token_count
int32 48
481k
| text
stringlengths 154
1.01M
| score
float64 3.66
5.34
|
|---|---|---|
919 |
Difference between revisions of "2021 AMC 10A Problems/Problem 2"
Problem
Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$
Solution 1
The following system of equations can be formed with $p$ representing the number of students in Portia's high school and $l$ representing the number of students in Lara's high school. $$p=3l$$ $$p+l=2600$$ Substituting $p$ with $3l$ we get $4l=2600$. Solving for $l$, we get $l=650$. Since we need to find $p$ we multiply $650$ by 3 to get $p=1950$, which is $\boxed{\text{C}}$
-happykeeper
Solution 2 (One Variable)
Suppose Lara's high school has $x$ students. It follows that Portia's high school has $3x$ students. We know that $x+3x=2600,$ or $4x=2600.$ Our answer is $$3x=2600\cdot\frac 34=650\cdot3=\boxed{\textbf{(C)} ~1950}.$$
~MRENTHUSIASM
Solution 3 (Arithmetic)
Clearly, $2600$ students is $4$ times as many students as Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{\textbf{(C)} ~1950}$ students.
~MRENTHUSIASM
Solution 4.1 (Quick Inspection)
The number of students in Portia's high school must be a multiple of $3.$ This eliminates $\textbf{(B)},\textbf{(D)},$ and $\textbf{(E)}.$ Since $\textbf{(A)}$ is too small (as it is clear that $600+\frac{600}{3}<2600$), we are left with $\boxed{\textbf{(C)} ~1950}.$
~MRENTHUSIASM
Solution 4.2 (Plug in the Answer Choices)
For $\textbf{(A)},$ we have $600+\frac{600}{3}=800\neq2600.$ So, $\textbf{(A)}$ is incorrect.
For $\textbf{(B)},$ we have $650+\frac{650}{3}=866\frac{2}{3}\neq2600.$ So, $\textbf{(B)}$ is incorrect.
For $\textbf{(C)},$ we have $1950+\frac{1950}{3}=2600.$ So, $\boxed{\textbf{(C)} ~1950}$ is correct. For completeness, we will check choices $\textbf{(D)}$ and $\textbf{(E)}.$
For $\textbf{(D)},$ we have $2000+\frac{2000}{3}=2666\frac{2}{3}\neq2600.$ So, $\textbf{(D)}$ is incorrect.
For $\textbf{(E)},$ we have $2050+\frac{2050}{3}=2733\frac{1}{3}\neq2600.$ So, $\textbf{(E)}$ is incorrect.
~MRENTHUSIASM
Video Solution (Very fast & Simple)
~ Education, the Study of Everything
Video Solution #1(Setting Variables)
https://youtu.be/qNf6SiIpIsk?t=119 ~ThePuzzlr
Video Solution #2(Solving by equation)
https://www.youtube.com/watch?v=aOpgeMfvUpE&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=1 ~North America Math Contest Go Go Go
- pi_is_3.14
~savannahsolver
~IceMatrix
~MathWithPi<|endoftext|>
| 4.875 |
1,024 |
The Shawnee Trail was the first major route used by the cattle trailing industry to deliver longhorns to the markets of the Midwest. Longhorns were collected around San Antonio, Texas, and taken northward through Austin, Waco, and Dallas, crossing the Red River near Preston, Texas, at Rock Bluff. There the outcroppings that provide the place-name formed a natural chute that forced the cattle together at the ford, and a gradual rise on the north bank made it easy to exit the river. North of the Red River the trail divided for a time, coming together near Boggy Depot in the Choctaw Nation, Indian Territory. Here some herds veered sharply eastward to pass through Fort Smith, Arkansas. The main trail led to the Canadian River directly below the confluence of the north and south branches and forded the Arkansas River between the mouths of the Verdigris and Neosho rivers, and followed the Neosho past Fort Gibson almost to the Kansas border. The trail then subdivided into various routes that, depending on the final destination, led to one of the following: Baxter Springs, Kansas, and Westport, Kansas City, Sedalia, and St. Louis, Missouri.
Throughout the 1830s settlers from the United States heading for Texas traveled across present Oklahoma along the Texas Road. When the first herds were taken north in the early 1840s, they reversed the trek, opening a trail to the railheads in Missouri. Newspapers referred to the route as the Sedalia Trail or simply the cattle trail. No one knows why it was called the Shawnee Trail; however, the route did pass by a Shawnee village in north Texas and near the Shawnee Hills in Indian Territory. By the late 1850s the name was in general use.
In the 1840s herds were taken up the trail primarily to Missouri, and during the Mexican War the trail was used almost constantly during the summer months. The gold rush in California increased demand for cattle after 1848, and for several years the Shawnee Trail was heavily used. By the mid-1850s Kansas City, Missouri, was the largest stock market in the west, and the Texas cattle trailing industry was well established.
The westward expansion of the farming frontier soon intervened. In 1853 farmers in Missouri turned back the drovers, fearing that the longhorns would infect their cattle with a tick-borne disease called Texas fever. Longhorns were immune to it, but they harbored ticks that spread it to local herds. Newly infected animals either died or required expensive treatment. Between 1853 and 1855 herds continued to use the trail, but resistance continued to grow. In December 1855 the Missouri legislature passed the first law banning diseased animals. Some drives avoided Missouri, staying on the eastern edge of Kansas Territory. Anxious farmers there pushed a bill through the territorial legislature in 1859 that limited access to the cattle drives. For a time the drovers were forced to run a gauntlet of angry farmers and justices of the peace to get the cattle to rail heads. Through 1859 and 1860 violence erupted when the drovers encountered the blockades. The outbreak of the Civil War in 1861 virtually stopped traffic on the Shawnee Trail north of Indian Territory.
The end of the Civil War signaled the rebirth of the cattle drives on the Shawnee Trail. More than two hundred thousand longhorns were taken up the trail in 1866. However, resistance grew stiffer and better organized. By 1867 six states had enacted laws limiting trail drives. Drovers attempted to avoid populated areas by turning to follow the Arkansas River westward or by grazing their herds in the Cherokee Strip until local quarantines were lifted. These delays and poor grazing in Indian Territory reduced profits, and the future of the trail driving industry seemed in peril. However, in 1867 Joseph G. McCoy, a young entrepreneur from Illinois, built stock pens and loading chutes on the railroad at Abilene, Kansas. Soon the majority of cattle were following the old Shawnee Trail from central Texas to Waco, but there they turned toward Fort Worth, following the Chisholm Trail. When the advancing frontier and barbed wire closed the Chisholm Trail, the trail drives turned to the Western or Dodge City Trail.
Wayne Gard, "The Impact of the Cattle Trails," Southwestern Historical Quarterly 71 (July 1967).
Wayne Gard, "The Shawnee Trail," Southwestern Historical Quarterly 56 (January 1953).
Jimmy M. Skaggs, The Cattle-Trailing Industry: Between Supply and Demand, 1866–1890 (Lawrence: University Press of Kansas, 1973).
The following (as per The Chicago Manual of Style, 17th edition) is the preferred citation for articles:
Carl N. Tyson, "Shawnee Trail," The Encyclopedia of Oklahoma History and Culture, https://www.okhistory.org/publications/enc/entry.php?entry=SH015.
© Oklahoma Historical Society.<|endoftext|>
| 3.875 |
1,481 |
Pest controllers have pondered it for a long time. In order to answer this question, we need to delve deeper in some of the crucial elements of ant family (or Formicidae, if you want to be more technical). How long ants live is determined by several different factors. Life expectancy is highly dependent on the species, with common representatives living for several years, while the members of others die off within a few weeks.
It also depends on the social role in the caste system. Ants that have a more perpetual role in the colony (such as the workers) tend to live longer than those with episodic roles. Males (or drones) for example, usually have the shortest life span because their only role is to fertilize the queen ant. On the other hand, queens usually have the longest life span, with some species extending over 30 years!
Ant Life Cycle
Like many other insects, an ant’s life begins in the form of an egg. Depending on fertilization, the egg may bear a male or a female. Because ants subscribe to the philosophy of biological determinism, their gender will play a crucial part of their “professional” development in the caste system.
Once the eggs hatch, they do not spawn adult ants right away. Instead, ants hatch in the form of a larva. Larvae are completely immobile and as such, totally defenseless. Their entire existence depends on the workers, who care for them, feed them, and protect them from invaders.
After the larva stage, ants “hit puberty” and rush into the pupal stage. They aren’t quite ants yet, but they’re starting to look the part. It is within this stage that the caste roles are determined in many species. These roles are infinitely complex and we’ve just started to find out more about them. Scientists need to do a lot more research before we can fully understand them.
Some believe that gene expression is dependent on current conditions. For example, if the queen is dead, a new pupa will be raised to take her place. However, such gene expression is discouraged if there is already a living queen.
Other entomologists postulate the caste roles are highly dependent on nutrition. Whatever the case may be, it is within this stage when the future of the ant and its occupation become known.
Finally, ants emerge from the pupae having reached maturity. They jump in their roles right away. Queens are cared for and protected by the colony. Young workers take on protective duties and caring for the young and the queen before moving on to something else. Males are used to mate with the queen to ensure the next generation of the colony. Everything is harmoniously balanced.
Different members have different life expectancy, depending on their role in the caste. Queens usually have the longest lifespan, with some species reaching over 30 years. This is not universal, however. In some species, there can be more than one queen and they live shorter lives. Some colonies don’t even have queens and instead, many females can be fertilized and give birth, without having a specialized role for the process. As a rule of thumb, however, the queen lives the longest.
Drones are male, winged ants with one main purpose – to fertilize the queen. As such, their functions are rather limited in nature and time. This negates the need for them to live any longer than a few weeks. That’s why they rarely survive any longer than that.
Ant soldiers are smaller than the queen, but are larger than the workers and usually have bigger mandibles. Their duty is to the protect the queen and the colony. They do so by either biting, stinging, or both. Usually an invader is chemically marked by the first defender that detects him. Hence, the rest know to attack.
They also sometimes help the workers with the more physically demanding tasks, such as moving a heavy piece of food. Due to the nature of their role, occupational hazards are implied. Because of that, they sometimes live less than the workers.
Workers have the second longest lifespan after the queen. Their role is essential for the survival of the colony. They take care of the queen and the offspring. They gather food and leave chemical trails for future expansions. And they make sure the colony is well-maintained. The last bit also includes ensuring enough space for the colony and expanding it, if necessary.
Workers’ duties are ingeniously delegated. Usually, the older members of the caste are tasked with the more difficult and dangerous responsibilities, whereas the younger tend to stay safer. After they emerge from their pupae, the first task they take on is caring for the queen and the young ones inside the colony. Usually, it is the oldest of ants who dig and forage for food.
Ants life expectancy widely varies on the species, as well. The lifespan of some species is quite short, while others will surprise you with their inherent longevity.
Black garden ants
The common black garden ant workers have a relative span of 1-2 years. The queen, however, can live up to 15 years, though accounts for even greater longevity are not uncommon.
Fire ant workers complete their life cycle in approximately 5-7 weeks. However, the queen can live up to 7 years.
Pharaoh ant workers have a longer lifespan than most of their cousins – 4 years. However, their queens are relatively short-lived at approximately 12 years.
Carpenter ants go through their entire life cycle in 6-12 weeks. Their queen, however, can survive up to the astounding 25 years.
Argentine ant queens are far smaller and more mobile than the queens of other species. They also take care of their own brood, allowing the workers to take better care of the colony and expand more. They sometimes even lead the expansion, themselves, which is why it’s particularly dangerous to suffer an argentine ant infestation as they are nimble to spread. To make matters worse, the queen can live up to 7 years. After her second year, she gives birth to other future queens to colonize other places. Workers live from 6 to 9 months, but since the colonies reproduce quickly, this is more than enough time for a huge infestation. In fact, the world’s largest known supercolony, spreading thousands of kilometres across over four different continents, was created by argentine ants.
Meat ants are carnivorous ants that are a polygyne species, which means they usually have more than one queen. That’s why the queen’s life expectancy is not as long as other species – 2-3 years. Workers usually live up to 9 months, but that is more than enough time to establish themselves as pests before they expire. However, they are also a very important part of the natural ecological balance in Australia as they lead the charge against the cane toads. This is still a relevant problem.
If you don’t feel like waiting for ants infesting your home to die from natural causes, you can always book professional ant control services. This way you can be sure they will leave you alone.<|endoftext|>
| 3.75 |
673 |
The SKA Layout
How the SKA Telescope will be spread out across two continents
A huge logistical challenge
The total collecting area of the SKA will be well over one square kilometre or 1,000,000 square metres. This will make the SKA the largest radio telescope array ever constructed, by some margin.
Rather than just clustered in the central core regions, the telescopes will be arranged in multiple spiral arm configurations, with the dishes extending to vast distances from the central cores. creating what is known as a long baseline interferometer array.
In such an array, physical distance separates the telescopes, and the distance between them is calculated precisely using the time difference between the arrival of radio signals at each receiver. Computers can then calculate how to combine these signals to synthesise something the equivalent size of a single dish measuring the width of the distance between the two scopes.
In doing so, the interferometry techniques enables astronomers to emulate a telescope with a size equal to the maximum separation between the telescopes in the array, or if needed, just the distance between a subset of telescopes, or indeed, multiple subsets of the main array.
This way, rather than build one gigantic dish, the capabilities of one huge dish are in some ways surpassed by the flexibility that this interferometry configuration brings. The system can act either as one gigantic telescope, or multiple smaller telescopes and any combination in between.
The spiral layout design has been chosen after detailed study by scientists into how best optimise the configuration to get the best possible results.
This spiral configuration gives many different lengths (baselines) and angles between antennas resulting in very high-resolution imaging capability.
The perfect layout would be a random arrangement that maximises the number of different baselines and angles between antennas. However, the practicalities of construction as well as linking the antennas together with cables mean that the spiral configuration is the best trade off between image resolution and cost.
Whilst the high-frequency dishes will be located over thousands of kilometres across Africa, the aperture array antennas will extend to about 200km from the core regions in Africa and Australia.
Every single telescope will be connected to a central core which will combine the data from each into via correlators into more manageable sized data packages. These will be then carried around the globe by high-speed links, to the computer screens of scientists working on the immense amounts of information being gathered by what will be the world’s largest radio telescope.
- At present, there are several long baseline networks dotted around the world. They are located in Europe, Canada, the United States, Russia, Japan and Australia
- The African Very Long Baseline Network (AVN) will modify existing but redundant large (30m) telecommunications dishes for astronomical use
- In Europe, the JIVE (Joint Institute for VLBI in Europe) is a SKA Pathfinder founded in 1993
- The Very Long Baseline Array (VLBA) uses ten dedicated, 25-meter telescopes spanning 5351 miles across the United States, and is the largest radio telescope array that operates all year round as an astronomical instrument
- The LOFAR telescope, built by ASTRON in the Netherlands, which is a pathfinder for the SKA is currently the largest connected radio telescope currently in existence.<|endoftext|>
| 3.9375 |
1,041 |
Apple banana blue walk tree happy sing. Surely you were able to read each of the words in that sentence and understand what they meant independently. An apple is a fruit that is usually round and red, green or yellow. A banana is another fruit that is yellow. Blue is a color…and so on and so forth. However, when you look at the sentence as a whole, does it make sense? Probably not. This nonsense sentence demonstrates the difference between being able to read words and comprehend text. As practiced readers we may take this distinction for granted since the acts of reading and comprehension occur almost simultaneously for us. For developing readers this relationship is not as apparent, but is essential for them to become strong, capable readers.
What exactly IS reading comprehension?
Simply put, reading comprehension is the act of understanding what you are reading. While the definition can be simply stated the act is not simple to teach, learn or practice. Reading comprehension is an intentional, active, interactive process that occurs before, during and after a person reads a particular piece of writing.
Reading comprehension is one of the pillars of the act of reading. When a person reads a text he engages in a complex array of cognitive processes. He is simultaneously using his awareness and understanding of phonemes (individual sound “pieces” in language), phonics (connection between letters and sounds and the relationship between sounds, letters and words) and ability to comprehend or construct meaning from the text. This last component of the act of reading is reading comprehension. It cannot occur independent of the other two elements of the process. At the same time, it is the most difficult and most important of the three.
There are two elements that make up the process of reading comprehension: vocabulary knowledge and text comprehension. In order to understand a text the reader must be able to comprehend the vocabulary used in the piece of writing. If the individual words don’t make the sense then the overall story will not either. Children can draw on their prior knowledge of vocabulary, but they also need to continually be taught new words. The best vocabulary instruction occurs at the point of need. Parents and teachers should pre-teach new words that a child will encounter in a text or aid her in understanding unfamiliar words as she comes upon them in the writing. In addition to being able to understand each distinct word in a text, the child also has to be able to put them together to develop an overall conception of what it is trying to say. This is text comprehension. Text comprehension is much more complex and varied that vocabulary knowledge. Readers use many different text comprehension strategies to develop reading comprehension. These include monitoring for understanding, answering and generating questions, summarizing and being aware of and using a text’s structure to aid comprehension.
How does reading comprehension develop?
As you can see, reading comprehension is incredibly complex and multifaceted. Because of this, readers do not develop the ability to comprehend texts quickly, easily or independently. Reading comprehension strategies must be taught over an extended period of time by parents and teachers who have knowledge and experience using them. It might seem that once a child learns to read in the elementary grades he is able to tackle any future text that comes his way. This is not true. Reading comprehension strategies must be refined, practiced and reinforced continually throughout life. Even in the middle grades and high school, parents and teachers need to continue to help their children develop reading comprehension strategies. As their reading materials become more diverse and challenging, children need to learn new tools for comprehending these texts. Content area materials such as textbooks and newspaper, magazine and journal articles pose different reading comprehension challenges for young people and thus require different comprehension strategies. The development of reading comprehension is a lifelong process that changes based on the depth and breadth of texts the person is reading.
Why is reading comprehension so important?
Without comprehension, reading is nothing more than tracking symbols on a page with your eyes and sounding them out. Imagine being handed a story written in Egyptian hieroglyphics with no understanding of their meaning. You may appreciate the words aesthetically and even be able to draw some small bits of meaning from the page, but you are not truly reading the story. The words on the page have no meaning. They are simply symbols. People read for many reasons but understanding is always a part of their purpose. Reading comprehension is important because without it reading doesn’t provide the reader with any information.
Beyond this, reading comprehension is essential to life. Much has been written about the importance of functional literacy. In order to survive and thrive in today’s world individuals must be able to comprehend basic texts such as bills, housing agreements (leases, purchase contracts), directions on packaging and transportation documents (bus and train schedules, maps, travel directions). Reading comprehension is a critical component of functional literacy. Think of the potentially dire effects of not being able to comprehend dosage directions on a bottle of medicine or warnings on a container of dangerous chemicals. With the ability to comprehend what they read, people are able not only to live safely and productively, but also to continue to develop socially, emotionally and intellectually.<|endoftext|>
| 3.75 |
634 |
Question
# If ${\text{A}} = \{ 10,15,20,25,30,35,40,45,50\}$,${\text{B}} = \{ 1,5,10,15,20,30\}$ and ${\text{C}} = \{ 7,8,15,20,35,45,48\} ,$ find ${\text{A}} - ({\text{B}} \cap {\text{C}})$.
Hint:- Draw Venn’ diagram. First find ${\text{B}} \cap {\text{C}}$.
As, we are given with three sets and that were,
$\Rightarrow {\text{A}} = \{ 10,15,20,25,30,35,40,45,50\} ,$
$\Rightarrow {\text{B}} = \{ 1,5,10,15,20,30\}$ and
$\Rightarrow {\text{C}} = \{ 7,8,15,20,35,45,48\}$
And asked to find ${\text{A}} - ({\text{B}} \cap {\text{C}})$.
And as we know that in set theory $\cap$ depicts intersection.
$\Rightarrow$An intersection of two sets gives us the common elements of both sets.
So, ${\text{B}} \cap {\text{C}}$ will give us common elements of sets B and C
$\Rightarrow$So, ${\text{(B}} \cap {\text{C) }} = {\text{ }}\{ 1,5,10,15,20,30\} \cap \{ 7,8,15,20,35,45,48\} = \{ 15,20\}$
As we know that if X and Y are some sets then,
$\Rightarrow$X-Y will give us a set having all elements of X excluding elements of Y.
So, ${\text{A}} - ({\text{B}} \cap {\text{C}})$ will give us those elements of set A which are not in set ${\text{(B}} \cap {\text{C)}}$
$\Rightarrow$${\text{A}} - ({\text{B}} \cap {\text{C}}) = \{ 10,15,20,25,30,35,40,45,50\} - \{ 15,20\}$
$\Rightarrow$Hence, ${\text{A}} - ({\text{B}} \cap {\text{C}}) = \{ 10,25,30,35,40,45,50\}$
Note:- Whenever we come up with this type of problem then first, we have to Draw Venn’s diagram because this will give proper clarity for the problem. Then remember that for any set X and Y, X-Y will give us a set having all elements of
X excluding elements of Y.<|endoftext|>
| 4.84375 |
1,802 |
10
Q:
# The distance of the college and home of Rajeev is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed (or increased) speed of Rajeev?
A) 28 km/h B) 30 km/h C) 40 km/h D) 20 km/h
Answer: D) 20 km/h
Explanation:
Let the normal speed be x km/h, then
$80x-80(x+4)=1$
$⇒$$x2+4x-320=0$
$⇒$x (x + 20) - 16 (x + 20) = 0
(x + 20 ) (x - 16) =0
x = 16 km/h
Therefore (x + 4) = 20 km/h
Therefore increased speed = 20 km/h
Q:
Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes?
A) 5 kmph B) 6 kmph C) 7.5 kmph D) 8 kmph
Answer & Explanation Answer: A) 5 kmph
Explanation:
Let p be the speed of man in kmph
According to the given data in the question,
Distance travelled by bus in 10 min with 20 kmph == Distance travelled by man in 8 min with (20 + p) kmph in opposite direction
=> 20 x 10/60 = 8/60 (20 + p)
=> 200 = 160 + 8p
=> p = 40/8 = 5 kmph.
1 44
Q:
With an average speed of 40 km/hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. Find the length of the total journey?
A) 70 kms B) 60 kms C) 45 kms D) 30 kms
Answer & Explanation Answer: A) 70 kms
Explanation:
Let the time taken by train be 't' hrs.
Then,
40t = 35t + 35/4
t = 7/4 hrs
Therefore, Required length of the total journey d = s x t
= 40 x 7/4
= 70 kms.
3 403
Q:
Ashwin fires two bullets from the same place at an interval of 15 minutes but Rahul sitting in a bus approaching the place hears the second sound 14 minutes 30 seconds after the first. If sound travels at the speed of 330 meter per second, what is the approximate speed of bus?
A) 330/29 m/s B) 330 x 30 m/s C) 330/14 m/s D) 330/900 m/s
Answer & Explanation Answer: A) 330/29 m/s
Explanation:
Second gun shot take 30 sec to reach rahul imples distance between two.
given speed of sound = 330 m/s
Now, distance = 330 m/s x 30 sec
Hence, speed of the bus = d/t = 330x30/(14x60 + 30) = 330/29 m/s.
2 302
Q:
Important Time and Distance formulas with examples.
1. How to find Speed(s) if distance(d) & time(t) is given:
Ex: Find speed if a person travels 4 kms in 2 hrs?
Speed = D/T = 4/2 = 2 kmph.
2. Similarly, we can find distance (d) if speed (s) & time (t) is given by
Distance (D) = Speed (S) x Time (T)
Ex : Find distance if a person with a speed of 2 kmph in 2 hrs?
Distance D = S X T = 2 x 2 = 4 kms.
3. Similarly, we can find time (t) if speed (s) & distance (d) is given by
Time (T) =
Ex : Find in what time a person travels 4 kms with a speed of 2 kmph?
Time T = D/S = 4/2 = 2 hrs.
4. How to convert km/hr into m/sec :
Ex : Convert 36 kmph into m/sec?
36 kmph = 36 x 5/18 = 10 m/sec
5. How to convert m/sec into km/hr :
.
Ex : Convert 10 m/sec into km/hr?
10 m/sec = 10 x 18/5 = 36 kmph.
6. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is b : a.
7. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km/hr . Then, the average speed during the whole journey is $2xyx+y$ km/hr.
295
Q:
Two trains are running with speeds 30 kmph and 58 kmph respectively in the same direction. A man in the slower train passes the faster train in 18 seconds. Find the length of the faster train?
A) 105 mts B) 115 mts C) 120 mts D) 140 mts
Answer & Explanation Answer: D) 140 mts
Explanation:
Speeds of two trains = 30 kmph and 58 kmph
=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s
Given a man takes time to cross length of faster train = 18 sec
Now, required Length of faster train = speed x time = 70/9 x 18 = 140 mts.
2 299
Q:
A train-A passes a stationary train B and a pole in 24 sec and 9 sec respectively. If the speed of train A is 48 kmph, what is the length of train B?
A) 200 mts B) 180 mts C) 160 mts D) 145 mts
Answer & Explanation Answer: A) 200 mts
Explanation:
Length of train A = 48 x 9 x 5/18 = 120 mts
Length of train B = 48 x 24 x 5/18 - 120
=> 320 - 120 = 200 mts.
2 420
Q:
Tilak rides on a cycle to a place at speed of 22 kmph and comes back at a speed of 20 kmph. If the time taken by him in the second case is 36 min. more than that of the first case, what is the total distance travelled by him (in km)?
A) 132 km B) 264 km C) 134 km D) 236 km
Answer & Explanation Answer: B) 264 km
Explanation:
Let the distance travelled by Tilak in first case or second case = d kms
Now, from the given data,
d/20 = d/22 + 36 min
=> d/20 = d/22 + 3/5 hrs
=> d = 132 km.
Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.
0 382
Q:
A train covers 180 km distance in 5 hours. Another train covers the same distance in 1 hour less. What is the difference in the distances covered by these trains in one hour if they are moving in the same direction?
A) 15 kms B) 9 kms C) 6 kms D) 18 kms
Answer & Explanation Answer: B) 9 kms
Explanation:
The first train covers 180 kms in 5 hrs
=> Speed = 180/5 = 36 kmph
Now the second train covers the same distance in 1 hour less than the first train => 4 hrs
=> Speed of the second train = 180/4 = 45 kmph
Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.<|endoftext|>
| 4.5 |
1,703 |
Some people encourage ketosis by following a diet called the ketogenic or low-carb diet. The aim of the diet is to try and burn unwanted fat by forcing the body to rely on fat for energy, rather than carbohydrates.
- Ketosis occurs when the body does not have sufficient access to its primary fuel source, glucose.
- Ketosis describes a condition where fat stores are broken down to produce energy, which also produces ketones, a type of acid.
- As ketone levels rise, the acidity of the blood also increases, leading to ketoacidosis, a serious condition that can prove fatal.
- People with type 1 diabetes are more likely to develop ketoacidosis, for which emergency medical treatment is required to avoid or treat diabetic coma.
- Some people follow a ketogenic (low-carb) diet to try to lose weight by forcing the body to burn fat stores.
What is ketosis?
In normal circumstances, the body's cells use glucose as their primary form of energy. Glucose is typically derived from dietary carbohydrates, including:
- sugar - such as fruits and milk or yogurt
- starchy foods - such as bread and pasta
The body breaks these down into simple sugars. Glucose can either be used to fuel the body or be stored in the liver and muscles as glycogen.
If there is not enough glucose available to meet energy demands, the body will adopt an alternative strategy in order to meet those needs. Specifically, the body begins to break down fat stores to provide glucose from triglycerides.
Ketones are a by-product of this process.
Ketones are acids that build up in the blood and are eliminated in urine. In small amounts, they serve to indicate that the body is breaking down fat, but high levels of ketones can poison the body, leading to a process called ketoacidosis.
Ketosis describes the metabolic state whereby the body converts fat stores into energy, releasing ketones in the process.
The ketogenic diet
Some people follow a ketogenic diet as a way to lose weight.
Due to the fact that ketosis breaks down fat stored within the body, some diets aim to create this metabolic state so as to facilitate weight loss.
Ketosis diets are also referred to as:
- ketogenic diets
- keto diets
- low-carbohydrate diets
The diet itself can be regarded as a high-fat diet, with around 75 percent of calories derived from fats. In contrast, around 20 percent and 5 percent of calories are gained from proteins and carbohydrates, respectively.
Adhering to the ketogenic diet can lead to short-term weight loss. A study conducted in 2008 and published in the American Journal of Clinical Nutrition found that obese men following a ketogenic diet for 4 weeks lost an average of 12 pounds during this time.
The participants were able to consume fewer calories without feeling hungry while following the diet.
Is ketosis healthy?
The ketogenic diet could have a healthful effect on serious health conditions such as:
- cardiovascular disease
- metabolic syndrome
It may also improve levels of HDL cholesterol (high-density lipoproteins, also known as "good" cholesterol) better than other moderate carbohydrate diets.
These health benefits could be due to the loss of excess weight and eating of healthier foods, rather than a reduction in carbohydrates.
The ketogenic diet has also been used under medical supervision to reduce seizures in children with epilepsy who do not respond to other forms of treatment. Some studies have suggested that the diet could also benefit adults with epilepsy, although more research is required to confirm these findings.
However, longer-term adherence to the ketogenic diet does not appear to yield great benefit.
The American Heart Association (AHA), American College of Cardiology, and the Obesity Society have concluded that there is not enough evidence to suggest that low-carbohydrate diets such as the ketogenic diet provide health benefits to the heart.
Other conditions are also being studied to see if a ketogenic diet might be beneficial; these include:
- metabolic syndrome
- Alzheimer's disease
- polycystic ovary disease (PCOS)
- Lou Gehrig's disease
Ketosis and diabetes
In diabetic patients, ketosis can occur due to the body not having enough insulin to process the glucose in the body. The presence of ketones in the urine is an indicator that a patient's diabetes is not being controlled correctly.
Some dietitians recommend a ketogenic diet for individuals with type 2 diabetes, also known as non-insulin dependent diabetes (NIDDM). With type 2 diabetes, the body still produces some insulin but is unable to properly use the insulin to transport glucose into cells for use as fuel.
The ketogenic diet focuses on the reduction of dietary carbohydrate intake. Individuals with type 2 diabetes are recommended to reduce carbohydrate intake as carbohydrates are converted to glucose and increase blood sugar levels.
Patients with diabetes who follow a ketogenic diet need to carefully monitor their ketone levels. A serious condition called ketoacidosis can occur if these levels get too high, and although it is most prevalent in individuals with type 1 diabetes, people with type 2 diabetes can also develop ketoacidosis.
Ketoacidosis is a condition where the levels of ketones in the body are abnormally high, poisoning the body. It is a serious and dangerous condition that can quickly develop, sometimes within the space of 24 hours.
There are several different potential triggers for ketoacidosis. It is most commonly caused by illnesses that lead to the production of higher levels of hormones that work against insulin.
It can also result from problems with insulin therapy, either through missing scheduled treatments or not being given enough insulin. Less common triggers of ketoacidosis include:
- drug abuse
- emotional trauma
- physical trauma
Ketoacidosis most commonly occurs in people with type 1 diabetes due to the body not producing any insulin. Ketoacidosis can also occur in people with type 2 diabetes, although it is much less common.
High levels of ketones in the urine and high blood sugar levels (hyperglycemia) are signs of ketoacidosis and can be detected with kits in the home.
Early symptoms of ketoacidosis include:
- abdominal pain
- confusion and difficulty concentrating
- dry or flushed skin
- excessive thirst and dry mouth
- fruity breath
- frequent urination
- nausea and vomiting
- shortness of breath or rapid breathing
Ketosis treatment and prevention
Ketone levels can be monitored using urine testing kits that are commonly available over the counter.
Ketosis does not usually occur in healthy individuals that eat balanced diets and regular meals. Drastically reducing the amount of calories and carbohydrates that are consumed, exercising for extended periods of time, or being pregnant can all trigger ketosis.
In patients with diabetes, ketosis and eventually ketoacidosis may occur if insufficient insulin is used to properly manage the condition, if meals are skipped, or if an insulin reaction occurs (often while asleep).
Diabetic ketoacidosis is considered an emergency as it can lead to diabetic coma and even death. Treatment is usually administered by emergency healthcare workers, followed by hospitalization in an intensive care unit.
For diabetic patients, the following measures are commonly taken:
- Fluid replacement - to rehydrate the body and dilute the excess sugar in the blood.
- Electrolyte replacement - these are needed to help keep the heart, muscles, and nerve cells functioning correctly. Levels in the blood often drop in the absence of insulin. Electrolyte supplements are available to purchase online.
- Insulin therapy - to reverse the processes that caused the episode of ketoacidosis.
Among otherwise healthy people, following a healthy, balanced diet and exercising regularly can prevent ketosis.
In addition, there are a number of measures that people with diabetes can take to help prevent ketoacidosis:
- Monitor blood sugar levels carefully and frequently - at least three to four times a day.
- Discuss insulin dosage with a specialist and follow a diabetes treatment plan.
- Keep an eye on ketone levels with a test kit, particularly when ill or under stress.<|endoftext|>
| 3.796875 |
166 |
Sounds can be made by twanging a string or an elastic band, blowing down a pipe, banging something together, and scraping or shaking something. There are three families of musical instruments - string, wind and percussion. When a string is plucked on an instrument such as a guitar, the vibration is passed into the air and you hear a sound. Blowing into panpipes or a horn makes the air inside the pipes vibrate. Bongos and other percussion instruments produce vibrations when you bang them.
Useful as an introduction to how sounds are made through vibration and could lead on to an activity where children make their own musical instruments using household items. Children could then use these home-made instruments to do a class performance. This could also be used within a Music lesson introducing the children to the types of instruments.<|endoftext|>
| 4.1875 |
302 |
An (XY, D)—CIRCULAR NUMBER is a number whose last two digits are X and Y, and multiplication of the number by the single digit D is equivalent to moving the last two digits (XY) to the beginning of the number. For example, the number 132832080200501253 is a (53, 4)—CIRCULAR NUMBER because 132832080200501253 * 4 = 531328320802005012. Write a program that asks for two-digit number XY and a single digit D, and prints the SMALLEST (XY, D)—CIRCULAR NUMBER or prints a message NONE if none exists. It is guaranteed that your number will not exceed 255 digits.
INPUTAn integer number, N, indicating the total number of cases. On each of the next N lines are the two-digit number XY and the single digit D, separated by a space.
OUTPUTThe smallest circular number (if one exists with less than 255 digits) that satisfies the given condition; otherwise output the message NONE.
1 53 4
Point Value: 10
Time Limit: 2.00s
Memory Limit: 16M
Added: Feb 24, 2009
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3<|endoftext|>
| 4.46875 |
1,032 |
# Chain Rule
• Dec 6th 2006, 02:06 PM
aaasssaaa
Chain Rule
Four questions in the Homework are bothering me:mad: :
see attachment cause I had to use word to put the equation
• Dec 6th 2006, 02:48 PM
galactus
For the most part, it's just differentiation. Find the derivative and enter in the given values.
• Dec 6th 2006, 03:34 PM
aaasssaaa
Yea this is what I tried to do except i kept getting the wrong answer
• Dec 6th 2006, 04:12 PM
topsquark
Quote:
Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :
see attachment cause I had to use word to put the equation
let g(x) = 27/(2x+1)^3 Determine the slope of the tangent to the curve at x=1
$\displaystyle g(x) = \frac{27}{(2x+1)^3}$
$\displaystyle g'(x) = \frac{-3 \cdot 27}{(2x+1)^4}$
$\displaystyle g'(1) = \frac{-3 \cdot 27}{(2 \cdot 1 +1)^4} =$ $\displaystyle -\frac{81}{81} = -1$
-Dan
• Dec 6th 2006, 04:21 PM
topsquark
Quote:
Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :
see attachment cause I had to use word to put the equation
Let $\displaystyle f(x) = \frac{6}{5 \sqrt{3x^4} - x}$. Find f'(1).
$\displaystyle f(x) = \frac{6}{5 \sqrt{3x^4} - x}$
$\displaystyle f'(x) = \frac{-6}{(5 \sqrt{3x^4} - x)^{2}} \cdot \left ( 5 \cdot \frac{1}{2\sqrt{3x^4}} \cdot 3 \cdot 4x^3 - 1 \right )$
$\displaystyle f'(x) = \frac{-6 \cdot (10\sqrt{3}x - 1)}{(5\sqrt{3}x^2 - x)^2}$
$\displaystyle f'(1) = \frac{-6 \cdot (10\sqrt{3} \cdot 1 - 1)}{(5\sqrt{3}(1)^2 - 1)^2} \approx -1.66878$
-Dan
• Dec 6th 2006, 04:26 PM
topsquark
Quote:
Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :
see attachment cause I had to use word to put the equation
$\displaystyle s = 5 - 3t - \left [ t^{-2} + (2t - 5)^5 \right ] ^4$
$\displaystyle s' = -3 - 4 \left [ t^{-2} + (2t - 5)^5 \right ] ^3 \cdot \left ( -2t^{-3} + 5(2t - 5)^4 \cdot 2 \right )$
$\displaystyle s' = -3 - 4 \left ( -2t^{-3} + 10(2t - 5)^4 \right ) \left [ t^{-2} + (2t - 5)^5 \right ] ^3$
-Dan
• Dec 6th 2006, 04:31 PM
topsquark
Quote:
Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :
see attachment cause I had to use word to put the equation
$\displaystyle V(t) = 50000 \left ( 1 - \frac{t}{30} \right ) ^2$
$\displaystyle V'(t) = 50000 \cdot 2 \left ( 1 - \frac{t}{30} \right ) \cdot \frac{-1}{30}$
$\displaystyle V'(t) = - \frac{10000}{3} \left ( 1 - \frac{t}{30} \right )$
$\displaystyle V'(10) = - \frac{10000}{3} \left ( 1 - \frac{10}{30} \right )$
$\displaystyle V'(10) = - \frac{20000}{9}$
Technically they want 20000/9 gal/min (ie not negative) because the question asked for the rate the water is flowing out.
-Dan<|endoftext|>
| 4.40625 |
443 |
Question #7cfc8
Oct 5, 2016
Proof below
Explanation:
First we will find the expansion of $\sin \left(3 x\right)$ separately (this will use the expansion of trig functions formulae):
$\sin \left(3 x\right) = \sin \left(2 x + x\right)$
$= \sin 2 x \cos x + \cos 2 x \sin x$
$= 2 \sin x \cos x \cdot \cos x + \left({\cos}^{2} x - {\sin}^{2} x\right) \sin x$
$= 2 \sin x {\cos}^{2} x + \sin x {\cos}^{2} x - {\sin}^{3} x$
$= 3 \sin x {\cos}^{2} x - {\sin}^{3} x$
$= 3 \sin x \left(1 - {\sin}^{2} x\right) - {\sin}^{3} x$
$= 3 \sin x - 3 {\sin}^{3} x - {\sin}^{3} x$
$= 3 \sin x - 4 {\sin}^{3} x$
Now to solve the original question:
$\frac{\sin 3 x}{\sin x} = \frac{3 \sin x - 4 {\sin}^{3} x}{\sin} x$
$= 3 - 4 {\sin}^{2} x$
$= 3 - 4 \left(1 - {\cos}^{2} x\right)$
$= 3 - 4 + 4 {\cos}^{2} x$
$= 4 {\cos}^{2} x - 1$
$= 4 {\cos}^{2} x - 2 + 1$
$= 2 \left(2 {\cos}^{2} x - 1\right) + 1$
$= 2 \left(\cos 2 x\right) + 1$<|endoftext|>
| 4.5 |
830 |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Can I Graph You, Too? | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Algebra I Student Edition Go to the latest version.
# 7.3: Can I Graph You, Too?
Created by: CK-12
0 0 0
This activity is intended to supplement Algebra I, Chapter 6, Lesson 6.
## Probmem 1 - Introduction to Disjunction and Conjunction
Consider the equation $|x|=5$. To solve, you would graph both sides of the equation as functions $(y =|x|$ and $y = 5$) and mark the solution as the area where the graphs intersect.
The same method can be applied to inequalities.
Press APPS and select the Inequalz app. Press any key to begin.
Example 1: $|x|<5$
• Using $Y1=$ graph the left side as $y = |x|$. The absolute value function is located by pressing MATH $\rightarrow$ and selecting abs(.
• Using $Y2=$ graph the right side as $y < 5$. On the equals sign, press ALPHA $[F2]$ for the $<$ sign. Press ZOOM and select ZoomStandard.
• Find the intersection points by pressing $2^{nd}$ [TRACE] and selecting intersect. Now just move the cursor to the intersection point and press ENTER three times. The solution is where the shading overlaps the graph of the absolute value function.
In this case, the solution is $-5 < x < 5$.
When an absolute value is less than a number, it is a conjunction because the solution is just one part of the graph.
$|ax+b| < c \quad \rightarrow \quad -c < ax + b < c.$
Example 2: $|x-4| \ge 8$
• Using $Y1=$ graph the left side as $y = |x-4|$.
• Using $Y2=$ graph the right side as $y \ge 8$. On the equals sign, press ALPHA$[F5]$ for the $\ge$ sign. Press WINDOW to choose appropriate window settings.
• Find the intersection points.
In this case, the solution is $x \le -4$ or $x \ge 12$.
When an absolute value is greater than a number, it is a disjunction because the solution is two separate parts of the graph.
$|ax + b| > c \quad \rightarrow \quad ax + b < -c \quad \text{or} \quad ax + b > c.$
## Problem 2 - Application of Disjunction and Conjunction
For the problems below, write the inequalities as either a conjunction or disjunction, then solve for $x$. Check your solution by graphing using the method described in Examples 1 and 2. Please use your graphing calculator to check your results.
#1: $|2x-3|>9$
#2: $\left |\frac{1}{3} x -10 \right | \le 11$
#3: $|3x|-1 \ge 5$
#4: $2 |4x-7| + 6 < 18$
## Problem 3 - Real World Application
One application of absolute value inequalities is engineering tolerance. Tolerance is the idea that an ideal measurement and an actual measurement can only differ within a certain range.
A bolt with a $10 \ mm$ diameter has a tolerance range of $9.965 \ mm$ to $10 \ mm$, while the hole that it fits into has a tolerance range of $10.05 \ mm$ to $10.075 \ mm$.
How can you express the tolerances of both the bolt and the hole in terms of an absolute value inequality?
Feb 22, 2012
Aug 19, 2014<|endoftext|>
| 4.625 |
508 |
How many comets are there?
Occasionally a comet makes the headlines, perhaps because it is visible from Earth with the unaided eye. This gives the impression that comets are rare. However, nothing could be further from the truth. There are thought to be so many comets that even astronomers can’t count them all…
Comets were born in the outer reaches of the Solar System, 4600 million years ago, when the planets were forming, from dust and ice.
Only when a comet swings in towards the Sun does it begin to evaporate and generate the tails for which comets are so famous. Observations of Comet Halley during its last appearance in 1986 show that the top one-metre layer of the comet’s surface was lost to form the tail.
A comet probably has enough surplus ice for a few hundred passes of the Sun. After that, it may become so weakened by the loss of material that it shatters, or its surface may become so choked with tar-like substances, left behind when the ice evaporates, that it forms a layer, insulating the remaining ice from further exposure to the Sun.
If this happens it transforms into a ‘stealth’ comet. It stops producing a tail and joins the army of other near-Earth asteroids (NEAs).
Only better observations or even visits by spacecraft will tell us how many of the NEAs are really extinct comet cores.
The fact that comets die in what, astronomically speaking, is a short period of time (around 10 000 years), suggests that there must be a great reservoir of extra comets to restock them.
By studying the orbits of comets, astronomers have come to the conclusion that two such reservoirs exist. The Kuiper belt, out beyond Pluto, is a flared disc of comets that supplies most of the short-period comets (those that orbit the Sun in less than a century).
The Oort cloud (named after Jan H. Oort) is much larger and supplies the long-period comets. It encloses the Solar System, with an outer edge that reaches almost a quarter of the way to the nearest star.
By taking the size of the Oort cloud into account, and the number of long-period comets that have been seen, astronomers estimate that a staggering one 'trillion' (12 zeros) comets may be out there!
Last update: 1 December 2017<|endoftext|>
| 3.796875 |
554 |
# How do you rewrite this Logarithmic problem in expanded form?
May 11, 2018
$= \frac{7}{2} \log x - \frac{5}{2} \log y - \frac{7}{2} \log z$
#### Explanation:
$\log \sqrt{{x}^{7} / \left({y}^{5} {z}^{7}\right)} = \log {\left({x}^{7} / \left({y}^{5} {z}^{7}\right)\right)}^{\frac{1}{2}}$
Recall that $\log \left({x}^{a}\right) = a \log x$, so
$\log {\left({x}^{7} / \left({y}^{5} {z}^{7}\right)\right)}^{\frac{1}{2}} = \frac{1}{2} \log \left({x}^{7} / \left({y}^{5} {z}^{7}\right)\right)$
Furthermore, recalling that $\log \left(\frac{a}{b}\right) = \log a - \log b ,$
$\frac{1}{2} \log \left({x}^{7} / \left({y}^{5} {z}^{7}\right)\right) = \frac{1}{2} \left[\log \left({x}^{7}\right) - \log \left({y}^{5} {z}^{7}\right)\right]$
Recalling that $\log \left(a b\right) = \log a + \log b ,$
$\frac{1}{2} \left[\log \left({x}^{7}\right) - \log \left({y}^{5} {z}^{7}\right)\right] = \frac{1}{2} \left[\log \left({x}^{7}\right) - \left(\log \left({y}^{5}\right) + \log \left({z}^{7}\right)\right)\right]$
Apply the exponent property to all remaining logarithms and distribute the negative through:
$= \frac{1}{2} \left[7 \log x - 5 \log \left(y\right) - 7 \log \left(z\right)\right]$
Distribute the $\frac{1}{2} :$
$= \frac{7}{2} \log x - \frac{5}{2} \log y - \frac{7}{2} \log z$<|endoftext|>
| 4.6875 |
337 |
A line that intersects a circle in exactly one point is called a tangent and the point where the intersection occurs is called the point of tangency. The tangent is always perpendicular to the radius drawn to the point of tangency.
A secant is a line that intersects a circle in exactly two points.
When a tangent and a secant, two secants, or two tangents intersect outside a circle then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.
