms, we can easily understand the outcome of the comparison. For this, we have framed a ratio formula for different techniques of comparison. In this section, we will study how a ratio between two quantities is determined and how we can use this concept to design various types of ratio formulas.
### What is a Ratio?
A ratio is the comparison of two quantities of the same kind that helps to compare them easily. This is how we express a fraction where a numerator is compared to a denominator. The definition of numerator and denominator depends on the formula of a ratio used in different aspects of a mathematical calculation. For instance, a simple ratio between the profit margin and the cost price of a product is (profit : cost). When we multiply it by 100, we get the profit percentage.
In this way, the quick ratio formula is also formulated by using the concept and the terms associated with it. The formula is:
Quick Ratio = (Current assets – Inventory)/Current liabilities
This is how the rest of the ratio formulas are determined. The definition is properly studied and the relation between the terms is determined to form a ratio formula.
### How a Ratio Formula is Determined?
The first method of comparison a student learns in mathematics is by forming a ratio. He is provided with a set of information where the same physical quantity is available in two different forms. For instance, Ram has 3 mangoes and Ashok has 5 mangoes. If we want to compare the number of mangoes owned by these two kids, we can form a ratio (3 : 5). In this way, other ratio formulas are also formed. You have already learned what the quick ratio formula stands for. Let us consider a few technical examples here.
### Return on Equity Formula
This ratio formula is used by the commerce students in their advanced subjects. The return on equity formula is:
Return on Equity (ROE) = Net income/shareholder’s equity
### Net Profit Margin Formula
This formula is one of the most common formulas you have studied in previous classes. Learn the definition of the terms used in the net profit margin formula.
Net Profit Margin = (Revenue – Cost)/Revenue
### Debt Ratio Formula
The debt ratio formula is also called the debt-to-equity formula. It is used in the commerce subjects to calculate the ratio of debts to assets. The formula is
Debt Ratio = total debts/total assets
### Gearing Ratio Formula
There are different kinds of gearing ratios used in the companies to understand the current financial conditions. In this section, you will study different types of gearing ratios such as debt to equity ratio, equity ratio, debt ratio, debt to capital ratio, etc.
FAQ (Frequently Asked Questions)
1. What is a Ratio Formula?
The term ratio formula is the expression of a numerator and a denominator considering the conditions of a physical quantity. This formula is used to compare a numerator with a denominator.
2. Why Should We Use Different Kinds of Ratio Formulas?
To compare different quantities with the same unit, a ratio operation is done. It helps to understand how a quantity changes and how it affects the ratio.
3. How Can You Remember the Ratio Formulas?
You will come across different kinds of ratio formulas. Understand the meaning of the ratio and the reason behind its formulation.
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## Perfect Shuffle
### March 13, 2020
A perfect shuffle, also known as a faro shuffle, splits a deck of cards into equal halves (there must be an even number of cards), then perfectly interleaves them. Eventually a series of perfect shuffles returns a deck to its original order. For instance, with a deck of 8 cards named (1 2 3 4 5 6 7 8), the first shuffle rearranges the cards to (1 5 2 6 3 7 4 8), the second shuffle rearranges the cards to (1 3 5 7 2 4 6 8), and the third shuffle restores the original order (1 2 3 4 5 6 7 8).
Your task is to write a program that performs a perfect shuffle and use it to determine how many perfect shuffles are required to return an n-card deck to its original order; how many perfect shuffles are required for a standard 52-card deck? When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 15 Responses to “Perfect Shuffle”
1. Zack said
Cool drill. Here is my take on it using Julia: https://pastebin.com/ppSpZ5BH
For a 52 card deck, it takes just n = 8 perfect shuffles, while for a 54 card one (one which includes 2 jokers) it takes many more.
Have a nice weekend and stay healthy!
2. Paul said
In Python.
```from itertools import chain
def pshuffle(seq):
half = len(seq) // 2
return list(chain.from_iterable(zip(seq[:half], seq[half:])))
for N in (8, 24, 52, 100, 1020, 1024, 10000):
seq = orig = list(range(N))
for i in range(100000):
seq = pshuffle(seq)
if seq == orig:
print(i+1, end=", ")
break
# -> 3, 11, 8, 30, 1018, 10, 300,
```
3. chaw said
Here is a simple solution in R7RS Scheme and a few popular helpers.
It includes a formulaic method for computing the cycle-length in
``` (import (scheme base) (scheme write) (only (srfi 1) fold-right split-at cons* iota) (only (srfi 8) receive) (only (org eip10 okmij myenv) assert)) (define (perfect-out-shuffle items) (assert (even? (length items))) (receive (top bot) (split-at items (quotient (length items) 2)) (fold-right (lambda (top-item bot-item rslt) (cons* top-item bot-item rslt)) '() top bot))) (write (perfect-out-shuffle (iota 8 1))) (newline) (define (iterate-perfect-out-shuffles items limit show?) (let loop ((i 0) (shuf items)) (and (< i limit) (let ((shuf (perfect-out-shuffle shuf))) (when show? (write shuf) (newline)) (or (equal? shuf items) (loop (+ i 1) shuf)))))) (write (iterate-perfect-out-shuffles (iota 8 1) 10 #t)) (newline) (define (perfect-out-shuffle-cycle-length num-items) (assert (> num-items 1) (even? num-items)) (if (= 2 num-items) 1 (let ((m (- num-items 1))) (let loop ((k 1) (e 2)) (if (= 1 e) k (loop (+ k 1) (modulo (* 2 e) m))))))) (write (map perfect-out-shuffle-cycle-length (iota 100 2 2))) (newline) (write (perfect-out-shuffle-cycle-length #e1e10)) (newline) ```
Output:
``` (1 5 2 6 3 7 4 8) (1 5 2 6 3 7 4 8) (1 3 5 7 2 4 6 8) (1 2 3 4 5 6 7 8) #t (1 2 4 3 6 10 12 4 8 18 6 11 20 18 28 5 10 12 36 12 20 14 12 23 21 8 52 20 18 58 60 6 12 66 22 35 9 20 30 39 54 82 8 28 11 12 10 36 48 30 100 51 12 106 36 36 28 44 12 24 110 20 100 7 14 130 18 36 68 138 46 60 28 42 148 15 24 20 52 52 33 162 20 83 156 18 172 60 58 178 180 60 36 40 18 95 96 12 196 99) 54540 ```
4. BW25 said
Looking at the pattern of numbers, it seems it is sometimes possible to calculate the number of perfect shuffles with an equation instead of actually doing the perfect shuffles.
For an even number N and N-1, they will both have the same answer.
E.g. 52 and 51 both take 8 perfect shuffles
I’ve also noticed for any number that is 2^n, the number of perfect shuffles is n.
E.g. 8 = 2^3, n=3, so the number of perfect shuffles is 3.
For any number that is (2^n) + 2 or 2^n + 1, the answer is 2n.
E.g. 10 = (2^3)+2, n = 3. The answer is 2n = 6.
Also: 9 = (2^3)+1, n = 3. The answer is 2n = 6.
Unfortunately, I can’t figure out the pattern beyond that, but in these special cases, the number of perfect shuffles can be found in O(1) time.
If there is an equation to describe the entire pattern, perhaps all cases could be solved in O(1) time.
Code:
import numpy as np
import math
#In some cases, we do not need to actually perform the perfect shuffle: we can determine the answer with simple math
def checkSpecialCases(arr):
#An even number N will have the same solution as N-1
#Add 1 to odd numbers to make checks easier
N = len(arr) + len(arr)%2
``````#If a number is 2^x, the solution is x
calc = math.log2(N)
if calc % 1 == 0:
return int(calc)
#If a number is (2^x)+2, the solution is 2x
calc = math.log2(N-2)
if calc % 1 == 0:
return int(calc*2)
#If none of these cases are met, we have no choice but to do the perfect shuffle
return perfectShuffle(arr)
``````
def perfectShuffle(arr):
orig = np.copy(arr)
``````n = 0 #Count the number of perfect shuffles
while True:
#Copy array
arr2 = np.copy(arr)
#Place upper half of array into position
for i in range(math.ceil((len(arr)/2))):
arr[i*2] = arr2[i]
#Place lower half of array into position
for i in range(len(arr)//2):
arr[(i*2)+1] = arr2[math.ceil(len(arr)/2)+i]
n+=1 #Increase shuffle counter
#print(arr)
if np.array_equal(orig,arr): #If the array matches the unshuffled array, we are finished.
break
return n
``````
if name == ‘main‘:
for i in range(2,9):
arr = np.arange(1,i+1)
print(str(i) + ‘ : ‘ + str(checkSpecialCases(arr)))
Output:
2 : 1
3 : 2
4 : 2
5 : 4
6 : 4
7 : 3
8 : 3
5. [email protected] said
Quoting Programming Praxis :
Further;If the aim of shuffling is to make the card order unpredicatable or at least difficuld to predict; how many “faro-shuffles” of a 52-card stack is ideal ?
> programmingpraxis posted: “A perfect shuffle, also known as a faro > shuffle, splits a deck of cards into equal halves (there must be an > even number of cards), then perfectly interleaves them. Eventually a > series of perfect shuffles returns a deck to its original order. For > instance,” > > > T
6. Daniel said
Here’s a solution in C.
```#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void shuffle(int* array, int n) {
int* shuffled = malloc(sizeof(int) * n);
int half = n / 2;
for (int i = 0; i < half; ++i) {
shuffled[i * 2] = array[i];
}
for (int i = half; i < n; ++i) {
shuffled[(i - half) * 2 + 1] = array[i];
}
memcpy(array, shuffled, sizeof(int) * n);
free(shuffled);
}
int main(int argc, char* argv[]) {
assert(argc == 2);
int n = atoi(argv[1]);
assert(n % 2 == 0);
int* array = malloc(sizeof(int) * n);
for (int i = 0; i < n; ++i) {
array[i] = i;
}
int count = 0;
while (1) {
shuffle(array, n);
++count;
for (int i = 0; i < n; ++i) {
if (array[i] != i) goto end_outer;
}
break;
end_outer: ;
}
printf("%d\n", count);
free(array);
return EXIT_SUCCESS;
}
```
Example:
```\$ for i in `seq 2 2 20`; do ./a.out \$i; done
1
2
4
3
6
10
12
4
8
18
```
As indicated on Wikipedia, this matches sequence A002326 on OEIS.
7. Globules said
```{-# LANGUAGE LambdaCase #-}
import Data.Maybe (listToMaybe)
import GHC.Natural (Natural, powModNatural)
import Text.Printf (printf)
-- The type of faro shuffle.
data Shuffle = In -- orig. top card to 2nd, bottom card to penultimate
| Out -- orig. top card to top, bottom card to bottom
deriving (Show)
-- The number of shuffles, of type s, required to restore an n card deck to its
-- original order.
numShuffles :: Shuffle -> Natural -> Maybe Natural
numShuffles s n | odd n = Nothing
| otherwise = go (modulusFrom s n)
where modulusFrom = \case In -> succ; Out -> pred
go m = listToMaybe [e | e <- [1..], let p = powModNatural 2 e m, p == 1]
--------------------------------------------------------------------------------
test :: [(Shuffle, Natural)] -> IO ()
test sns = do
printf "Deck Size Shuffle Type Num Shuffles\n"
printf "--------- ------------ ------------\n"
forM_ sns \$ \(s, n) -> do
let shufs = maybe "-" show (numShuffles s n)
printf "%5d %7s %6s\n" n (show s) shufs
main :: IO ()
main = test [( In, 7), (Out, 7),
( In, 8), (Out, 8),
( In, 52), (Out, 52),
( In, 64), (Out, 64),
( In, 87654), (Out, 87654)]
```
```\$ ./faro
Deck Size Shuffle Type Num Shuffles
--------- ------------ ------------
7 In -
7 Out -
8 In 6
8 Out 3
52 In 52
52 Out 8
64 In 12
64 Out 6
87654 In 8556
87654 Out 6732
```
8. matthew said
I had fun with this one:
```# Cut pack into two halves, deal alternately from each
# half. An outshuffle starts from the half with the original
# top card, an inshuffle starts from the other half (so the
# top card always stays the same after an outshuffle). The
# procedure makes sense for both odd and even numbers of cards.
# The next() function describes effect of the shuffle:
# type is outshuffle = 0, inshuffle = 1
def next(i,n,type):
# A card originally at index 'i' goes to index 'next(i,n,t)'
# Need special case for last element of an even outshuffle
if n%2 == 0 and type == 0 and i == n-1: return i
if n%2 == 0 and type == 0: return (2*i) % (n-1)
if n%2 == 1 and type == 0: return (2*i) % n
if n%2 == 0 and type == 1: return (2*i+1) % (n+1)
if n%2 == 1 and type == 1: return (2*i+1) % n
assert False
print([next(i,10,0) for i in range(10)])
print([next(i,10,1) for i in range(10)])
print([next(i,11,0) for i in range(11)])
print([next(i,11,1) for i in range(11)])
# [0, 2, 4, 6, 8, 1, 3, 5, 7, 9]
# [1, 3, 5, 7, 9, 0, 2, 4, 6, 8]
# [0, 2, 4, 6, 8, 10, 1, 3, 5, 7, 9]
# [1, 3, 5, 7, 9, 0, 2, 4, 6, 8, 10]
# 'shuffle()' constructs the entire permutation, both
# as an array map and as a set of cycles.
#
# The length of the longest cycle is the number of shuffles
# needed to restore the original pack order.
def shuffle(n, type = 0):
# Go through 0..n-1, finding the next index, until a cycle is
# found. Record the cycle and move on to next unhandled index.
a = [-1]*n
cycles = []
for i in range(n):
c = []
while a[i] < 0:
# Add i to the current cycle
c += [i]
# and find the next index.
j = next(i,n,type)
a[i],i = j,j
# Collect non-empty cycles
if len(c) > 0: cycles.append(c)
return (a,cycles)
# inshuffle length 2n is like outshuffle length 2n+2 (imagine discarding both
# end cards), and both are like inshuffle and outshuffles of length 2n+1 and
# all four shuffles have similar cycle structure:
print(shuffle(20,1)[1])
print(shuffle(21,1)[1])
print(shuffle(21,0)[1])
print(shuffle(22,0)[1])
# [[0, 1, 3, 7, 15, 10], [2, 5, 11], [4, 9, 19, 18, 16, 12], [6, 13], [8, 17, 14]]
# [[0, 1, 3, 7, 15, 10], [2, 5, 11], [4, 9, 19, 18, 16, 12], [6, 13], [8, 17, 14], [20]]
# [[0], [1, 2, 4, 8, 16, 11], [3, 6, 12], [5, 10, 20, 19, 17, 13], [7, 14], [9, 18, 15]]
# [[0], [1, 2, 4, 8, 16, 11], [3, 6, 12], [5, 10, 20, 19, 17, 13], [7, 14], [9, 18, 15], [21]]
# Here, the longest cycle is 6 and this is how many shuffles are needed
# to restore the original pack order.
# The simplest form is the odd outshuffle, where index i goes to 2*i%n:
for n in range(3,22,2): print(n,[len(c) for c in shuffle(n,0)[1]])
# 3 [1, 2]
# 5 [1, 4]
# 7 [1, 3, 3]
# 9 [1, 6, 2]
# 11 [1, 10]
# 13 [1, 12]
# 15 [1, 4, 4, 2, 4]
# 17 [1, 8, 8]
# 19 [1, 18]
# 21 [1, 6, 3, 6, 2, 3]
# and we can see that the smallest k > 0 with 2**k%n == 1 will be the
# longest cycle (the cycle beginning 1 will be of this length) and in
# fact we have: 2**4%15 = 16%15 = 1; 2**6%21 == 64%21 == 1 etc.
# In general, for k outshuffles of n items, index i moves to:
def outshuffle(n,k,i):
if n%2 == 0 and i == n-1: return n-1 # Special case
m = n-1 if n%2 == 0 else n
return 2**k*i%m
# and for k inshuffles of n items, index i moves to:
def inshuffle(n,k,i):
m = n+1 if n%2 == 0 else n
return (2**k*(i+1)-1)%m
print ([inshuffle(8,1,i) for i in range(8)])
print ([inshuffle(9,1,i) for i in range(9)])
print ([outshuffle(9,1,i) for i in range(9)])
print ([outshuffle(10,1,i) for i in range(10)])
# [1, 3, 5, 7, 0, 2, 4, 6]
# [1, 3, 5, 7, 0, 2, 4, 6, 8]
# [0, 2, 4, 6, 8, 1, 3, 5, 7]
# [0, 2, 4, 6, 8, 1, 3, 5, 7, 9]
# and the longest cycle here is 6, so:
print ([inshuffle(8,6,i) for i in range(8)])
print ([inshuffle(9,6,i) for i in range(9)])
print ([outshuffle(9,6,i) for i in range(9)])
print ([outshuffle(10,6,i) for i in range(10)])
# [0, 1, 2, 3, 4, 5, 6, 7]
# [0, 1, 2, 3, 4, 5, 6, 7, 8]
# [0, 1, 2, 3, 4, 5, 6, 7, 8]
# [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for k in range(7): print ([outshuffle(9,k,i) for i in range(9)])
# [0, 1, 2, 3, 4, 5, 6, 7, 8]
# [0, 2, 4, 6, 8, 1, 3, 5, 7]
# [0, 4, 8, 3, 7, 2, 6, 1, 5]
# [0, 8, 7, 6, 5, 4, 3, 2, 1]
# [0, 7, 5, 3, 1, 8, 6, 4, 2]
# [0, 5, 1, 6, 2, 7, 3, 8, 4]
# [0, 1, 2, 3, 4, 5, 6, 7, 8]
# Finally, for a full 52-card pack:
for c in shuffle(52,0)[1]: print(c)
# [0]
# [1, 2, 4, 8, 16, 32, 13, 26]
# [3, 6, 12, 24, 48, 45, 39, 27]
# [5, 10, 20, 40, 29, 7, 14, 28]
# [9, 18, 36, 21, 42, 33, 15, 30]
# [11, 22, 44, 37, 23, 46, 41, 31]
# [17, 34]
# [19, 38, 25, 50, 49, 47, 43, 35]
# [51]
# The longest cycle is 8 and we have:
print([outshuffle(52,8,i) for i in range(52)] == list(range(52)))
# True
```
9. Steve said
Klong version
```
shuffle::{[d n n1 n2 n3]; d::x; n::_(#d)%2; n1::n#d; n2::(-n)#d; n{n3::(*n1),*n2; n1::1_n1; n2::1_n2; x,n3}:*[]}
shuffle(1+!20)
[1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20]
shuffle(shuffle(1+!20))
[1 6 11 16 2 7 12 17 3 8 13 18 4 9 14 19 5 10 15 20]
nshuffle::{[cnt d f]; d::x; cnt::f::0; {x; ~f}{[s]; cnt::cnt+1; f::(#d)=+/d=s::shuffle(x); s}:~d; cnt}
{.p((\$x)," --> ",\$nshuffle(1+!x))}'2*1+!26;:ok
2 --> 1
4 --> 2
6 --> 4
8 --> 3
10 --> 6
12 --> 10
14 --> 12
16 --> 4
18 --> 8
20 --> 18
22 --> 6
24 --> 11
26 --> 20
28 --> 18
30 --> 28
32 --> 5
34 --> 10
36 --> 12
38 --> 36
40 --> 12
42 --> 20
44 --> 14
46 --> 12
48 --> 23
50 --> 21
52 --> 8
:ok
```
10. Steve said
Klong version
```
shuffle::{[d n n1 n2 n3]; d::x; n::_(#d)%2; n1::n#d; n2::(-n)#d; n{n3::(*n1),*n2; n1::1_n1; n2::1_n2; x,n3}:*[]}
shuffle(1+!20)
[1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20]
shuffle(shuffle(1+!20))
[1 6 11 16 2 7 12 17 3 8 13 18 4 9 14 19 5 10 15 20]
nshuffle::{[cnt d f]; d::x; cnt::f::0; {x; ~f}{[s]; cnt::cnt+1; f::(#d)=+/d=s::shuffle(x); s}:~d; cnt}
{.p((\$x)," --> ",\$nshuffle(1+!x))}'2*1+!26;:done
2 --> 1
4 --> 2
6 --> 4
8 --> 3
10 --> 6
12 --> 10
14 --> 12
16 --> 4
18 --> 8
20 --> 18
22 --> 6
24 --> 11
26 --> 20
28 --> 18
30 --> 28
32 --> 5
34 --> 10
36 --> 12
38 --> 36
40 --> 12
42 --> 20
44 --> 14
46 --> 12
48 --> 23
50 --> 21
52 --> 8
:done
```
11. Sam Claflin said
Sorry for lack of formatting, new to the site. This is a vanilla Python solution; I was stumped for a bit but figured it out pretty quickly after drawing out a couple of cards and seeing how they moved.
Number of shuffles for 52 cards: 8
for i in range(2, 14, 2):
# Initialize deck size
deck_size = i
if deck_size % 2 != 0:
msg = “Deck size must be even”
raise ValueError(msg)
``````# Create deck function
def create_deck(deck_number):
card_values = []
for i in range(1, deck_number + 1):
card_values.append(i)
return card_values
# Shuffle function
def shuffle(deck):
# Split deck and initialize list
half_1 = deck[0: int(len(deck)/2)]
half_2 = deck[int(len(deck)/2): len(deck)]
shuffled_deck = []
# Main shuffle loop
for i in range(len(half_1)):
shuffled_deck.append(half_1[i])
shuffled_deck.append(half_2[i])
return shuffled_deck
# Initial shuffle
next_shuffle = shuffle(create_deck(deck_size))
shuffle_count = 1
# Mainloop
while True:
if next_shuffle == create_deck(deck_size):
break
next_shuffle = shuffle(next_shuffle)
shuffle_count += 1
# Print output
print(f"For a deck of {deck_size} cards, it takes {shuffle_count} perfect shuffles to return to "
f"its original permutation.")
print("=" * 40)
``````
=====================================
OUTPUT
For a deck of 2 cards, it takes 1 perfect shuffles to return to its original permutation.
For a deck of 4 cards, it takes 2 perfect shuffles to return to its original permutation.
For a deck of 6 cards, it takes 4 perfect shuffles to return to its original permutation.
For a deck of 8 cards, it takes 3 perfect shuffles to return to its original permutation.
For a deck of 10 cards, it takes 6 perfect shuffles to return to its original permutation.
For a deck of 12 cards, it takes 10 perfect shuffles to return to its original permutation.
12. matthew said
@Sam – to format code, put ‘[code lang=”python”]’ on a line at the start of the block of code, and ‘[/code]’ on a line at the end.
13. matthew said
Just to be clear – don’t include the single quotes there, just the square brackets and the stuff inside:
```[code lang="python"]
fib = lambda n: 1 if n <= 1 else fib(n-1)+fib(n-2)
[/code]
```
14. mclrhn said
Further;If the aim of shuffling is to make the card order unpredictable or at least as unpredictable as possible ;
how many “faro-shuffles” of a 52-card stack is ideal ?….sorry about such a late comment , it was stimulated by the current array shuffling problem even though it has no relation to it !
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https://colincrain.com/2020/09/05/wear-a-wig-to-play-the-goldbach-variations/?like_comment=226&_wpnonce=da5a176e77
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Wear a Wig to Play the Goldbach Variations!
Wherein we search the texts for hidden meaning, adding up the indivisible holes in the master pattern. Wish us luck.
THE WEEKLY CHALLENGE – PERL & RAKU #76
episode 1
“The Goldbach Variations”
Reviewed by: Ryan Thompson
You are given a number `\$N`. Write a script to find the minimum number of prime numbers required, whose summation gives you `\$N`.
For the sake of this task, please assume `1` is not a prime number.
Example:
``````Input:
\$N = 9
Ouput:
2 as sum of 2 prime numbers i.e. 2 and 7 is same as the input number.
2 + 7 = 9.``````
ANALYSIS
How many prime numbers are required to create any given positive integer? Before really thinking about it very much my initial feeling said “two or three”, but I wasn’t at all sure why I thought that.
How can you generalize about numbers that defy predictability? There is no pattern to the prime numbers, there are only an infinite number of patterns for numbers that are not prime. By applying the Sieve of Eratosthenes, we can view the patterns behind the excluded numbers; the primes are just the holes left afterwords, their very existence defined by the absence of pattern.
Goldbach’s Conjecture (not to be confused with the Goldberg Variations), states that any even integer can be reduced to the sum of two primes. While remaining one of the great unproven problems in mathematics, it has been verified for all numbers up to to 4 × 1018. Thus although it still retains the most minute chance to render any algorithm using it to be flawed, by constricting our data space to values less than that very, very large number we should be safe. In any case with that caveat, there’s no reason not to let the conjecture guide our analysis.
So that’s half the field, even numbers, right there. For those, the answer is two. What about the odd numbers, though?
One quality about prime numbers is that they are, save 2, odd. So the sum of any two primes will always be even, the sum of any three, odd. Ok, yes, there is an edge case where when one of the primes is 2; an odd number can be created by adding an odd prime and the sole even prime 2. We’ll have to watch for that. But by subtracting any smaller odd prime from an odd integer we will end up with a even integer, and as assumed by the Conjecture, any even integer can be constructed from two primes.
Which leads to the conclusion that any integer can, if even, be constructed by the sum of two primes, and if odd, by a maximum of three primes.
There’s one more loose end out there, though, which is what if the number given is itself prime? Can it be said to be “constructed” from one prime, itself? Concluding “no” immediately leads to an existential crisis of identity, so I am disinclined to pursue that line of reasoning. One is already commonly accepted to be the loneliest number, and the primes in their pride stand alone, so I see no need to further the primal Weltschmerz. They have enough problems. Look around — there’s no shortage of prime problems. In Number Theory they’re everywhere.
So in an absolute sense, we have answered the challenge right here: if a number is prime itself the answer is one, any even integer results in two primes, and for odd integers we can, by subtracting 2 and checking the result for primeness, determine whether the edge case applies and report either two or three primes for those. The challenge never actually asks what those primes are, only how many are required. The examples, however, suggest that last step might be a better interpretation of the task, to provide at least one example, so we’ll do that. Even if, perchance, no one actually cares.
Perl Solution
To find our primes, we’ll start with a list of every prime less than the target number. To do that we’ll bring in our trusty `make_prime_list()` routine, which does exactly what we need by dividing out up to the square root. Once we have an array of primes, we can make a parallel hash structure from to allow quick and easy verification of the elements.
Armed with these, we can take the largest prime less than the target, subtract it and then do a hash lookup to see whether the difference is also prime. If not, the next largest prime is tried, then the next, until a hit is found; as per the Goldbach Conjecture it will eventually be so. This is an easy way to determine sum pairs.
For the triplets, the action is quite similar, only first we remove the value of the largest prime, then proceed to the summing pair for the remaining even number. By subtracting the largest prime we reduce the difference as much as we can, leaving the smallest number remaining to find a prime pair for. The only reason we choose to do it this way is that smaller numbers are just simpler numbers to work with, having fewer ways to decompose them.
To facilitate these actions we will park the prime pair sums over in their own routine, to be available for computing either pairs or triplets. Further, a little effort is made to avoid having to replicate the prime list, being the most computationally expensive aspect of the task. If we hand in a precomputed reference that is used, but if we don’t, as when we partition the smaller difference in a triplet, a new prime list is created.
``````[colincrain:~/PWC]\$ perl 76_1_goldbach_variations.pl 222227
222227 = ( 222199 + 23 + 5 )``````
``````use warnings;
use strict;
use feature ":5.26";
use DDP;
## ## ## ## ## MAIN:
my \$num = shift @ARGV // 57;
my (\$primes, \$lookup, \$result);
\$primes = make_prime_list(\$num);
\$lookup = { map { \$_ => 1 } \$primes->@* };
## descend through the cases, in increasing complexity
if (\$num == \$primes->[-1]) {
\$result = [\$num];
}
elsif (\$num % 2 == 0 ) {
\$result = get_prime_pair(\$num, \$primes, \$lookup);
}
elsif (exists \$lookup->{ \$num-2 }) { ## edge case for odd numbers
\$result = [\$num-2, 2];
}
else {
my \$diff = \$num - @{\$primes}[-1];
\$result = [ \$primes->[-1], get_prime_pair(\$diff)->@* ];
}
say "\$num = ( ", (join ' + ', @\$result), " )";
## ## ## ## ## SUBS:
sub make_prime_list {
## creates a list of all primes less than or equal to a given number
my \$max = shift;
my @output = (2);
CANDIDATE: for( my \$candidate = 3; \$candidate <= \$max; \$candidate += 2 ) {
my \$sqrt_candidate = sqrt( \$candidate );
for( my \$test = 3; \$test <= \$sqrt_candidate; \$test += 2 ) {
next CANDIDATE if \$candidate % \$test == 0;
}
push @output, \$candidate;
}
return \@output;
}
sub get_prime_pair {
## given an even number returns two primes that sum to it
## if \$primes and \$lookup are absent, makes new ones for \$num
my ( \$num, \$primes, \$lookup ) = @_;
if (not defined \$primes) {
\$primes = make_prime_list(\$num);
\$lookup = { map { \$_ => 1 } \$primes->@* };
};
my \$i = @\$primes;
while (--\$i >= 0) {
my \$diff = \$num - \$primes->[\$i];
if ( exists \$lookup->{ \$diff } ) {
return [\$primes->[\$i], \$diff];
last;
}
}
return undef;
}``````
Raku Solution
In Raku, we get a prime test out of the box, so constructing a list of prime candidates is quick and easy. The \$result assignment is also a fine example to demonstrate the
``\$result = do given \$num { when {} ... }``
construct. Which in turn highlights that instead of using a Hash to check for existence of a value in the prime list, we are using a Set, and the set operator “is an element of” that comes with it.
I’m well aware that just comparing with `.is-prime` is presumably a better way of going about this, but using explicit sets is just cool, so I’m in. Likewise creating the entire list of lesser primes is probably overkill as well; we might just count down from the target and check as we go to find the highest prime, then check the difference for primeness to find a pair. And once we eliminate the prime list, we no longer need a `multi`, so that’s out too.
Wow. We seem to be losing a lot of the cool Raku tricks here. When this goes into production I’ll have to change that.
But this works, and I’ve not used `Set` yet, and `multi` is cool too, so that was all fun and I’m keeping it as it is.
``````unit sub MAIN (Int \$num where \$num > 1 = 51) ;
## generate prime list and set
my @primes = (2..\$num).grep: { .is-prime };
my \$p = @primes.Set;
## descend through the cases, in increasing complexity
my \$result = do given \$num {
when @primes.tail { \$num.List }
when \$_ %% 2 { get_prime_pair(\$num) }
when \$_ - 2 ∈ \$p { \$num-2, 2 }
default { @primes.tail, |get_prime_pair(\$num - @primes.tail) }
};
## output
my \$quan = \$result.elems;
say "\$quan\n\n\$num can be summed from the \$quan primes " ~ \$result.join: ' + ';
multi get_prime_pair (\$num, @primes) {
## calculates prime pairs that add to make number
## give it a prime list to avoid regenerating it
my \$p = @primes.Set;
my \$i = @primes.end;
while (\$i > -1) {
return (@primes[\$i], \$num - @primes[\$i]) if (\$num - @primes[\$i] ∈ \$p);
\$i--;
}
}
multi get_prime_pair (\$num) {
## without a prime list it will make a new one
my @p = (2..\$num).grep: { .is-prime };
get_prime_pair(\$num, @p);
}``````
episode 2
“Where in the World is Wigged?”
Submitted by: Neil Bowers
Reviewed by: Ryan Thompson
Write a script that takes two file names. The first file would contain word search grid as shown below. The second file contains list of words, one word per line. You could even use local dictionary file.
Print out a list of all words seen on the grid, looking both orthogonally and diagonally, backwards as well as forwards.
Search Grid
``````B I D E M I A T S U C C O R S T
L D E G G I W Q H O D E E H D P
U S E I R U B U T E A S L A G U
N G N I Z I L A I C O S C N U D
T G M I D S T S A R A R E I F G
S R E N M D C H A S I V E E L I
S C S H A E U E B R O A D M T E
H W O V L P E D D L A I U L S S
R Y O N L A S F C S T A O G O T
I G U S S R R U G O V A R Y O C
N R G P A T N A N G I L A M O O
E I H A C E I V I R U S E S E D
S E T S U D T T G A R L I C N H
H V R M X L W I U M S N S O T B
A E A O F I L C H T O D C A E U
Z S C D F E C A A I I R L N R F
A R I I A N Y U T O O O U T P F
R S E C I S N A B O S C N E R A
D R S M P C U U N E L T E S I L
``````
Output
Found 54 words of length 5 or more when checked against the local dictionary. You may or may not get the same result but that is fine.
``aimed, align, antes, argos, arose, ashed, blunt, blunts, broad, buries, clove, cloven, constitution, constitutions, croon, depart, departed, enter, filch, garlic, goats, grieve, grieves, hazard, liens, malign, malignant, malls, margo, midst, ought, ovary, parted, patna, pudgiest, quash, quashed, raped, ruses, shrine, shrines, social, socializing, spasm, spasmodic, succor, succors, theorem, theorems, traci, tracie, virus, viruses, wigged``
Method
What an interesting puzzle, with what turns out to be a lot of parts and moving pieces. Checking every possible substring against every possible word seems like a tall order. But after examining the problem a little closer its not so bad.
There’s two halves of the word search, traversing the matrix for candidates, suitable and unsuitable, and the cross-referencing to a dictionary of valid words to verify them. Of the two, it seemed the lookup to be the more formidable, as the dictionary on my system has about 279,000 words in it.
It seemed like a trie would serve well for a lookup, hashing the words as a linked list of characters. This in itself isn’t too difficult a task or anything, but it does involve a bit of thinking through and a pair of accessor functions to add and verify words in the data structure. Each word check would be reduced to a half to a dozen or so small hash lookups. Sounds fast. Perl hashes are extremely fast, all in all. But when considering speed, it occurred to me that Perl hashes are fast, and also don’t scale. So why, again, do I need a trie? Dumping the entire word list into look-up hash is a trivial operation. In Raku it’s one line from filename to hash. Until we found it to be wanting, that was going to be just fine.
Curious, I ran some benchmarks on hash sizes vs time to fetch, by the way, and found a remarkable lack of linear scaling, with negligible difference overall, over a range of a hundred keys to ten million.
``````[colincrain:~/PWC]\$ perl hash-speedtest-linear.pl
Rate tthou hthou thous milli hundr tmill
tthou 1316991/s -- -1% -1% -1% -2% -2%
hthou 1323859/s 1% -- -0% -1% -1% -2%
thous 1325748/s 1% 0% -- -0% -1% -2%
milli 1331858/s 1% 1% 0% -- -0% -1%
hundr 1338024/s 2% 1% 1% 0% -- -1%
tmill 1349578/s 2% 2% 2% 1% 1% --``````
A “mere” couple of hundred thousand words is nothing; a lookup takes what a lookup takes, no matter the size of the hash. If you notice, the ten million key hash is fastest for this run, although the actual difference is insignificant.
The other side of the search, finding candidate letter groupings, again became less daunting once it was broken down.
Each letter in the grid is potentially the beginning of a word. Each word can extend in one of eight directions from a starting letter: forwards, backwards, up, down, and diagonally on the four angles between those cardinal points. Each direction will have words of length 5, 6, 7 etc until we run out of grid, and each of those combinations will need to be considered a different word. A little back of the envelope counting led me to conclude that for a starting letter not close to any edge there would be about 37 letter combinations to look at, give or take. The grid given is 16 × 19, comprising 304 letters. Multiplying this by 37 yields only 11248 possible words. I had thought it would be a lot larger. In fact, because proximity to the edges eliminated some possibilities, the real number, for words 5 or more letters long, was only 10340.
All there was left was to determine the letter strings to check.
Not long ago we visited the classic 8-queens chess problem in 3D, in what I called the Spacequeen Problem. Having thought out all that geometry in three dimensions left me well prepared to computing it in an easy two. No triagonal unicorns here to confound us here! The result is a series of interrelated equations that calculate out the letter vectors in eight directions.
Once we have the vectors, we iterate over them from the base letter in increasing lengths to produce candidate words until we run out of vector. Cross-referencing these against the dictionary lookup filters these 10000 or so letter combinations down to the 68 words found.
Perl Solution
In Perl, the individual letter vectors are made by a series of small `for` loops, iterating over the index values in play and adding the letter found. For the diagonal vectors, we need an additional iterator to shift the other index simultaneously; this iterator is reset back to the starting x value between runs. An additional check must be made as well for the diagonals, this one within the loop to make sure the vector does not run out of bounds.