$\m\angle A=\frac{1}{2}(m\overline{DE}-m\overline{BC} )$
When two chords intersect inside a circle, then the measures of the segments of each chord multiplied with each other is equal to the product from the other chord:
$AB\cdot EB=CE\cdot ED$
If two secants are drawn to a circle from one exterior point, then the product of the external segment and the total length of each secant are equal:
$AB\cdot AD=AC\cdot AE$
If one secant and one tangent are drawn to a circle from one exterior point, then the square of the length of the tangent is equal to the product of the external secant segment and the total length of the secant:
$AB^{2}=AC\cdot AD$
If we have a circle drawn in a coordinate plane, with the center in (a,b) and the radius r then we could always describe the circle with the following equation:
$(x-a)^{2}+(y-b)^{2}=r^{2}$
## Video lesson
Find the value of t in the figure<|endoftext|>
| 4.40625 |
217 |
While the small dinosaur Ornitholestes is long since extinct, her basic biology was solid enough in her close genetic cousins, for instance, to adapt to flight or even lose their teeth completely in favor of a beak. It it is not difficult to imagine, when looking at a modern Cassowary for example, to really get an idea of just how these small theropods looked or even moved around and behaved during the Jurassic Period. Feathers as a flock of fur (a type of scale actually) would have come around first to provide warmth and dryness....and would then further be adapted for flight. Close examination of fossil evidence now confirms that virtually ALL theropod dinosaurs were covered in some form of insulating feather/fur. The Triassic, Jurassic and Cretaceous Periods were a large chunk of the Earth's history and had plenty of cold and wet climates, feathers would have been nearly tantamount to long term survival.
Having chicken for dinner tonight? Well, you're eating a dinosaur.
High resolution stock illustration available for commercial use Here<|endoftext|>
| 4.03125 |
1,028 |
# Derivative of a function defined by conditionals
The question is:
Which values $a$ and $b$ can assume to be possible to derive the function $f$ at $x = 1$
$f(x) = \left\{ \begin{array}{rl} x^2 &\mbox{$x < 1$} \\ ax + b &\mbox{$x \geq 1$} \end{array} \right.$
Here is my progress
The one way i can wonder to solve this is using the definition of derivatives. So i started by verifying the following limit
$$\lim_{x \to 1^{+}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{+}} \frac{ax + b - (a + b)}{x - 1}$$ $$= \lim_{x \to 1^{+}} \frac{ax - a}{x - 1} = \lim_{x \to 1^{+}} \frac{a(x - 1)}{x - 1} = \lim_{x \to 1^{+}} a = a$$
And then i verify the limit by the left side
$$\lim_{x \to 1^{-}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{-}} \frac{x^2 - (a + b)}{x - 1}$$
From this point i can't go ahead.
-
You're almost there! If $a+b=1$, then your left limit is... If $a+b\neq 1$, then... – 1015 Jan 25 '13 at 19:24
Very roughly: you want $f$ continuous and differentiable at $x=1$. Continuity means setting the two expressions equal to each other (why?):
$$1^2 = a(1) + b$$
differentiability means matching the value of the derivative of each expression at $x=1$; that is
$$2 (1) = a$$
-
I guess i found the answer
If the function is differentiable, then it's continuous. So, let's verify if it's continuos
To be continuos it must satisfy the following conditions
- the limit must exists when $x$ tends to $a$
- $f(a)$ must be equal to the limit when $x$ tends to $a$
So, lets begin
$$f(1) = a + b$$ $$\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} x^2 = 1$$ $$\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} ax + b = a + b$$
So, we need the equality $\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = f(1)$ to be true. We get $a + b = 1$
Verifying if the function is differentiable
$$\lim_{x \to 1^{-}} \frac{f(x) - f(1)}{x - 1}$$ We have $f(1) = a + b = 1$ , so
$$\lim_{x \to 1^{-}} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^{-}} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1^{-}} x + 1 = 2$$
Now the limit by the right side
$$\lim_{x \to 1^{+}} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^{+}} \frac{ax + b - (a + b)}{x - 1} = \lim_{x \to 1^{+}} \frac{ax - a}{x - 1} = \lim_{x \to 1^{+}} \frac{a(x - 1)}{x - 1} = a$$
As we need this limits to be equal, so $a = 2$
We had the condition $a + b = 1$, and $a = 2$, so we can substitute $2$ in $a$, then $2 + b = 1 \leftrightarrow b = 1 - 2 \leftrightarrow b = -1$
Finally we have that if $a = 2$ and $b = -1$ the function $f$ is differentiable at $x = 1$
-<|endoftext|>
| 4.5625 |
1,840 |
# Completing the Square Instructor: Dr. Jo Steig
Completing the square is a process that can be used to solve an equation, find the vertex on a parabola, or to put an equation of a conic in standard form. On this page I am going to give several examples of completing the square that will allow us to find the vertex of a parabola.
BACKGROUND: The basic idea behind completing the square is that we are going to build an expression that will factor into a perfect square. Note that in the following,
x2 + 2bx + b2 = (x + b)2
x2 + 2bx + b2 is called a perfect square trinomial because it factors into (x + b)2, a squared binomial. In the process of completing the square we will build a perfect square trinomial that will factor into a squared binomial.
Note that in the above example the following is true:
(1) The coefficient of the squared term, x2, is 1
(2) The second term of (x + b)2 , b, is (1/2) of the coefficient of the linear term of the trinomial, x2 + 2bx + b2. That is,
The linear term of x2 + 2bx + b2 is 2bx
The coefficient of the linear term is 2b
(1/2) of 2b = b, the second term of (x + b)2
We will use this information in the process of completing the square.
First, I will give and example of completing the square and will then follow that with detailed steps of the process. In the following examples we will change the equation of a parabola (y = ax2 + bx + c) into the form y = a( x - h) 2 + k. Determining the vertex from this form will be discussed at the end of the discussion of completing the square.
EXAMPLE 1: Complete the square on x: y = x2 + 6x + 2
Add and subtract the square of (1/2) of the coefficient of the linear term
The linear term is 6x
The coefficient of the linear term is 6
(1/2) of the coeffient of the linear term is (1/2) 6 = 3
The square of (1/2) of the coeffient of the linear term is 32
y = x2 + 6x + 9 - 9+ 2
y = (x2 + 6x + 9) - 9+ 2
y = (x + 3)2 - 7
Now, here is a summary of the steps of the process:
## Completing the Square
(1) The coefficient of the squared term, x2, must be 1
If the coefficient is not 1 then factor it out of the two variable terms (example will follow)
(2) Add and subtract the square of (1/2) of the coefficient of the linear term
(3) Separate the perfect square trinomial from the constant terms
(4) Factor the perfect square trinomial
EXAMPLE 2: Complete the square on x: y = 2x2 + 12x- 5
(1) The coefficient of the squared term, x2, must be 1
If the coefficient is not 1 then factor it out of the two variable terms
y = 2(x2 + 6x) - 5
(2) Add and subtract the square of (1/2) of the coefficient of the linear term
y = 2(x2 + 6x + 32 - 32) - 5
(3) Separate the perfect square trinomial from the constant terms
y = 2(x2 + 6x + 32) - 32(2) - 5
(4) Factor the perfect square trinomial
y = 2(x + 3)2 - 32(2) - 5
y = 2(x + 3)2 - 23
You should take particular notice that in step (3) above, the constant term that came out of the parentheses had to be multiplied by 2.
Here is another example, but without the commentary:
EXAMPLE 3: Complete the square on t: x = -3t 2+ 3t - 1
x = -3( t 2- t) - 1
x = -3( t 2- t + .5 2 - .5 2 ) - 1
x = -3( t 2- t + .5 2 ) - .5 2 (-3) - 1
x = -3( t - .5 )2 - .5 2 (-3) - 1
x = -3( t - .5 )2 + .75 - 1
x = -3( t - .5 )2 - .25
This process can be used even with numbers that would be considered unwiedy without a calculator - you just follow the same steps and pattern shown above. Here is such an example.
EXAMPLE 4: Complete the square on x: y = 42.94x 2+ 62.34x - 12.31
y = 42.94 (x 2+ 1.45x) - 12.31 [1.45 = 62.34/42.94]
y = 42.94 (x 2+ 1.45x + .73 2 - .73 2 ) - 12.31[.73 = 1.45/2]
y = 42.94 (x 2+ 1.45x + .73 2 ) - .73 2 (42.94) - 12.31
y = 42.94 (x + .73 ) 2 - .73 2 (42.94) - 12.31
y = 42.94 (x + .73 ) 2 - 22.88 - 12.31
y = 42.94 (x + .73 ) 2 - 35.19
DETERMINE THE VERTEX OF A PARABOLA
Recall that in the above examples we changed the equation of a parabola (y = ax2 + bx + c) into the form y = a( x - h) 2 + k.In this latter form the vertex is identified as the point with coordinates (h,k). Here is why:
A parabola that opens upward or downward is either going to have a minimum value of y (if the parabola opens upward) or a maximum value of y (if the parabola opens downward).
If a > 0 then the parabola opens upward and there is a minimum value for y. This will occur when a( x - h) 2 = 0 for that is the smallest value that a (x - h) 2 can take on.
a( x - h) 2 = 0
( x - h) 2 = 0
( x - h) = 0
x = h, so the x-coordinate of the vertex is h.
When x = h, y = k so the coordinates of the vertex are (h,k)
If a < 0 then the parabola opens downward and there is a maximum value for y. This will occur when a( x - h) 2 = 0 for that is the largest value that a(x - h) 2 can take on (recall that a < 0).
a ( x - h) 2 = 0
( x - h) 2 = 0
( x - h) = 0
x = h, so the x-coordinate of the vertex is h.
When x = h, y = k so the coordinates of the vertex are (h,k)
Now let's go back to each of the previous four examples and identify the vertex.
FROM EXAMPLE 1: y = (x + 3)2 - 7
Vertex ( - 3, - 7)
FROM EXAMPLE 2: y = 2(x + 3)2 - 23
Vertex ( - 3, - 23)
FROM EXAMPLE 3: x = -3( t - .5 )2 - .25
Vertex (.5, - . 25) [assuming that the ordered pairs are in the form, (t,x)]
FROM EXAMPLE 4: y = 42.94 (x + .73 ) 2 - 35.19
Vertex: ( - .73, - 35.19)<|endoftext|>
| 5.03125 |
4,182 |
# 2.3 Converting Units
### Learning Objective
1. Convert from one unit to another unit of the same type.
In Section 2.2 “Expressing Numbers”, we showed some examples of how to replace initial units with other units of the same type to get a numerical value that is easier to comprehend. In this section, we will formalize the process.
Consider a simple example: how many feet are there in 4 yards? Most people will almost automatically answer that there are 12 feet in 4 yards. How did you make this determination? Well, if there are 3 feet in 1 yard and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards.
This is correct, of course, but it is informal. Let us formalize it in a way that can be applied more generally. We know that 1 yard (yd) equals 3 feet (ft):
1 yd = 3 ft
In math, this expression is called an equality. The rules of algebra say that you can change (i.e., multiply or divide or add or subtract) the equality (as long as you don’t divide by zero) and the new expression will still be an equality. For example, if we divide both sides by 2, we get
We see that one-half of a yard equals 3/2, or one and a half, feet—something we also know to be true, so the above equation is still an equality. Going back to the original equality, suppose we divide both sides of the equation by 1 yard (number and unit):
The expression is still an equality, by the rules of algebra. The left fraction equals 1. It has the same quantity in the numerator and the denominator, so it must equal 1. The quantities in the numerator and denominator cancel, both the number and the unit:
When everything cancels in a fraction, the fraction reduces to 1:
We have an expression, 3 ft1 yd, that equals 1. This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. The expression 3 ft1 yd is called a conversion factor, and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.)
To see how this happens, let us start with the original quantity:
4 yd
Now let us multiply this quantity by 1. When you multiply anything by 1, you don’t change the value of the quantity. Rather than multiplying by just 1, let us write 1 as 3 ft1 yd:
The 4 yd term can be thought of as 4 yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard:
That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:
Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.
How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter, which is
1 mm = 1/1,000 m
The 1/1,000 is what the prefix milli- means. Most people are more comfortable working without fractions, so we will rewrite this equation by bringing the 1,000 into the numerator of the other side of the equation:
1,000 mm = 1 m
Now we construct a conversion factor by dividing one quantity into both sides. But now a question arises: which quantity do we divide by? It turns out that we have two choices, and the two choices will give us different conversion factors, both of which equal 1:
Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Therefore, we will use the second conversion factor. Canceling units and performing the mathematics, we get
Note how m cancels, leaving mm, which is the unit of interest.
The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses.
### Example 2.6
1. Convert 35.9 kL to liters.
2. Convert 555 nm to meters.
Solution
1. We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1,000 L/1 kL. Applying this conversion factor, we get
2. We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: 1 m/109 nm. Applying this conversion factor, we get
In the final step, we expressed the answer in scientific notation.
Test Yourself
1. Convert 67.08 μL to liters.
2. Convert 56.8 m to kilometers.
1. 6.708 × 10−5 L
2. 5.68 × 10−2 km
What if we have a derived unit that is the product of more than one unit, such as m2? Suppose we want to convert square meters to square centimeters? The key is to remember that m2 means m × m, which means we have two meter units in our derived unit. That means we have to include two conversion factors, one for each unit. For example, to convert 17.6 m2 to square centimeters, we perform the conversion as follows:
### Example 2. 7
How many cubic centimeters are in 0.883 m3?
Solution
With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have
You should demonstrate to yourself that the three meter units do indeed cancel.
Test Yourself
How many cubic millimeters are present in 0.0923 m3?
9.23 × 107 mm3
Suppose the unit you want to convert is in the denominator of a derived unit; what then? Then, in the conversion factor, the unit you want to remove must be in the numerator. This will cancel with the original unit in the denominator and introduce a new unit in the denominator. The following example illustrates this situation.
### Example 2.8
Convert 88.4 m/min to meters/second.
Solution
We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1 min/60 s. Apply and perform the math:
Notice how the 88.4 automatically goes in the numerator. That’s because any number can be thought of as being in the numerator of a fraction divided by 1.
Test Yourself
Convert 0.203 m/min to meters/second.
0.00338 m/s or 3.38 × 10−3 m/s
Sometimes there will be a need to convert from one unit with one numerical prefix to another unit with a different numerical prefix. How do we handle those conversions? Well, you could memorize the conversion factors that interrelate all numerical prefixes. Or you can go the easier route: first convert the quantity to the base unit, the unit with no numerical prefix, using the definition of the original prefix. Then convert the quantity in the base unit to the desired unit using the definition of the second prefix. You can do the conversion in two separate steps or as one long algebraic step. For example, to convert 2.77 kg to milligrams:
Alternatively, it can be done in a single multistep process:
You get the same answer either way.
### Example 2.9
How many nanoseconds are in 368.09 μs?
Solution
You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-,
Test Yourself
How many milliliters are in 607.8 kL?
6.078 × 108 mL
When considering the significant figures of a final numerical answer in a conversion, there is one important case where a number does not impact the number of significant figures in a final answer—the so-called exact number. An exact number is a number from a defined relationship, not a measured one. For example, the prefix kilo- means 1,000—exactly 1,000, no more or no less. Thus, in constructing the conversion factor
neither the 1,000 nor the 1 enter into our consideration of significant figures. The numbers in the numerator and denominator are defined exactly by what the prefix kilo- means. Another way of thinking about it is that these numbers can be thought of as having an infinite number of significant figures, such as
The other numbers in the calculation will determine the number of significant figures in the final answer.
### Example 2.10
A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.
Solution
Area is defined as the product of the two dimensions, which we then have to convert to square meters and express our final answer to the correct number of significant figures, which in this case will be three.
The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.
Test Yourself
What is the volume of a block in cubic meters whose dimensions are 2.1 cm × 34.0 cm × 118 cm?
0.0084 m3
### Chemistry Is Everywhere: The Gimli Glider
On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed “the Gimli Glider.”
The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.
What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.
View this video on unit conversion by Dr. Jessie A. Key / Vancouver Island University for a lecture on unit conversion.
### Key Takeaways
• Units can be converted to other units using the proper conversion factors.
• Conversion factors are constructed from equalities that relate two different units.
• Conversions can be a single step or multistep.
• Unit conversion is a powerful mathematical technique in chemistry that must be mastered.
• Exact numbers do not affect the determination of significant figures.
### Exercises
1. Write the two conversion factors that exist between the two given units.
a) milliliters and liters
b) microseconds and seconds
c) kilometers and meters
2. Write the two conversion factors that exist between the two given units.
a) kilograms and grams
b) milliseconds and seconds
c) centimeters and meters
3. Perform the following conversions.
a) 5.4 km to meters
b) 0.665 m to millimeters
c) 0.665 m to kilometers
4. Perform the following conversions.
a) 90.6 mL to liters
b) 0.00066 ML to liters
c) 750 L to kiloliters
5. Perform the following conversions.
a) 17.8 μg to grams
b) 7.22 × 102 kg to grams
c) 0.00118 g to nanograms
6. Perform the following conversions.
a) 833 ns to seconds
b) 5.809 s to milliseconds
c) 2.77 × 106 s to megaseconds
7. Perform the following conversions.
a) 9.44 m2 to square centimeters
b) 3.44 × 108 mm3 to cubic meters
8. Perform the following conversions.
a) 0.00444 cm3 to cubic meters
b) 8.11 × 102 m2 to square nanometers
9. Why would it be inappropriate to convert square centimeters to cubic meters?
10. Why would it be inappropriate to convert from cubic meters to cubic seconds?
11. Perform the following conversions.
a) 45.0 m/min to meters/second
b) 0.000444 m/s to micrometers/second
c) 60.0 km/h to kilometers/second
12. Perform the following conversions.
a) 3.4 × 102 cm/s to centimeters/minute
b) 26.6 mm/s to millimeters/hour
c) 13.7 kg/L to kilograms/milliliters
13. Perform the following conversions.
a) 0.674 kL to milliliters
b) 2.81 × 1012 mm to kilometers
c) 94.5 kg to milligrams
14. Perform the following conversions.
a) 6.79 × 10−6 kg to micrograms
b) 1.22 mL to kiloliters
c) 9.508 × 10−9 ks to milliseconds
15. Perform the following conversions.
a) 6.77 × 1014 ms to kiloseconds
b) 34,550,000 cm to kilometers
16. Perform the following conversions.
a) 4.701 × 1015 mL to kiloliters
b) 8.022 × 10−11 ks to microseconds
17. Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.
a) 88 ft/s to miles/hour (Hint: use 5,280 ft = 1 mi.)
b) 0.00667 km/h to meters/second
18. Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.
a) 3.88 × 102 mm/s to kilometers/hour
b) 1.004 kg/L to grams/milliliter
19. What is the area in square millimeters of a rectangle whose sides are 2.44 cm × 6.077 cm? Express the answer to the proper number of significant figures.
20. What is the volume in cubic centimeters of a cube with sides of 0.774 m? Express the answer to the proper number of significant figures.
21. The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square centimeters if its base is 1.007 m and its height is 0.665 m? Express the answer to the proper number of significant figures.
22. The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square meters if its base is 166 mm and its height is 930.0 mm? Express the answer to the proper number of significant figures.
1.
a) 1,000 mL/1 L and 1 L/1,000 mL
b) 1,000,000 μs/1 s and 1 s/1,000,000 μs
c) 1,000 m/1 km and 1 km1,000 m
3.
a) 5,400 m
b) 665 mm
c) 6.65 × 10−4 km
5.
a) 1.78 × 10−5 g
b) 7.22 × 105 g
c) 1.18 × 106 ng
7.
a) 94,400 cm2
b) 0.344 m3
9.
One is a unit of area, and the other is a unit of volume.
11.
a) 0.75 m/s
b) 444 µm/s
c) 1.666 × 10−2 km/s
13.
a) 674,000 mL
b) 2.81 × 106 km
c) 9.45 × 107 mg
15.
a) 6.77 × 108 ks
b) 345.5 km
17.
a) 6.0 × 101 mi/h
b) 0.00185 m/s
19.
1.48 × 103 mm2
21.
3.35 × 103 cm2
Image Credits
Example 2.8
Figure 2.2<|endoftext|>
| 5 |
1,314 |
# Case Based Questions Test: Real Numbers - 2
## 10 Questions MCQ Test NCERT Mathematics for CAT Preparation | Case Based Questions Test: Real Numbers - 2
Description
Attempt Case Based Questions Test: Real Numbers - 2 | 10 questions in 20 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study NCERT Mathematics for CAT Preparation for Class 10 Exam | Download free PDF with solutions
QUESTION: 1
### Direction: Read the following text and answer the following questions on the basis of the same:A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.108 can be expressed as a product of its primes as
Solution: All even numbers except 2 are composite numbers, so 108 is a composite number.
Therefore, we can conclude that 108 can be expressed as a product of prime factors by 22 × 33.
QUESTION: 2
### Direction: Read the following text and answer the following questions on the basis of the same:A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.What is the minimum number of rooms required during the event?
Solution: Minimum no. of rooms required are total number of students divided by number of students in each room.
= 21
QUESTION: 3
### Direction: Read the following text and answer the following questions on the basis of the same:A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.The product of HCF and LCM of 60,84 and 108 is
Solution: 60 = (22 × 3 × 5)
84 = (22 × 3 × 7)
108 = (22 × 33)
HCF(60, 84,108) = (22 × 3) = 12
LCM =
LCM of 60, 84, and 108 is 3780.
Product of HCF x LCM = 3780 x 12 = 45360
QUESTION: 4
Direction: Read the following text and answer the following questions on the basis of the same:
A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.
In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are
Solution: No. of participants seated in each room would be HCF of all the three values above.
60 = 2 × 2 × 3 × 5
84 = 2 × 2 × 3 × 7
108 = 2 × 2 × 3 × 3 × 3
Hence, HCF = 12.
QUESTION: 5
Direction: Read the following text and answer the following questions on the basis of the same:
A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.
The LCM of 60, 84 and 108 is
Solution:
LCM = 2 x 2 x 3 x 3 x 3 x 5 x 7
LCM = 3780
Therefore, LCM of 60, 84, and 108 is 3780.
QUESTION: 6
Direction: Read the following text and answer the following questions on the basis of the same:
A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
The prime factorisation of 13915 is
Solution:
Using the prime factorization tree, we have
∴ 13915 = 5 × 11 × 11 × 23 = 5 × 112 × 23
QUESTION: 7
Direction: Read the following text and answer the following questions on the basis of the same:
A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
What will be the value of y?
Solution: 2783 = y × 253
y = 2783/253
y = 11
QUESTION: 8
Direction: Read the following text and answer the following questions on the basis of the same:
A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
According to Fundamental Theorem of Arithmetic 13915 is a
Solution: The prime factorization of 13,915 is 5 × 112 × 23.
Since it has a total of 4 prime factors, 13,915 is a composite number.
QUESTION: 9
Direction: Read the following text and answer the following questions on the basis of the same:
A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
What will be the value of x?
Solution: x = 2783 × 5
x = 13915
QUESTION: 10
Direction: Read the following text and answer the following questions on the basis of the same:
A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
What will be the value of z?
Solution: 253 = 11 × z
z = 253/11
z = 23
Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code<|endoftext|>
| 4.625 |
155 |
Pre-Algebra Practice Questions: Finding the Volume of Prisms and Cylinders
To find the volume of a prism or cylinder, you can use the following formula, where Ab is the area of the base and h is the height:
- Find the volume of a prism with a base that has an area of 6 square centimeters and a height of 3 centimeters.
- Figure out the approximate volume of a cylinder whose base has a radius of 7 millimeters and whose height is 16 millimeters.
Answers and explanations
- 18 cubic centimeters
- Approximately 2,461.76 cubic millimeters
First, use the area formula for a circle to find the area of the base:
Plug this result into the formula for the volume of a cylinder:<|endoftext|>
| 3.953125 |
1,042 |
Many students do not have adequate rest. We notice students struggling to concentrate and lagging in their responses to questions. We notice students yawning during lessons, both in school and during tuition. Some students even doze off, especially when the room is air-conditioned, for example during computer-assisted learning sessions.
Students need to learn to manage and prioritize their time. Hanging out at the mall after school, playing computer games, watching television (or online video), catching a movie, going out with parents or friends on weekends, and excessive CCA commitments not only take up precious time, but also distract and tire him/her out. Even a machine gets worn out after running non-stop for a period of time.
Here is one more reason to prioritize sleep.
A New Scientist article summarized a research paper showing an association between chronic sleep deprivation (lack of sleep for many nights, which is a source of stress for the brain) and increased phagocytosis (“eating”) by microglia (innate immune cells resident in the brain). The research paper suggested that a lack of sleep may increase the risk of neurological disorders later in life. The article is reproduced below.
The brain starts to eat itself after chronic sleep deprivation
Sleep loss in mice sends the brain’s immune cells into overdrive. This might be helpful in the short term, but could increase the risk of dementia in the long run.
Burning the midnight oil may well burn out your brain. The brain cells that destroy and digest worn-out cells and debris go into overdrive in mice that are chronically sleep-deprived.
In the short term, this might be beneficial – clearing potentially harmful debris and rebuilding worn circuitry might protect healthy brain connections. But it may cause harm in the long term, and could explain why a chronic lack of sleep puts people at risk of Alzheimer’s disease and other neurological disorders, says Michele Bellesi of the Marche Polytechnic University in Italy.
Bellesi reached this conclusion after studying the effects of sleep deprivation in mice. His team compared the brains of mice that had either been allowed to sleep for as long as they wanted or had been kept awake for a further eight hours. Another group of mice were kept awake for five days in a row – mimicking the effects of chronic sleep loss.
The team specifically looked at glial cells, which form the brain’s housekeeping system. Earlier research had found that a gene that regulates the activity of these cells is more active after a period of sleep deprivation.
One type of glial cell, called an astrocyte, prunes unnecessary synapses in the brain to remodel its wiring. Another type, called a microglial cell, prowls the brain for damaged cells and debris.
Bellisi’s team found that after an undisturbed sleep, astrocytes appeared to be active in around 6 per cent of the synapses in the brains of the well-rested mice. But astrocytes seemed to be more active in sleep-deprived mice – those that had lost eight hours of sleep showed astrocyte activity in around 8 per cent of their synapses, while the cells were active in 13.5 per cent of the synapses of the chronically sleep-deprived animals.
This suggests that sleep loss can trigger astrocytes to start breaking down more of the brain’s connections and their debris. “We show for the first time that portions of synapses are literally eaten by astrocytes because of sleep loss,” says Bellesi.
For all we know, this may be a good thing. Much of the remodeling was of the largest synapses, which are more mature and used more intensively. “They are like old pieces of furniture, and so probably need more attention and cleaning,” says Bellesi.
But the team also found that microglial cells were more active after chronic sleep deprivation.
This is a more worrying find, says Bellesi; excessive microglial activity has been linked to a range of brain disorders. “We already know that sustained microglial activation has been observed in Alzheimer’s and other forms of neurodegeneration,” he says.
The finding could explain why a lack of sleep seems to make people more vulnerable to developing such dementias, says Agnès Nadjar of the University of Bordeaux in France.
It’s not yet clear whether getting more sleep could protect the brain or rescue it from the effects of a few sleepless nights. The researchers plan to investigate how long the effects of sleep deprivation last.
Original research paper:
Journal of Neuroscience: Sleep Loss Promotes Astrocytic Phagocytosis and Microglial Activation in Mouse Cerebral Cortex. Michele Bellesi, Luisa de Vivo, Mattia Chini, Francesca Gilli, Giulio Tononi and Chiara Cirelli. 24 May 2017, 37 (21) 5263-5273<|endoftext|>
| 3.671875 |
383 |
Did you know, in the recorded history, the average temperature of the earth’s climate system has been rising at the fastest pace over the past 50 years? Owing to this rise in temperature, the average sea level has risen around seven inches in the course of the past century.
Here you will learn about what is global warming in Urdu and English.
How will you define global warming? Here is a comprehensive definition. Alternatively known as climate change, global warming can be defined as the observed century-scale increase in the average temperature of the global climate system and the effects caused by such a rise.
Global warming occurs due to the accumulation of air pollutants and greenhouse gases, like carbon dioxide, methane, chlorofluorocarbons, ozone, nitrous oxide and water vapors, in the earth’s atmosphere. Such elements absorb the solar radiations that have bounced off the surface of earth and prevent them from escaping into the space. In other words, the greenhouse gases and pollutants remain in the atmosphere for years to centuries and trap the heat radiations which causes the planet earth to get warmer and harsher. You can also read detail of global warming causes here.
The scientists observe a close link between global warming and extreme weather conditions. According to them, the rising global temperatures lead to hotter and longer heat waves, more frequent droughts, more powerful hurricanes and heavier rainfall.
As the temperature of oceans on earth is rising, the tropical storms are able to pick up more energy. It indicates that global warming has the potential to turn a category 3 storm into a more powerful and devastating category 4 storm.
Other effects of global warming are: early snowmelt, dramatic water shortages, increased flooding, outbreak of infectious diseases and extinction of many animal and plant species due to disruption in their habitats. Here you can also learn about what is global warming in Urdu.<|endoftext|>
| 3.921875 |
1,192 |
2018 AIME I Problems/Problem 7
Problem
A right hexagonal prism has height $2$. The bases are regular hexagons with side length $1$. Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).
Solution 1
We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.
Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \sqrt{3}$ with 6 cases or $2, 2, 2$ with 2 cases. This can be repeated on the other base for $16$ cases.
Case 2: The vertices span two bases. WLOG call the only vertex on one of the bases $X$. Call the closest vertex on the other base $B$, and label clockwise $C, D, E, F, G$. We will multiply the following scenarios by $12$, because the top vertex can have $6$ positions and the top vertex can be on the other base. We can have $XCG, XDF$, but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is $2$ and the height is $2$, so $XBE$ is also correct! Those are the only three cases, so there are $12*3=36$ cases for this case.
In total there's $\boxed{052}$ cases.
Solution 2
If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \cdot 2\cdot 6 = 28$ ways. Then, if one length was $\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\sqrt{3}$ side, and two ways to put the other point. $2 \cdot 12 = 24$ for four ways to put the third point. Adding the number up, the final answer is $24+28 = \boxed{052}.$
~kevinmaths~
Solution 3
To start, let's find the distances from any vertex (call it A, doesn't matter since the prism is symmetrical) to all the other 11 vertices. Using Pythagorean relations, we find that the distances are $1$, $\sqrt{3}$, $2$, $\sqrt{3}$, $1$, $\sqrt{7}$, $\sqrt{5}$, $2$, $\sqrt{5}$, $\sqrt{7}$, and $\sqrt{8}$.
We can clearly form an isosceles triangle using any vertex and any two neighboring edges that are equal. There are 12 total vertices, all of which are symmetrical to the vertex A. Thus, for each vertex, we can form 5 isosceles triangles (with neighboring edges $1$-$1$, $2$-$2$, $\sqrt{3}$-$\sqrt{3}$, $\sqrt{5}$-$\sqrt{5}$, and $\sqrt{7}$-$\sqrt{7}$) , so have so far $12 \times 5 = 60$ isosceles triangles.
To check for overcounting, note that isosceles triangles can be split into two major categories: equilateral isosceles, and non-equilateral isosceles. We only counted non-equilateral isosceles triangles once since there is only one vertex whose two neighboring edges are equal. But equilateral triangles were counted three times (namely once for each vertex).
By observation, there are only 4 equilateral triangles (1-1-1 side lengths), two on each hexagonal face. Since we counted each two more times than we should of (three instead of once), we will subtract ($4$ eq. triangles) $\times$ ($2$ overcounts per eq. triangle) = $8$ overcounts.
Finally, we have $60 - 8$ overcounts = $\boxed{052}$ isoceles triangles.
Solution 4
Start by drawing the hexagonal prism. Start from one of the points, let's call it $A$. From simple inspection, we can see that each point we choose creates $4$ isosceles triangles; $A$ and the two adjacent points, $A$ and the two points adjacent to those, $A$ and the adjacent point on the other face, and $A$ and the two points adjacent to those. Here, we have $12\cdot4\Rightarrow48$ isosceles triangles.
Next, notice that you can create an isosceles triangle with the bases of the hexagonal prism. Let's say that our base, starting from the very left-most point of the hexagon, has points $A, B, C, D, E, F$, rotating counter-clockwise. We can create a triangle with either points $\bigtriangleup{BCE}$ or $\bigtriangleup{BCF}$. WLOG, assuming we use $\bigtriangleup{BCE}$, we can use the Law of Cosines and find that $\overline{CE}$ = $\sqrt{3}$. Thus, this would mean $\overline{CF}$ is $2$, which is the exactly the same length as the height of the hexagonal prism. There are $4$ possibilities of this case, so our final solution is $48+4=\boxed{52}$.
Solution by IronicNinja~<|endoftext|>
| 4.78125 |
506 |
Fractions, Decimals, and Percents
Remember that when you are dealing with decimals, the first place to the right of the decimal is the tenths place, the second place is the hundredths place, and the third place is the thousandths place, and so on. Most statistics are reported to at least the hundredths place. If there is no whole number before the decimal, it is customary to report a zero before the decimal. This calls attention to the fact that you are dealing with a number smaller than one.
Many formulas in statistics contain fractions. To understand what these formulas are doing, it is very helpful to recall the basics of fractions. Fractions can be expressed two primary ways: With a horizontal line dividing the numerator (the top of a fraction) and denominator (the bottom of a fraction), or a slash dividing the numerator and denominator.
We can also divide the numerator by the denominator and get a proportion, which is a fraction expressed in decimal form. We can easily convert that proportion to a percentage by multiplying it by 100 (which moves the decimal two places to the right).
```Examples:
2/3 = = 0.6667 = 66.67%
1/2 = = 0.5000 = 50.00%```
Remember the following characteristics of fractions:
• As the numerator increases, the value of the faction increases.
• As the denominator increases, the value of the fraction decreases.
• With complex fractions, perform all math operations in both the numerator and the denominator before dividing.
A percentage is a special fraction where the denominator is always 100, regardless of how many subjects we are describing. If we say 50% of our subjects have a particular characteristic, we are saying that 50 out of 100 have it. It may be helpful to recall that cent comes from Latin and means 100. By using a common denominator, percentages are very useful in allowing us to compare groups of different sizes. It also aids us in understanding the characteristics of large groups that we can’t intuitively grasp.
`To convert a fraction to a percent, divide the numerator by the denominator and multiply by 100.`
[ Back | Contents | Next ]
`Last Modified: 06/28/2018`
This site uses Akismet to reduce spam. Learn how your comment data is processed.
Doc's Things and Stuff uses Accessibility Checker to monitor our website's accessibility.<|endoftext|>
| 4.6875 |
176 |
Mountain top boiling
If you were on a mountain top you would find the water in a kettle would start boiling at less than 100 °C.
Why? Because the air pressure is much lower the higher up you are – and the lower the pressure the lower the temperature at which liquids boil.
You can change the temperature at which any liquid boils by changing the pressure. Boiling at a particular pressure is known as Vaporization or Evaporation.
For example, in your fridge at home a liquid (known as a refrigerant) is circulated through pipework at the back. The space inside the fridge is cooled when this liquid refrigerant evaporates at low pressure drawing heat from the space inside.
The evaporated gas is re-pressurised by a motor and turns back to a liquid in a continuous circuit.
This is how your fridge works.<|endoftext|>
| 4.0625 |
199 |
What do zebras and ostriches have in common with clownfish and sea anemones? They're both examples of interdependent relationships! Students will learn about helpful partnerships in the animal kingdom as they multiply their way through fractions. By integrating math and literacy skills, this math reader makes multiplying fractions simple, relevant, and fun, and the real-world examples of problem solving allow students to explore the concepts in meaningful ways. With intriguing full-color images, the book includes text features such as a glossary, index, captions, and a table of contents to increase understanding and build academic vocabulary. The Let's Explore Math sidebars, the extensive Problem Solving section, and the clear mathematical charts and diagrams provide numerous opportunities for students to practice what they have learned. The DOK-leveled Math Talk section includes questions that facilitate mathematical discourse with activities that students can respond to at school or home. Explore the world of strange animal partnerships with this engaging grade 4 math reader!<|endoftext|>
| 3.734375 |
411 |
This week in precalc 11, we finished our whole unit on rational expressions and equations. One of my favorite things we learned was solving rational equations.
There are many different ways you can solve a rational equation such as moving the terms to one side, cross multiplying, multiply each fraction by a common denominator, etc. Whatever question you are presented with, you must choose the best option so you can solve your question efficiently.
Here is one way to solve an equation with cross multiplying : $\frac{5}{x+1} = \frac{2}{x+2}$
Since it is called cross multiplying you take whatever is on the top left term and the bottom right term then combine together (only allowed to cross multiply when an = sign is between two terms), then proceed to do the same with the rest which will give us this $5(x+2) = 2(x+1)$
Now we finish this question off with basic algebra, you distribute then solve for x !
$5x +10 = 2x +2$
$3x = -8$
$x = -\frac{8}{3}$
Another way to solve is by multiplying each fraction by a common denominator $\frac{5}{x} + 2 = \frac{3}{x}$
For this equation, we know that the common denominator is x so we can multiply everything by x
Since there is an x on the top and the bottom for 2 of the 3 terms, you can cancel them out because one number divided by the same number = 1
It will leave us with $5 + 2x = 3$, we can now solve algebraically as usual
$2x = -2$
$x = -1$
As for the more complicated questions with quadratics, we must know how to factor then find the common denominators. Solving is fairly simple as long as you do not mess up the first few steps, after then its all algebra that we have been doing since middle school.<|endoftext|>
| 4.59375 |
2,530 |
# Search by Topic
#### Resources tagged with Creating expressions/formulae similar to Triangle Inequality:
Filter by: Content type:
Stage:
Challenge level:
### There are 85 results
Broad Topics > Algebra > Creating expressions/formulae
### Christmas Chocolates
##### Stage: 3 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Marbles in a Box
##### Stage: 3 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Seven Squares
##### Stage: 3 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Lower Bound
##### Stage: 3 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Partly Painted Cube
##### Stage: 4 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### Pinned Squares
##### Stage: 3 Challenge Level:
The diagram shows a 5 by 5 geoboard with 25 pins set out in a square array. Squares are made by stretching rubber bands round specific pins. What is the total number of squares that can be made on a. . . .
### Sum Equals Product
##### Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Always a Multiple?
##### Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Your Number Is...
##### Stage: 3 Challenge Level:
Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
### How Big?
##### Stage: 3 Challenge Level:
If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Magic Sums and Products
##### Stage: 3 and 4
How to build your own magic squares.
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Boxed In
##### Stage: 3 Challenge Level:
A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box?
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
### Quick Times
##### Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Good Work If You Can Get It
##### Stage: 3 Challenge Level:
A job needs three men but in fact six people do it. When it is finished they are all paid the same. How much was paid in total, and much does each man get if the money is shared as Fred suggests?
### Fibs
##### Stage: 3 Challenge Level:
The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?
### Days and Dates
##### Stage: 3 Challenge Level:
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Generating Triples
##### Stage: 4 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### ' Tis Whole
##### Stage: 4 and 5 Challenge Level:
Take a few whole numbers away from a triangle number. If you know the mean of the remaining numbers can you find the triangle number and which numbers were removed?
### Your Number Was...
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Hike and Hitch
##### Stage: 4 Challenge Level:
Fifteen students had to travel 60 miles. They could use a car, which could only carry 5 students. As the car left with the first 5 (at 40 miles per hour), the remaining 10 commenced hiking along the. . . .
### Pareq Calc
##### Stage: 4 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.<|endoftext|>
| 4.53125 |
1,829 |
Objectives
1. Learn two main criteria for a matrix to be diagonalizable.
2. Develop a library of examples of matrices that are and are not diagonalizable.
3. Recipes: diagonalize a matrix, quickly compute powers of a matrix by diagonalization.
4. Pictures: the geometry of diagonal matrices, why a shear is not diagonalizable.
5. Theorem: the diagonalization theorem (two variants).
6. Vocabulary words: diagonalizable, algebraic multiplicity, geometric multiplicity.
Diagonal matrices are the easiest kind of matrices to understand: they just scale the coordinate directions by their diagonal entries. This section is devoted to the question: “When is a matrix similar to a diagonal matrix?” We will see that the algebra and geometry of such a matrix is relatively easy to understand.
Subsection5.4.1Diagonalizability
First we make precise what we mean when we say two matrices are “similar”.
Definition
Two matrices and are similar if there exists an invertible matrix such that
If two matrices are similar, then their powers are similar as well.
Proof
First note that
Next we have
The pattern is clear.
In this chapter, we will determine when a matrix is similar to a diagonal matrix. This property is important enough to deserve its own name.
Definition
An matrix is diagonalizable if it is similar to a diagonal matrix: that is, if there exists an invertible matrix and a diagonal matrix such that
Example
Any diagonal matrix is is diagonalizable because it is similar to itself. For instance,
Example
If a matrix is diagonalizable, and if is similar to then is diagonalizable as well. Indeed, if for diagonal, and then
so is similar to
Powers of diagonalizable matrices
Multiplying diagonal matrices together just multiplies their diagonal entries:
Therefore, it is easy to take powers of a diagonal matrix:
By this fact, if then so it is also easy to take powers of diagonalizable matrices. This is often very important in applications.
Recipe: Compute powers of a diagonalizable matrix
If where is a diagonal matrix, then
A fundamental question about a matrix is whether or not it is diagonalizable. The following is the primary criterion for diagonalizability. It shows that diagonalizability is an eigenvalue problem.
Proof
By this fact in Section 5.1, if an matrix has distinct eigenvalues then a choice of corresponding eigenvectors is automatically linearly independent.
An matrix with distinct eigenvalues is diagonalizable.
Non-Uniqueness of Diagonalization
We saw in the above example that changing the order of the eigenvalues and eigenvectors produces a different diagonalization of the same matrix. There are generally many different ways to diagonalize a matrix, corresponding to different orderings of the eigenvalues of that matrix. The important thing is that the eigenvalues and eigenvectors have to be listed in the same order.
There are other ways of finding different diagonalizations of the same matrix. For instance, you can scale one of the eigenvectors by a constant
you can find a different basis entirely for an eigenspace of dimension at least etc.
In the above example, the (non-invertible) matrix is similar to the diagonal matrix Since is not invertible, zero is an eigenvalue by the invertible matrix theorem, so one of the diagonal entries of is necessarily zero. Also see this example below.
Here is the procedure we used in the above examples.
Recipe: Diagonalization
Let be an matrix. To diagonalize
1. Find the eigenvalues of using the characteristic polynomial.
2. For each eigenvalue of compute a basis for the -eigenspace.
3. If there are fewer than total vectors in all of the eigenspace bases then the matrix is not diagonalizable.
4. Otherwise, the vectors in the eigenspace bases are linearly independent, and for
where is the eigenvalue for
We will justify the linear independence assertion in part 4 in the proof of this theorem below.
The following point is often a source of confusion.
Diagonalizability has nothing to do with invertibility
Of the following matrices, the first is diagonalizable and invertible, the second is diagonalizable but not invertible, the third is invertible but not diagonalizable, and the fourth is neither invertible nor diagonalizable, as the reader can verify:
Subsection5.4.2The Geometry of Diagonalizable Matrices¶ permalink
A diagonal matrix is easy to understand geometrically, as it just scales the coordinate axes:
A daigonalizable matrix is not much harder to understand geometrically. Indeed, if are linearly independent eigenvectors of an matrix then scales the -direction by the eigenvalue in other words, Since the vectors form a basis of this determines the action of on any vector in
Example
Consider the matrices
One can verify that see this example. Let and the columns of These are eigenvectors of with corresponding eigenvalues and
The matrix is diagonal: it scales the -direction by a factor of and the -direction by a factor of
If we write a vector in terms of the basis say, then it is easy to compute
Here we have used the fact that are eigenvectors of Since the resulting vector is still expressed in terms of the basis we can visualize what does to the vector it scales the -coordinate” by and the -coordinate” by
For instance, let We see from the grid on the right in the picture below that so
The picture illustrates the action of on the plane in the usual basis, and the action of on the plane in the -basis.
Now let We see from the grid on the right in the picture below that so
This is illustrated in the picture below.
In the following examples, we visualize the action of a diagonalizable matrix in terms of its dynamics. In other words, we start with a collection of vectors (drawn as points), and we see where they move when we multiply them by repeatedly.
Subsection5.4.3Algebraic and Geometric Multiplicity
In this subsection, we give a variant of the diagonalization theorem that provides another criterion for diagonalizability. It is stated in the language of multiplicities of eigenvalues.
In algebra, we define the multiplicity of a root of a polynomial to be the number of factors of that divide For instance, in the polynomial
the root has multiplicity and the root has multiplicity
Definition
Let be an matrix, and let be an eigenvalue of
1. The algebraic multiplicity of is its multiplicity as a root of the characteristic polynomial of
2. The geometric multiplicity of is the dimension of the -eigenspace.
Since the -eigenspace of is its dimension is the number of free variables in the system of equations i.e., the number of columns without pivots in the matrix
We saw in the above examples that the algebraic and geometric multiplicities need not coincide. However, they do satisfy the following fundamental inequality, the proof of which is beyond the scope of this text.
In particular, if the algebraic multiplicity of is equal to then so is the geometric multiplicity.
If has an eigenvalue with algebraic multiplicity then the -eigenspace is a line.
We can use the theorem to give another criterion for diagonalizability (in addition to the diagonalization theorem).
Proof
The first part of the third statement simply says that the characteristic polynomial of factors completely into linear polynomials over the real numbers: in other words, there are no complex (non-real) roots. The second part of the third statement says in particular that for any diagonalizable matrix, the algebraic and geometric multiplicities coincide.
Let be a square matrix and let be an eigenvalue of If the algebraic multiplicity of does not equal the geometric multiplicity, then is not diagonalizable.
The examples at the beginning of this subsection illustrate the theorem. Here we give some general consequences for diagonalizability of and matrices.
Diagonalizability of 2 × 2 Matrices
Let be a matrix. There are four cases:
1. has two different eigenvalues. In this case, each eigenvalue has algebraic and geometric multiplicity equal to one. This implies is diagonalizable. For example:
2. has one eigenvalue of algebraic and geometric multiplicity To say that the geometric multiplicity is means that i.e., that every vector in is in the null space of This implies that is the zero matrix, so that is the diagonal matrix In particular, is diagonalizable. For example:
3. has one eigenvalue of algebraic multiplicity and geometric multiplicity In this case, is not diagonalizable, by part 3 of the theorem. For example, a shear:
4. has no eigenvalues. This happens when the characteristic polynomial has no real roots. In particular, is not diagonalizable. For example, a rotation:<|endoftext|>
| 4.53125 |
5,431 |
Allergies, including allergic rhinitis, affect an estimated 40 to 50 million people in the United States. Some allergies may interfere with day-to-day activities or lessen the quality of life.
Your allergist-immunologist, with his or her specialized training and expertise in managing allergies and asthma, can develop a treatment plan for your individual condition. The goal will be to enable you to lead a life that is as normal and symptom-free as possible.
What is rhinitis?
Rhinitis is a term describing the symptoms produced by nasal irritation or inflammation.
Symptoms of rhinitis include runny nose, itching, sneezing and stuffy nose due to blockage or congestion. These symptoms are the nose's natural response to inflammation and irritation.
Arbitrarily, rhinitis lasting less than six weeks is called acute rhinitis, and persistent symptoms are called chronic rhinitis. Acute rhinitis is usually caused by infections or chemical irritation. Chronic rhinitis may be caused by allergy or a variety of other factors.
The nose normally produces mucus, which traps substances like dust, pollen, pollution, and germs such as bacteria and viruses. Mucus flows from the front of the nose and drains down the back of the throat.
When mucus production is excessive, it can flow from the front, as a runny nose, or become noticeable from the back, as post-nasal drip. Nasal mucus, normally a thin, clear liquid, can become thick or colored, perhaps due to dryness, infection or pollution.
When post-nasal drip is excessive, thick, or contains irritating substances, cough is the natural response for clearing the throat.
Itching and sneezing are also natural responses to irritation caused by allergic reactions, chemical exposures including cigarette smoke, or temperature changes, infections and other factors.
The nasal tissues congest and decongest periodically. In most people, nasal congestion switches back and forth from side to side of the nose in a cycle several hours long.
Some people, especially those with narrow nasal passages, notice this nasal cycle more than others. Strenuous exercise or changes in head position can affect nasal congestion. Severe congestion can result in facial pressure and pain, as well as dark circles under the eyes.
What is sinusitis?
Sinusitis is inflammation or infection of any of the four groups of sinus cavities in the skull, which open into the nasal passages. Sinusitis is not the same as rhinitis, although the two may be associated and their symptoms may be similar.
The terms "sinus trouble" or "sinus congestion" are sometimes wrongly used to mean congestion of the nasal passage itself. Most cases of nasal congestion, though, are not associated with sinusitis.
What is allergic rhinitis?
Known to most people as hay fever, allergic rhinitis is a very common medical problem affecting more than 15 percent of the population, both adults and children.
Allergic rhinitis takes two different forms seasonal and perennial. Symptoms of seasonal allergic rhinitis occur in spring, summer and/or early fall and are usually caused by allergic sensitivity to pollens from trees, grasses or weeds, or to airborne mold spores.
Other people experience symptoms year-round, a condition called "perennial allergic rhinitis." It's generally caused by sensitivity to house dust, house dust mites, animal dander and/or mold spores. Underlying or hidden food allergies are considered a possible cause of perennial nasal symptoms.
Some people may experience both types of rhinitis, with perennial symptoms worsening during specific pollen seasons. As will be discussed later, there are also other causes for rhinitis.
What causes the sneezing, itchy eyes and other symptoms?
When a sensitive person inhales an allergen (allergy-causing substance) like ragweed pollen, the body's immune system reacts abnormally with the allergen. The allergen binds to allergic antibodies (immunoglobulin E) that are attached to cells that produce histamine and other chemicals.
The pollen "triggers" these cells in the nasal membranes, causing them to release histamine and the other chemicals. Histamine dilates the small blood vessels of the nose and fluids leak out into the surrounding tissues, causing runny noses, watery eyes, itching, swelling and other allergy symptoms.
Antibodies circulate in the blood stream, but localize in the tissues of the nose and in the skin. This makes it possible to show the presence of these antibodies by skin testing, or less commonly, by a special blood test. A positive skin test mirrors the type of reaction going on in the nose.
No hay, no fever so why "Hay Fever?"
"Hay fever" is a turn-of-the-century term which has come to describe the symptoms of allergic rhinitis, especially when it occurs in the late summer. However, the symptoms are not caused by hay (ragweed is one of the main culprits) and are not accompanied by fever. So physicians prefer the term "allergic rhinitis" because it's more accurate.
Similarly, springtime symptoms are sometimes called "rose fever," but it's just coincidental that roses are in full-bloom during the grass-pollinating season. Roses and other sweet-smelling, showy flowers rely on bees, not the wind, for pollination, so not much of their pollen gets into the air to cause allergies.
Is there any escape?
A common question from allergic rhinitis sufferers is: Can I move someplace where my allergies will go away?
Some allergens are tough to escape. Ragweed which affects 75 percent of allergic rhinitis sufferers blankets most of the United States. Less ragweed is found in a band along the West Coast, the southern-most tip of Florida and northern Maine, but it is still present. Even Alaska and Hawaii have a little ragweed.
Allergist-immunologists seldom recommend moving to another locale as a cure for allergies. A move may be of questionable value because a person may escape one allergy to ragweed, for example only to develop sensitivity to grasses or other allergens in the new location.
Since moving can have a disrupting effect on a family financially and emotionally, relocation should be undertaken only after consultation with an allergist-immunologist.
Is allergic rhinitis ever the cause of other problems?
Some known complications include ear infections, sinusitis, recurrent sore throats, cough, headache, fatigue, irritability, altered sleep patterns and poor school performance.
Occasionally, children may develop altered facial growth and orthodontic problems. Allergy treatment can eliminate or alleviate most of these problems.
Are all cases of rhinitis caused by allergies?
Rhinitis may result from many causes other than allergic reaction. Not all rhinitis symptoms are the result of allergies. Listed below are the three most common Rhinitis types and their characteristics.
Allergic. Common name: Hay fever. Allergic sensitivity: Yes. Causes may include dust, foods, animals, pollens and molds. Duration of symptoms are perennial and/or seasonal.
Infectious. Common name: Colds or flu. Allergic sensitivity: No. Causes may include viruses and bacteria. Duration of symptoms are 3-7 days.
Non-allergic. Common name: Irritant. Allergic sensitivity: No. Causes may include Smoke, air pollution, exhaust fumes, aerosol sprays, fragrance, paint fumes, etc. Duration of symptoms are Perennial and/or following exposure.
The most common condition causing rhinitis is the common cold, an example of infectious rhinitis. Most infections are relatively short-lived, lasting from three to seven days. Colds can be caused by any one of more than 200 viruses.
Children, particularly young children in school or day care centers, may have from eight to 12 colds each year. Fortunately, the frequency of colds lessens after immunity has been produced from exposure to many viruses.
Colds usually begin with a sensation of congestion, rapidly followed by runny nose and sneezing. Over the next few days, congestion becomes more prominent, the nasal mucus may become colored, and there may be a slight fever and cough.
Cold symptoms resolve within a couple of weeks, although a cough may sometimes persist. Cold symptoms that last longer may be due to other causes, such as chronic rhinitis or sinusitis.
What are other causes of rhinitis?
Not all symptoms in the nasal passage are caused by allergy or infection. Similar symptoms can be caused by mechanical blockage, use of certain medications, irritants, temperature changes or other physical factors. Rhinitis can also be a feature of other diseases and medical conditions.
Drug-induced nasal congestion can be caused by birth control pills and other female hormone preparations, certain blood pressure medications and prolonged use of decongestant nasal sprays.
Decongestant nasal sprays work quickly and effectively, but they alter how the nasal passages normally work. After a few weeks of use, nasal tissues swell after the medication wears off. The only thing that seems to relieve the obstruction is more of the medicine, and the medication's effect lasts shorter lengths of time. Permanent damage to the nasal tissues may result. Consultation with a physician to "get off" the medication is often necessary.
Cocaine also alters how the nasal passages normally work, causing a condition identical to, or even more severe than that produced by decongestant nasal sprays. If you use cocaine, it is important to tell your physician so that appropriate therapy can be prescribed.
Irritant rhinitis, or "vasomotor rhinitis" describes a group of poorly understood causes of rhinitis, with symptoms not caused by infection or allergy. Many people have recurrent or chronic nasal congestion, excess mucus production, itching, and other nasal symptoms similar to those of allergic rhinitis, but the disorder is not caused by allergy.
What triggers vasomotor rhinitis?
Irritants that can trigger vasomotor rhinitis include cigarette smoke, strong odors and fumes including perfume, hair spray, other cosmetics, laundry detergents, cleaning solutions, pool chlorine, car exhaust and other air pollution. Other irritants are spices used in cooking, alcoholic beverages (particularly beer and wine), aspirin, and certain blood pressure medications. Some people are very sensitive to abrupt changes in weather or temperature.
Skiers often develop a runny nose, but in some people any cold exposure may cause a runny nose. Others start sneezing when leaving a cold, air conditioned room. These agents are not allergens, do not induce formation of allergic antibodies and do not produce positive skin test reactions. Occasionally, one or two positive skin tests may be observed, but they do not match with the history and are not relevant or significant.
The cause of vasomotor rhinitis is not well understood. In a sufficiently high concentration, many odors will cause nasal irritation in almost anyone. Some people are unusually sensitive to irritation and will have significant nasal symptoms even when exposed to low concentrations of irritants. Thus, vasomotor rhinitis seems to be an exaggeration of the normal nasal response to irritation, occurring at levels of exposure which don't bother most people. It occurs more often in smokers and older individuals.
As is the case with allergic rhinitis, vasomotor rhinitis often can't be cured. Fortunately, symptoms can be kept under control by avoiding or reducing exposure to substances that cause symptoms and by taking medication when needed. Patients with vasomotor rhinitis should not smoke or permit smoking in their homes.
Dryness of the nasal tissues can be a normal effect of aging, or a characteristic of a nasal condition associated with a foul smelling nasal discharge. Rhinitis can also be a feature of endocrine disease, like hypothyroidism, or can occur during pregnancy. Rhinitis can be made worse or even improved during pregnancy. Alcoholic beverages can cause the blood vessels in the nose to enlarge temporarily and produce significant nasal congestion.
How do you know what kind of rhinitis you have?
Consult your physician. Sometimes several conditions can coexist in the same person. In a single individual, allergic rhinitis could be complicated by vasomotor rhinitis, septal deviation (curvature of the bone separating the two sides of the nose) or nasal polyps. Use of spray decongestants for chronic sinusitis, septal deviation or vasomotor rhinitis may cause rhinitis medicamentosa.
Any of these conditions will be made worse by catching a cold. Nasal symptoms caused by more than one problem can be difficult to treat, often requiring the cooperation of an allergist-immunologist and an otolaryngologist (ear, nose and throat specialist).
How is allergic rhinitis diagnosed?
Your allergist-immunologist may begin by taking a detailed history, looking for clues in your lifestyle that will help pinpoint the cause of your symptoms. You'll be asked about your work and home environments, your eating habits, your family's medical history, the frequency and severity of your symptoms, and miscellaneous matters, such as if you have pets.
Then, you may require some tests. Your allergist-immunologist may employ skin testing, in which small amounts of suspected allergen are introduced into the skin. Skin-testing is the easiest, most sensitive and generally least expensive way of making the diagnosis. Another advantage is that results are available immediately. In rare cases, it also may be necessary to do a special blood test for allergens, using the RAST or other methods.
How is rhinitis treated?
When no specific cure is available, options are ignoring your symptoms, avoiding or decreasing exposure to irritants or allergens to the extent practical, and taking medications for symptom relief.
Once allergic rhinitis is diagnosed, treatment options include avoidance, medication and immunotherapy (allergy shots).
Avoidance - A single ragweed plant may release one million pollen grains in just one day. The pollen from ragweed, grasses and trees is so small and buoyant that the wind may carry it miles from its source. Mold spores, which grow outdoors in fields and on dead leaves, also are everywhere and may outnumber pollen grains in the air even when the pollen season is at its worst.
While it's difficult to escape pollen and molds, here are some ways to lessen exposure.
- Keep windows closed and use air-conditioning in the summer, if possible. A HEPA (High Energy Particulate Air) filter or an electrostatic precipitator may help clean pollen and mold from the indoor air. Automobile air conditioners help, too.
- Don't hang clothing outdoors to dry. Pollen may cling to towels and sheets.
- The outdoor air is most heavily saturated with pollen and mold between 5 and 10 a.m., so early morning is a good time to limit outdoor activities.
- Wear a dust mask when mowing the lawn, raking leaves or gardening, and take appropriate medication beforehand.
Medication - When avoidance measures don't control symptoms, medication may be the answer. Antihistamines and decongestants are the most commonly used medications for allergic rhinitis. Newer medications, such as cromolyn, inhibit the release of chemicals that cause allergic reactions.
Nasal corticosteroid sprays reduce the inflammation from the allergic trigger. Medications help to alleviate nasal congestion, runny nose, sneezing and itching. They are available in many forms, including tablets, nasal sprays, eye drops and liquids. Some medications may cause side effects, so its best to consult your allergist-immunologist if there's a problem.
Immunotherapy - Allergen immunotherapy, known as "allergy shots," may be recommended for persons who don't respond well to treatment with medications, experience side-effects from medications or have allergen exposure which is unavoidable. Immunotherapy does not cure allergies but can be very effective in controlling allergic symptoms. Allergy injections are usually given at variable intervals over a period of three to five years.
An immunotherapy treatment program consists of injections of a diluted allergy extract, administered frequently in increasing doses until a maintenance dose is reached. Then, the injection schedule is changed so that the same dose is given with longer intervals between injections.
Immunotherapy helps the body build resistance to the effects of the allergen, reduces the intensity of symptoms caused by allergen exposure, and sometimes can actually make skin test reactions disappear. As resistance develops, symptoms should improve, but the improvement from immunotherapy will take several months to occur. Immunotherapy does not help the symptoms produced by non-allergic rhinitis.
There are many ways of treating allergies, and each person's treatment must be individualized based on the frequency, severity and duration of symptoms and on the degree of allergic sensitivity. If you have more questions, your allergist-immunologist will be happy to answer them.
Antihistamines are the most inexpensive and commonly used treatment for rhinitis. These medications counter the effects of histamine, the irritating chemical released within your body when an allergic reaction takes place. Although other chemicals are involved, histamine is primarily responsible for causing the symptoms.
Antihistamines do not cure, but help relieve: nasal allergy symptoms, such as sneezing, itching and discharge; eye symptoms, such as itching, burning, tearing, and clear discharge; skin conditions, such as hives, eczema, itching and some rashes; and other allergic conditions as determined by your physician.
There are dozens of different antihistamines and wide variations in how patients respond to them. Some are available over-the-counter and others require a prescription.
Generally, they work well and produce only minor side effects. The body tends to build up resistance to some antihistamines over time. This tendency varies from individual to individual. If you find that an antihistamine loses its "strength," notify your physician who may then recommend an antihistamine of a different class or strength.
Persons with nasal dryness or thick nasal mucus should avoid taking antihistamines without consulting a physician. Contact your physician for advice if an antihistamine causes drowsiness or other side effects.
Proper use. Short-acting antihistamines can be taken every four to six hours, while timed-release antihistamines are taken every 24 hours. The short-acting antihistamines are often most helpful taken 30 minutes before anticipated allergic exposure (picnic during ragweed season). Timed-release antihistamines are better suited to chronic (long-term) use for those who need daily medications.
Proper use of these drugs is just as important as their selection. The most effective way to use them is before symptoms develop. A dose taken early can eliminate the need for many later to reduce established symptoms. Many times a patient will say that he "took one, and it didn't work." If he or she had taken the antihistamine regularly for three to four days, and built up blood levels, it might have been effective.
Side effects. The most common side effect is sedation or drowsiness. For this reason, it is important that you do not drive a car or work with dangerous machinery the first time you take an antihistamine. You should take the antihistamine for the first time at home, several hours before bedtime.
When you are sure that the medicine will not cause sedation, you then can take it any time as prescribed during the day. In persons who experience drowsiness, the sedation effect usually lessens over time. Some of the newer antihistamines have no drowsiness side effects.
Another frequently encountered side effect is excessive dryness of the mouth, nose, and eyes. Less common side effects include restlessness, nervousness, over excitability, insomnia, dizziness, headaches, euphoria, fainting, visual disturbances, decreased appetite, nausea, vomiting, abdominal distress, constipation, diarrhea, increased or decreased urination, high or low blood pressure, nightmares (especially in children), sore throat, unusual bleeding or bruising, chest tightness or palpitations. Consult your allergist-immunologist should these reactions occur.
Alcohol and tranquilizers increase the sedation side effects of antihistamines.
- Never take anyone else's medication.
- Do not use more than one antihistamine at a time, unless prescribed.
- Keep these medications out of the reach of children.
- Know the effect of the medication on you before working with heavy machinery or driving.
- Follow your physician's instructions.
- There have not been enough studies to determine absolute safety of antihistamines in pregnancy. Again, consult your allergist-immunologist or obstetrician if antihistamines must be taken.
While antihistamines have been taken safely by millions of people in the last 50 years, don't take antihistamines before telling your allergist-immunologist if you are allergic to or intolerant of any medicine; are pregnant or intend to become pregnant while using this medication; are breast feeding; have glaucoma or enlarged prostate; have any medical illness.
What other medications are effective in treating rhinitis?
Decongestants help relieve the stuffiness and pressure caused by allergic, swollen nasal tissue. They do not contain antihistamines, so do not cause antihistamine side effects. They do not relieve the other symptoms of allergic rhinitis, such as runny nose, post-nasal drip and sneezing.
Decongestants are available as prescription and non-prescription medications and are often seen in combination with antihistamines or other medications. It is not uncommon for patients using decongestants to experience insomnia if taking the medication in the afternoon or evening. If this occurs, a dose reduction may be needed.
At times, men with prostate enlargement may encounter urinary problems while on decongestants. Patients using medications for the management of emotional or behavioral problems should discuss this with their physicians before using decongestants. Pregnant patients should also check with their physician before starting decongestants.
Non-prescription decongestant nasal sprays work within minutes and last for hours, but should not be used for more than a few days at a time without a physician's order.
Oral decongestants are found in many over-the-counter and prescription medications, and may be the treatment of choice for nasal congestion. They don't cause rhinitis medicamentosa, but need to be avoided by some patients with high blood pressure. If you have high blood pressure, you should check with your physician before using them.
Non-prescription saline nasal sprays will help counteract symptoms of dry nasal passages or thick nasal mucus. Unlike decongestant nose sprays, a saline nose spray can be used as often as needed. Sometimes, your physician may recommend washing (douching) of the nasal passage.
Corticosteroids counteract the inflammation caused by the body's release of allergy-causing substances, as well as that caused by other non-allergic factors. Thus, they generally work for many causes of rhinitis symptoms and are sometimes useful for chronic sinusitis.
Corticosteroids are sometimes injected or taken orally, but usually on a short-term basis for extremely severe symptoms. Physicians warn that injected or oral steroids may produce severe side effects when used for long periods or used repeatedly and, for this reason, they should be used with extreme caution.
In rhinitis, a corticosteroid is much safer when used by spraying it into the nose. Side effects are less common, but may include nasal ulceration, nasal fungal infection, or bleeding.
Cromolyn is a medication that blocks the body's release of allergy-causing substances. It does not work in all patients. The full dosage is four times daily, and improvement may take several weeks to occur.
Atropine and the related drug ipratropium bromide are sometimes used to relieve the runny nose of rhinitis; in fact, most antihistamines have a slight atropine-like effect. Atropine can be taken orally and as a nasal spray. It is a component of some antihistamine decongestant preparations.
Antibiotics are for the treatment of bacterial infections. They do not affect the course of uncomplicated common colds, and are of no benefit for non-infectious rhinitis, including allergic rhinitis. In chronic sinusitis, antibiotics may help only temporarily, and surgery may be needed.
Eye allergy preparations are used when the eyes are affected by the same allergens that trigger rhinitis, causing redness, watery eyes and itching. Eye preparations are available as prescription and non-prescription medications. Check with your physician or pharmacist about these medications.
Nasal surgery will usually cure or improve symptoms caused by mechanical blockage or chronic sinusitis not responsive to prolonged antibiotics and nasal steroid sprays. Stopping the use of offending medications will cure rhinitis medicamentosa, providing that there is no underlying disorder.
Check with your physician or pharmacist if you are unsure about a specific drug or formula.
Advice from your allergist: rhinitis
Allergies, including allergic rhinitis, affect an estimated 40 to 50 million people in the United States. Some allergies may interfere with day-to-day activities or lessen the quality of life.
The allergist-immunologist, with his or her specialized training and expertise in managing allergies and asthma, can develop a treatment plan for your individual condition. The goal will be to enable you to lead a life that is as normal and symptom-free as possible.<|endoftext|>
| 3.8125 |
1,749 |
# Calculator search results
Formula
Expand the expression
Factorize the expression
$3 \left( 2x-y \right) ^{ 2 } -4x+2y-5$
$12 x ^ { 2 } - 12 x y - 4 x + 3 y ^ { 2 } + 2 y - 5$
Organize polynomials
$3 \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \right ) ^ { \color{#FF6800}{ 2 } } - 4 x + 2 y - 5$
Expand the binomial expression
$3 \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \right ) - 4 x + 2 y - 5$
$\color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 4 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \right ) - 4 x + 2 y - 5$
Organize the expression with the distributive law
$\color{#FF6800}{ 12 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ y } + \color{#FF6800}{ 3 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } - 4 x + 2 y - 5$
$\color{#FF6800}{ 12 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
Sort the polynomial expressions in descending order
$\color{#FF6800}{ 12 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\left ( 2 x - y + 1 \right ) \left ( 6 x - 3 y - 5 \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ 3 } \left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
Expand the expression
$\color{#FF6800}{ 12 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
$\color{#FF6800}{ 12 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 12 } \color{#FF6800}{ x } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ y } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 5 }$
Do factorization
$\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ y } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \right )$
Solution search results<|endoftext|>
| 4.4375 |
1,442 |
The term peripheral vascular disease (or PVD) refers to any obstruction of large arteries in the limbs, most commonly in the legs and feet. The condition is also known as peripheral artery disease (PAD), and the two terms are often used interchangeably.
Peripheral vascular disease can cause pain, weakness, numbness, and changes in color of the affected limb.
Peripheral vascular disease eventually causes narrowing and hardening of the arteries that carry blood to the feet, and the decrease in blood flow can cause injury to the nerves. PVD is estimated to affect 12 percent of the general population, and up to 20 percent or more of Americans over the age of 70.
Most sufferers are men, although woman can become more susceptible during and after menopause.
Symptoms of Peripheral Vascular Disease
Some cases of peripheral vascular disease are mild, and in about 20 percent of cases there are no symptoms at all. In more severe cases, however, patients may experience:
- Claudication—pain in the legs that occurs with walking and impairs mobility
- Tingling sensation in the feet and toes
- Change in color (the limb or digits may appear pale or even blue)
- Chronic discomfort in the thigh or calf muscles
- Varicose veins
- Hair loss on the legs and feet
- Sores or wounds that heal very slowly or do not heal at all
In the early stages of peripheral vascular disease, sufferers may find that vigorous exercise has begun to seem more difficult. As the disease progresses, even mild exercise tends to aggravate the symptoms, and eventually those symptoms even present themselves when the sufferer is at rest.
The Causes of Peripheral Vascular Disease
The most common cause of PVD is arteriosclerosis. This is a buildup of plaque on the walls of the arteries, which stiffens them and narrows the space through which blood needs to flow.
The problem is exacerbated by the stiffening of the artery walls, which prevents the arteries from dilating to allow more blood flow. In some cases blood clots may also form in the arteries, further inhibiting blood flow.
While arteriosclerosis is the most common cause of peripheral artery disease, the condition can also be caused by trauma or infection. Patients who suffer from coronary artery disease also tend to suffer from peripheral artery disease.
Categories and Stages of Peripheral Vascular Disease
There are two categories of PVD—functional and occlusive. Occlusive PVD involves blockage of the artery (usually by plaque). In cases of functional PVD, the artery spasms and constricts. Raynaud’s Phenomenon, a condition that makes sufferers unusually sensitive to cold, especially in their extremities, is thought to be a type of functional peripheral vascular disease.
Raynaud’s Phenomenon causes many of the same symptoms as other forms of peripheral vascular disease, and sufferers may also experience brittle nails.
In 1954 the progress of peripheral vascular disease was divided into four stages by René Fontaine; these stages are now called the Fontaine Stages:
- Stage I: The arterial obstruction is not yet significant, and the subject is not experiencing symptoms
- Stage II: Mild pain when walking
- Stage III: Pain is present even when the subject is at rest
- Stage IV: Gangrene begins to set in
Risk Factors for Peripheral Vascular Disease To Be Aware Of
Peripheral artery disease is most common among men over the age of 50, especially those who are are heavy smokers. Anyone who has ever had a stroke is also at elevated risk. Additional risk factors include:
- High Cholesterol
- High blood pressure
- Heart or kidney disease
When Should You Seek Medical Attention?
Never let anyone tell you that pain and numbness in your legs is just a normal part of the aging process. It is not normal; it is a potentially serious medical condition that requires treatment.
If peripheral vascular disease is left untreated it will progress, and there is even the possibility—however small—that you might lose a foot, or even a leg.
Contact your healthcare provider immediately if you have leg pain while walking, or if you have numbness or tingling in your limbs.
Even if you are not experiencing any of the symptoms of PVD, you should be screened for it periodically if you are over the age of 70, or if you are over the age of 50 and have risk factors such as diabetes, obesity, or tobacco use.
Complications of untreated peripheral vascular disease can include:
- Blood clots
- Coronary artery disease
- Heart attack
- Deep vein thrombosis (a blood clot in a vein travels to the lungs and blocks a vein, causing a life-threatening pulmonary embolism)
- Gangrene, possibly requiring amputation (this usually appears first in the form of sores that do not heal)
- Ulcerations or non-healing wounds of the feet or legs
Prevention and Treatment of Peripheral Vascular Disease
While peripheral vascular disease is very common, there are many lifestyle choices that can help to prevent it, or at least to reduce your risk. Tobacco cessation is the first and simplest of these choices.
If you have diabetes or hypertension, work with your primary care doctor to keep it under control. Watch your diet, and ask your doctor for his or her advice on controlling your cholesterol intake.
Lastly, it is important to exercise regularly; if you can manage 30 minutes of vigorous exercise daily, you should do so. At the very least, make a habit of going for a 20-minute walk each day, if you can.
Treatment of peripheral vascular disease may involve surgery such as angioplasty (surgical widening of obstructed or narrowed arteries) or plaque excision, in which the plaque that is blocking the artery is scraped away.
Your doctor may attempt to treat your peripheral vascular disease with medications rather than surgery (or in addition to it).
A variety of medications are available for the control of blood sugar, cholesterol, high blood pressure, and blood clots, and your doctor may want to try a medication-oriented approach before resorting to surgery.
Talking to Your Doctor
Here are some questions you can ask your doctor about peripheral vascular disease:
- What is the cause of the numbness and tingling in my feet?
- Do I need to change my eating habits?
- Should I take medication for high blood pressure?
- What kind of an exercise regimen do you recommend?
- Are there any non-surgical ways we can handle this condition?
- Why are my feet cold all the time?
- Why does the hair on my legs not grow anymore?
- Why won’t these sores on my feet and legs heal?
- What does my blood sugar have to do with blood supply to my feet?<|endoftext|>
| 3.65625 |
807 |
#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 57 maths textbook solution
Answer: $\frac{2 \log 2}{\log \frac{11}{10}}$
Hint: Separate the terms of x and y and then integrate them.
Given: In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. We have to find, in how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
Solution: Let the count of bacteria be N
As the rate of growth of bacteria is proportional to the no. present
\begin{aligned} &\therefore \frac{d N}{d t} \alpha N \\\\ &\Rightarrow \frac{d N}{d t}=K N \\\\ &\Rightarrow \frac{d N}{N}=K d t \end{aligned}
Integrating both sides
\begin{aligned} &\int \frac{d N}{N}=K \int d t\\\\ &\Rightarrow \log |N|=K t+c \end{aligned} .................(1)
Initially$N=100000;\; t=0$
By (1)
\begin{aligned} &\log 100000=K(0)+c \\\\ &\Rightarrow c=\log 100000 \end{aligned}
Put in (1) we get
$\log N=K t+\log 100000$ ...................(2)
According to given
When
\begin{aligned} \mathrm{t}=2, \mathrm{~N} &=10 \% \text { of } 100000+100000 \\ &=10000+100000=110000 \end{aligned}
\begin{aligned} &\operatorname{By}(2) \log 110000=\mathrm{K}(2)+\log 100000 \\\\\ &\log 110000=K(2)+\log 100000 \\\\ &\Rightarrow \log \frac{110000}{100000}=K(2) \\\\ &{\left[\log m-\log n=\log \frac{m}{n}\right]} \end{aligned}
\begin{aligned} &{\left[\log m-\log n=\log \frac{m}{n}\right]} \\\\ &\Rightarrow K=\frac{1}{2} \log \frac{11}{10} \\\\ &\therefore \log N=\frac{1}{2} \log \frac{11}{10} t+\log 100000 \end{aligned} ....................(3)
Now we have to find in how many hours i.e t1 ; N=200000
By (3)
\begin{aligned} &\log 200000=\frac{1}{2}\left(\log \frac{11}{10}\right) t^{1}+\log 100000 \\\\ &\Rightarrow \log \left|\frac{200000}{100000}\right|=\frac{1}{2} \log \left(\frac{11}{10}\right) t^{1} \\\\ &\Rightarrow \log 2=\frac{1}{2} \log \left(\frac{11}{10}\right) t^{1} \end{aligned}
\begin{aligned} &\Rightarrow 2 \log 2=\log \left(\frac{11}{10}\right) t^{1} \\\\ &\Rightarrow \frac{2 \log 2}{\log \left(\frac{11}{10}\right)}=t^{1} \\\\ &\Rightarrow t^{1}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \end{aligned}<|endoftext|>
| 4.59375 |
637 |
Night vision goggles are immensely useful to hunters, soldiers and law enforcement agencies, but despite being used since World War II, there are still many people who don’t know how these devices function. Thanks to military, action and spy movies almost everyone knows what night vision goggles do, but how it is done remains a mystery to many.
Night Vision Principles [Source – NiviTech]
What happens when light particles, called photons, enter the goggles is that they hit the photocathode . The photocathode is light sensitive and immediately converts these photons into subatomic particles called electrons. Using a photomultiplier, the goggles then amplify these electrons. This means that there are much more electrons leaving the photomultiplier than what enters it. The final step in the process is a phosphor screen which causes the electrons to emit tiny flashes of light as they hit the phosphor. Since the whole process results in much more photons than what there was originally the result is that the original scene appears might brighter on the screen.
The process also explains why the night vision goggles displays the original screen in green colors. The reason is that photons that are converted to electrons lose all the color information, effectively turning the scene black and white. Since human eyes are actually more sensitive towards green light this color was deliberately chosen for the night vision goggles to display. The other advantage of using the green color is that it is more comfortable to look at for extended periods of time as black and white cause eye fatigue sooner.
Of course, the above described process produces a lower resolution and blurrier version of the original scene. Other drawbacks to first generation devices include a high-pitched noise which occurs when the goggles are switched on. The unity usually continues to glow green for a short time after it has been turned off as well. Second generation night vision goggles, which are more advanced, but also comes with a heftier price tag, forces the electrons through a so called micro channel plate for multiplication.
The MCP is made up of millions of tiny glass tubes which releases thousands more electrons when the original electrons pass through them. The end result is much more electrons reaching the phosphor screen and a better image. For even better, albeit more expensive results, third and fourth generation night vision goggles make use of chemical layers, consisting of gallium arsenide, to cover the phosphor screen which produces a brighter image and improves the contrast.
Despite the geometric distortion first generation night vision goggles are some of the most popular devices amongst hunters due to the lower price. Thanks to the increased price of second generation devices they are primarily in use by law enforcement agencies. Third and fourth generation devices are mostly confined to professional or military use only because of the high cost.
Since some light is required for these scopes to operate they come with infrared illuminators which produces infra-red light. This light is invisible to the naked eye, but is picked up by the goggles and allows for use in situations where there is total darkness.
Visit PointOptics for more military and huntings optics, as well as a guide to night vision trail cameras.<|endoftext|>
| 3.6875 |
445 |
Every glacier melts. Liquified glacier water trickles down rivers and streams during warm seasons; the cold seasons’ falling snow usually replaces the loss and, eventually, compresses into ice. But the last century’s changing climate has helped melt to outpace snowfall, and glaciers around the world are dying.
Doomed glaciers are thought to melt in place, like an ice cube abandoned on the counter; to retreat instead of advancing like a river of ice; or, in the case of tropical glaciers, to sublimate directly from ice to water vapor. (The last case might not be due to global warming — the ice loss on Kilimanjaro is actually caused by less snowfall and more sun.) At least one glacier, however, has surprised scientists by not following any of these patterns.
The Falljökull glacier of Skaftafell National Park in Iceland has abandoned its rotting, dying lower portion. The upper portion is once again advancing forward. Emrys Phillips, a researcher with the British Geological Survey, compared the glacier to a lizard that has detached its tail to escape a predator. "In the case of the glacier, the predator is our warming climate," he told Climate Central.
While acknowledging that “it’s strange to anthropomorphise a glacier,” Phillips says it’s “as if the glacier is trying to save itself by shortening its active length to adapt to the increasingly warmer summers and less snowfall during the winter months....
In effect, Phillips says, “The glacier has downsized.”
Phillips and his colleagues deployed several different forms of reconnaissance technologies — satellites, ground-penetrating radar and remote-sensing LiDAR — to reveal the steep mountain glacier’s survival strategy. They published their findings in the Journal of Geophysical Research last week.
If other mountain glaciers are staying alive in this way, research measuring the effects of climate change should look beyond the total length of glaciers and focus on the size of active portions, the researchers say. Over 90 percent of alpine glaciers in the world are retreating, but perhaps a few might have life in them yet.<|endoftext|>
| 4 |
413 |
If gravity is uniform - the force has the same magnitude and direction everywhere, the trajectory is a parabola. This is a very good approximation for trajectories that don't go very far.
But in fact the force is not perfectly uniform. It actually does point to the center of the earth. It is stronger nearer the center. The trajectory for this case is an ellipse.
A typical parabolic trajectory hits the ground before it gets very far. If it didn't, it would be a very long skinny ellipse.
A parabola is an infinitely long ellipse.
For typical trajectories that hit the ground quickly, parabolic and elliptical trajectories are almost identical.
Edited to respond to comments.
The rotation of the earth does have an effect. From the point of view of an inertial observer floating in space, the initial velocity of a thrown rock is about the speed of rotation of the earth's surface at that latitude.
The earth rotates $360$ degrees in $1$ sidereal day = $85604.1$ sec, or about $0.0042$ degrees/sec. So gravity isn't quite uniform. It has tilted a bit by the end of the trajectory. But in a flight lasting only a couple seconds, it isn't enough to notice.
The observer in space sees an observer on the ground moving sideways at the speed of the earth's surface. The ground observer is following a circular path. In those few seconds, he deviates from a straight line at $0.0042$ degrees/sec. To a good approximation, he is moving at uniform speed on a straight line. This gives the same result as if he wasn't moving.
If the rock fell through the earth and orbited, it would follow an ellipse as seen by the space observer. It would not be as skinny as I had thought. To the observer on earth, the motion would look complicated.
So thanks to Peter for pointing this out.<|endoftext|>
| 3.8125 |
1,192 |
Smartick is an online platform for children to master math in only 15 minutes a day
Oct27
Grade 5 Math Word Problems with Smartick
Math word problems are a really important part of Smartick. In this entry, we’re going to look specifically at some of the fifth-grade problems you can find in Smartick.
Math word problems with one operation
Yesterday a lot of us went to the school cafeteria for breakfast, and we drank all the juice. If they had served 470 liters less of juice, they would have served the same number of liters as they did today. Today they served 910 liters of juice. How many liters of juice did they serve yesterday?
Yesterday they served an unknown quantity of liters. If we take 470 liters away from this quantity, we get the number of liters that they have served today. To work out how many liters they served yesterday, we have to add:
910 + 470 = 1380
They served 1380 liters yesterday.
Subtraction
Charlie invited us to his house to try his strawberry and raspberry cake. When he started making it, he realized that he needed 400 more strawberries in order to have as many strawberries as the 680 raspberries he had. Work out how many strawberries he had.
Charlie had 400 fewer strawberries than raspberries. Since he had 680 raspberries, we need to subtract:
680 – 400 = 280
Multiplication
It’s sales season, and there are 15 bags of potato chips in Cecilia’s shop. Alfred told me that there are 5 times fewer bags of potato chips in Cecilia’s shop than there are in his. Work out how many bags of potato chips there are in Alfred’s shop.
There are 5 times fewer bags of potato chips in Cecilia’s shop than there are in Alfred’s. This means that there are 5 times more bags in Alfred’s than there are in Cecilia’s. To work out how many bags there are in Alfred’s shop, we have to multiply:
15 x 5 = 75
There are 75 bags of potato chips in Alfred’s shop.
Division
A supermarket sold 1000 kilos of food in total on Monday, between vegetables, fruit, meat, fish, and bread. On Tuesday, they sold much less, exactly 5 times less food than they sold on Monday. How many kilos of food did they sell on Tuesday?
They sold 5 times less food on Tuesday than they sold on Monday. Since we know that they sold 1000 kilos on Monday, we have to divide to work out how many kilos they sold on Tuesday:
1000 ÷ 5 = 200
They sold 200 kilos of food on Tuesday.
Math word problems with two operations
Since I like fish, I’ve put a fish tank in my room which contains 8 orange fish and 5 green fish. I’ve also added some things to decorate the fish tank, like sand, shells, and snails. Anthony also likes fish, but he only has 6 in his fish tank. How many fish would Anthony have to buy to have the same number of fish as me?
Before we can subtract the fish that I have with the fish Anthony has, we have to know how many fish I have. To do this, we have to add:
8 + 5 = 13
13 is the number of fish I have. Now we have to find the difference between the number of fish that I have and the number of fish that Anthony has. To do this, we need to subtract.
13 – 6 = 7
Anthony would have to buy 7 fish to have the same number of fish as me.
Fractions
Valentina is a nature photographer and all the photographs in her albums are of countryside and animals. 1/9 of the album that she just started is made up of photos of the countryside, and 1/6 is photos of animals. How much of the album has she filled?
She’s filled one part of the album with photos of the countryside, and another part with photos of animals. So, to work out the total amount of the album that she’s filled, we need to add:
1/9 + 1/6 = 15/54
She has filled 15/54 of the album.
Tables
Tomorrow is the premiere of the film “Alan vs. Aliens” and it’s going to play on all the screens at the Space Cinema. There are just 14 tickets left for the 5 p.m. screening, which is half of the number of tickets left for the 3 p.m. screening. Complete the table of available tickets.
We have to look at the data in the table and fill in the gaps.
There are 14 tickets left for the 5 p.m. screening. If there are 5 tickets left for Screen 1 and 4 left for Screen 3, how many are left for Screen 2?
14 – 5 – 4 = 5
There are 5 tickets left for the 5 p.m. screening in Screen 2.
There are 28 tickets left for the 3 p.m. screening (double the number of tickets left for the 5 p.m. screening). If there are 11 tickets left for Screen 2 and 4 left for Screen 3, how many tickets are there left for Screen 1?
28 – 11 – 4 = 13
There are 13 tickets left for the 3 p.m. screening in Screen 1.
The completed table will look like this:
These are some examples of the math word problems that you can find in Smartick. If you want to see more math word problems, sign up to Smartick and try it free.<|endoftext|>
| 4.53125 |
148 |
Or download our app "Guided Lessons by Education.com" on your device's app store.
Sight Word Puzzle
Year two students benefit from learning to read words quickly and automatically. This is especially true of high-frequency words (words that appear frequently in written language, and which often do not follow phonetic rules). The key to learning these words, of course, is having lots of exposure to them with repeated practise. But if you’re puzzled as to how to go beyond the basic flashcard approach to sight word practise, here’s an activity that can help! In this “out-of-the-box” approach to practising sight words, practise looks a lot like play!<|endoftext|>
| 3.828125 |
2,805 |
Kony 2012: Using Technology for Empathy
The term empathy has recently been on the minds of philosophers (Krznaric 2012), neurologists (Decety and Ickes 2009), health care professionals (Spiro 1992; VanHooft 2011), social workers (Gerdes, Importance of Empathy for Social Work Practice: Integrating New Science 2011), educators (Brooks 2009; Endacott 2010; Yilmaz 2007), and in the media (First Read Minute 2014). Empathy, though simply defined as one’s ability to comprehend the situation of another, is a complex process of the mind and heart (Gerdes 2011; Davis 1980). Findings connect it to a myriad of pro-social outcomes such as helping behavior (Graziano, Habashi and Sheese 2007), cooperating skills, self-esteem, and civic engagement (Noddings 2002). It also has inverse associations with bullying, racism, depression, and aggression (Legault, Gutsell and Inzlict 2011). In education particularly, there has been an emerging focus on building “21st Century Skills,” increasing the spotlight on empathy building in schools (Framework for 21st Century Learning 2010).
These 21st Century Skills require students to cooperate and communicate across cultural lines, to have a command of technology, and to demonstration imaginative problem solving skills, all of which relate to empathy. They are also represented in the engagement model of Kony 2012 and its parent organization, Invisible Children(Russell 2012). For this reason Invisible Children was a fixture in many schools prior to the release of Kony 2012. Educators identified Invisible Children’s engagement programs as a way to teach 21st Century skills and empathy (Kligler-Vilenchik and Shresthova 2012).
This viral phenomenon provides an opportunity to examine the opportunities when technology, film, and sound educational pedagogy are used to engage millennials’ empathic potential.
Listed are the top 15 viral videos, represented by millions of views. Kony 2012 is the only of the videos of serious content, and also at least 20 minutes longer than each of the other videos represented. This offers evidence that millennials can engage in content-rich media consumption when that media is strategically engaging. Educational or instructional content can be used to impart curricular goals such as empathy.
To view Invisible Children and Kony 2012 as an empathic case study, the term “empathy” must be further clarified. Researchers agree that empathy occurs both intellectually and emotively (Davis 1980; Preston and DeWaal 2002). This dual-process function combines a neurological reflex (mirror neurons) (Gutsell and Inzlicht 2010), and learned skill (Spiro 1992). Though some people are born with empathic predispositions, empathy can be fostered intellectually and emotionally (Mehrabian, Young, and Sato 1998). The tool often used to measure empathy levels is the Interpersonal Reactivity Index (IRI) (Davis 1983). The IRI is based on four subscales, two that are cognitive (c) and two emotional (e); imagination (c), perspective taking (c), affective connection (e), and personal distress (e). These subscales offer the theoretical framework from which to examine the case of Invisible Children and Kony 2012. Through the use of technology, social media, and film, this organization inadvertently put millennials in an empathic pipeline that aligns with these subscales.
For most, Kony 2012 was the first experience with Invisible Children, though the decade-old organization has made hundreds of podcasts and films. The story structure of their film is simple: a westerner is introduced on-screen; the viewer would affectively connect to him/her. he westerner would make a connection with a war-affected peer in northern Uganda, and the audience would be gently guided into that connection as well. This method of reporting and storytelling is not unique, especially for those telling the stories of people in distant places. In a youtube question and answer session, Nicholas Kristof discusses a “white arc” that he uses to emotionally scaffold his readership into a connection to a far off humanitarian crisis (Kristof, Youtube Question on Africa Coverage 2010). Normally, distance is a barrier to empathic connection (Ginzburg 1994), so Invisible Children’s the use of visuals and Kristof’s “arc” help overcome that geographical barrier. Next, the open sourcing of podcasts, webinars, school assembly and classroom screenings offered a multi-sensory entry point, which was both engaging and easy to virtually share. This method of story telling helps forge an empathic connection with a world issue that many are overwhelmed by. In this way an “affective connection” is formed, marking the first step in a viewers’ empathic engagement.
Invisible Children capitalizes on audiences’ emotional investment to impart the context and the scale of the issue. The affective connection creates a motivational effect so statistics that might otherwise be boring are now interesting (Kristof, New York Times Op Ed 2007). The use of animation sequencing, re-enactments, and real-time footage, Invisible Children kept engagement high and compelling, helping viewers through the “perspective-taking” phase of empathy. After that, the films’ story arc reaches a climax—the viewer is guided toward a feeling of “personal distress.” Upon feeling the affective connection to a victim in crisis, gaining an understanding of context and scale (“perspective taking”), audiences begin to sense an urgency of the situation in real time. At this point viewers are still only partially through an empathy building process. Viewer empowerment is the next necessary step in that process, and required to set viewers up to engage in “helping behavior” outcomes. At this point of experiencing “personal distress,” one’s brain seeks relief in one of two ways: constructive action, or avoidance/redirection. Avoidance often looks like “compassion fatigue” and burn out (Batson, Fultz and Schoenrade 1987).
It is here that Invisible Children’s strategy is to first cast a vision to end the crisis, and then to invite the viewer toward constructive engagement (relieving personal distress). The empowerment message was made even more effective through the medium of film because a movie literally offers a vision of what can be, igniting the “imagination” aspect of empathy (Davis 1983). Audiences, therefore, are carefully guided through the four clinical requirements of empathy in an artful and pedagogically sound way. Kony 2012 mirrored this empathic journey, and viewers of the film responded just as empathy research would suggest, by participating in helping behaviors (Preston and DeWaal 2002). Though controversial, this 30-minute film about a remote war was the most shared video in history. It broke expectations of millennials’ apathetic nature, and gave insight on how to tap into the empathy and desirable pro-social behaviors of this generation.
This is the empathic pipeline is represented throughout Invisible Children’s overall brand engagement strategy. After every film or online experience, the audience is individually connected with a peer mentor who can guide them toward customized involvement, and can encourage the empathic process along in a nuanced manner (Dillenburg 2009). Local supporter communities are formed and are connected through online platforms to motivate each other. Webinars and curriculum enhance their cognitive understanding and perspective of the problem, as well as real time updates on crisis were available through their innovative crisis tracker (LRA Crisis Tracker n.d.) (therefore sustainable level urgency, or “personal distress”) (Cameron and Payne 2011; Kony 2012 Campaign Analysis 2012). All of these elements combine to positively effect the empathy skills of millennials, and as a bonus, they made it all fun.
Batson, C. D., J. Fultz, and P. A. Schoenrade. 1987. "Distress and Empathy: Two Qualitatively Distinct Vicarious Emotions with Different Motivational Consequences." Journal of Personality 55 (1): 19-39.