There’s a temptation to refactor these somewhat repetitive loops, perhaps functionally with a `map` construct, but Perl maps won’t allow an early exit with loop control statements like `next` and `last`. But no mind, this has the benefit of clarity and discrete action. To see how this might work in practice, take a look down at the Raku version in all its first-class functional glory.
``````[colincrain:~/PWC]\$ perl 76_2_where_is_wigged.pl 'wordsearch.txt' '/Users/colincrain/dictionaries/scrabble.txt'
B I D E M I A T S U C C O R S T
L D E G G I W Q H O D E E H D P
U S E I R U B U T E A S L A G U
N G N I Z I L A I C O S C N U D
T G M I D S T S A R A R E I F G
S R E N M D C H A S I V E E L I
S C S H A E U E B R O A D M T E
H W O V L P E D D L A I U L S S
R Y O N L A S F C S T A O G O T
I G U S S R R U G O V A R Y O C
N R G P A T N A N G I L A M O O
E I H A C E I V I R U S E S E D
S E T S U D T T G A R L I C N H
H V R M X L W I U M S N S O T B
A E A O F I L C H T O D C A E U
Z S C D F E C A A I I R L N R F
A R I I A N Y U T O O O U T P F
R S E C I S N A B O S C N E R A
D R S M P C U U N E L T E S I L
found 68 words of minimum length 5:
AIMED
ALIGN
ANTES
AROSE
ASHED
BLUNT
(...)
VIRUS
VIRUSES
WIFIE
WIGGED``````
``````use warnings;
use strict;
use feature ":5.26";
## ## ## ## ## MAIN:
my \$file = shift @ARGV // 'wordsearch.txt';
my \$dict = shift @ARGV // '/usr/share/dict/words';
my \$MINWORD = shift @ARGV // 5;
print_matrix(\$matrix);
my \$words = build_word_hash(\$dict);
my @possibles;
my \$height = @\$matrix - 1;
my \$width = \$matrix->[0]->@* - 1 ;
for my \$y (0..\$height) {
for my \$x (0..\$width) {
push @possibles, word_vectors( \$x, \$y, \$matrix)->@*;
}
}
my @output = grep { exists \$words->{\$_} } @possibles;
say '';
say "found ", scalar @output, " words of minimum length \$MINWORD: \n";
say for sort @output;
## ## ## ## ## SUBS:
sub word_vectors {
my (\$x, \$y, \$matrix) = @_;
my \$height = @\$matrix - 1;
my \$width = \$matrix->[0]->@* - 1 ;
my @words;
my @vec ;
my \$i;
## horz forward
push \$vec[0]->@*, \$matrix->[\$y][\$_] for (\$x..\$width);
## horz back
push \$vec[1]->@*, \$matrix->[\$y][\$_] for reverse (0..\$x);
## vert down
push \$vec[2]->@*, \$matrix->[\$_][\$x] for (\$y..\$height);
## vert up
push \$vec[3]->@*, \$matrix->[\$_][\$x] for reverse (0..\$y);
## diag down forward
\$i = \$x;
for (\$y..\$height) { ## y to height index
last if \$i > \$width;
push \$vec[4]->@*, \$matrix->[\$_][\$i++];
}
## diag down back
\$i = \$x;
for (\$y..\$height) { ## y to height index
last if \$i < 0;
push \$vec[5]->@*, \$matrix->[\$_][\$i--];
}
## diag up forward
\$i = \$x;
for (reverse (0..\$y)) { ## 0 to y
last if \$i > \$width;
push \$vec[6]->@*, \$matrix->[\$_][\$i++];
}
## diag up back
\$i = \$x;
for (reverse (0..\$y)) { ## 0 to y
last if \$i < 0;
push \$vec[7]->@*, \$matrix->[\$_][\$i--];
}
## turn vectors into strings \$MINWORD letters or longer
for my \$v (@vec) {
next if @\$v < \$MINWORD;
my \$stem = join '', @\$v[0..\$MINWORD-2];
push @words, map { \$stem .= \$_ } @\$v[\$MINWORD-1..@\$v-1];
}
return \@words;
}
my \$file = shift;
open my \$fh, '<', \$file
or die "cannot open file \$file: \$!\n";
my @search;
while (my \$line = <\$fh>) {
push @search, [split /\s/, \$line];
}
close \$fh;
return \@search;
}
sub print_matrix {
my \$matrix = shift;
for (@\$matrix) {
say join ' ', @\$_;
}
}
sub build_word_hash {
my \$dict = shift;
my %hash;
open my \$fh, "<", \$dict
or die "can't open \$dict to read: \$!";
while (my \$word = <\$fh>) {
\$word =~ s/[\n\r]//g;
\$word = uc(\$word);
\$hash{\$word} = 1;
}
return \%hash;
}``````
Raku Solution
In Raku, everything is a first-class citizen, and every loop block, including those in `map` and `grep`, allows control statements like `next` and `last`. As such, the vector definitions can be rewritten as mappings that can be exited should they stray out of bounds. But why stop there? Assignment to a master list of vectors is another repetitive task, and the eight functions that create the vectors can be collected in a list and processed all at once with
``@vectors.push( \$_() ) for { block }, { block }, ...``
which looks very clean to me. Sweet.
Bughunt
One aspect of all this functional looping, one that caught me several times in this challenge, is how `map` does not produce a list or array after processing, but rather a sequence, or `Seq`.
This has caught me before, but I think I worked around it without a thorough understanding of why what I did worked. In short, the difference being that a Seq has lazy evaluation, so in a sense doesn’t exist until you look at it. When I first refactored the Perl `for` loops as mapped functions, several of them broke. Not completely, mind you, but the x axis was way off on the down-back and up-back diagonals. I explicitly reset the iterator variables between runs, but it made no difference, it wouldn’t stick. As it was, I determined the extra iterator was only being manipulated within the maps, not from the surrounding code. The iterator was incrementing in one map, then going down again in the next. As blocks are closures, this should not be the case, and in fact as we’re handing in the \$i variable at the first of the diagonals, it obviously isn’t. So what on Earth was happening?
One thing I noticed was that when debugging, the print statements I was interspersing to peek at the values for `\$i` were coming out out-of-order. Adding a few more to nail down exactly when the code was being executed finally gave me what I needed to know, and it dawned on me. The @vectors were not a list of lists, a very perlish thing to think, but rather they were an array of sequences, and the individual vectors were only evaluated when they were looked at a half-dozen lines lower, when they were iterated through to produce word candidates. By placing the `\$i = \$x` reset code there, everything worked as expected.
But sequences weren’t done with me. When I had first read in the letter grid file, I printed it out to make sure I had gotten things right. I had, and it was handy see it there for quick reference as I worked out the rest, so I left it. Fast-forward to the end of the story; I want the output all gathered together at the end and move the line. And everything breaks.
``````The iterator of this Seq is already in use/consumed by another Seq
(you might solve this by adding .cache on usages of the Seq, or
by assigning the Seq into an array)
in sub word_vectors at /Users/colincrain/Code/PWC/76-2-where-is-wigged.raku line 47
in block at /Users/colincrain/Code/PWC/76-2-where-is-wigged.raku line 32
in sub MAIN at /Users/colincrain/Code/PWC/76-2-where-is-wigged.raku line 30
in block <unit> at /Users/colincrain/Code/PWC/76-2-where-is-wigged.raku line 1``````
All from just moving the line
``.put for @matrix;``
which prints the sub-arrays. What?
After a bit of hair-pulling, failure and drama, after searching the web for the error message, I ended up on PerlMonks with a nearly identical problem broached by Athanasius. Why would a print statement fix this error? It turns out that the print requires evaluation of the Seq, of course, but also assumes you want to keep those values for later, so it calls `.cache` on the Seq. A Seq by default is expected to be evaluated once and moved on from, so it comes with an iterator that moves forward and is expended. Calling .cache works around this by caching the results so they can be reviewed, as this behavior isn’t on by default. Before I had fixed this by casting to an Array, which also works, but now I have a deeper understanding of what’s really going on. Having scores of built-in data structures is nice, but takes a bit of getting used to shifting over from only three.
Like most problems, its impossible until you know the answer, then its trivial. Oh, that little thing? Of course it’s a sequence. Pshaw.
``````unit sub MAIN (Str \$file = 'wordsearch.txt', Str \$dict = '/usr/share/dict/words') ;
## cfg
my \$MINWORD = 5; ## val > 1
## in
my @matrix = \$file.IO.lines
.map( { .comb(/\w/).cache; } ); ## <-- need to cache the internal Seqs
my %lookup = \$dict.IO.lines
.map( { my \$copy = \$_; \$copy.=uc; \$copy => 1} );
## work
my @possibles;
for [email protected] -> \$y { ## height
for [email protected] -> \$x { ## width
@possibles.append: |word_vectors( \$x, \$y, @matrix);
}
}
my @output = @possibles.grep:{ %lookup{\$_}:exists }
## out
.put for @matrix;
say '';
say "found ", @output.elems, " words of minimum length \$MINWORD: \n";
.say for sort @output;
# + + + + + + + + + + + + + + + + + + + + + +
sub word_vectors (\$x, \$y, @mat) {
my \$height = @mat.end;
my @words;
my @vectors ;
my \$i = \$x;
## formulae for the 8 vectors
@vectors.push( \$_() ) for
{ (\$x..\$width).map: { @mat[\$y][\$_]} }, ## horz forward
{ (0..\$x).reverse.map: { @mat[\$y][\$_]} }, ## horz back
{ (\$y..\$height).map: { @mat[\$_][\$x]} }, ## vert down
{ (0..\$y).reverse.map: { @mat[\$_][\$x]} }, ## vert up
{ (\$y..\$height).map: { last if \$i > \$width; @mat[\$_][\$i++] } }, ## diag down forward
{ (\$y..\$height).map: { last if \$i < 0; @mat[\$_][\$i--] } }, ## diag down back
{ (0..\$y).reverse.map: { last if \$i > \$width; @mat[\$_][\$i++] } }, ## diag up forward
{ (0..\$y).reverse.map: { last if \$i < 0; @mat[\$_][\$i--] } } ; ## diag up back
## turn vectors into strings \$MINWORD letters or longer
for @vectors -> @v {
\$i = \$x; ## <-- lazy evaluation in map Seqs above need the reset to be here!
next if @v.elems < \$MINWORD;
my \$stem = ( @v[0..\$MINWORD-2] ).join;
my @vec_words = ( @v[\[email protected]] ).map: { \$stem ~= \$_ } ;
@words.push: |@vec_words;
}
return @words;
} ``````
4 thoughts on “Wear a Wig to Play the Goldbach Variations!”
1. Myoungjin Jeon says:
the perl code is so clean as usual.
your raku solution is awesome as well.
TBH I never expect that raku solution can be so fast as yours.
but I’d like to suggest that you would better to add `List::MoreUtils::uniq’
e.g.)
my @output = grep { exists \$words->{\$_} } uniq @possibles;
otherwise you will get some duplicated words when under 5 characters.
Like
1. Thank You! Yes I had considered that, it’s especially apparent for 2 letter words: if the same word is found in multiple locations it is listed twice. But it isn’t counting the same word twice, it’s counting different instances. It depends on whether you’re counting different words or the act of circling words, the way this puzzle would traditionally be solved in a newspaper. The real problem (unsolved) is what happens if a word is a palindrome? Here it gets counted twice, once forward and once backwards. But then again one might reasonably consider them to be, read this way, different words comprised of the same letters, making the same word. Pathological edge cases are fun.
Like
2. Myoungjin Jeon says:
oh, yes. it’s not duplicated word. just duplicated count. so you could get how many times the same word occurs in the quiz.
and interesting ideas. I’m never good at solving word quiz so couldn’t think about that.
but I thought about a quiz generator. my solution only generating all the possible indices only and compile possible strings from grid search file so I could make some random quiz file by randomly chosen alphabets several times (of course it could take very long time to make as size of matrix is growing)
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# DSA Part 2: Mathematics
Posted on Dec 25, 2020
## Number System
### Types of number
TypeValue
Natural Numbers$N=1,2,3,4, \ldots$
Prime Number$P=2,3,5,7,11,13,17, \ldots$
Composite Number$4,6,8,9,10,12, \ldots$
Whole Numbers$W=0,1,2,3,4, \ldots$
Integers$Z= \ldots,−3,−2,−1,0,1,2,3, \ldots$
Rational Numbers$Q= \frac{1}{2} ,0.33333 \ldots,52,1110, \ldots$
Irrational Numbers$F= \ldots ,π, \sqrt{2} ,0.121221222 \ldots$
Real Numbers$R= \ldots ,−3,−1,0, \frac{1}{5},1.1, \sqrt{2} ,2,3,π, \ldots$
Complex Number$C= \ldots ,−3+2i,0,1+3i, \ldots$
### Types of number system
Number system consists values from 0 to N-1
Number SystemValue
Binary0 and 1
Octal0 to 7
Decimal0 to 9
HexaDecimal0 to 9 and A to F
## Number Series
SeriesFormula
Sum of first n numbers$\frac{n \cdot (n+1)}{2}$
Squares of first n numbers$\frac{n \cdot (n+1) \cdot (2n+1)}{6}$
Cube of first n numbers$(\frac{n \cdot (n+1)}{2})^2$
Sum of first n even numbers$n \cdot (n+1)$
Sum of first n odd numbers$n^2$
## Number Sequences
### Arithmetic progression
$2, 4, 6, 8, \ldots$
$\rightarrow a_{1} = a + 0.d$
$\rightarrow a_{2} = a + 1.d$
$\rightarrow a_{3} = a + 2.d$
$\vdots$
$\rightarrow T_{n} = a + (n-1) \cdot d$
We know that
$\rightarrow avg = \frac{sum}{n}$
$\rightarrow sum = avg \times n$
And in case of evenly distributed numbers
$\rightarrow avg = \frac{first + last}{2}$
$\rightarrow sum = \frac{(a + a + (n-1).d)n}{2}$
$\rightarrow S_n = \frac{n(2a + (n-1)*d)}{2}$
### Geometric progression
2, 4, 16, 32, 64 ……
a, $a.r^{1}$, $a.r^{2}$, $a.r^{3}$, … $r^{n-1}$
Nth term $\rightarrow$ $T_{n} = a*r^{n-1}$
Sum till N terms $\rightarrow$ $S_{n} = a*(r^{n} - 1)/ r-1$
Sum of infinite terms $\rightarrow$ $S_{\infty} = \frac{a}{1-r}$
### Harmonic Progression
Harmonic progression (or harmonic sequence) is a progression formed by taking the reciprocals of an arithmetic progression. $\rightarrow \frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \ldots$
Nth term $\rightarrow$ $T_{n} = \frac{1}{a + (n-1) \cdot d}$
## Average
### Mean
The central of middle value in a set of data. Commons ways to calculate average are mean, median, mode.
Suppose
$a_{1}, a_{2}, a_{3}, …. , a_{n}$
Mean is the arithmetic average of a given data.
$\rightarrow mean = \frac{a_{1} + a_{2} + a_{3} + …. + a_{n}}{n}$
#### Types of Mean
NumberTest
Arithmetic Mean(AM)$\frac{a+b}{2}$
Geometric Mean(GM)$\sqrt{ab}$
Harmonic Mean(HM)$(2ab) \cdot (a+b)$
$\rightarrow GM^2 = AM \times GM$
### Median
Median is the middle value
$\rightarrow median = A[\frac{n+1}{2}] if n is odd else \frac{A[\frac{n}{2}] + A[\frac{n+1}{2} + 1]}{2}$
### Mode
Mode is the number with the maximum frequency
$\rightarrow 3 Median = 2 Mean + Mode$
$ax^{2} + by + c = 0$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
$D = b^{2} - 4ac$
• D < 0 : Imaginary roots
• D = 0 : Two equal roots
• D > 0 : Two distinct roots
### Prime numbers
2,3,5,7,11, ….
All the numbers having exactly two factors.
Can be represented as 6n+1 or 6n-1, except 2 and 3.
2 and 3 are only consecutive prime numbers.
## Test Of Divisibility
NumberTest
2Last one digit is divisible by 2 (Unit digist is 0, 2, 4, 6, 8)
3Sum is divisble by 3
4Last two digit is divisible by 4
5Unit digit is either 0 or 5
6Divisible by both 2 and 3
7Difference between twice the unit digit of the given number and the remaining part is divisible by 7
8Last three digit is divisible by 8
9Sum is divisble by 9
10Unit digit is 0
11Difference between sum of digits at odd places and even places is either 0 or is divisible by 11
## HCF and LCM
There are three ways to find HCM and LCM
1. Listing factors/multiple
2. Prime fractorization
3. Division method
$\rightarrow a\times b = HCF(a,b) \times LCM(a,b)$
HCF of co-primes is 1.
For fractions
$HCF = \frac{HCF(Numerators)}{LCM(Denominators)}$
$LCM = \frac{LCM(Numerators)}{HCF(Denominators)}$
## Area
### Rectangle
$\rightarrow area = length \times bread$
$\rightarrow perimeter = 2 ( length + bread)$
### Square
$\rightarrow area = side^{2}$
$\rightarrow perimeter = 4 \times side$
### Rhombus
$\rightarrow area = \frac{product of diagonals}{2}$
### Equilateral Triangle
$\rightarrow area = \frac{ \sqrt{n} \times side^{2}}{4}$
$\rightarrow radius of incircle = \frac{side}{2 \times \sqrt{3}}$
$\rightarrow radius of circle = \frac{side}{\sqrt{3}}$
### Isosceles Triangle
$\rightarrow area = \frac{base \times length}{2}$
### Heron’s Formula
$\rightarrow area = \sqrt{s(s-a)(s-b)(s-c)}$
where s is semi-perimeter
$\rightarrow s = \frac{a + b + c}{2}$
### Circle
$\rightarrow area = \pi \times R^{2}$
$\rightarrow circumference = 2 \times \pi \times R$
### Parallelogram
$\rightarrow area = base \times height$
## Volume
### Cuboid
$\rightarrow Volume = L \ast B \ast H$
$\rightarrow Surface area = 2 \cdot (LB + BH + LH)$
$\rightarrow Diagonal = \sqrt{L^2 + B^2 + H^2}$
### Cube
$\rightarrow Volume = a^3$
$\rightarrow Surface area = 6 \cdot a^2$
$\rightarrow Diagonal = \sqrt{3}a$
### Cylinder
$\rightarrow Volume = \pi r^2h$
$\rightarrow Curved surface area = 2 \pi rh$
$\rightarrow Total surface area = 2 \pi rh + 2 \pi r^2$
### Cone
$\rightarrow Slant height = \sqrt{h^2 + r^2}$
$\rightarrow Volume = \frac{\pi r^2 h}{3}$
$\rightarrow Curved surface area = \pi rL$
$\rightarrow Total surface area = \pi rL + \pi r^2$
### Sphere
$\rightarrow Volume = \frac{4 \pi r^3}{3}$
$\rightarrow Surface area = 4 \pi r^2$
## Trignometry
$Radian = \frac{\pi}{180} \times θ$
### Trignometric Ratios
$sin θ = \frac{Perpendicular}{Hypotenuse}$
$cos θ = \frac{Base}{Hypotenuse}$
$tan θ = \frac{Perpendicular}{Base}$
$sec θ = \frac{Hypotenuse}{Base}$
$cosec θ = \frac{Hypotenuse}{Perpendicular}$
$cot θ = \frac{Base}{Perpendicular}$
### Fundamental Trignometric identities
1. $sin^2 A + cos^2 A = 1$
2. $1 + tan^2 A = sec^2 A$
3. $1 + cot^2 A = cosce^2 A$
### Reciprocal Identities
$sin θ = \frac{1}{cosec θ}$
$cos θ = \frac{1}{sec θ}$
$tan θ = \frac{1}{cot θ}$
### Trigonometry Table
Angle0°, 030°, π/645°, π/460°, π/390°, π/2
$sin θ$01/21/√2√3/21
$cos θ$1√3/21/√21/20
$tan θ$01/√31√3
$cot θ$√311/√30
$sec θ$12/√3√22
$cosec θ$2√22/√31
### Periodicity Identities
• $sin (π/2 – θ) = cos θ$
• $cos (π/2 – θ) = sin θ$
• $sin (π/2 + θ) = cos θ$
• $cos (π/2 + θ) = – sin θ$
• $sin (3π/2 – θ) = – cos θ$
• $cos (3π/2 – θ) = – sin θ$
• $sin (3π/2 + θ) = – cos θ$
• $cos (3π/2 + θ) = sin θ$
• $sin (π – θ) = sin θ$
• $cos (π – θ) = – cos θ$
• $sin (π + θ) = – sin θ$
• $cos (π + θ) = – cos θ$
• $sin (2π – θ) = – sin θ$
• $cos (2π – θ) = cos θ$
• $sin (2π + θ) = sin θ$
• $cos (2π + θ) = cos θ$
### Co-function Identities
• sin(90° − x) = cos x
• cos(90° − x) = sin x
• tan(90° − x) = cot x
• cot(90° − x) = tan x
• sec(90° − x) = cosec x
• cosec(90° − x) = sec x
### Inverse Formulas
• $sin^{-1} (-x) = -sin^{-1} x$
• $cos^{-1} (-x) = π - cos^{-1} x$
• $tan^{-1} (-x) = -tan^{-1} x$
• $cosec^{-1} (-x) = -cosec^{-1} x$
• $sec^{-1} (-x) = π - sec^{-1} x$
• $cot^{-1} (-x) = π - cot^{-1} x$
## Logarithm
Logarithm is the inverse function to exponentiation.
$\log_a{b} = c \implies a^c = b$
$\log_2{8} = 3 \implies 2^3 = 8$
### Types of logs
• Common log $\rightarrow$ base 10 $\rightarrow log_10{x}$
• Natural log $\rightarrow$ base e $\rightarrow log_e{x}$ or $lnx$
• Binary log $\rightarrow$ base 2 $\rightarrow log_2{x}$
### Laws of logarithms
$\log_ab + \log_ac = \log_a{bc}$
#### Subtraction Property
$\log_ab - \log_ac = \log_a{\frac{b}{c}}$
#### Argument-Power Property
$\log_ab^n = n\log_ab$
#### Inverse Property
$\log_ab = \frac{1}{\log_ba}$
$\log_a{\frac{1}{b}} = - \log_ab$
#### Power Property
$\log_{a^n}b^n = \log_ab$
#### Change-of-Base
$\frac{\log_cb}{\log_ca} = \log_ab$
• $log_a{1} = 0$
• $log_aa = 1$
## Problems
### Number of digits in an integer
#### Iterative
count(n){
int cnt = 0;
while(n){
cnt += n%10;
n /= 10;
}
return cnt;
}
#### Recursive
count(n){
if(!n) return n;
return 1+count(n)l
}
#### Logarithmic
count(n){
return floor(log(n) + 1);
}
### Check if the given integer is a palindrome or not.
bool isPal(int n){
int rev = 0;
int tmp = n;
while(tmp){
int ones = tmp % 10;
rev = rev * 10 + ones;
tmp /= rmp;
}
return rev == n;
}
### Factorial of a number
#### Iterative
int fact(int n);
int res = 1;
for(int i=2; i<=n; i++)
res *= i;
return res;
}
#### Recursive
int fact(int n){
if(!n) return 1;
return n * fact(n-1);
}
### Count trailing zeros in the factorial of a number
int ctz(int n){
int res = 0;
for(int i=5; i<=n; i *= 5)
res += n/i;
return res;
}
TC : O(log5n)
### GCD of two numbers
#### Naive
int gcd(int a, int b){
int res = min(a, b);
while(res){
if(a%res == 0 and b%res == 0)
break;
res--;
}
return res;
}
Time Complexity: O(min(a,b))
#### Euclidean Algorithm
int gcd(int a, int b){
while(a != b)
if(a>b)
a = a-b;
else
b = b-a;
return a;
}
#### Optimized Euclidean Algorithm
int gcd(int a, int b){
if(b==0)
return a;
return gcd(b, a%b);
}
Time Complexity: O(log max(a,b))
#### Extended Euclidean Algorithm
int gcd(int a, int b, int *x, int *y)
{
if (a == 0)
{
*x = 0;
*y = 1;
return b;
}
int res = gcd(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return res;
}
### LCM of two numbers
#### Naive
int gcd(int a, int b){
int res = max(a, b);
while(res){
if(res%a == 0 and res%a == 0)
break;
res++;
}
return res;
}
Time Complexity: O(a*b - max(a,b))
#### Efficient Approach
$\rightarrow a\times b = gcd(a,b) \times lcm(a,b)$
$\rightarrow lcm(a,b) = \frac{a\times b}{gcd(a,b)}$
Time Complexity: O(log max(a,b))
### Prime number
#### Naive approach
bool isPrime(int n)
{
if (n <= 1)
return false;
for (int i = 2; i < n; i++)
if (n % i == 0)
return false;
return true;
}
Time Complexity: O(n)
#### Efficient approach
bool isPrime(int n)
{
if (n <= 1)
return false;
for (int i = 2; i*i <= n; i++)
if (n % i == 0)
return false;
return true;
}
Time Complexity: O($\sqrt{n}$)
#### Optimized approach
bool isPrime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
### Prime factorization of a number
#### Naive
bool isPrime(int n)
{
if (n <= 1)
return false;
if (n <= 3)
return true;
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
void primeFactors(int n)
{
for (int i = 2; i <= n; i = i + 2)
{
if(isPrime(i))
int x = i;
while (n % x == 0)
{
cout << i << " ";
x = x*i;
}
}
}
Time Complexity: O($nlogn$)
#### Efficient
void primeFactors(int n)
{
if (n <= 1) return;
for (int i = 2; i*i <= n; i = i + 2)
{
while (n % i == 0)
{
cout << i << " ";
n = n/i;
}
}
if (n > 1)
cout << n << " ";
}
Time Complexity: O($\sqrt{n} logn$)
#### Using Sieve
void primeFactors(int n)
{
int c=2;
while(n>1)
{
if(n%c==0){
cout<<c<<" ";
n/=c;
}
else c++;
}
}
Time Complexity: O(n)
#### Naive
void printDivisors(int n)
{
for (int i = 1; i <= n; i++)
if (n % i == 0)
cout <<" " << i;
}
Time Complexity: O(n)
#### Efficient
void printDivisors(int n)
{
for (int i=1; i*i<=n; i++)
{
if (n%i == 0)
cout << i << " ";
}
for (; i>=1; i--)
{
if (n%i == 0)
cout << n/i << " ";
}
}
Time Complexity: O($\sqrt{n}$)
### N Prime Numbers
#### Naive
void nPrimes(){
for(int i=2; i<=n; i++)
if(isPrime(i))
cout << i << " ";
}
Time Complexity: O($n \times \sqrt{n}$)
#### Sieve of Eratosthenes
void manipulated_seive(int n)
{
vector<bool> isPrime(n+1, true);
for (int i=2; i*i<=n ; i++)
{
if (isprime[i])
{
cout << i << " "
for(int j=i*i; j<=n; j+= i)
isPrime[j] = false;
}
}
}
Time Complexity: O($log log n$)
### Computating Pow(x,n)
#### Naive
FOR(1->n) : x = x*x;
Time Complexity: O(n)
#### Optimized
double myPow(double x, int n) {
if(n==0) return 1.0;
double y = myPow(x, n/2);
if(n % 2 == 0)
return y*y;
else return n < 0 ? (y*y)/x : x*y*y;
}
Time Complexity: O(logn)
#### Constant Space
$\rightarrow$ Every natural number can be written as the sum of distinct powers of 2.
double myPow(long long int x, long long int n) {
long long int res = 1;
while(n>0){
if(n&1)
res = res * x;
x *= x;
n = n >> 1;
}
return res;
}
Time Complexity: O(logn)
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# How many days can it be achieved?
I was studying optimization problems and came up with a problem. __ How long does it take to travel all over Japan in the shortest distance? __ In order to answer the question, I calculated a little with python.
Since the purpose is to hands-on the optimization problem with python, please forgive the sweetness of the problem setting. Also, as I wrote in this article, I couldn't solve the problem of creating a circular route other than the favorite route. It is an approximate solution, not an optimal solution. If you are familiar with the solutions around here, please comment! !!
# Problem setting
This time I will use Takashikun, which everyone loves (I don't know if it's true) that often appears in math problems.
## Takashi-kun specs
・ __ Takashi-kun is strong __ so you can cross the sea. ・ __ Takashi-kun is strong __ so you can run at a constant speed regardless of whether there are mountains or valleys. ・ __ Takashi-kun is strong __ so you can run for 24 hours.
## Configuration
・ __ Takashi-kun has a lot of intellectual curiosity, so I thought about going around Japan at a speed of 10km / h. ・ "All over Japan" is defined as "the location of all prefectures" this time. ・ Since __ Takashi-kun is impatient, all movements will be done in a straight line.
# I tried it
## Acquisition of prefectural office location data
The longitude and latitude of the prefectural capital are quoted from this article. After downloading Excel, it will be molded only to the location name, longitude, and latitude data.
``````import pandas as pd
import numpy as np
display(df)
``````
The data looks like this. lati is latitude and long is longitude.
## Distance calculation between each location
Calculate the distance between 47 locations. Originally, I think it would be good to narrow down the combinations, thinking, for example, "Okinawa-Hokkaido will not move." This time, ~~ the selection work was troublesome, and ~~ __ I had a faint expectation that something miracle might happen __, so I calculated all the combinations.
This article was used for distance calculation using longitude and latitude. It is a function that takes the longitude and latitude of two points as arguments (note that each is a list type!) And returns the distance between them in km.
``````def cal_rho(lon_a,lat_a,lon_b,lat_b):
F=(ra-rb)/ra # flattening of the earth
c1=(np.sin(xx)-xx)*(np.sin(pa)+np.sin(pb))**2/np.cos(xx/2)**2
c2=(np.sin(xx)+xx)*(np.sin(pa)-np.sin(pb))**2/np.sin(xx/2)**2
dr=F/8*(c1-c2)
rho=ra*(xx+dr)
return rho
``````
After that, I will put all the combinations into this function.
``````#Prepare a container for the list as an argument
loc_A = []
lon_A = []
lat_A = []
loc_B = []
lon_B = []
lat_B = []
#Intently
for i in range(47):
for j in range(47):
loc_A.append(i)
lon_A.append(df["long"][i])
lat_A.append(df["lati"][i])
loc_B.append(j)
lon_B.append(df["long"][j])
lat_B.append(df["lati"][j])
#Calculate!
rho=cal_rho(lon_A,lat_A,lon_B,lat_B)
#Put the result in a data frame
combi_df = pd.DataFrame([loc_A,loc_B,lon_A,lat_A,lon_B,lat_B,rho]).T
combi_df.columns = ["loc_A","loc_B","long_A","lati_A","long_B","lati_B","Dist/km"]
combi_df = combi_df.fillna(0)
``````
As a result, we were able to calculate the distance between each point.
## Calculation of shortest path
This is almost the same implementation as Previous article and Original article, but only the size of drawing by matplotlib is changed (Reference article). ) Click here for the drawing function.
``````import random, numpy as np, pandas as pd, networkx as nx, matplotlib.pyplot as plt
from itertools import chain, combinations
from pulp import *
def draw(g):
plt.figure(dpi=150)
rn = g.nodes() #Node list
pos = nx.spring_layout(g) #Node position
"""drawing"""
nx.draw_networkx_labels(g, pos=pos)
nx.draw_networkx_nodes(g, node_color='w', pos=pos)
nx.draw_networkx_edges(g, pos=pos)
plt.show()
``````
``````#Define 47 points
n = 47 #Number of nodes
g = nx.random_graphs.fast_gnp_random_graph(n, 1, 8)
#Input the distance between 47 points
for a,b in g.edges():
col_idx = a * 47 + b
g.edges[a, b]["dist"] = combi_df.loc[col_idx,"Dist/km"]
#Calculation
source, sink = 0, 46 #start point,end point
r = list(enumerate(g.edges()))
m = LpProblem() #Mathematical model
x = [LpVariable('x%d'%k, lowBound=0, upBound=1,cat="Integer" ) for k, (i, j) in r] #variable(Whether to enter the road)
m += lpSum(x[k] * g.edges[i,j]["dist"] for k, (i, j) in r) #Objective function
for nd in g.nodes():
m += lpSum(x[k] for k, (i, j) in r if i == nd) \
+ lpSum(x[k] for k, (i, j) in r if j == nd) == {source:1, sink:1}.get(nd, 2) #Constraint
m.solve()
print([(i, j) for k, (i, j) in r if value(x[k]) == 1])
print(LpStatus[m.status])
print(value(m.objective))
``````
The first print is the display of the route, the second is the display of the optimization calculation, and the third is the display of the optimum solution. Click here for the output result. Since it is optimal, it seems that the optimal solution could be found! !! And the shortest route is __4304km! !! __ So Takashi-kun seems to be able to travel all over Japan in __430 hours, that is, about 17 days and 22 hours! __
__ I did it! I was able to calculate! !! !! __ It was a moment when I thought it was __. __
# Circular path problem
What kind of route did you take? I thought, I tried drawing. Click here for the code to check the optimization result.
``````G = nx.Graph()
for k,(i,j) in r:
if value(x[k]) ==1:
draw(G)
``````
Here is the drawing result. The __circle generation problem __, which was also mentioned in the previous article (https://qiita.com/canonno/items/711201febd5b2bf746c4), occurred. This is a problem that arises from the lack of constraint that "when focusing on a certain point, two roads extend at the waypoint and one road extends at the start and end points."
I borrowed a blank map from this site and plotted it. There are places in the Tohoku, Kanto, and Chugoku regions that go round and round.
From here, I will play with the route a little and make it a single route. Since arbitrary operations are added, it is difficult to know whether it will be the optimum solution, but let's think that it will be close.
## Akita-Niigata
Try to make the length of the path you want to cut extremely long. It's like this.
``````pair_list = [(4,14),(14,4)]
for a,b in pair_list:
col_idx = a * 47 + b
combi_df.loc[col_idx,"Dist/km"] = 1000000
``````
Now let's do the optimization calculation. The result is only a map. Successful integration of the Tohoku region! !! I did it. Even in this case, it is about 4309km, so changing only one route does not seem to have that much effect.
## Cut between Niigata and Nagano
I was worried about what to do with the integration of the Kanto region. So, I thought about a route that goes from Niigata to Gunma without going to Nagano, joins the Kanto circle, and then goes to Nagano. So I cut the route between Niigata and Nagano.
The result is here.
I went from Niigata to Tochigi and was able to integrate well. The distance has increased a little, but it is now 4326km.
## Cut between Shimane and Okayama
Finally, what should we do with the integration of the Chugoku region? You can see the movement between Shimane and Okayama, but if you want to move between Okayama and Shimane, it seems better to go via Tottori or Hiroshima. So I will cut between Shimane and Okayama.
__ I got involved in Kagawa ... __ __ How much do you want to be small there? __
## Cut between Shimane and Kagawa
So I will cut the Shimane-Kagawa section. The result is here! Although it is warping from Shimane to Kochi, it is finally one! !! The distance is 4361km.
I was a little worried about the warp between Shimane and Kochi, so I cut it off here as well. This time it warps from Kochi to Oita, and Yamaguchi seems to go to Kyushu and then pick it up. The distance is about the same as 4361km. It seems difficult to write a single stroke around Chugoku and Shikoku.
# Conclusion
__ Distance to run: Approximately 4361km__ __ Running time: Approximately 18 days 4 hours __
# at the end
Optimization problem interesting! I will continue to experiment. Thank you for reading to the end, and it will be encouraging to have LGTM!
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# Fermat Factoring Factoid
When C is odd, (C mod 4) reveals if m needs to be even or odd, to be greater than n
(+2) [vote for, against]
It is possible that this Idea is known by many mathematicians, but it is also possible (see 2nd link) that it has previously been overlooked. Since it certainly isn't widely known outside of mathematics, ....
I've described Fermat Factoring in another Idea (first link) so I don't need to repeat it here. What matters here is the result that if C is a composite number, its factors can be described as combinations of two values, m+n and m-n, such that m-squared minus n-squared equals C.
When C is an even number, we all know that one of its factors is 2, which means that either C/2 is another even number (and also divisible by 2) or that C is an odd number (and might be a prime, not a composite number). In this presentation we only want to pay attention to values of C that are both composite and odd.
Now note that if C is a composite odd number, then all its factors must be odd numbers. NOW note that if two of its factors can be described as m+n and m-n, the only way that the results can be odd numbers is if one of the two, m OR n, is an even number, while the other is an odd number. Because of the subtraction, it is normal for m to be assigned the larger of the two values (prevents computing a negative factor of C).