Brooks, Sarah. 2009. "Historical empathy in the social studies classroom: A review of the literature." The Journal of Social Studies Research 33(2).
Cameron, D., and K. Payne. 2011. "Escaping Affect: How Motivational Emotional Regulation Creates Insensitivity to Mass Suffering." Journal of Personality and Social Psychology 100 (1): 1-15.
Davis, M. H. 1980. "A multidimensional approach to individual differences in empathy." JSAS Catalog of Selected Documents in Psychology 10 (85).
Davis, M. H. 1983. "Measuring individual differences in empathy: Evidence for a multidimensional approach." Journal of Personality and Social Psychology 44 : 113-126.
Decety, J., and W. Ickes. 2009. The Social Neuroscience of Empathy. Cambridge, MA: The MIT Press.
Dillenburg, Margie. 2009. "Invisible Children Curriculum." Invisible Children, Inc. . Invisible Children. http://invisiblechildren.com/get-involved/start-a-club/ (accessed June 30, 2014).
Endacott, J. L. 2010. "Reconsidering affective engagement in historical empathy." Theory & Research in Social Education 38: 6-49.
First Read Minute. 2014. NBC. January 7. http://firstread.nbcnews.com/_news/2014/01/07/22217507-first-read-minute-what-gop-empathy-gap-means-for-party-2014 (accessed July 20, 2014).
"Framework for 21st Century Learning." Partnership for 21st Century Skills. 2010. http://www.p21.org/about-us/p21-framework (accessed June 30, 2014).
Gerdes, K. 2011. "Empathy, Sympathy, and Pity: 21st Century Definitions and Implications for Practice and Research." Journal of Social Services Research 37.
Gerdes, K. 2011. "Importance of Empathy for Social Work Practice: Integrating New Science." Social Work 56 (2).
Ginzburg, Carlo. 1994."Killing a Chinese Mandarin; The Moral Implications of Distance." Critical Inquiry 21 (1): 46-60.
Graziano, W., M. Habashi, and B. Sheese. 2007. "Agreeableness, Empathy, and Helping: A Person X Situation Perspective." Journal of Personality and Social Psychology 93 (4): 583-599.
Gutsell, Jennifer, and M. Inzlicht. 2010. "Empathy constrained: Prejudice predicts reduced mental simulations of actions during observations of outgroups." Journal of Experimental Social Psychology 46 (5): 841-845.
Invisible Children Youtube Channel. 2009. http://invisiblechildren.com/program/schools-for-schools/ (accessed June 30, 2014).
Kligler-Vilenchik, Neta, and Sangita Shresthova. 2012. "Learning Through Practice: Participatory Culture Civics." Los Angeles, CA: Annenberg School for Communcation and Journalism.
Kristof, Nicholas. 2007. "New York Times Op Ed." New York Times. May 10. http://www.nytimes.com/2007/05/10/opinion/10kristof.html?_r=0 (accessed July 20, 2014).
Kristof, Nicholas. Youtube Question on Africa Coverage New York Times, (July 9, 2010).
Krznaric, Roman. RSA Animate. December 12, 2012. https://www.youtube.com/watch?v=BG46IwVfSu8 (accessed July 20, 2014).
Legault, Lisa, Jennifer Gutsell, and Michael Inzlict. 2011. "Ironic Effects of Antiprejudice Message: How Motivational Interventions Can Reduce (but also increase) Prejudice." Psychological Science 22 (12): 1472-1477.
"LRA Crisis Tracker." Invisible Children; Resolve. http://www.lracrisistracker.com/ (accessed June 30, 2014).
Mehrabian, A., A. L. Young, and S. Sato. 1998. "Emotional empathy and associated individual differences." Current Psychology: Research & Reviews 7.
Noddings, Nell. 2002. Educating Moral People: A Caring Alternative to Character Education. New York, NY: Teachers College Press.
Preston, S., and F. DeWaal. 2002. "Empathy: Its ultimate and proximate bases." Behavioral and Brain Science 25: 1-72.
Kony 2012. Directed by Jason Russell. Produced by Invisible Children. 2012.
Spiro, H. 1992. "What is Empathy and Can it be Taught." Annals of Internal Medicine 116 (10).
VanHooft, S. 2011. "Caring, objectivity and justice: An integrated view." Nursing Ethics 18 (2): 149-160.
Yilmaz, K. 2007. "Historical empathy and its implication for classroom practices in schools." The History Teacher 40 (3).<|endoftext|>
| 3.78125 |
585 |
## What is the next value of 1 i 2 l 3/f 4?
Related Posts
Looking at the design of the letters, “I” is made up of 1 line, “L” is made up of 2 traces, and “F” is made up of 3 lines. Following this pattern, the next letter would consist of 4 traces. Hence the next value is “E”, which is made up of 4 traces.
## What is the 7th term in the sequence of square numbers?
Answer: 49. Step-by-step explanation: To remedy, you can merely checklist the square numbers so as…
(*2*)
How do you in finding the fourth time period in a sequence?
Such sequences may also be expressed in phrases of the nth term of the sequence. In this situation, the nth term = 2n. To find the 1st time period, put n = 1 into the method, to find the 4th time period, change the n’s by means of 4’s: 4th time period = 2 × 4 = 8.
What is the fourth time period of the series a1 k an 2an 1?
The fourth time period a4 = 2a4 -1 = 2a3. So we now have 2(4k) = 8k. Hence the fourth term is 8k.
### How do you work out quadratic equations?
1. Put all phrases on one facet of the equal sign, leaving zero on the different aspect.
2. Factor.
3. Set each issue equal to 0.
4. Solve each and every of these equations.
5. Check through putting your resolution in the authentic equation.
### What are the similarities and variations between linear equations and arithmetic sequences?
The similarity between linear functions and mathematics sequences is that the slope of linear function is constant and difference between any two consecutive terms of an arithmetic sequence is consistent.
What is the nth term of a Fibonacci series?
Binet’s Formula: The nth Fibonacci quantity is given by the following formula: fn=[(1+√52)n−(1−√52)n]√5. Binet’s components is an example of an explicitly defined series. This signifies that terms of the sequence aren’t dependent on earlier phrases.
How do you to find the nth unusual quantity?
(*4*)
The nth atypical quantity is given by the components 2*n-1.
#### Can sequences have detrimental terms?
Geometric sequences wherein each term is got from the previous one by means of multiplying by means of a relentless, known as the not unusual ratio and steadily represented by means of the symbol r. Note that r will also be sure, damaging or zero. The terms in a geometrical series with negative r will oscillate between sure and unfavorable.
(*1*)
(*3*)<|endoftext|>
| 4.5625 |
175 |
Students should read pages 14 - 27 (from the revelation of Eva's suicide to the end of Sheila's questioning).
Common Core Objectives
CCSS.ELA-Literacy.CCRA.R.4 Interpret words and phrases as they are used in a text, including determining technical, connotative, and figurative meanings, and analyze how specific word choices shape meaning or tone.
CCSS.ELA-LITERACY.CCRA.R.5 Analyze the structure of texts, including how specific sentences, paragraphs, and larger portions of the text (e.g., a section, chapter, scene, or stanza) relate to each other and the whole.
CCSS.ELA-Literacy.CCRA.R.6 Assess how point of view or purpose shapes the content and style of a text.<|endoftext|>
| 3.859375 |
264 |
Leonardo Da Vinci is known to have popularized mirror writing. This involves writing in a way that it can only be read when its reflected in a mirror.
Mirror Reading involves having the ability to decode messages that are presented initially in mirror-reverse. The skills required for reading mirrored text are different from those required for normal reading. Such as spatial transformation and mental rotation.
While most people in general are able to read texts that have been reflected into a mirror slowly with some degree of effort, there have been a team of scientists in the Basque Centre on Cognition, Brain and Language for the first time ever have been able to show that we can turn these visuals around and understand them unconsciously and automatically.
The results of their research have shown that a short-term and intensive practice of mirror-reverse words can significantly improve reading performance not just for trained typographers, but for other mirrored typographies and mirror-reversed Arabic digits.*
We have created a tool to convert text into mirrored text and enhanced it with images to help. Kids have to read all three of the possible answers in the mirror and find the words that match the image. This is a perfect game for building children’s vocabulary, as well as acting like a gym for the brain and just a fun activity.<|endoftext|>
| 3.6875 |
543 |
# which statement is true if the refractive index of a medium a is greater than that of medium b
Question
which statement is true if the refractive index of a medium a is greater than that of medium b
in progress 0
1 year 2021-09-04T22:13:23+00:00 2 Answers 4 views 0
b.
Explanation:
B) total internal reflection is possible when light travels from medium a to medium b.
Explanation:
The index of refraction of a medium is the ratio between the speed of light in a vacuum (c) and the speed of light in that medium (v):
$$n=\frac{c}{v}$$
When a ray of light crosses the interface between two mediums, it undergoes refraction, which means it is is bent and its speed changes, according to Snell’s law:
$$n_1 sin \theta_1 = n_2 sin \theta_2$$
where
$$n_1,n_2$$ are the index of refraction of medium 1 and 2
$$\theta_1, \theta_2$$ are the angle of incidence and of refraction
We can rewrite the equation as
$$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$$
From this equation, we observe that if $$n_1>n_2$$, then the ratio $$\frac{n_1}{n_2}>1$$; however, $$sin \theta_2$$ cannot be larger than 1. This means that in this case, there will be a maximum value of $$\theta_1$$ above which refraction no longer occurs: in that situation, the light coming from the 1st medium is entirely reflected inside the 1st medium, and this phenomenon is called total internal reflection.
The value of the critical angle above which total internal reflection occurs is
$$\theta_c = sin^{-1}(\frac{n_2}{n_1})$$
In this situation, we have:
$$n_A > n_B$$ (index of refraction of A is larger than index of refraction of B)
Also, the index of refraction of air is 1.00, so it is lower than the index of refraction of every medium:
$$n_A > n_B > n_{air}$$
Since total internal reflection occurs only if the index of refraction of medium 1 is higher than that of medium 2, the correct statements are:
B) total internal reflection is possible when light travels from medium a to medium b.<|endoftext|>
| 4.5 |
257 |
Ancient Egyptian art is the painting, sculpture, architecture and other arts produced by the civilization of ancient Egypt in the lower NileValley from about 3000 BC to 100 AD.
Ancient Egyptian art reached a high level in painting and sculpture, and was both highly stylized and symbolic. Much of the surviving art comes from tombs and monuments and thus there is an emphasis on life after death and the preservation of knowledge of the past.
Ancient Egyptian architects used sun-dried and kiln-baked bricks, fine sandstone, limestone and granite. Architects carefully planned all their work. The stones had to fit precisely together, since there was no mud or mortar. When creating the pyramids, ramps were used to allow workmen to move up as the height of the construction grew. When the top of the structure was completed, the artists decorated from the top down, removing ramp sand as they went down. Exterior walls of structures like the pyramids contained only a few small openings. Hieroglyphic and pictorial carvings in brilliant colors were abundantly used to decorate Egyptian structures, including many motifs, like the scarab, sacred beetle, the solar disk, and the vulture. They described the changes the Pharaoh would go through to become a god.<|endoftext|>
| 4 |
1,012 |
Carbon nanotubes are cylindrical structures consisting of carbon molecules. Although their diameter is in the order of a few nanometers, they may reach several millimeters in length. They are mostly valued for their extraordinary mechanical, thermal, and electrical properties. In particular, the discrete separation of the carbon nanotube energy levels is the reason for their excellent semiconducting properties. They also exhibit unique stiffness and strength, and can be used in a variety of applications such as the manufacturing of electronic circuits, materials, optics, batteries, solar cells, textiles, and many more.
In general, nanotubes can be found in two types: single-walled (SWNTs), consisting of only one graphitic layer, and multi-walled (MWNTs) consisting of multiple concentric graphitic layers.
The Process of Self-Assembly
There are various methods for the synthesis of nanotubes in large quantities, including arc discharge, laser ablation, chemical vapor deposition, and others. However, there is also a smaller scale process that can take place without any external stimulation. It is known as molecular self-assembly, where molecules can be arranged in well-defined patterns without human intervention. The same process can be applied to nanotubes as well.
Researchers have achieved the right experimental conditions and mix of catalysts to reproduce the self-assembling of nanotubes, in the same way that the deoxyribonucleic acid or DNA does. Carbon molecules are then linking in groups of six to form rosette-shaped rings and the rings, in their turn, assemble to create rod-like nanotubes. The result is the formation of self-replicating nanotubes that can be regarded as a superior material for multiple applications.
Such experiments have already taken place in recent years, regarding the creation of artificial joints, body implants and electronic device applications. The results, although at an experimental level, were very promising.
Fabrics and Other Applications
Apart from the wide application potential in biomedical fields, a recent development also involves the production of a perfectly uniform layer of self-assembled nanotubes known as "nanocarpet." When magnified a million times, the nanocarpet structure resembles the fibers of a shag rug. These tiny carpets are used as thermal interface materials in chips to protect them from overheating. They also have the ability to sense their environment by modifying their color according to the environmental conditions and even detect and kill bacteria (such as E. coli). Other possible applications include the detection and decontamination of chemical and biological weapons.
Electronic textiles or fabrics are another major field where nanotubes could be used more extensively. The process is simple: a cotton thread is being coated with conductive carbon nanotubes and then can be woven into fabric. These threads can conduct a significant amount of electric current, so that a LED attached to them, can shine brightly enough. It has been observed that these nanotube textiles demonstrate interesting biosensing abilities, such as sensing human blood.
Fears, Speculation, and Reality
An undeniable number of benefits are likely to be acquired through the use of self-assembling nanotubes as a raw material. They are inexpensive (as they are made of carbon), and since they can replicate in room-temperature conditions within a few hours, they have the additional advantage of low production costs.
According to popular belief however, these developments may prove quite risky. (See "Is Nanotechnology in Fuel Cells Ultimately Bad for the Environment?") Self-assembling nanotube structures could "go wild," as the fearful theory goes, and replicate uncontrollably in the field. The possible consequences are only limited by our imagination.
Although concerns like this are always expected when new technologies are tested, there is no solid proof yet that could possibly justify them. The self-assembling phenomenon has only been observed under certain conditions- even without human help- and are limited to the nano-scale. Despite the concerns, this is a first and important step toward the design of novel nanostructured materials with unlimited application potential.
- "Self assembly of carbon nanotubes (CNTs)" , "Carbon Nanotubes for electronic textile applications" by NanotechnologyDevelopment.Com
- "Self-assembling 'nanotubes' offer promise for future artificial joints" by NanotechWire.Com
- "Carbon nanotube" ScienceDaily.Com
- "Self-Assembling Nanotubes Change Colors, Form ‘Nanocarpet’ and Even Kill Bacteria" mirm.pitt.edu
- This is an STM Atomically Resolved Image Showing the Structure of a Nanotube by Taner Yildirim
- The Structure of a Carbon Nanotube by Guillaume Paumier
- Aligned Carbon Nanotubes by Argonne National Laboratory<|endoftext|>
| 3.796875 |
1,383 |
# What is Whole Number? Definition and Whole Number Example
Whole numbers (W) form the foundation of arithmetic, encompassing all non-negative integers from zero onwards {W =(0,1,2,3,4,…)}. Understanding their properties and operations is fundamental to mastering mathematics. In this blog, we will explore the important characteristics of whole numbers, including their properties such as closure under addition and multiplication, the concept of identity elements, and the role they play in everyday calculations. Whether you’re a student looking into the basics or a curious learner, this blog will provide you with the knowledge to learn more about whole numbers confidently.
## What are Whole Numbers?
Whole numbers are a set of numbers that include all the non-negative integers from zero upwards. In mathematical notation, they are represented as W ={0,1,2,3,4,…}. Whole numbers are essentially natural numbers (positive integers) including zero.
Important Properties:
1. Closure under Addition and Multiplication: When you add or multiply two whole numbers, the result is always another whole number. For example, 2+3=5 and 2×3=6, both results are whole numbers.
2. Identity Elements:
• Additive Identity: The number 0 is the additive identity for whole numbers. Adding zero to any whole number n gives n itself: n+0=n.
• Multiplicative Identity: The number 1 is the multiplicative identity for whole numbers. Multiplying any whole number nnn by 1 gives n: n×1=n.
3. Non-negativity: Whole numbers are always non-negative, meaning they are greater than or equal to zero.
4. Counting and Cardinality: Whole numbers are commonly used for counting objects and determining cardinality (size or quantity).
5. Basic Operations:
• Addition and Subtraction: Whole numbers can be added or subtracted following basic arithmetic rules.
• Multiplication and Division: Multiplication involves repeated addition, and division is the inverse operation of multiplication.
6. Application in Mathematics and Real Life: Whole numbers are foundational in mathematics, and used in various calculations, measurements, and scenarios where counting and discrete quantities are involved.
Examples:
• W = 0,1,2,3,4,5,6,………..
## The Power of Whole Numbers
Whole numbers aren’t just for counting objects. They empower us to perform various mathematical operations:
• Addition: We combine whole numbers to find a total quantity (e.g., 3 apples + 2 apples = 5 apples).
• Subtraction: We take away a whole number from another to find the difference (e.g., 10 cookies – 5 cookies = 5 cookies left).
• Multiplication: We repeatedly add a whole number to itself to find the total (e.g., 4 x 3 = 4 added to itself 3 times, equaling 12).
These basic operations form the building blocks of more complex math, allowing us to solve problems in various fields like finance, engineering, and even computer science.
## Properties of Whole Numbers
Here are all the key properties of whole numbers:
Closure under Addition and Multiplication Property:
• When you add two whole numbers, the result is always another whole number. For example, 2+3=5.
• When you multiply two whole numbers, the result is always another whole number. For example, 2×3=6.
Associative Property:
• The addition and multiplication of whole numbers are associative operations. This means that the way numbers are grouped in an operation does not affect the result.
• For example, (2+3)+4=2+(3+4) and (2×3)×4=2×(3×4).
Commutative Property:
• The addition and multiplication of whole numbers are commutative operations. This means that the order of numbers does not affect the result.
• For example, 2+3=3+2 and 2×3=3×2.
Identity Elements:
• Additive Identity: The number 0 is the additive identity for whole numbers. Adding zero to any whole number n gives n itself: n+0=n.
• Multiplicative Identity: The number 1 is the multiplicative identity for whole numbers. Multiplying any whole number nnn by 1 gives nnn: n×1=n.
Distributivity:
• Multiplication distributes over addition in whole numbers. For example, 2×(3+4)=2×3+2×4.
Non-negativity:
• Whole numbers are always non-negative, meaning they are greater than or equal to zero.
Closure under Subtraction:
• Subtraction of whole numbers may not always result in a whole number unless the subtraction does not lead to a negative number.
## Whole Numbers Solved Examples
Here are five solved examples involving whole numbers:
1. Example 1: Addition Calculate the sum of 25 and 17.
25+17=42
2. Example 2: Multiplication Find the product of 8 and 6.
8×6=48
3. Example 3: Subtraction Subtract 12 from 30.
30−12=18
4. Example 4: Division Divide 45 by 9.
45/9=5
5. Example 5: Application Problem A shopkeeper had 85 pencils. If she sold 42 pencils, how many pencils does she have left?
85−42=43
Answer: She has 43 pencils left.
Also Read: Questions of Logical Problems Reasoning
## FAQs
Is 1 to 100 are all whole numbers?
You would say all the numbers from 1 to 100 if you were counting them. Starting with 1, go through 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and so on, until you reach 100. In addition to zero, whole numbers are all positive values from one to infinity.
Why is 0 a whole number?
It is important to note that whole numbers are real numbers and do not have any fractions, decimals, or negative numbers. Not a part or decimal of any number, zero is just zero. This is neither good nor bad. As a whole number, zero follows the rule.
What is the largest whole number?
It’s clear that 0 is the smallest whole number and 1 is the smallest natural number. There is no biggest whole number, though, because every number comes after it. So, there isn’t a biggest whole number.
This was all about “Whole Numbers”. For more such informative blogs, check out our Study Material Section, or you can learn more about us by visiting our Indian exams page.<|endoftext|>
| 4.84375 |
1,921 |
September 21, 2016
INTRODUCTION: WHY TRANSPARENCY?
The concepts of transparency, disclosure and accountability are deeply entwined. Transparency refers to the principle of creating an environment where information on existing conditions, decisions, actions or other matters are made visible, accessible and understandable to all interested parties. Closely related to transparency is disclosure. Disclosure is the process and methodology of providing information and decisions related to an organization through timely dissemination and openness. Accountability refers to the need for those in an organization to justify their actions and policies and accept responsibility for their decisions and results.
Transparency is necessary for the concept of accountability to take hold among the major “players” both within and outside an organization. Transparency requires organizations to face the reality of a situation and makes members of that organization more responsible, particularly if they know they will have to explain their views, decisions and actions. Transparency is also a means of fostering accountability, internal discipline and better governance. Transparency and accountability improve the quality of decision making.
Transparency and accountability are mutually reinforcing. Transparency enhances accountability by making it easier to monitor an organization’s activities, and accountability enhances transparency by incentivizing participants within an organization to explain their actions and ensure that such explanations are disseminated and understood. The goal is for transparency and accountability to improve current decision making, as well as the public’s understanding of the key factors in current and future decisions.
A Transparency Policy, therefore, seeks to ensure that sufficient information is disclosed to interested parties and the public in general, while holding the decision makers accountable.
There are a number of obvious advantages to a Transparency Policy, particularly for a panel of experts that oversees scientific research such as the Expert Panel for Fragrance Safety (“Expert Panel”).
First, by its very nature, safety and science is meant to be “open” and transparent. Accordingly, it is contrary to the essence of a scientific expert panel not to have a transparency policy.
Second, many of the Expert Panel’s actions with respect to safety assessments are already subject to public scrutiny through peer reviewed publications. A Transparency Policy will do nothing more than expand this process.
Third, the Expert Panel has nothing to hide and should encourage the dissemination of information regarding its decision making process. For example, the current Expert Panel’s Conflict of Interest Statement, as well as its Operating Guidelines, demonstrate the Expert Panel’s strict ethical standards and openness and should be readily available to any interested parties.
Fourth, the Expert Panel may well benefit by knowing that it will be held accountable for its actions and policies when such actions and policies are more accessible to interested parties pursuant to a Transparency Policy.
There is, however, a tension between transparency and confidentiality. The existence of proprietary information may deter the implementation of a transparency program in certain instances. For example, releasing preliminary “raw” data and results on fragrance material safety may cause undue and misplaced concern or give those with knowledge of that information a perceived or actual competitive advantage. Also, legal advice protected by the attorney-client privilege should be maintained as confidential to preserve the privilege and encourage frank discussion among Expert Panel Members. And any third party information provided to the Expert Panel in confidence should be maintained as confidential. Accordingly, a Transparency Policy must balance the goals of transparency with – in certain circumstances – the need for confidentiality. On the whole though, transparency should be the general rule, and confidentiality the exception.
During its January 18, 2016 meeting, the Expert Panel discussed establishing a formal Transparency Policy and the key elements of such a policy. The following arises from these discussions and forms the basis of a final Transparency Policy that will govern and guide the Expert Panel in its actions and deliberations.
ELEMENTS OF THE EXPERT PANEL’S TRANSPARENCY POLICY
1. General Recommendation
The Expert Panel should adopt and periodically update a formal Transparency Policy.
Safety and science initiatives – the bread and butter of the Expert Panel – are meant to be open and transparent. Additionally, many of the Panel’s actions with respect to safety assessments are already subject to public scrutiny through peer reviewed publications. A Transparency Policy is well advised and would facilitate accountability in the fragrance industry and allow continued public scrutiny of the Expert Panel’s actions.
2. Exceptions to Expert Panel Transparency Policy
Third party proprietary and confidential information provided to the Panel in confidence, as well as Attorney-Client communications, are exempt from the Transparency Policy.
The existence of proprietary and confidential information will affect any transparency policy. Accordingly, the Expert Panel’s Transparency Policy should exclude third party proprietary and confidential information provided to the Panel in confidence, as well as attorney-client communications.
3. The Expert Panel’s Presence on the Internet
The Expert Panel will establish and maintain a separate website under the domain name “fragrancesafetypanel.com/.net/.org.”
Information regarding the Expert Panel can be found on the RIFM website by clicking the “About Us” tab on the RIFM website, and then clicking “Expert Panel” to arrive at a single webpage that briefly explains the Expert Panel and then lists its members and their affiliations. A separate, stand-alone Expert Panel website is essential to promote transparency in the fragrance community. The following recommendations (Items 4 through 12 below) address the nature and content of the Expert Panel website so as to achieve the goal of transparency.
4. Biographies of Expert Panel Members
In addition to listing the panelists and their affiliations, the Expert Panel website will also provide biographies and more detailed CVs on Panel members.
It is insufficient merely to identify the name and affiliation of each Panel member without providing any information regarding the knowledge, expertise and skill set of the panelist. The Expert Panel website, therefore, will provide a biography on each Panel Member setting forth the panelist’s background and expertise. The biography will then provide a hyperlink to a more detailed CV should additional information be desired.
5. Conflict of Interest Statement and Operating Procedures
The Expert Panel Conflict of Interest Statement and its Operating Procedures should be publicly available.
Every Expert Panel member must sign a Conflict of Interest Statement to ensure impartiality and objectiveness. Additionally, the Expert Panel has promulgated Operating Procedures to govern its actions. These documents will be made available to the general public and posted on the Expert Panel website.
6. Dates and Locations of Expert Panel Meetings
The dates and locations of all Expert Panel Meetings that have been scheduled should be publicly available.
The Expert Panel typically schedules its meetings up to one (1) year ahead of time. This information will be made available, not only to the Expert Panel, but also the general public. Accordingly, the meeting schedule and locales for the Expert Panel will be posted and regularly updated on the Expert Panel website.
7. Agendas for Expert Panel Meetings
Agendas of Expert Panel Meetings should be publicly available.
Typically, Agendas for each Expert Panel Meeting are provided to Panelists approximately 10 – 14 days prior to meetings. Upon distribution to the Expert Panel, the current Agenda for the upcoming meeting will be posted on the Expert Panel website.
8. List of Fragrance Materials to be Discussed at Expert Panel Meetings
The list of all fragrance materials to be discussed and/or assessed by the Expert Panel at its next meeting should be publicly available.
In addition to making Expert Panel Meeting Agendas available to the general public, it makes sense to identify all fragrance materials that will be discussed and/or assessed by the Expert Panel during the upcoming meeting. This information should be posted on the Expert Panel’s website, in addition to RIFM’s website.
9. Soliciting and Allowing Written Submissions/Data for Expert Panel Consideration with Respect to Fragrance Materials to be Discussed at Panel Meetings
The Expert Panel should solicit from the public and allow written submissions and data with respect to fragrance materials that will be discussed at Panel Meetings.
To ensure that the Expert Panel receives and considers all available information with respect to fragrance materials that will be considered by the Panel in its assessments, the Panel will solicit and allow written submissions and data that are relevant to the fragrance materials being considered. Such solicitations will appear on the Expert Panel’s website in conjunction with notices of upcoming meetings.
10. Minutes of Expert Panel Meetings
The Minutes of Expert Panel Meetings should be publicly available.
The actions and decisions of the Expert Panel should be available to the public. Making minutes publicly available is not uncommon and complies with best practices in scientific organizations. Accordingly, the Minutes of Panel Meetings will be posted on the Expert Panel website.
11. Expert Panel’s Vision and Mission
The Expert Panel should adopt “Vision” and “Mission” statements and post them on the Expert Panel website.
The Expert Panel’s Vision Statement and Mission Statements should be made available to the general public and posted on the Panel’s website.
12. Availability of Expert Panel’s Transparency Policy
The current Expert Panel Transparency Policy should be made available to the general public.
The Expert Panel should be transparent about its own Transparency Policy. Accordingly, the current policy should be posted on the web for everyone to see.<|endoftext|>
| 4 |
3,238 |
You are on page 1of 10
# Chapter 2:
## Random Variables and Probability Distributions
Lesson 3: Random Variables
TIME FRAME: 1 hour session
OVERVIEW OF LESSON
In this lesson, the concept of a random variable will be discussed. The notion of a statistical
experiment will be defined. Then, random variables that relate to experiments. Finally, two types
of random variables, discrete and continuous, will be described.
LEARNING OUTCOME(S): At the end of the lesson, the learner should be able to:
## illustrate/provide examples of random variables
distinguish between discrete and continuous random variables
find the possible values of a random variable.
## PRE-REQUISITE LESSONS: Types of Data (in particular, classifications of numerical
variables), Probability
LESSON OUTLINE:
1. Introduction/Motivation: The coin toss and breath holding activity.
2. Introduce the concepts of a statistical experiment and a random variable
3. Distinguish between discrete and continuous random variables
4. Give examples of random variables
KEY CONCEPTS: Statistical Experiment, Outcomes, Random Variables, Discrete Random
Variables, Continuous Random Variables
MATERIALS NEEDED:
one 1 peso coin per student, one stop timer per group.
DEVELOPMENT OF THE LESSON
(A) Motivation The coin toss and breath holding
Look at the current 1- peso coin in circulation. It has Jose Rizal on one side, which we will
call Head (H), and the other side Tail (T). Ask students to toss a 1-peso coin three times and
record in their Activity Sheet the results of the three tosses (Use H for heads, and T for tails.
If needed, define first which is the heads side of the coin and the tails side of the coin). For
example, a student tosses heads, tails, heads, then the student should write HTH on his/her
notebook. Ask them to count the number of heads that appeared and write it also in their
Activity Sheets.
Next, have each student hold their breath and record the time. This is best done if the students
time it as accurately as possible (if possible, use a cell phone timer and record up to the
nearest hundredth of a second). If there is limited number of timers, do it one at a time with
one student holding the timer, as the other one is holding his/her breath. Ask students to
record the time in their Activity Sheets.
Then, record all the possible answers on the board for both activities. For the first activity
write all eight possible outcomes, and then list down which one had 0, 1, 2, or 3 heads. If you
have time, tally the results. You can do this systematically so that you do not get confused
TTT
TTH
THT
HTT
THH
HTH
HHT
HHH
The second activity is a little bit more challenging. Expect to have a lot of possible values.
Just write about 10, then just tell the students if they have different values, they can just raise
their hand. Notice that the first one had only four possible values to take, while the second
one is almost unique for each individual. What could help might be getting the lowest value
and the highest value recorded by students.
Emphasize the difference in the number of possible values in these two activities as this is
important in the discussion.
## Experiments and Random Variables
Begin the discussion with the definition of a Statistical Experiment: an activity that will
produce outcomes, or a process that will generate data. The outcomes have a corresponding
chance of occurrence. Examples of which are (a) tossing three coins and counting the number
of heads, (b) recording the time a person can hold his/her breath, (c) counting the number of
students in the classroom who are present today, (d) obtaining the height of a student, etc.
Say that the two activities are examples of statistical experiments. You can come up with
several examples, such as recording the results of an examination, asking the weekly baon
(or allowance) of students, identifying the waist line of students.
Emphasize that Statistical Experiments can have a few or a lot of possible outcomes. In the
coin toss example, there are eight possible outcomes. In the breath example, there can be a
lot of possible outcomes. However, we can indicate that the possible values are in the range
of 10 seconds to 60 seconds (you can ask the shortest and the longest times in class and use
that as the limits for this example).
Suppose you will give a student candy based on the number of heads that appear in the coin
toss experiment (actually giving a candy is optional). List down all the possible number of
candies that can be given. Notice that it should only be 0, 1, 2, or 3. Then, you can list down
all the outcomes of the experiment under each value:
0 TTT
1 TTH, THT, HTT
2 THH, HTH, HHT
3 HHH
## Next, define a Random Variable: it is a way to map outcomes determined by chance to a
number. It is typically denoted by a capital letter, usually X.
X: outcome number
It is actually neither random nor a variable in the traditional sense defined in an algebra class
(where we solve for the value of a variable). It is technically a function from the space of all
possible events to the set of real numbers.
Tell students that a random variable must take exactly one value for each random outcome.
As with functions generally, a number of possible outcomes may have the same value of the
random variable, and in practice this occurs frequently. For instance, three outcomes above
for tossing a coin thrice would have 1 candy, and three outcomes would have 2 candies.
Students need to understand that random variables are conceptually different from the
mathematical variables that they have met before in math classes. A random variable is
linked to observations in the real world, where uncertainty is involved.
Students should be told that random variables are central to the use of probability in practice.
They help model random phenomena, that is, random variables are relevant to a wide range
of human activity, and disciplines, including agriculture, biology, ecology, economics,
medicine, meteorology, physics, psychology, computer science, engineering and others. They
are used to model outcomes of random processes that cannot be predicted deterministically in
advance (but the range of numerical outcomes may however be viewed).
In the coin example, we can define the random variable X to be the number of heads that
appears from tossing a coin three times. While we do not know what will be the specific
outcome resulting, but we know the possible values of X in this case are 0, 1, 2, or 3. You can
also define another random variable Y to be the time a person can hold his/her breath. The
possible values for this variable can be one of so many possible values.
In the second example, the possible values range between the lowest and the highest value
recorded by students. Notice that it is really difficult to list down all the possible values. That
is why in this example, it is better to state the possible values as an interval, such as
10 Y 60 , if the lowest and highest values are 10 and 60 respectively.
## Types of Random Variables:
Then distinguish the two types of random variables, viz., discrete and continuous.
(a) Discrete Random Variables are random variables that can take on a finite (or countably
infinite) number of distinct values. Examples are number of heads obtained when tossing a
coin thrice, the number of siblings a person has, the number of students present in a
classroom at a given time, the number of crushes a person has at a particular time, etc.
Categorical variables can be considered discrete variables, Example: whether a person has
normal BMI or not, you can assign 1 as the value for normal BMI and 0 for not normal BMI.
You can also put numbers to represent certain categorical variables with more than two
categories. You can also use ordinal variables, like how much they like adobo on a scale of 1
to 10 (where 1 means favorable, and 10 unfavorable).
Continuous Random Variables, on the other hand, are random variables that take an
infinitely uncountable number of possible values, typically measurable quantities. Examples
are the time a person can hold his/her breath, the height or weight or BMI of a person (if
measured very accurately), the time a person takes for a person to bathe. The values that a
continuous random variable can take on lie on a continuum, such as intervals.
Extra Notes:
You can modify the experiment to just tossing a coin twice instead of thrice to
make things simpler. Here, the outcomes will be only four: HH, HT, TH, TT. and
the possible values of X are 0, 1, and 2.
You may use other examples of continuous variables such as height, weight,
lengths, age)
Feel free to add more examples, or get examples from the seatwork that is in the
next section.
## (C) Group Discussion
Group students into threes. Given the following experiments and random variables, ask the
groups to identify what are the possible values of the random variable? Also, for each random
variable, identify whether the variable is discrete or continuous. (Answers in bold are
Discrete, while answers in italics are Continuous)
1. Experiment: Roll a pair of dice
Random Variable: Sum of numbers that appears in the pair of dice
2. Experiment: Ask a friend about preparing for a quiz in statistics
Random Variable: How much time (in hours) he/she spends studying for this quiz
3. Experiment: Record the sex of family members in a family with four children
Random Variable: The number of girls among the children
4. Experiment: Buy an egg from the grocery
Random Variable: The weight of the egg in grams
5. Experiment: Record the number of hours watching tv from 7pm to 11pm for the past
five nights.
Random Variable: The number of hours watching tv from 7 pm to 11 pm
(D) Enrichment
In tossing a coin four times, how many outcomes correspond to each value of the random
variable?
What if coin would be tossed five times? six times? seven times? eight times?
## Try to relate the outcomes to the numbers in Pascals triangle.
For tossing the coin four times, there will be five possible values,
0, 1, 2, 3, 4, with
1, 4, 6, 4, 1 outcomes, respectively.
## For five coins there are six possible values,
0, 1, 2, 3, 4, and 5, with
1, 5, 10, 10, 5, 1 outcomes, respectively.
## In general, for n tosses of a coin, there are n+1 possible values, 0, 1, 2, 3, , n. If k is a
possible value, then there are
n!
nCx = (nk )= k ! ( nk )!
## outcomes associated with x.
Next, possibly read on probability distributions, which will be covered in the next lesson.
KEY POINTS
## A Random Variable may be viewed as a way to map outcomes determined by chance to
numbers.
There are two types of random variables:
o Discrete: takes on a finite (or countably infinite) number of values
o Continuous: takes an infinitely uncountable number of possible values, typically
measurable quantities
REFERENCES
De Veau, R. D., Velleman, P. F., and Bock, D. E. (2006). Intro Stats. Pearson Ed. Inc.
Workbooks in Statistics 1: 11th Edition, Institute of Statistics, UP Los Banos, College Laguna
4031
topic/random_variables_prob_dist/v/random-variables
ACTIVITY SHEET 02-02
1. Toss a coin three times and record results of the three tosses below
(Use H for heads, and T for tails. .
Outcome
First Toss
Second Toss
Third Toss
## 2. Count the number of heads that appeared.
3. Write all possible outcomes for tossing a coin three times, and then list count the number of
4. Hold your breath and accurately record the time you held your breath. (If possible, use a cell
phone timer and record up to the nearest hundredth of a second). Record the time below:
________ seconds
ASSESSMENT
1. Identify a possible random variable (or if possible two random variables) given the following
statistical experiments. If possible, identify whether the variable is Discrete or Continuous:
a. Take a quiz (score of students, whether a student passed or failed the quiz, how
long it took a student to answer the quiz)
long students ate breakfast, the time students had breakfast, how many calories they
consumed)
c. Ask a neighbor about television shows (how many shows he/she watches every
night, what tv channel he/she prefers the most, how long does he/she watch TV per
week)
number of Facebook friends he/she has, the amount of time he/she spends per week
e. Run 100m on the track (whether students were able to complete it in under 15s or
not, time to finish running 100m)
f. Ask a classmate about musical instruments (whether he/she plays an instrument or
not , how many instruments he/she can play, length of time he/she plays the
instrument per week)
g. Visit the nearest market and look for poultry, such as chickens (how many stalls sell
chickens, whether the first stall sells chickens or not, total weight of chickens sold
in a certain day)
time of the EDSA revolution or not, whether she was there or not, what her age
was during the EDSA revolution)
2. During a game of Tetris, we observe a sequence of three consecutive pieces. Each Tetris
piece has seven possible shapes labeled here by the letters , , , , , and . So in this
random procedure, we can observe a sequence such as STT, JJ , SOL, JJJ and so on. Define:
## X to be the number of occurrences of `J' in a sequence of three pieces. Then X
can take the value 0, 1, 2 or 3.
Y to be the number of different shapes in a sequence of three pieces. Then Y can
take the value 1, 2 or 3.
T to be the time it takes a randomly selected Tetris gamer to end a game
## Identify whether X, Y and T are discrete or continuous.
Explanatory Notes:
Teachers have the option to just ask this assessment orally to the entire class, or to
group students and ask them to identify answers, or to give this as homework, or to use
some questions for a chapter examination.
The answers here are some of the possible answers and are not limited to these only.
Should the students think of other examples, that would be better.<|endoftext|>
| 4.4375 |
844 |
Our curriculum focuses on the development of basic skills, providing children with a range of stimulating and exciting learning experiences.
Through our teaching we aim to:
- enable children to become confident, resourceful, enquiring and independent learners;
- foster children’s self-esteem and help them build positive relationships with other people;
- develop children’s self-respect and encourage children to respect the ideas, attitudes, values and feelings of others;
- show respect for all cultures and, in so doing, to promote positive attitudes towards other people;
- enable children to understand their community and help them feel valued as part of this community;
- help children grow into reliable, independent and positive citizens for the 21st centuryTop of Form
The Foundation Stage
Children in the Nursery (FS1) and Reception (FS2) classes access the Early Years Foundation Stage Curriculum. They engage in learning that is primarily through first hand experiences. The Foundation Curriculum has seven areas of learning. These are: Personal, Social and Emotional Development; Communication and Language; Literacy; Physical Development; Mathematics, Expressive Arts; and Understanding the World.
The children’s progress is regularly assessed by observational assessment so that staff can plan effectively to support the development of each of the children.
Personal, Social and Emotional Development is vital in the Foundation Stage. We support the children in making relationships, building their self-confidence and in managing their feelings and behaviour. This is part of everyday learning but we teach specific skills in circle time.
Children access phonics daily and use the Read Write Inc programme. Children are taught in small groups appropriate to their level of development. By the end of the foundation stage, many children can blend simple words and read simple sentences. In literacy, children apply these skills in an environment rich in language. Activities include role play, drama, small world activities and the reading area. There is a balance between adult and child led learning with the indoors and outdoors used equally to support learning. Children use their phonic skills when learning to read and write. They are supported in small groups in order to develop their basic skills and learn to label pictures, write simple sentences and make lists.
In Physical Development, children learn to move with control and coordination; to balance and climb; to move with agility and self -expression. They learn to hold one- handed tools such as paint brushes and scissors with control and to hold and manipulate a pencil with care. They also are supported in managing their own personal hygiene and are taught what they need to do to stay fit and healthy
Mathematics in the early years focuses on practical skills-for example: counting, ordering, addition and subtraction, doubling, halving and sharing. In the Shape, Space and Measures element there is also an emphasis on play based learning-for example : using simple shapes, measuring and weighing, filling and emptying, whilst using the correct language to describe and compare.
Children are given the opportunity to use their skills imaginatively in Expressive Arts and Design. They are given the opportunity to explore and use media and materials such as paint, clay and construction equipment. Children are also encouraged to be imaginative in areas such as role play, art and dance.
In Understanding the World, children are encouraged to be curious and explore natural materials and learn about living things and compare different places. They learn about technology, using computers and programmable toys and the wider application of technology in everyday life. Moreover they learn about people and communities; how they are different or similar to their friends and they learn about and celebrate other cultures.
Children in the Foundation Stage engage in high quality learning experiences both indoors and outdoors.
Key Stage One & Two
We follow the National Curriculum, which is skills based and appropriate for the age of the children. Our curriculum is topic based, encouraging cross-curricular links between the subjects. Each topic is planned to engage and stimulate the children’s s curiosity and interest through a variety of activities where children learn and apply new skills. There are also learning opportunities provided through out of school trips and visitors.<|endoftext|>
| 3.75 |
976 |
Review question
# Are Alice, Bob and Charlie telling the truth, or lying? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5157
## Solution
1. Alice, Bob, and Charlie make the following statements:
Alice: Bob is lying.
Bob: Charlie is lying.
Charlie: 1 + 1 = 2.
We take ‘lying’ here as meaning, ‘saying something untrue’.
Certainly Charlie is telling the truth.
Bob says Charlie is lying, so he must himself be lying.
Alice says that Bob is lying, which we have just shown to be true.
So Alice and Charlie are telling the truth, and Bob is lying.
A more condensed version of this could read:
Charlie is clearly telling the truth, so Bob is lying, which means Alice is telling the truth.
1. Now Alice, Bob, and Charlie make the following statements:
Alice: Bob is telling the truth.
Bob: Alice is telling the truth.
Charlie: Alice is lying.
What are the possible numbers of people telling the truth? Explain your answer.
Let’s introduce some notation: write, for example, TFT to describe the situation where Alice is telling the truth, Bob is lying, and Charlie is telling the truth.
Looking at the statements above, Bob and Charlie contradict one another, but one of them must be telling the truth (Alice is either telling the truth or lying!).
So either Bob or Charlie is telling the truth, and the other is lying. This narrows down our possible situations to TFT,FFT, TTF, FTF.
From the first and second statements, Alice and Bob must both be telling the truth, or both lying (if, for example, Alice is telling the truth and Bob is not, we reach a contradiction).
This narrows down possible situations further to FFT and TTF. In each of these cases, all three are consistent, so either one or two people are lying.
#### An alternative version without notation
They cannot all be telling the truth, and they cannot all be lying. Bob and Charlie must be opposites.
Alice and Bob telling the truth and Charlie lying is consistent, as is Alice and Bob lying and Charlie telling the truth.
So there can be either one or two people telling the truth.
1. They now make the following statements:
Alice: Bob and Charlie are both lying.
Bob: Alice is telling the truth or Charlie is lying (or both).
Charlie: Alice and Bob are both telling the truth.
Who is telling the truth and who is lying on this occasion? Explain your answer.
Looking at Charlie’s statement, either Charlie is telling the truth, and so everyone is telling the truth, or Charlie is lying, and either Alice or Bob (or both) is lying.
So we have narrowed down our possibilities to TTT, FTF, TFF, FFF.
Looking at Alice’s statement, if Alice is telling the truth, Bob and Charlie are not, ruling out TTT.
If Alice is lying, either Bob or Charlie (or both) are telling the truth, ruling out FFF. So we are only left with FTF and TFF.
Looking at Bob’s statement, if Bob is lying then Alice is lying and Charlie is telling the truth. This rules out TFF.
This leaves FTF, which is consistent with Bob’s statement, and is therefore the correct solution. Hence Bob is telling the truth, and Alice and Charlie are lying.
A condensed version of this argument could read, if Alice is telling the truth, then Bob’s statement is also true, contradicting Alice.
So Alice is lying, which makes Bob’s statement true, and Charlie’s statement a lie.
The form of “or” used in Bob’s statement here, which includes the possibility of “both”, is the usual mathematical/logical definition of the term “or”.
If we have a statement $S$ of the form “$A$ is true or $C$ is true”, and we decide $S$ is false, this is the same as saying that the statement $T$ which says that “$A$ is false and $C$ is false” IS true.
If you know a little about logical notation, we can write this rule as $\neg(A\vee C) \iff (\neg A) \wedge (\neg C),$ where $\neg =$ NOT, and $\wedge =$ AND, and $\vee =$ OR.
This is known as one of de Morgan’s laws—the other is $\neg(A \wedge C) \iff (\neg A)\vee (\neg C).$
What this rule tells us in this case is that if Bob’s statement is false, then both Alice is not telling the truth and Charlie is not lying.<|endoftext|>
| 4.5 |
1,435 |
There were some events earlier in history that are related to matters of infection control, but it was in the early 1800s that things really began to pop. That’s when it began to be recognized that clean healthcare worker (HCW) hands equaled lower infection rates.
As early as 1822, a French pharmacist noted that chloride solutions got rid of the nasty smells that lingered on hands after handling corpses. He also urged HCWs who came in contact with patients who had infectious diseases to use liquid chloride solutions to disinfect their hands. In 1843, Oliver Wendell Holmes touted the benefits of handwashing after he concluded childbed (puerperal) fever was spread by the hands of HCWs. In 1846, Ignaz Semmelweis observed that students and physicians carried with them a foul odor picked up in the autopsy suite that persisted despite hand washing. When Semmelweis insisted that students and physicians disinfect their hands with a chlorine solution between working patient to patient and patient to cadaver, mortality rates dropped, providing evidence that use of an antiseptic agent lowered rates of infection more effectively than handwashing alone.
Discovery of microorganisms and the acceptance of the germ theory in the late 1800s only reinforced the reasonableness of the trend toward clean HCW hands. Clean HCW hands, to this day, continue to be at the core of infection control and prevention.
Discovery of the importance of clean hands and discovery of antibiotics are two of the most important factors in infection control as we know it today. The discovery of antibiotics to control spread of pathogenic organisms was considered nothing short of a miracle; in fact, antibiotics were often referred to as miracle drugs early on.
Penicillin, while discovered in 1928, did not begin being used as a therapeutic agent to treat infections in humans until the 1940s. Initially, penicillin was held by the government and became known to the public following a devastating fire at the Cocoanut Grove in Boston where hundreds of people died. The 200 survivors were, essentially, members of an historic clinical trial for this unique drug. Penicillin was so successful in treating infections such as Staphylococcus, common in skin wounds of burn patients, that it launched a quest to find more antibiotics.
Unfortunately, with antibiotics arose the issue of resistance, an enormous problem in the healthcare environment today and one predicted by penicillin’s discoverer, British bacteriologist Alexander Fleming. In a 1945 interview with The New York Times, Fleming predicted that misuse of penicillin could lead to selection and propagation of mutant resistant bacteria. In the time since antibiotic use has become common, most bacteria that were previously susceptible to antibiotics are now resistant to at least some antibiotics, and in some cases, to many different ones. We will talk about more about antibiotic resistance in an upcoming blog.
Another milestone in infection control was the organization of the Communicable Disease Center (CDC) in Atlanta, Georgia, in 1946. It sprang from the Malaria Control in War Areas program, which was responsible for keeping the southeastern states malaria-free during World War II and, for approximately 1 year, from murine typhus fever. In 1970, it was once again renamed, the Centers for Disease Control and Prevention, to reflect its mission and activities more accurately.
When CDC was born, the world was under threat of infections we hardly give a passing thought to these days. CDC played a large role in the eradication of smallpox, for instance. Later, in the 1970s and 1980s, CDC gained attention by tracking down emerging infectious diseases such as Legionnaire’s disease, toxic-shock syndrome, and HIV.
In the 1980s, gloves became used increasingly in response to the establishment of Universal Precautions, now called Standard Precautions, in the Bloodborne Pathogen Standard (OSHA 29CFR 1910.1030 ). With the prevalence of HIV, hepatitis B, and hepatitis C came increased need for HCWs to protect themselves against exposure to bloodborne infections during procedures such as blood draws and insertion of central venous catheters.
In the 1970s and 1980s we began to see formal guidelines published by organizations such as CDC, the Society for Healthcare Epidemiology of America, established in 1980, and the Association for Professionals in Infection Control and Epidemiology, Inc, established in 1972, on a wide variety of issues including but not limited to handwashing, disinfection and sterilization, prevention of intravascular catheter-related Infections, isolation precautions to prevent transmission of infectious diseases, and environmental infection control in healthcare facilities.
It seems incredible to think that, much as the importance of hand washing was once unrecognized, the environment as a source of cross-transmission was very much underappreciated until fairly recently. In a 2013 study published in the Current Opinion in Infectious Diseases, Weber et al noted, “Contact with the contaminated environment by healthcare personnel is equally as likely as direct contact with a patient to lead to contamination of the healthcare provider's hands or gloves that may result in patient-to-patient transmission of nosocomial pathogens.”
Over the last decade, a new cross-contamination point has been discovered: computer keyboards. In fact, it’s become such an issue that specialized medical keyboards are made for hospitals, and HCWs typically have a policy in place to disinfect them. The need for safe and easy keyboard disinfection is met by Seal Shield. Seal Shield medical keyboards, medical mice, and remotes are 100% waterproof. They can be cleaned with hospital-grade disinfectants and even soaked in a bleach solution. Seal Shield waterproof technology, by enabling the user to clean the keyboard, helps protect the user from cross contamination infections. Read more about medical keyboards.
In addition to computer keyboards, high-touch electronics such as touch screens, mobile tablets, and cell phones need disinfection, too; however, these delicate electronics are extremely difficult to clean without harming them. One hi-tech solution is UVC disinfection technology. The germicidal wavelength of UVC light disrupts the DNA of bacteria, viruses, and protozoa, rendering them incapable of reproducing and infecting. Read more about UV disinfection.
Products such as these contribute to the trend toward prevention rather than just control of infection. They help further the goal of “zero tolerance.” It was once thought that the best healthcare facilities could do was to hold down the number of infections. That is no longer acceptable. Facilities are aiming for zero infections.
No doubt, the Centers for Medicare and Medicaid Services have contributed to the goal of zero infections—and attention directed toward developing new products designed to thwart infection—by refusing to pay for hospital-acquired infections. Infections are so very costly, and, if you want to get a medical facility’s attention, one sure way is to hit the pocketbook. More on that in our next blog.
Susan Cantrell, ELS
Infection Control Corner<|endoftext|>
| 3.953125 |
393 |
# Angle of Rotational Symmetry
In this chapter we will look at the concept of Rotational symmetry with solved examples.
Let us first review the basics of Rotational Symmetry.
## What is Rotational Symmetry ?
An object which after rotation produce the same image and can coincide with each other is said to be in rotational symmetry.
For example, consider the below rectangle.
Let us rotate the rectangle by 90 degree clockwise.
Note that image (ii) and (iv) match the original object.
The above image shows that when we rotate the rectangle by 180 degree we get the same image.
Hence, rectangle has rotational symmery.
## What is Angle of Rotational symmetry ?
It’s the angle of rotation at which the identical image of the given object is produced.
Hence, it is the smallest angle of rotation in which the object can be rotated to coincide with itself.
For example, consider the rectangle used above.
We have seen that of we rotate the rectangle by 180 degree, we will get the same image.
Let us now look at some other example of angle of rotational symmetry.
Example 02
Rotational symmetry of equilateral triangle
Consider the below equilateral triangle.
If we rotate the equilateral triangle by 120 degree, we will get the same image.
Hence, the rotational symmetry for equilateral triangle is 120 degree.
You can consecutively rotate the above equilateral triangle and all the time you will get the same image.
Example 03
Rotational symmetry for Hexagon.
Consider the below regular hexagon with all equal sides.
If we rotate the hexagon by 60 degree we get the same image which can coincide each other.
Hence, the angle of rotational symmetry for hexagon is 60 degree.
Example 04
Consider the below triangle ABC
The given triangle does not have any angle of rotational symmetry as we can’t get the same image after the rotation.<|endoftext|>
| 4.75 |
197 |
in this topic you will learn about what a solution is and be able to name examples of solutions. You will find out about what substances are soluble in different solvents. You will do experiments to separate substances from their solvents.
Here are the learning objectives for this topic
o Name soluble and insoluble substances
o Use the words solute, solvent, solution
o Make a saturated solution
o Separate inks
o Separate sand and water
o Draw the particles of a solution
o That when a solvent cannot dissolve any more solute it is saturated
o That saturated solutions can form crystals, like in kidney stones
o How chromatography works
o When we can use distillation and evaporation to separate solutes and solvents
o That mass does not change when we add solute and solvent together
o That the volume can change when a solution is made
useful websites for 7h
Click here to edit.<|endoftext|>
| 4.3125 |
703 |
What’s Normal And What’s Not
As adults, we often feel that it is easy to say all of the speech sounds correctly, and that young children who don’t do so might be lazy, or have just developed bad habits. However speech sound production is a very complex process. It requires very specific coordination of movements of the tongue, teeth, lips and other muscles. It also requires the brain to send the correct messages to these speech organs.
Speech sounds develop gradually as children grow and develop in other areas. Many sounds are difficult for young children to produce and they may have difficulty with some sounds, even after they enter junior kindergarten. Children might be able to produce some sounds on their own, but will have difficulty, if these sounds are produced together with other sounds. The order of sounds in a word will affect how easy or difficult they are for children. This can sometimes be frustrating for parents who feel that the child can produce the sound but doesn’t always do so.
The most common challenging sounds for children under 3, include, /th/, /l/, /k/, /f/, /s/, /r/. At this age, difficulty with sounds is normal and some children may have difficulty with sounds not listed above.
After the age of three, we might see continued difficulties with the production of /s/, /r/, /th/, /sh/, and /ch/. After three, we do expect that children will not drop sounds, but might substitute one sound for another. Therefore, we might hear the word “htop” for “stop”, but would not expect the child to say “top” for “stop”, dropping the first sound in the word. This reflects an earlier stage of development.
At three, in spite of some of these speech sound difficulties, all children should be understandable to even unfamiliar listeners. If this is not the case, and if they are dropping sounds rather than substituting sounds, a referral to tykeTALK would be appropriate. You can refer your child to tykeTALK by completing the Online Referral Form or by calling 519-663-0273.
How can you help your child when they are having trouble with speech sounds?
When your child produces a sound incorrectly in a word, repeat the word back, saying the sound correctly while emphasizing it (eg. Your child says: “I am free years old.” . You reply: “Yes, you are three years old.” . Your child says: “There’s a top sign.” You reply: “Yes, I see the stop sign.”).
Don’t ask your child to repeat sounds after you, unless your child is seeing a Speech-Language Pathologist and you have been shown how to do this. For some children, this may work, but for many it causes frustration and may make them feel self-conscious about their speech. Remember, speech production is very complex.
Always say the sound correctly for your child. At any age, children should not be hearing “baby talk”. They should always hear speech produced clearly and correctly. Although some of the speech of very young children may sound cute, we do not want to encourage this, either by imitating it or reinforcing it. Your ultimate goal is for your child to speak clearly and confidently.<|endoftext|>
| 3.828125 |
936 |
2001 IMO Shortlist Problems/G6
Problem
Let $ABC$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ (or extensions thereof) in $D$, $E$, $F$, respectively. Suppose further that the areas of triangles $PBD$, $PCE$, $PAF$ are all equal. Prove that each of these areas is equal to the area of triangle $ABC$ itself.
Solution
Solution 1 by Mewto55555:
We use barycentric coordinates.
So $A$ is $(1,0,0)$, $B$ is $(0,1,0)$, $C$ is $(0,0,1)$, and $P$ is $(p,q,r)$, with $p+q+r=1$.
Now, the equation of line $AP$ is just the line $qz=ry$, $BP$ is just $pz=rx$, and $CP$ is $qx=py$.
Also, $AB$ is just $z=0$, $BC$ is $x=0$, and $AC$ is $y=0$.
Thus, the coordinates of $D$ is $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$. Similarly, $E$ is at $\left(\frac{p}{p+r},0,\frac{r}{p+r}\right)$ and $F$ is at $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$
Now, the ratio $[PBD]$ to $[ABC]$ is just
$\begin{vmatrix} p & 0 & 0 \\ q & 1 & \frac{q}{q+r}\\ r & 0 & \frac{r}{q+r} \end{vmatrix}= \frac{pr}{q+r}$
The other ratios are similarly $\frac{pq}{p+r}$ and $\frac{qr}{p+q}$
Since $p+q+r=1$, we have $\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K$ and we want to show that $|K|=1$.
Thus, we have $\frac{pqr}{p(1-r)}=\frac{pqr}{r(1-q)}=\frac{pqr}{q(1-p)}$.
Since none of $p,q,r=0$ (else $P$ would be on one of the sides of $ABC$):
$p(1-r)=r(1-q)=q(1-p)$.
We know $r=1-p-q$. Substuting:
$p^2+pq=1-p-2q+pq+q^2=q-pq$.
From the first and third, we get that $q(1-2p)=p^2 \implies q=\frac{p^2}{1-2p}$
Now consider first and second;
$p^2+p-1=q^2-2q$
Subbing back in $q$:
$(p^2+p-1)(1-2p)^2=p^4-2p^2(1-2p)$
which rearranges to
$0=3p^4-4p^3-5p^2+5p-1=(3p-1)(p^3-p^2-2p+1)=0$
If $p=\frac{1}{3}$, then $q=r=\frac{1}{3}$, so $P$ is in the triangle (as all of $p,q,r>0$) contradiction.
Thus, we have $p^3-p^2-2p+1=0$
So, $1-2p=p^2(1-p) \implies q=\frac{p^2}{1-2p}=\frac{1}{1-p}$
Thus, $K=\frac{pq}{1-q}=\frac{\frac{p}{1-p}}{1-\frac{1}{1-p}}=\frac{p}{1-p-1}=-1$
Therefore, if $[PBD]=[PCE]=[PAF]$, necessarily $[PBD]=[PCE]=[PAF]=[ABC]$.<|endoftext|>
| 4.5 |
586 |
Addition practice and critical thinking skills can be a tough sell for kids. But these detective themed Math Mysteries – FUN Addition and Subtraction Math Problems are a great way to work on both! Figure out the mystery numbers using addition and subtraction skills as well as knowledge of even and odd math worksheets
- 34 Math Games for Kids
- Make Kindergarten Math FUN
- 1st Grade Math Worksheets
- 2nd Grade Math Worksheets
- Deck of Cards Addition Worksheets
- Free Deck of Cards Subtraction Worksheets
- Gumball Math Worksheets
- Solve & Stamp Mystery Pictures (Addition & Subtraction)
- Addition Tic Tac Toe Math
Math Mysteries – FUN Addition and Subtraction Math Problems
What Number Am I Detective Math
It’s time to pretend to be a detective and find these mystery numbers! Use the clues to figure out which number belongs on each card! This Math Mysteries – FUN Addition and Subtraction Math Problems is sure to make math fun.
There are two sets of these cards – one for addition under 10 and one for addition under 20. Each card includes a clue based on even and odd numbers and 2 clues that include either addition or subtraction.
There is also a recording sheet included for use with a group of kids.
How to Set Up the Mystery Math Cards
If you’re going to use the set in a group, I’d recommend laminating the cards. This way kids can use dry erase markers on the cards but they can still be reused. Have them write their final answer on a recording sheet.
Otherwise, the cards can be used on plain paper for just one child.
How to Use the Detective Math Mystery Cards
These cards rely on logic to figure out the answer, but the answer can be found purely through using math.
Go through each clue and eliminate as many numbers as possible until the final number is revealed.
The first clue is always indicating whether the number is even or odd. So start by crossing off any number that is excluded.
Keep moving through the cards until there is just one number left. That is the mystery number!
My kids enjoyed acting like detectives while doing their math (and I totally keep magnifying glasses on hand just for occasions like this!)
For more detective math ideas: Detective Math with Invisible Ink
Download Math Mysteries – FUN Addition and Subtraction Math Problems
Before you download your free pack you agree to the following:
- This set is for personal and classroom use only.
- This printable set may not be sold, hosted, reproduced, or stored on any other website or electronic retrieval system.
- Graphics Purchased and used with permission from
- All downloadable material provided on this blog is copyright protected.<|endoftext|>
| 3.78125 |
538 |
Samsung Advanced Institute of Technology (SAIT), the core R&D incubator for Samsung Electronics, has developed a new transistor structure utilizing graphene. Current semiconductors consist of billions of silicon transistors. To increase the performance of a semiconductor used today, you need to either reduce the size of individual transistors to shorten the traveling distance of electrons, or use a material with higher electron mobility which allows for faster electron velocity.
Graphene possesses electron mobility about 200 times higher then silicon, and a switch to this material could be an alternative path for making faster and more energy efficent devices as it allows faster electron velocity. An issue with using graphene is however that unlike the conventional semiconducting materials, current cannot be switched off because the material is semi-metallic. This has long been a key issue, preventing graphene transistors to be realized.
The industry have for a long time tried to increasing performance in existing semiconductors by reducing the size, but as experts now believe we are nearing the potential limits of scaling down a new solution is required. Alternative materials like graphene has therefore been explored for a long time. Since both on and off flow of current is required in a transistor to represent “1” and “0” of digital signals, using graphene as a solution have been difficult to accomplish because of the materials semi-metallic properties.
Previous research have tried to convert graphene into a semi-conductor, but this approach radically decrease the mobility of graphene and in large remove it’s major benefit over silicone. By re-engineering the basic operating principles of digital switches, Samsung Advanced Institute of Technology has managed to develop a device that can switch off the current in graphene without degrading its mobility.
The institute has solved the most difficult problem in graphene device research and has opened the door to new directions for future studies. The solution uses a graphene-silicon Schottky barrier that can switch current on or off by controlling the height of the barrier. The device demonstrating this approach was named Barristor, after its barrier-controllable feature.
A Schottky Barrier is a potential energy barrier formed at a metal-semiconductor interface. It prevents an electric charge to flow from metal to silicon. The Schottky Barrier is named after a German physicist Walter H Schottky. Generally a metal-semiconductor junction would have fixed work function and Schottky barrier height, but as for graphene the Schottky barrier height can be controlled through the work function.
Samsung Advanced Institute of Technology owns 9 major patents related to the structure and the operating method of the Graphene Barristor.<|endoftext|>
| 3.671875 |
380 |
Terms in this set (19)
The building blocks of rocks - a solid, inorganic substance that occurs in nature (not man-made). Has a definite chemical composition and atom structure.
The color of a mineral in its powdered form.
The way light reflects off a mineral.
The ability to scratch an object or be scratched by an object
A property where minerals break into pieces that have smooth, flat surfaces.
A property where minerals break into pieces that have uneven or curved surfaces.
Mohs Hardness Scale
A scale used to rank a mineral from 1 to 10.
A substance that does not contain matter that was once alive.
An expert in the field of geology, the study of what the Earth is made of and how it was formed.
A mineral that can be cut and polished for use in jewelry.
A round rock that contains a hollow cavity lined with crystals on the inside.
The process by which atoms form a solid and arrange themselves in an orderly, repeating pattern (6 main types)
A piece of white, hard plate (called porcelain)
A solid in which the atoms are arranged in a certain pattern that repeats over and over again
A characteristic in some minerals that is very unique; for example, fluorite is green, sulfur is yellow - not always reliable because the same mineral can come in different colors.
Two substances mixed together evenly
Substance that gets dissolved
Substance that does the dissolving
the process of dissolving
YOU MIGHT ALSO LIKE...
MCAT Organic Chemistry | Kaplan Guide
chapter 3 minerals
ALL OF CHAPTER 4
OTHER SETS BY THIS CREATOR
Figurative Language, Figurative Language
Tech Systems no pix
THIS SET IS OFTEN IN FOLDERS WITH...
Chapter 5 Weathering and Soil<|endoftext|>
| 3.828125 |
14,478 |
Chapter 6. Linear Transformation. 6.1 Intro. to Linear Transformation
Save this PDF as:
Size: px
Start display at page:
Transcription
1 Chapter 6 Linear Transformation 6 Intro to Linear Transformation Homework: Textbook, 6 Ex, 5, 9,, 5,, 7, 9,5, 55, 57, 6(a,b), 6; page 7- In this section, we discuss linear transformations 89
2 9 CHAPTER 6 LINEAR TRANSFORMATION Recall, from calculus courses, a funtion f : X Y from a set X to a set Y associates to each x X a unique element f(x) Y Following is some commonly used terminologies: X is called the domain of f Y is called the codomain of f If f(x) = y, then we say y is the image of x The preimage of y is preimage(y) = {x X : f(x) = y} 4 The range of f is the set of images of elements in X In this section we deal with functions from a vector sapce V to another vector space W, that respect the vector space structures Such a function will be called a linear transformation, defined as follows Definition 6 Let V and W be two vector spaces A function T : V W is called a linear transformation of V into W, if following two prperties are true for all u,v V and scalars c T(u+v) = T(u)+T(v) (We say that T preserves additivity) T(cu) = ct(u) (We say that T preserves scalar multiplication) Reading assignment Read Textbook, Examples -, p 6- Trivial Examples: Following are two easy exampes
3 6 INTRO TO LINEAR TRANSFORMATION 9 Let V,W be two vector spaces Define T : V W as T(v) = for all v V Then T is a linear transformation, to be called the zero transformation Let V be a vector space Define T : V V as T(v) = v for all v V Then T is a linear transformation, to be called the identity transformation of V 6 Properties of linear transformations Theorem 6 Let V and W be two vector spaces Suppose T : V W is a linear transformation Then T() = T( v) = T(v) for all v V T(u v) = T(u) T(v) for all u,v V 4 If v = c v + c v + + c n v n then T(v) = T (c v + c v + + c n v n ) = c T (v )+c T (v )+ +c n T (v n ) Proof By property () of the definition 6, we have T() = T() = T() =
4 9 CHAPTER 6 LINEAR TRANSFORMATION So, () is proved Similarly, T( v) = T(( )v) = ( )T(v) = T(v) So, () is proved Then, by property () of the definition 6, we have T(u v) = T(u + ( )v) = T(u) + T(( )v) = T(u) T(v) The last equality follows from () So, () is proved To prove (4), we use induction, on n For n = : we have T(c v ) = c T(v ), by property () of the definition 6 For n =, by the two properties of definition 6, we have T(c v + c v ) = T(c v ) + T(c v ) = c T(v ) + c T(v ) So, (4) is prove for n = Now, we assume that the formula (4) is valid for n vectors and prove it for n We have T (c v + c v + + c n v n ) = T (c v + c v + + c n v n )+T (c n v n ) = (c T (v ) + c T (v ) + + c n T (v n )) + c n T(v n ) So, the proof is complete 6 Linear transformations given by matrices Theorem 6 Suppose A is a matrix of size m n Given a vector v = v v Rn define T(v) = Av = A v v v n v n Then T is a linear transformation from R n to R m
5 6 INTRO TO LINEAR TRANSFORMATION 9 Proof From properties of matrix multiplication, for u,v R n and scalar c we have and T(u + v) = A(u + v) = A(u) + A(v) = T(u) + T(v) The proof is complete T(cu) = A(cu) = cau = ct(u) Remark Most (or all) of our examples of linear transformations come from matrices, as in this theorem Reading assignment Read Textbook, Examples -, p 65-6 Projections along a vector in R n Projections in R n is a good class of examples of linear transformations We define projection along a vector Recall the definition 56 of orthogonal projection, in the context of Euclidean spaces R n Definition 64 Suppose v R n is a vector Then, for u R n define proj v (u) = v u v v Then proj v : R n R n is a linear transformation Proof This is because, for another vector w R n and a scalar c, it is easy to check proj v (u+w) = proj v (u)+proj v (w) and proj v (cu) = c (proj v (u)) The point of such projections is that any vector u R n can be written uniquely as a sum of a vector along v and another one perpendicular to v: u = proj v (u) + (u proj v (u))
6 94 CHAPTER 6 LINEAR TRANSFORMATION It is easy to check that (u proj v (u)) proj v (u) Exercise 65 (Ex 4, p 7) Let T(v,v,v ) = (v + v, v v,v v ) Compute T( 4, 5, ) Solution: T( 4, 5, ) = ( ( 4)+5, 5 ( 4), 4 ) = (,, 5) Compute the preimage of w = (4,, ) Solution: Suppose (v,v,v ) is in the preimage of (4,, ) Then (v + v, v v,v v ) = (4,, ) So, v +v = 4 v v = v v = The augmented matrix of this system is 4 its Gauss Jordan f orm So, Preimage((4,, )) = {(574, 8574, 574)}
7 6 INTRO TO LINEAR TRANSFORMATION 95 Exercise 66 (Ex p 7) Determine whether the function T : R R T(x,y) = (x,y) is linear? Solution: We have T((x,y) + (z,w)) = T(x + z,y + w) = ((x + z),y + w) (x,y) + (z,w) = T(x,y) + T(z,w) So, T does not preserve additivity So, T is not linear Alternately, you could check that T does not preserve scalar multiplication Alternately, you could check this failure(s), numerically For example, T((, ) + (, )) = T(, ) = (9, ) T(, ) + T(, ) Exercise 67 (Ex 4, p 7) Let T : R R be a linear transformation such that T(,, ) = (, 4, ), T(,, ) = (,, ), T(,, ) = (,, ) Compute T(, 4, ) Solution: We have (, 4, ) = (,, ) + 4(,, ) (,, ) So, T(, 4, ) = T(,, )+4T(,, ) T(,, ) = (, 4, )+(,, )+(,, ) = (, 5, ) Remark A linear transformation T : V V can be defined, simply by assigning values T(v i ) for any basis {v,v,,v n } of V In this case of the our problem, values were assigned for the standard basis {e,e,e } of R
8 96 CHAPTER 6 LINEAR TRANSFORMATION Exercise 68 (Ex 8, p 7) Let 4 A = Let T : R 5 R be the linear transformation T(x) = Ax Compute T(,,,, ) Solution: T(,,,, ) = 4 = 7 5 Compute preimage, under T, of (, 8) Solution: The preimage consists of the solutions of the linear system 4 x x x x 4 = 8 The augmented matrix of this system is 4 8 x 5 The Gauss-Jordan form is
9 6 INTRO TO LINEAR TRANSFORMATION 97 We use parameters x = t,x 4 = s,x 5 = u and the solotions are given by x = 5 + t + 5s + 4u,x = t,x = 4 + 5s,x 4 = s,x 5 = u So, the preimage T (, 8) = {(5+t+5s+4u, t, 4+5s, s, u) : t,s,u R} Exercise 69 (Ex 54 (edited), p 7) Let T : R R be the linear transformation such that T(, ) = (, ) and T(, ) = (, ) Compute T(, 4) Solution: We have to write (, 4) = a(, )+b(, ) Solving (, 4) = 5(, ) 5(, ) So, T(, 4) = 5T(, ) 5T(, ) = 5(, ) 5(, ) = (, 5) Compute T(, ) Solution: We have to write (, ) = a(, )+b(, ) Solving (, ) = 5(, ) 5(, ) So, T(, ) = 5T(, ) 5T(, ) = 5(, ) 5(, ) = (, ) Exercise 6 (Ex 6 (edited), p 7) Let T : R R the projection T(u) = proj v (u) where v = (,, )
10 98 CHAPTER 6 LINEAR TRANSFORMATION Find T(x,y,z) Solution: See definition 64 proj v (x,y,z) = v (x,y,z) v v = (,, ) (x,y,z) (,, ) ( x + y + z = Compute T(5,, 5) (,, ) = x + y + z (,, ), x + y + z, x + y + z Solution: We have ( x + y + z T(5,, 5) =, x + y + z, x + y + z ) ( =,, ) Compute the matrix of T ) Solution: The matrix is given by A =, because T(x,y,z) = x y z
11 6 KERNEL AND RANGE OF LINEAR TRANSFORMATION99 6 Kernel and Range of linear Transformation We will essentially, skip this section Before we do that, let us give a few definitions Definition 6 Let V,W be two vector spaces and T : V W a linear transformation Then the kernel of T, denoted by ker(t), is the set of v V such that T(v) = Notationally, ker(t) = {v V : T(v) = } It is easy to see the ker(t) is a subspace of V Recall, range of T, denoted by range(t), is given by range(t) = {v W : w = T(v) for some v V } It is easy to see the range(t) is a subspace of W We say the T is isomorphism, if T is one-to-one and onto It follows, that T is an isomorphism if ker(t) = {} and range(t) = W
12 CHAPTER 6 LINEAR TRANSFORMATION 6 Matrices for Linear Transformations Homework: Textbook, 6, Ex 5, 7,,, 7, 9,,, 5, 9,,, 5(a,b), 7(a,b), 9, 4, 45, 47; p 97 Optional Homework: Textbook, 6, Ex 57, 59; p 98 (We will not grade them) In this section, to each linear transformation we associate a matrix Linear transformations and matrices have very deep relationships In fact, study of linear transformations can be reduced to the study of matrices and conversely First, we will study this relationship for linear transformations T : R n R m ; and later study the same for linear transformations T : V W of general vector spaces In this section, we will denote the vectors in R n, as column matrices Recall, written as columns, the standard basis of R n is given by B = {e,e,,e n },,, Theorem 6 Let T : R n R m be a linear transformation Write T(e ) = a a,t(e ) = a a,,t(e n) = a n a n a m a m a mn
13 6 MATRICES FOR LINEAR TRANSFORMATIONS These columns T(e ),T(e ),T(e n ) form a m n matrix A as follows, a a a n A = a a a n a m a m a mn This matrix A has the property that T(v) = Av for all v R n This matrix A is called the standard matrix of T Proof We can write v R n as v v = v = v e + v e + + v n e n v n We have, Av = a a a n a a a n a m a m a mn v v v n = v a a a m +v a a a m + +v n = v T (e )+v T (e )+ +v n T (e n ) = T (v e + v e + + v n e n ) = T(v) The proof is complete Reading assignment: Read Textbook, Examples,; page 89-9 In our context of linear transformations, we recall the following definition of composition of functions Definition 6 Let a n a n a mn T : R n R m, T : R m R p
14 CHAPTER 6 LINEAR TRANSFORMATION be two linear transformations Define the composition T : R n R p of T with T as T(v) = T (T (v)) for v R n The composition T is denoted by T = T ot Diagramatically, R n T R m T=T ot T R p Theorem 6 Suppose T : R n R m, T : R m R p are two linear transformations Then, the composition T = T ot : R n R p is a linear transformation Suppose A is the standard matrix of T and A is the standard matrix of T Then, the standard matrix of the composition T = T ot is the product A = A A Proof For u,v R n and scalars c, we have T(u+v) = T (T (u+v)) = T (T (u)+t v)) = T (T (u))+t T (v)) = T(u)+T(v) and T(cu) = T (T (cu)) = T (ct (u)) = ct (T (u)) = ct(u) So, T preserves addition and scalar multiplication Therefore T is a linear transformation and () is proved To prove (), we have T(u) = T (T (u)) = T (A u) = A (A u) = (A A )u
15 6 MATRICES FOR LINEAR TRANSFORMATIONS Therefore T(e ) is the first column of A A, T(e ) is the second column of A A, and so on Therefore, the standard matrix of T is A A The proof is complete Reading assignment: Read Textbook, Examples ; page 9 Definition 64 Let T : R n R n, T : R n R n be two linear transformations such that for every v R n we have T (T (v)) = v and T (T (v)) = v, then we say that T is the inverse of T, and we say that T is invertible Such an inverse T of T is unique and is denoted by T (Remark Let End(R n ) denote the set of all linear transformations T : R n R n Then End(R n ) has a binary operation by composition The identity operation I : R n R n acts as the identitiy under this composition operation The definition of inverse of T above, just corresponds to the inverse under this composition operation) Theorem 65 Let T : R n R n be a linear transformations and let A be the standard matrix of T Then, the following are equivalent, T is invertible T is an isomorphism A is invertible And, if T is invertible, then the standard matrix of T is A
16 4 CHAPTER 6 LINEAR TRANSFORMATION Proof (First, recall definition 6, that T is isomorphism if T is to and onto) (Proof of () () :) Suppose T is invertible and T be the inverse Suppose T(u) = T(v) Then, u = T (T(u)) = T (T(v)) = v So, T is to Also, given u R n we have So, T is an isomorphism u = T(T (u)) So, T onto R n (Proof of () () :) So, assume T is an isomorphism Then, Ax = T(x) = T() x = So, Ax = has an unique solution Therefore A is invertible and () follows from () (Proof of () () :) Suppose A is invertible Let T (x) = A x, then T is a linear transformation and it is easily checked that T is the inverse of T So, () follows from () The proof is complete Reading assignment: Read Textbook, Examples 4; page 9 6 Nonstandard bases and general vector spaces The above discussion about (standard) matrices of linear transformations T had to be restricted to linear transformations T : R n R m This wa sbecause R n has a standard basis {e,e,,e n, } that we could use Suppose T : V W is a linear transformation between two abstract vector spaces V,W Since V and W has no standard bases, we cannot associate a matrix to T But, if we fix a basis B of V and B of W we can associate a matrix to T We do it as follows
17 6 MATRICES FOR LINEAR TRANSFORMATIONS 5 Theorem 66 Suppose T : V W is a linear transformation between two vector spaces V,W Let B = {v,v,,v n } be basis of V and B = {w,w,,w m } be basis of w We can write T(v ) = w w w m a a a m Writing similar equations for T(v ),,T(v n ), we get T(v ) T(v ) T(v n ) = w w w m a a a n a a a n a m a m a mn Then, for v = x v + x v + + x n v n, We have a a a n T(v) = w w w m a a a n a m a m a mn Write A = a ij Then, T is determined by A, with respect to bases B,B x x x n Exercise 67 (Ex 6, p 97) Let T(x,y,z) = (5x y + z, z + 4y, 5x + y)
18 6 CHAPTER 6 LINEAR TRANSFORMATION What is the standard matrix of T? Solution: We have T : R R We write vectors x R as columns x x = y instead of (x, y, z) Recall the standard basis e = We have T(e ) = z, e = 5 5, T(e ) =, e = So, the standard matrix of T is 5 A = Exercise 68 (Ex, p 97) Let of R, T(e ) = T(x,y,z) = (x + y, y z) Write down the standard matrix of T and use it to find T(,, ) Solution: In this case, T : R R With standard basis e,e,e as in exercise 67, we have T(e ) =, T(e ) =, T(e ) =
19 6 MATRICES FOR LINEAR TRANSFORMATIONS 7 So, the standard matrix of T is A = Therefore, T(,, ) = A = = 4 We will write our answer in as a row: T(,, ) = (, 4) Exercise 69 (Ex 8, p 97) Let T be the reflection in the line y = x in R So, T(x,y) = (y,x) Write down the standard matrix of T Solution: In this case, T : R R With standard basis e = (, ) T,e = (, ) T, we have T(e ) =, T(e ) = So, the standard matrix of T is A = Use the standard matrix to compute T(, 4) Solution: Of course, we know T(, 4) = (4, ) They want use to use the standard matrix to get the same answer We have 4 T(, 4) = A = = or T(, 4) = (4, ) 4 4
20 8 CHAPTER 6 LINEAR TRANSFORMATION Lemma 6 Suppose T : R R is the counterclockwise rotation by a fixed angle θ Then T(x,y) = cos θ sin θ x sin θ cos θ y Proof We can write x = r cos φ, y = r sin φ, where r = x + y, tanφ = y/x By definition T(x,y) = (r cos(φ + θ),r sin(φ + θ)) Using trig-formulas r cos(φ + θ) = r cosφcos θ r sin φ sin θ = x cos θ y sin θ and r sin(φ + θ) = r sin φ cos θ + r cos φ sin θ = y cos θ + x sin θ So, cos θ sin θ T(x,y) = sin θ cos θ The proof is complete x y Exercise 6 (Ex, p 97) Let T be the counterclockwise rotation in R by angle o Write down the standard matrix of T Solution: We use lemma 6, with θ = So, the standard matrix of T is A = cosθ sin θ cos sin = sin θ cos θ sin cos = 5 5
21 6 MATRICES FOR LINEAR TRANSFORMATIONS 9 Compute T(, ) Solution: We have T(, ) = A = 5 5 = + We write our answer as rows: T(, ) = (, + ) Exercise 6 (Ex, p 97) Let T be the projection on to the vector w = (, 5) in R : T(u) = proj w (u) Find the standard matrix Solution: See definition 64 T(x,y) = proj w (x,y) = w (x,y) (, 5) (x,y) x 5y w = (, 5) = (, 5) w (, 5) 6 ( ) x 5y 5x + 5y =, 6 6 So, with e = (, ) T,e = (, ) T we have (write/think everything as columns): T(e ) = So, the standard matrix is Compute T(, ) A =, T(e ) = Solution: We have T(, ) = A = We write our answer in row-form: T(, ) = ( 5, 65 6) =
22 CHAPTER 6 LINEAR TRANSFORMATION Exercise 6 (Ex 6, p 97) Let T(x,y,z) = (x y + z, x y,y 4z) Write down the standard matrix of T Solution: with e = (,, ) T,e = (,, ) T,e = (,, ) T we have (write/think everything as columns): T(e ) =, T(e ) = So, the standard matrix is A = Compute T(,, ) 4, T(e ) = 4 Solution: We have T(,, ) = A = 4 = 7 7 Exercise 64 (Ex 4, p 97) Let T : R R, T (x,y) = (x y, x + y) and T : R R, T (x,y) = (y, ) Compute the standard matrices of T = T ot and T = T T Solution: We solve it in three steps:
23 6 MATRICES FOR LINEAR TRANSFORMATIONS Step-: First, compute the standard matrix of T With e = (, ) T,e = (, ) T we have (write/think everything as columns): T (e ) =, T (e ) = So, the standard matrix of T is A = Step-: Now, compute the standard matrix of T With e = (, ) T,e = (, ) T we have : T (e ) =, T (e ) = So, the standard matrix of T is A = Step- By theorem 6, the standard matrix of T = T T is A A = = So, T(x,y) = (x+y, ) Similarly, the standard matrix of T = T T is A A = = So, T (x,y) = (y, y) Exercise 65 (Ex 46, p 97) Determine whether T(x,y) = (x + y,x y)
24 CHAPTER 6 LINEAR TRANSFORMATION is invertible or not Solution: Because of theorem 65, we will check whether the standard matrix of T is invertible or not With e = (, ) T,e = (, ) T we have : T(e ) =, T(e ) = So, the standard matrix of T is A = Note, deta = 4 So, T is invertible and hence T is invertible Exercise 66 (Ex 58, p 97) Determine whether T(x,y) = (x y,,x + y) Use B = {v = (, ),v = (, )} as basis of the domain R and B = {w = (,, ),w = (,, ),w = (,, )} as basis of codomain R Compute matrix of T with respect to B,B Solution: We use theorem 66 We have T(u ) = T(, ) = (,, ), T(u ) = T(, ) = (,, ) We solve the equation: (,, ) = aw + bw + cw = a(,, ) + b(,, ) + c(,, ) and we have (,, ) = (,, ) (,, )+(,, ) = w w w
25 64 TRANSITION MATRICES AND SIMILARITY Similarly, we solve (,, ) = aw + bw + cw = a(,, ) + b(,, ) + c(,, ) and we have (,, ) = (,, ) (,, ) + (,, ) = w w w So, the matrix of T with respect to the bases B,B is A = 64 Transition Matrices and Similarity We will skip this section I will just explain the section heading You know what are Transition matrices of a linear transformation T : V W They are a matrices described in theorem 66 Definition 64 Suppose A,B are two square matrices of size n n We say A,B are similar, if A = P BP for some invertible matrix P
26 4 CHAPTER 6 LINEAR TRANSFORMATION 65 Applications of Linear Trans Homework: Textbook, 65 Ex (a), (a), 5, 7, 9, 5, 7, 9, 4, 49, 5, 5, 55, 6, 65; page In this section, we discuss geometric interpretations of linear transformations represented by elementary matrices Proposition 65 Let A be a matrix and x T(x,y) = A y We will write the right hand side as a row, which is an abuse of natation If A =, then T(x,y) = A x y = ( x,y) T represents the reflection in y axis See Textbook, Example (a), p47 for the diagram If A =, then T(x,y) = A x y = (x, y) T represents the reflection in x axis See Textbook, Example (b), p47 for the diagram
27 65 APPLICATIONS OF LINEAR TRANS 5 If A =, then T(x,y) = A x y = (y,x) T represents the reflection in line y = x See Textbook, Example (c), p47 for the diagram 4 If A = k, then T(x,y) = A x y = (kx,y) T If k >, then T represents expansion in horizontal direction and < k <, then T represents contraction in horizontal direction See Textbook, Example Fig 6, p49 for diagrams 5 If A = k, then T(x,y) = A x y = (x,ky) T If k >, then T represents expansion in vertical direction and < k <, then T represents contraction in vertical direction See Textbook, Example Fig 6, p49 for diagrams 6 If A = k, then T(x,y) = A x y = (x + ky,y) T Then T represents horizontal shear (Assume k > ) The upperhalf plane are sheared to right and lower-half plane are sheared to left The points on the x axis reamain fixed See Textbook, Example, fig 64, p49 for diagrams
28 6 CHAPTER 6 LINEAR TRANSFORMATION 7 If A = k, then T(x,y) = A x y = (x,kx + y) T Then T represents vertical shear (Assume k > ) The righthalf-plane are sheared to upward and left-half-plane are sheared to downward The points on the y axis reamain fixed See Textbook, Example, fig 65, p49 for diagrams 65 Computer Graphics Linear transformations are used in computer graphics to move figures on the computer screens I am sure all kinds of linear (and nonlinear) transformations are used Here, we will only deal with rotations by an angle θ, around () x axis, () y axis and () z axis as follows: Proposition 65 Suppose θ is an angle Suppose we want to rotate the point (x, y, z) counterclockwise about z axis through an angle θ Let us denote this transformation by T and write T(x,y,z) = (x,y,z ) T Then using lemma 6, we have T(x,y,z) = x y z = cos θ sin θ x sin θ cos θ y z = x cos θ y sin θ x sin θ + y cos θ Similarly, we can write down the linear transformations corresponding to rotation around x axis and y axis We write down the transition matrices for these three matrices as follows: z The standard matrix for this transformation of counterclockwise
29 65 APPLICATIONS OF LINEAR TRANS 7 rotation by an angle θ, about x axis is cosθ sin θ sinθ cosθ The standard matrix for this transformation of counterclockwise rotation by an angle θ, about y axis is cos θ sinθ sin θ cosθ The standard matrix for this transformation of counterclockwise rotation by an angle θ, about z axis is cos θ sin θ sin θ cos θ Reading assignment: Read Textbook, Examples 4 and 5; page 4-4 Exercise 65 (Ex 6, p 44) Let T(x,y) = (x, y) (This is a vertical expansion) Sketch the image of the unit square with vertices (, ), (, ), (, ), (, ) Solution: We have T(, ) = (, ), T(, ) = (, ), T(, ) = (, ), T(, ) = (, )
30 8 CHAPTER 6 LINEAR TRANSFORMATION Diagram:,,,,,, Here, the solid arrows represent the original rectangle and the brocken arrows represent the image Exercise 654 (Ex, p 44) Let T be the reflection in the line y = x Sketch the image of the rectangle with vertices (, ), (, ), (, ), (, ) Solution: Recall, (see Proposition 65 ()) that T(x, y) = (y, x) We have T(, ) = (, ),T(, ) = (, ),T(, ) = (, ),T(, ) = (, ) Diagram:,,,,,,, Here, the solid arrows represent the original rectangle and the brocken arrows represent the image
31 65 APPLICATIONS OF LINEAR TRANS 9 Exercise 655 (Ex 8, p 44) Suppose T is the expansion and contraction represented by T(x,y) = ( x, ) y Sketch the image of the rectangle with vertices (, ), (, ), (, ), (, ) Solution: Recall, (see Proposition 65 ()) that (x, y) = (y, x)we have T(, ) = (, ),T(, ) = (, ),T(, ) = (, ),T(, ) = (, ) Diagram:,,,,, 5,,, Here, the solid arrows represent the original rectangle and thebrocken arrows represent the image Exercise 656 (Ex 44, p 44) Give the geometric description of the linear transformation defined by the elementary matrix A = Solution: By proposition 65 (6) this is a horizontal shear Here, T(x,y) = (x + y,y)
32 CHAPTER 6 LINEAR TRANSFORMATION Exercise 657 (Ex 5 and 54, p 45) Find the matrix of the transformation T that will produce a 6 o rotation about the x axis Then compute the image T(,, ) Solution: By proposition 65 () the matrix is given by A = cos θ sin θ = cos 6 o sin 6 o = sinθ cos θ So, T(,, ) = A = sin 6 o cos 6 o = + Exercise 658 (Ex 64, p 45) Determine the matrix that will produce a 45 o rotation about the y axis followed by 9 o rotation about the z axis Then also compute the image of the line segment from (,, ) to (,, ) Solution: We will do it in three (or four) steps Step- Let T be the rotation by 45 o about the y axis By proposition 65 () the matrix of T is given by A = cos θ sinθ cos 45 sin 45 = sin θ cos θ sin 45 cos 45 = Step- Let T be the rotation by 9 o rotation about the z axis By proposition 65 () the matrix of T is given by B = cos θ sin θ cos 9 o sin 9 o sin θ cos θ = sin 9 o cos 9 o =
33 65 APPLICATIONS OF LINEAR TRANS Step- So, thematrix of the composite transformation T = T T is matrix BA = = The Last Part: So, T(,, ) = BA = =
34 CHAPTER 6 LINEAR TRANSFORMATION
LINEAR ALGEBRA W W L CHEN
LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,
0.1 Linear Transformations
.1 Linear Transformations A function is a rule that assigns a value from a set B for each element in a set A. Notation: f : A B If the value b B is assigned to value a A, then write f(a) = b, b is called
Linear Equations in Linear Algebra
1 Linear Equations in Linear Algebra 1.8 INTRODUCTION TO LINEAR TRANSFORMATIONS LINEAR TRANSFORMATIONS A transformation (or function or mapping) T from R n to R m is a rule that assigns to each vector
α = u v. In other words, Orthogonal Projection
Orthogonal Projection Given any nonzero vector v, it is possible to decompose an arbitrary vector u into a component that points in the direction of v and one that points in a direction orthogonal to v
Lectures notes on orthogonal matrices (with exercises) 92.222 - Linear Algebra II - Spring 2004 by D. Klain
Lectures notes on orthogonal matrices (with exercises) 92.222 - Linear Algebra II - Spring 2004 by D. Klain 1. Orthogonal matrices and orthonormal sets An n n real-valued matrix A is said to be an orthogonal
MATH2210 Notebook 1 Fall Semester 2016/2017. 1 MATH2210 Notebook 1 3. 1.1 Solving Systems of Linear Equations... 3
MATH0 Notebook Fall Semester 06/07 prepared by Professor Jenny Baglivo c Copyright 009 07 by Jenny A. Baglivo. All Rights Reserved. Contents MATH0 Notebook 3. Solving Systems of Linear Equations........................