If the factors of C are known, then it is easy to compute m and n. For example, consider 77, which most folks know equals 7*11. To find m, guaranteed to be the larger than n, simply add 7+11 to get 18, and divide by 2 to get 9. m is 9. It is midway between 7 and 11 by a quantity of 2, so n is 2. (And 9-squared minus 2-squared is, indeed, 77.)
For another example, consider 51, which equals 3*17. If we add them to get 20 and divide by 2, then m=10. And since 10 is midway between 3 and 17 by a quantity of 7, n is 7. (And 10- squared minus 7-squared is, indeed, 51.)
Now note that in the first example m is an odd number, while in the second example, m is an even number. This Idea is about, whenever you don't know the factors of C, you can still be certain that m is an even number, OR you can be certain that m is an odd number.
At the 3rd link I have a little table for your edification. It consists of the results of Even Squares minus Odd Squares, the first part of all possible combinations (the RESULTS are listed, not the squares, except for the first column, the result of Zero squared minus odd squares). As you can see, some of the results are negative, simply because sometimes an Even Square is smaller than an Odd Square. If we temporarily ignore the minus signs we can say that all possible values of C-as-odd must be in this table somewhere. (Even primes are in it, having factors of p and 1, which can still lead to computations of m and n.)
But here's the thing: In that table EVERY positive number in the table has the form of 4x+3. And EVERY negative number in the table has the form of 4x+1. The proof that the observation continues for numbers not in the table is simple: If you follow the numbers diagonally from upper-left to lower-right, you can see that the difference between any two numbers is always a multiple of 4. That kind of pattern means that the descriptions 4x+1 and 4x+3 NEVER change. (Also, the differences between adjacent vertical values, and between adjacent horizontal values, are multiples of 4, too. Only when crossing from negative numbers to positive numbers, or vice-versa, do the remainders change, after dividing by 4.)
Anyway, the observation also means that if you divide any odd composite C by 4, and look at the remainder (it will be either 1 or 3), then that can tell you the part of the table where C can be found (the positive numbers or the negative numbers). If the remainder is 1, then C is in the negative part of the table, and if the remainder is 3, then C is in the positive part of the table.
Keeping in mind that the table consists of Even Squares minus Odd Squares, the preceding also tells us that any negative result means that an Odd Square was greater than an Even Square. So when C is shown to be in the negative part of the table, we know that m will be an odd number; if C is shown to be in the positive part of the table, we know that m will be an even number.
Remember we WANT m to be greater than n, so that when n- squared is subtracted from m-squared, the result will be a positive C. And simply dividing C by 4 tells us how to know whether or not m will be even, or odd.
— Vernon, Mar 25 2016
Fermat Factoring described (among other things) Factoring_20Algorithm
As mentioned in the main text. [Vernon, Mar 25 2016]
Simple math discoveries are likely still out there http://www.israelha...rticle.php?id=32345
As mentioned in the main text. Note this article describes the theorem poorly; the lines inside the circle all need to be the same length. [Vernon, Mar 25 2016]
Table of Even Squares minus Odd Squares http://www.nemitz.net/vernon/SqrDiffs.png
As mentioned in the main text. The table can fit reasonably well on a piece of "legal" size paper (landscape orientation), if you want to print it out. [Vernon, Mar 25 2016]
The image file was derived from a screen shot of the first spreadsheet; the other two sheets in the file are about studying the diagonals [Vernon, Mar 30 2016]
There is a more general proof of the factoid of this Idea.
Any odd number can be described as having the GENERIC form of either 4x+1 or 4x+3
If we multiply 4x+1 by itself, we get 16x^2 + 8x + 1 Since both 16 and 8 are divisible by 4, the net effect is that the square is a number having the GENERIC form of 4y+1 (the fact that y can represent a more complicated expression, such as (4x^2 + 2x), makes no difference when talking about generic stuff).
If we multiply 4x+3 by itself, we get 16x^2 + 24x + 9 Well! 9=4*2 +1, and that means we could rewrite the overall expression as 4y+1, where y=(4x^2 + 6x +2) That means the square of ANY odd number has the GENERIC form of 4y+1 (!)
Therefore we need to see what happens when we subtract 4y+1 from an even square. First, though, we should note that since any even number has a factor of 2, the square of that number will have two factors of 2, which means the square of any even number will be divisible by 4, and thus has the generic form of 4z.
If the even square is smaller than the odd square that we want to subtract from the even square, then the Rule For Subtraction is, "subtract the lesser-magnitude value from the greater-magnitude value, and mark the result as negative".
So 4z minus a greater-magnitude 4y+1 means we actually do 4y+1 minus 4z, and get -(4(y-z) +1) --and all the negative values in the linked table did indeed each have the generic form of 4-times-something, plus 1
If the even square is larger than the odd square, then we need to do a little trick, to prepare for subtraction:
4z equals 4z-4+4 equals 4(z-1) + 4
We can now subtract 4(z-1)+4 minus 4y+1, and get this
4(z-y-1) +3 --and all the positive numbers in the linked table did indeed each have the generic form of 4-times-something, plus 3.
— Vernon, Mar 28 2016
Your excitement makes me wish I understood this better. (+)
— 2 fries shy of a happy meal, Mar 28 2016
super nifty [Vernon]
I think this could be used as gene logic to figure out things like Did I have sex (even or odd chromosome thing, meiosis, mitosis)
also at other technologies I am reminded of fly with wire, or the space shuttle two out of three computers fly the spacecraft. Is it possible there is a super obvious mathematics bettter than of two of three to guide?
like you could do mod 4 on computer guided pathwayizing so rather than just knowing it is 2/3 you could know which 2/3 or even notice it was 3/3.
from a voltage rather than a digital perspective it would be easy to note three capacitors were together, or three LEDS were modifying a voltage. a mod 4 voltage square wave or ripple could say which 2/3 or say 3/3
I thought about a math approach to distributing data at a computer parallel processing computer with n processors. like um, there is a square wave bits (bytes) then if some of these have slightly elevated voltage, the |_|-|_| with the elevated voltage could make it through a capacitor that accumulated a different bnary datastream than the main datastream. the different bnary datastream could be a compressed data thing, then mathematically there is an n, a quantity of rpocessors, where expanding the comrpessed data was actually more efficient than sending the entire decoded data bnary stream along the main datastream. Your modulo 4 even oddness could build that possibly.
like um, say you have a a really compressed thing that takes an entire hundredth of a second to decompress. decompressing that at client would be faster than sending it from the opposite side of the earth, rapidifying the internet. kind of like a multivoltage coprocessor could expand an MP\$ movie faster than sending it, yet the ultracompressed mp4 would be just slight voltage variations at the data stream that was travelling regardless. so modulo4 to note odd or even could possibly lift out bits to make the compressed data out of.
suddenly I conceptualize saying, "photons are different than voltage levels, the internet is photons. hmmm. electrons at microsd cards having multilevel voltage perhaps thus amplifying storage capacity of flash memory to be really ultra high data amounts at a multiprocessor computer possibly...
— beanangel, Mar 28 2016
Your excitement also makes me wish I understood what you said better.
"computer guided pathwayizing " [marked-for-tagline]
— 2 fries shy of a happy meal, Mar 28 2016
computer guided pathwayizing is um, a way to say computer guided spacecraft trajectory producing as a verb.
I think Simple math discoveries are likely still out there is likely
I heard about a big bang theory, the thing is I also read that the duodish shape of an erythrocyte actually has higher surface area to volume than a sphere, also the mathematics of that squiggle thing that might be called an antisphere is another continuous lnked curve. the idea that a big bang could be mathematically from something more energy or matter parsimonious than a sphere could be new physics or new math.
Also regarding the big bang theory wikipedia says that the blue glow of nuclear reactor comes from cerenkov radiation, which is produced when neutrinos travel faster than light through water or other material with a sufficient refractive index. thus if you think of ultradense matter emitting neutrinos, then the neutrinos actually expand at a radius faser than phtons, so the absorbability of neutrinos at prematter actually effects universe shape as much or more than photon absorption effects of prematter. neutrino shaped universe asymmetry is new to me. yet when you think about the possibility of something more parsimonious than a sphere that would be mathematics that is new to me.
— beanangel, Mar 28 2016
Wait a minute. I thought *everything* had a higher surface-to-volume ratio than a sphere - that a sphere is what you built if you wanted to *minimize* surface area per unit volume. No?
— pertinax, Apr 02 2016
It turns out that a slightly simpler way to identify if m is even or odd is this: Just add 1 to the odd number c, and divide by 2. If the result is even, then m is even. If the result is odd, then m is odd.
Note that if c has the form of 4x+3, then adding 1 creates a number having the form of 4x+4, which if divided by 2 yields 2x+2, which *must* be an even number. And if c has the form of 4x+1, then adding 1 creates a number having the form of 4x+2, which when divided by 2 will yield an odd number: 2x+1.
Just for fun, consider the composite number c=1155, which is the product of (3)(5)(7)(11). It has all these combinations of pairs of factors: 1 and 1155 3 and 385 5 and 231 7 and 165 11 and 105 15 and 77 21 and 55 33 and 35
To get m and n (when a pair of factors is known), such that m-squared minus n-squared equals c, the larger number to square (m) is always equal to the sum of a factor- pair divided by 2. And n is simply m minus the smaller factor (OR the larger factor minus m --OR the larger factor minus the smaller factor, which is then divided by two). So: 1 and 1155 yield m=578 and n=577 3 and 385 yield m=194 and n=191 5 and 231 yield m=118 and n=113 7 and 165 yield m=86 and n=79 11 and 105 yield m=58 and n=47 15 and 77 yield m=46 and n=31 21 and 55 yield m=38 and n=17 33 and 35 yield m=34 and n=1
Feel free to square m and subtract the square of n get 1155 (that is, c) in every case. As guaranteed by the algebra presented at the start of this anno (along with the algebra in the first anno above), since in this example (c+1)/2 yields an even value for m, the larger number to square, ALL the possible values of m are also even numbers.
— Vernon, Apr 11 2016
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1. rationalize
$\displaystyle \displaystyle\ A=\frac{6}{\sqrt{3-\sqrt{3}}}=\sqrt{\frac{36}{3-\sqrt{3}}}=\sqrt{\frac{36(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}$
$\displaystyle \displaystyle\sqrt{\frac{36(3+\sqrt{3})}{9+3\sqrt{ 3}-3\sqrt{3}-3}}=\sqrt{\frac{36(3+\sqrt{3})}{9-3}}$
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A360078 Moebius function for the floor quotient poset. 2
1, -1, -1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, -1, -1, -1, -1, -1, -1, 0, 0, 0, -1, -1, -1, -2, -2, -2, -2, -2, -2, -1, -1, -1, -1, 0, 0, -1, -1, -1, -2, -2, -2, -2, -3, -3, -3, -3, -3, -1, -1, -1, -1, -1, -1, 1, 1, 1, 0, -1, -1, -1, -1, -1, -1, -1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,30 COMMENTS Say d is a "floor quotient" of n if d = [n/k] for some positive integer k. This defines a partial order relation on the positive integers. This sequence records the Moebius function values of this poset. LINKS Andrew Howroyd, Table of n, a(n) for n = 1..10000 J.-P. Cardinal, Symmetric matrices related to the Mertens function, arXiv:0811.3701 [math.NT], 2008-2009. J. C. Lagarias and D. H. Richman, The floor quotient partial order, Adv. Appl. Math., 153 (2024); arXiv:2212.11689 [math.NT], 2022-2023. EXAMPLE For n = 9, the set of floor quotients of 9 are Q(9) = {1, 2, 3, 4, 9} with Moebius values a(1) = 1, a(2) = -1, a(3) = -1, and a(4) = 0. The Moebius recursion requires that the Moebius values summed over Q(9) must equal zero, so a(9) = 1. MATHEMATICA LinearSolve[Table[If[Floor[i/j] > Floor[i/(j + 1)], 1, 0], {i, n}, {j, n}], UnitVector[n, 1]] PROG (PARI) seq(n)={my(v=vector(n)); v[1]=1; for(n=2, #v, my(S=Set(vector(n-1, k, n\(k+1)))); v[n]=-sum(i=1, #S, v[S[i]])); v} \\ Andrew Howroyd, Jan 24 2023 CROSSREFS Cf. A002321, A008683, A360079. Sequence in context: A163529 A283655 A262742 * A027354 A192227 A102673 Adjacent sequences: A360075 A360076 A360077 * A360079 A360080 A360081 KEYWORD sign,look AUTHOR Harry Richman, Jan 24 2023 STATUS approved
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Messages - Tzu-Ching Yen
Pages: 1 [2]
16
Quiz-2 / Re: Q2 TUT 0701
« on: October 05, 2018, 05:42:24 PM »
\begin{gather*}
N_x = \frac{2x}{y} - \frac{3y}{x^2}, M_y = -\frac{6}{y^2}, xM = 3x^2 + \frac{6x}{y}, yN = x^2 + 3\frac{y^2}{x}, \\\frac{N_x - M_y}{xM - yN} = \frac{\frac{2x}{y} - \frac{3y}{x^2} + \frac{6}{y^2}}{2x^2 - \frac{3y^2}{x} + \frac{6x}{y}} = R = \frac{1}{xy}
\end{gather*}
From assignment
\begin{gather*}
\mu (z) = \exp(\int \frac{1}{z} d(z)) = z,\\
\mu M = 3x^2y + 6x, \mu N = x^3 + 3y^2,\\
\int \mu M dx = x^3y + 3x^2 + f(y) + c_0,\\
\int \mu N dy = x^3y + y^3 + g(x) + c_1.
\end{gather*}
Combine the previous two result gives
$$\phi(x, y) = x^3y + 3x^2 + y^3 = c,$$
where
$\frac{\partial\phi}{\partial x} = uM, \frac{\partial\phi}{\partial x} = uN$
We do not need this
17
Quiz-1 / Re: Q1: TUT 0201, TUT 5101 and TUT 5102
« on: September 29, 2018, 09:23:56 PM »
Rephrase equation
$y' + \frac{2}{t}y = \frac{sin(t)}{t}$
Find integrating factor
$u(t) = e^{\int \frac{2}{t}} = t^2$
the constant from integration is chosen to be zero. Now
$y = \frac{1}{u(t)}\int u(t)\frac{sin(t)}{t}$
Use int by parts,
$y = -\frac{cos(t)}{t} + \frac{sin(t)}{t^2} + \frac{c_1}{t^2}$
Since $t$ is in denominator and the nominators are bounded, as $t \to \infty$, $y \to 0$
18
Quiz-1 / Re: Q1: TUT 0701
« on: September 28, 2018, 06:00:02 PM »
First divide by $t^3$ on both side of the equation, we get
$$y' + \frac{4}{t}y = \frac{e^{-t}}{t^3}$$
Using the method of integrating factor we have equation for $u(t)$
$$u(t) = e^{\int \frac{4}{t}dt} = e^{4\ln(t) + c} = t^4$$
where constant $c$ is arbitrary, it's chosen to be 0 here. Then
$$\bigl(y u(t)\bigr)' = u(t)\frac{e^{-t}}{t^3}$$
rearranging gives equation
$$y = \frac{1}{u(t)}\int u(t)\frac{e^{-t}}{t^3}$$
substitute in $u(t) = t^4$
$$y = \frac{1}{t^4}\int te^{-t}$$
use integration by parts
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4} + \frac{c_1}{t^4}$$
to check $c_1$, plug in condition $y(-1) = 0$
$$y(-1) = e - e + c_1 = c_1= 0$$
Plug in $c_1 = 0$ gets
$$y = -\frac{e^{-t}}{t^3} - \frac{e^{-t}}{t^4}$$
Pages: 1 [2]
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# Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 10
Divergent
#### Work Step by Step
The p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\sum_{n=3}^{\infty}n^{-0.9999}=\sum_{n=3}^{\infty}\frac{1}{n^{0.9999}}$ The series $\sum_{n=1}^{\infty}\frac{1}{n^{0.9999}}$is a p-series with $p= 0.9999 \leq 1$ and it is divergent. But the given series is not exactly a p-series, it is because the summation starts from $n=3$ But adding or subtracting finite number of terms will not change the convergence/divergence nature of the series. Hence, the given series is divergent.
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SOLUTION 22: $$\displaystyle{ \lim_{x \to \infty} \ (1 + 3/x)^x } = \displaystyle{ \ (1+ 3/\infty)^{ \ \infty } \ " } = \displaystyle{ \ (1+0)^{ \ \infty } \ " } = \displaystyle{ \ 1^{ \ \infty } \ " }$$ (Rewrite the problem to circumvent this indeterminate form. Recall that $\displaystyle{ e^{\ln z} = z }.$) $$\displaystyle{ \lim_{x \to \infty} \ (1 + 3/x)^x } = \displaystyle{ \lim_{x \to \infty } \ e^{ \ \ln (1 + 3/x)^x } }$$ $$= \displaystyle{ \lim_{x \to \infty } \ e^{ \ x \cdot \ln (1 + 3/x) } }$$ $$= \displaystyle{ e^{ \ \displaystyle{ \lim_{x \to \infty } \ x \cdot \ln (1 + 3/x) } } } = \displaystyle{ \ e^ { \ \displaystyle{ \ \infty \cdot \ln 1 \ " } } } = \displaystyle{ \ e^ { \ \displaystyle{ \ \infty \cdot 0 \ " } } }$$ (Flip" $x$ to circumvent this indeterminate form.) $$\displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { \ln (1 + 3/x) \over {1/x} } } } } = \displaystyle{ \ e^ { \ \displaystyle{ { \ \ln (1+3/\infty) / (1/ \infty) \ " } } } } = \displaystyle{ \ e^ { \ \displaystyle{ { \ (\ln (1+0)) / 0 \ " } } } } = \displaystyle{ \ e^ { \ \displaystyle{ { \ (\ln 1)/ 0 \ " } } } } = \displaystyle{ \ e^ { \ \displaystyle{ { \ 0 / 0 \ " } } } }$$ (Apply Theorem 1 for l'Hopital's Rule. Recall that $\displaystyle{ D \{ \ln f(x) \} = \frac{1}{f(x)} \cdot f'(x) }.$) $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 1 \over 1 + 3/x } \cdot (0-{3/x^2}) \over { -1/x^2 } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 1 \over 1 + 3/x } \cdot { -3 \over x^2 } \cdot { x^2 \over -1 } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 3 \over 1 + 3/x } } } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{{3}/(1+0) } } }$$ $$= \displaystyle{ \ e^ { \ \displaystyle{ 3 } } }$$
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## Fluid Flow: Water Pipe with Vertical Pipes
1. The problem statement, all variables and given/known data
Consider a water pipe that tapers down from a diameter dA= 5.0 cm at
end A to a diameter dB = 2.5 cm at end B. At each end a vertical pipe that is open to the air at the
top is attached to the pipe as shown in the Figure. (Not to scale, the pipes are much taller than
shown.) Assume that water flows through the pipe at high enough pressure that the vertical pipes
1 and 2 are partially filled with water. You may take g=10 m/s2
(a) In which pipe will the water level be higher or will it be at the same height in both pipes?
Explain. Does this depend on the direction of the water flow?
(b) If the water enters the pipe at point A with a velocity vA=2.0 m/s, what is its velocity when it
exits at point B?
(c) If the water enters the pipe at point A with a velocity vA=2.0 m/s, what is the height
difference between the levels in the two vertical pipes? If the difference is not zero, please
indicate which pipe has the higher level.
2. Relevant equations
A1v1=A2v2
Bernoulli's Equation
P=ρgh
3. The attempt at a solution
(a) My answer is that because end A will have higher pressure than end B due to Bernoulli's principle and the equation above, the water level in column 1 will be higher than column 2.
I'm just not entirely sure that it's correct.
(b) A1v1=A2v2
vA=2.0 m/s
rA=0.025 m
rB=0.0125 m
vB=πrA2 * vA / (πrB2)
= 8.0 m/s
From Bernoulli's equation, I derived
(yA2-yA1)=h1=[(Po-P1) - (1/2)*ρvA2]/(ρg)
and
(yB2-yB1)=h2=[(Po-P2) - (1/2)*ρvB2]/(ρg)
and
P1=P2+ρgΔy
where
P0 = atmospheric pressure above water in vertical pipe
P1 = pressure right below vertical pipe #1
P1 = pressure right below vertical pipe #2
Δy = difference in height between A and B
I then found the difference in height between the two openings, A and B, to be 0.4625,
and, still assuming that column #1 has a higher water level, subtracted h2 from h1:
h1-h2 = [[(Po-P2 - ρg(0.4625)) - (1/2)*ρvA2] - [(Po-P2) - (1/2)*ρvB2]] / (ρg)
=[[- ρg(0.4625) - (1/2)*ρvA2] - [-(1/2)*ρvB2]] / (ρg)
=[(1/2)*vB2 - (1/2)*vA2 - g(0.4625)] / g
Plugging in
vA=2.0
vB=8.0
g=10
I get Δh=2.5375 m
As I said, I'm very uncertain about the way I approached this last part, so help would be greatly appreciated!
--Johan
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
I think part (c) is a little off. You should get a nice clean expression for the column height difference. Try writing Bernoulli's equation: Pa + (1/2)ρva^2 = Pb + (1/2)ρvb^2 Solve for Pa - Pb and note this is equal to ρg(h1 - h2).
edgepflow: Could you explain how Pb-Pa=ρg(h1 - h2)? Solving for Pb-Pa in Bernoulli's equation, I get Pb-Pa=(1/2)ρ(va^2-vb^2) I'm not sure how this ties to the height of the water in the columns. To me it seems to only relate the velocities, pressures, and heights of the two ends, A and B.
## Fluid Flow: Water Pipe with Vertical Pipes
Quote by JohanM edgepflow: Could you explain how Pb-Pa=ρg(h1 - h2)? Solving for Pb-Pa in Bernoulli's equation, I get Pb-Pa=(1/2)ρ(va^2-vb^2) I'm not sure how this ties to the height of the water in the columns. To me it seems to only relate the velocities, pressures, and heights of the two ends, A and B.
The expression Pa-Pb=ρg(h1 - h2) is basically the same as P=ρgh you listed in Relevant Equations. It relates static pressure difference in a fluid column (Pa - Pb) to the height difference of the fluid column (h1 - h2). So the value of Pb - Pa is the pressure differerence due to the velocity difference. And now you can find h1 - h2.
edgepflow: Thanks! It's so much clearer now, I appreciate your quick response. PA-PB=(1/2)ρ(vB2-vA2) PA-PB=ρg(h1-h2) (1/2)ρ(vB2-vA2) = ρg(h1-h2) (h1-h2)=Δh=(vB2-vA2)/(2g) Δh=3.0 m EDIT: I'm assuming my answers to (a) and (b) are correct since you didn't comment on them. Is that right?
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# Optimal strategy for determining unknown permutation given the number of correctly placed elements after each guess
I was scrolling on TikTok the other day when I came across a video similar to this one. They're playing a game with bottles. Initially, $$6$$ bottles, each of a different color, are hidden behind a box and the goal is to guess the correct order of the bottles. The player takes turns guessing the order of the bottles and after each guess it is revealed how many bottles the player got right, i.e. exactly how many bottles are in their correct position.
A question naturally arises - Given $$n$$ bottles, each of a different color out of $$n$$ colors, what is the optimal strategy for this game?
I've tried to answer the question with no success. An alternative question is what is the optimal strategy if we allow for repetition (instead of guessing permutations of $$\{1, 2, ..., n\}$$, we guess $$n$$-tuples from $$\{1, 2, ..., k\}^n$$) - Given $$n$$ bottles, each being one of $$k$$ colors, what is the optimal strategy for this game?
Note: when I say optimal strategy, I mean what is the smallest $$m$$ such that there exists a strategy that guarantees a win in $$m$$ guesses or less.
• I have edited the title to something more descriptive of the game and its mathematical content. Some additional tags may also be helpful but these seem fine for now Commented Aug 4 at 23:44
• You may want to describe "optimal" in more detail. Are you trying to minimise the expected number of guesses or minimise the worst-case number of guesses? (For the similar Mastermind game they give different results.) Commented Aug 4 at 23:47
• The "information-theoretic lower bound" is that you need at least $\frac{\log n!}{\log (n+1)} \approx n - \frac{n}{\log n}$ guesses, which for $n = 6$ is $4$. So it would be natural to try to shoot for a strategy that wins using $n$ or so guesses. Commented Aug 4 at 23:55
• This question seems to be the similar to this one or this one . Maybe some of it is also useful here. Commented Aug 5 at 0:34
• Actually, $\ln n!/\ln n$ is an information theoretic lower bound as there are only $n$ different outcomes of each round: the number of matching bottles cannot be $n-1$. Commented Aug 5 at 4:17
Here is an improved lower bound for the permutation case.
Assuming there are $$n$$ bottles of $$n$$ different colours, and in each round the guess has to be a permutation of the colours, there is a lower bound on the number of rounds as has been pointed out in the comments. Since in each round there are at most $$n$$ different outcomes (since $$n-1$$ correct colours is not possible), the minimum number of rounds is $$\ln n!/\ln n$$. However, this is a weak bound since, at least in the first rounds, the number of correct colours is likely to be small.
Let $$A_k$$ be the number of different $$n$$-permutations with exactly $$k$$ fixed points. This can be expressed as $$A_k=\binom{n}{k}\,!(n-k)$$ where $$!m$$ is the number of fixed point free $$m$$-permutations, also known as derangements.
At the start, there are $$M_0=n!$$ possible permutations. In round $$r$$, we make a guess, and worst case is a result $$k$$ leaving at least $$M_r$$ possible permutations where $$M_r$$ is the smallest number so that $$\sum_{k=0}^n\min(A_k,M_r)\ge M_{r-1}$$.
This gives stronger lower bound, $$R_n$$, on the number of guesses required to get the right answer: ie after $$R_n-1$$ guesses, you know the right answer, and in round $$R_n$$ you provide it.
I found the lower bound on $$R_n$$ by this method to be $$R_n\ge n$$ for $$n=1,\ldots,164$$, and after that it was $$R_n\ge n-1$$ for as far as I tested.
The actual values of $$R_n$$ for low $$n$$ are $$R_1=1$$, $$R_2=2$$, $$R_3=4(?)$$, beyond which I would not solve by hand.
My guess is that $$R_n\approx n$$, perhaps even $$R_n\le n+C$$ for some constant $$C$$.
• How did you get $\sum_{k=0}^n\min(A_k,M_r)\ge M_{r-1}$? Commented Aug 12 at 14:06
• @Natrium: After $r-1$ rounds there are $M_{r-1}$ possible permutations. You then make a guess that splits these into groups depending on the number of fixed points $k$ (ie matching bottles) in you guess. If the biggest of these groups has size $M_r$, that is the worst case: all other groups have at most $M_r$ permutations. However, as there are only $A_k$ permutations with $k$ fixed points, so the maximal size of group $k$ is $\min(A_k,M_r)$. Combined, these groups must contain all $M_{r-1}$ permutations, hence the inequality. Commented Aug 12 at 16:44
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# 129.4 pounds in stones and pounds
## Result
129.4 pounds equals 9 stones and 3.4 pounds
You can also convert 129.4 pounds to stones.
## How to convert 129.4 pounds to stones and pounds?
In order to convert 129.4 pounds to stones and pounds we first need to convert 129.4 pounds into stones.
We know that 1 pound is equal to 1/14 stones, therefore to convert 129.4 pounds to stones we simply multiply 129.4 pounds by 1/14 stones:
129.4 pounds × 1/14 stones = 9.242857 stones
We already know the amount of stones is 9. Now we have to find out the amount of pounds, to do so we take the decimal part of 9.242857 stones and convert it into pounds. In this case we need to convert 0.242857 stones into pounds. To convert 0.242857 stones to pounds we simply multiply 0.242857 stones by 14 pounds.
0.242857 stones × 14 pounds = 3.4 pounds
Finally, we can say that 129.4 pounds in stones and pounds is equivalent to 9 stones and 3.4 pounds:
129.4 pounds = 9 stones and 3.4 pounds
One hundred twenty-nine point four pounds is equal to nine stones and three point four pounds.
## Conversion table
For quick reference purposes, below is the pounds and stones to pounds conversion table:
pounds(lbs) stones(st) pounds(lb)
130.4 pounds 9 stones 4.4 pounds
131.4 pounds 9 stones 5.4 pounds
132.4 pounds 9 stones 6.4 pounds
133.4 pounds 9 stones 7.4 pounds
134.4 pounds 9 stones 8.4 pounds
135.4 pounds 9 stones 9.4 pounds
136.4 pounds 9 stones 10.4 pounds
137.4 pounds 9 stones 11.4 pounds
138.4 pounds 9 stones 12.4 pounds
139.4 pounds 9 stones 13.4 pounds
## Units definitions
The units involved in this conversion are stones and pounds. This is how they are defined:
### Stones
The pound or pound-mass is a unit of mass used in the imperial, United States customary and other systems of measurement. A number of different definitions have been used; the most common today is the international avoirdupois pound, which is legally defined as exactly 0.45359237 kilograms, and which is divided into 16 avoirdupois ounces. The international standard symbol for the avoirdupois pound is lb; an alternative symbol is lbm (for most pound definitions), # (chiefly in the U.S.), and ℔ or ″̶ (specifically for the apothecaries' pound). The unit is descended from the Roman libra (hence the abbreviation "lb"). The English word pound is cognate with, among others, German Pfund, Dutch pond, and Swedish pund. All ultimately derive from a borrowing into Proto-Germanic of the Latin expression lībra pondō ("a pound by weight"), in which the word pondō is the ablative case of the Latin noun pondus ("weight"). Usage of the unqualified term pound reflects the historical conflation of mass and weight.
### Pounds
The stone or stone weight (abbreviation: st.) is an English and imperial unit of mass now equal to 14 pounds (6.35029318 kg). England and other Germanic-speaking countries of northern Europe formerly used various standardised "stones" for trade, with their values ranging from about 5 to 40 local pounds (roughly 3 to 15 kg) depending on the location and objects weighed. The United Kingdom's imperial system adopted the wool stone of 14 pounds in 1835. With the advent of metrication, Europe's various "stones" were superseded by or adapted to the kilogram from the mid-19th century on. The stone continues in customary use in Britain and Ireland used for measuring body weight, but was prohibited for commercial use in the UK by the Weights and Measures Act of 1985.
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Question Video: Finding the Relation between the Unknown Components of Two Parallel Vectors | Nagwa Question Video: Finding the Relation between the Unknown Components of Two Parallel Vectors | Nagwa
# Question Video: Finding the Relation between the Unknown Components of Two Parallel Vectors Mathematics • First Year of Secondary School
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Given that ๐ = โฉ๐ฅ, โ19โช, ๐ = โฉโ19, ๐ฆโช, and ๐โฅ๐, find the relationship between ๐ฅ and ๐ฆ.
03:27
### Video Transcript
Given that ๐ is the vector ๐ฅ, negative 19 and ๐ is the vector negative 19, ๐ฆ and the vector ๐ is parallel to the vector ๐, find the relationship between ๐ฅ and ๐ฆ.
In this question, weโre given two vectors, the vector ๐ and the vector ๐. And in fact, weโre told some information about vectors ๐ and vector ๐. For example, weโre told that ๐ and ๐ are parallel. This is represented by the two vertical lines between our vectors in the question. We need to use all of this information to determine the relationship between ๐ฅ and ๐ฆ, where ๐ฅ and ๐ฆ are given as components of the vectors ๐ and ๐, respectively.
To do this, letโs start by recalling what we mean when we say that two vectors are parallel. We say that two vectors are parallel if they point in the same direction or in exactly opposite directions. For vectors ๐ฎ and ๐ฏ, this is exactly the same as saying that there is some scalar constant ๐ which is not equal to zero such that ๐ฎ is equal to ๐ times ๐ฏ. In other words, we need our vectors to be a nonzero scalar multiple of each other.
So because weโre told that vector ๐ and the vector ๐ are parallel in the question, we know that ๐ is equal to ๐ times ๐ for some scalar constant ๐ not equal to zero. And weโre given the components of ๐ and ๐. So we can write this in our equation. By writing our vectors ๐ and ๐ out component-wise, we have the vector ๐ฅ, negative 19 should be equal to ๐ times the vector negative 19, ๐ฆ.
Now remember, when we multiply a vector by a scalar, we do this component-wise. In other words, we need to multiply every single component of our vector by ๐. Multiplying every component in our vector ๐ by ๐, we get the vector ๐ฅ, negative 19 is equal to the vector negative 19๐, ๐๐ฆ. So for our vectors ๐ and ๐ to be parallel, these two vectors need to be equal.
We can use this to find the value of ๐. Remember, for two vectors to be equal, they must have the same number of components and all of their components have to be equal. So for these two vectors to be equal, their horizontal components must be equal and their vertical components must be equal. Setting these to be equal, we get two equations which must be true: ๐ฅ must be equal to negative 19๐ and negative 19 must be equal to ๐ times ๐ฆ.
We can rearrange both of these equations to solve for ๐. Dividing our first equation through by negative 19, we get ๐ is equal to negative ๐ฅ over 19. And dividing our second equation through by ๐ฆ, we get that ๐ should be equal to negative 19 over ๐ฆ.
And before we continue, there is one thing worth pointing out here. We know our value of ๐ฆ and our value of ๐ฅ cannot be equal to zero. If ๐ฅ was equal to zero, then ๐ would only point in the vertical direction. It would have no horizontal component, so it could not be parallel to vector ๐. Something very similar is true if ๐ฆ was equal to zero. So ๐ฅ and ๐ฆ are both not equal to zero. So we donโt need to worry about dividing through by ๐ฆ.
Now we see we have two equations for our constant ๐. Since both of these are equal to ๐, we can set them equal to each other. In other words, we know that negative ๐ฅ over 19 is equal to ๐ and negative 19 over ๐ฆ is also equal to ๐. So these two are equal. And this is in fact a relationship between ๐ฅ and ๐ฆ. However, we can simplify this even further. We could multiply through by negative ๐ฆ and we could also multiply through by 19. Doing this and simplifying, we get the equation ๐ฅ times ๐ฆ should be equal to 361, which is our final answer.
Therefore, we were able to show if the vector ๐ ๐ฅ, negative 19 and the vector ๐ negative 19, ๐ฆ are parallel, then we must have that ๐ฅ times ๐ฆ is equal to 361.
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##### In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Given:
Radius of circle = 21 cm
Angle subtended by the arc = 60°
Length of arc = × 2πr
= × 2 × × 21 = 22 cm
Hence, the length of the arc is 22 cm.
14
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# Proportions and Similar Figures
## If 𝒂𝒃 = 𝒄𝒅, then The ratio 𝟐𝟒: 𝟑𝟔 𝒄 −𝟏 𝟏 𝟏𝟎
= =
the proportion is: simplified is: 𝟐 𝟒 𝒄+𝟐 𝟓
𝒂 𝒄 𝟏
= 𝟖: 𝟗 𝒄=−
𝒃 𝒅 𝟐
𝒂 𝒄 False 𝟑
= 𝟐: 𝟑 𝒄=−
𝒅 𝒃 𝟐
Two squares are A proportion is
𝟐 𝒄 𝟐 𝟒 always similar. an equation having two
= =
𝟗 𝟏𝟖 𝟐𝒄 𝟏𝟐 ratios ________.
## The similar The corresponding The corresponding
𝟑 𝒄
If = 𝟔 ,then 𝟐𝒄 is: angles of the similar sides of the similar
figures have: 𝟐
figures are: shapes are:
Same shape but
different size
𝟐𝒄 = 𝟏𝟖 Zero
## Different shape 𝟏 𝒂𝒄 = 𝒃𝒅 Proportional
𝟐𝒄 =
but same size 𝟐
𝒂 𝒄
The ratio 𝟎. 𝟑: 𝟏. 𝟐 𝟏 𝒃 If 𝒃
= 𝒅 , then
simplified is: =
𝒂 𝒄
𝟏: 𝟒 𝒂+𝒅 =𝒄+𝒅 𝒂𝒅 = 𝒄𝒅
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# What is the number of integers from 1 to 1000 inclusive that
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What is the number of integers from 1 to 1000 inclusive that [#permalink] 17 Jul 2007, 14:34
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What is the number of integers from 1 to 1000 inclusive that are not divisible by 11 nor by 35?
a) 884
b) 890
c) 892
d) 910
e) 945
Senior Manager
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GK_Gmat wrote:
dahcrap wrote:
I got A
How?
Lets see.
Number of numbers between 1 and 1000 divisible by 11 is 90. Similarly there are 28 numbers divisible by 35. Now remember to consider numbers that are divisible by both 11 and 35.There are 2 such numbers. So total is 1000 - 90 - 28 + 2 = 884
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how do you figure out in a very timely fashion the numbers that are both divisible by 11 and 35??
Intern
Joined: 14 Jul 2007
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The best and easiest way to figure out the numbers divisible by 11 and 35 is to split it into its prime factors
Prime Factors of 11 = 11
Prime Factors of 35 = 7 and 5
Since there is no overlap, the Least Common Multiple is 11 *35 = 385
Now there are 2 such multiples of 385 below 1000 (i.e 385 and 770)
The rest of the reasoning is the same as that provided earlier
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Can some still provide a better way . I am still confused.
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View more editions
# Vector Mechanics for Engineers Dynamics (10th Edition)
• 1706 step-by-step solutions
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 1
7949-B-1P AID: 1825 | 30/11/2013
Show the dimensions of the triangular plate as in Figure (1).
Calculate the height of the triangle (h).
Here, side of the triangle is a.
Express the formula for area of the triangle (A).
Substitute for h.
Calculate the distance between point E and centroid C.
Substitute for h.
Express the volume of triangle (V).
Here, thickness of triangle plate is t.
Calculate the mass of triangular plate (m).
Here, density of triangular plate is.
Substitute At for V.
Express the formula for mass moment of inertia.
Here, moment of inertia is I.
a)
Calculate the moment of inertia of triangular plate with respect to axis .
Substitute for h.
Now, calculate the mass moment of inertia of triangular plate with respect to axis .
Substitute for , for , and for A.
Hence, the mass moment of inertia of triangular plate with respect to axis is.
Calculate the moment of inertia of triangular plate with respect to axis .
Substitute for h.
Now, calculate the mass moment of inertia of triangular plate with respect to axis .
Substitute for , for , and for A.
Hence, the mass moment of inertia of triangular plate with respect to axis is.
b)
Express the formula for mass moment of inertia of triangular plate with respect to axis .
Substitute for both and .
Hence, the mass moment of inertia of triangular plate with respect to axis is.
Corresponding Textbook
Vector Mechanics for Engineers Dynamics | 10th Edition
9780077402327ISBN-13: 0077402324ISBN:
Alternate ISBN: 9780077531348, 9780077573263, 9780077889715
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Are projective measurement bases always orthonormal?
Are projective measurement bases always orthonormal?
Yes.
Remember that you require several properties of a projective measurement including $$P_i^2=P_i$$ for each projector, and $$\sum_iP_i=\mathbb{I}.$$ The first of these show you that the $$P_i$$ have eigenvalues 0 and 1. Now take a $$|\phi\rangle$$ that is an eigenvector of eigenvalue 1 of a particular projector $$P_i$$. Use this in the identity relation: $$\left(\sum_jP_j\right)|\phi\rangle=\mathbb{I}|\phi\rangle$$ Clearly, this simplifies to $$|\phi\rangle+\sum_{j\neq i}P_j|\phi\rangle=|\phi\rangle.$$ Hence, $$\sum_{j\neq i}P_j|\phi\rangle=0.$$ The $$P_j$$ are all non-negative, so the only way that this can be 0 is if $$P_j|\phi\rangle=0$$ for all $$j\neq i$$. (To expand upon this, assume there's a $$P_k$$ such that $$P_k|\phi\rangle=|\psi\rangle\neq 0$$. This means that $$\sum_{j\neq i,k}\langle\psi|P_j|\phi\rangle=-\langle\psi|P_k|\phi\rangle,$$ so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)
Here is another way to see this.
A projection $$P$$ is an operator such that $$P^2=P$$.
This directly implies that we can attach to each projector $$P$$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $$P_i$$ has trace $$\operatorname{tr}(P_i)=n$$, then we can represent $$P_i$$ as a set of orthonormal states $$\{\lvert\psi_{i,j}\rangle\}_{j=1}^n$$. Note in particular that if $$\operatorname{tr}(P_i)=1$$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.
The projector $$P_i$$ and the corresponding states are connected through $$P_i=\sum_{j=1}^n \lvert\psi_{ij}\rangle\!\langle \psi_{ij}\rvert.$$ In the simpler case of $$\operatorname{tr}(P_i)=1$$ this reads $$P_i=\lvert\psi_i\rangle\!\langle\psi_i\rvert$$.
Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state. This condition is expressed mathematically by requiring $$\sum_i P_i=I,$$ which in terms of the associated ket states reads $$\sum_{ij}\lvert\psi_{ij}\rangle\!\langle\psi_{ij}\rvert=I,$$ which is the completeness relation for the vectors $$\{\lvert\psi_{ij}\rangle\}_{ij}$$. This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $$\lvert\psi_{ij}\rangle$$).
Orthogonality of $$P_i$$ is equivalent to orthogonality of the corresponding $$\lvert\psi_{ij}\rangle$$, thus the conclusion.
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# prime numbers and composite numbers
This category only includes cookies that ensures basic functionalities and security features of the website. A composite number has more than two factors. All Rights Reserved.
It is a composite numbers. 24 / 1 = 24. But opting out of some of these cookies may have an effect on your browsing experience.
51 is composite.
3.Smallest composite number is also 2. Solution: The factors of 2 are 1 x 2.
12 is a composite number because it can be divided by 1,2,3,4,6 and 12.
For example: The number 28 has six factors which are 1, 2, 4, 7, 14, and 28. The number 113 is odd, doesn’t end in 5 or 0, and has a digital root of 5, so it’s not divisible by 2, 5, or 3. Example 3: Find all prime numbers between 2 and 9. The next number is 6, but since we already have 6 as a factor of 24, we are finished calculating the factors of 24. It is mandatory to procure user consent prior to running these cookies on your website.
Prime and composite numbers worksheets have a variety pdf exercises to understand recognize prime and composite numbers. A Prime number is a whole number that has two factors, one and itself, A prime number can be divided, without a remainder, only by itself and by 1. Both 1 and 24 are its factors.
So, the number ‘12’ has 6 factors. Example 4: Find all prime numbers between 10 and 19. Required fields are marked *. That is, composite numbers can be divided by 1, themselves, and some other numbers also. The next three odd numbers are prime — 3, 5, and 7.
When the area of a rectangle is a composite number, there are two or more sets of possible dimensions for that rectangle.
The table shows a whole bunch of composite numbers up to 100, but there’s a problem.
A composite number has more than two factors, which means apart from getting divided by number 1 and itself, it can also be divided by at least one integer or number. Directions: Read each question below. In this lesson, the details about prime numbers and composite numbers are given in detail along with examples and chart. The number 23 isn’t even, doesn’t end in 5 or 0, has a digital root of 5, and isn’t a multiple of 7. b.
A composite number is a positive integer, greater than 1, that has factors other than just 1 and itself.
The number 1 is neither prime nor composite.
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We have determined if a single number is prime or composite. The whole-number dimensions, 1, 2, 4 and 8, of the rectangular gardens in Problem 2, are the factors of the number 8. Thank you, and keep up the great work here. Following are the answers to the practice questions: a. Notice how prime numbers only have 1 and itself as its factors?
a. It only has the factors 1 and 11.
Take a look at our chart for more examples. Necessary cookies are absolutely essential for the website to function properly. In today's post, we're going to learn how to find prime numbers using the Sieve of Eratosthenes.
Here also, we have infinite composite numbers.
A prime number has exactly two factors — 1 and the number itself. The number 12 can be written as the 1 x 12, 3 x 4, and 2 x 6.
Solution: a) 43 is only divisible by 1 and 43. Since there is a remainder, 5 is not a factor of 24. Most people are able to quickly identify the prime numbers up to 50.
From the given chart you can see the smallest prime number is 2 and the smallest composite number is 4. 169 is composite. If you make a purchase on one of these sites, I may receive a small commission at no cost to you.
Solution: The prime numbers between 10 and 19 are 11, 13, 17 and 19. Since 12 is divisible by numbers other than 1 and itself, 12 is a composite number. It isn’t divisible by 7, because 263 / 7 = 37 r 2.
Not ready to subscribe? +1 – 8 + 7 = 0, so 187 is a multiple of 11. d. 283 is prime. Solution: The prime numbers between 20 and 29 are 23 and 29. What is a Prime Number?
They have “factors.” Factors are the numbers that divide into it.
Figure out which of the following are prime numbers and which are composite numbers: a.
In this lesson, the details about prime numbers and composite numbers are given in detail along with examples and chart. 3 is prime. Of the following numbers, tell which are prime and which are composite. The number 1 in natural numbers is an exception as this number cannot be categorized as a prime or composite number.
This realization leads us into our A prime number is a positive integer, greater than 1, whose only factors are 1 and itself.. Composite Numbers. This is because the number 31 has only two factors: 1 and 31. +1 – 4 + 3 = 0, so 143 is divisible by 11. b. If a number less than 289 isn’t divisible by 2, 3, 5, 7, 11, or 13, it’s prime; otherwise, it’s composite.
Prime Numbers or Composite Numbers | Worksheet #2.
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Because 65 ends in 5, it’s divisible by 5. b.
Eleven’s only factors are 1 and 11. d. 14 is composite.
The next prime number is number 3. 1 and the number itself.
To determine if a number is prime or composite, follow these steps: Example 1: Is the number 2 prime or composite? What is a Prime Number?
is all very important.
Students can also practice by creating charts for prime and composite numbers from 1 to 1000. Feedback to your answer is provided in the RESULTS BOX. But if we consider another number say, 21, whose factors are 1, 3 and 7. ‘1’ and number itself. The next prime number is 7. For both the numbers, we can see, the common factor is 1.
Summary: 1.Prime numbers have 1 and themselves as their factor while composite numbers can have more factors than 1 and themselves. Hence it is prime number. It’s also not a multiple of 7: 113 / 7 = 16 r 1. When testing to see whether a number is prime or composite, perform divisibility tests in the following order (from easiest to hardest): 2, 5, 3, 11, 7, and 13. Apart from these two, there is also a similar category of number which are coprime numbers. In today’s entry, we’re going to talk about length, width, and height as tools to find the dimensions of an object. Factors are the numbers you multiply together to getanother number: Example1: 1 and 5 are factors of 5 1x5=5example2: 3 and 4 are factors of 12 3x4=12, Note: 1 is not prime or composite. 2.Smallest prime number is 2. 185 is composite.
Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. The prime numbers between 2 and 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 since each of these numbers has only two factors, itself and 1. They are numbers (greater than 1) that are not prime. Composite Number Definition. It isn’t divisible by 11, because it fails the + and – test (+2 – 6 + 3 = –1, which isn’t 0 or divisible by 11). Solution 3: I divided the number 31 by all numbers between 1 and 31 and found no factors other than one and thirty-one.
definitions of prime and composite numbers. Here’s how you know which tests to perform: If a number less than 121 isn’t divisible by 2, 3, 5, or 7, it’s prime; otherwise, it’s composite. The following are
We don’t consider ‘1’ as a composite number.
c. 91 is composite. 3 is a prime number because 3 can be divided by only two number’s i.e.
And it isn’t divisible by 13, because 263 / 13 = 20 r 3. Example 1: 6 can be divided evenly by 2, so 6 is a composite number. Solution : To find the primality of 5 and 6, first we need to find the factors of the numbers 5 = 1x5 6 = 1x6, 2x3 , 3x2 5 has only 1 and 5 as its factors. When the area of a rectangle is a prime number, there is only one set of possible dimensions for that rectangle.
Solution 2: Thirty-one is a prime number. Therefore, 14 and 15 are coprime numbers. Make sure that you have a clear understanding of Number 2 is a prime number but all multiples of 2 will be composite numbers because they are divisible by 2. The number 263 ends in an odd number, so it isn’t divisible by 2.
No — unfortunately, it doesn’t. Then, 21 is neither a coprime for 14 nor for 15.
As you've begun your study of factors, These two types of numbers are prime numbers and composite numbers.
The coprime-numbers or mutually primes or relatively primes are the two numbers which have only one common factor, which is 1.
Prime and composite number worksheets use stimulating concepts to get your child thinking. The first six prime numbers are 2, 3, 5, 7, 11, and 13. Get access to hundreds of video examples and practice problems with your subscription! I think it reinforces the concept that ‘factors’ are always integers.
The prime numbers between 2 and 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 since each of these numbers has only two factors, itself and 1. All the numbers other than prime numbers, except 1, are composite numbers because they have more than two factors.
A Prime number is a wholenumber that has two factors, one and itself,A prime number can be divided, without a remainder, only by itself and by 1.Example: 11 is a prime number because the only numbers it can be divided byevenly is 1 and 11What is Composite Number: Composite numbers hasmore than two factors. Identifying the Greatest Common Factor (GCF). About Us | Contact Us | Advertise With Us | Facebook | Recommend This Page. These numbers are also called composites.. All the natural numbers which are not prime numbers are composite numbers as they can be divided by more than two numbers. It doesn’t end in 5 or 0, so it isn’t divisible by 5.
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As you have understood already about the composite number, now let us know the smallest number which is composite in nature. What are Variables in Coding?
Robby’s New Laboratory.
The number 1 is neither prime nor composite.
you've probably realized that some numbers have a lot of factors and The number 7 is prime.
send us a message to give us more detail!
The number 91 is a multiple of 7: 7 x 13 = 91. d. 113 is prime. A composite number is a whole number that can be divided evenly by numbers other than 1 or itself. You do not need to learn them by heart, but if you’d like, remember the smaller ones, like 2, 3, 5, 7, 11, and 13: A factor of a number is a value that divides evenly into the number—that is, with no remainder.
The next prime number is 11. Select your answer by clicking on its button. The number 73 isn’t even, doesn’t end in 5 or 0, and isn’t a multiple of 7. c. 111 is composite. Also amusing display charts which list the prime and composite numbers from 1 to 100 and extremely engaging activities like coloring, cutting, pasting and mazes are here for your children in grade 4 through grade 7. Each set of dimensions is a pair of factors. The number 283 is odd, doesn’t end in 5 or 0, and has a digital root of 4; therefore, it’s not divisible by 2, 5, or 3. A prime number is the one which has exactly two factors, which means, it can be divided by only “1” and itself.
If you find that a number is divisible by one of these, you know that it’s composite and you don’t have to perform the remaining tests. Example 5: Find all prime numbers between 20 and 29.
Since 12 is divisible by numbers other than 1 and itself, 12 is a composite number. Prime Composite Numbers.
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# Second order differential equation
• Linder88
In summary: The orthogonality conditions are \int_{-L/2}^{L/2} \cos(k_n x)\cos(k_m x)\,dx = L/2 \delta_{nm} and similar for the sines.
#### Linder88
Homework Statement
I need help with finding the approximate solution of a second order differential equation.
Relevant Equations
I am given the differential equation as in
$$\lambda \frac{d^2u}{dx^2} + q = 0$$
which is defined over $$x \in [-L/2,L/2]$$ and where $$q = a+bx$$. I am supposed to find an approximative solution using a spectral method, i.e. using cosine and sine terms that fulfills the essential boundary conditions given by
$$u \bigg ( -\frac{L}{2} \bigg ) = 0, u \bigg (\frac{L}{2} \bigg ) = 0$$
The appoximative solution should be represented by a Fourier series and by noticing that the terms in the integral are orthogonal, the matrix of the system of equations should be diagonal.
We choose an approximative solution given by
$$u_N(x) = \frac{a_0}{2} + \sum_{n=1}^N a_n \cos nx + b_n \sin nx$$
Comparing this approximative solution with the differential equation yields that
$$\frac{a_0}{2} = a$$
and the boundary conditions yields the equation system
$$a + \sum_{n=1}^N a_n \cos \bigg ( \frac{nL}{2} \bigg ) + b_n \sin \bigg ( \frac{nL}{2} \bigg ) = 0 \\ a + \sum_{n=1}^N a_n \cos \bigg ( -\frac{nL}{2} \bigg ) + b_n \sin \bigg ( -\frac{nL}{2} \bigg ) = 0$$
So at least I got a system of two equations but I do not know where to go from here. How should I use the orthogonality to set up a matrix from these two equations?
Linder88 said:
Comparing this approximative solution with the differential equation yields that
##{a_0\over 2}=a##
How ?
and the boundary conditions yields the equation system
I miss the ## +bx## from
$$\lambda \frac{d^2u}{dx^2} + a + bx = 0$$
(and I find the use of ##a_n## and ##b_n## in your notation confusing, since ##a## and ##b## are allready used).
Linder88 said:
Homework Statement:: I need help with finding the approximate solution of a second order differential equation.
Relevant Equations:: I am given the differential equation as in
$$\lambda \frac{d^2u}{dx^2} + q = 0$$
which is defined over $$x \in [-L/2,L/2]$$ and where $$q = a+bx$$.
I am supposed to find an approximative solution using a spectral method, i.e. using cosine and sine terms that fulfills the essential boundary conditions given by
$$u \bigg ( -\frac{L}{2} \bigg ) = 0, u \bigg (\frac{L}{2} \bigg ) = 0$$
The appoximative solution should be represented by a Fourier series and by noticing that the terms in the integral are orthogonal, the matrix of the system of equations should be diagonal.
We choose an approximative solution given by
$$u_N(x) = \frac{a_0}{2} + \sum_{n=1}^N a_n \cos nx + b_n \sin nx$$
Comparing this approximative solution with the differential equation yields that
$$\frac{a_0}{2} = a$$
No. It's the second derivative of $u$ that is equal to $\frac{-q(x)}{\lambda}$, not $u$ itself. When you differentiate the constant term you get zero, and in any case a non-zero constant does not satisfy the boundary condition.
and the boundary conditions yields the equation system
$$a + \sum_{n=1}^N a_n \cos \bigg ( \frac{nL}{2} \bigg ) + b_n \sin \bigg ( \frac{nL}{2} \bigg ) = 0 \\ a + \sum_{n=1}^N a_n \cos \bigg ( -\frac{nL}{2} \bigg ) + b_n \sin \bigg ( -\frac{nL}{2} \bigg ) = 0$$
So at least I got a system of two equations but I do not know where to go from here. How should I use the orthogonality to set up a matrix from these two equations?
You have $u(x) \approx \sum_n u_n\phi_n(x)$ where $$\phi_n(-L/2) = \phi_n(L/2) = 0.$$ Therefore $$\lambda \sum_n u_n \phi_n''(x) \approx -q(x)$$ and we determine the $u_n$ by requiring that the inner product with $\phi_m$ should be exact, ie. $$\lambda \sum_n u_n \int_{-L/2}^{L/2} \phi_n''(x) \phi_m(x)\,dx = -\int_{-L/2}^{L/2} q(x)\phi_m(x)\,dx.$$ The matrix you are looking for is the matrix of coefficients $$M_{mn} = \int_{-L/2}^{L/2} \phi_n''(x) \phi_m(x)\,dx.$$
Now the choice of a Fourier basis means that $$\phi_n(x) = A_n\cos(k_n x) + B_n\sin(k_n x)$$ and $k_n$ is such that the boundary conditions are satisfied by something other than $A_n = B_n = 0$.
Last edited:
etotheipi
## What is a second order differential equation?
A second order differential equation is a mathematical equation that involves the second derivative of a function. It is used to model various physical phenomena, such as motion, heat transfer, and electrical circuits.
## What is the general form of a second order differential equation?
The general form of a second order differential equation is: y'' = f(x, y, y'), where y'' is the second derivative of y with respect to x, and f(x, y, y') is a function of x, y, and the first derivative of y with respect to x, y'.
## What is the difference between a homogeneous and non-homogeneous second order differential equation?
A homogeneous second order differential equation is one where the right-hand side of the equation is equal to zero, while a non-homogeneous second order differential equation has a non-zero right-hand side. Homogeneous equations have simpler solutions compared to non-homogeneous equations.
## What is the role of initial conditions in solving a second order differential equation?
Initial conditions are necessary in solving a second order differential equation because they provide the values of the function and its first derivative at a specific point. These conditions are used to find the particular solution to the equation.
## What methods can be used to solve a second order differential equation?
There are several methods for solving a second order differential equation, including separation of variables, substitution, and the method of undetermined coefficients. Other methods include Laplace transforms, power series, and numerical methods such as Euler's method and the Runge-Kutta method.
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# Round to Nearest Multiple (Up or Down or Both)
Welcome to our blog post on rounding to the nearest multiple in Excel. In this tutorial, we will explore three different functions that can assist you in rounding numbers up or down to the nearest multiple: FLOOR, CEILING, and MROUND. Whether you need to calculate the number of product bundles or adjust sales targets, these functions will come in handy. So let’s dive right in and learn how to use them!
## Walkthrough
In this tutorial, we will learn how to:
1. Use the FLOOR function to calculate the number of product bundles needed based on order quantities.
2. Utilize the CEILING function to determine the number of bundles required to cover an entire order.
3. Apply the MROUND function to round sales targets up or down to the nearest five.
Let’s take it one exercise at a time.
### Exercise 1: Rounding Down to the Nearest Multiple
Let’s begin by working on the task of calculating the number of product bundles needed based on order quantities. For this example, let’s assume that products are only sold (a) individually or (b) in discounted five-pack bundles. We have received a bunch of orders. The order quantities are shown below:
Since the FLOOR functions rounds down to the nearest multiple, we can use it to round the order quantity down to the nearest multiple of 5. Since we sell 5-pack bundles, this will tell us how many items we can deliver with 5-pack bundles.
`=FLOOR(B13,5)`
• B13 is the number to round
• 5 is the significance (multiple)
• Note: Depending on your Excel version, you may have FLOOR or FLOOR.MATH. Use FLOOR.MATH if available, otherwise use FLOOR.
Drag the formula down the column to apply it to all order quantities.
Now we see the order quantities that can be filled with the 5-pack bundles. Computing the number of singles needed to fulfill the order is straightforward. We use a subtraction formula like this:
`=B13-C13`
We fill that formula down:
Finally, we need to convert the FLOOR column quantites to the actual number of 5-packs. To do so, we can use a basic division formula like this:
`=C13/5`
Fill the formula down, and bam:
Now we can see how many bundles and how many singles are needed to fulfill the order quantity!
But, what if we only sold 5-pack bundles? How many bundles would be needed to fill the entire order? Well, that leads us to our next exercise.
### Excel Exercise 2: Rounding Up to the Nearest Multiple
For this exercise, let’s say we only sell the product in five-pack bundles, and we need to calculate the number of bundles required to cover the entire order quantity.
Same order quantities as before:
This time, we’ll use the CEILING function to round the order quantity up to the nearest 5. This will tell us the quantity needed to cover the order. So, we write the following:
`=CEILING(B13, 5)`
• B13 is the quantity
• 5 is the significance (the multiple)
• Note: if you have a version of Excel with the CEILING.MATH function, use it; otherwise, use the legacy CEILING function.
Now, we just need to convert the CEILING quantity to the actual number of 5-pack bundles. We can do that with the following:
`=C13/5`
Fill it down, and got it:
So, we know how to force Excel to round up with CEILING, and we know how to force it to round down with FLOOR, but what if we wanted it to round up or down as needed? Well, that takes us to the next exercise.
### Exercise 3: Rounding to the Nearest Multiple with MROUND
In this exercise, we want to round sales targets to the nearest multiple of five, based on a percent increase.
The prior quarter sales and growth target % are provided:
First, let’s compute the new sales target with the following formula:
`=B12*(1+C12)`
This formula multiplies the sales q by 110% and the results are shown below:
But, rather than use this result as the target, let’s clean it up by having Excel round it up or down to the nearest multiple of 5. For this, we can use the MROUND function as follows:
`=MROUND(D12,5)`
• D12 is the number to round
• 5 is the multiple
Fill it down, and got it:
Yay … we did it!
## Conclusion
Congratulations! You have learned how to round numbers to the nearest multiple in Excel by utilizing the FLOOR, CEILING, and MROUND functions. Whether you need to calculate product bundles or adjust sales targets, these functions provide a simple and efficient solution. If you have any improvements, questions, suggestions, or alternatives … please post a comment!
## FAQs
Q: What if I have an older version of Excel that does not have the newer functions (CEILING.MATH or FLOOR.MATH)?
A: Don’t worry! You can still use the Legacy CEILING and FLOOR functions, which serve the same purpose.
Q: Can I change the multiple values in the exercise examples?
A: Absolutely! You can adjust the multiple values to match your specific requirements. For example, instead of rounding to the nearest five, you can round to the nearest ten or any other desired multiple.
Q: How can I round to a specific decimal place instead of a whole number multiple?
A: To round to a specific decimal place, you can use the ROUNDUP, ROUNDDOWN, or ROUND functions instead of CEILING, FLOOR, or MROUND. Simply adjust the formula to include the desired number of decimal places.
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# Find the angle between vector A = i + j - 2k and vector B = i + j - k
8 views
in Physics
closed
Find the angle between $\vec A=\hat i+\hat j-2\hat k \,and\, \vec B=\hat i+\hat j-\hat k$
+1 vote
by (9.6k points)
selected by
We have
$\vec A.\vec B$= | A | | B | cos θ
Here | A | = $\sqrt{1^2+1^2+(-2)^2}=\sqrt6$
| B | =$\sqrt{1^2+1^2+(-1)^2}=\sqrt3$
$\vec A.\vec B=(\hat i+\hat j-\hat k).(\hat i+\hat j-\hat k)$
= 1 × 1 + 1 × 1 + (-2)(-1)
= 1 + 1 + 2 = 4
∴ cos θ = $\frac{\vec A.\vec B}{|A||B|}=\frac{4}{\sqrt6\sqrt3}$
cos θ = $\frac{4}{\sqrt{18}}$
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+0
# Algebra 2
0
82
1
Given V = 2πr2h, solve for h
A)h = V/2r2
B)h = V2πr2
C)h = V - 2πr2
D)h = V/2πr2
f(x) = x2 - 5
For the function shown, what is the range of the function when the domain is {2, 4, 7}?
A){-1, 44}
B){-1, -5}
C){6, 21, 54}
D){-1, 11, 44}
Guest Aug 15, 2017
Sort:
#1
+1364
+1
1)
When the question asks to solve for in the equation of $$V=2\pi r^2h$$, it means get h, alone, on one side of the equation. This only requires one step, in this case:
$$V=2\pi r^2h$$ Divide by $$2\pi r^2$$ on both sides of the equation. $$h=\frac{V}{2\pi r^2}$$ We're done! h is by itself.
The appropriate answer choice, therefore, is D
2) If the domain is only {2, 4, 7}, just plug them in the original equation of $$f(x)=x^2-5$$.
$$f(2)=2^2-5=-1$$ $$f(4)=4^2-5=11$$ $$f(7)=7^2-5=44$$
Now, find the answer choice that has all three of these in its domain. That would be D again.
TheXSquaredFactor Aug 15, 2017
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# Cross five
The figure on the picture is composed of the same squares and has a content of 45cm². What's its perimeter?
o = 36 cm
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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# Ticknor's Mensuration, Or, Square and Triangle: Being a Practical and Concise System of Geometry and Mensuration : Adapted to the Use of Schools and Academies in the American Republic
B. Bannan, 1849 - Measurement - 144 pages
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### Popular passages
Page 13 - ... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained...
Page 80 - To find the solidity of a cylinder. RULE. — Multiply the area of the base by the altitude, and the product will be the solidity.
Page 28 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight hue BC. In BC take any point D, and join AD; and at the point A, in the straight line AD, make (I.
Page 63 - To find the circumference of a circle when the diameter is given, or the diameter when the circumference is given. RULE. — Multiply the diameter by 3.1416...
Page 29 - BD the two triangles ABD, BDC, have all the sides of the one equal to the corresponding...
Page 32 - The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the other two sides.
Page 68 - From eight times the chord of half the arc, subtract the chord of the whole arc, and divide the remainder by 3, and the quotient will be the length of the arc, nearly.
Page 22 - If equals be added to equals, the sums will be equal. 3. If equals be taken from equals, the remainders will be equal. 4. If equals be added to unequals, the sums will be unequal.
Page 89 - To find the solidity of a sphere or globe.. RULE.* "... Multiply the cube of the diameter by .5236, and the product will be the solidity.
Page 10 - RULE. Divide as in whole numbers, and from the right hand of the quotient point off as many places for decimals as the decimal places in the dividend exceed those in the divisor.
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# Solving of Equations with Method of Introducing New Variable
It is easier to explain essence of the method of introducing new variable on example.
Example 1. Solve equation (x^2-3x)^2+3(x^2-3x)-28=0.
Let x^2-3x=y, then initial equation can be rewritten as y^2+3y-28=0. This equation has two roots: y_1=-7,y_2=4.
Thus, we obtained set of two equations: x^2-3x=-7,x^2-3x=4, i.e. x^2-3x+7=0,x^2-3x-4=0.
First equation doesn't have roots because its discriminant D=(-3)^2-4*1*7=-19<0. Second equation has roots x=4 and x=1. These roots are also roots of initial equation.
Example 2. Solve equation 24/(x^2+2x-8)-15/(x^2+2x-3)=2.
Let x^2+2x=y, then equation can be rewritten as (24)/(y-8)-(15)/(y-3)=2.
Common denominator is (y-8)(y-3). So, (24(y-3))/((y-8)(y-3))-(15(y-8))/((y-3)(y-8))=(2(y-3)(y-8))/((y-3)(y-8)). This can be rewritten as (24(y-3)-15(y-8)-2(y-3)(y-8))/((y-8)(y-3))=0.
Domain of this equation is all y, except y=3 and y=8.
Now, fraction equals zero, when numerator equals zero: 24(y-3)-15(y-8)-2(y-3)(y-8)=0 or y(2y-31)=0. Therefore, either y=0 or y=31/2. Both of these roots are in domain of the equation.
But y=x^2+2x, so we obtain set of equations: x^2+2x=0,x^2+2x=31/2.
First equation has solutions 0 and -2. Second equation has solutions x=-1+-sqrt(33/2).
All these solutions are solutions of initial equation: 0,-2,-1+sqrt(33/2),-1-sqrt(33/2).
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Custom math worksheets at your fingertips
# Details for problem "Slope triangle to be drawn for a given line"
Quickname: 8003
Elementary School, Primary School, Junior High School, Middle School, High School.
## Summary
In a coordinate system with a given straight line a slope triangle is to be drawn.
## Description
A straight line is given in a coordinate system. Add a slope triangle to this drawing. The task can be extended by the labeling of the gradient triangle, and also by the derivation of the slope.
The straight line equation and the dimensions of the coordinate system are designed in such a way that a gradient triangle can be drawn without any particular difficulty. The intersection of the line and the Y-axis can be chosen to lie on only full or also half coordinate plane squares (vertical divisions).
With respect to the coordinate system, it can be set whether only quadrant one is to be covered with positive X and Y coordinates, or also the quadrants containing coordinates with negative signs. The coordinate system can be plotted in three different sizes.
The slope is chosen randomly, but can also be restricted to be positive or negative. As an additional task a labeling of the gradient triangle can be requested, either with
• "dx"="run" and "dy"="rise" or
• with the actual lengths of dx and dy
can be requested.
If values are required, the derivation of the gradient via the corresponding fraction can also be requested.
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## Customization options for this problem
Parameter
Possible values
Number of problems
1, 2
Yes, No
Slope
pos & neg, positive, negative
Sub-square rises
Yes, No
Size
small, medium, large
none, actual lengths, run=dx/rise=dy, derive values,slope
Remark
Description
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https://www.coursehero.com/file/6775242/HW3/
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# HW3 - ECE 302 Homework#3 due date...
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Unformatted text preview: ECE 302, Homework #3, due date: 2/2/2011 http://cobweb.ecn.purdue.edu/ ∼ chihw/11ECE302S/11ECE302S.html Review of Calculus: Question 1: Compute the following integrals. Z 2 π a cos( ωt + θ ) dθ Z 2 π a cos( ωt + θ ) da Question 2: Define a 2-D function f ( x,y ) as follows. f ( x,y ) = ( x/y 2 if y ∈ [1 , 2] and x ∈ [0 ,y ] otherwise Compute the values of the following 2-dimensional integrals. Z 4 / 3 y =2 / 3 Z 3 / 2 x =1 / 2 f ( x,y ) dxdy Z 4 / 3 y =2 / 3 Z ∞ x =-∞ f ( x,y ) dxdy Z ∞ y =-∞ Z 3 / 2 x =1 / 2 f ( x,y ) dxdy. Question 3: Define a 1-D function f X ( x ) as follows. f X ( x ) = x if x ∈ [0 , 1] 1 2 if x ∈ (1 , 2] otherwise Another function F ( x ) can be defined based on the integral of f X ( x ) as follows: F ( x ) = Z x s =-∞ f X ( s ) ds. 1. Find the expression of F ( x ) for the case of x < 0. 2. Find the expression of F ( x ) for the case of x ∈ [0 , 1]. 3. Find the expression of F ( x ) for the case of x ∈ (1 , 2]. 4. Find the expression of F ( x ) for the case of x > 2. 5. Write down the complete expression of F ( x ) by considering the above four different cases. Your answer is simply a piecewise function that considers four different cases. Question 4: [Basic] Problem 2.62. Please assume the die is fair. Question 5: [Basic] Consider a best-of-three series between teams A and B. The condi- tional distributions are as follows. P ( A wins the next game | B is leading in the series ) = 0 . 7 P ( A wins the next game | A and B are tied in the series ) = 0 . 5 P ( A wins the next game | A is leading in the series ) = 0 . 4 1. Construct the weight assignment for the sample space1....
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## This note was uploaded on 02/12/2012 for the course ECE 302 taught by Professor Gelfand during the Spring '08 term at Purdue.
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# Linear Motion under a variable force
### Linear Motion under a Variable Force
Essentials of Linear Motion under a Variable Force
• The force acting on an object is variable, rather than constant as in basic mechanics. When force changes over time, it alters object’s velocity and position.
• The force is often expressed as a function of time, position, or velocity. This complicates the force-equation but also adds greater detail and flexibility to the problem.
• Newton’s Second Law of Motion still applies: F = ma. Here however, a, the acceleration, is the derivative of the velocity which again is the derivative of position. Since the force is also changing, these properties interlock in new ways which require differential equations to solve.
Understanding Variable Forces
• Drag Forces: These act in the opposite direction to the motion of the particle. They can depend on speed, the square of speed, or be constant.
• Spring Forces: According to Hooke’s law, these are proportional to the extension or compression of a spring. The force is given as F = -kx where k is the spring constant and x the position.
• Gravitational Forces: Act vertically downwards. These can become variable in problems involving extremely great heights (as in problems involving spacecraft), where the gravitational field strength diminishes with altitude.
Key Equations
• The definition of force and acceleration holds firm: F = ma. Both sides of this equation can be variable.
• For drag forces: F = -kv or F = -k(v^2) where k is the drag constant, and v is velocity.
• For spring forces: F = -kx where x is the displacement from the rest or equilibrium position and k is the spring constant.
• The gravitational force is F = mg where m is mass and g is the gravitational field strength. However, for large altitudes (as in space mechanics) it becomes F = GMm/r² with M the mass of Earth, m the mass of the particle, r the distance from the centre of Earth, and G the gravitational constant.
Analyzing Problems with Variable Forces
• You generally need calculus to solve these problems, specifically differential equations. These allow for the study of how one quantity changes in relation to another, opening up the mechanics to variables changing over time.
• The wording in the problem can often indicate which type of force you’re being asked to consider. Look for words like “drag”, “resistance”, “spring”, or “gravity”.
• Sketching force diagrams is still very useful. Being able to visualize how the forces at work often leads to clearer understanding of the problem.
• The Analytic approach requires step-by-step integration, and often gives a solution in terms of an indefinite integral.
• The Numeric approach uses numerical approximations, particularly useful where an analytic approach is challenging or impossible. Examples include problems involving air resistance proportional to velocity squared.
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# Indeterminate system
An indeterminate system is a system of simultaneous equations (especially linear equations) which has more than one solution. The system may be said to be underspecified. If the system is linear, then the presence of more than one solution implies that there are an infinite number of solutions, but that property does not extend to nonlinear systems.
An indeterminate system is consistent, the latter implying that there exists at least one solution. For a system of linear equations, the number of equations in an indeterminate system could be the same as the number of unknowns, less than the number of unknowns (an underdetermined system), or greater than the number of unknowns (an overdetermined system). Conversely, any of those three cases may or may not be indeterminate.
## Examples
The following examples of indeterminate systems have respectively fewer, the same, and more equations than unknowns:
$x+y=2$
$x+y=2, \,\,\,\,\, 2x+2y=4$
$x+y=2, \,\,\,\,\, 2x+2y=4, \,\,\,\,\, 3x+3y=6$
## Conditions giving rise to indeterminacy
In linear systems, indeterminacy occurs if and only if the number of independent equations (the rank of the augmented matrix of the system) is less than the number of unknowns and is the same as the rank of the coefficient matrix. For if there are at least as many independent equations as unknowns that will eliminate any stretches of overlap of the equations' surfaces in the geometric space of the unknowns (aside from possibly a single point); and if the rank of the augmented matrix exceeds (necessarily by one if at all) the rank of the coefficient matrix then the equations jointly contradict each other.