MAT188H1S Lec0101 Burbulla
Winter 206 Linear Transformations A linear transformation T : R m R n is a function that takes vectors in R m to vectors in R n such that and T (u + v) T (u) + T (v) T (k v) k T (v), for all vectors u
Linear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007)
MAT067 University of California, Davis Winter 2007 Linear Maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) As we have discussed in the lecture on What is Linear Algebra? one of
4.6 Null Space, Column Space, Row Space
NULL SPACE, COLUMN SPACE, ROW SPACE Null Space, Column Space, Row Space In applications of linear algebra, subspaces of R n typically arise in one of two situations: ) as the set of solutions of a linear
Chapter 17. Orthogonal Matrices and Symmetries of Space
Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length
Inner Product Spaces and Orthogonality
Inner Product Spaces and Orthogonality week 3-4 Fall 2006 Dot product of R n The inner product or dot product of R n is a function, defined by u, v a b + a 2 b 2 + + a n b n for u a, a 2,, a n T, v b,
Vectors, Gradient, Divergence and Curl. 1 Introduction A vector is determined by its length and direction. They are usually denoted with letters with arrows on the top a or in bold letter a. We will use
5.3 The Cross Product in R 3
53 The Cross Product in R 3 Definition 531 Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] Then the vector given by [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] is called the cross product (or
Mathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
Images and Kernels in Linear Algebra By Kristi Hoshibata Mathematics 232
Images and Kernels in Linear Algebra By Kristi Hoshibata Mathematics 232 In mathematics, there are many different fields of study, including calculus, geometry, algebra and others. Mathematics has been
Review: Vector space
Math 2F Linear Algebra Lecture 13 1 Basis and dimensions Slide 1 Review: Subspace of a vector space. (Sec. 4.1) Linear combinations, l.d., l.i. vectors. (Sec. 4.3) Dimension and Base of a vector space.
Systems of Linear Equations
Systems of Linear Equations Beifang Chen Systems of linear equations Linear systems A linear equation in variables x, x,, x n is an equation of the form a x + a x + + a n x n = b, where a, a,, a n and
B such that AB = I and BA = I. (We say B is an inverse of A.) Definition A square matrix A is invertible (or nonsingular) if matrix
Matrix inverses Recall... Definition A square matrix A is invertible (or nonsingular) if matrix B such that AB = and BA =. (We say B is an inverse of A.) Remark Not all square matrices are invertible.
2.1 Functions. 2.1 J.A.Beachy 1. from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair
2.1 J.A.Beachy 1 2.1 Functions from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 21. The Vertical Line Test from calculus says that a curve in the xy-plane
T ( a i x i ) = a i T (x i ).
Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b)
THREE DIMENSIONAL GEOMETRY
Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,
Orthogonal Matrices. u v = u v cos(θ) T (u) + T (v) = T (u + v). It s even easier to. If u and v are nonzero vectors then
Part 2. 1 Part 2. Orthogonal Matrices If u and v are nonzero vectors then u v = u v cos(θ) is 0 if and only if cos(θ) = 0, i.e., θ = 90. Hence, we say that two vectors u and v are perpendicular or orthogonal
( ) which must be a vector
MATH 37 Linear Transformations from Rn to Rm Dr. Neal, WKU Let T : R n R m be a function which maps vectors from R n to R m. Then T is called a linear transformation if the following two properties are
1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number
ISOMETRIES OF R n KEITH CONRAD
ISOMETRIES OF R n KEITH CONRAD 1. Introduction An isometry of R n is a function h: R n R n that preserves the distance between vectors: h(v) h(w) = v w for all v and w in R n, where (x 1,..., x n ) = x
1 VECTOR SPACES AND SUBSPACES
1 VECTOR SPACES AND SUBSPACES What is a vector? Many are familiar with the concept of a vector as: Something which has magnitude and direction. an ordered pair or triple. a description for quantities such
Lecture 14: Section 3.3
Lecture 14: Section 3.3 Shuanglin Shao October 23, 2013 Definition. Two nonzero vectors u and v in R n are said to be orthogonal (or perpendicular) if u v = 0. We will also agree that the zero vector in
A matrix over a field F is a rectangular array of elements from F. The symbol
Chapter MATRICES Matrix arithmetic A matrix over a field F is a rectangular array of elements from F The symbol M m n (F) denotes the collection of all m n matrices over F Matrices will usually be denoted
NOTES ON LINEAR TRANSFORMATIONS
NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all
Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis
Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis 2. Polar coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers (r, θ). If r >
Solving Linear Systems, Continued and The Inverse of a Matrix
, Continued and The of a Matrix Calculus III Summer 2013, Session II Monday, July 15, 2013 Agenda 1. The rank of a matrix 2. The inverse of a square matrix Gaussian Gaussian solves a linear system by reducing
Orthogonal Projections
Orthogonal Projections and Reflections (with exercises) by D. Klain Version.. Corrections and comments are welcome! Orthogonal Projections Let X,..., X k be a family of linearly independent (column) vectors
Dot product and vector projections (Sect. 12.3) There are two main ways to introduce the dot product
Dot product and vector projections (Sect. 12.3) Two definitions for the dot product. Geometric definition of dot product. Orthogonal vectors. Dot product and orthogonal projections. Properties of the dot
Rotation Matrices and Homogeneous Transformations
Rotation Matrices and Homogeneous Transformations A coordinate frame in an n-dimensional space is defined by n mutually orthogonal unit vectors. In particular, for a two-dimensional (2D) space, i.e., n
MATH1231 Algebra, 2015 Chapter 7: Linear maps
MATH1231 Algebra, 2015 Chapter 7: Linear maps A/Prof. Daniel Chan School of Mathematics and Statistics University of New South Wales [email protected] Daniel Chan (UNSW) MATH1231 Algebra 1 / 43 Chapter
Cross product and determinants (Sect. 12.4) Two main ways to introduce the cross product
Cross product and determinants (Sect. 12.4) Two main ways to introduce the cross product Geometrical definition Properties Expression in components. Definition in components Properties Geometrical expression.
Lecture 11. Shuanglin Shao. October 2nd and 7th, 2013
Lecture 11 Shuanglin Shao October 2nd and 7th, 2013 Matrix determinants: addition. Determinants: multiplication. Adjoint of a matrix. Cramer s rule to solve a linear system. Recall that from the previous
Some Basic Properties of Vectors in n
These notes closely follow the presentation of the material given in David C. Lay s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation
Recall that two vectors in are perpendicular or orthogonal provided that their dot
Orthogonal Complements and Projections Recall that two vectors in are perpendicular or orthogonal provided that their dot product vanishes That is, if and only if Example 1 The vectors in are orthogonal
Section 5.3. Section 5.3. u m ] l jj. = l jj u j + + l mj u m. v j = [ u 1 u j. l mj
Section 5. l j v j = [ u u j u m ] l jj = l jj u j + + l mj u m. l mj Section 5. 5.. Not orthogonal, the column vectors fail to be perpendicular to each other. 5..2 his matrix is orthogonal. Check that
Recall the basic property of the transpose (for any A): v A t Aw = v w, v, w R n.
ORTHOGONAL MATRICES Informally, an orthogonal n n matrix is the n-dimensional analogue of the rotation matrices R θ in R 2. When does a linear transformation of R 3 (or R n ) deserve to be called a rotation?
Linear Transformations
a Calculus III Summer 2013, Session II Tuesday, July 23, 2013 Agenda a 1. Linear transformations 2. 3. a linear transformation linear transformations a In the m n linear system Ax = 0, Motivation we can
1 Orthogonal projections and the approximation
Math 1512 Fall 2010 Notes on least squares approximation Given n data points (x 1, y 1 ),..., (x n, y n ), we would like to find the line L, with an equation of the form y = mx + b, which is the best fit
Matrices: 2.3 The Inverse of Matrices
September 4 Goals Define inverse of a matrix. Point out that not every matrix A has an inverse. Discuss uniqueness of inverse of a matrix A. Discuss methods of computing inverses, particularly by row operations.
Lecture Notes: Matrix Inverse. 1 Inverse Definition
Lecture Notes: Matrix Inverse Yufei Tao Department of Computer Science and Engineering Chinese University of Hong Kong [email protected] Inverse Definition We use I to represent identity matrices,
Matrix Algebra 2.3 CHARACTERIZATIONS OF INVERTIBLE MATRICES Pearson Education, Inc.
2 Matrix Algebra 2.3 CHARACTERIZATIONS OF INVERTIBLE MATRICES Theorem 8: Let A be a square matrix. Then the following statements are equivalent. That is, for a given A, the statements are either all true
Lecture L3 - Vectors, Matrices and Coordinate Transformations
S. Widnall 16.07 Dynamics Fall 2009 Lecture notes based on J. Peraire Version 2.0 Lecture L3 - Vectors, Matrices and Coordinate Transformations By using vectors and defining appropriate operations between
a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.
Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given
UNIT 2 MATRICES - I 2.0 INTRODUCTION. Structure
UNIT 2 MATRICES - I Matrices - I Structure 2.0 Introduction 2.1 Objectives 2.2 Matrices 2.3 Operation on Matrices 2.4 Invertible Matrices 2.5 Systems of Linear Equations 2.6 Answers to Check Your Progress
7 - Linear Transformations
7 - Linear Transformations Mathematics has as its objects of study sets with various structures. These sets include sets of numbers (such as the integers, rationals, reals, and complexes) whose structure
Linear Algebra Notes for Marsden and Tromba Vector Calculus
Linear Algebra Notes for Marsden and Tromba Vector Calculus n-dimensional Euclidean Space and Matrices Definition of n space As was learned in Math b, a point in Euclidean three space can be thought of
Lecture notes on linear algebra
Lecture notes on linear algebra David Lerner Department of Mathematics University of Kansas These are notes of a course given in Fall, 2007 and 2008 to the Honors sections of our elementary linear algebra
Math 241, Exam 1 Information.
Math 241, Exam 1 Information. 9/24/12, LC 310, 11:15-12:05. Exam 1 will be based on: Sections 12.1-12.5, 14.1-14.3. The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/241fa12/241.html)
Math 312 Homework 1 Solutions
Math 31 Homework 1 Solutions Last modified: July 15, 01 This homework is due on Thursday, July 1th, 01 at 1:10pm Please turn it in during class, or in my mailbox in the main math office (next to 4W1) Please
Cofactor Expansion: Cramer s Rule
Cofactor Expansion: Cramer s Rule MATH 322, Linear Algebra I J. Robert Buchanan Department of Mathematics Spring 2015 Introduction Today we will focus on developing: an efficient method for calculating
Geometry of Vectors. 1 Cartesian Coordinates. Carlo Tomasi
Geometry of Vectors Carlo Tomasi This note explores the geometric meaning of norm, inner product, orthogonality, and projection for vectors. For vectors in three-dimensional space, we also examine the
v w is orthogonal to both v and w. the three vectors v, w and v w form a right-handed set of vectors.
3. Cross product Definition 3.1. Let v and w be two vectors in R 3. The cross product of v and w, denoted v w, is the vector defined as follows: the length of v w is the area of the parallelogram with
Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula:
Chapter 7 Div, grad, and curl 7.1 The operator and the gradient: Recall that the gradient of a differentiable scalar field ϕ on an open set D in R n is given by the formula: ( ϕ ϕ =, ϕ,..., ϕ. (7.1 x 1
x1 x 2 x 3 y 1 y 2 y 3 x 1 y 2 x 2 y 1 0.
Cross product 1 Chapter 7 Cross product We are getting ready to study integration in several variables. Until now we have been doing only differential calculus. One outcome of this study will be our ability
Geometric Transformations
Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted
Vector Math Computer Graphics Scott D. Anderson
Vector Math Computer Graphics Scott D. Anderson 1 Dot Product The notation v w means the dot product or scalar product or inner product of two vectors, v and w. In abstract mathematics, we can talk about
1.3. DOT PRODUCT 19. 6. If θ is the angle (between 0 and π) between two non-zero vectors u and v,
1.3. DOT PRODUCT 19 1.3 Dot Product 1.3.1 Definitions and Properties The dot product is the first way to multiply two vectors. The definition we will give below may appear arbitrary. But it is not. It
Similarity and Diagonalization. Similar Matrices
MATH022 Linear Algebra Brief lecture notes 48 Similarity and Diagonalization Similar Matrices Let A and B be n n matrices. We say that A is similar to B if there is an invertible n n matrix P such that
Interpolating Polynomials Handout March 7, 2012
Interpolating Polynomials Handout March 7, 212 Again we work over our favorite field F (such as R, Q, C or F p ) We wish to find a polynomial y = f(x) passing through n specified data points (x 1,y 1 ),
1 Eigenvalues and Eigenvectors
Math 20 Chapter 5 Eigenvalues and Eigenvectors Eigenvalues and Eigenvectors. Definition: A scalar λ is called an eigenvalue of the n n matrix A is there is a nontrivial solution x of Ax = λx. Such an x
Topic 1: Matrices and Systems of Linear Equations.
Topic 1: Matrices and Systems of Linear Equations Let us start with a review of some linear algebra concepts we have already learned, such as matrices, determinants, etc Also, we shall review the method
Algebra and Linear Algebra
Vectors Coordinate frames 2D implicit curves 2D parametric curves 3D surfaces Algebra and Linear Algebra Algebra: numbers and operations on numbers 2 + 3 = 5 3 7 = 21 Linear algebra: tuples, triples,...
Harvard College. Math 21b: Linear Algebra and Differential Equations Formula and Theorem Review
Harvard College Math 21b: Linear Algebra and Differential Equations Formula and Theorem Review Tommy MacWilliam, 13 [email protected] May 5, 2010 1 Contents Table of Contents 4 1 Linear Equations
2.1: MATRIX OPERATIONS
.: MATRIX OPERATIONS What are diagonal entries and the main diagonal of a matrix? What is a diagonal matrix? When are matrices equal? Scalar Multiplication 45 Matrix Addition Theorem (pg 0) Let A, B, and
MATH 240 Fall, Chapter 1: Linear Equations and Matrices
MATH 240 Fall, 2007 Chapter Summaries for Kolman / Hill, Elementary Linear Algebra, 9th Ed. written by Prof. J. Beachy Sections 1.1 1.5, 2.1 2.3, 4.2 4.9, 3.1 3.5, 5.3 5.5, 6.1 6.3, 6.5, 7.1 7.3 DEFINITIONS
Using determinants, it is possible to express the solution to a system of equations whose coefficient matrix is invertible:
Cramer s Rule and the Adjugate Using determinants, it is possible to express the solution to a system of equations whose coefficient matrix is invertible: Theorem [Cramer s Rule] If A is an invertible
Chapter 4: Binary Operations and Relations
c Dr Oksana Shatalov, Fall 2014 1 Chapter 4: Binary Operations and Relations 4.1: Binary Operations DEFINITION 1. A binary operation on a nonempty set A is a function from A A to A. Addition, subtraction,
NON SINGULAR MATRICES. DEFINITION. (Non singular matrix) An n n A is called non singular or invertible if there exists an n n matrix B such that
NON SINGULAR MATRICES DEFINITION. (Non singular matrix) An n n A is called non singular or invertible if there exists an n n matrix B such that AB = I n = BA. Any matrix B with the above property is called
ex) What is the component form of the vector shown in the picture above?
Vectors A ector is a directed line segment, which has both a magnitude (length) and direction. A ector can be created using any two points in the plane, the direction of the ector is usually denoted by
WHICH LINEAR-FRACTIONAL TRANSFORMATIONS INDUCE ROTATIONS OF THE SPHERE?
WHICH LINEAR-FRACTIONAL TRANSFORMATIONS INDUCE ROTATIONS OF THE SPHERE? JOEL H. SHAPIRO Abstract. These notes supplement the discussion of linear fractional mappings presented in a beginning graduate course
5.3 ORTHOGONAL TRANSFORMATIONS AND ORTHOGONAL MATRICES
5.3 ORTHOGONAL TRANSFORMATIONS AND ORTHOGONAL MATRICES Definition 5.3. Orthogonal transformations and orthogonal matrices A linear transformation T from R n to R n is called orthogonal if it preserves
1 Sets and Set Notation.
LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most
Modern Geometry Homework.
Modern Geometry Homework. 1. Rigid motions of the line. Let R be the real numbers. We define the distance between x, y R by where is the usual absolute value. distance between x and y = x y z = { z, z
1.5 Elementary Matrices and a Method for Finding the Inverse
.5 Elementary Matrices and a Method for Finding the Inverse Definition A n n matrix is called an elementary matrix if it can be obtained from I n by performing a single elementary row operation Reminder:
Vector Spaces 4.4 Spanning and Independence
Vector Spaces 4.4 and Independence October 18 Goals Discuss two important basic concepts: Define linear combination of vectors. Define Span(S) of a set S of vectors. Define linear Independence of a set
(a) The transpose of a lower triangular matrix is upper triangular, and the transpose of an upper triangular matrix is lower triangular.
Theorem.7.: (Properties of Triangular Matrices) (a) The transpose of a lower triangular matrix is upper triangular, and the transpose of an upper triangular matrix is lower triangular. (b) The product
Homework: 2.1 (page 56): 7, 9, 13, 15, 17, 25, 27, 35, 37, 41, 46, 49, 67
Chapter Matrices Operations with Matrices Homework: (page 56):, 9, 3, 5,, 5,, 35, 3, 4, 46, 49, 6 Main points in this section: We define a few concept regarding matrices This would include addition of
Section 9.5: Equations of Lines and Planes
Lines in 3D Space Section 9.5: Equations of Lines and Planes Practice HW from Stewart Textbook (not to hand in) p. 673 # 3-5 odd, 2-37 odd, 4, 47 Consider the line L through the point P = ( x, y, ) that
Geometric Transformation CS 211A
Geometric Transformation CS 211A What is transformation? Moving points (x,y) moves to (x+t, y+t) Can be in any dimension 2D Image warps 3D 3D Graphics and Vision Can also be considered as a movement to
1 2 3 1 1 2 x = + x 2 + x 4 1 0 1
(d) If the vector b is the sum of the four columns of A, write down the complete solution to Ax = b. 1 2 3 1 1 2 x = + x 2 + x 4 1 0 0 1 0 1 2. (11 points) This problem finds the curve y = C + D 2 t which
MAT 200, Midterm Exam Solution. a. (5 points) Compute the determinant of the matrix A =
MAT 200, Midterm Exam Solution. (0 points total) a. (5 points) Compute the determinant of the matrix 2 2 0 A = 0 3 0 3 0 Answer: det A = 3. The most efficient way is to develop the determinant along the
Math 018 Review Sheet v.3
Math 018 Review Sheet v.3 Tyrone Crisp Spring 007 1.1 - Slopes and Equations of Lines Slopes: Find slopes of lines using the slope formula m y y 1 x x 1. Positive slope the line slopes up to the right.
LINEAR ALGEBRA W W L CHEN
LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2008. This chapter originates from material used by author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals,
Linear Equations and Inverse Matrices
Chapter 4 Linear Equations and Inverse Matrices 4. Two Pictures of Linear Equations The central problem of linear algebra is to solve a system of equations. Those equations are linear, which means that
1. For each of the following matrices, determine whether it is in row echelon form, reduced row echelon form, or neither.
Math Exam - Practice Problem Solutions. For each of the following matrices, determine whether it is in row echelon form, reduced row echelon form, or neither. (a) 5 (c) Since each row has a leading that
Linear transformations Affine transformations Transformations in 3D. Graphics 2011/2012, 4th quarter. Lecture 5: linear and affine transformations
Lecture 5 Linear and affine transformations Vector transformation: basic idea Definition Examples Finding matrices Compositions of transformations Transposing normal vectors Multiplication of an n n matrix
LECTURE 1 I. Inverse matrices We return now to the problem of solving linear equations. Recall that we are trying to find x such that IA = A
LECTURE I. Inverse matrices We return now to the problem of solving linear equations. Recall that we are trying to find such that A = y. Recall: there is a matri I such that for all R n. It follows that
is in plane V. However, it may be more convenient to introduce a plane coordinate system in V.
.4 COORDINATES EXAMPLE Let V be the plane in R with equation x +2x 2 +x 0, a two-dimensional subspace of R. We can describe a vector in this plane by its spatial (D)coordinates; for example, vector x 5
Chapter 19. General Matrices. An n m matrix is an array. a 11 a 12 a 1m a 21 a 22 a 2m A = a n1 a n2 a nm. The matrix A has n row vectors
Chapter 9. General Matrices An n m matrix is an array a a a m a a a m... = [a ij]. a n a n a nm The matrix A has n row vectors and m column vectors row i (A) = [a i, a i,..., a im ] R m a j a j a nj col
Name: Section Registered In:
Name: Section Registered In: Math 125 Exam 3 Version 1 April 24, 2006 60 total points possible 1. (5pts) Use Cramer s Rule to solve 3x + 4y = 30 x 2y = 8. Be sure to show enough detail that shows you are
Section 2.1. Section 2.2. Exercise 6: We have to compute the product AB in two ways, where , B =. 2 1 3 5 A =
Section 2.1 Exercise 6: We have to compute the product AB in two ways, where 4 2 A = 3 0 1 3, B =. 2 1 3 5 Solution 1. Let b 1 = (1, 2) and b 2 = (3, 1) be the columns of B. Then Ab 1 = (0, 3, 13) and
Sec 4.1 Vector Spaces and Subspaces
Sec 4. Vector Spaces and Subspaces Motivation Let S be the set of all solutions to the differential equation y + y =. Let T be the set of all 2 3 matrices with real entries. These two sets share many common
Math 54. Selected Solutions for Week Is u in the plane in R 3 spanned by the columns
Math 5. Selected Solutions for Week 2 Section. (Page 2). Let u = and A = 5 2 6. Is u in the plane in R spanned by the columns of A? (See the figure omitted].) Why or why not? First of all, the plane in<|endoftext|>
| 4.65625 |
137 |
# How do you find three consecutive integers such that twice the smallest is 12 more than the largest?
Jan 22, 2016
Convert the conditions into an equation and solve to find that the three integers are: $14 , 15 , 16$
#### Explanation:
Suppose the three integers are $n$, $n + 1$ and $n + 2$
We are given:
$2 n = \left(n + 2\right) + 12$
Subtract $n$ from both sides to find:
$n = 2 + 12 = 14$
So the three integers are $14 , 15 , 16$.<|endoftext|>
| 4.40625 |
654 |
Some LED lights can change color, but it is not normally the LED that changes color. When you see an LED light change color, what you actually see is a different colored diode inside emitting the light.
Interested in how this is all possible? Here is a quick explanation on how LED lights change color (click here to learn how LEDs work).
Method 1: Color Changing LED Lights Using Several LED Colors
A Light Emitting Diode is essentially a semiconductor that releases energy in the form of light. The light color depends on the material of the semiconductor.
Most color changing LEDs contain three separate light emitting diodes in the same casing. Each LED emits its own specific color but by controlling the energy levels of these three LEDs, other colors can be created. This is known as the RGB color model.
Take for example a standard color changing LED with a red, blue and green LED inside the same casing (there are many other different types of LED lights on the market). Inside the casing is also a microcontroller that manages which LED receives a current.
A current to a single LED in this casing gives off either red, blue or green while a current to two LEDs results in a color combination like purple, pink, yellow and so on. This is commonly referred to as LED color temperature.
Method 2: Color Changing LED Lights Through Alternating Currents
Another way to control the color that an LED emits is by using alternating currents. In this case, a lower current is sent to the colored LED so that the shade changes.
For example, sending current at 100% brightness through a red LED gives off the color red. Then, lowering the brightness to only 50% gives the color pink.
This color effect is specific to LED technology. The same mechanism does not give the same results with halogen lights or a CFL for example.
You have probably noticed that other types of light bulbs need a few seconds to reach full brightness. An LED reaches full brightness instantly so there is no lag time, this is one of the many benefits of LED lighting.
The human eye cannot detect the rapid speed at which an LED turns on and off. Instead, humans detect the net result of the LED turning on and off at this extreme pace.
In other words, when an LED switches on and off at a steady fast pace, we humans only detect the light as being dimmer. This is why the red LED can appear as pink.
What you actually see is the red LED turning on and off so fast that your brain detects the color at 50% brightness, in this case, pink. Depending on the set pace of the LED, this shade that we register changes.
Common LED Colors and Their Materials
Different materials are used for different LED colors. Below is a quick overview of the common semiconductor materials used for color changing LED lights.
- Aluminium gallium indium phosphate (AIGaInP): yellow, orange and red
- Gallium Phosphide (GaP): yellow and green
- Indium Gallium Nitride (InGaN): green, blue and white
- Aluminium Gallium Arsenide (AIGaAs): red and infrared<|endoftext|>
| 4.03125 |
1,584 |
+0
# Math Trivia - Proving in Mathematics
+1
683
8
+349
We all know what a proof is, right? We proved that humans can't fly. We proved that slavery is immoral. We use reasoning to show that a statement is either true or false (in rare cases, neither, but I'll get to that later). But proving in mathematics can get pretty complex. Some things, like the irrationality of $$\sqrt{2}$$, can be proven in a few simple steps. Some, like Andrew Wiles' proof of Fermat's Last Theorem, get extreme. Furthermore, there are different kinds of proofs, not just in mathematics, but anywhere. Here, I will enumerate three types of proofs that are most popular, and why they are used. So, let us begin, shall we?
1. Algebraic proof
An algebraic proof is a simple, straight-forward, brute-force approach. You use algebra to prove that a fact is true. Here is an example.
Proof that negative*negative=positive
$$-a(b-b)=0$$ because $$b-b=0$$.
$$-ab+(-a)(-b)=0$$
$$(-a)(-b)=ab$$. Q.E.D.
This type of proof is very useful because it uses elementary algebra, so it can be understood by most of the community (who are mathematically inclined). But usually this type of proof fails to get you anywhere in problems like, oh, I don't know, the Fold-and-Cut theorem (xD). But in all seriousness, elementary algebra sometimes just isn't enough. Anyway, on to the next one.
This should be easy to understand. Proof by contradiction assumes the statement is false, then proves that it cannot be false, therefore the statement is true. An example of this is the famous proof that $$\sqrt{2}$$ is irrational, but I won't put that here. Instead, I'll give you another example.
Proof that ALL roots greater than 2 of 2 are irrational
Assuming $$\sqrt[n]{2}$$ is rational for $$n>2$$, $$\sqrt[n]{2}={a\over b}$$ such that $$a$$ and $$b$$ are integers and $${a\over b}$$ is fully simplified.
$$2=({a\over b})^n$$
$$2={a^n\over b^n}$$
$$2b^n=a^n$$
$$b^n+b^n=a^n$$
By Fermat's Last Theorem, this statement is impossible, therefore $$\sqrt[n]{2}$$ is irrational.
Q.E.D.
Did you like that? Proof by contradiction is a very powerful tool. It's easy to understand, and usually you get to an answer right away. But, this proof fails too. An example is "This statement is false." Prove it.
3. Proof by induction
Some of you probably have not heard of this before, but it is exetremely useful. To prove by induction, first show that the statement is true for 0 or 1, then show that if the statement is true for $$n$$, it is also true for $$n+1$$. Doing this, you create a domino effect that proves the statement for all natural numbers. Here is an example:
Proof that $$3^n-1$$ is always a multiple of 2
Show that it is true for 1:
$$3^1-1=2$$
Show that if it is true for $$n$$, it is true for $$n+1$$:
Assume $$3^n-1$$ is a multiple of 2.
$$3^{n+1}-1$$
$$=3*3^n-1$$
$$=(2*3^n+3^n)-1$$
$$=2*3^n+3^n-1$$
$$2*3^n$$ is a multiple of 2 (multiplying by 2), and $$3^n-1$$ is also a multiple of 2 (our assumption), therefore $$3^{n+1}-1$$ is a multiple of 2, therefore the statement is true for all natural numbers.
Nice! Proof by induction can save us a lot of time, right? It pretty much does the work for us so we don't have to prove it infinitely many times. But of course, it has its own downsides. If you are not careful, you can end up with ridiculous claims. I'd give you an example, but this question is getting long.
These three are just a few ways you can prove things in mathematics, and they are all very useful indeed, if we use them correctly. Now, this wouldn't be a web2.0calc post without a question, eh? So...
Using any type of proof, prove the Pythagorean identity: $$\sin^2\theta+\cos^2\theta=1$$.
Oct 19, 2017
#1
+95859
+1
Thanks, Mathhemathh !!!!!!
Oct 19, 2017
#2
0
For your "Proof by Contradiction", you should have used the famous proof of Euclid of "Infinity of Primes!" That is one of the most elegant proofs in the history of Math.
Oct 19, 2017
#3
+96956
+1
Why don't you demonstrate :)
Melody Oct 19, 2017
#4
+1
Good Point! I thought everybody interested in Math knew it! Here is a brief proof of it.
https://math.stackexchange.com/questions/1017218/euclids-proof-for-infinitely-many-prime-numbers
Oct 19, 2017
#6
+96956
+1
You 'assumed' all people interested in mathematics knew.
As someone interested in mathematics you should know that assumptions usually lead to errors.
That is why in mathematics all things must be proven. :)
And proofs must be demonstrated :)
Melody Oct 19, 2017
edited by Melody Oct 19, 2017
edited by Melody Oct 19, 2017
#7
+1
True what you say! Here is my take on it:
Suppose we ONLY had these Primes: 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19=9,699,690. Now, we add 1 to it and we have: 9,699,690 + 1=9,699,691. Now, this new number is either a prime or has Prime Factors that are NOT included in our list, because if you divide it by any of them, it will always leave a remainder of 1. It, therefore, must either be a Prime Number or has Prime Factors > 19 listed above. And if we check it, we see that 9,699,691 = 347 * 27953, which are its Prime factors and both, in this case, are larger than our listed largest prime, or 19!. And that proves there are infinitely MANY PRIMES!!.
Guest Oct 19, 2017
edited by Guest Oct 19, 2017
#8
+96956
+1
Yes, I did accept your website proof but it is good that you explained it in your own words :)
Melody Oct 19, 2017<|endoftext|>
| 4.59375 |
635 |
In the heart of the jungle in northern Guatemala and surrounded by lush vegetation, lies one of the major sites of Mayan civilization. The ceremonial center contains superb temples and palaces. Remains of dwellings are scattered throughout the surrounding countryside.
Tikal National Park is located in northern Guatemala's Petén Province within a large forest region often referred to as the Maya Forest, which extends into neighboring Mexico and Belize. Embedded within the much larger Maya Biosphere Reserve, exceeding 2,000,000 ha (5,000,000 acres) and contiguous with additional conservation areas, Tikal National Park is one of the few World Heritage properties inscribed according to both natural and cultural criteria for its extraordinary biodiversity and archaeological importance.
The park comprises 57,600 ha (142,330 acres) of wetlands, savannah, tropical broadleaf and palm forests with thousands of architectural and artistic remains of the Mayan civilization from the Preclassic Period (600 B.C.) to the decline and eventual collapse of the urban center around 900 AD.
The diverse ecosystems and habitats harbor a wide spectrum of neotropical fauna and flora. Five cats, including Jaguar and Puma, several species of monkeys and anteaters and more than 300 species of birds are among the notable wildlife. The forests comprise more than 200 tree species and over 2000 higher plants have been recorded across the diverse habitats.
Tikal, a major pre-Columbian political, economic and military center, is one of the most important archaeological complexes left by the Maya civilization. An inner urban zone of around 400 hectares contains the principal monumental architecture and monuments which include palaces, temples, ceremonial platforms, small and medium sized residences, ball-game courts, terraces, roads, large and small squares.
Many of the existing monuments preserve decorated surfaces, including stone carvings and mural paintings with hieroglyphic inscriptions, which illustrate the dynastic history of the city and its relationships with urban centers as far away as Teotihuacán and Calakmul in Mexico, Copán in Honduras or Caracol in Belize. A wider zone of key archaeological importance, around 1,200 ha (3,000 acres), covers residential areas and historic water reservoirs, today known as "aguadas."
The extensive peripheral zone features more than 25 associated secondary sites, historically serving protective purposes and as check-points for trade routes. The peripheral areas also played a major role for agricultural production for the densely populated center.
Research has revealed numerous constructions, carved monuments and other evidence bearing witness to highly sophisticated technical, intellectual and artistic achievements that developed from the arrival of the first settlers (800 B.C.) to the last stages of historic occupation around the year 900.
Tikal has enhanced our understanding not only of an extraordinary bygone civilization but also of cultural evolution more broadly. The diversity and quality of architectonical and sculptural ensembles serving ceremonial, administrative and residential functions are exemplified in a number of exceptional places, such as the Great Plaza, the Lost World Complex, the Twin Pyramid Complexes, as well as in ball courts and irrigation structures.<|endoftext|>
| 3.703125 |
773 |
# Video: Using Proportions to Solve Problems in a Real-World Context
Tim Burnham
The amount of stain needed to cover a wooden surface is directly proportional to the area of the surface. If 4 pints are required to cover a square deck of side 3 feet, how many pints of stain are needed to paint a square deck of side 8 feet 3 inches?
03:24
### Video Transcript
The amount of stain needed to cover a wooden surface is directly proportional to the area of its surface. If four pints are required to cover a square deck of side three feet, how many pints of stain are needed to paint a square deck of side eight feet three inches?
Well first up, let’s define a couple of variables. Let’s let 𝑆 be the amount of stain in pints, and we’ll let 𝐴 be the area of the surface in square feet. Now the question tells us that the amount of stain needed is directly proportional to the area of the surface. So we can write that as 𝑆 is directly proportional to 𝐴. And this just means that 𝑆 is equal to some constant times 𝐴. That’s what direct proportionality means.
Now we know that four pints are required to cover a square deck of side three feet. So if we’ve got a square deck of side three feet, that means each of its sides are three feet. So the area of that deck is gonna be three times three is nine square feet. And we can put that into our equation. We know that the amount of stain is four pints. We don’t know what the constant is yet. That’s what we’re trying to work out here. And that goes with an area of nine square feet. So four is equal to 𝑘 times nine. If I divide both sides of that by nine, the nines cancel from the right-hand side, and it tells me that 𝑘 is equal to four-ninths. So now I’ve got a grand formula that tells me the amount of stain that I need to cover a wooden deck of a certain area. The amount of stain in pints is equal to four-ninths times the area of the deck in square feet.
Now let’s use that formula to answer this last part of the question. We’ve got a square deck of side eight feet three inches. So we need to work out the area of that deck with sides of eight feet three inches. So the calculation is gonna be eight feet three by eight feet three. Well, a foot has got twelve inches in it, so eight feet and three inches means eight and three-twelfths of a foot. So our calculation becomes eight and three-twelfths times eight and three-twelfths. Well the more observant among you will notice that three-twelfths is the same as a quarter. So an even simpler form of that calculation would be eight and a quarter times eight and a quarter.
Turning those into top heavy fractions or improper fractions, that’s 33 over four times 33 over four, which is 1089 over 16 square feet. So let’s just write that into our diagram here and make a little bit of space for some more calculation. Then when the area is 1089 over 16, the volume of stain that we need is four-ninths times 1089 over 16. Now I can cancel top and bottom a little bit. Fours go into 16 four times and nines go into 1089 121 times. So the volume is 121 over four pints. Now we can change that into a mixed number, which would be a really nice way to present our final answer of 30 and a quarter pints.<|endoftext|>
| 4.8125 |
794 |
# University of Florida/Egm4313/s12.team8.dupre/R5.4
Problem 5.4
## Problem Statement
(1) Show that:
${\displaystyle \displaystyle y_{p}(x)=\sum _{i=0}^{n}y_{p,i}(x)}$ (5.1)
is indeed the overall particular solution of the L2-ODE-VC:
${\displaystyle \displaystyle y''+p(x)y'+q(x)y=r(x)}$ (5.2)
with the excitation:
${\displaystyle \displaystyle r(x)=r_{1}(x)+r_{2}(x)+...+r_{n}(x)=\sum _{i=0}^{n}r_{i}(x)}$ (5.3)
(2) Discuss the choice of ${\displaystyle \displaystyle y_{p}(x)}$ in the above table , e.g., for:
${\displaystyle \displaystyle r(x)=kcos(\omega x)}$
Why would you need to have both ${\displaystyle \displaystyle cos(\omega x),sin(\omega x)}$ in ${\displaystyle \displaystyle y_{p}(x)}$?
### Solution (1)
Using the following equation:
${\displaystyle \displaystyle r_{i}(x)=y_{p,i}''+p(x)y_{p,i}'+q(x)y_{p,i}}$ (5.4)
for different r and y values gives us the following:
${\displaystyle \displaystyle r_{1}(x)=y_{p,1}''+p(x)y_{p,1}'+q(x)y_{p,1}}$ (5.5)
${\displaystyle \displaystyle r_{2}(x)=y_{p,2}''+p(x)y_{p,2}'+q(x)y_{p,2}}$ (5.6)
${\displaystyle \displaystyle r_{3}(x)=y_{p,3}''+p(x)y_{p,3}'+q(x)y_{p,3}}$ (5.7)
Now, adding (5.4),(5.5), and (5.6), gives us:
${\displaystyle \displaystyle r_{1}(x)+r_{2}(x)+r_{3}(x)=(y_{p,1}+y_{p,2}+y_{p,3})''+p(x)(y_{p,1}+y_{p,2}+y_{p,3})'+q(x)(y_{p,1}+y_{p,2}+y_{p,3})}$ (5.8)
Equation (5.8) shows us that the overall particular solution of (5.2) with excitation (5.3), is in fact, equation (5.1).
### Solution (2)
We know that the given example for an excitation is the periodic excitation:
${\displaystyle \displaystyle r(x)=kcos(\omega x)}$
When we decompose a periodic excitation into a Fourier trigonometric series, we find:
${\displaystyle \displaystyle r(x)=a_{0}+\sum _{n=0}^{\infty }[a_{n}cos(n\omega x)+b_{n}sin(\omega x)]}$
Since we know that the particular solution should depend on the excitation, we know that for a periodic excitation ${\displaystyle \displaystyle r(x)}$, we would need both ${\displaystyle \displaystyle cos(\omega x),sin(\omega x)}$ in ${\displaystyle \displaystyle y_{p}(x)}$ to obtain the correct particular solution.<|endoftext|>
| 4.375 |
358 |
# Mental Math Workbook - Adding two digit numbers to another two digit number.
### Class Room
The secret to being able to add two digit numbers in your head is to simply the problem.
## Adding numbers when there is no carry from the units place.
For example, conside the problem of adding two two-digit numbers 83 and 25.
We can simplify the problem mentally by splitting the problem into two simpler sums, like this:
Which essentially breaks the problem into two smaller additions:
Step 1
And then to the next (simpler) step:
Note that there is no magic here. We are simply adding the tens digits together and the units digits together. The above works if there is no carry involved from adding the the tens digit.
## Adding numbers when there is a carry from the units place
If there is a carry from the lower digits, then we can round up the second number to the next multiple of 10 and add in two steps. For example, consider the following problem:
Adding 8 to 4 involves a carry and it is cumbersome to remember the carry. Instead, we can simplify the addition as:
And calculate like this:
Once again, note that there is no magic here. We are simply pre-adding the carry to the ten's digit by rounding it up to the next multiple of 10.
Also, by rounding the number to the next multiple of 10, the units addition in the first step is trivial.
We could round up either the first number or the second number. In the above examples, we always rounded the second number simply because it is easier to grasp the concept. However, you could round either of the numbers. In general, round the number which is closer to the next multiple of 10.<|endoftext|>
| 4.59375 |
654 |
Resilience: What it is & How to get more of it!
What is resilience?
Resilience is a set of skills, attributes, and abilities that enable individuals to adapt well to difficulties and challenges. This is something parents can help nurture in their children for greater success across all spheres of life: social, emotional, & academic.
Children seem to understand the meaning of resilience best with this analogy: Stress placed on a rubber band might stretch it to the breaking point. Ideally, however, the rubber band will have the flexibility to withstand the stretching and slowly return to its original size and shape. Learning how to make an active effort and be flexible is fundamental to building resilience skills (Alford, Zucker, & Grados, 2011).
A person’s ability to take the initiative, believe in his or her effectiveness, and think realistically but positively also contributes to resilience. For example, children who can learn to compromise in their relationships with others are more apt to develop successful friendships. Students who are proactive – who aren’t afraid to seek out a teacher and ask for extra help – are more likely to turn a poor grade into an A or a B. Being proactive means setting goals, planning and problem solving, thinking optimistically, and building a more positive sense of self.
Listening to what we say to ourselves and others provides a clue as to whether we are being proactive, being reactive, or being passive. Things we say to ourselves are called our “self-talk.” For example, if we are told that screen time has been reduced by half, our thinking and self-talk would be:
- Positive and Proactive if we ask ourselves, “What are my choices for how to best use the available time?”
- Negative and Reactive if we say, “That’s terrible! You’re a jerk! I hate you!”
- Negative and Passive if we respond with, “I can’t do anything fun in that time” and then just give up.
A positive proactive approach involves taking responsibility and ownership for our thoughts, feelings, and actions so that we can work on changing them.
As parents, it is helpful to teach children about the private conversations that they have with themselves and how they are always thinking! Sometimes the thoughts are negative and sometimes the thoughts are positive. We can encourage positive, proactive self-talk as well as behavior.
As children face a challenge or a problem, these 5 STEPS can help them deal with difficulty in a POSITIVE AND PROACTIVE WAY:
- Acknowledge the problem.
- Keep perspective that it is a specific problem,
- Keep in mind that the problem won’t last forever.
- Come up with a plan to make the situation better.
- Act on the plan.
Help children practice! As they do, this will build confidence and resiliency!
Written by Victoria L. Norton, Psy.D. at Linden BP. If you are interested in receiving Linden Blog updates with original articles about parenting, families, mental health, and wellness, subscribe using the field below.<|endoftext|>
| 4.34375 |
1,127 |
DNA is normally found as a loosely contained structure called chromatin within the nucleus, where it is wound up and associated with a variety of histone proteins. The two complementary strands are separated, much like unzipping a zipper.
DISCUSSION QUESTIONS FOR CHAPTERS 1, 2, 3, 4, 5, 6, 7, 8 'There is no such thing as a specific origin of DNA replication in eukaryotes'. Discuss. 9. The Nucleus and DNA Replication . DNA replication is the copying of DNA that occurs before cell division can take place. After a . Critical Thinking Questions.
How does the replication machinery know where to begin? The following section will explore the structure of the nucleus and its contents, as well as the process of DNA replication.
Brush up on biological concepts and practice critical thinking skills that 3 questions Explore DNA's replication and repair processes, the significance of its. in knowledge, develop critical thinking skills, learn how to work together to develop . will need to address several essential questions about DNA. Each student.
Congrats you bunch of brainiacs! In prokaryotes, three main types of polymerases are known: The kidneys cannot make urine salty enough to remove the excess salt you consume in the question water. The genetic instructions that are used to build and maintain an organism are arranged in an orderly manner in strands of DNA.
Clicking on any base in D. Compare and contrast the major pathways involved in repairing damage in human DNA.
Together, this spliting and splicing, the critical and replication of forms, comprise dna dialectic method: The differentiated state is generally stable and can be inherited from one somatic cell to another. What is a "codon"?
Critical Thinking Questions. Resource ID:[email protected] . DNA replication is bidirectional and discontinuous; explain your understanding of those concepts. Critical thinking questions dna replication. Genetics Graphing Practice with Critical Thinking Questions. $ Pinterest. Explore Life Science, Science Lessons, and.
Discuss the evidence for and against the existence of an analogous skeleton within the nucleus. Discuss the relative importance of cis- and trans-acting factors in the control of transcription.
Outline how you would test experimentally the possible role of this octamer sequence in regulating the expression of these genes.
The result is billions of new cells being created each day. Why do mitotic chromosomes have the shape they do?
Describe the mechanisms that ensure that parental and daughter duplexes have the same DNA sequences. Estimates dna replication critical thinking questions gene numbers suggest that mammals have four times more genes than flies, and ten times more than yeast. How is the structure of the nuclear pore related to its function?
- Critical Thinking Questions | Texas Gateway
- Solved: Date:. Critical Thinking Questions 1. Make A Simpl | thefireworkshoplist.com
- Client manager cover letter sample why the divorce rate increased. essay
- DNA Replication in Prokaryotes – Biology 2e
- Writing essay on interview curriculum vitae fashion model
How do cells make accurate copies of DNA? Illustrate how the study of human disease has helped us to understand the different pathways involved in repairing damage in DNA.
The Nucleus and DNA Replication
Discuss how the scientific community learned that DNA replication takes place in a semi- conservative fashion. If helicase is mutated, the DNA strands will not be joined together at the beginning of replication.
The expression of bacterial genes is controlled by the action of diffusible repressors and activators. Why do mitotic chromosomes have the shape they do? Two strains of Vibrio cholerae were used for the experiment.
How are primary transcripts processed and what roles do such modifications play? Review the evidence supporting current models for the initiation of DNA replication in eukaryotic cells.
Discuss telomeres in terms of their discovery, location, universality, duplication, and relationship with ageing and cancer. How has the study of developmental biology impinged upon our understanding of cancer? The two sides of the ladder are not identical, but are complementary.
answered a question related to DNA Replication “Dear Ru-Jeng Teng, I will try to stimulate the discussion by the proposal for giving attention to the approach . Start studying DNA Critical Thinking Questions. DNA polymerase adds complementary bases 3. What happens when there's an error in DNA Replication?.
Explain the events taking place at the replication fork. The strand with the Okazaki fragments is known as the lagging strand. The nucleolus is a region of the nucleus that is responsible for manufacturing the RNA necessary for construction of ribosomes.
dna replication practice q
Multiple nucleosomes along the entire molecule of DNA appear like a beaded necklace, in which the string is the DNA and the beads are the associated histones. Comparison of the promoter sequences of a family of mammalian genes reveals that all share a sequence of eight nucleotides.
One zone of each strand dna replication critical thinking questions made up of identical repeating units, while another zone is made up of differing units. Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorous.
Stage 2: RNA polymerases make mistakes. I will insert a DNA sequence herewhat would be the amino acids in the protein? Interestingly, some cells in the body, such as muscle cells, contain more than one nucleus [link]which is known as multinucleated.<|endoftext|>
| 4.0625 |
349 |
Differences Between the Flu and a Cold
Diana Rosenberg, M.D.