## Finding the solution set of an indeterminate linear system
Let the system of equations be written in matrix form as
$Ax=b$
where A is the coefficient matrix, x is the vector of unknowns, and b is a vector of constants. Then if the system is indeterminate, the infinite solution set is the set of all x vectors generated by
$x=A^+b + [I-A^+A]w$
where $A^+$ is the Moore-Penrose pseudoinverse of A and w is any appropriately dimensioned vector.
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https://byjus.com/question-answer/suppose-you-have-fuses-of-rating-5a-and-15a-what-will-be-the-rating-of/
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Question
# suppose you have fuses of rating 5A and 15A .what will be the rating of the combination if they are connected 1)in series 2)in parallel
Solution
## According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated using the equation V = IR, where I equals the current in amps (A) and R is the resistance in ohms (Ω). Another way to think of this is that V is the voltage necessary to make a current I flow through a resistance R. So the voltage drop across R1 is V1 = IR1, that across R2 is V2 = IR2, and that across R3 is V3 = IR3. The sum of these voltages equals the voltage output of the source; that is, V = V1 + V2 + V3. This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation PE = qV, where q is the electric charge and V is the voltage. Thus the energy supplied by the source is qV, while that dissipated by the resistors is qV1 + qV2 + qV3.
Suggest corrections
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https://www.jiskha.com/search?query=Palmer+and+teller+go+to+the+local+ice+cream+parlor+after+work+that+offers+40+flavors+of+ice+cream.+Palmer+wants+a+cone+with+chocolate+mousse+on+top%2C+cherry+truffle+in+the+middle%2C+and+double+Dutch+chocolate+on+the+bottom&page=76
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# Palmer and teller go to the local ice cream parlor after work that offers 40 flavors of ice cream. Palmer wants a cone with chocolate mousse on top, cherry truffle in the middle, and double Dutch chocolate on the bottom
32,255 results, page 76
1. ## statistics
If the probability of running out of gas is .03 and the probability the electronic starting system will not work is .01 a.) what is the probability there will be enough gas and that the starting system will work? assume the two events are independent b.)
asked by kkg on September 2, 2011
2. ## Physics
A car with mass m require 5.0 KJ of work to move from rest to a final spee v. If this sam amount of work is performed during the same amount of time on a car with a mass of 2m, what is the final speed of the second car.
asked by Cristiao on May 20, 2012
3. ## mathematics
The basic wage earned by a truck driver for a 40 hr work week is \$560.00. A)Calculate his hourly rate. Is the ans \$14 per hour 2) For overtime work, the driver is paid one and a half time the basic hourly rate. B)Calculate his overtime wage for 10 hrs of
asked by MONIKA on June 14, 2009
4. ## ETH 125
2. Final Project: Diversity and Your Community • Resources: Appendix A. • Write a 1,400- to 1,750-word autobiographical research paper that analyzes the influences of diversity as it relates to your community. In your paper, write your first-person
asked by Amanda on January 20, 2011
5. ## Physics
A 58kg skier is coasting down 25degrees slope at an initial velocity of 3,6m per second.A kinetic frictional force of 70N appores the motion.Calculate the net force acting on the skeir, (b)the work done by net force, (c)the final kinetic of the skier using
asked by Somila mdunyelwa on March 28, 2012
6. ## English
Think About the work you completed in your reading role. Determine the ideas that would be most worthy to share in a literary discussion about Johnny Tremain. Provide an explanation for your choices. How did the role you select and the work you completed
asked by Ouch on January 11, 2016
7. ## scientist
Mrs. leuenberger ran the same marathon but since her husband was hunting and she couldn't find a babysitter she had to carry her two daughters in her arms they weigh 40pounds and 55pounds Mrs Leuenberger weighs 145 pound (but you didn't hear that from
asked by Barb on October 25, 2009
8. ## Physics
this is the information i was given, it is confusing since the teacher didnt explain it at all... - speed of light = 3x10^8 m/s - 1nm= 10^-9 m the problem i have to solve is... A laser has a wavelength of 2.5 nm. What is the frequency in Hz? (Type your
asked by Anonymous on June 13, 2018
9. ## Critical Thinking (philosophy)
1. 21{1[2(24)24]}> The average height of members of the high school basketball team is six feet, three inches. Jerry is on the high school basketball team, so Jerry must be taller than six feet. The argument above is flawed because it confuses (Points : 1)
asked by Jean on April 30, 2011
10. ## Economics
The public debt is the amount of money that A. Americans owe to foreigners. B. state and local governments owe to the federal government. C. the federal government owes to holders of U.S. securities. D. the federal government owes to taxpayers. I chose C
asked by Fatima on March 27, 2018
11. ## Personal Finance-Mathe
Sue and Tom Wright are assistant professors at the local university. They each take home about \$40,000 per year after taxes. Sue is 37 years of age, and Tom is 35. Their two children, Mike and Karen are 13 and 11. Were either one to die, they estimate that
asked by Jamie on May 6, 2007
12. ## Law
for my CSE 1010 course I solved every single problem in my text book. Sense it's a electronic file I could easily publish it to the internet. My only concern is copyrights behind all of it. All work in the file is all my work, except for a few exceptions
asked by George on July 5, 2011
13. ## Environmental Science
I have been studying for several hours and am having problems with finding the answer to this question. Can anyone help please? Conservationists argue that watershed protection and other ecological functions of forget are more economically valuable than
asked by Sandy on April 19, 2011
14. ## Geography
Regarding tropical rainforests, the term evapotranspiration refers to the fact that A. deforestation causes a decline in atmospheric oxygen. B. the humidity of rain forests can produce local and regional precipitation C. deforestation causes a loss of soil
asked by Suzanne on July 11, 2014
15. ## physics
A pendulum of length L=26.0 cm and mass m= 168 g is released from rest when the cord makes an angle of 65.2 degrees with the vertical. (a) How far (m) does the mass fall before reaching its lowest point. (b) How much work (J) is done by gravity as it falls
asked by Sarah on December 11, 2011
16. ## physics
A pendulum of length L = 20.8cm and mass m = 168 g is released from rest when the cord makes an angle of 65.2 with the vertical. (a) How far (m) does the mass fall before reaching its lowest point? (b) How much work (J) is done by gravity as it falls to
asked by waqas on August 9, 2011
17. ## Math
Jordan decides to start a garden care service. He charges 20\$ for the first hour and 10\$ for each additional hour. He works 5 days a week and takes care of one different garden each day. How many hours will he need to work to earn at least 250\$. Write and
asked by Anonomyous on January 15, 2014
18. ## art
Using Internet resources, find one example of installation art. Describe the artwork you have found, including the title, artist, year created, and where it is installed. What is the meaning of the work? How does the meaning relate to what the artist has
asked by Anonymous on August 17, 2012
19. ## Physics
Two friends are at the local high school track, a circle measuring 440 yards for one complete lap. Abe can jog at 8.2 miles per hour while Bob’s jogging speed is 4.6 miles per hour. If they both start at the same point and jog in the same direction (say
asked by Bob on April 30, 2012
20. ## History
The second amendment protects the right of the people to keep and bear arms, yet it is illegal to owns some weapons. how this happen A.)A right to public safety is guaranteed in the constitution B.)The united states is not threatened by invasion in modern
asked by YRN DJ on November 6, 2015
21. ## math
Charles work a total of 30 hours last week. he worked 10 hours on Monday. what fraction of his work hours for the week did he complete on Monday?
asked by leah on January 20, 2016
22. ## Dianamic sckul
Tom takd 3 days to do a piece work while jerry take only one day for the same . Together. They both can finish the job in 15 days. In how many days tom will finish the work.
23. ## Statistics
The probability of Joe is at work is 49%. The probability Joe is at home is 25%. What is the probability Jow is neither at Work nor at home? A) 39% B)74% C)52% D)26% I get the answer at about 38%,but doesn't round to 39%, can anyone help, I'm stuck.
asked by Luke on December 8, 2013
24. ## Statistics/Math
1. A magazine is considering the launch of an online edition. The magazine plans to go ahead only if it's convinced that more than 25% of the current readers would subscribe. The magazine contacted a simple random sample of 500 current subscribers and
asked by Michael on October 30, 2013
25. ## math
20 men can do a piece of work in 24 days. After working for 6 days, additional men are employed to finish the work in 21 days from the beginning. Find the number of additional men
asked by saurav on March 19, 2015
26. ## math
What is the square root of 121? show work thanx I memorized this long ago in the fourth grade as 11. If you want to show work, show that 11x11 is 121. 11x11=121
asked by lida on March 30, 2007
27. ## American History
1. Believing that truth is what works for actual real people, Harvard psychologist William James developed the philosophy of (A. pragmatism.) B. social progressivism. C. radical progressivism. D. social idealism. 2. Which of the following was true of
asked by Andrea on November 14, 2012
28. ## physics
A flight attendant pulls her 70 N flight bag a distance of 253 m along a level airport floor at a constant speed. The force she exerts is 40 N at an angle of 53° above the horizontal A) Find the work that she does on the flight bag B) Find the work done
asked by zheane on March 7, 2013
29. ## prealgebra
a local pool charges a \$6 sign up fee plus \$2 per visit as as summer special. Another pool runs a summer special that only charges \$3 per visit. Write a system of equations to represent this situation. Is it 6+ 2x=y 3x=y
asked by mike on February 23, 2016
Consider a local business whose services you frequently use. What are the main components of this company's business model? I regualrly use a gas station so would my 3 main components be 1. gas prices 2.location 3.prices on product
asked by Tracie on February 19, 2010
31. ## Physics
Thank you for answering my previous questions! I have another one now in regards to the following question... An unknown potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to
asked by Sandra on February 28, 2010
32. ## science
Why do scientists use models? a. Scientists use models to learn about things that are too small, too large, or too complex to observe directly. b. Scientists use models because doing so is always part of the scientific method. c. Scientists use models
asked by Anonymous on September 19, 2013
33. ## Physics
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +13.0 m/s^2. At t1 the rocket shut down and the sled moves with constant velocity v until t2. The total distance traveled by the sled is 5.30 x 10^3m and the
asked by Kahlei on September 1, 2010
34. ## physics
Ian is firmly attached to a snowboard which is on a flat surface of ice, which you can assume to be frictionless. He and the snowboard are initially at rest. He then throws a 138 g ball which travels at a speed of 19.2 m/s, and finds himself moving
asked by Lydia on November 8, 2011
35. ## Physics
A closed box is filled with dry ice at a temperature of -81.5 C, while the outside temperature is 27.4 C. The box is cubical, measuring 0.307 m on a side, and the thickness of the walls is 4.47 × 10-2 m. In one day, 3.37 × 106 J of heat is conducted
asked by <3 on April 21, 2013
36. ## Chemistry
Copper (I) chloride, CuCL, has Ksp = 1.9 x 10^ -7.calculate the molar solubility of CuCL in A)pure water, B)0.0200M HCl solution C) 0.200M HCl solution, and d) 0.150M CaCl2 solution. I used the ice table and for a) i got 0.000435 M sort of lost for the
asked by priya on November 12, 2016
37. ## physics
The following forces act on a hockey puck (round, rubber disk) sitting on a frictionless surface: F_1 = 12 N at 15 degrees; F_2 = 28 N at 125 degrees; and F_3 = 33.3 N at 235 degrees. All the forces are in the plane of the ice. Determine the net force on
asked by Jacob on November 16, 2009
38. ## statics
Suppose we want to estimate the average number of purchases in a given hour in a local convenience store. Using this example, describe in your own words how you would go about determining this average through: a. Drawing a random sample b. Drawing a
asked by Anonymous on March 23, 2012
39. ## Math
Can you please help me set up this equation? Copies at the local library cost \$.15 easch. A new copier costs \$200 and it costs \$.02 to make a copy. How many copies would have to be made for the cost of making your own copies equal to the cost of copying at
asked by Kate on December 20, 2011
40. ## writing
Can you check to see if this looks and sounds better please? Dear Jordyn, I am thrilled you’re moving back to town and looking for a job. My boss, Mrs. Cartucci, is seeking a personal assistant to help manage the sales department at JR’s Fashions. The
asked by jacqui on March 30, 2011
41. ## Science
I don't understand what I am supposed to do?? 1)you are a climatologist who has been given the task of finding out how much glaciers and ice caps have receded over the past 50 years. 2) you have at your disposal satellite image data from this entire time
asked by Hannah on May 3, 2018
42. ## drawing 5,
What type of resume emphasizes the continuity and growth of your career? A. Targeted B. Chronological C. Functional D. Creative 2. Suppose you're an artist who has been doing projects on your own and, at the same time, working at an art supply store.
asked by liliy on July 7, 2011
43. ## Physics
Is this right? When a 5 kg box is lifted 3 meters from the ground, calculate the gravitational potential energy stored in the box. Calculate the work done by a Force to lift the box at a constant speed. Compare this answer to the amount of gravitational
asked by A on March 22, 2016
44. ## english
In the story Animal Farm how is the use of allegory as a rhetorical device different from simply laying out a non-fictional account, or an historical or statistical analysis of the period and the rise of the Communist Party? Answer: Within an allegory
asked by sue on May 6, 2018
45. ## Math
Five employees at the Wild Fowl Publishing Company each come to work by a different one of five means: car, railroad, subway, and walking. They hold these positions at the company: art director, editor, publisher, receptionist, and secretary. Two of them
asked by Mike on December 16, 2011
46. ## Science
How sediments,fossil, and ice core record Earth's geologic history? It record Earth's geologic history by giving scientist an idea of when certain climates or events happen.(Is this a good answer?) Thanks
47. ## Chemistry
queston: the pH of a white vinegar solution is 2.45. this vinager is an aquous solution of acetic acid with a density of 1.09g/mL. What is the mass percentage of acetic acid in the soulution. What exactly do i have to do? do i set up an Ice table. What do
asked by Anonymous on April 16, 2009
48. ## Gen Chem
Which provides the greatest increase in entropy for a mole of substance? a)Water melts at 0°C. b)Liquid water is heated from 0°C to 25°C. c)Gaseous water is converted to solid ice at 0.1°C. d)Water is converted from liquid to vapor at 100°C. e)Cannot
asked by Sarah H on March 28, 2008
49. ## Chemistry
How do I start an ICE table from the following information: BrCl3(g)+ Cl2(l) -> BrCl5(g) where Kp=7.8*10^-6 and there is originally 0.215 atm BrCl3, 725g Cl2, 0.115 atm BrCl5 - Can pressures be used as initial concentrations?
asked by Ken on October 3, 2012
50. ## science
350g ofwater at 30 degree celcius is contained in a copper vessel of mass 50g. Calculate the mass of ice required to bring down the temperature of the vessel and its content at 5 degree celcius.
asked by manoj on March 16, 2014
51. ## chem
The reaction is Br2 (g) 2Br (g). It occurs at T = 1600 C. When 1.05 mol Br2 are placed in a 2 Liter flask, 2.50% of Br2 undergoes dissociation. Calculate the Kp for the reaction. Okay so i did the ICE table for this, Br2 (g) 2Br (g) I : 1.05/2 ..... 0 C :
asked by Katie on October 14, 2013
52. ## employment law
The supervisor comes to your office and wants to see the employee’s file. You provide the supervisor the file. The supervisor asks for the results from the drug test, the medical follow-up for the work-related injury a year ago, and the employee’s
asked by Crystal on September 28, 2011
53. ## Spanish
How exactly will I do this? I seriously need help yet,I can't even attempt this. My teacher hasn't taught me this stuff as yet. So help would be appreciated. This is what I'm working on: Command form(imperative mood)negative familiar/polite. I have to
asked by John on May 10, 2008
54. ## computer science
Use the Library and other resources to write a "work-from-home" policy manual for an insurance company, educational institution, or another organization concerned about protecting an individual's privacy. Your manual should include policies to cover remote
asked by shutera on February 6, 2012
55. ## Physics 2414
An airplane pilot falls 355 m after jumping from an aircraft without his parachute opening. Fortunately, he lands in a snowbank, creating a crater 1.5 m deep, and survives with only minor injuries. Assuming the pilot's mass is 77 kg and his terminal
asked by Aestas on March 31, 2011
56. ## American Government
What is the purpose of city planning? (2 answers) a. to fulfill federal government requirements*** b. to correct past mistakes*** c. to create an opportunity for state and local governments to cooperate d. to gain federal funding for city planning efforts
asked by Elise Bedford on April 24, 2017
57. ## Chemistry
I saw that article on Google - that is actually how I got the idea to sublimate the naphthalene. But thank you for the suggestion! It makes me feel much more confident to know that I am on the right track! Thank you! We have a lab where we have to separate
asked by Summer on February 28, 2007
Question is what do the terms race and ethnicity man to you? Question is Why are these concepts important to United States Society? Answer: The term race to be means the way a person distinguishes how one looks at ones skin color, facial features, ancestry
asked by ann on July 1, 2007
59. ## Writing Evaluation (Writeacher!)
What do you think? Have I written according to the directions? What improvements do you suggest? Write 2 paragraphs describing Annie Johnson's sitation ("New Directions" by Maya Angelou.) A: Annie Johnson’s situation shows the familiar struggle that many
asked by Anonymous on February 5, 2014
60. ## Chemistry
Jack occasionally goes to work by bicycle where he exhales once per 3.00 sec. If in Jack’s every exhale is 4.50% CO2 (by volume), he exhales a total volume of 4.6 L each time, and it takes Jack 20.0 minutes to bike to work, how many grams of CO2 has he
asked by Kitty on January 16, 2012
According to "Life Without Gravity" why can astronauts' muscles become weak? a. Without gravity, the blood is rerouted from the legs to the head. b. Without gravity, the body "grows" as much as several inches. c. Without gravity, eating is difficult, so
asked by Dani on October 12, 2016
62. ## Math 11 Foundations
Last year, a local hockey association charged \$10 a ticket and averaged 12000 in attendance in its 2000 capacity area. The committee calculated that for every \$0.50 decreases in the price of the ticket, there would be 100 more tickets sold. what is the
asked by Jason Keller on January 25, 2017
63. ## English
I left out the following questions. Thank you very much, Writeacher. 1) First, the organization collects information. When it has proof that a person is a prisoner of conscience, it hands over that person’s case to a local group. 2)This group send (or
asked by Matthew on April 15, 2012
64. ## home inspection
Which of the following is the main purpose of home inspection societies and professional associations? A.To license and regulate home inspectors B.To "self-regulate" the home inspection industry C.To certify people starting out in the home inspection
asked by sib on August 24, 2010
65. ## math
speakeasy company offers customers a choice between 3 schemes:1st,a monthly charge of \$15 + 5cents/ min.2nd,a monthly charge of \$5 for the line rental +20 cents/min,3rd a(pay as you go) charge of 35cents/min with no monthly line rental charge.
asked by rima on May 10, 2010
66. ## Chemistry
I don't understand ice table at all..can anyone please guide me in how to solve the following question? : The pH of a 0.00250 mol/L solution of benzoic acid is 3.65. Calculate the Ka for the benzoic acid. (This question does not state that the given
asked by Jessica on June 15, 2010
67. ## Physics
A 95.1-g ice cube is initially at 0.0°C. (a) Find the change in entropy of the cube after it melts completely at 0.0°C. Hint: The latent heat of fusion for water is 3.33 105 J/kg. answer in J/kg (b) What is the change in entropy of the environment in
asked by Sam on November 26, 2012
68. ## chemistry
if 27 grams of ice at 0 degrees celsius is added to 123 grams of water at 100 degrees celsius in an insulated container, calculate the final temperature. assume specific heat of water is 4.184 J/K g
asked by Gabriel on October 5, 2012
69. ## Physics
How much ice at 0 degrees celsius can be melted with 1kg of steam at 100 degrees celsius, assuming no loss of energy to the surroundings? Use specific latent heat values. The specific heat of water is 4200 J kg-1 K-1
asked by Mariya on June 26, 2013
70. ## Math calculus
An ice cube is 3 by 3 by 3 inches is melting in such a way that the length of one of its side is decreasing at a rate of half an inch per minute. Find the rate at which its surface area is decreasing at the moment when the volume of the cube is 8 cubic
asked by Ashley on August 7, 2016
71. ## chemistry
How do I find out how many ml of NaOH i need to titrate H3PO4. I used the ice table to knock off the two protons but im using a .2200M NaOH and do not Know how to determine the volume needed before hand. In other words i need some pre-calculations for
asked by ted on July 10, 2011
72. ## Physics
an ice skater with a mass of 80kg pushes off against a second skater with a mass of 32kg. both skaters are initially at rest. a. what is the total momentum of the system after the push off? b. if the larger skater moves off with a speed of 3m/s, what is
asked by Titi on November 10, 2009
73. ## geometry
you are to design, draw and label the dimensions of 3 boxes that will hold enough detergent to do 18 loads of laundry. Each scoop of detergent does one load. The size of the scoop is 3in.x2in.x2in. all work must be shown on the work page. At the bottom of
asked by Aaron on May 18, 2011
74. ## legal and ethical issues
Which element do you think plays the greatest role in influencing employee behavior? Explicit policy or implicit policy. First find out what explicit and implicit mean. Then think about a couple of job situations. Pretend you are the employer, then the
asked by Anonymous on August 8, 2007
75. ## Algebra 1
suppose you can work a total of no more than 20 hours per week at your two jobs. Babysitting pays \$5 per hour, and your cashier job pays \$6 per hour. youneed to earn at least \$90per week to cover your expenses. Write a system of inequalities that show s
asked by molly on December 16, 2014
76. ## physics
A stewardess pulls her 70 N flight bag a distance of 200 m along an airport floor at constant speed. The force she exerts is 40 N at an angle of 50 degrees above the horizontal. Find (a) the work she does, (b) the work done by the force of friction, and
asked by Anonymous on October 23, 2007
77. ## College Chemistry
Given 400.0 g of hot tea at 80.0C, what mass of ice at 0C must bea added to obtain iced tea at 10.0C? The specific heat of water is 4.18 J/(g*C), The specific heat of water is 4.184 kJ/mol.
asked by Ashley on December 6, 2010
78. ## physics
A 5.0 MULTIPLIED BY 10 TO THE POWER OF 2g aluminium block is heated to 350 degrees celcius. determine the number of kilograms of ice at 0 degrees celcius that the aluminium block will melt as it cools.
asked by audrey on January 21, 2008
79. ## physics
A 2.50m -long, 450g rope pulls a 10.0kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.50m/s2 . How much force pulls forward on the rope?
asked by Matt on March 27, 2015
80. ## Physics
A 10.3 kg block of ice slides without friction down a long track. The start of the track is 4.2 m higher than the end of the track, and the path traveled is 8.0 m. Find the speed of the block when it reaches the end of the track.
asked by Breezerlea on February 22, 2011
81. ## Physics
How many grams of ice must be added to 100 grams of water at 60 degrees so that its temperature drops to 20 degrees celcius? As far as i know I have to use: (Mass)*(specific heat of water) *(finaltemperature - initial temperature = ??
asked by Jose on January 7, 2010
82. ## physics
A hockey puck slides across the ice at a con- stant speed. Which of the following is true? 1. The puck is moving and thus not in equi- librium. 2. The puck can be considered neither at rest nor in equilibrium. 3. It is in equilibrium. 4. None of these 5.
asked by BOB on October 18, 2011
83. ## Physics
When a block slides a certain distance down an incline the work done by gravity is 300 J. What is the work done by gravity if this block slides the same distance up the incline? a) 300 J b) zero c) -300 J d) It cannot be determined without knowing the
asked by Tammy on March 25, 2007
84. ## physics/math
Joe is a former basketball player who is now a work-out junkie, and especially loves to do push-ups. He was a good athlete, but not so good at science in school, especally not good in Physics. He wants you, yes you, to calculate how much work he does, in
asked by kristin on August 12, 2008
85. ## algebraic expression
Whitney is speaking to four different classes at the local grade school . She will spend m minutes with each class, and she will have to wait 30 minutes before seeing the last group. How many minutes will Whitney spend at the school? Thank you.
asked by S. Fuentes on July 13, 2014
86. ## intermediate algebra
A local electronics store will sell seven AC adapters and two rechargeable flashlights for \$86, or three AC adapters and four rechargeable flashlights for \$84. What is the price of a single AC adapter and a single rechargeable flashlight?
asked by amanda on September 30, 2013
87. ## math
at the local clothing store, 3 similar shirts and 4 similar jackets cost \$360 and 1 shirt and 3 jackets cost \$220. find the cost of each shirt.
asked by shohanur on March 31, 2013
88. ## math
at the local clothing store, 3 similar shirts and 4 similar jackets cost \$360 and 1 shirt and 3 jackets cost \$220. find the cost of each shirt.
asked by shohanur on March 31, 2013
89. ## Physics
A 855 N crate rests on the floor. a) How much work is required to move it at constant speed 7.7 m along the floor against a friction force of 180 N? b) How much work is required to move it at constant speed 7.7 m vertically? I really NEED help I am very
asked by Jen on October 15, 2009
90. ## American Government
Which of the following is an example of a company trying to influence government to hurt its competition? A. a toy manufacturer seeking help from the Consumer Product Safety Commission B. local chambers of commerce holding fund-raisers for congressional
asked by Sharon on August 24, 2014
91. ## English Improving writing rough draft
please check his for me thank you very much :) There are on going problems that need to be addressed.First there is uneven allocation of tasks. While Bonnie and Molly are working overtime at least twice a month, Jack and Rachel often have no work to do.
asked by Kaleigh-Anne on June 18, 2008
92. ## Math
At a local baseball game, the concession stand has two different meal choices. You can get 3 hotdogs and 1 drinks for \$18.00 or you can get 2 hotdogs and 4 drinks for \$22.00. If the price of hotdogs and drinks is the same for each meal, find how much it
asked by Help on November 18, 2017
My brother and I are trying to start our own wireless indirect agent. We both have experience in managing a wireless store and the kowledge in rate plan and wireless market, but becasue we are a new business it is hard to get fiancing. We have dealed with
asked by Will on April 5, 2008
94. ## physics
A driver of a 7.50 N car passes a sign stating "Bridge Out 30 Meters Ahead." She slams on the brakes, coming to a stop in 10.0 s. How much work must be done by the brakes on the car if it is to stop just in time? Neglect the weight of the driver, and
asked by Anonymous on October 22, 2007
95. ## Criminal justice
Write a 700- to 1,050-word paper in which you describe the issues facing police departments in today’s society. · Include a description of how local, state, and federal law enforcement agencies currently interact with the U.S. Department of Homeland
asked by Michele on July 15, 2012
96. ## TX HISTORY; HELP ASAP
How did fighting in the civil war change the lives of Texas soldiers ? a) They came to support the cause of secession b) They traveled far from home to participate in bloody battles c) They earned enough money to buy cotton plantations and slaves d) They
asked by Natalie on December 6, 2017
97. ## criminal justice
· Write a 700- to 1,050-word paper in which you describe the issues facing police departments in today’s society. · Include a description of how local, state, and federal law enforcement agencies currently interact with the U.S. Department of Homeland
asked by asia on May 18, 2012
98. ## SAT Essay
Please critique and let me know how it is and what I can improve. Thank you! Assignment: Do you think that ease does not challenge us and that we need adversity to help us discover who we are? Plan and write an essay in which you develop your point of view
asked by Sanya on November 30, 2012
99. ## buisness
Which sentences highlight a unilateral contract? a. Monique goes to a supermarket and picks up a shampoo. She then goes to the counter and pays for the shampoo. b. John offers a ride back home to Jack in exchange for his favorite music DVD for a day. Jack
asked by Judy on April 25, 2017
100. ## English
For homework I have to answer the following questions regarding Puerto Rican obituary by Pedro pietri but I am horrible at interpreting poems if someone can please help thank you "Puerto Rican Obituary", by Pedro Pietri: What has the American dream turned
asked by Jessica on May 23, 2013
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$$\require{cancel}$$
# 5.1: Introduction- Sources in a Uniform Permeable Material
The equations of magnetostatics are given by Equation (4.1.2)
$\operatorname{div}(\overrightarrow{\mathrm{B}})=0, \nonumber$
and Equation (4.1.3)
$\operatorname{curl}(\overrightarrow{\mathrm{B}})=\mu_{0}\left(\overrightarrow{\mathrm{J}}_{f}+\operatorname{curl}(\overrightarrow{\mathrm{M}})\right). \nonumber$
( refer to section(4.1)). For a linear, isotropic, magnetic medium $$\vec B$$ is proportional to $$\vec H$$ where the factor of proportionality is called the permeability.
$\overrightarrow{\mathrm{B}}=\mu \overrightarrow{\mathrm{H}}=\mu_{0}(\overrightarrow{\mathrm{H}}+\overrightarrow{\mathrm{M}}) , \nonumber$
so that
$\overrightarrow{\mathrm{M}}=\left(\frac{\mu}{\mu_{0}}-1\right) \overrightarrow{\mathrm{H}} , \nonumber$
or
$\overrightarrow{\mathrm{M}}=\left(\mu_{r}-1\right) \overrightarrow{\mathrm{H}}. \label{5.1}$
In Equation (\ref{5.1}) µr = µ/µ0 is the relative permeability. The second of the above Maxwell’s equations can be re-written in the form
$\operatorname{curl}(\overrightarrow{\mathrm{H}})=\overrightarrow{\mathrm{J}}_{f} , \nonumber$
or
$\operatorname{curl}(\overrightarrow{\mathrm{B}})=\mu \overrightarrow{\mathrm{J}}_{f} . \label{5.2}$
The substitution $$\vec B$$ = curl($$\vec A$$) ensures that Equation (4.1.2) will be satisfied since the divergence of any curl is zero. Using this substitution in Equation (\ref{5.2}) gives
$\operatorname{curlcurl}(\overrightarrow{\mathrm{A}})=\mu \overrightarrow{\mathrm{J}}_{f} . \label{5.3}$
$\operatorname{div}(\overrightarrow{\mathrm{A}})=0, \label{5.4}$
then
$\nabla^{2} \overrightarrow{\mathrm{A}}=-\mu \overrightarrow{\mathrm{J}}_{f} , \label{5.5}$
and this equation has the particular solution
$\overrightarrow{\mathrm{A}}(\overrightarrow{\mathrm{R}})=\frac{\mu}{4 \pi} \, iiint_{S p a c e} \mathrm{d} \tau \frac{\overrightarrow{\mathrm{J}}_{f}(\overrightarrow{\mathrm{r}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}, \label{5.6}$
where d$$\tau$$ is an element of volume. This development exactly follows the procedure described in Chpt.(4); the only difference is that the integration in Equation (\ref{5.6}) is carried out over the free current density distribution, and the fields due to the effective current density curl($$\vec M$$) are taken into account through the permeability µ that multiplies the integral. It should be noted that this procedure only works if µ does not depend upon position in space. If there are regions characterized by different values of µ the problem of calculating the magnetic field distribution becomes much more difficult. This is because at the boundaries between regions having different permeabilities there are discontinuities in the normal and tangential components of $$\vec M$$ that act as field sources.
In the usual situation the current density is zero except within a finite number of thin wires. For a current of I Amps carried in a wire of negligible cross-section Equation (\ref{5.6}) becomes
$\overrightarrow{\mathrm{A}}_{P}=\frac{\mu \mathrm{I}}{4 \pi} \int_{W i r e} \frac{| \mathrm{d} \overrightarrow{\mathrm{L}}}{|\overrightarrow{\mathrm{r}}|} , \label{5.7}$
where $$\vec r$$ is the vector from the element of length d$$\vec L$$ to the point P where the vector potential $$\vec A$$ is to be calculated. From $$\vec B$$ = curl($$\vec A$$) one obtains
$\overrightarrow{\mathrm{B}}(\overrightarrow{\mathrm{r}})=\frac{\mu \mathrm{I}}{4 \pi} \int_{W i r e} \frac{\mathrm{d} \overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|^{3}} . \label{5.8}$
These formulae are very similar to Equations (4.2.1) and (4.17) of Chpt.(4). The fields corresponding to the standard problems of a long straight wire, the field along the axis of a circular loop, and along the axis of a finite solenoid are given by Equations (4.3.3),(4.3.4), and (4.3.5) where the permeability of free space, µ0, is replaced by the permeability µ. In particular, the field of an infinite solenoid that is filled with a magnetic material is given by
$\overrightarrow{\mathrm{B}}=\mu \mathrm{NI} , \label{5.9}$
where N is the number of turns per meter.
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# Lines and Line Segments
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## Objective
SWBAT identify and measure a line segment using the appropriate tool, recording their outcome using the correct units.
#### Big Idea
Common Core expectations for second grade are that students will be able to measure a line segment, record the length in appropriate units, and add and subtract to compare the lengths of the lines.
## Warm Up
10 minutes
We start today by generating a list of measurement tools. I ask students to list all the tools they can think of to measure with. I record the answers on my computer projected on the White Board. (You can do this on a board or easel, but I want to be able to classify the list by the kinds of measuring it does ie length, weight, volume).
After students have given their ideas, I put the 3 headings on the board LENGTH, WEIGHT, VOLUME (LIQUID). I ask students one at a time to tell me where to place each tool. We discuss how different tools measure in different ways. This will set the foundation for measuring in this unit, and also open the discussion for appropriate units of measure, as well as selecting the appropriate tools for measurement (MP 5).
## Teaching the Lesson
20 minutes
I want to help students understand the distinction between a line and a line segment. I start drawing a line across the board. I keep going to the end of the board. At the end of the board I pretend to continue my line along the wall. I ask students what I am making? (A line). I go back to the board and draw a short line with a dot at each end. I ask what I am making now? Most students will reply that I have made a line. I tell them that this is just a piece of a line. It has a special name, does anyone know what the name is?
I show the students an orange that I have broken apart into segments. Does anyone know what the pieces are called? They are not the whole orange, but pieces with a special name. Does anyone know what the name is? I give hints using a "hangman strategy" where I put up S __ __ __ __ __ __S. We work together to fill in the missing sounds and get the word segments. I say that a segment is a piece of the whole.
I return to my line segment. I point out that it is not as big as my first line it is just a piece so it is called a line ________________. (segment)
I tell students that today we will be measuring line segments. What tool do they think would be good for measuring line segments? I help them work towards using a ruler, centimeter stick, meter stick or a yardstick.
We talk about the unit we will measure with. We discuss inches, centimeters, miles, feet, and come to a decision about a unit that would measure a small length (inches or centimeters) and how we would record each one (in, or cm).
## Measurement Practice
20 minutes
I tell students that today they will be measuring a series of line segments on a worksheet Line Segment Practice and then solving problems using those lengths.
Before students begin I point out how the line segments on their papers have dots and letters at each end. A line segment is named by the points at each end. They will know which lines to compare by the letter names at each end.
We practice identifying the lines on their papers by the letter names. When students are comfortable with this I explain the directions and ask them to work independently to measure the lengths of the lines to the nearest inch and then to answer the questions at the bottom of the page. Using Line Segment Distances to Solve Addition and Subtraction Problems
I circulate around supporting the students who may need help.
## Closing
10 minutes
When students complete their papers, I ask them to report out on the sizes of the lines, and how they solve the problems presented (MP1). Students are reminded that as their peer shares, they look at his/her own work to see if their answer matches the answer of others, or if they have a different answer.
If their answer is different, we share the solutions and work together to decide which answer fits the problem that is presented.
Students learn from correcting their own work. They can see if their strategy worked, they can see other strategies, and they can develop new strategies for similar problems that they may face in the future.
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• # question_answer A block ${{M}_{1}}$ of mass $2.0\text{ }kg$ is moving on a frictionless horizontal surface with a velocity of $1.0\text{ m/s}$ towards another block ${{M}_{2}}$ of equal mass kept at rest as shown. The spring constant of the spring fixed at one end of ${{m}_{2}}$ is $100N/m$. The maximum compression of the spring is: A) $2\,cm$ B) $5\,cm$ C) $10\,cm$ D) $20\,cm$
At the time of maximum compression velocities of both the blocks will be same (say V) By conservation of linear momentum. $2\times 1=4V$ $V=0.5m/s$ By conservation of energy. $\frac{1}{2}\times 2\times {{(1)}^{2}}=\frac{1}{2}\times 4\times {{(0.5)}^{2}}+\frac{1}{2}\times 100\times {{(X)}^{2}}$ $X=0.1m=10cm$
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2 Vectors
2.1 Atomic vectors
1. Q: How do you create scalars of type raw and complex? (See ?raw and ?complex)
A: In R scalars are represented as vectors of length one. For raw and complex types these can be created via raw() and complex(), i.e.:
raw(1)
#> [1] 00
complex(1)
#> [1] 0+0i
For raw vectors it’s easy to coerce numeric or character scalars to raw.
as.raw(42)
#> [1] 2a
charToRaw("A")
#> [1] 41
For complex numbers real and imaginary parts can be provided directly.
complex(length.out = 1, real = 1, imaginary = 1)
#> [1] 1+1i
2. Q: Test your knowledge of vector coercion rules by predicting the output of the following uses of c():
c(1, FALSE) # will be coerced to numeric -> 1 0
c("a", 1) # will be coerced to character -> "a" "1"
c(TRUE, 1L) # will be coerced to integer -> 1 1
3. Q: Why is 1 == "1" true? Why is -1 < FALSE true? Why is "one" < 2 false?
A: These comparisons are carried out by operator-functions, which coerce their arguments to a common type. In the examples above these cases will be character, double and character: 1 will be coerced to "1", FALSE is represented as 0 and 2 becomes "2" (in ASCII numerals preceed letter).
4. Q: Why is the default missing value, NA, a logical vector? What’s special about logical vectors? (Hint: think about c(FALSE, NA_character_).)
A: It is a practical thought. When you combine NAs in c() with other atomic types they will be coerced like TRUE and FALSE to integer (NA_integer_), double (NA_real_), complex (NA_complex_) and character (NA_character_). Recall that in R there is a hierarchy of recursion that goes logical >> integer >> double >> character. If NA was a character and provided as part of a set of other values, all of these would be coerced into character as well. Making NA a logical means that involving an NA in a dataset (which happens often) will not affect coercion.
5. Q: Precisely what do is.atomic(), is.numeric(), and is.vector() test for?
A: The documentation informs, that:
• is.atomic() tests if an object has one of these types: "logical", "integer", "double", "complex", "character", "raw" or "NULL" (!).
• is.numeric() tests if an object has integer or double type and is not of "factor", "Date", "POSIXt" or "difftime" class.
• is.vector() tests if an object has no attributes, except of names and if its mode() is atomic ("logical", "integer", "double", "complex", "character", "raw"), "list" or "expression".
2.2 Attributes
1. Q: How is setNames() implemented? How is unname() implemented? Read the source code.
A: setNames() is implemented as:
setNames
#> function (object = nm, nm)
#> {
#> names(object) <- nm
#> object
#> }
#> <bytecode: 0x4881f48>
#> <environment: namespace:stats>
As the data comes first setNames() also works well with the magrittr-pipe operator. When no first argument is given, the result is a named vector:
setNames( , c("a", "b", "c"))
#> a b c
#> "a" "b" "c"
However, the implemention also means that setNames() only affects the names attribute and not any other more specific naming-attributes like the dimnames attribute for matrices and arrays.
unname() is implemented in the following way:
unname
#> function (obj, force = FALSE)
#> {
#> if (!is.null(names(obj)))
#> names(obj) <- NULL
#> if (!is.null(dimnames(obj)) && (force || !is.data.frame(obj)))
#> dimnames(obj) <- NULL
#> obj
#> }
#> <bytecode: 0x2dddbf8>
#> <environment: namespace:base>
If set, unname() removes the names and dimnames attributes from its input object. Note that the dimnames attribute (names and row names) won’t be affected for data frames, even if the documentation currently (R 3.5.1) mentions this in the case where force == TRUE is supplied. Instead the line dimnames(obj) <- NULL setting NULL as the replacement value in the underlying dimnames<-.data.frame function, raises the first error condition in the underlying source code:
dimnames<-.data.frame
#> function (x, value)
#> {
#> d <- dim(x)
#> if (!is.list(value) || length(value) != 2L)
#> stop("invalid 'dimnames' given for data frame")
#> value[[1L]] <- as.character(value[[1L]])
#> value[[2L]] <- as.character(value[[2L]])
#> if (d[[1L]] != length(value[[1L]]) || d[[2L]] != length(value[[2L]]))
#> stop("invalid 'dimnames' given for data frame")
#> row.names(x) <- value[[1L]]
#> names(x) <- value[[2L]]
#> x
#> }
#> <bytecode: 0x3a10418>
#> <environment: namespace:base>
2. Q: What does dim() return when applied to a 1d vector? When might you use NROW() or NCOL()?
A: From ?nrow:
dim() will return NULL when applied to a 1d vector.
One might want to use NROW() or NCOL() to handle atomic vectors, lists and NULL values (1 column, 0 rows) analog to a 1-column matrix / data frame. In these cases the alternatives nrow() and ncol() return NULL (consistently to the behaviour of dim()). When subsetting data frames (interactively) this might be convenient as it is not affected by and hence more robust regarding the default drop = TRUE(-idiom):
NROW(iris[, 1, drop = TRUE])
#> [1] 150
nrow(iris[, 1, drop = TRUE])
#> NULL
NCOL(iris[, 1, drop = TRUE])
#> [1] 1
ncol(iris[, 1, drop = TRUE])
#> NULL
3. Q: How would you describe the following three objects? What makes them different to 1:5?
x1 <- array(1:5, c(1, 1, 5)) # 1 row, 1 column, 5 in third dimension
x2 <- array(1:5, c(1, 5, 1)) # 1 row, 5 columns, 1 in third dimension
x3 <- array(1:5, c(5, 1, 1)) # 5 rows, 1 column, 1 in third dimension
A: They are of class array and so they have a dim attribute.
4. Q: An early draft used this code to illustrate structure():
structure(1:5, comment = "my attribute")
#> [1] 1 2 3 4 5
But when you print that object you don’t see the comment attribute. Why? Is the attribute missing, or is there something else special about it? (Hint: try using help.)
A: From the help of comment (?comment):
Contrary to other attributes, the comment is not printed (by print or print.default).
Also from the help of attributes (?attributes):
Note that some attributes (namely class, comment, dim, dimnames, names, row.names and tsp) are treated specially and have restrictions on the values which can be set.
Apart of that, we can get it easily when we are more specific, i.e.:
bla <- structure(1:5, comment = "my attribute")
attributes(bla)
#> \$comment
#> [1] "my attribute"
attr(bla, "comment")
#> [1] "my attribute"
2.3 S3 atomic vectors
1. Q: What sort of object does table() return? What is its type? What attributes does it have? How does the dimensionality change as you tabulate more variables?
A: table() returns a crosstabulation of its input. The result is an S3 table object, which is an array (implicit class) of integers (type) under the hood. Attributes are dim (dimension of the underlying array) and dimnames (one for each input column). The dimensionality equals to the number of unique values (accordingly factor levels) of the input arguments, i.e.:
dim(table(iris))
#> [1] 35 23 43 22 3
sapply(iris, function(x) length(unique(levels(as.factor(x)))))
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 35 23 43 22 3
2. Q: What happens to a factor when you modify its levels?
f1 <- factor(letters)
levels(f1) <- rev(levels(f1))
A: Both, the entries of the factor and also its levels are being reversed:
f1
#> [1] z y x w v u t s r q p o n m l k j i h g f e d c b a
#> Levels: z y x w v u t s r q p o n m l k j i h g f e d c b a
3. Q: What does this code do? How do f2 and f3 differ from f1?
f2 <- rev(factor(letters)) # changes only the entries of the factor
f3 <- factor(letters, levels = rev(letters)) # changes only the levels of the factor
A: Unlike f1 f2 and f3 change only one thing. They change the order of the factor or its levels, but not both at the same time.
2.4 Lists
1. Q: List all the ways that a list differs from an atomic vector.
A:
• Atomic vectors are homogeneous (all contents must be of the same type). Lists are heterogeneous (all contents can be of different types).
• Atomic vectors point to one value, while lists contain references which point to one value each:
lobstr::ref(1:3)
#> [1:0x2e961d0] <int>
lobstr::ref(list(1:3,2,3))
#> █ [1:0x61b6658] <list>
#> ├─[2:0x62078c0] <int>
#> ├─[3:0x2e75460] <dbl>
#> └─[4:0x2e75498] <dbl>
• Subsetting with out of bound values and NAs leads to NAs for atomics and NULL values for lists:
(1:2)[3]
#> [1] NA
as.list(1:2)[3]
#> [[1]]
#> NULL
(1:2)[NA]
#> [1] NA NA
as.list(1:2)[NA]
#> [[1]]
#> NULL
#>
#> [[2]]
#> NULL
2. Q: Why do you need to use unlist() to convert a list to an atomic vector? Why doesn’t as.vector() work?
A: To get rid of (flatten) the nested structure as.vector() doesn’t work, because a list is already a vector.
3. Q: Compare and contrast c() and unlist() when combining a date and date-time into a single vector.
A: Date and date-time objects are build upon doubles. Dates are represented as days, while date-time-objects (POSIXct) represent seconds (counted in regard to the reference date 1970-01-01, also known as “The Epoch”).
Let’s define date and date-time objects:
date <- as.Date("1970-01-02")
dttm_ct <- as.POSIXct("1970-01-01 01:00", tz = "UTC")
When combining these objects method-dispatch leads to suprising output:
c(date, dttm_ct) # equal to c.Date(date, dttm_ct)
#> [1] "1970-01-02" "1979-11-10"
c(dttm_ct, date) # equal to c.POSIXct(date, dttm_ct)
#> [1] "1970-01-01 01:00:00 UTC" "1970-01-01 00:00:01 UTC"
The generic function dispatches based on the class of its first argument. When c.Date() is executed, dttm_ct is converted to a date, but the 3600 seconds are mistaken for 3600 days! When c.POSIXct() is called on date, one day counts as one second only, as illustrated by the following line:
unclass(c(date, dttm_ct)) # internal representation
#> [1] 1 3600
date + 3599
#> [1] "1979-11-10"
Some of these problems may be avoided via explicit conversion of the classes:
c(as.Date(dttm_ct, tz = "UTC"), date)
#> [1] "1970-01-01" "1970-01-02"
Let’s look at unlist(), which operates on list input.
# attributes are stripped
unlist(list(date, dttm_ct))
#> [1] 1 3600
We see that internally dates(-times) are doubles. Unfortunately this is all we are left with, when unlist strips the attributes of the list. (This wouldn’t happen for vector input, but then we would have to combine the different classes into one vector, which is tricky as seen above.)
To summarise: c() coerces types and errors may occur because of inappropriate method dispatch. unlist() strips attributes.
2.5 Data frames and tibbles
1. Q: Can you have a data frame with 0 rows? What about 0 columns?
A: Yes, you can create one easily and in many ways. Also both dimensions can be 0. The fastest way is to subset the regarding dimension with one of 0, NULL or a valid 0-length atomic (logical(0), character(0), integer(0), double(0)). Also a negative integer sequence would work. Here we use the recycling rules for logical subsetting:
iris[FALSE,]
#> [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> <0 rows> (or 0-length row.names)
iris[ , FALSE] # or iris[FALSE]
#> data frame with 0 columns and 150 rows
iris[FALSE, FALSE] # or just data.frame()
#> data frame with 0 columns and 0 rows
2. Q: What happens if you attempt to set rownames that are not unique?
A For matrices this will work without any problems. For data frames it is not possible and what happens depends on the approach. When using the row.names<- replacement function, no further arguments can be set and the underlying .rowNamesDF<- will throw an error (and an additional warning):
row.names(mtcars) <- rep(1, nrow(mtcars))
#> Warning: non-unique value when setting 'row.names': '1'
#> Error in .rowNamesDF<-(x, value = value): duplicate 'row.names' are not allowed
However, by calling .rowNamesDF<- directly one can set the make.names argument to NA or TRUE. When set to NA, any non unique row name will trigger the new row names to become seq_len(nrow(x)). When make.names = TRUE, row names will automatically converted into unique ones via make.names(value, unique = TRUE). The same behaviour is caused, when a matrix with non unique row names is converted into a data frame.
3. Q: If df is a data frame, what can you say about t(df), and t(t(df))? Perform some experiments, making sure to try different column types.
A Both will return matrices with dimensions regaring the typical transposition rules. As t() uses as.matrix.data.frame() for the preprocessing infront of applying t.default() and elements of matrices need to be of the same type, all elements get coerced in the usual order (logical << integer << double << character), while factors, dates and datetimes are treated as characters during coercion.
4. Q: What does as.matrix() do when applied to a data frame with columns of different types? How does it differ from data.matrix()?
A: From ?as.matrix:
The method for data frames will return a character matrix if there is only atomic columns and any non-(numeric/logical/complex) column, applying as.vector to factors and format to other non-character columns. Otherwise the usual coercion hierarchy (logical < integer < double < complex) will be used, e.g., all-logical data frames will be coerced to a logical matrix, mixed logical-integer will give a integer matrix, etc.
To illustrate this, we create an easy example where the data frame gets coerced to a character matrix:
a <- c("a", "b", "c")
b <- c(TRUE, FALSE, FALSE)
c <- c("TRUE", "FALSE", "FALSE")
d <- c(1L, 0L, 2L)
e <- c(1.5, 2, 3)
f <- c("one" = 1, "two" = 2, "three" = 3)
g <- c("first" = 1, "second" = 2, "third" = 3)
h <- factor(c("f1", "f2", "f3"))
df_cols <- data.frame(a, b, c, d, e, f, g, h, stringsAsFactors = FALSE)
# Note that format is applied to the characters, which can complicate
# inverse conversion (back to the previous type).
# For example TRUE in the b variable becomes " TRUE" (starting with a space)
as.matrix(df_cols)
#> a b c d e f g h
#> one "a" " TRUE" "TRUE" "1" "1.5" "1" "1" "f1"
#> two "b" "FALSE" "FALSE" "0" "2.0" "2" "2" "f2"
#> three "c" "FALSE" "FALSE" "2" "3.0" "3" "3" "f3"
From ?as.data.matrix:
Return the matrix obtained by converting all the variables in a data frame to numeric mode and then binding them together as the columns of a matrix. Factors and ordered factors are replaced by their internal codes.
So for data.matrix we’ll get a numerix matrix containing NAs for original character columns:
data.matrix(df_cols)
#> Warning in data.matrix(df_cols): NAs introduced by coercion
#> Warning in data.matrix(df_cols): NAs introduced by coercion
#> a b c d e f g h
#> one NA 1 NA 1 1.5 1 1 1
#> two NA 0 NA 0 2.0 2 2 2
#> three NA 0 NA 2 3.0 3 3 3
2.6 Old exercises
1. Q: What does dim() return when applied to a vector?
A: NULL
2. Q: If is.matrix(x) is TRUE, what will is.array(x) return?
A: TRUE, as also documented in ?array:
A two-dimensional array is the same thing as a matrix.
3. Q: What attributes does a data frame possess?
A: names, row.names and class.
4. Q: What are the six types of atomic vector? How does a list differ from an atomic vector?
A: The six types are logical, integer, double, character, complex and raw. The elements of a list don’t have to be of the same type.
5. Q: What makes is.vector() and is.numeric() fundamentally different to is.list() and is.character()?
A: The first two tests don’t check for a specific type.
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Guided Lessons
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# A litre Please
This lesson introduces students to measuring and estimating liquid volume, using liters (l) as the standard unit of measurement.
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Students will be able to measure and estimate liquid volume, using liters as the standard unit of measurement.
(15 minutes)
• Tell your students that they will learn about the Litre(l), which is a metric unit of measurement, as opposed to a standard unit of measurement.
• On the board, write the word LitreWith its abbreviation L.
• Show the students the container with the Capacity, or the space a solid, liquid, or gas takes up, of a litre.
• Ask your students where they might see the litre as a unit of measurement. For example, it is found on some grocery products such as canned goods.
(15 minutes)
• Demonstrate to your students what they will do in groups later.
• Collect everything you need for this activity: two medium-sized containers, one 1 litre container, a container of water, a plastic container or sheet to shield objects from water, and the worksheet.
• Give the containers names on the worksheet so that you can keep track of which container you're using.
• In front of the class, estimate if the medium-sized container can hold less than one litre, more than litre, or exactly one litre.
• Pour water into the medium-sized container until it is full.
• Then, to measure how much water you actually poured in, pour that water from the medium-sized container into the one litre container.
• Check the estimate and compare it to the actual result.
(5 minutes)
• Group the students in three to four groups of three to four students.
• Give each group two different medium-sized containers, one 1 litre container, a container full of water, and a plastic container or sheet to cover objects. Give each student one worksheet.
(15 minutes)
• Instruct your students to complete the activity and the worksheet as you modeled. Make sure they measure both of the medium-sized containers.
• Have them fill out the bottom of the worksheet.
• Walk around and observe the students as they work on completing the activity.
• Monitor the students as they work, making sure the activity is done appropriately.
• During this time, ask questions to make sure the students understand the concept of liquid measurement.
• Enrichment:Have more than two medium-sized containers available for your students.
• Support:Give your students a a smaller container that is less than one litre. Make sure your students understand "less than" before moving on.
(5 minutes)
• At the end of the independent working time, ask your students to identify the medium-sized containers as less than, equal to, or more than one litre.
(5 minutes)
• Ask for a spokesperson from each group to share the group's findings.
• After each group has shared the results of their activity, ask the class if there are any questions or comments.
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# How many ways can we distribute $7$ different color pencils to $10$ drawers so there are at least $2$ pencils in the $10$th drawer? [closed]
Тhere are $10$ drawers and $7$ different color pencils. How many ways we can arrange them, so there are at least $2$ pencils in the $10$th drawer?
So i tried doing it this way:
1.If all 7 pencils are in the 10 drawer this is 1.
2.If 6 pencils are in the 10 drawer, this are Variations with rep. $\to 9^1=9$
3.If 5 pencils are in the 10 drawer, this are Variations with rep. $\to 9^2=81$
4.If 4 pencils are in the 10 drawer, this are Variations with rep. $\to 9^3=729$
5.If 3 pencils are in the 10 drawer, this are Variations with rep. $\to 9^4=6561$
6.If 2 pencils are in the 10 drawer, this are Variations with rep. $\to 9^5=59049$
So when i sum them i get $66430$ but it's far from the answer.
Any hints? Thank you!
## closed as off-topic by Austin Mohr, Magdiragdag, Claude Leibovici, Juniven, user223391 Apr 22 '17 at 13:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Austin Mohr, Magdiragdag, Claude Leibovici, Juniven, Community
If this question can be reworded to fit the rules in the help center, please edit the question.
• Hint: Substract the conditions with 0 pencil or 1 pencil on the 10th drawer from the all combinations. – offret Apr 15 '17 at 17:16
• A different approach from Burak's: how many ways can you distribute 5 pencils to 10 drawers, without additional restrictions? And how many ways can you pick 2 pencils out of 7? – Paul Sinclair Apr 15 '17 at 22:52
• @PaulSinclair: That can work, but OP would have to be careful of overcounting. For instance, if pencils 1, 2, and 3 are in the tenth drawer, then without being careful, that would be counted as selecting pencils 1 and 2 to go in the tenth drawer, plus selecting pencils 1 and 3 to go in the tenth drawer, plus selecting pencils 2 and 3 to go in the tenth drawer. I think it would be harder to keep track of the overcounting than some more direct method. – Brian Tung Apr 21 '17 at 14:48
There are $10^7$ ways of distributing seven pencils into ten drawers. Of course, many of those ways don't have at least two pencils in the tenth drawer. So let's subtract those out.
How many ways are there to have no pencils in the tenth drawer? That's equal to the number of ways to distribute seven pencils into the first nine drawers, which is $9^7$. So we subtract that out.
Now, how many ways are there to have exactly one pencil in the tenth drawer? Well, let's take it slowly: How many ways are there to have only the first pencil in the tenth drawer? That means that the remaining six pencils have to all be distributed into the first nine drawers, which is $9^6$. So we have to subtract that out. But that's just the first pencil. Each of the other six pencils can be the unique pencil in the tenth drawer, too. So that's six more subtractions of $9^6$.
Altogether, then, we have
$$10^7-9^7-(7 \times 9^6)$$
which is, indeed, quite a ways from $66430$.
The reason why you didn't get the right answer is that you didn't account for all the different combinations of pencils that could be in the tenth drawer. For instance, you write, for $4$ pencils in the tenth drawer, a total count of $729$. But that only takes care of the ways to put the other $3$ pencils in the other $9$ drawers. You forgot that there are many ways to select the $4$ pencils that will go into the tenth drawer—$\binom{7}{4} = 35$, in fact.
If you take that into account in your approach, you will have
$$1+\binom76\times9+\binom75\times9^2+\binom74\times9^3+\binom73\times9^4+\binom72\times9^5$$
and you will end up with the same answer as above.
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# Convergence of an alternating improper integral.
I'm trying to determine the convergence/divergence of two improper integrals, both with the same attribute:
$$\int_2^\infty f(x)\,dx=\int_2^\infty \frac{x-\lfloor x\rfloor-\frac{1}{2}}{\ln{x}}dx$$
$$\int_1^\infty g(x)\,dx=\int_1^\infty \frac{(-1)^{\lfloor x \rfloor}}{x\ln{x}}dx$$
The common attribute these two have is that their numerators are alternating between two constant values as the functions progresses twoards infinity.
Note the numerators are bounded by some constant values, and the denominators are monotonically decreasing. Thus, I thought I could use Dirichlet's test for convergence of improper integrals to determine convergence.
In $f(x)$ I can do that, since both the numerator and denominator are continuous as that's what the test requires for convergence. However, in $g(x)$ I can't, since the numerator isn't continuous.
For the first integral $$\int_2^{\infty} \frac{\{x\}- 1/2}{\ln(x)} dx$$ Define $\displaystyle B_1(x) = \frac{\{x\}^2}{2} - \frac{\{x\}}{2} + \frac1{12}$. Then we get that $$\int_2^{\infty} f(x) dx = \int_2^{\infty} \frac{d(B_1(x))}{\ln(x)} = \left. \frac{B_1(x)}{\ln(x)} \right \rvert_2^{\infty} + \int_2^{\infty} \frac{B_1(x)}{x \ln^2(x)} dx = \frac1{12 \ln(2)} + \int_2^{\infty} \frac{B_1(x)}{x \ln^2(x)} dx$$ which converges clearly since $$\left \lvert \int_2^{\infty} \frac{B_1(x)}{x \ln^2(x)} dx \right \rvert \leq \frac1{12} \left \lvert \int_2^{\infty} \frac{1}{x \ln^2(x)} dx \right \rvert < \infty$$
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# RCL Series Circuit Question
Discussion in 'Homework Help' started by wildnixon, May 27, 2008.
1. ### wildnixon Thread Starter Member
May 1, 2008
12
0
I am having trouble figuring out the phasor voltage and phasor current of an RCL series circuit. I am given the following values: v(t)=10sin(wt) (that w is a small Omega) and that the value of w is 1000 rad/sec. The inductance is 120mH, the capacitance= 1 micro Farad, and the resistance =100 ohms. I have calculated the frequency given the omega value, but I end up with a really small number : 6.3x10-3, and then I end up with a really small XL number and a really huge XC!! I know the math once I get the value of the frequency right, but I am just beating my head against the wall here!!
Thanks for any help!
D
2. ### mrmount Active Member
Dec 5, 2007
59
7
I am not sure how you came up with that value for w. w=2*Pi*f; f=w/2*pi. So you will be getting 159.23 Hz! The value you have found is 'T' or the time taken for one complete cycle of the sine wave.
3. ### wildnixon Thread Starter Member
May 1, 2008
12
0
I will proceed to use the 159 Hz...I see now where I botched it...it's always something simple...
Thanks!
4. ### silvrstring Active Member
Mar 27, 2008
159
0
wildnixon,
omega = 2*pi*f. So f = omega/(2*pi) = 159.2Hz.
X(L) = 2*pi*f*L, and will be at 90 degrees (or * j).
X(c) = 1/(2*pi*f*C), and will be at -90 degrees (or * -j).
Z = R + jX(l) - jX(c) = sqrt(R^2 + X^2) at angle tan^(-1) (X/R).
If you are keeping with peak voltage values, your pk V value is 10V at 0 degrees.
I = V / Z.
example (10V at 0 degrees/500 ohms at 60 degrees = 20 mA at -60 degrees). Not the answer by the way!
Hope this helps.
5. ### zamansabbir Member
May 27, 2008
15
0
XL=w*L (w=small omega)
Xc=1/(w*c)
R= 100 ohm
Hence Z=R+J(XL-Xc) ohm
in phasor form Z= sqrt(R^2+(XL-Xc)^2)< arc tangent of((XL-Xc)/R)
then use V= I*Z;
to get the current through every component and then use voltage divider rule or same ohms law to get the voltage phasor
Hope u understand
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# Exploring reflections and translations (Part 2)
In this lesson, we will look at more examples of how a shape can be reflected or translated wihlst preserving its appearance.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.How many types of transformations are there?
1/5
Q2.Which statement below is correct?
2/5
Q3.Which statement below is correct?
3/5
Q4.Which statement below is correct?
4/5
Q5.If the vertices of triangle A are (-5,3), (-5,1) and (-2,1) and the vertices of triangle B are (-5,-1), (-5,-3) and (-2,-3), which of the below best describes the transformation from A to B?
5/5
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.How many types of transformations are there?
1/5
Q2.Which statement below is correct?
2/5
Q3.Which statement below is correct?
3/5
Q4.Which statement below is correct?
4/5
Q5.If the vertices of triangle A are (-5,3), (-5,1) and (-2,1) and the vertices of triangle B are (-5,-1), (-5,-3) and (-2,-3), which of the below best describes the transformation from A to B?
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Exploring Reflections and Translations Quiz
Quiz about the similarities of translating and reflecting shapes
Q1.Which of the statements below is correct?
1/5
Q2.Which of the statements below is correct?
2/5
Q3.What would happen if we reflected a shape onto the x-axis and then onto the y-axis and then back onto the x-axis and then onto the y-axis?
3/5
Q4.Which of the statements below is correct?
4/5
Q5.Which of the statements below is NOT correct?
5/5
Quiz:
# Exploring Reflections and Translations Quiz
Quiz about the similarities of translating and reflecting shapes
Q1.Which of the statements below is correct?
1/5
Q2.Which of the statements below is correct?
2/5
Q3.What would happen if we reflected a shape onto the x-axis and then onto the y-axis and then back onto the x-axis and then onto the y-axis?
3/5
Q4.Which of the statements below is correct?
4/5
Q5.Which of the statements below is NOT correct?
5/5
# Lesson summary: Exploring reflections and translations (Part 2)
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Dance
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# How do you find the derivative of x^2 lnx?
$y ' = 2 x \cdot \ln \left(x\right) + {x}^{2} \cdot \frac{1}{x} = 2 x \ln \left(x\right) + x = x \left[2 \ln \left(x\right) + 1\right]$
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# Jee Mains Year 2015 Maths Solutions
## Maths Year 2015 Maths Jee Mains Solutions
Free Jee Mains video solutions with Jee Mains 2015 question paper for Maths. Doubtnut provides Jee Mains solution of 2015 for Maths with Jee Mains 2015 question paper.
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### JEE MAINS Class 11 | SETS
Let `A and B` be two sets containing four and two elements respectively. Then the number of subsets of the set `A xx B,` each having at least three elements is : (1) `219` (2) `256` (3) `275` (4) `510`
### JEE MAINS Class 11 | COMPLEX NUMBERS
A complex number z is said to be unimodular if . Suppose `z_1` and `z_2` are complex numbers such that `(z_1-2z_2)/(2-z_1z_2)` is unimodular and `z_2` is not unimodular. Then the point `z_1` lies on a : (1) straight line parallel to x-axis (2) straight line parallel to y-axis (3) circle of radius 2 (4) circle of radius `sqrt(2)`
### JEE MAINS Class 11 | QUADRATIC EQUATIONS
Let `alpha` and `beta` be the roots of equation `x^2-6x-2""=""0` . If `a_n=alpha^n-beta^n ,""for""ngeq1` , then the value of `(a_(10)-2a_8)/(2a_9)` is equal to: (1) 6 (2)`-6` (3) 3 (4) `-3`
### JEE MAINS Class 12 | MATRICES
If `A=[(1,2,2),(2,1,-2),(a,2,b)]` is a matrix satisfying the equation `A A^T=9I`, where `I` is `3xx3` identity matrix, then the ordered pair `(a,b)` is equal to : (A) `(2,-1)` (B) `(-2,1)` (C) `(2, 1)` (D) `(-2,-1)`
### JEE MAINS Class 12 | MATRICES
The set of all values of `lambda` for which the system of linear equations : `2x_1-2x_2+""x_3=lambdax_1` `2x_1-3x_2+""2x_3=lambdax_2` `-x_1+""2x_2=lambdax_3` has a non-trivial solution, (1) is an empty set (2) is a singleton (3) contains two elements (4) contains more than two elements
### JEE MAINS Class 11 | PERMUTATIONS AND COMBINATIONS
The number of integers greater than `6,000` that can be formed, using the digits `3, 5, 6, 7 and 8,` without repetition, is :
### JEE MAINS Class 11 | BINOMIAL THEOREM
The sum of coefficients of integral powers of x in the binomial expansion of `(1-2sqrt(x))^(50)` is: (1) `1/2(3^(50)+1)` (2) `1/2(3^(50))` (3) `1/2(3^(50)-1)` (4) `1/2(2^(50)+1)`
### JEE MAINS Class 11 | SEQUENCES AND SERIES
If m is the A.M. of two distinct real numbers `l` and `n""(""l ,""n"">""1)` and` G_1, G_2` and `G_3` are three geometric means between `l` and n, then `G_1^4+2G_2^4+G_3^4` equals, (1) `4l^2` mn (2) `4l^m^2` mn (3) `4l m n^2` (4) `4l^2m^2n^2`
### JEE MAINS Class 11 | SEQUENCES AND SERIES
The sum of first 9 terms of the series `(1^3)/1+(1^3+2^3)/(1+3)+(1^3+2^3+3^3)/(1+2+3)+. . . . .` is (1) 71 (2) 96 (3) 142 (4) 192
### JEE MAINS Class 11 | LIMITS AND DERIVATIVES
`(lim)_(xvec0)((1-cos2x)(3+cosx)/(xtan4x)` is equal to: (1) 4 (2) 3 (3) 2 (4) `1/2`
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# Project #22673 - Mathematics Problems
Please do following 5 questions. thanks.
1- John earns an hourly wage of \$ 10.00 an hour at a local retail store. If he works a 35 hour week, how much will he earn in one year?
2- Shauna earns an hourly wage of \$ 12.00. If she works an 8 hour shift with two hours at time and a half, how much will she earn in total for the shift?
3- Martina has graduated from school and has an office job that pays her \$36,000 starting. What would be her hourly income? Assume 40 hr work week. Assume 52 weeks in a year.
4- The employees of a manufacturing plant received a 6 % increase in pay. What is the amount of the increase for an employee who makes \$1240 a week? What is the weekly wage for the employee after the wage increase?
5- A supervisor’s salary this year is \$ 32,000. This salary will increase by 8 % next year.
a. a- What is the amount of increase?
b. b- What will the salary be next year?
Subject Mathematics Due By (Pacific Time) 02/11/2014 10:01 pm
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# Suppose ABCD is a parallelogram in which angle A = 108'. Calculate angles B, C and D.
2
by AishwaryaRangan
2014-05-30T15:35:46+05:30
If ABCD is a parallelogram
In parallelogram ABCD
m/_A=108·
m/_C=m/_A=108· (opposite angle of a parallelogram are congruent)
m/_A+m/_B=180 (adjacent angle of a parallelogram are congruent)
108+m/_B=180
m/_B=180-108
m/_B=72
m/_B=m/_D=72 (opposite angle of a parallelogram are congruent)
• Brainly User
2014-05-30T15:46:26+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Consider a,b,c,d to be the vertices of the parallelogram taken in cyclic order.
A = 108 degree
opposite sides are equal
C = 108 degree
base angles are supplementary
A+B = 180
B =180-108=72
B and D are opposite . Hence D = 72 degree
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Application Center - Maplesoft
# Cauchy problems for wave equations (updated to Maple 7)
You can switch back to the summary page by clicking here.
wave_eqc.mws
Solving Cauchy problems for wave equations
by Aleksas Domarkas
Vilnius University, Faculty of Mathematics and Informatics,
Naugarduko 24, Vilnius, Lithuania
[email protected]
NOTE: In this session we solve Cauchy problems for wave equations.
Cauchy problem for 1d wave equation (1)
Problem
from M.Kawski
ftp://math.la.asu.edu/pub/kawski/MAPLE/362/dAlembert2hats.mws
> restart;
> eq:=diff(u(x,t),t,t)=4*diff(u(x,t),x,x);#x>-infinity,x0;
> ic:=u(x,0)=u0(x);ict:=D[2](u)(x,0)=u1(x);#x>-infinity,x
> hat2:=x->1-abs(abs(x)-2);
> ramp:=x->(x+abs(x))/2;
> u0:=ramp@hat2;
> u1:=0;
> plot(u0,-4..4,-1..2,scaling=constrained,axes=boxed);
> convert(u0(x),piecewise);
> u0:=unapply(convert(u0(x),Heaviside),x);
>
Solving(1 method)
We use d'Alembert's formula for solution:
> a:=2:
> sol1:=(u0(x-a*t)+u0(x+a*t))/2;
>
Solving(2 method)
> assume(t>0);
> with(inttrans):
> fourier(eq,x,w);
> ode:=subs(fourier(u(x,t),x,w)=s(t),%);
> dsolve({ode,s(0)=fourier(u0(x),x,w), D(s)(0)=fourier(u1(x),x,w)},s(t));
> sol2:=simplify(invfourier(rhs(%),w,x));
>
Checking the Solution
sol1=sol2 ???
> simplify(sol1-sol2);
> u:=unapply(sol1,(x,t)):
> simplify(rhs(eq)-lhs(eq));
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
Plots
> plot3d(sol1,x=-10..10,t=0..5,orientation=[-60,20],axes=framed);
> with(plots):with(plottools):
```Warning, the name changecoords has been redefined
```
```Warning, the name arrow has been redefined
```
Animation, created by M.Kawski:
> display([ seq( display([ textplot([1.4,1.4,cat(`t = `,convert(tt,string))], 'align=RIGHT',font=[TIMES,ROMAN,18]), plot(subs(t=tt,sol1),x=-6..6,-1..2, thickness=3) ]), tt=[seq(evalf(k/20,3),k=0..100)]) ],insequence=true, titlefont=[TIMES,BOLD,16], title=`Animation: d'Alembert's solution`);
>
Cauchy problem for 1d wave equation (2)
Problem
> restart;
> eq:=diff(u(x,t),t,t)=a^2*diff(u(x,t),x,x);
> ic:=u(x,0)=u0(x);ict:=D[2](u)(x,0)=u1(x);
> u0:=unapply(randpoly([x]),x);
> u1:=unapply(randpoly([x]),x);
> assume(a>0,t>0);
Solving(1 method)
We use d'Alembert's formula for solution:
> (u0(x-a*t)+u0(x+a*t))/2+int(u1(xi),xi = x-a*t .. x+a*t)/(2*a);
> sol1:=simplify(%);
>
Solving(2 method)
> L:=f->diff(f,x,x);
> d:=max(degree(u0(x)),degree(u1(x)));
> Sum(a^(2*k)*t^(2*k)/(2*k)!*(L@@k)(u0(x))+ a^(2*k)*t^(2*k+1)/(2*k+1)!*(L@@k)(u1(x)),k=0..d);
> sol2:=simplify(value(%));
> is(sol1=sol2);
>
Solving(3 method)
> with(inttrans):
> fourier(eq,x,w);
> ode:=subs(fourier(u(x,t),x,w)=s(t),%);
> dsolve({ode,s(0)=fourier(u0(x),x,w), D(s)(0)=fourier(u1(x),x,w)},s(t));
> sol3:=invfourier(rhs(%),w,x);
> is(sol1=sol3);
Checking the Solution
> u:=unapply(sol1,(x,t)):
> simplify(rhs(eq)-lhs(eq));
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
>
Cauchy problem for 2d wave equation
Problem( 2d wave equation)
> restart;with(linalg,laplacian):with(student,Doubleint):
> eq:=diff(u(x,y,t),t,t)=a^2*laplacian(u(x,y,t),[x,y]);
> ic:=u(x,y,0)=u0(x,y);ict:=D[3](u)(x,y,0)=u1(x,y);
> u0:=unapply(randpoly([x,y]),x,y);
> u1:=unapply(randpoly([x,y]),x,y);
> assume(a>0,t>0);
Solving(1 method)
We use Poisson formula for solution:
where B={( ): }.