Family Medicine, Rush Copley Medical Group
Many people can mistake the common cold for the flu, simply because they become ill during “flu season.” It is true that the flu and the common cold are both respiratory illnesses, however different viruses cause each illness. Both flu and cold viruses are transmitted the same way, but because their symptoms are similar, it can be difficult to distinguish between them. Influenza or “the flu” develops when a flu virus infects your respiratory system, including your nose, throat, bronchial tubes and possibly the lungs. A cold virus commonly infects only your upper respiratory tract. What is referred to as “stomach flu” is actually just another virus that causes diarrhea and vomiting. Primary symptoms of the flu include fever, fatigue, aches, chills and a cough. One important thing to know about flu viruses is how to prevent from passing them on. Here are some things you can do to prevent the spreading of cold and flu viruses:
- Wash your hands frequently and use alcohol-based gel
- After you cough and sneeze into a tissue or into your hands, wash them
- Avoid touching your eyes, nose or mouth excessively
- Keep shared surfaces like phones and keyboards clean
- Avoid crowds during cold and flu season
Tamiflu and the other prescription drugs were developed to cut short a case of the flu. In order to be effective, these drugs must be taken within 48 hours after symptoms are detected. If you are unsure whether your symptoms are the flu or a cold, it is always a good idea to contact your physician.<|endoftext|>
| 3.78125 |
1,141 |
## Wednesday, January 2, 2013
### What are the prime years in the 21st Century?
Yesterday, we checked to see if 2013 was a prime number. Because the digits sum to 2+0+1+3 = 6, we can tell immediately that it is not prime because it is divisible by 3, 2013 = 671 × 3. It turns out that 671 isn't prime, either, since it factors to 61 × 11, both of which are prime.
This means the prime factorization of 2013 = 3 × 11 × 61. I put the primes in order from smallest to largest, but that's just me being tidy. The order of the primes doesn't matter. We say that every number has a unique prime factorization up to order.
Not everyone knows this rule for divisibility by 3, but the rule for divisibility by 5 is much simpler and better known. It shows us quickly that 2015 can't possibly be prime, since 2015 = 5 × 403. (To get the complete prime factorization, we have to check to see if 403 is prime. It turns out it isn't, 403 = 13 × 31.)
So let's ask a slightly more general problem. What numbers between 2000 and 2100 are prime? This sounds like a lot of work, which is why I decided to use Excel to answer the question.
(If you are new to Excel, you can skip down to the answer.)
Step 1: What is the square root of 2100?
In cell A1, type =sqrt(2100). The answer is 45.8257... The reason to find this is because we need to check each of our numbers for divisibility by the primes less than their square root. If a × b = n, the smaller of a and b has to be less than or equal to the square root of n and the larger must be greater than or equal to the square root of n.
Step 2: Across row 1 starting in B1, list the primes less than 45.
Starting in B1, type 2[tab] 3[tab] 5[tab] 7[tab] 11[tab] 13[tab] 17[tab] 19[tab] 23[tab] 29[tab] 31[tab] 37[tab] 41[tab] 43[tab].
Step 3: Put the numbers from 2001 to 2099 in column A.
The easiest way to do this is to type 2001 in A2, drag the value down to A101 and change the method of filling the values in to "Fill Series".
Step 4: How to tell if a number is evenly divisible by a prime.
2001/3 = 667, but 2001/5 = 400.2, which means 2001 is divisible by 3, but not by 5. The Excel formula to type will use the Excel built-in formulas IF and INT.
IF asks for a statement that is true or false, what to do if it is true and what to do if it is false.
INT takes a number x and rounds down to the largest whole number less than x. For example, =INT(2001/3) will give us 667, but =INT(2001/5) will give us 400 instead of 400.2, a difference we will be using in our logical statement to test.
Here is the formula to type into cell B2.
=IF(INT(\$A2/B\$1)=\$A2/B\$1, "yes"," ")
What this does is check if INT(2001/2) = 2001/2. In this case it does not, so Excel will put a blank in this cell, which is what " " means.) If I click and drag this formula down the column and across the rows, The spreadsheet will now have a pattern of the word "yes" scattered about among the blank cells.
Step 5: Use an IF statement to find the numbers that weren't divisible by any of the primes we used.
We now use IF and COUNTIF in the column just beyond N to see if the word "yes" showed up in that row. If it didn't, the number in column A is prime and we will copy that number into column P.
=IF(COUNTIF(B2:N2,"yes")=0, A2, " ")
Yet again, click and drag the formula down the column. Column P now has the primes between 2000 and 2100 listed. Here they are.
2003
2011
2017
2027
2029
2039
2053
2063
2069
2081
2083
2087
2089
2099
Notice that some primes have a difference of only two, like 2081 and 2083. These are called twin primes. Mathematicians have known since the time of Euclid that there are an infinite number of primes, which means that whatever big number n you choose, there have to be primes larger than n. It is unknown if the twin primes are infinite. It is not of vital importance, but mathematicians are fascinated by problems that are easy to state and difficult to solve.
Tomorrow: A practical application of number theory known as casting out nines.<|endoftext|>
| 4.40625 |
908 |
# 3-5: Linear Programming.
## Presentation on theme: "3-5: Linear Programming."— Presentation transcript:
3-5: Linear Programming
Vocabulary constraints: conditions given to variables, often expressed as linear inequalities feasible region: the intersection of the graphs in a system of constraints bounded: a region is bounded when the graph of a system of constraints is a polygonal region vertices: the maximum or minimum value that a linear function has for the points in a feasible region unbounded: a system of inequalities that forms a region that is open
Linear Programming Businesses use linear programming to find out how to maximize profit or minimize costs. Most have constraints on what they can use or buy.
Find the minimum and maximum value of the function f(x, y) = 3x - 2y.
We are given the constraints: y ≥ 2 1 ≤ x ≤5 y ≤ x + 3
Linear Programming Find the minimum and maximum values by graphing the inequalities and finding the vertices of the polygon formed. Substitute the vertices into the function and find the largest and smallest values.
1 ≤ x ≤5 8 7 6 5 4 3 y ≥ 2 2 y ≤ x + 3 1 1 2 3 4 5
Linear Programming The vertices of the quadrilateral formed are: (1, 2) (1, 4) (5, 2) (5, 8) Plug these points into the function f(x, y) = 3x - 2y
Linear Programming f(x, y) = 3x - 2y
Linear Programming f(1, 4) = -5 minimum f(5, 2) = 11 maximum
Find the minimum and maximum value of the function f(x, y) = 4x + 3y
We are given the constraints: y ≥ -x + 2 y ≤ x + 2 y ≥ 2x -5
y ≥ 2x -5 6 5 4 y ≥ -x + 2 3 2 1 1 2 3 4 5
Vertices f(x, y) = 4x + 3y f(0, 2) = 4(0) + 3(2) = 6
Linear Programming f(0, 2) = 6 minimum f(4, 3) = 25 maximum
Lisa has an online jewelry shop where she sells earrings and necklaces
Lisa has an online jewelry shop where she sells earrings and necklaces. She sells earrings for \$30 and necklaces for \$40. It takes 30 minutes to make a pair of earrings and 1 hour to make a necklace, and, since Lisa is a math tutor, she only has 10 hours a week to make jewelry. In addition, she only has enough materials to make 15 total jewelry items per week. She makes a profit of \$15 on each pair of earrings and \$20 on each necklace. How many pairs of earrings and necklaces should Lisa make each week in order to maximize her profit, assuming she sells all her jewelry?
Define the variables, write an inequality for this situation, and graph the solutions to the inequality.
Then take the vertices (corner points) and plug in the x and y values into the objective function.
So, in order to maximize her profits, Lisa should make 10 pair of earrings and 5 necklaces per week, and her weekly profit is \$250.
Identify the vertices of the feasible region
Using the vertices, find the minimum and maximum values of the function f(x, y) = 3x + y
Identify the vertices of the feasible region
Using the vertices, find the minimum and maximum values of the function f(x, y) = 5x + 4y
Veterinary medicine As a receptionist for a veterinarian, one of Dolores Alvarez’s tasks is to schedule appointments. She allots 20 minutes for a routine office visit and 40 minutes for a surgery. The veterinarian cannot do more than 6 surgeries per day. The office has 7 hours available for appointments. If an office visit costs \$55 and most surgeries cost \$125, find a combination of office visits and surgeries that will maximize the income the veterinarian practice receives per day.
Veterinary medicine<|endoftext|>
| 4.71875 |
1,654 |
# Introduction To Trigonometry Class 10 Notes
## Class 10 Maths Chapter 8 Introduction to Trigonometry Notes
The notes for trigonometry class 10 Maths is provided here. Get the complete concept on trigonometry which is covered in Class 10 Maths. Also, get the various trigonometric ratios for specific angles, the relationship between trigonometric functions, trigonometry tables, various identities given here.
Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:
## Trigonometric Ratios
### Opposite & Adjacent Sides in a Right Angled Triangle
In the ΔABC right-angled at B, BC is the side opposite to A, AC is the hypotenuse and AB is the side adjacent to A.
### Trigonometric Ratios
For the right ΔABC, right-angled at B, the trigonometric ratios of the A are as follows:
• sin A=opposite side/hypotenuse=BC/AC
• cosec A=hypotenuse/opposite side=AC/BC
To know more about Trigonometric Ratios, visit here.
### Visualization of Trigonometric Ratios Using a Unit Circle
Draw a circle of the unit radius with the origin as the centre. Consider a line segment OP joining a point P on the circle to the centre which makes an angle θ with the x-axis. Draw a perpendicular from P to the x-axis to cut it at Q.
• sinθ=PQ/OP=PQ/1=PQ
• cosθ=OQ/OP=OQ/1=OQ
• tanθ=PQ/OQ=sinθ/cosθ
• cosecθ=OP/PQ=1/PQ
• secθ=OP/OQ=1/OQ
• cotθ=OQ/PQ=cosθ/sinθ
### Relation between Trigonometric Ratios
• cosec θ =1/sin θ
• sec θ = 1/cos θ
• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ=1/tan θ
## Trigonometric Ratios of Specific Angles
### Range of Trigonometric Ratios from 0 to 90 degrees
For 0θ90,
• 0≤sinθ≤1
• 0≤cosθ≤1
• 0≤tanθ<∞
• 1≤secθ<∞
• 0≤cotθ<∞
• 1≤cosecθ<∞
tanθ and secθ are not defined at 90.
cotθ and cosecθ are not defined at 0.
### Variation of trigonometric ratios from 0 to 90 degrees
As θ increases from 0 to 90
• siθ increases from 0 to 1
• coθ decreases from 1 to 0
• taθ increases from 0 to
• coseθ decreases from to 1
• seθ increases from 1 to
• coθ decreases from to 0
### Standard values of Trigonometric ratios
∠A 0o 30o 45o 60o 90o sin A 0 1/2 1/√2 √3/2 1 cos A 1 √3/2 1/√2 1/2 0 tan A 0 1/√3 1 √3 not defined cosec A not defined 2 √2 2/√3 1 sec A 1 2/√3 √2 2 not defined cot A not defined √3 1 1/√3 0
To know more about Trigonometric Ratios of Standard Angles, visit here.
## Trigonometric Ratios of Complementary Angles
### Complementary Trigonometric ratios
If θ is an acute angle, its complementary angle is 90θ. The following relations hold true for trigonometric ratios of complementary angles.
• si(90− θcoθ
• co(90− θsiθ
• ta(90− θcoθ
• co(90− θtaθ
• cose(90− θseθ
• se(90− θcoseθ
To know more about Trigonometric Ratios of Complementary Angles, visit here.
## Trigonometric Identities
• sin2θ+cos2θ=1
• 1+cot2θ=coesc2θ
• 1+tan2θ=sec2θ
To know more about Trigonometric Identities, visit here.
### Trigonometry for Class 10 Problems
Example 1:
Find Sin A and Sec A, if 15 cot A = 8.
Solution:
Given that 15 cot A = 8
Therefore, cot A = 8/15.
We know that tan A = 1/ cot A
Hence, tan A = 1/(8/15) = 15/8.
Thus, Side opposite to ∠A/Side Adjacent to ∠A = 15/8
Let BC be the side opposite to ∠A and AB be the side adjacent to ∠A and AC be the hypotenuse of the right triangle ABC respectively.
Hence, BC = 15x and AB = 8x.
Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.
(i.e) AC2 = AB2 + BC2
AC2 = (8x)2+(15x)2
AC2 = 64x2+225x2
AC2 = 289x2
AC = 17x.
Therefore, the hypotenuse AC = 17x.
Finding Sin A:
We know Sin A = Side Opposite to ∠A / Hypotenuse
Sin A = 15x/17x
Sin A = 15/17.
Finding Sec A:
To find Sec A, find cos A first.
Thus, cos A = Side adjacent to ∠A / Hypotenuse
Cos A = 8x/17x
We know that sec A = 1/cos A.
So, Sec A = 1/(8x/17x)
Sec A = 17x/8x
Sec A = 17/8.
Therefore, Sin A = 15/17 and sec A = 17/8.
Example 2:
If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that 0° <A+B ≤ 90°; A>B ]
Solution:
Given that
Tan (A+B) = √3.
We know that tan 60 = √3.
Thus, tan (A+B) = tan 60° = √3.
Hence A+B= 60° …(1)
Similarly, given that,
Tan (A-B) = 1/√3.
We know that tan 30° = 1/√3.
Thus, tan (A-B) = tan 30° = 1/√3.
Hence, A-B = 30° …(2)
Now, adding the equations (1) and (2), we get
A+B+A-B = 60° + 30°
2A = 90°
A = 45°.
Now, substitute A = 45° in equation (1), we get
45° +B = 60°
B = 60°- 45°
B = 15°
Hence, A = 45 and B = 15°.
Stay tuned with BYJU’S – The Learning App and download the app to learn all Maths-related concepts easily by exploring more videos.<|endoftext|>
| 4.8125 |
431 |
+0
# Pls help thks
+1
137
1
Danny has 2/3 as many toy cars as Jack. They have a total of 75 toy cars. How many toy cars does Jack have at first? How many toy cars must Jack give to Danny so that Danny will have 35 more toy cars than Jack?
Nov 2, 2021
#1
+121
+1
Step 1- Given Informations : -
Danny has 2/3 as many toy cars as Jack. They have a total of 75 toy cars. How many toy cars does Jack have at first? How many toy cars must Jack give to Danny so that Danny will have 35 more toy cars than Jack?
According to Given Problem : -
Danny has 2/3 as many toy cars as Jack.
total toy cars = 75
2/3J + J = 75 (According to Question)
5J/3 = 75
J $$= {75 * 3\over 5}$$
J = 45
So, (Jack have = 45 toy cars at first)
So, Danny have = 2/3 * J
= 2/3 * 45
= 30
(Danny have = 30 toy cars at First)
Step 2- Now,
We have to Find out how many cars (toy)
must Jack Give to Danny so that Danny
will have 35 more toy cars than Jack?
Jack will Give (45-x) (∴x = Number of toy cars )
Jack Give to Danny
Now, Danny will have (30+x)
So, According to Guide Situation : -
(45-x) + 35 = 30 + x
-x - x = 30 x 35 - 45
-2x = 30 - 80
-2x = -50
x = 25
(25 toy cars must Jack give to Danny, so
that Danny will have 35 more toy cars
than Jack)
Dec 4, 2021<|endoftext|>
| 4.53125 |
4,373 |
# Volumes and Surface Areas
1. The volume of a sphere is $\displaystyle\frac{{88}}{{21}} \times {14^3}c{m^3}$ The curved surface of its sphere is :
a. 2424 $c{m^2}$
b. 2446 $c{m^2}$
c. 2464 $c{m^2}$
d. 2484 $c{m^2}$
Correct Option: C
Volume of the sphere = $\displaystyle\frac{4}{3}\pi {r^3}$
$\displaystyle\frac{4}{3} \times \displaystyle\frac{{22}}{7} \times {r^3} = \displaystyle\frac{{88}}{{21}} \times {(14)^3} \Rightarrow r = 14$
Curved Surface area = $4\pi {r^2}$ = $4 \times \displaystyle\frac{{22}}{7} \times {14^2}$ = 2464 $c{m^2}$
2. If the radius of a sphere is doubled, then its volume is increased by :
a. 100%
b. 200%
c. 700%
d. 800%
Correct Option: D
Explanation:
Original volume = $\displaystyle\frac{4}{3}\pi {r^3}$
New volume = $\displaystyle\frac{4}{3}\pi {(2r)^3} = \displaystyle\frac{{32}}{3}\pi {r^3}$
Increase % = ${\left( {\displaystyle\frac{{32}}{3}\pi {r^3} \times \displaystyle\frac{3}{{4\pi {r^3}}}} \right)\% = 800\% }$
3. If the height of a cone is doubled, then its volume is increased by :
a. 100%
b. 200%
c. 300%
d. 400%
Correct Option: A
Explanation:
Original volume = $\displaystyle\frac{1}{3}\pi {r^2}h$
New volume $\displaystyle\frac{1}{3}\pi {r^2}(2h) = \displaystyle\frac{2}{3}\pi {r^2}h$
Increase % $\left( {\displaystyle\frac{{\displaystyle\frac{2}{3}\pi {r^2}h}}{{\displaystyle\frac{1}{3}\pi {r^2}h}} \times 100} \right)\% = 200\%$
4. The cost of painting the four walls of a room is Rs.350. The cost of painting a room three times in length, breadth and height will be :
a. Rs.1050
b. Rs.1400
c. Rs.3150
d. Rs.4200
Correct Option: C
Explanation:
Area of 4 walls of the room = $[2(l + b) \times h]{m^2}$
Area of 4 walls of new room = $[2(3l + 3b) \times 3h]{m^2}$ = $9[2(l + b) \times h]{m^2}$
Cost of painting the 4 walls of new room = Rs.$(9 \times 350) = Rs.3150$
5. The area of the base of a right circular cone is 154 $c{m^2}$ and its height is 14 cm. The curved surface of the cone is :
a. $(154 \times \sqrt 5 )c{m^2}$
b. 11 $c{m^2}$
c. $(154 \times \sqrt 7 )c{m^2}$
d. 5324 $c{m^2}$
Correct Option: A
Explanation:
$\displaystyle\frac{{22}}{7} \times {r^2} = 154 \Rightarrow {r^2} = \left( {154 \times \displaystyle\frac{7}{{22}}} \right)$ = 49 or r = 7 cm
Now, r = 7 and h = 14.
So, l = $\sqrt {{{(7)}^2} + {{(14)}^2}}$
= $\sqrt {245} = 7\sqrt 5$ cm
Area of curved surface = $\pi rl$ = $\left( {\displaystyle\frac{{22}}{7} \times 7 \times 7\sqrt 5 } \right)c{m^2} = 154\sqrt 5 {\rm{ }}c{m^2}$
6. The material of a cone is converted into the shape of a cylinder of equal radius. If the height of the cylinder is 5 cm, the height of the cone is :
a. 10 cm
b. 15 cm
c. 18 cm
d. 24 cm
Correct Option: B
Explanation:
$\displaystyle\frac{1}{3}\pi {r^2} \times h = \pi {h^2} \times 5$ or h = 15 cm
7. A sold consists of a circular cylinder with an exact fitting right circular cone placed on the top. The height of the cone is h. If the total volume of the solid is three times the volume of the cone, then the height of the cylinder is :
a. 2 h
b. 4 h
c. $\displaystyle\frac{{2h}}{3}$
d. $\displaystyle\frac{{3h}}{3}$
Correct Option: C
Explanation:
Let the height of the cylinder be H and its radius = r.
Then, $\pi {r^2}H + \displaystyle\frac{1}{3}\pi {r^2}h$ = $3 \times \displaystyle\frac{1}{3}\pi {r^2}h$
$\Rightarrow \pi {r^2}H = \displaystyle\frac{2}{3}\pi {r^2}h$ or H = $\displaystyle\frac{2}{3}h$
8. If the volumes of two cones are in the ratio 1:4 and their diameters are in the ratio 4:5, then the ratio of their height is :
a. 1:5
b. 5:4
c. 5:16
d. 25:64
Correct Option: D
Explanation:
Since the diameters are in the ratio 4:5, it follows that their radii are in the ratio 4:5. Let them be 4 r and 5 r.
Let the height be h and H.
Ratio of volumes = $\displaystyle\frac{{\displaystyle\frac{1}{3}\pi \times {{(4r)}^2} \times h}}{{\displaystyle\frac{1}{3}\pi \times {{(5r)}^2} \times H}} = \displaystyle\frac{{16h}}{{25H}}$
$\displaystyle\frac{{16h}}{{25H}} = \displaystyle\frac{1}{4}{\rm{ or }}\displaystyle\frac{h}{H}{\rm{ = }}\left( {\displaystyle\frac{1}{4} \times \displaystyle\frac{{25}}{{16}}} \right){\rm{ = }}\displaystyle\frac{{25}}{{64}}{\rm{ = 25:64}}$
9. If a rightg circular cone of vertical height 24 cm has a volume of 1232 $c{m^3}$, then the area of its curved surface is :
a. 1254 $c{m^2}$
b. 704 $c{m^2}$
c. 550 $c{m^2}$
d. 154 $c{m^2}$
Correct Option: C
Explanation :
$\displaystyle\frac{1}{3} \times \displaystyle\frac{{22}}{7} \times {r^2} \times 24 = 1232$ or
$\left( {1232 \times \displaystyle\frac{7}{{22}} \times \displaystyle\frac{3}{{24}}} \right) = 49$
$\Rightarrow {r^2} = \left( {1232 \times \displaystyle\frac{7}{{22}} \times \displaystyle\frac{3}{{24}}} \right) = 49$
$\Rightarrow$ r = 7 cm
Now r = 7 and h = 24, So, l = $\sqrt {{7^2} + {{(24)}^2}} = \sqrt {625} = 25$ cm
Curved surface area = $\pi {\rm{rl}}$ = $\left( {\displaystyle\frac{{22}}{7} \times 7 \times 25} \right)c{m^2} = 550c{m^2}$
10. A cylindrical piece of metal of radius 2cm and height 6 cm is shaped into a cone of same radius. The height of the cone is :
a. 18cm
b. 14cm
c. 12cm
d. 8 cm
Correct Option: A
Explanation:
${\displaystyle\frac{1}{3}\pi \times {{(2)}^2} \times h = \pi \times {{(2)}^2} \times 6 \Rightarrow h = 18}$ cm
11. The radii of two cylinder are in the ratio of 2:3 and their heights are in the ratio 5:3. The ratio of their volumes is :
a. 27:20
b. 20:27
c. 4 : 9
d. 9 : 4
Correct Option: B
Explanation:
Let their radii be 2 r and 3r and heights 5h and 3h respectively. Ratio of their volumes.
=$\displaystyle\frac{{\pi {{(2r)}^2} \times 5h}}{{\pi {{(3r)}^2} \times 3h}} = \displaystyle\frac{{20}}{{27}} = 20:27$
12. If the volume and surface area of a sphere are numerically the same, then its radius is :
a. 1 unit
b. 2 units
c. 3 units
d. 4 units
Correct Option: C
Explanation:
$\displaystyle\frac{4}{3}\pi {r^3} = 4\pi {r^2} \Rightarrow r = 3$ units
a. 800
b. 125
c. 400
d. 8000
Correct Option: D
Explanatioin:
Number of balls = $\displaystyle\frac{{{\rm\text{Volume of big ball}}}}{{{\rm\text{Volume of small ball}}}}$ $= \displaystyle\frac{{\displaystyle\frac{4}{3} \times \pi \times 10 \times 10 \times 10}}{{\displaystyle\frac{4}{3} \times \pi \times 0.5 \times 0.5 \times 0.5}} = 800$
14. The radii of two spheres are in the ratio 1:2. The ratio of their surface areas is :
a. 1 : 2
b. 1 : 4
c. 1 : $\sqrt 2$
d. 3 : 8
Correct Option: B
Explanation:
Let their radii be x and 2 x.
Ratio of their surface areas = $\displaystyle\frac{{4\pi {x^2}}}{{4\pi {{(2x)}^2}}} = \displaystyle\frac{1}{4} = 1:4$
15. A right cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is :
a. 3 : 5
b. 2 : 5
c. 3 : 1
d. 1 : 3
Correct Option: D
Explanation :
Let the height of cylinder = h
and height of cone = H
Then, ${\pi ^2}h = \displaystyle\frac{1}{3}\pi {r^2}H$ or $\displaystyle\frac{h}{H} = \displaystyle\frac{1}{3} = 1:3$
16. The radius of a circular cylinder is the same as that of a sphere. Their volumes are equal. The height of the cylinder is :
a. $\displaystyle\frac{4}{3}$ times its radius
b. $\displaystyle\frac{2}{3}$ times its radius
d. equal to its diameter
Correct Option: A
Explanation:
$\displaystyle\frac{4}{3}\pi {r^3} = \pi {r^2}h \Rightarrow h = \displaystyle\frac{4}{3}r$
Height = $\displaystyle\frac{4}{3}$ times of its radius.
17. The number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is :
a. 3
b. 4
c. 5
d. 6
Correct Option: C
Explanation:
Let the number of spheres be x.
Then, $x \times \displaystyle\frac{4}{3}\pi \times {(3)^3} = \pi \times {(2)^2} \times 45$
or 36x = 180 or x = $\displaystyle\frac{{180}}{{36}} = 5$
18. The length of the wire of 0.2 mm radius that can be drawn after melting a solid copper sphere of diameter 18 cm is :
a. 24.3 m
b. 243 m
c. 2430 m
d. 24300m
Correct Option: D
Explanation:
Radius of sphere = 9 cm
Volume of sphere = $\left[ {\displaystyle\frac{4}{3} \times \pi {{(9)}^3}} \right]c{m^3} = (972\pi )c{m^3}$ = $\displaystyle\frac{1}{{50}}$ cm
Let the length be = x cm
Then, $972\pi = \pi \times {\left( {\displaystyle\frac{1}{{50}}} \right)^2} \times x$ $\Rightarrow x = (972 \times 50 \times 50)$ cm
Length of wire = $\left( {\displaystyle\frac{{972 \times 50 \times 50}}{{100}}} \right)$ m = 24300 m
19. The radius of a wire is decreased to one-third. If volume remains the same, length will increase:
a. 1 time
b. 3 times
c. 6 times
d. 9 times
Correct Option: D
Explanation:
Let original radius = r and original length = h
Now radius = $\displaystyle\frac{1}{3}r$, Let new length = H
Then, $\pi {r^2}{\rm{ h = }}\pi {\left( {\displaystyle\frac{1}{3}r} \right)^2} \times H = \displaystyle\frac{{\pi {r^2}H}}{9}$ H = 9h
Thus, the length becomes 9 times.
20. If the volumes of two cubes are in the ratio 8:1, the ratio of their edges is :
a. 8 : 1
b. 2$\sqrt 2$ : 1
c. 2 : 1
d. None of these
Correct Option: C
Explanation:
Let their volumes be $8{x^2}$ and ${x^3}$ respectively
Then, their edges are
2x and x respectively
Ratio of their edges = 2 : 1
21. A metal sheet 27 cm long 8 cm broad and 1 cm thick is melted into a cube. The difference between the surface areas of two solids is :
a. 284 $c{m^2}$
b. 286 $c{m^2}$
c. 296 $c{m^2}$
d. 300 $c{m^2}$
Correct Option: B
Explanatioin:
Volume of sheet = $(27 \times 8 \times 1){\rm{ }}c{m^2} = 216{\rm{ }}c{m^2}$
Volume of cube formed = 216 $c{m^2}$
Edge of the cube = ${(6 \times 6 \times 6)^{1/3}} = 6$ cm
Surface area of the cube formed = $\left[ {6 \times {{(6)}^2}} \right]$ $c{m^2}$ = 216 $c{m^2}$
Difference in areas = (502-216)$c{m^2}$=286 $c{m^2}$
22. A wooden box of dimensions $8m \times 7m \times 6m$ is to carrry rectangular boxes of dimensions $8cm \times 7cm \times 6cm$ . The maximum number of boxes that can be carried in 1 wooden box is :
a. 1200000
b. 1000000
c. 9800000
d. 7500000
Correct Option: B
Explanation:
Number of boxes = $\displaystyle\frac{{{\rm\text{Volume of wooden box in c}}{{\rm{m}}^{\rm{3}}}}}{{{\rm\text{Volume of 1 small ball}}}}$ = $\displaystyle\frac{{800 \times 700 \times 600}}{{8 \times 7 \times 6}} = 1000000$
23. If the length, breadth and the height of a cuboid are in the ratio 6:5:4 and if the total surface area is 33300 $c{m^2}$, then the length , breadth and height in cms, are respectively.
a. 90,85,60
b. 85,75,60
c. 90,75,70
d. 90,75,60
Correct Option: D
Explanation:
Let length = 6x, breadth = 5x and height =4x in cm
$2(6x \times 5x + 5x \times 4x + 6x \times 4x) = 33300$
$148{x^2} = 33300 \Rightarrow {x^2}$=$\displaystyle\frac{{33300}}{{148}} = 225 \Rightarrow x = 15$
Length = 90 cm, Breadth = 75 cm and Height = 60 cm
24. The maximum length of a pencil that can be kept in a rectangular box of dimensions $8cm \times 6cm \times 2cm$, is :
a. $2\sqrt {13}$ cm
b. $2\sqrt {14}$ cm
c. $2\sqrt {26}$ cm
d. $10\sqrt 2$ cm
Correct Option: C
Explanation:
Length of pencil = $\sqrt {{{(8)}^2} + {{(6)}^2} + {{(2)}^2}}$ cm = $\sqrt {104}$ cm = 2$\sqrt {26}$ cm
25. The surface area of a cube is 726 ${m^2}$.Its volume is:
a. 1300 ${m^3}$
b. 1331 ${m^3}$
c. 1452 ${m^3}$
d. 1542 ${m^3}$
Correct Option: B
Explanation:
$6{a^2} = 726 \Rightarrow {a^2} = 121 \Rightarrow a = 11$ cm
Volume of the cube = $(11 \times 11 \times 11)c{m^3}$ = 1331$c{m^3}$<|endoftext|>
| 4.46875 |
662 |
Videos
Choose from over 80 mathematical videos.
Our team of top teacher mathematicians have put together all the important topics and explain them with the types of examples that you would find in a normal textbook or exam.
• Fractions - basic ideas
What are fractions? What do fractions look like? What are equivalent fractions? What are the different types of fractions? This video explores all these concepts.
Length: 26 minutes
• Fractions - Multiplying and Dividing
This video shows how to multiply fractions and how to divide fractions by turning the second fraction upside down.
Length: 33 minutes
• Hyperbolic functions
Hyperbolic functions. The hyperbolic functions have similar names to the trigonometric functions, but they are defined in terms of the exponential function. This unit defines the three main hyperbolic functions and sketches their graphs. Inverse functions and reciprocal functions are also considered.
Length: 29 minutes
• Implicit Differentiation
Implicit differentiation . Sometimes functions are not given in the form y = f(x) but in a more complicated form where it is difficult to express y explicitly in terms of x. These are called implicit functions, and they can be differentiated to give dy/dx.
Length: 30 minutes
• Integration as a summation
Integration as summation . Integration may be introduced as a means of finding areas using summation and limits. This process gives rise to the definite integral of a function.
Length: 41 minutes
• Integration as the reverse of differentiation
Integration as the reverse of differentiation. Integration can be seen as differentiation in reverse; that is we start with a given function f(x), and ask which functions, F(x), would have f(x) as their derivative. The result is called an indefinite integral. A definite integral can be obtained by substituting values into the indefinite integral.
Length: 34 minutes
• Integration by Parts
Integration by parts. A special rule, integration by parts, can often be used to integrate the product of two functions. It is appropriate when one of the functions forming the product is recognised as the derivative of another function. The result still involves an integral, but in many cases the new integral will be simpler than the original one.
Length: 26 minutes
• Integration by substitution
Integration by substitution . There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable and the integrand. With definite integrals the limits of integration can also change.
Length: 36 minutes
• Integration leading to log functions
Integration leading to log functions . The derivative of x is 1/x. As a consequence, if we reverse the process, the integral of 1/x is ln x + c. In this unit we generalise this result and see how a wide variety of integrals result in logarithm functions.
Length: 8 minutes
• Integration using a table of anti-derivatives
Integration using a table of anti-derivatives. Integration may be regarded as the reverse of differentiation, so a table of derivatives can be read backwards as a table of anti-derivatives. The final result for an indefinite integral must, however, include an arbitrary constant.
Length: 15 minutes<|endoftext|>
| 4.625 |
744 |
Measurement While Drilling (MWD), also known as Logging While Drilling (LWD), is a measurement taken of the wellbore (the hole) inclination from vertical, and also magnetic direction from north. Using basic trigonometry, a three-dimensional plot of the path of the well can be produced.
Essentially, a MWD Operator measures the trajectory of the hole as it is drilled (for example, data updates arrive and are processed every few seconds or faster). This information is then used to drill in a pre-planned direction into the formation which contains the oil, gas, water or condensate. Additional measurements can also be taken of natural gamma ray emissions from the rock; this helps broadly to determine what type of rock formation is being drilled, which in turn helps confirm the real-time location of the wellbore in relation to the presence of different types of known formations (by comparison with existing seismic data).
Density and porosity, rock fluid pressures and other measurements are taken, some using radioactive sources, some using sound, some using electricity, etc.; this can then be used to calculate how freely oil and other fluids can flow through the formation, as well as the volume of hydrocarbons present in the rock and, with other data, the value of the whole reservoir and reservoir reserves.
An MWD downhole tool is also “lined-up” with the bottom hole drilling assembly, enabling the wellbore to be steered in a chosen direction in 3D space known as directional drilling. Directional drillers rely on receiving accurate, quality tested data from the MWD engineer to allow them to keep the well safely on the planned trajectory.
Directional survey measurements are taken by three orthogonally mounted accelerometers to measure inclination, and three orthogonally mounted magnetometers which measure direction (azimuth). Gyroscopic tools may be used to measure Azimuth where the survey is measured in a location with disruptive external magnetic influences, inside “casing”, for example, where the hole is lined with steel tubulars (tubes). These sensors, as well as any additional sensors to measure rock formation density, porosity, pressure or other data, are connected, physically and digitally, to a logic unit which converts the information into binary digits which are then transmitted to surface using “mud pulse telemetry” (MPT, a binary coding transmission system used with fluids, such as, combinatorial, Manchester encoding, split-phase, among others).
This is done by using a downhole “pulser” unit which varies the drilling fluid (mud) pressure inside the drill-string according to the chosen MPT: these pressure fluctuations are decoded and displayed on the surface system computers as wave-forms; voltage outputs from the sensors (raw data); specific measurements of gravity or directions from magnetic north, or in other forms, such as sound waves, nuclear wave-forms, etc.
Surface (mud) pressure transducers measure these pressure fluctuations (pulses) and pass an analogue voltage signal to surface computers which digitize the signal. Disruptive frequencies are filtered out and the signal is decoded back into its original data form. For example, a pressure fluctuation of 20psi (or less) can be “picked out” of a total mud system pressure of 3,500psi or more.
Downhole electrical and mechanical power is provided by downhole turbine systems, which use the energy of the “mud” flow, battery units (lithium), or a combination of both.<|endoftext|>
| 3.765625 |
1,120 |
10 Key Accounting Terms
Knowing the correct accounting terms and what they mean can make a world of difference when you’re deciphering financial statements and reports and determining profits and losses. It’s easy to get debits and credits confused, and it’s a must to know which documents make up a complete financial report. A ton of cash could depend on your understanding of the following basic accounting terms:
- Accounting: The methods and procedures for identifying, analyzing, recording, accumulating, and storing information and data about the activities of an entity that has financial results and for preparing summary reports of these activities internally for managers and externally for those entitled to receive financial reports about the entity. A business’s managers, investors, and lenders depend on accounting reports called financial statements to make informed decisions. Accounting also encompasses preparing tax returns that the entity must file with government tax authorities and facilitating day-to-day operating functions.
- Balance sheet: This financial statement summarizes the assets, liabilities, and owners’ equity of a business at a moment in time. It’s prepared at the end of every profit period (and whenever else it’s needed). The main elements of a balance sheet are called accounts — such as cash, inventory, notes payable, and capital stock. Each account has a dollar amount, which is called its balance. But be careful: The fact that the accounts have balances is not the reason this financial statement is called a balance sheet; rather, the name refers to the equality (or balance) of assets with the total of liabilities and owners’ equity. This financial statement is also called the statement of financial condition and the statement of financial position.
- Cash flow: An ambiguous term that can refer to several different sources of or uses of cash. This term is often shorthand for cash flow from earning profit or from operating activities. Some friendly advice: When using this term, always make clear the particular source or use of cash you have in mind!
- Debits and credits: Accounting jargon for decreases and increases recorded in accounts according to the centuries-old scheme based on the accounting equation (Assets = Liabilities + Owners’ equity, or Assets = Sources of assets). An increase in an asset is a debit, and the ingenious twist of the scheme is that a decrease in a liability or an owners’ equity is also a debit. Conversely, a decrease in an asset is a credit, and an increase in a liability or an owners’ equity is a credit.
Revenue is recorded as a credit, and expenses are recorded as debits. In recording transactions, the debit or sum of debits must equal the credit or sum of credits. The phrase “the books are in balance” means that the total of accounts with debit balances equals the total of accounts with credit balances.
- Financial reports: The periodic financial communications from a business (and other types of organizations) to those entitled to know about the financial performance and position of the entity. Financial reports of businesses include three primary financial statements (the balance sheet, income statement, and statement of cash flows) as well as footnotes and other information relevant to the owners of the business. Public companies must file several types of financial reports and forms with the Securities and Exchange Commission (SEC), which are open to the public. The financial reports of a private business are generally sent only to its owners and lenders.
- Financial statement: Generally refers to one of the three primary accounting reports of a business: the balance sheet, statement of cash flows, and income statement. Sometimes financial statements are simply called financials. Internal financial statements and other accounting reports to managers contain considerably more detail, which is needed for decision-making and control.
- Fixed assets: The shorthand term for the variety of long-life physical resources used by a business in conducting its operations, which include land, buildings, machinery, equipment, furnishings, tools, and vehicles. These resources are held for use, not for sale. Please note that fixed assets is an informal term; the more formal term used in a balance sheet is property, plant, and equipment.
- Generally accepted accounting principles (GAAP): The authoritative standards and approved accounting methods that should be used by profit-motivated businesses and private not-for-profit organizations domiciled in the United States to measure and report their revenue and expenses; to present their assets, liabilities, and owners’ equity; and to report their cash flows in their financial statements. GAAP are not a straitjacket; these official standards are loose enough to permit alternative interpretations.
- Income statement: This financial statement summarizes sales revenue (and other income) and expenses (and losses) for a period and reports one or more different profit lines. Also, any unusual gains and losses are reported separately in this financial statement. The income statement is one of the three primary financial statements of a business included in its financial report and is also called the earnings statement, operating statement, or similar titles.
- Profit: A very general term that is used with different meanings. It may mean gains minus losses or other kinds of increases minus decreases. In business, the term means sales revenue (and other sources of income) minus expenses (and losses) for a period of time, such as one year. In an income statement, the preferred term for final or bottom-line profit is net income. For public companies, net income is put on a per-share basis, called earnings per share.<|endoftext|>
| 3.84375 |
1,596 |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 5.4: Bipartite Graphs
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
We have already seen how bipartite graphs arise naturally in some circumstances. Here we explore bipartite graphs a bit more.
It is easy to see that all closed walks in a bipartite graph must have even length, since the vertices along the walk must alternate between the two parts. Remarkably, the converse is true. We need one new definition:
Definition: distance between vertices
The distance between vertices $$v$$ and $$w$$, $$\d(v,w)$$, is the length of a shortest walk between the two. If there is no walk between $$v$$ and $$w$$, the distance is undefined.
Theorem 5.4.2
$$G$$ is bipartite if and only if all closed walks in $$G$$ are of even length.
Proof
The forward direction is easy, as discussed above.
Now suppose that all closed walks have even length. We may assume that $$G$$ is connected; if not, we deal with each connected component separately.
Let $$v$$ be a vertex of $$G$$, let $$X$$ be the set of all vertices at even distance from $$v$$, and $$Y$$ be the set of vertices at odd distance from $$v$$. We claim that all edges of $$G$$ join a vertex of $$X$$ to a vertex of $$Y$$. Suppose not; then there are adjacent vertices $$u$$ and $$w$$ such that $$\d(v,u)$$ and $$\d(v,w)$$ have the same parity. Then there is a closed walk from $$v$$ to $$u$$ to $$w$$ to $$v$$ of length $$\d(v,u)+1+\d(v,w)$$, which is odd, a contradiction.
$$\square$$
The closed walk that provides the contradiction is not necessarily a cycle, but this can be remedied, providing a slightly different version of the theorem.
Corollary 5.4.3
$$G$$ is bipartite if and only if all cycles in $$G$$ are of even length.
Proof
Again the forward direction is easy, and again we assume $$G$$ is connected. As before, let $$v$$ be a vertex of $$G$$, let $$X$$ be the set of all vertices at even distance from $$v$$, and $$Y$$ be the set of vertices at odd distance from $$v$$. If two vertices in $$X$$ are adjacent, or two vertices in $$Y$$ are adjacent, then as in the previous proof, there is a closed walk of odd length.
To finish the proof, it suffices to show that if there is a closed walk $$W$$ of odd length then there is a cycle of odd length. The proof is by induction on the length of the closed walk.
If $$W$$ has no repeated vertices, we are done. Otherwise, suppose the closed walk is
$$v=v_1,e_1,\ldots,v_i=v,\ldots,v_k=v=v_1.$$
Then
$$v=v_1,\ldots,v_i=v \quad\hbox{and}\quad v=v_i,e_i,v_{i+1},\ldots, v_k=v$$
are closed walks, both are shorter than the original closed walk, and one of them has odd length. By the induction hypothesis, there is a cycle of odd length.
$$\square$$
It is frequently fruitful to consider graph properties in the limited context of bipartite graphs (or other special types of graph). For example, what can we say about Hamilton cycles in simple bipartite graphs? Suppose the partition of the vertices of the bipartite graph is $$X$$ and $$Y$$. Because any cycle alternates between vertices of the two parts of the bipartite graph, if there is a Hamilton cycle then $$|X|=|Y|\ge2$$. In such a case, the degree of every vertex is at most $$n/2$$, where $$n$$ is the number of vertices, namely $$n=|X|+|Y|$$. Thus the Ore condition (\)\d(v)+\d(w)\ge n\) when $$v$$ and $$w$$ are not adjacent) is equivalent to $$\d(v)=n/2$$ for all $$v$$. This means the only simple bipartite graph that satisfies the Ore condition is the complete bipartite graph $$K_{n/2,n/2}$$, in which the two parts have size $$n/2$$ and every vertex of $$X$$ is adjacent to every vertex of $$Y$$. The upshot is that the Ore property gives no interesting information about bipartite graphs.
Of course, as with more general graphs, there are bipartite graphs with few edges and a Hamilton cycle: any even length cycle is an example.
We note that, in general, a complete bipartite graph $$K_{m,n}$$ is a bipartite graph with $$|X|=m$$, $$|Y|=n$$, and every vertex of $$X$$ is adjacent to every vertex of $$Y$$. The only such graphs with Hamilton cycles are those in which $$m=n$$.<|endoftext|>
| 4.6875 |
230 |
Everyone knows that eating fruit is an important part of a well balanced diet, but many fall short of eating their minimum 2 servings of fruit and 3 servings of vegetables per day.
Many fruits are low in sodium, have no cholesterol, have few calories, and are also low fat.
Fruits are packed with nutrients:
- Potassium: Potassium in fruit helps maintain a healthy blood pressure. Bananas, dried peaches, dried apricots, cantaloupe, honeydew melon, orange juice, prunes, and prune juice are some common fruits that contain Potassium.
- Fiber: Fiber is not only great for digestion, but more specifically, fiber from fruits helps reduce blood cholesterol and may help reduce risks of chronic heart disease. The fiber in fruits also helps the stomach feel fuller and provides other metabolic benefits.
- Vitamin C: Vitamin C is important for the growth and repair of all body tissues. It also helps heal cuts and wounds, and also keeps teeth and gums healthy.
- Folate: Folate or folic acid found in fruits helps the body form red blood cells.<|endoftext|>
| 4.03125 |
679 |
### Problem Statement
The usual longhand scheme for multiplying two numbers A and B is to multiply the last digit of B by A, shift left by one digit, multiply the second-to-last digit of B by A, and so on. This process is illustrated below:
``` 36
x 15
-----
180
+ 36
-----
540
```
But let's say we didn't multiply in the usual way. Let us define a new method of multiplication called "NC-Multiplication", where the "NC" stands for "No Carry". It is called this because we do not carry when numbers exceed 9, no matter what. To multiply 36 by 15 in this manner, we would do:
``` 3 6
x 1 5
---------
15 30
+ 3 6
---------
3 21 30
```
and so the result would be {3, 21, 30}.
You will be given a int[] digits, that represents the result of NC-multiplying two numbers together. You wish to factor this result by finding the two numbers that multiplied together to form the result. There may be multiple pairs of numbers that work. If we call the larger number A and the smaller B, then we want the pair such that A - B is minimized. Of this pair, your method should return a long that is equal to A. If no such A and B exist that NC-multiply to digits, your method should return -1.
### Definition
Class: NCMultiplication Method: findFactors Parameters: int[] Returns: long Method signature: long findFactors(int[] digits) (be sure your method is public)
### Constraints
-digits will contain between 1 and 15 elements, inclusive.
-All elements of digits will be between 0 and 2000, inclusive.
-At least one element in digits will be nonzero.
-The number represented by digits will be less than 1014 = 100000000000000.
-There will be no leading or trailing zeros in digits.
### Examples
0)
`{3,21,30}`
`Returns: 36`
36 and 15 NC-Multiply together to make {3,21,30}, as seen above.