For computing integrals we use polar coordinates:
> T:={xi=x+r*cos(phi),eta=y+r*sin(phi)};
> subs(T,1/(2*Pi*a)* Doubleint(u1(xi,eta)/sqrt((a*t)^2-r^2)*r, phi=0..2*Pi,r=0..a*t)+1/(2*Pi*a)*Diff( Doubleint(u0(xi,eta)/sqrt((a*t)^2-r^2)*r, phi=0..2*Pi,r=0..a*t),t));
> sol1:=simplify(value(%));
Solving(2 method)
> L:=f->laplacian(f,[x,y]);
> d:=max(degree(u0(x,y)),degree(u1(x,y)));
> Sum(a^(2*k)*t^(2*k)/(2*k)!*(L@@k)(u0(x,y))+ a^(2*k)*t^(2*k+1)/(2*k+1)!*(L@@k)(u1(x,y)),k=0..d);
> sol2:=simplify(value(%));
> is(sol1=sol2);
>
Solving(3 method)
> with(inttrans);
> fourier(eq,x,w1);
> fourier(%,y,w2);
> subs(fourier(fourier(u(x,y,t),x,w1),y,w2)=s(t),%);
> dsolve({%,s(0)=fourier(fourier(u0(x,y),x,w1),y,w2), D(s)(0)=fourier(fourier(u1(x,y),x,w1),y,w2)},s(t));
> invfourier(rhs(%),w2,y);
> sol3:=invfourier(%,w1,x);
> is(sol1=sol3);
Checking the Solution
> u:=unapply(sol1,(x,y,t)):
> simplify(rhs(eq)-lhs(eq));
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
>
Cauchy problem for 3d wave equation
Problem
> restart;with(linalg):with(student):
```Warning, the protected names norm and trace have been redefined and unprotected
```
> eq:=diff(u(x,y,z,t),t,t)=a^2*laplacian(u(x,y,z,t),[x,y,z]);
> ic:=u(x,y,z,0)=u0(x,y,z);ict:=D[4](u)(x,y,z,0)=u1(x,y,z);
> u0:=unapply(randpoly([x,y,z]),x,y,z);
> u1:=unapply(randpoly([x,y,z]),x,y,z);
> assume(a>0,t>0);
Solving(1 method)
We use Kirchhoff formula for solution:
where S={( ): }.
> sol:=Int(u1,S)/(4*Pi*a^2*t)+Diff(Int(u0,S)/t,t)/(4*Pi*a^2);
For computing integrals we use spherical coordinates:
> T:={xi[1]=x+r*cos(phi)*sin(theta),xi[2]=y+r*sin(phi)*sin(theta), xi[3]=z+r*cos(theta)};
> linalg[jacobian]([x+r*cos(phi)*sin(theta),y+r*sin(phi)*sin(theta), z+r*cos(theta)],[phi,r,theta]);
> J:=simplify(det(%));
> subs(Int(u1,S)=Doubleint(subs(T,u1(xi[1],xi[2],xi[3]))*J, theta=0..Pi,phi=0..2*Pi), Int(u0,S)=Doubleint(subs(T,u0(xi[1],xi[2],xi[3]))*J, theta=0..Pi,phi=0..2*Pi), r=a*t,sol);
> sol1:=simplify(value(%));
Solving(2 method)
> L:=f->laplacian(f,[x,y,z]);
> d:=max(degree(u0(x,y,z)),degree(u1(x,y,z)));
> Sum(a^(2*k)*t^(2*k)/(2*k)!*(L@@k)(u0(x,y,z))+ a^(2*k)*t^(2*k+1)/(2*k+1)!*(L@@k)(u1(x,y,z)),k=0..d);
> sol2:=simplify(value(%));
> is(sol1=sol2);
>
Solving(3 method)
> with(inttrans):
```Warning, the name hilbert has been redefined
```
> fourier(eq,x,w1);
> fourier(%,y,w2);
> fourier(%,z,w3);
> ode:=subs(fourier(fourier(fourier(u(x,y,z,t),x,w1),y,w2),z,w3)=s(t),%);
> dsolve({ode,s(0)=fourier(fourier(fourier(u0(x,y,z),x,w1),y,w2),z,w3), D(s)(0)=fourier(fourier(fourier(u1(x,y,z),x,w1),y,w2),z,w3)},s(t));
> invfourier(rhs(%),w3,z):
> invfourier(%,w2,y):
> sol3:=invfourier(%,w1,x);
> is(sol1=sol3);
>
Checking the Solution
> u:=unapply(sol1,(x,y,z,t)):
> simplify(rhs(eq)-lhs(eq));
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
>
Non-homogenous wave equation
Problem
> restart;with(linalg,laplacian):
> eq:=diff(u(x,y,z,t),t,t)-a^2*laplacian(u(x,y,z,t),[x,y,z])=F(x,y,z,t);#-infty0
> ic:=u(x,y,z,0)=0;#-infty
> ict:=D[4](u)(x,y,z,0)=0;#-infty
The function F(x,y,z,t) may be polynomial respect to x,y,z .
> randpoly([x,y,z,t,t*sin(t),cos(t),exp(t)]):
> F:=unapply(%,[x,y,z,t]);
procedure
> wsol:=proc(u) option `Copyright Aleksas Domarkas, 2002`; local dsol; dsol:=proc(F) local g,f,d,L; if has(F, t) and not has(F,[x,y,z]) then g:=unapply(F,t); f:=1; elif not has(F,t) then g:=1;f:=F; else f:=select(has,F,[x,y,z]); g:=unapply(F/f,t); end if; d:=max(degree(f),1); L:=f->linalg[laplacian](f,[x,y,z]); sum(a^(2*k)/(2*k+1)!*(L@@k)(f)*Int((t-tau)^(2*k+1)*g(tau),tau=0..t),k=0..d);value(%); end proc: if type(u,`+`) then map(dsol,u) else dsol(u) end if: end proc:
solution
> sol:=wsol(F(x,y,z,t));
>
test
> u:=unapply(sol,(x,y,z,t)):
> simplify(rhs(eq)-lhs(eq)):if %<>0 then combine(%,trig) else % fi;
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
>
Decomposition method
> restart;with(linalg,laplacian):with(inttrans,fourier,invfourier):
Proble m
From V.S.Vladimirov(ed.), Exercises book on Equations of Mathematical Physics, Nauka, Moscow, 1982(in Russian) 12.38.7.
> eq:=diff(u(x,y,z,t),t,t)-a^2*laplacian(u(x,y,z,t),[x,y,z])=exp(z)*sin(y)*cos(x);#-infty0
> ic:=u(x,y,z,0)=x^2*exp(y+z);#-infty
> ict:=D[4](u)(x,y,z,0)=sin(x)*exp(y+z);#-infty
I
> eq1:=eq;
> ic1:=u(x,y,z,0)=0;
> ict1:=D[4](u)(x,y,z,0)=0;
Solving of I
> subs(u(x,y,z,t)=v(t)*rhs(eq1),eq1);
> simplify(%/rhs(eq1));
> dsolve({%,v(0)=0,D(v)(0)=0},v(t));
> sol1:=rhs(%)*rhs(eq1);
>
II
> eq2:=diff(u(x,y,z,t),t,t)-a^2*laplacian(u(x,y,z,t),[x,y,z])=0;
> ic2:=u(x,y,z,0)=0;
> ict2:=D[4](u)(x,y,z,0)=sin(x)*exp(y+z);
Solving of II
> subs(u(x,y,z,t)=v(t)*rhs(ict2),eq2);
> simplify(%/rhs(ict2));
> dsolve({%,v(0)=0,D(v)(0)=1},v(t));convert(%,trig): combine(%,trig);
> sol2:=rhs(%)*rhs(ict2);
>
III
> eq3:=eq2;
> ic3:=u(x,y,z,0)=x^2*exp(y+z);
> ict3:=D[4](u)(x,y,z,0)=0;
Solving of III
> subs(u(x,y,z,t)=v(x,t)*exp(y+z),eq3);
> eqn:=simplify(%/exp(y+z));
> fourier(eqn,x,w);
> subs(fourier(v(x,t),x,w)=s(t),%);
> dsolve({%,s(0)=fourier(x^2,x,w),D(s)(0)=0},s(t));
> subs(s(t)=fourier(v(x,t),x,w),%);
> invfourier(%,w,x);
> convert(%,trig): combine(%,trig);
> sol3:=rhs(%)*exp(y+z);
>
Solution
> sol:=sol1+sol2+sol3;
Checking the Solution
> u:=unapply(sol,(x,y,z,t)):
> simplify(rhs(eq)-lhs(eq));
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
>
Radial-symmetric solution
For radial-symmetric solutions of linear elliptic equations see radsymell.mws
Problem
> restart;with(linalg,laplacian):
[vlad] 12.38.9.
> eq:=diff(u(x,y,z,t),t,t)-a^2*laplacian(u(x,y,z,t),[x,y,z])=0;
> ic:=u(x,y,z,0)=cos((x^2+y^2+z^2)^(1/2));
> ict:=D[4](u)(x,y,z,0)=cos((x^2+y^2+z^2)^(1/2));
Solving problem
> eqn:=diff(u(r,t),t,t)-a^2*expand(laplacian(u(r,t),[r,theta,phi],coords=spherical));
Solution we search in form :
> subs(u(r,t)=v(r,t)/r,eqn);
> eq1:=simplify(%*r);
> ic1:=v(r,0)=r*cos(r);
> ict1:=D[2](v)(r,0)=r*cos(r);
> phi:=unapply(rhs(ic1),r);
> psi:=unapply(rhs(ict1),r);
By d'Alembert's formula :
> v(r,t)=(phi(r+a*t)+phi(r-a*t))/2+Int(psi(xi),xi=r-a*t .. r+a*t)/2/a;
> value(rhs(%))/r;
> expand(%);
> collect(%,{sin(r),cos(r)});
> applyop(collect,[1,1],%,r):applyop(collect,[2,1],%,r);
Solution
> sol:=subs(r=sqrt(x^2+y^2+z^2),%);
>
Checking the Solution
> u:=unapply(sol,(x,y,z,t)):
> simplify(rhs(eq)-lhs(eq));
> simplify(rhs(ic)-lhs(ic));
> simplify(rhs(ict)-lhs(ict));
>
While every effort has been made to validate the solutions in this worksheet, Waterloo Maple Inc. and the conevibutors are not responsible for any errors contained and are not liable for any damages resulting from the use of this material.
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# Question #36d12
Dec 15, 2014
${C}_{3} {H}_{8} + 5 {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$
$\Delta {H}_{c} = - 2220 k J . m o {l}^{- 1} .$
${C}_{8} {H}_{18} + 8 \frac{1}{2} {O}_{2} \rightarrow 8 C {O}_{2} + 9 {H}_{2} O$
$\Delta {H}_{c} = - 5530 k J . m o {l}^{- 1}$
${M}_{r} {C}_{3} {H}_{8} = \left(3 \times 12 + 8 \times 1\right) = 44$
So 44g gives 2220kJ
So 1g gives 2220/44 = 50.45 kJ
${M}_{r} {C}_{8} {H}_{18} = \left(8 \times 12 + 18 \times 1\right) = 114$
So 114g gives 5530 kJ
So 1g gives 5530/114 = 48.51 kJ.
So we can see that propane gives slightly more energy per gram.
From an environmental point of view the amount of $C {O}_{2}$ produced per gram is about the same.
The other big factor is cost. The cost of crude oil, hence gasoline, is falling due a drop in demand and the increase in shale gas production in the US. I'm not sure how propane compares currently. If anyone knows perhaps they could update.
Dec 15, 2014
Propane
The heat of combustion of propane is -2202 kJ/mol.
Energy per gram = $\text{2202 kJ"/"1 mol" × "1 mol"/"44.10 g}$ = 49.9 kJ/g
Octane
The heat of combustion of octane is -5430 kJ/mol.
Energy per gram = $\text{5430 kJ"/"1 mol" × "1 mol"/"114.2 g}$= 47.5 kJ/g
a. Efficiency
You get about 5 % more heat energy from a gram of propane than from a gram of octane.
b. Cost
In American units, the international cost of propane at the time of writing is $0.51/gal. The average international cost of gasoline is$4.43/gal. Propane is much cheaper on a volume basis.
a. Difficulty in handling
Propane must be handled as a pressurized liquid.
b. Danger
Propane vapour is denser than air. If there is a leak, the gas will sink into any enclosed area and cause a risk of explosion and fire.
c. Limited availability
Propane is less readily available to consumers than gasoline. Buses can refuel at the terminal, but consumers have fewer options.
Most personal vehicles need expensive conversions to handle propane
d. About 18 % less driving distance per litre of fuel
Because liquid propane is less dense than gasoline.
e. Fuel tanks can be filled to only 80 % of capacity
Because liquid propane expands as it warms from -42 °C to ambient temperature.
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: 136.330 Intro to Algebra: Answers to problems
By Anonymous on Tuesday, September 17, 2002 - 09:05 pm: Edit Post
Hi! i'm wondering how to do 2.8., 2.21,2.22
By Sasho on Tuesday, September 17, 2002 - 11:11 pm: Edit Post
Please write down the statements of the problems (for the sake of the people who want to see both the statement and a solution or a hint, and because I do not drag the textbook with me and so I do not know what, say, 2.8 is).
By Anonymous on Thursday, September 19, 2002 - 12:57 pm: Edit Post
2.8 Describe the inverse of the mapping Beta in Example 2.1
Example 2.1
Let S={x,y,z} T={1,2,3} and U={a,b,c}
Define theta:S-T by theta(x)=2 theta(y)=1 theta(z)=3
beta(1)=b, beta(2)=c beta(3)=a
2.21 give an example of sets s,t, and and mappings theta: s-t and beta: t-u such beta*theta is onto, but theta is not onto.[compare theorem 2.1(b)}
2.22 give an example of sets s,t, and and mappings theta: s-t and beta: t-u such beta*theta is onto, but beta is not one to one.[compare theorem 2.1(b)}
By Sasho on Thursday, September 19, 2002 - 01:49 pm: Edit Post
2.8.
The inverse b-1 of b: T ® U is the mapping U ®T that works as follows: b-1(a)=3, b-1(b)=1 and b-1(c)=2 (just follow in the opposite direction the arrows in the textbook describing b).
2.21.
Take S={1}, T={a,b}, U={x} and a(1)=a, b(a)=x=b(b). Then ba is certainly onto, but a is not onto.
2.22.
The example in 2.21 will do the work for 2.22 as well ! (check !)
By Paul on Tuesday, October 01, 2002 - 12:05 pm: Edit Post
There are a couple of examples in the book that I am having problems with. If you could please help me out it would be greatly appreciated.
page 14 #1.21a),b)
Assume that S and T are finite sets containing m and n elements, respectively.
a) How many mappings are there form S to T?
B) How many one-to-one mappings are there from S to T?
#1.29
Prove that a mapping alpha: S-T is one-to-one iff alpha(A intersection B) = alpha(A) intersection alpha(B) for every pair of subsets A and B of S.
By Sasho on Tuesday, October 01, 2002 - 01:58 pm: Edit Post
14. (a) the first element of S could go to any of the n elements of T (n choices); given where the first element of S is mapped, the second could still be mapped to any of the n elements of T; we have n.n=n2 many choices for the images of the first two elements of S. Keep going untill we exhaust all of the elements of S, to get the final ansewer of nm mappings S®T.
(b) requires a similar argument. The final answer is either 0 if m>n, or n(n-1)(n-2)...(n-m+1) if n is at least as large as m.
1.29. I wil show a part of one of the two implications: if a:S®T is one-to-one, then a(A intersect B) = a(A) intersect a(B).
So, suppose a: is 1-1. Take an element x in a(A intersect B). Then x=a(y) for some y in A intersect B. But since y is in A intersect B, we have that y is both in A and in B. So, x=a(y) is in a(A) (since y is in A), and x=a(y) is in a(B) (since y is in B). So, since we got that x is in both a(A) and a(B), it must be in a(A) intersect a(B). This shows that a(A intersect B) is a subset of a(A) intersect a(B). You then prove that a(A) intersect a(B) is a subset of a(A intersect B) by taking an element in the first set and showing it must be in the second. That would do this implication. [Note that I have not used the assumption that a is 1-1; it is needed in the part that is omitted (that a(A) intersect a(B) is a subset of a(A intersect B))].
Then you do the converse (if a(A intersect B) = a(A) intersect a(B) then a is 1-1).
Here is something to make your life easier: I will not give you that question for the test - it needs too much of manual work. But, a small portion of it might be in.
By Paul on Monday, October 28, 2002 - 10:35 pm: Edit Post
I am having a few difficulties with equivalence classes. If you could please answer the following questions, I should be able to figure the rest out.
q. 9.8 b)
Define a relation ~ on the set N of natural numbers by a~b iff a = (b)(10^k) for some k in Z.
Give a complete set of equivalence class representatives.
q. 9.10
Give a complete set of equivalence class representatives for the equivalence relation in Example 9.2. (Let L denote the set of all lines in a plane with rectangular coordinate system. For l1, l2 in L, let l1~l2 mean that l1 and l2 have equal slopes or that both slopes are undefined. This is an equivalence relation on L. The set of lines equivalent to a line l consists of l and all lines in L that are parallel to l.)
Thanks
By Sasho on Tuesday, October 29, 2002 - 09:57 am: Edit Post
9.8(b) Observe that a~b if a=b00000 (a bunch of 0-s) or if b=a00000 (a bunch of 0-s). So, the equivalence classes are {1,10,100,1000,...}, {2,20,200,...}, ....{9,90,900,...}, {11,110,1100,...}, {12,120,...},.... (representatives of these classes are numbers that do NOT end with 0).
9.10. The last sentence in the statement almost gives the answer to the question: one equivalence class is the set of lines of a fixed slope r; as r ranges over the set of all real numbers we get almost all equivalence classes; the only one that is missing and should also be accounted for is the equivalence class of all vertical lines (of slope "infinity").
By Paul on Tuesday, October 29, 2002 - 04:14 pm: Edit Post
Hello again,
You gave us a proof to do in class a while ago, order of An is 1/2(n!). I took a look in the book but I don't understand what they are doing with theta. If you could please clear that up for me, thanks.
By Sasho on Tuesday, October 29, 2002 - 04:42 pm: Edit Post
I dont have the textbook with me; so you need to write down the part that is unclear to you.
And by the way - please start a new topic; this one is becoming too long. (use "create new conversation" button at the bottom of the page listing the topics so far)
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# Calculate the sum.
There is given positive decreasing sequence of real numbers $$a_{1},a_{2},a_{3},...$$
• Calculate the sum in nice form: $$SUM=\sum_{i=1}^{N}a_{i}-2\sum_{1\le i
• If it is impossible to do it in general then let try it in case when $$a_{n}=\frac{(-1)^{n}}{n^\sigma}$$ where $$0<\sigma<1$$.
• Find lower bound of the $$SUM$$
After few transformations i got $$SUM=(\sum_{n=1}^{N}(-1)^{n}a_{n})^{2}-4\sum_{n=1}^{N}a_{n}\sum_{m=1}^{N-n}a_{n+2m}$$
But is this going to help me?
Second idea is about considering these elements of sequence as a variables of polynomial(which degree is equal to $$2$$)
Any hint?
Regards.
• Huh? $a_n=(-1)^n/n^\sigma$ does not satisfy $(a_n)$ decreasing. – user10354138 May 17 at 17:17
• Hint: $2\sum_{i<j} a_i a_j = \sum_{i\ne j}a_i a_j = \left(\sum_i a_i\right)^2 - \sum_i a_i^2$ If you sum stop at $\infty$ instead of finite $N$, the sum you have can be expressed in terms of zeta functions. – achille hui May 17 at 17:22
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# The magnitude of (|E| / |H|) in a uniform plane wave is:
1. $$\sqrt {\mu \epsilon}$$
2. Infinity
3. $$\sqrt {\dfrac{\mu}{ \epsilon}}$$
4. 1
## Answer (Detailed Solution Below)
Option 3 : $$\sqrt {\dfrac{\mu}{ \epsilon}}$$
## Parallel Plane Waveguide MCQ Question 1 Detailed Solution
Concept:
• Electric field $$\left( {\vec E} \right)$$ & magnetic field $$\left( {\vec H} \right)$$ are both orthogonal/transverse to each other as well as for the direction of propagation, called transverse electromagnetic wave (TEM).
• E & H combination called uniform plane wave because E & H has some magnitude through any transverse plane.
Intrinsic Impedance $$\left( \eta \right) = \frac{E}{H}$$
And also $$\eta = \sqrt {\frac{{j\omega \mu }}{{\sigma + j\omega \varepsilon }}}$$
$$\Rightarrow \eta = \frac{E}{H} = \sqrt {\frac{\mu }{\varepsilon }}$$
Pointing vector is a vector whose direction is the direction of wave propagation pointing vector $$= \vec E \times \vec H$$ (Hence it is a direction of wave propagation)
Where,
E = Electric filed
B = magnetic field
H = Magnetic field
μ0 = Permeability of free space = 4π x 10-7 H / m
ϵ0 = Permittivity of free space = 8.85 x 10-12 F/m
# The dispersion equation of a waveguide, which relates the wavenumber k to the frequency ω, is $$k\left( \omega \right) = \left( {1/\;c} \right)\;\sqrt {{\omega ^2} - \omega _o^2}$$ where the speed of light c = 3 × 108 m/s, and ωo is a constant. If the group velocity is 2 × 108 m/s, then the phase velocity is
1. 1.5 × 108 m/s
2. 2 × 108 m/s
3. 3 × 108 m/s
4. 4.5 × 108 m/s
## Answer (Detailed Solution Below)
Option 4 : 4.5 × 108 m/s
## Parallel Plane Waveguide MCQ Question 2 Detailed Solution
Concept:
Phase velocity:
The rate at which the phase of the wave propagates in space
$${V_p} = \frac{\omega }{\beta }$$
Group Velocity:
The velocity with which overall envelope of the wave travels
$${V_g} = \frac{{d\omega }}{{d\beta }} = \frac{{d\omega }}{{dk}}$$
$$K\left( \omega \right) = \frac{1}{c}\sqrt {\omega - \omega _0^2} \ldots \left( 1 \right)$$
Vg = 2 × 108 m/s (green)
$${V_g} = \frac{{d\omega }}{{dk}}$$
$$\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}}$$
differentiating (1)
$$\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}} = \frac{{1 \times 2\omega }}{{2C\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{{V_g}}}$$
$$\Rightarrow \frac{\omega }{{3 \times {{10}^8}\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{2 \times {{10}^8}}}$$
$$\sqrt {{\omega ^2} - \omega _0^2} = \frac{{2\omega }}{3}$$
$${V_p} = \frac{\omega }{K} = \frac{{\omega c}}{{\sqrt {{\omega ^2} - \omega _0^2} }}$$
$$= \frac{{\omega c}}{{\left( {\frac{{2\omega }}{3}} \right)}}$$
$${V_p} = \frac{{3c}}{2} = \frac{{3 \times {{10}^8} \times 3}}{2} = 4.5 \times {10^8}m/s$$
Shortcut:
Group velocity × Phase velocity = c2
2 × 108 × phase velocity = (3 × 108)2
$${V_p} = \frac{{9 \times {{10}^{16}}}}{{2 \times {{10}^8}}}$$
= 4.5 × 108 m/s
# The velocity of electromagnetic waves in a dielectric medium with relative permittivity of 4 will
1. 3 × 108 m/s
2. 6 × 108 m/s
3. 1.5 × 108 m/s
4. 0.75 × 108 m/s
## Answer (Detailed Solution Below)
Option 3 : 1.5 × 108 m/s
## Parallel Plane Waveguide MCQ Question 3 Detailed Solution
Concept:
The velocity (Vp) of a plane electromagnetic wave is given as-
$${V_p} = \frac{1}{{\sqrt {μ \varepsilon } }}$$
Here, μ = μoμr
ε = εo εr
$${v_p} = \frac{1}{{\sqrt {{μ _o}{μ _r}{\varepsilon _o}{\varepsilon _r}} }}$$
$${v_p} = \frac{c}{{\sqrt {{μ _r}{\varepsilon _r}} }}$$-----(1)
$$c = \frac{1}{{\sqrt {{μ _o}{\varepsilon _o}} }} = 3 \times {10^8}\;m/sec$$
μ0 = permeability in free space 4π × 10-7 H/m
εo = permittivity in free space 8.854 × 10-12 C2/Nm2
Calculation:
Given:
μr = 4
From Equation (1):
$$v_p=\frac{3\times10^8}{\sqrt4}$$
vp = 1.5 x 108 m/sec
# For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of spacing between the plates by a factor of two results in
1. halving of v and no change in Z
2. no change in v and halving of Z
3. no change in both v and Z
4. halving of both v and Z
## Answer (Detailed Solution Below)
Option 2 : no change in v and halving of Z
## Parallel Plane Waveguide MCQ Question 4 Detailed Solution
Concept:
Consider the following parallel plate line:
Assuming w ≫ h, the fringing effect can be ignored.
Where:
The bold line represents the direction of the Electric Field, and the dotted lines represent the direction of the Magnetic Field.
V = Ey h
I = Hx w
The Characteristic Impedance is given by:
$$Z = \frac{V}{I} = \frac{{{E_y}}}{{{H_x}}} \cdot \frac{h}{w} = \eta \frac{h}{w}$$
$$Z = \eta \frac{h}{w}$$ ---(1),
η = intrinsic impedance of dielectric medium.
If ‘h’ is reduced by a factor of 2 then Z will be halved.
Alternate Method:
$$Z = \sqrt {\frac{L}{C}}$$
$$Z = \sqrt {\frac{{\frac{{\mu h}}{w}}}{{\frac{{Ew}}{h}}}}$$
$$Z = \sqrt {\frac{\mu }{E}} \;\frac{h}{w}$$
We observe that If ‘h’ is halved, z is halved.
Important Point:
If the student has learned the formula for L and C then this problem will be easily solved by:
$$Z = \sqrt {\frac{L}{C}} \;\;and\;\;v = \frac{1}{{\sqrt {LC} }}$$
Also, the formula for L and C for the following are also important:
Coaxial cable:
$$C = \frac{{2\pi El}}{{In\;\left( {\frac{b}{a}} \right)}}$$
$$h = \frac{{\mu l}}{{2\pi }}In\;\left( {\frac{b}{a}} \right)$$
Parallel wires:
$$C = \frac{{\pi El}}{{In\left( {\frac{D}{r}} \right)}}$$
r = radius of the wire
$$L = \frac{{ul}}{\pi }In\left( {\frac{D}{r}} \right)$$
Important Points:
For lossless lines, the three quantities Z, L, and C are related as:
$$h = \mu \frac{Z}{\eta },\;\;C =\epsilon \frac{\eta }{Z}$$
where,
η = Intrinsic Impedance of the dielectric medium
Putting the value of Z from equation (1) in L and C, we get:
$$L = \frac{\mu }{\eta }\eta \frac{h}{w} = \frac{{\mu h}}{w}$$
$$L = \frac{{\mu h}}{w}$$
L represents the inductance per unit length of parallel plate lines.
$$C = \frac{{E\;\eta \;w}}{{\eta \;h}} = \frac{{E\;w}}{h}$$
$$C = \frac{{E\;w}}{h}$$
C represents the Capacitance per unit length of parallel plate lines.
Speed of propagation is given by
$$v = \frac{1}{{\sqrt {LC} }}$$
$$v = \frac{1}{{\sqrt {\frac{{\mu h}}{w}\frac{{E\;w}}{h}} }} = \frac{1}{{\sqrt {\mu E} }}$$
v does not depend on the spacing between plates, hence v will not change
# A rectangular waveguide of width w and height h has cut-off frequencies for TE10 and TE11 modes in the ratio 1 : 2. The aspect ratio w/h, rounded off to two decimal places, is __________.
## Parallel Plane Waveguide MCQ Question 5 Detailed Solution
Concept:
The cutoff frequency of TEmn mode is given by:
$${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$$
Application:
for TE10 mode
$${f_{{c_{10}}}} = \frac{c}{2}\left( {\frac{1}{a}} \right)$$ ----(1)
for TE11 mode
$${f_{{c_{11}}}} = \frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}$$ ----(2)
$$\frac{{eqn\left( 1 \right)}}{{eqn\left( 2 \right)}} = \frac{1}{2}$$
$$\frac{1}{2} = \frac{{\frac{c}{2}\left( {\frac{1}{a}} \right)}}{{\frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}$$
$$= \frac{1}{2} = \frac{{\frac{c}{{2a}}}}{{\frac{c}{2}\sqrt {\frac{{{a^2} + {b^2}}}{{ab}}} }} = \frac{b}{{\sqrt {{b^2} + {a^2}} }}$$
$$\frac{1}{2} = \frac{1}{{\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} }}$$
$${\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{w}{h}} \right)^2} = 4 - 1 = 3$$
$$\frac{w}{h} = \sqrt 3 = 1.732$$
# A rectangle waveguide of internal dimensions (a = 3 cm and b = 1 cm) is to be operated at TE11 mode. The minimum operating frequency is:
1. 6.25 GHz
2. 10.5 GHz
3. 31.6 GHz
4. 15.8 GHz
## Answer (Detailed Solution Below)
Option 4 : 15.8 GHz
## Parallel Plane Waveguide MCQ Question 6 Detailed Solution
Concept:
Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:
$${f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}}$$
a = length of the waveguide
b = height of the waveguide
m,n = modes of operation
Calculation:
Given, a = 3 cm
b = 1 cm
The minimum frequency in TE11 is nothing but the cut-off frequency calculated as:
$${f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {1 \times {{10}^{ - 2}}} \right)}^2}}}}$$
$${f_c} = 150 \times {10^6}\sqrt {11111.11} \;\;$$
$${f_c} = 150 \times {10^6} \times 105.41$$
fC = 15.81 GHz
# As the wave frequency approaches to cut-off frequency of wave guide, the correct statement is:
1. The phase velocity of waves tends to zero.
2. The phase velocity of waves tends to velocity of light.
3. The phase velocity of waves tends to infinite.
4. None of these.
## Answer (Detailed Solution Below)
Option 3 : The phase velocity of waves tends to infinite.
## Parallel Plane Waveguide MCQ Question 7 Detailed Solution
Concept:
Phase velocity:
$${{v}_{p}}=\frac{c}{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}$$
Where c is the free space velocity, f is the operating frequency and fc is the cut-off frequency.
1. We can readily see that the phase velocity is a non-linear function of the operating frequency.
2. We can also write phase velocity as:
$${v_p} = \frac{c}{{\cos \theta }};$$
Where θ is the angle with which the wave enters the waveguide as shown:
Calculation:
Given, f = fc
$${v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}$$
$$v_p = \frac{c}{{\sqrt {1 - 1} \;}} = \frac{c}{0} = \infty$$
vp = ∞ means that,
cos θ = 0, i.e. θ = 90°.
This situation is shown as below:
This implies that at f = fc, the wave will oscillate between the walls as shown.
Hence option (3) is the correct answer.
# Transverse Electro-magnetic waves are characterized by
1. During wave propagation in Z-direction, the components of H and E are transverse 60° to the direction of propagation of the waves
2. During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.
3. During wave propagation in Z-direction, the components of H and E are transverse 120° to the direction of propagation of the waves
4. None of the above
## Answer (Detailed Solution Below)
Option 2 : During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.
## Parallel Plane Waveguide MCQ Question 8 Detailed Solution
Transverse Electro-magnetic waves:
• In an electromagnetic wave, electric and magnetic field vectors are perpendicular to each other and at the same time are perpendicular to the direction of propagation of the wave.
• This nature of electromagnetic wave is known as Transverse nature.
• Maxwell proved that both the electric(E) and magnetic fields(H) are perpendicular to each other in the direction of wave propagation. He considered an electromagnetic wave propagating along positive z-axis
• The electric and magnetic fields propagate sinusoidal with the z-axis.
# Standard air-filled rectangular waveguides of dimensions a = 2.29 cm and b = 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominant TE10 mode with the operating frequency at least 25% above the cutoff frequency of the TE10 mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is
1. 8.19 GHz ≤ f ≤ 13.1 GHz
2. 8.19 GHz ≤ f ≤ 12.45 GHz
3. 6.55 GHz ≤ f ≤ 13.1 GHz
4. 1.64 GHz ≤ f ≤ 10.24 GHz
## Answer (Detailed Solution Below)
Option 2 : 8.19 GHz ≤ f ≤ 12.45 GHz
## Parallel Plane Waveguide MCQ Question 9 Detailed Solution
Concept:
Cut-off frequency of waveguide (rectangular)
$$\Rightarrow {f_c} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \;\;\;\; \ldots \left( 1 \right)$$
Dominant mode in rectangular waveguide (a > b) is TE10 and the cutoff frequency is defined as;
$${f_{{c_{\left( {10} \right)}}}} = \frac{C}{2}\sqrt {\frac{1}{{{a^2}}}} = \frac{C}{{2a}}$$
Calculation:
$${f_{{c_{\left( {10} \right)}}}} = \frac{C}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 2.29}} = 6.554\;Hz$$
Desired operating frequency = 25% more than $${f_{{c_{\left( {10} \right)}}}}$$
$$\left( {1 + 0.25} \right){f_{{c_{\left( {10} \right)}}}} = 1.25{f_{{c_{\left( {10} \right)}}}} = 1.25 \times 6.55$$
= 8.199 Hz
Using equation (1), we calculate different cut-off frequency for higher modes to find the next dominant mode,
fc for TE01 mode = 14.74 Hz
fc for TE11 mode = 16.4 Hz
and
fc for TE20 mode = 13.144 Hz
So, TE20 is the next higher cut-off frequency,
95% of 13.144 Hz
⇒ 12.484 GHz
# A standard air-filled rectangular waveguide with dimensions a = 8 cm, b = 4cm, operates at 3.4 GHz, For the dominant mode of wave propagation, the phase velocity of the signal is vp. The value (rounded off two decimal places) of vp / C, where C denotes the velocity of light, is _____
## Parallel Plane Waveguide MCQ Question 10 Detailed Solution
Concept:
Phase velocity is given by:
$${v_p} = \frac{\omega }{\beta }$$
Where, ω = angular frequency
β = phase constant, calculated for a parallel waveguide as:
$$\beta = \sqrt {{\omega ^2}\mu - {{\left( {\frac{{m\pi }}{a}} \right)}^2}}$$ [for parallel waveguide]
The Phase velocity can now be written as:
$${v_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}$$
$${v_p} = \frac{1}{{\sqrt {\mu - {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}$$
$${v_p} = \frac{{\frac{1}{{\sqrt {\mu } }}}}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu } }}} \right)}^2}} }}$$
With $$c = \frac{1}{{\sqrt {\mu } }}$$
$${v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{m\pi c}}{{a\omega }}} \right)}^2}} }}\;$$
Also, the cut-off frequency is defined as:
$${\omega _c} = \frac{{m{\pi _c}}}{a}$$
$${v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}$$
$$\sin \theta = \frac{{{\omega _c}}}{\omega }$$
$${v_p} = \frac{c}{{\sqrt {1 - {{\sin }^2}\theta } }}$$
$${v_p} = \frac{C}{{\cos \theta }}$$
$${v_p} = \frac{c}{{\cos \theta }}$$
Note: This is valid for the Rectangular waveguide also.
Calculation:
Given
a = 8 cm
b = 4 cm
f = 3.4 GHz
The cut-off frequency will be:
$${f_c} = \frac{c}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 8}}$$
$$\sin \theta = \frac{{{f_c}}}{f}$$
$$\sin \theta = \frac{{1.875}}{{3.4}} = 0.5514$$
∴ cos θ = 0.83419
$$\frac{{{v_p}}}{c} = \frac{1}{{\cos \theta }}$$
$$\frac{{{v_p}}}{c} = 1.198$$
# The phase velocity of waves propagating in a hollow metal waveguide is
1. equal to the group velocity
2. equal to the velocity of light in free space
3. less than the velocity of light in free space
4. greater than the velocity of light in free space
## Answer (Detailed Solution Below)
Option 4 : greater than the velocity of light in free space
## Parallel Plane Waveguide MCQ Question 11 Detailed Solution
Concept:
Phase velocity is defined as:
$${V_p} = \frac{\omega }{\beta }$$
β is the phase constant defined as:
$$\beta = \sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}}$$
$${V_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}$$
$${V_p} = \frac{1}{{\sqrt {\mu \epsilon - {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}$$
$${V_p} = \frac{{\frac{1}{{\sqrt {\mu \epsilon } }}}}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu \epsilon} }}} \right)}^2}} }}$$
Using $$c = \frac{1}{{\sqrt {\mu C} }}$$ where c = speed of light, the above expression becomes:
$${V_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{m\pi C}}{{a\omega }}} \right)}^2}} }};$$
Also $${\omega _c} = \frac{{m\pi c}}{a}$$
$${V_p} = \frac{C}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}$$
Using $$\sin \theta = \frac{{{\omega _c}}}{\omega }$$, we get:
$${V_p} = \frac{c}{{\sqrt {1 - {{\sin }^2}\theta } }}$$
$${V_p} = \frac{c}{{\cos \theta }};$$
Since -1 ≤ cos θ ≤ 1
Vp > c
Extra Information:
Group velocity is given by:
$${V_g} = \frac{{d\omega }}{{d\beta }}$$
$$\beta = \sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}}$$
$$\frac{{d\beta }}{{d\omega }} = \frac{{2\omega \mu\epsilon }}{{2\sqrt {{\omega ^2}\mu\epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}$$
$$\frac{{d\beta }}{{d\omega }} = \frac{{\sqrt {\mu\epsilon } }}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu\epsilon } }}} \right)}^2}} }}$$
$$\frac{{d\beta }}{{d\omega }} = \frac{1}{{C\sqrt {1 - {{\left( {\frac{{{\omega _C}}}{\omega }} \right)}^2}} }}$$
$${V_g} = c\sqrt {1 - {{\left( {\frac{{{\omega _C}}}{\omega }} \right)}^2}}$$
Vg = c cos θ
Vg < c
Conclusion: ∴ The phase velocity is always greater than the speed of light and group velocity is always less than the speed of light.
# For transverse electric waves between parallel plates, the lowest value of m, without making all the field components zero, is equal to
1. 3
2. 2
3. 1
4. 0
Option 3 : 1
## Parallel Plane Waveguide MCQ Question 12 Detailed Solution
Concept:
For parallel plate waveguide, the propagation constant is defined as:
$$\beta = \frac{{m\pi }}{a}$$
a = distance between the plates
Where the minimum value of m = 1
Cut-off frequency for the parallel plate waveguide is given by:
$${f_c} = \frac{m}{{2a}}$$
The wavelength will be:
$${\lambda _c} = \frac{{2a}}{m}$$
# The cut-off frequency of TE01 mode of an air-filled rectangular waveguide having inner dimensions a cm × b cm (a > b) is twice that of the dominant TE10 mode. When the waveguide is operated at a frequency which is 25% higher than the cut-off frequency of the dominant mode, the guide wavelength is found to be 4 cm. The value of b (in cm, correct to two decimal places) is _______.
## Parallel Plane Waveguide MCQ Question 13 Detailed Solution
Concept:
Dominant mode of rectangular waveguide is TE10
The cutoff frequency is given as:
$${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$$
Calculations:
Given: $${\left( {{f_c}} \right)_{TE01}} = 2{\left( {f_c} \right)_{TE10}}$$ ---(1)
Operating frequency:
f = (fc)TE10 + 25% of (fc)TE10
f = 1.25(fc)TE10
For air filled wavelength C, we can write:
$$f_c = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$$
For TE01 mode
m = 0, n = 1
$${\left( {fc} \right)_{TE10}} = \frac{c}{{2a}}$$
For TE10 mode
m = 1; n = 0
$${\left( {f_c} \right)_{TE10}} = \frac{c}{{2a}}$$ ----(2)
From equation (1)
(fc)TE01 = 2(fc)TE10
$$\frac{c}{{2b}} = 2.\frac{c}{{2a}}$$
a = 2b
operating frequency:
f = (1.25) (fc)TE01 = 2(fc)TE10
$$f = \left( {1.25} \right)\left[ {\frac{c}{{2a}}} \right]$$
$$= \frac{{1.25\;c}}{{4b}}$$
Operating wavelength
$$\lambda = \frac{c}{f} = \frac{{4b}}{{1.25}} = 3.2b$$
Guide wavelength is given by:
$${\lambda _g} = \frac{\lambda }{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}$$
$$4 = \frac{{3.2b}}{{\sqrt {1 - {{\left( {\frac{1}{{1.25}}} \right)}^2}} }}$$
$$4 = \frac{{3.2b}}{{0.6}}$$
$$\Rightarrow b = \frac{{4 \times 0.6}}{{3.2}}$$
b = 0.75 cm
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OCR MEI Statistics 2 (complete)
This bundle contains all the Powerpoints and question resources that, between them, cover all of the S2 module for OCR MEI A-level Maths. Lots relevant to other courses also.
Contingency Tables and Chi Squared (OCR MEI Statistics 2 : Chapter 3 Part 2)
This presentation introduces the idea of contingency tables, the Chi-squared hypothesis test. Also included are a collection of relevant past paper questions (and answers) drawn from MEI papers. It is not therefore specific to OCR or MEI and could be used for any maths/stats course where such an introduction is needed.
Contingency Tables
A powerpoint presentation explaining the use of contingency tables to carry out a chi-squared test. Used for maths AQA S2 students.
Chi Squared on Excel Guide
Created for my IB Maths Studies group to help them with their IA. This is a guide on how to do Chi Squared on Excel.
BTEC Assignment Briefs (1 - 4) Unit 8: Using Statistics for Science
This unit enables the learner to use statistical techniques that are essential for the handling, collection and interpretation of scientific data. This contains BTEC assignments for Unit 8: Using Statistics for Science. All of them have been screened for internal verification. Should you need to adapt this for your teaching you may ask your internal verifier to double check them. They have been written in alignment with the learning outcomes of Unit 8. After download you may change the front cover to your own schools logo.
Quantitative skills: Spearman's Rank and Chi Squared
This lesson is designed to meet requirements of the Geography AQA A-Level specification on quantitative and qualitative skills. As per my usual format this lesson consists of; - Starter activity to get them thinking about prior knowledge and using this to link to this lessons learning (the learning linked to this is found on slides 12-16 as I set this as for students to work on in study time/homework). - Learning objectives and success criteria that are clearly linked to the specification. - New information on Spearman's rank and chi squared with examples designed to work through as a class or in teams, with supporting worksheet at the very end of the PPT. - Plenary/application task - to consolidate learning from the lesson. - Further reading with hyperlinked websites. - Further study - hyperlink and consolidating exam style question on fieldwork techniques that would be used in coastal areas.
S3 Goodness of Fit and Contingency Tables Topic Mat
A mat to summarise key concepts from the S3 topic Goodness of Fit and Contingency Tables (Edexcel). This one is quite big so I put it onto A3 paper! I also include a suggestion for how the mat could be filled out. I usually use it as a plenary – ask the pupils for the key ideas from the lesson and collectively fill in sections over a sequence of lessons. At the end of a topic they have a topic mat from which to revise from. I also photocopy them on different coloured paper so they act as dividers in students’ folders.
Chi – Square testing.
For BTEC: UNIT 8 – Using Statistics for Science LO2: Chi – Square lesson notes (P2, M2, D2) In this resource, students will learn - About the χ2 distribution. - How to use χ2 tables and workout the number of degrees of freedom, ν (pronounced, “Nu���) - How to perform a χ2 test for independence between variables, using a contingency table.
Higher Statistics Module (SCQF Level 6) - Extended Notes
These notes complement my Course Notes for this SQA course. They include more further examples, more complicated statistical tests and links to Excel for examples.
Limitations of the chi-squared test
Power Point presentation, 9 slides, Explaining what are the limitations when performing the Chi squared test of independence, using examples to show how to overcome those limitations, based on IB Standard Level Mathematical Studies Syllabus. For a preview of the power point copy the following link on your browser: https://drive.google.com/file/d/0B8z00qLZV7Olb2YwcUlqb1g3UGs/view?usp=sharing
Chi squared test of independence
Power Point presentation, 29 slides, Explaining what is Chi squared test of independence, using examples to show how to perform the test, and show how to use GDC to calculate chi-squared, based on IB Standard Level Mathematical Studies Syllabus. For a preview of the power point copy the following link on your browser: https://drive.google.com/file/d/0B8z00qLZV7OlV1cxZFBGQjZXZG8/view?usp=sharing
Statistics and Probability
Discrete and continuous random variables, expectation algebra, chi-squares and goodness of fit, hypothesis testing, confidence levels CLT and distribution of sample means..
Goodness of Fit and Chi Squared.
Introduction to the concept of testing for independence, goodness of fit for discrete and continuous distributions combined with hypothesis testing.
A Level Biology- Stats
A collection of resources I have picked up and adapted along the way when teaching statistic elements of A Level Biology. These resources focus on Chi Squared and Students T Test. The stats booklet I refer to is an unedited version from another user which can be found via a TES search.
Statistics 2
This bundle includes all the topics needed to cover the S2 AQA course including; Discrete and Continuous Random Variables Poisson Distribution Estimation Hypothesis Testing Chi-squared
Hypothesis testing and Chi-squared
This presentation introduces the students to hypothesis testing using normal distribution. It starts by explaining what a hypothesis test is, the key language used and also covers type 1 and type 2 errors. Also included is a presentation that shows students how to use Chi-squared test to find a goodness of fit. It covers all parts of chi-squared from a basic introduction to very clear examples, including using the standard results tables (don't forget to move them to show the answers underneath). It also goes on to contingency tables including Yates' correction. SMART Notebook Express presentations can be used for free by using SMART online or by downloading SMART software, also free, I use it because it allows a more interactive presentation than a powerpoint. Just search 'SMART Notebook Express' and try it with my free SMART resource.
Chi squared worksheet- A level geography- Rivers
Chi squared worksheet- A level geography- Rivers. Contains an introduction to the purpose of chi squared and then gives a partially blank table to fill in, complete with a writing frame with gaps to fill in the analysis from the critical value table.
Environmental Sciences: Inferential Statistics
A resource for pupils studying at SCQF Level 8, (Further and Higher Education) which details the use of inferential statistics in the field of geoscience. May be of use to teachers of geography, the sciences or even as an application in a mathematics or statistics course. Of use for helping to achieve problem solving, data handling and numeracy learning outcomes. All content released under a CC BY license. Authors: Kay Douglas and Sophie Flack © University of Edinburgh Keywords: Biology/data and statistics/data handling, Environmental science/data and statistics/data handling, Chemistry/data and statistics/data handling University of Edinburgh, EdUniOERGeo
Guide to Inferential Statistics in Geosciences
A resource for pupils studying at SCQF Level 7, (Scottish Advanced Higher, A2-Level) which details the use of inferential statistics in the field of geoscience. May be of use to teachers of geography, the sciences or even as an application in a mathematics or statistics course. Of use for helping to achieve problem solving, data handling and numeracy learning outcomes. All content released under a CC BY license. Authors: Kay Douglas and Sophie Flack © University of Edinburgh Keywords: Biology/data and statistics/data handling, Environmental science/data and statistics/data handling, Chemistry/data and statistics/data handling University of Edinburgh, EdUniOERGeo
Statistics 3 AQA End of Topic Tests
These are varying mark test sets for each topic within the Statistics 3 AQA Maths syllabus. Students write their solutions in the boxes provided. Any feedback would be very welcome. Enjoy! Em
MEI stats 2 power point presentation
This presentation covers the Poisson distribution, The normal distribution, samples and hypothesis testing and Bivariate data. Including approximations and hints and tips. The resource is also extremely useful just before exams if printed out as flash cards to induce mathematical thinking before students enter an exam
Chi squared calculations
Introduction to chi squared calculations by hand and using technology. IB Studies. For a tns file please visit my website www.mathexams.com.au
AQA S3 Past Paper Revision
A differentiated, PowerPoint-formatted version of all the 2006-2014 S3 past paper questions for AQA A-Level Statistics, listed by topics. Green - questions that appear towards the beginning of the paper (easy) Amber - questions that appear in the middle of the paper (medium) Red - questions that appear towards the end of the paper (hard) Any feedback would be appreciated and please let me know if there are any mistakes with the animations etc. Enjoy! Emma
Statistics Tables Test - S2 looking up tables
Give your A level students a tables test! How quickly and accurately can they find the correct values from the statistical tables? Covers Poisson, Chi-squared and critical values for hypothesis testing with the normal distribution and Student's t-distribution. Presented in simple, no-frills ppt format.
Chi squared test and contingency table (Biology and maths focus)
Powerpoint and examples on turning maggots and fingerprint types. Plus questions, notes and answers on simple chi squared test
S3 - Chapter 4 - Goodness of Fit and Contingency Tables
Based on Edexcel syllabus.
Interactive graphs of the Binomial and Poisson Distributions - spreadsheet
An interactive Excel spreadsheet that graphs the Binomial and Poisson distributions, with sliders to affect the probability for the Binomial, and lambda for the Poisson. The graphs update instantly so that you can see how the distributions change as the parameters change. Two further sheets overlays the normal distribution (with the relevant mean and standard deviation), showing how well the normal curve models these distributions. I suggest that it is shown via projector/SWB, and the teacher manipulates the slider to give students a more intuitive grasp of the distributions.
Edexcel S2 Revision Cards
These S2 revision cards can be used as an individual or class activity. Individuals can ask friends/relatives to test them using these cards. For use in class, hand out one laminated card per pupil. They circulate round the class, pairing up to ask and answer their questions, then swapping cards and pairing up with someone else. It's a nice active revision lesson. Please do let me know if I&'ve made any errors on these cards.
Dr Who themed Statistics 2 test
Binomial Poisson crv cdf pdf
S2 Treasure Hunt
A fun way of revising S2
Statistics 2 powerpoints
A set of powerpoints covering all topics in S2. Examples labelled WB correspond to the separately attached 'Workbook&' (I give this as a single booklet so pupils have a clear model answer to each topic). References to Exercises are from the Pearson Edexcel S2 textbook.
Complex Numbers Applications of de Moivres theorem
Designed for the Edexcel spec but applicable to AQA, OCR,MEI and WJEC.
Statistics 2 (S2) - Var(X)
Statistics 2 (S2) - Var(X) - Continuous Uniform Distribution Edexce. Designed for the Edexcel spec but applicable to AQA, OCR,MEI and WJEC
Binomial Distribution: S2 Edexcel January 2012 Q3
This video from ExamSolutions looks at binomial distribution : S2 Edexcel January 2012 Q3(d)
S2 Edexcel January 2012 Q3(c) : ExamSolutions
This video from ExamSolutions looks at binomial distribution : S2 Edexcel January 2012 Q3(c)
Binomial Expansion: Maths C4 June 2012 Q3(a)
This video from ExamSolutions looks at binomial expansion : Edexcel Core Maths C4 June 2012 Q3(a)
Binomial Distribution : S2 Edexcel January 2013 Q3
This video from ExamSolutions looks at Binomial Distribution : S2 Edexcel January 2013 Q3
Binomial Distribution : S2 Edexcel January 2012 Q3
This video from ExamSolutions looks at Binomial Distribution : S2 Edexcel January 2012 Q3(a)(b)
Induction to prove that the sum formula works.
video tutorial on induction/
Video: Write the Equation of the Polynomial
Video tutorial on polynomials. Given three real zeros write the equation of the polynomial .
Video: Remainder Theorem with Polynomials .
Video tutoriaol on how the remainder theorem work with polynomials .
Video: Factoring a Trinomial to the Fourth Power
Video tutorial on factoring a trinomial to the fourth power .
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# Resources tagged with: Spheres, cylinders & cones
Filter by: Content type:
Age range:
Challenge level:
### There are 21 results
Broad Topics > 3D Geometry, Shape and Space > Spheres, cylinders & cones
### Tin Tight
##### Age 14 to 16Challenge Level
What's the most efficient proportion for a 1 litre tin of paint?
### Conical Bottle
##### Age 14 to 16Challenge Level
A right circular cone is filled with liquid to a depth of half its vertical height. The cone is inverted. How high up the vertical height of the cone will the liquid rise?
### Funnel
##### Age 14 to 16Challenge Level
A plastic funnel is used to pour liquids through narrow apertures. What shape funnel would use the least amount of plastic to manufacture for any specific volume ?
### Cola Can
##### Age 11 to 14Challenge Level
An aluminium can contains 330 ml of cola. If the can's diameter is 6 cm what is the can's height?
### Peeling the Apple or the Cone That Lost Its Head
##### Age 14 to 16Challenge Level
How much peel does an apple have?
### Ball Packing
##### Age 14 to 16Challenge Level
If a ball is rolled into the corner of a room how far is its centre from the corner?
### Volume of a Pyramid and a Cone
##### Age 11 to 14
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Efficient Cutting
##### Age 11 to 14Challenge Level
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Mouhefanggai
##### Age 14 to 16
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### Which Solid?
##### Age 7 to 16Challenge Level
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen.
### When the Angles of a Triangle Don't Add up to 180 Degrees
##### Age 14 to 18
This article outlines the underlying axioms of spherical geometry giving a simple proof that the sum of the angles of a triangle on the surface of a unit sphere is equal to pi plus the area of the. . . .
### In a Spin
##### Age 14 to 16Challenge Level
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### Packing 3D Shapes
##### Age 14 to 16Challenge Level
What 3D shapes occur in nature. How efficiently can you pack these shapes together?
### Fill Me up Too
##### Age 14 to 16Challenge Level
In Fill Me Up we invited you to sketch graphs as vessels are filled with water. Can you work out the equations of the graphs?
### Three Balls
##### Age 14 to 16Challenge Level
A circle has centre O and angle POR = angle QOR. Construct tangents at P and Q meeting at T. Draw a circle with diameter OT. Do P and Q lie inside, or on, or outside this circle?
### There and Back Again
##### Age 11 to 14Challenge Level
Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives?
### Immersion
##### Age 14 to 16Challenge Level
Various solids are lowered into a beaker of water. How does the water level rise in each case?
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Gym Bag
##### Age 11 to 16Challenge Level
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Make Your Own Pencil Case
##### Age 11 to 14Challenge Level
What shape would fit your pens and pencils best? How can you make it?
### Witch's Hat
##### Age 11 to 16Challenge Level
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
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# Correlation and squared variables
According to my textbooks if two variables are uncorrelated, they are not necessarily independent (unless they are normally distributed).
My question is, are 2 variables still not independent if they are not correlated, but their squares are correlated? I believe that they are still not independent, but I am not sure since at some point I thought that what if the squares of those 2 variables have Chi-distribution (form a new varialbe Y) and then the variables are normally distributed and independent. So I am quite confused now.
I would be very grateful to you for your help. Thank you very much.
-
If $X$ and $Y$ are independent random variables, then so are $g(X)$ and $h(Y)$ independent random variables for all (measurable) functions $g(\cdot)$ and $h(\cdot)$. Thus, if $X$ and $Y$ are independent (and hence uncorrelated), then $X^2$ and $Y^2$ cannot be correlated random variables; they too must be independent (and hence uncorelated) random variables. Now, since $$X ~\text{and}~ Y ~ \text{independent} \Rightarrow X^2 ~\text{and}~ Y^2 ~ \text{independent}$$ it follows that $$X^2 ~\text{and}~ Y^2 ~ \text{not independent} \Rightarrow X ~\text{and}~ Y ~ \text{not independent}.$$ Since $X^2$ and $Y^2$ are correlated, they are not independent, and so $X$ and $Y$ are not independent either. Whether $X$ and $Y$ are correlated or not has no bearing on the matter.
-
Thank you very much! – James Sep 29 '12 at 20:20
Normal distribution doesnt explicitely imply independence. If two jointly normally distributed random variables are uncorrelated, then they are independent.
In your case you know that two random variables $X$ and $Y$ are dependent and uncorrelated. This means
$$P(X=a|X=b)\neq P(X=a)$$
for at least one $a$ and $b$ and $$E[XY]=E[X]E[Y]$$
Given these two conditions, extra conditions defined on $X^2$ or $Y^2$ will not change the independency condition.
-
Thank you so much! – James Sep 29 '12 at 20:20
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# ALEX Lesson Plan Resources
Back
ALEX Lesson Plans
Subject: Mathematics (4 - 5)
Title: Fraction Action
Description: The students will participate in a hands-on-lesson where they will understand and be able to write equivalent forms of fractions. The students will learn and apply the concept of using fractions in everyday life.
Subject: Mathematics (5)
Title: Gallon Man Meets Fraction Friend: An Introductory Lesson on Adding Fractions with Unlike Denominators
Description: This hands-on, minds-on activity helps students use what they already know about customary measurement (CCRS 2010 #18 [5.MD.1]: Convert among different-sized standard measurement units within a given measurement system.) to help them add and subtract fractions with like and unlike denominators.
Subject: Mathematics (5 - 6), or Technology Education (6 - 8)
Title: Aquarium Project
Description: The class is given money to purchase fish for a class aquarium. Students must use the size and lifestyle requirements for the fish to decide which type to buy. Students must present their proposal to the principal in a letter.This lesson plan was created as a result of the Girls Engaged in Math and Science University, GEMS-U Project.
Subject: Mathematics (4 - 5), or Technology Education (3 - 5)
Title: Adding and Subtracting Mixed Numbers
Description: During this technology based lesson, students learn how to add and subtract mixed numbers by exploring websites on the Internet. Students demonstrate their knowledge of fractions by participating in practice activities on the Internet.
Subject: Mathematics (4 - 5), or Technology Education (3 - 5)
Title: Kids and Money
Description: This lesson is used to show students how a bank pays interest to people who deposit money into it and charges interest to people who borrow money from it.
Thinkfinity Lesson Plans
Subject: Mathematics
Title: Fun with Fractions Add Bookmark
Description: In this seven-lesson unit from Illuminations, students explore relationships among fractions through work with the set model. This early work with fraction relationships helps students make sense of basic fraction concepts and facilitates work with comparing and ordering fractions and working with equivalency.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Title: Investigating Equivalent Fractions with Relationship Rods Add Bookmark
Description: In this lesson, one of a multi-part unit from Illuminations, students investigate the length model by working with relationship rods to find equivalent fractions. Students develop skills in reasoning and problem solving as they explain how two fractions are equivalent (the same length). Relationship rods are wooden or plastic rods in ten different colors, ranging in length from one to ten centimeters.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Title: Inch by Inch Add Bookmark
Description: In this lesson, one of a multi-part unit from Illuminations, students use an actual ruler to represent various fractions as lengths. This lesson builds on work done in a previous lesson with nonstandard measurement as students use a standard instrument to measure a variety of items. Several pieces of literature appropriate for use with this lesson are suggested.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Description: This collection of Web links, reviewed and presented by Illuminations, offers teachers and students information about and practice in concepts related to arithmetic. Users can read the Illuminations Editorial Board's review of each Web site, or choose to link directly to the sites.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Description: Mathematical games can foster mathematical communication as students explain and justify their moves to one another. In addition, games can motivate students and engage them in thinking about and applying concepts and skills. This e-example from Illuminations contains an interactive version of a game that can be used in the grades 3-5 classroom to support students' learning about fractions. e-Math Investigations are selected e-examples from the electronic version of the Principles and Standards of School Mathematics (PSSM). The e-examples are part of the electronic version of the PSSM document. Given their interactive nature and focused discussion tied to the PSSM document, the e-examples are natural companions to the i-Math investigations.
Thinkfinity Partner: Illuminations
Subject: Mathematics
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# 平行四边形的面积 PowerPoint PPT Presentation
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## 平行四边形的面积
6
4
24
6
4
24
1、拼出的长方形与原来的平行四边形比面积变了没有？
2、拼出的长方形的长和宽与原来的平行四边形的底和高有什么关系？
3、能否根据长方形的面积计算公式推导出平行四边形面积计算公式？
4m
8m
2
S=ah =8×4=32（m ）
4米
4米
6米
6米
(1) 平行四边形的底是7米,高是4米,面积是28米。( )
(2)两个平行四边形的底相等，它们的面积相等。 ( )
×
×
7分米
8分米
6分米
×
?
h=
6cm
8cm
4cm
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# â€؛ gan.1 â€؛ teaching â€؛ fall08 â€؛ lec4.pdf lecture 4: r-l-c circuits and...
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• RLC series circuit: ● What's VR?
◆ Simplest way to solve for V is to use voltage divider equation in complex notation:
◆ Using complex notation for the apply voltage Vin = V0cosωt = Real(V0 e jωt ):
■ We are interested in the both the magnitude of VR and its phase with respect to Vin.
■ First the magnitude:
K.K. Gan 1
Lecture 4: R-L-C Circuits and Resonant Circuits
RL C V = V
X
in cosω0
XL C
t
VR = VinR
R+ XC + XL
= VinR
R+ 1 jωC
+ jωL
VR = V0e
jωt R
R+ j ωL − 1 ωC
VR = V0e
jωt R
R+ j ωL − 1 ωC
= V0R
R2 + ωL − 1 ωC
2
K.K. Gan 1 L4: RLC and Resonance Circuits
• K.K. Gan 2
■ The phase of VR with respect to Vin can be found by writing VR in purely polar notation. ❑ For the denominator we have:
❑ Define the phase angle φ :
❑ We can now write for VR in complex form:
■ Finally, we can write down the solution for V by taking the real part of the above equation:
R+ j ωL − 1 ωC
= R2 + ωL −
1 ωC
2 exp j tan−1
ωL − 1 ωC
R
VR = VoR e
jωt
e jφ R2 + ωL − 1 ωC
2
= VR e j(ωt−φ)
tanφ = Imaginary X Real X
= ωL − 1
ωC R
VR = Real V0R e
j(ωt−φ)
R2 + ωL − 1 ωC
2
= V0R cos(ωt −φ)
R2 + ωL − 1 ωC
2
L4: RLC and Resonance Circuits
Depending on L, C, and ω, the phase angle can be positive or negative! In this example, if ωL > 1/ωC, then VR(t) lags Vin(t).
• K.K. Gan 3
◆ VR
• K.K. Gan 4
1e-1 1e-7
100
V = V cos t0in ω
Vout
K.K. Gan 4
1e-1 1e-7
100
inV = Vcosω t
Vout
φ(- )
L4: RLC and Resonance Circuits
Bode plot of magnitude of VR/Vin vs. frequency
• K.K. Gan 5
◆ In general VC(t), VR(t), and VL(t) are all out of phase with the applied voltage.
◆ I(t) and VR(t) are in phase in a series RLC circuit.
◆ The amplitude of VC, VR, and VL depend on ω.
◆ The table below summarizes the 3 cases with the following definitions:
● RLC circuits are resonant circuits ◆ energy in the system "resonates" between the inductor and capacitor ◆ “ideal" capacitors and inductors do not dissipate energy ◆ resistors dissipate energy i.e. resistors do not store energy
Gain Magnitude Phase VR/Vin R/Z -φ VL/Vin ωL/Z π/2 - φ VC/Vin 1/ωCZ -π/2 - φ
K.K. Gan 5
Z = R2 + (ωL −1/ωC)2[ ] 1/2
tanφ = (ωL −1/ωC) /R
L4: RLC and Resonance Circuits
• K.K. Gan 6
● Resonant Frequency: ◆ At the resonant frequency the imaginary part of the impedance vanishes. ◆ For the series RLC circuit the impedance (Z) is:
◆ At resonance (series, parallel etc):
◆ At the resonant frequency the following are true for a series RLC circuit: ■ |VR| is maximum (ideally = Vin) ■ φ = 0 ■
☞ The circuit acts like a narrow band pass filter.
● There is an exact analogy between an RLC circuit and a harmonic oscillator (mass attached to spring):
◆ x ⇔ q (electric charge), L ⇔ m, k ⇔ 1/C ◆ B (coefficient of damping) ⇔ R
K.K. Gan
Z = R+ XL + XC = R+ j(ωL −1/ωC)
Z = R2 + (ωL −1/ωC)2[ ] 1/2
ωL =1/ωC
ωR = 1 LC
VC Vin
= VL Vin
= L
R C (VC or VL can be > Vin!)
m d 2x dt2
+B dx dt
+ kx = 0 damped harmonic oscillator
L d 2q dt2
+ R dq dt
+ q C
= 0 undriven RLC circuit
6 L4: RLC and Resonance Circuits
• K.K. Gan 7
● Q (quality factor) of a circuit: determines how well the RLC circuit stores energy ◆ Q = 2π (max energy stored)/(energy lost) per cycle ◆ Q is related to sharpness of the resonance peak:
K.K. Gan 7 L4: RLC and Resonance Circuits
• K.K. Gan 8
◆ The maximum energy stored in the inductor is LI2/2 with I = IMAX. ■ no energy is stored in the capacitor at this instant because I and VC are 900 out of phase.
■ The energy lost in one cycle:
■ There is another popular, equivalent expression for Q
❑ ωU (ωL) is the upper (lower) 3 dB frequency of the resonance curve. ❍ Q is related to sharpness of the resonance peak. ❑ Will skip the derivation here as it involves a bit of algebra. ❍ two crucial points of the derivation:
K.K. Gan 8
power × (time for cycle) = IRMS 2 R× 2π
ωR = 12 Imax
2 R× 2π ωR
Q = 2π LIMax
2
2
2π ωR
RIMax 2
2
= ωRL R
Q = ωR ωU −ωL
VR Vin
= 1
1+Q2 ω ωR
− ωR ω
2
Q ω ωR
− ωR ω
= ±1
L4: RLC and Resonance Circuits
at the upper and lower 3 dB points
• K.K. Gan 9
● Q can be measured from the shape of the resonance curve ■ one does not need to know R, L, or C to find Q!
● Example: Audio filter (band pass filter)
■ Audio filter is matched to the frequency range of the ear (20-20,000 Hz).
K.K. Gan 9
Q = ωR ωU −ωL
L4: RLC and Resonance Circuits
• K.K. Gan 10
● Let's design an audio filter using low and high pass RC circuits.
◆ Ideally, the frequency response is flat over 20-20,000 Hz, and rolls off sharply at frequencies below 20 Hz and above 20,000 Hz. ■ Set 3 dB points as follows: ❑ lower 3 dB point : 20 Hz = 1/2πR1C1
❑ upper 3 dB point: 2x104 Hz = 1/2πR2C2
■ If we put these two filters together we don't want the 2nd stage to affect the 1st stage. ❑ can accomplish this by making the impedance of the 2nd (Z2) stage much larger than R1. ❑ Remember R1 is in parallel with Z2.
■ In order to insure that the second stage does not "load" down the first stage we need: R2 >> R1 since at high frequencies Z2 ⇒ R2
■ We can now pick and calculate values for the R's and C's in the problem. ❑ Let C1 = 1 µF ⇒ R1 = 1/(20Hz 2πC1) = 8 kΩ ❑ Let R2 > 100R1 ⇒ R2 = 1 MΩ, and C2 = 1/(2x104 Hz 2πR2) = 8 pf
☞ R1 = 8 kΩ, C1 = 1 µF R2 = 1 MΩ, C2 = 8 pf
K.K. Gan 10
Z1 = R1+1/ jωC1 Z2 = R2 +1/ jωC2
L4: RLC and Resonance Circuits
• ◆ Exact derivation for above filter: ◆ In the above circuit we treated the two RC filters as independent. ◆ Why did this work? ◆ We want to calculate the gain (|Vout/Vin|) of the following circuit:
■ Working from right to left, we have:
❑ ZT is the total impedance of the circuit as seen from the input. ❑ Z1 is the parallel impedance of R1 and R2, in series with C2.
■ Finally we can solve for the gain G = |Vout/Vin|:
K.K. Gan 11 K.K. Gan 11
Vout =VaX2 /(X2 + R2 ) Va =VinZ1 /ZT
Z1 = R1 R2 + X2( ) R1 + R2 + X2
ZT = X1 + Z1
Va = VinR1 R2 + X2( )
X1 R1 + R2 + X2( )+ R1 R2 + X2( )
Vout Vin
= R1X2
X1 R1 + R2 + X2( )+ R1 R2 + X2( )
L4: RLC and Resonance Circuits
• K.K. Gan 12
■ We can relate this to our previous result by rewriting the above as:
■ If we now remember the approximation (R1
• K.K. Gan 13
● Another Example: Calculate |I| and the phase angle between Vin and I for the following circuit:
■ First calculate |I|. ◆ The total current out of the input source (I) is related to Vin and the total impedance (ZT) of the circuit by Ohm’s law: ◆ The total impedance of the circuit is given by the parallel impedance of the two branches:
◆ Putting in numerical values for the R's and X's we have:
◆ We can now find the magnitude of the current:
K.K. Gan 13
I =Vin /ZT
1/ZT =1/Z1+1/Z2 Z1 = R1+ X1 Z2 = R2 + X2
| 2,710 | 7,031 |
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| 3.78125 | 4 |
CC-MAIN-2020-24
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| 0.643725 |
https://www.archivemore.com/why-inclined-manometer-is-used-in-measuring-pressure/
| 1,713,815,080,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00744.warc.gz
| 567,166,200 | 7,967 |
## Why inclined manometer is used in measuring pressure?
The advantage of the inclined manometer is that the differential reading scales along the tube can be made large compared to a vertical manometer for a given pressure difference, hence improving the accuracy in reading the scale.
## What does an inclined manometer measure?
Inclined Manometer in Range 0-12 MM Defination: inInclined Manometer Range from 0-12 MM The inclined manometer is a scientific device, used to measure a very low pressure of liquid & gases. It is more suitable for measuring, very low pressure measurements or where greater accuracy is required.
## Which of the following is used to measure the difference in pressure between two points in a pipe or in two different pipes?
Differential manometers
## What type of manometer is best for measuring low pressures?
Low pressure and low differentials are better handled with an inclined-tube manometer, where 1 in. of vertical liquid height can be stretched to 12 in. of scale length. Liquid manometers measure differential pressure by balancing the weight of a liquid between two pressures.
## What is the manometer equation?
Manometers. One of the most important classes of pressure gauges applies the property that pressure due to the weight of a fluid of constant density is given by p = hρg.
## What is the formula for calculating pressure?
Pressure and force are related, and so you can calculate one if you know the other by using the physics equation, P = F/A. Because pressure is force divided by area, its meter-kilogram-second (MKS) units are newtons per square meter, or N/m2.
Solution: Pressure is per unit thrust on a surface, hence pressure is directly proportional to thrust. Greater the thrust, greater is the pressure and smaller the thrust, smaller is the pressure.
## How do I calculate thrust?
1. Thrust is the force which moves an aircraft through the air. Thrust is generated by the engines of the airplane.
2. F = ((m * V)2 – (m * V)1) / (t2 – t1)
3. F = m * a.
4. m dot = r * V * A.
5. F = (m dot * V)e – (m dot * V)0.
6. F = (m dot * V)e – (m dot * V)0 + (pe – p0) * Ae.
## Is thrust the same as force?
A force is a push or pull on an object. Picture pushing open a door or pulling the door closed. Both are forces on the door. Thrust is the force that pushes aircraft forward or upwards.
## Is thrust a push or pull force?
Thrust is actually a force! A Force is a push or pull on an object. Scientists name these forces so that they’re easier to deal with. You may have heard of weight: that’s the force of gravity pulling you down!
| 608 | 2,602 |
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https://www.jiskha.com/questions/77757/if-sin-4-5-with-a-in-qii-and-sin-b-3-5-with-b-in-qiv-find-cos-a-b
| 1,624,043,097,000,000,000 |
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| 772,037,073 | 5,185 |
# trig
If sin = 4/5 with A in QII, and sin B = -3/5 with B in QIV find cos(A-B)
1. 👍
2. 👎
3. 👁
1. cos(A-B)= cosAcosB+sinAsinB
If sinA = 4/5 in QII
draw you triangle in QII, and find the third side.
The third side is 3. (Do the same thing for sin B= -3/5
Plug in:
cosAcosB-sinAsinB
=(-3/5)*(4/5)+(4/5)*(-3/5)
=(-12/25) -(-12/25)
=-24/25
1. 👍
2. 👎
2. 53. the longest side is 55 cm. find the length of the shortest side
1. 👍
2. 👎
## Similar Questions
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2. ### calculus
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Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are in quadrant 1. =Sin(s)cos(t) + Cos(s)Sin(t) =Sin(1/5)Cos(3/5) + Cos(-1/5)Sin(3/5) = 0.389418 Sin(s-t) =sin(s)cos(t) - cos(s)sin(t) =sin(-3/5)cos(1/5) -
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1. ### Math:)
Explain why sin^-1[sin(3pi/4)] does not = 3pi/4 when y=sin(x) and y=sin^-1(x) are inverses. Any help on this question is greatly appreciated. Thank you!
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Stumped on this question - please explain! If sin A = 0.35, cos A = 0.94, sin B = 0.58, and cos B = 0.81, what is sin(A + B)? Am I just plugging the values into sin(A + B) or do I have to do more than that? Thank you.
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Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal. I am not sure, but would I find the derivative first: y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 But then I don't know what to
| 1,153 | 2,950 |
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| 4.34375 | 4 |
CC-MAIN-2021-25
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