1)
`{15,3,6}`
`Returns: 512`
2)
`{4,20,25}`
`Returns: 25`
25 NC-Multiplied by 25.
3)
`{6,61,124,129,90,27}`
`Returns: 6773`
4)
`{8,14,22,95,125,120,73,9,9}`
`Returns: -1`
5)
`{6, 5, 32, 68, 113, 143, 143, 124, 100, 75, 48, 23, 7, 1}`
`Returns: 65864431`
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=2416
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=5075&pm=2416
antimatter
#### Testers:
lbackstrom , brett1479<|endoftext|>
| 4.375 |
2,370 |
Axoneme: A bundle of MICROTUBULES and MICROTUBULE-ASSOCIATED PROTEINS forming the core of each CILIUM or FLAGELLUM. In most eukaryotic cilia or flagella, an axoneme shaft has 20 microtubules arranged in nine doublets and two singlets.Sperm Tail: The posterior filiform portion of the spermatozoon (SPERMATOZOA) that provides sperm motility.Flagella: A whiplike motility appendage present on the surface cells. Prokaryote flagella are composed of a protein called FLAGELLIN. Bacteria can have a single flagellum, a tuft at one pole, or multiple flagella covering the entire surface. In eukaryotes, flagella are threadlike protoplasmic extensions used to propel flagellates and sperm. Flagella have the same basic structure as CILIA but are longer in proportion to the cell bearing them and present in much smaller numbers. (From King & Stansfield, A Dictionary of Genetics, 4th ed)Cilia: Populations of thin, motile processes found covering the surface of ciliates (CILIOPHORA) or the free surface of the cells making up ciliated EPITHELIUM. Each cilium arises from a basic granule in the superficial layer of CYTOPLASM. The movement of cilia propels ciliates through the liquid in which they live. The movement of cilia on a ciliated epithelium serves to propel a surface layer of mucus or fluid. (King & Stansfield, A Dictionary of Genetics, 4th ed)Dyneins: A family of multisubunit cytoskeletal motor proteins that use the energy of ATP hydrolysis to power a variety of cellular functions. Dyneins fall into two major classes based upon structural and functional criteria.Chlamydomonas: A genus GREEN ALGAE in the order VOLVOCIDA. It consists of solitary biflagellated organisms common in fresh water and damp soil.Microtubules: Slender, cylindrical filaments found in the cytoskeleton of plant and animal cells. They are composed of the protein TUBULIN and are influenced by TUBULIN MODULATORS.Chlamydomonas reinhardtii: A species of GREEN ALGAE. Delicate, hairlike appendages arise from the flagellar surface in these organisms.Sperm Motility: Movement characteristics of SPERMATOZOA in a fresh specimen. It is measured as the percentage of sperms that are moving, and as the percentage of sperms with productive flagellar motion such as rapid, linear, and forward progression.Axonemal Dyneins: Dyneins that are responsible for ciliary and flagellar beating.Spermatozoa: Mature male germ cells derived from SPERMATIDS. As spermatids move toward the lumen of the SEMINIFEROUS TUBULES, they undergo extensive structural changes including the loss of cytoplasm, condensation of CHROMATIN into the SPERM HEAD, formation of the ACROSOME cap, the SPERM MIDPIECE and the SPERM TAIL that provides motility.Tubulin: A microtubule subunit protein found in large quantities in mammalian brain. It has also been isolated from SPERM FLAGELLUM; CILIA; and other sources. Structurally, the protein is a dimer with a molecular weight of approximately 120,000 and a sedimentation coefficient of 5.8S. It binds to COLCHICINE; VINCRISTINE; and VINBLASTINE.Sea Urchins: Somewhat flattened, globular echinoderms, having thin, brittle shells of calcareous plates. They are useful models for studying FERTILIZATION and EMBRYO DEVELOPMENT.Spermatids: Male germ cells derived from the haploid secondary SPERMATOCYTES. Without further division, spermatids undergo structural changes and give rise to SPERMATOZOA.Kartagener Syndrome: An autosomal recessive disorder characterized by a triad of DEXTROCARDIA; INFERTILITY; and SINUSITIS. The syndrome is caused by mutations of DYNEIN genes encoding motility proteins which are components of sperm tails, and CILIA in the respiratory and the reproductive tracts.Microtubule Proteins: Proteins found in the microtubules.Centrioles: Self-replicating, short, fibrous, rod-shaped organelles. Each centriole is a short cylinder containing nine pairs of peripheral microtubules, arranged so as to form the wall of the cylinder.Microscopy, Electron: Microscopy using an electron beam, instead of light, to visualize the sample, thereby allowing much greater magnification. The interactions of ELECTRONS with specimens are used to provide information about the fine structure of that specimen. In TRANSMISSION ELECTRON MICROSCOPY the reactions of the electrons that are transmitted through the specimen are imaged. In SCANNING ELECTRON MICROSCOPY an electron beam falls at a non-normal angle on the specimen and the image is derived from the reactions occurring above the plane of the specimen.Ciliary Motility Disorders: Conditions caused by abnormal CILIA movement in the body, usually causing KARTAGENER SYNDROME, chronic respiratory disorders, chronic SINUSITIS, and chronic OTITIS. Abnormal ciliary beating is likely due to defects in any of the 200 plus ciliary proteins, such as missing motor enzyme DYNEIN arms.Protozoan Proteins: Proteins found in any species of protozoan.Eukaryota: One of the three domains of life (the others being BACTERIA and ARCHAEA), also called Eukarya. These are organisms whose cells are enclosed in membranes and possess a nucleus. They comprise almost all multicellular and many unicellular organisms, and are traditionally divided into groups (sometimes called kingdoms) including ANIMALS; PLANTS; FUNGI; and various algae and other taxa that were previously part of the old kingdom Protista.Spermatogenesis: The process of germ cell development in the male from the primordial germ cells, through SPERMATOGONIA; SPERMATOCYTES; SPERMATIDS; to the mature haploid SPERMATOZOA.Algal Proteins: Proteins found in any species of algae.Microscopy, Interference: The science and application of a double-beam transmission interference microscope in which the illuminating light beam is split into two paths. One beam passes through the specimen while the other beam reflects off a reference mirror before joining and interfering with the other. The observed optical path difference between the two beams can be measured and used to discriminate minute differences in thickness and refraction of non-stained transparent specimens, such as living cells in culture.Microscopy, Electron, Transmission: Electron microscopy in which the ELECTRONS or their reaction products that pass down through the specimen are imaged below the plane of the specimen.Microtubule-Associated Proteins: High molecular weight proteins found in the MICROTUBULES of the cytoskeletal system. Under certain conditions they are required for TUBULIN assembly into the microtubules and stabilize the assembled microtubules.Trypanosoma brucei brucei: A hemoflagellate subspecies of parasitic protozoa that causes nagana in domestic and game animals in Africa. It apparently does not infect humans. It is transmitted by bites of tsetse flies (Glossina).Movement: The act, process, or result of passing from one place or position to another. It differs from LOCOMOTION in that locomotion is restricted to the passing of the whole body from one place to another, while movement encompasses both locomotion but also a change of the position of the whole body or any of its parts. Movement may be used with reference to humans, vertebrate and invertebrate animals, and microorganisms. Differentiate also from MOTOR ACTIVITY, movement associated with behavior.Paramecium: A genus of ciliate protozoa that is often large enough to be seen by the naked eye. Paramecia are commonly used in genetic, cytological, and other research.Mutation: Any detectable and heritable change in the genetic material that causes a change in the GENOTYPE and which is transmitted to daughter cells and to succeeding generations.Testis: The male gonad containing two functional parts: the SEMINIFEROUS TUBULES for the production and transport of male germ cells (SPERMATOGENESIS) and the interstitial compartment containing LEYDIG CELLS that produce ANDROGENS.Kinesin: A microtubule-associated mechanical adenosine triphosphatase, that uses the energy of ATP hydrolysis to move organelles along microtubules toward the plus end of the microtubule. The protein is found in squid axoplasm, optic lobes, and in bovine brain. Bovine kinesin is a heterotetramer composed of two heavy (120 kDa) and two light (62 kDa) chains. EC 3.6.1.-.Infertility, Male: The inability of the male to effect FERTILIZATION of an OVUM after a specified period of unprotected intercourse. Male sterility is permanent infertility.Microscopy, Immunoelectron: Microscopy in which the samples are first stained immunocytochemically and then examined using an electron microscope. Immunoelectron microscopy is used extensively in diagnostic virology as part of very sensitive immunoassays.Molecular Motor Proteins: Proteins that are involved in or cause CELL MOVEMENT such as the rotary structures (flagellar motor) or the structures whose movement is directed along cytoskeletal filaments (MYOSIN; KINESIN; and DYNEIN motor families).Molecular Sequence Data: Descriptions of specific amino acid, carbohydrate, or nucleotide sequences which have appeared in the published literature and/or are deposited in and maintained by databanks such as GENBANK, European Molecular Biology Laboratory (EMBL), National Biomedical Research Foundation (NBRF), or other sequence repositories.Amino Acid Sequence: The order of amino acids as they occur in a polypeptide chain. This is referred to as the primary structure of proteins. It is of fundamental importance in determining PROTEIN CONFORMATION.Models, Biological: Theoretical representations that simulate the behavior or activity of biological processes or diseases. For disease models in living animals, DISEASE MODELS, ANIMAL is available. Biological models include the use of mathematical equations, computers, and other electronic equipment.Drosophila melanogaster: A species of fruit fly much used in genetics because of the large size of its chromosomes.<|endoftext|>
| 3.875 |
245 |
Desertification is the spread or encroachment of a desert environment into arid or semiarid regions. It is caused by climatic changes, human influence or a combination of both. Climatic factors include temporary but severe drought periods and long term climatic changes resulting in aridity. Artificial changes in climate caused by human interferences such as the removal of vegetation (which can cause unnaturally high erosion), over-cultivation of land and excessive use of surface and groundwater are human factors that can result in desertification.
Desertification is the draining of the life-supporting capabilities of arid or semiarid land. Characteristic of this process is the declining of the groundwater table and depletion of surface water supplies, the salinisation of water and topsoil, increasing erosion and diminution of natural vegetation. Drought or misuse by humans makes land susceptible to desertification which spreads into arid and semiarid areas. Desertification can also occur within deserts when the ecological balance is disturbed, for example in the Sonoran and Chihuahuan deserts of the American Southwest.
Public awareness of desertification has increased in the late 20th century as more and more people and land are affected.<|endoftext|>
| 3.96875 |
1,362 |
Search 72,882 tutors
0 0
## Writing a system of equations and solving using elimination or substitution
Pam has \$6.35. She has only 35 coins, comprised of dimes and quarters. How many dimes and how many quarters does she have?
1. Write a system of equations that will solve the problem.
2. Solve the above problem by substitute or elimination method.
(1.) In writing a system of equations, first note what you are looking to solve for. In this problem, you are asked to find how many dimes and how many quarters are needed to make up a total amount of \$6.35 given that you have a total of 35 coins.
Let D represent the number of dimes and Q represent the number of quarters. From this, you can generate the following equation:
D + Q = 35
Since the value of a dime is \$0.10 and that of a quarter is \$0.25, then the sum of the # of dimes times the value of a dime (0.10) and the # of quarters times the value of a quarter (0.25) equals the total amount of \$6.35. That is,
(0.10)D + (0.25)Q = 6.35
Thus, the system of equations that will solve for the problem contains the following equations:
D + Q = 35
(0.10)D + (0.25)Q = 6.35
(2.) When solving a system of equations with 2 variables (i.e., D and Q), your goal is to eliminate one of the variables in order to solve for the other variable. Once you've solved for one of these variables, you can use it to solve for the other variable by plugging it into one of the original equations.
First, pick a variable to eliminate. I will choose to eliminate Q to demonstrate. If I multiply the second equation by -4 and add the 2 equations together, I will have eliminated Q and, therefore, be able to solve for D:
-4*[(0.10)D + (0.25)Q = 6.35] ==> (-0.40)D + (-1) Q = -25.4
==> (-0.40)D - Q = -25.4
Add the new equation to the first equation:
D + Q = 35
+ (-0.40)D - Q = -25.4
_______________________
D - (0.40)D + Q - Q = 9.60 ==> (0.60)D = 9.60
Divide both sides by 0.60 to solve for D:
(0.60)D / (0.60) = 9.60 / 0.60 ==> D = 16
Plug 16 in for D in the first equation to solve for Q:
D + Q = 35 ==> 16 + Q = 35
Subtract 16 from both sides of the equation:
16 + Q = 35
-16 -16
_________________
Q = 19
You can check your work by plugging in these values for D and Q into the second equation:
(0.10)D + (0.25)Q = (0.10)16 + (0.25)19 = 1.60 + 4.75 = 6.35
Thus, Pam has 16 dimes and 19 quarters.
Label the dimes as and the quarters as q. Since Pam has 35 coins in all, and 635 cents in dimes worth 10 cents and quarters worth 25 cents, we can set up two equations:
10d + 25q = 635
d + q = 35
__________________
Now multiply the second equation all the way across by -10, and add that result to the first equation.
10d + 25q = 635
-10d -10q = -350
__________________
Add the two equations to eliminate the d variable.
15q = 385
Divide by 15; q = 19.
Substitute 19 back into the first equation to solve for d:
d + 19 = 35
Thus d= 16.
Pam has 19 quarters and 16 dimes.
Hi Tracy,
The first step when converting a word problem to equations is to assign variables to the unknown quantities. In this case, we don't know how many of EACH type of coin she has, although we do know that she has 35 coins total.
So let's assign some variables - we'll call "d" the number of dimes she has, and "q" the number of quarters she has.
Since we know that she has 35 coins, we can write:
d + q = 35
Then, since we know that if she has "d" dimes, then the value of those dimes is 0.1 * d (because each dime is worth \$0.10). Also, if she has "q" quarters, then the value of those quarters is 0.25 * q (because each quarter is worth \$0.25). Then we can write a second equation, because we know that all of her coins put together are worth \$6.35:
0.1d + 0.25q = 6.35
So our two equations are:
d + q = 35
0.1d + 0.25q = 6.35
We can now solve. I like using the elimination method, so I will multiply the second equation by -10. Why -10? Because I see that 10 * 0.1 = 1, which is the coefficient of d in the first equation, so if we multiply by -10, then it will be -1, and then we can add the two equations to eliminate "d":
d + q = 35
-1d - 2.5q = -63.5
---------------------
- 1.5 q = 28.5
q = 19
Then we plug q back in to our d+q=35 equation to solve for d:
d + 19 = 35
d = 16
Thus, she has 16 dimes and 19 quarters.<|endoftext|>
| 4.65625 |
832 |
The authors do not work for, consult, own shares in or receive funding from any company or organisation that would benefit from this article, and have disclosed no relevant affiliations beyond their academic appointment.
Republish our articles for free, online or in print, under Creative Commons licence. Policymakers are tasked with addressing climate change in the face of uncertainty: The uncertainty is compounded by the fact that the consequences of any temperature change are unknown, including how something as basic as human fertility might be affected.
Understanding how climate change will affect fertility is an important economic concern.
According to World Bank estimates, in the United States and many European countries, a woman has fewer than two children on average by the end of her reproductive life. Any additional decline in births due to climate change could only make this worse.
First, hot weather could affect sexual behavior. After all, physically demanding activities are more difficult at high temperatures. Second, temperature could negatively influence reproductive health factors such as sperm motility and menstruation.
Expert colds sperm and 90 days naked pictures
There are some pretty compelling experimental studies on mammals to support this possibility. It is Colds sperm and 90 days two potential links that led us to hypothesize that global warming might be a threat to human reproduction, something that had yet to be thoroughly investigated by scientists and policymakers.
To isolate the effects of temperature, our study relies on a natural experiment: We tested to see if births in Louisiana changed after an unusually hot August. Our study also controls for many social and economic factors that are changing over time, including economic opportunities for women and access to birth control.
Colds sperm and 90 days Because we averaged the minimum and maximum temperature, the daytime temperature on these days is usually above 90F 32Cwhich most of us would find to be very hot.
The core finding is that hot days lead to a reduction in birth rates eight to 10 months later. The effect size is largest at nine months: Importantly, the data also show that air conditioning played a major role in minimizing the impact that hot days pose for fertility.
Our study also explores whether the initial decline in birth rates is offset by an increase in the following months.
This suggests that these shocks could reduce the number of children a woman has over the course of her reproductive life — a growing concern for the United States and many countries. As one limitation of our Colds sperm and 90 days, we tested for a rebound in births for only up to one year after the initial decline, so there could be some longer-term rebound for which we do not account.
While sexual behavior could certainly be influenced by hot weather, we present some novel evidence to suggest that reproductive health is especially vulnerable.
If the story were just about temperature making sex uncomfortable, then we would only see a fall in births eight to nine months later. Instead, we find that birth rates also fall 10 months later, suggesting that hot days have lasting health effects.
However, more research is needed to definitively verify this hypothesis. Currently, the United States experiences nearly 30 hot days per year.
A prominent global circulation model projects that the United States will experience a tripling of the number of hot days to about 90 by the end of the 21st century. We project that this warming will cause the number of births to fall by aboutper year by then. There will also be more summer births, due to the rebound, which will expose pregnancies to considerably hotter days during the third trimester and will threaten infant health.
As a caveat, these projections focus exclusively on the fertility cost of heat stress and do not offer insight into the costs of natural disasters or other major social changes resulting from climate change. Should nothing be done to mitigate climate change, our study indicates that air conditioning can lower the fertility costs.
But, we caution that in order to avoid exacerbating climate change, any increase in energy use for air conditioning must be offset with decreases in emissions in other parts of the economy. While our study offers lessons from the United States, it is uncertain how global warming might impact fertility elsewhere in the world.<|endoftext|>
| 3.703125 |
787 |
Bullying is one of the most prevalent and widely discussed topics pertaining to school safety and security. Incidents of bullying can affect the school environment, the community, and most importantly the psychological and developmental state of youth. By working together and educating one another, the most effective prevention efforts can be discovered and implemented.
Increasing attention has been given to school violence and the prevention measures utilized by schools. Research has shown that violence prevention efforts by teachers, administrators, parents, community members, and students can positively impact the school environment/climate. Prevention efforts should aim to reduce the risks and perceptions of violence on the campus, while creating a safe and supportive learning environment. As attention and concern surrounding school violence increases the need to create positive school environments free of violence is evident.
Mental health plays a substantial role in the safety and overall climate of a school. According to the U.S. Department of Health and Human Services, 1 in 5 adolescents has a diagnosable mental health disorder and even more show signs of depression. Educators also struggle with mental health, with recent surveys suggesting half or more describe themselves as under high levels of stress and in a “not good” mental health condition.These issues can disrupt the educational environment and lead to negative outcomes for students.
The health risks associated with tobacco and illicit drug use are well researched and documented. Cigarette smoking is the number one cause of preventable disease and mortality in the United States and has shown to be a path to illicit substances such as marijuana and cocaine. School administrators, teachers, resource officers, and school staff play an important role in the health and safety of youth. If students can be educated on the dangers of tobacco and drug use earlier in life, it saves not only our students but strengthens our communities.
Over the past decade, technology has altered the way youth communicate and interact with their peers. With over 90% of teenagers engaging in online use, the internet serves as a dominant medium of information gathering and sharing. As forms of internet communication evolve, it is the responsibility of youth, parents, and educators to understand the benefits and, when not used appropriately, the downfalls of these technologies.
The unfortunate reality is that school and junior college districts in Texas have the potential to be affected by any number of adverse events that can occur with little warning. Therefore, educational institutions must be self-reliant until help arrives and prepare to continue to participate in the emergency process after help arrives. Self-reliance can be achieved through the development of a comprehensive emergency management program that is mapped to the four phases of emergency management and includes a Multi-Hazard Emergency Operation Plan, regular training, drilling, and exercising, coordination with state and local partners, ongoing safety and security assessments, and the establishment of a district Safety and Security Committee.
School-Based Law Enforcement
The increased presence of school law enforcement during the 1990s has been attributed to the public’s perception that juvenile crime was increasing. The federal government has passed legislation and given millions of dollars in grant money to fund school-based law enforcement as a response. School-based law enforcement officers face challenges that are unique to serving in the school setting. The role these officers play has changed over time is not always clear. Research is beginning to uncover some of these issues and develop the best ways to utilize police to increase safety and security in schools.
Youth are making an impact across the nation by getting involved in policy change and by developing relationships with adults who value their feedback on important topics. Through a youth-led, adult assisted model, youth become the decision makers, while adults act in a supportive role, providing assistance when needed. Involving youth in leadership roles builds sustainability because young people who lead today are building the skills to continue leading tomorrow. Youth are valuable partners because they reflect genuine concern for their generation and project a powerful voice in advocating for community change.<|endoftext|>
| 3.828125 |
1,759 |
My name is Alexander FufaeV and here I will explain the following topic:
# Voltage: How it is created by charge separation!
Explanation
## Video
This lesson is also available as a YouTube video: Voltage: The Most Basic Explanation From a Physics Point of View
Have you ever wondered what the $$1.5 \, \text{V}$$ on an AA battery means? Or, what it means that an outlet provides $$230 \, \text{V}$$? Or, what it means that a USB port needs $$5 \, \text{V}$$ to function? These are all questions that have to do with voltage. Let's try to understand what these numbers mean physically.
With this basic knowledge it will be easier for you later on
• to analyze simple electrical circuits,
• to carry out calculations with electrical quantities,
• or simply get a feeling for voltage numbers on your everyday devices.
## Necessary for the voltage: Positive and negative charges
Particles can carry negative or positive electric charge. Let us denote this charge with the letter small $$q$$. Charge is measured in the unit Coulomb and abbreviated with the letter $$\text{C}$$:
Formula anchor
Particles, like the protons that make up our universe, carry a very small positive charge of $$q = 10^{-19} \, \text{C}$$. That is 18 zeros with a 1 behind the decimal point:
Formula anchor
This is a quite tiny amount of charge for our standards. Only if we take a lot of particles, they will carry a noticeable amount of charge in sum. So let's take a lot of charges.
In order that different particles don't get mixed up, we put the positive particle cluster in one box. And the negative particles into another box. We add up the individual charges small $$q$$ together and get a total charge, that we denote with a large $$Q$$.
Formula anchor
In the positively charged box, the total charge $$\class{red}{Q}$$ is positive because positively charged particles carry a charge with a positive sign. And in the negatively charged box, $$\class{blue}{Q}$$ is negative because negatively charged particles carry a charge with a negative sign.
Our oppositely charged boxes will attract each other and move towards each other. Because eventually, opposite charges will attract each other. Our boxes are just like two big charges. We say: they exert an electric force on each other. The more charges in the boxes, the greater the attracting electric force.
## Charge separation generates voltage
Let's do a thought experiment. We fix the two boxes so that they cannot move towards each other. In this way, the opposite charges will remain separate from each other.
The accumulation of negative charges in the one box is called minus pole and the accumulation of positive charges in the other box is called plus pole.
The charges from both poles attract each other, they want to move towards each other. But they can't do that because we have fixed them each in two places. The separation of the positive and negative poles creates a voltage between the poles. We denote the voltage with the symbol $$U$$.
• The voltage $$U$$ is smaller when less positive and negative charges are separated. In this case, the total charge $$Q$$ in the boxes is small.
• The voltage $$U$$ is large when many positive and negative charges are separated. In this case, the total charge $$Q$$ in the boxes is large.
The voltage arises only when positive and negative charges are separated. In our case, there are two charged boxes separated from each other. The voltage always refers to two different points. So always keep the question in mind: Voltage between which two points?
So always keep the question in mind: Voltage between which two points? Between the points where the two boxes are located.
So when you read or hear something about voltage, always remember that somewhere positive and negative charges are separated from each other and there is in principle the possibility to release them to generate an electric current. But more about that a little later.
## Voltage is energy per charge
So far, you have only learned that charge separation is the cause of voltage. Now let's try to understand what voltage means physically. With this you can understand what a number like $$5 \, \text{V}$$ means.
For this purpose we take a small positively charged particle with the charge $$q$$. Let's call this particle a test charge. With this we can test how big the charge separation is and therefore how big the voltage between the poles is.
Let's place the test charge directly at the positive pole. Then the test charge is attracted to the negative pole. It therefore experiences an electric force. This force causes the test charge to accelerate. Just before it hits the negative box, the test charge reaches a certain speed because it has been accelerating the whole time.
But if a test charge has a velocity, then it also has motion energy (kinetic energy). The test charge has thus gained kinetic energy by moving from the positive pole to the negative pole. The test charge had no kinetic energy at the beginning, but has received kinetic energy by the acceleration to the negative pole!
Let us call this gained energy $$W$$. The energy is measured in Joule and the unit is abbreviated with the letter $$\text{J}$$:
Formula anchor
Let's hold on to an important insight:
If we now divide this gained energy $$W$$ by the charge $$q$$ of the test particle, we get a quantity that gives gained energy per charge: $$\frac{W}{q}$$. And this quantity $$\frac{W}{q}$$ corresponds to the voltage $$U$$:
Formula anchor
## Voltage generates current
If we conductively connect the two poles, the positive charges can move to the negative pole. We assume here that the negative charges are firmly anchored in the negative box and cannot move. Otherwise, they would also move along the junction to the positive pole. That would be okay, but it would make our thought experiment unnecessarily complicated. Therefore, we only consider the current of positive charges.
The electric current $$I$$ is the amount of charge per second that travels through this connection. It is clear that the charges only travel through this connection if there is any charge separation at all. Without charge separation, there is neither a voltage nor a current here.
We can exploit this current to make a small lamp light up. The charges move and move... But with time, the number of positive charges in the box decreases. The charge separation decreases. The current decreases. The light shines dimmer and dimmer. As fewer charges are separated, the voltage naturally decreases as well, until there are no more charges in the positive box. Then the voltage is zero. The current stops flowing. The light stops shining.
The charges are no longer separated from each other, but are all in one box. The opposite charges have neutralized each other. The total charge in the box is now zero: $$Q = 0$$.
How can we make it so that the light continues to shine constantly? We must keep the charge in the box constant, so that the voltage and the current remain constant. Then the lamp will keep the same brightness all the time. If we supply charges, by whatever means, to the boxes, then we have thereby created what is called a voltage source with which we can then generate a non-decaying current which in turn causes the little light to glow steadily.
## Is current or voltage dangerous?
Voltage in itself is not dangerous, whether it be $$10 \, \text{V}$$ or $$10\,000 \, \text{V}$$. AS LONG as the charges are separated, nothing special happens. The positive charges are resting in one box. And the negative charges in the other box. Although the voltage between the poles is not zero.
Only when we conductively connect the two poles an electric current starts to flow, and that current in turn can become dangerous. The current becomes more dangerous...
• the greater the voltage $$U$$
• and the better the connection betweeen the poles.
Of course, the connection between the poles should not be your body! Even if your body is not such a good connection, such as a copper cable, a sufficiently large voltage can also 'push' the charges through your body. The consequences are for example: burns in the areas where the current has passed through.
In the next lesson, we will look at the Ohm's Law, which allows us to describe exactly how current and voltage are related.<|endoftext|>
| 4.5 |
1,145 |
All About Readability By Cheryl Stephens Why are we looking at readability tests? The use of readability tests in the plain language process is a controversial topic. Now that readability scores are easy to obtain by using computerized grammar and style checking software programs, there is new pressure to adopt them.
The Art of Learning Tips for Students Videos for Students Links for Students This page will develop into a collection of all sorts of things - tips, videos, links to great sites - all designed to help you study. At the moment I have made links to some great websites which will help you with studying all your school subjects, no matter where you are.
ReadStrong - Neurological Impress Method Neurological Impress Method (NIM) The Neurological Impress Method is a form of paired reading in which a student and tutor read the same text almost simultaneously. Sitting side-by-side, the tutor reads a text slightly faster and louder than the student while both follow the text with their fingers. Reading along with a more fluent reader is thought of as "an impress, an etching in of word memories on the natural process" (Heckelman, 1969). Biteslide is the engaging way to research, create, and present school projects The Common Core State Standards Initiative demands that students be able to read and understand more complex texts, write arguments and persuasive pieces with greater dexterity, and present their findings in short-term projects mirroring the tasks that will be asked of students in both college and career. Listening and public speaking skills, as well as the incorporation of media into presentations, are also important key points to take away from the Common Core State Standards Initiative for teachers to embrace and include in their lesson planning. Let’s accomplish all of this by incorporating Biteslide into a New York State 6th grade English Language Arts lesson within a larger unit on personal identity and personal choices with challenging texts from contemporary figures. The lessons will last for 7 Instructional days (not including student oral presentations).
11th Grade English Course: 11th Grade English Kristin Damo 260 Views 7 Favorites Readability Tests The main function of readability tests is to give you a quick assessment about the density of your writing. Readability tests cannot tell you how easily a reader can understand the information in the text. You can perform readability tests manually by counting and doing a mathematical calculation, or by using word-processing software.
Bloom's and ICT tools Many teachers use Bloom's Taxonomy and Bloom's Revised Taxonomy in developing and structuring their teaching & learning experiences. Bloom's Digital taxonomy is an attempt to marry Bloom's revised taxonomy and the key verbs to digital approaches and tools. This is not a replacements to the verbs in the revised taxonomy, rather it suppliments and supports these by including recent developments, processes and tools. This page looks at some specific examples of tools and match them to Bloom's Digital Taxonomy Many of these tools that are FOSS (Free or Open Source Software). These are in italics.
Performance Assessment Performance Assessment By Samuel J. Meisels, Ed.D. Most standardized tests are not designed to evaluate the individualized growth and development taking place in your classroom. But there are assessment tools that do! Many early childhood educators are uncomfortable with the idea of testing the young children they work with.
Place Value Charts, Number Grids, and more The links below list additional charts that have been developed to date including decimal/ fraction equivalency, percentage, number grids, and place value charts. Check out the Customizable Hundreds Number Chart. You can set the start number and the interval, change the number of rows, have negative numbers, decimals, and you can start the numbers from the bottom or the top. This is an example of a customizable math resource that lets you change the values to make the charts and tables just the way you need them. How I Teach English in the 45 Minute Timeframe One of the question I am asked the most is how do you teach English in 45 minutes? Not just reading, not just writing, but everything that English encompasses. And I can tell you; it is not easy, nor is it perfect, nor do I have everything figured out. The 45 minute block of time is the bane of my English existence. Yet as I have figured out it is within our biggest problems that we find our biggest inspiration, and that is very true for this situation.
Ultimate list of online content readability tests “In my younger and more vulnerable years my father gave me some advice that I’ve been turning over in my mind ever since. ‘Whenever you feel like criticizing any one,’ he told me, ‘just remember that all the people in this world haven’t had the advantages that you’ve had.’” Can you read the above paragraph easily? Scaffolding Definition In education, scaffolding refers to a variety of instructional techniques used to move students progressively toward stronger understanding and, ultimately, greater independence in the learning process. The term itself offers the relevant descriptive metaphor: teachers provide successive levels of temporary support that help students reach higher levels of comprehension and skill acquisition that they would not be able to achieve without assistance. Like physical scaffolding, the supportive strategies are incrementally removed when they are no longer needed, and the teacher gradually shifts more responsibility over the learning process to the student. Scaffolding is widely considered to be an essential element of effective teaching, and all teachers—to a greater or lesser extent—almost certainly use various forms of instructional scaffolding in their teaching.<|endoftext|>
| 4.03125 |
632 |
Yang-Mills theory, in physics, a generalization of Scottish physicist James Clerk Maxwell’s unified theory of electromagnetism, also known as Maxwell’s equations, used to describe the weak force and the strong force in subatomic particles in terms of a geometric structure, or quantum field theory. The Yang-Mills theory relies on a quantum mechanical property called the “mass gap.” The theory was introduced in 1954 by Chinese-born American physicist Chen Ning Yang and American physicist Robert L. Mills, who first developed a gauge theory, using Lie groups (see mathematics: Mathematical physics and the theory of Lie groups), to describe subatomic interactions. The current state of Yang-Mills theory has been compared to the early days of the calculus, when undeniably accurate and useful results were being obtained but before the formal development of analysis added rigorous definitions that eliminated logical fallacies. For Yang-Mills theory, one of the most important questions is to mathematically explain the mass gap, or nonzero mass, in quantum applications of the formulas. Evidence for the mass gap has been demonstrated in physical experiments and computer-based mathematical models, and it is believed to be the reason that the strong force operates only at very small distances (within atomic nuclei).
In 2000 the Yang-Mills theory was designated a Millennium Problem, one of seven mathematical problems selected by the Clay Mathematics Institute of Cambridge, Mass., U.S., for a special award. The solution for each Millennium Problem is worth $1 million.
Learn More in these related Britannica articles:
Millennium Problem>Yang-Mills theory, and Poincaré conjecture.…
James Clerk Maxwell
James Clerk Maxwell, Scottish physicist best known for his formulation of electromagnetic theory. He is regarded by most modern physicists as the scientist of the 19th century who had the greatest influence on 20th-century physics, and he is…
electromagnetism: Maxwell’s unified theory of electromagnetismThe final steps in synthesizing electricity and magnetism into one coherent theory were made by Maxwell. He was deeply influenced by Faraday’s work, having begun his study of the phenomena by translating Faraday’s experimental findings into mathematics. (Faraday was self-taught and had never mastered mathematics.)…
Maxwell’s equations, four equations that, together, form a complete description of the production and interrelation of electric and magnetic fields. The physicist James Clerk Maxwell in the 19th century based his description of electromagnetic fields on these four equations, which express experimental laws. The statements of these four equations are, respectively:…
Weak force, a fundamental force of nature that underlies some forms of radioactivity, governs the decay of unstable subatomic particles such as mesons, and initiates the nuclear fusion reaction that fuels the Sun. The weak force acts upon all known fermions—i.e., elementary particles with half-integer values of intrinsic angular momentum,…
More About Yang-Mills theory1 reference found in Britannica articles
- Millennium Problem<|endoftext|>
| 3.703125 |
222 |
Formation of Oil and Gas Deposits
Oil and natural gas are formed out of organic matter from dead plants and animals decomposing deep within the Earth. Under very specific pressure and temperature, the biologic material is first converted to kerogen, which is then transformed into bitumen and eventually into oil or gas. Hydrocarbon molecules then rise via migration paths – permeable rock – to the surface where they form oil and gas deposits.
For hydrocarbon deposits to exist, the entire system is essential – i.e. source rocks, migration paths and cap rocks or traps. All of these elements were formed by historical geological processes.
Source rocks are represented by claystone and marlstone rich in organic matter (long-decomposed animal and plant material), while migration paths are tectonic disturbances and porous horizons deep under the Earth’s surface.
Petroleum traps consist of porous rocks (reservoirs) that either form vaulted structures or are lithologically limited. These traps are sealed by impermeable rock which prevents hydrocarbons from escaping to the surface.<|endoftext|>
| 4.375 |
357 |
# A card is drawn from a standard deck of 52 playing cards. How do you find the probability that the card is a jack or a spade. Express the probability as a simplified fraction?
Oct 22, 2016
The probability is $\frac{4}{13}$
#### Explanation:
If we denote the events as:
$A$- a jack is chosen and
$B$ - a spade is chosen
Then the evemt we are looking for can be described as the alternative A or B ($A \cup B$).
The probability can be calculated using the formula:
$P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$
Now we have to calculate the probabilities on the right side:
$P \left(A\right) = \frac{4}{52}$ because there are $4$ jacks in a deck of cards
$P \left(B\right) = \frac{13}{52}$ because there are $13$ spades
$P \left(A \cap B\right) = \frac{1}{52}$ because there is one card which is jack and a spade (a jack of spades).
Finally if we substitute calculated values we get:
$P \left(A \cup B\right) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$
This result can be simplified because both numerator and denominator can be divided by $4$.
So the answer can be written as:
The probability is $\frac{4}{13}$<|endoftext|>
| 4.8125 |
734 |
For the complexity of the background of my research, and what I wanted to achieve, my lesson was very simple. So simple, that I didn't create a big fancy lesson plan around it. I already had the standards I wanted to cover, see below, so it was just how to structure the activity.
My students had already been introduced to the vocabulary words in the Read 180 unit, and completed the workbook page where the filled in the definition of the word, and wrote it in a sentence. That was when they took their pre-test where I gathered the base-line data. My project was to simply have my students make a Google slides presentation where each slide was dedicated to a word. They included the definition and the sentence, along with some similar meaning words, and a picture that could go with the word. When they presented them, they were encouraged to do some sort of action with each word to add the kinesthetic aspect to it.
Napa Valley Unified School District has adopted the "4 C's" to focus on in our teaching and learning. They include Communication, Collaboration, Creativity, and Critical Thinking. When I planned out how I wanted to execute this project, I kept these in mind.
Using technology as the main resource, students used creativity to utilize multiple modalities of learning to assist in their study of the words. They were challenged to use critical thinking skills to dive deeper into the meanings of words by looking at definitions, sentences, and other ways words could be defined. They were given opportunities to communicate and collaborate with their classmates in both the building of these projects, and in presenting them to each other.
Common Core State Standards- College and Career Readiness Standards for Reading: 4
Interpret words and phrases as they are used in a text, including determining technical, connotative, and figurative meanings, and analyze how specific word choices shape meaning or tone.
Common Core state Standards- College and Career Readiness Standards for Language: 6
Acquire and use accurately a range of general academic and domain-specific words and phrases sufficient for reading, writing, speaking, and listening at the college and career readiness level; demonstrate independence in gathering vocabulary knowledge when encountering an unknown term important to comprehension or expression.
Common Core State Standards- English Language Arts- Literacy- 6.4
Determine or clarify the meaning of unknown and multiple-meaning words and phrases based on grade 6 reading and content, choosing flexibly from a range of strategies.
Common Core State Standards- English Language Arts- Literacy- 6.6
Acquire and use accurately grade-appropriate general academic and domain-specific words and phrases; gather vocabulary knowledge when considering a word or phrase important to comprehension or expression.
Q1: Can technology resources be utilized to make student learning more efficient?
Between the pre-task test and post-task test, the participating students only engaged with the target vocabulary through the technology resources and displayed significant improvement. Therefore, as proven by this study, technology resources can be utilized to make student learning more efficient.
Q2: Can the use of technology resources and multiple modalities of learning help improve students’ vocabulary recollection?
In this study, adding multiple modalities of learning proved to be beneficial to improving students’ vocabulary recollection. The work the students did with the targeted vocabulary words before the task was very simple in the workbooks and on the computers. This was not enough for them to have mastery of the words. When visual aides and physical actions were incorporated into the task, multiple learning style needs were met, therefore, vocabulary recollection increased.<|endoftext|>
| 4.125 |
628 |
Home » Summation Formulas
# Category Archives: Summation Formulas
## Total Number of Triangles Puzzle
How many triangles are there in the figure below?
Solution to this Geometry Triangle Puzzle problem is given in the video below!
## Total Number of Rectangles Puzzle
How many rectangles are there in the figure below?
Solution to this Geometry Rectangle Puzzle is given in the video below!
## Total Number of Upright SQUARES in a Square
How many upright Squares are there in the following square?
The key is to understand vertices of any given upright square and a diagonal line of the largest square to see how many translations of that given upright square you can have. As such, there is a summation-derived FORMULA you can actually use for any largest size square to find the total number of possible upright squares contained within it!
Solution to this Geometry Square Puzzle is given in the video below!
## Total Number of Upright RECTANGLES in a Square (or Rectangle)
How many upright Rectangles are there in the following square?
The key is to understand vertices of any given upright rectangle to see how many unique translations of that rectangle are possible. As such, there is actually a summation-derived FORMULA you can utilize for the largest square or rectangle of any size to find the total number of possible upright rectangles within it!
Solution to this Geometry Rectangle Puzzle is given in the video below!
## Total Number of Upright SQUARES in a Rectangle
How many upright Squares are there in the following rectangle?
The key is to understand vertices of any given upright square to see how many translations of that given square you can have. As such, there is a summation-related FORMULA you can actually derive for any size rectangle to find the total number of possible upright squares contained within it!
Solution to this Square Rectangle Geometry Puzzle is given in the video below!
## Total Number of Upright & Tilted SQUARES in a Square
How many upright and tilted Squares are there in the following square?
The key is to use vertices of any given largest square to see how many translations of a given smaller square you can have. As such, there is a summation-derived FORMULA you can actually use for any size square to find the total number of possible upright and tilted squares contained within it!
Solution to this Square Geometry Puzzle example is given in the video below!
## Total Number of Upright & Tilted SQUARES in a Rectangle
How many upright and tilted Squares are there in the following rectangle?
The key is to use vertices of any given rectangle to see how many translations of a given square you can have. As such, there is a summation-derived FORMULA you can actually use for any size rectangle to find the total number of possible upright and tilted squares contained within it!
Solution to this Rectangle Geometry Puzzle example problem is given in the video below!
## Infinite Series Proof example question
Show that
when
Solution to this Calculus & Precalculus Infinite Series Proof practice problem is given in the video below!<|endoftext|>
| 4.53125 |
774 |
The path of space exploration is made of rocks. Unveiling the secrets harbored by minerals in other planets and moons requires testing advanced technologies right here on Earth. One of the experiments carried out during the latest PANGAEA-X test campaign in 2017 discovered water-related minerals in the volcanic landscapes of Lanzarote, Spain.
“We are paving the way to learn how and where to search for life, but also for new worlds to live in. Expect the unexpected,” says Jesús Martínez-Frías, professor of planetary geochemistry at the Geosciences Institute in Spain. As an expert in mineralogy, Jesús helped to collect and analyse hundreds of samples of rock, soil and dust around dormant volcanoes.
The results collected by the PANalytical experiment confirmed his best expectations. “We have detected for the first time previously unidentified water-related minerals. The list is long and promising: we have found oxides, oxy-hydroxides, clay minerals, zeolites, carbonates and sulphates,” he explains.
These minerals are crucial to determining what type of biomarkers tell us about habitability conditions – past and future – not only on our planet, but also on the Moon and Mars.
The results confirm the relevance of Lanzarote as an exceptional site for field campaigns such as PANGAEA-X and further astrobiology research studies. “We have discovered new pieces of the geological puzzle of the island, we can better understand its geological context. The basalts we found make it a good Moon analogue, while the water-related minerals are useful for Mars habitability studies,” adds Jesús.
The collection of high quality spectra of volcanic minerals was made possible by two portable devices: the TerraSpec-Halo spectrometer and Epsilon 1 X-ray fluorescence spectrometer.
“We are pleased that our technology allowed the team to identify yet unknown spectral signatures,” says Lieven Kempenaers, principal investigator from Malvern PANalytical, a scientific instrumentation company with worldwide customers in many industries such as mining, building materials, petrochemicals and pharmaceuticals.
While the Halo spectrometer has been screening rocks with PANGAEA since 2016, the Epsilon 1 device was a newbie to the test campaign.
“These two complementary devices could rapidly screen the minerals and identify its chemical elements. They told us the story of the rocks in-situ,” adds Lieven.
The PANalytical instruments allowed PANGAEA participants to measure hundreds of samples and multiple spots in the same sample. All procedures were automated, and they took from seconds to 30 minutes, according to the precision required.
ESA astronaut Matthias Mauer used the devices throughout the campaign. “These tools are a huge asset for the crew to gain quick non-destructive measurements on the spot. They are easy to use, and even without being a professional geologist I could perform autonomous activities with scientific relevance,” he says.
The astronaut’s feedback resulted in an upgrade to the Epsilon 1 device for the upcoming PANGAEA 2018 campaign – the device will now be equipped with a small camera to take pictures of the sample. This new feature adds valuable visual information for the identification of minerals.
The versatility of the instruments does not end here. “We are thinking about mounting components of the Epsilon X-ray instrument on a scouting rover in the future,” explains Lieven, who sees the campaign as an ideal scenario to test operational strategies for planetary surface geology.
“We are not lacking in ambition. We dream of sending our technology to the Moon,” Lieven says.<|endoftext|>
| 3.671875 |
574 |
#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.27 Question 5
Answer: $I=\frac{e^{2 x}}{8}[\sin 2 x-\cos 2 x]+c$
Hint: $[\sin 2 x-\cos 2 x]$
Given: $\int e^{2 x} \sin x \cos x d x$
Solution: $I=\frac{1}{2} \int e^{2 x} \sin x \cos x d x$
\begin{aligned} &=\frac{1}{2} \int e^{2 x} \sin 2 x d x \\ &=\frac{1}{2}\left[\sin 2 x \int e^{2 x}-\int\left[\frac{d}{d x} \sin 2 x \int e^{2 x} d x\right] d x\right] \\ &=\frac{1}{2}\left[\sin 2 x \frac{e^{2 x}}{2}-\int 2 \cos 2 x \frac{e^{2 x}}{2} d x\right] \end{aligned}
\begin{aligned} &=\frac{1}{4} \sin 2 x e^{2 x}-\frac{1}{2} \int e^{2 x} \cos 2 x d x \\ &=\frac{1}{4} \sin 2 x e^{2 x}-\frac{1}{2}\left[\cos 2 x \frac{e^{2 x}}{2}+\int 2 \sin 2 x \frac{e^{2 x}}{2} d x\right] \\ &=\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x-I+c \end{aligned}
\begin{aligned} &I+I=\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x+c \\ &2 I=\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x+c \\ &I=\frac{1}{2}\left[\frac{1}{4} e^{2 x} \sin 2 x-\frac{1}{4} e^{2 x} \cos 2 x\right]+c \\ &I=\frac{e^{2 x}}{8}[\sin 2 x-\cos 2 x]+c \end{aligned}<|endoftext|>
| 4.40625 |
856 |
# How to fill up the numbers in a set of empty discs drawing a pentagon?
The following pentagon is composed of 10 discs. Each disc must be filled with one number. The first 10 even positive numbers are allowed. The sum of the numbers to be placed in each side of the pentagon must be the same and the maximum possible. Find the sum of the numbers adjacent to the discs filled with the numbers 2 and 18.
The figure is shown below:
The choices given are:
1. 54
2. 60
3. 50
4. 64
I tried several ways to fill up this pentagon but did not find a clear solution. Does a strategy to solve this kind of puzzle exist? I'm stuck at the very beginning. I believe this requires mathematics. Please try to include some visual aid and the most detailed way to solve this problem.
• Need some clarification: In the question, you said to fill the pentagon with first ten positive even numbers, but at the end of first paragraph you ask the pentagon to be filled by only 2 and 18? Oct 13, 2019 at 13:42
• @SamRoy Fill the circles with each of the even numbers $2$ to $20$ as described. Then take the sum of the numbers that are adjacent to the numbers $2$ and $18$ in the filled diagram. Oct 13, 2019 at 18:31
The puzzle asks for a maximal sum for the numbers on the edges, so...
The sum of all five edges will include each number at a vertex twice, once for each edge that it is part of. We need to put the largest numbers at the vertices in order to maximize the sum. This will give us a maximum total sum of $$2(20+18+16+14+12)+10+8+6+4+2=190$$. Dividing the among the five edges gives us a maximum equal edge sum of $$\frac{190}{5}=38$$. In order to obtain equal sums on all edges, we need to balance the high and low numbers. Starting with $$20$$ at a vertex, we add $$2$$ and $$4$$ to either side of the $$20$$. Given the target sum of $$38$$, we complete the first two edges as $$20+2+\color{blue}{16}=38$$ and $$20+4+\color{blue}{14}=38$$. Now consider the other two vertices. We need to put $$12$$ and $$18$$ in those discs. If we put $$18$$ on the edge with $$16$$, we would need $$16+4+18=38$$, but we have already placed the $$4$$. So we place $$18$$ with $$14$$ and $$12$$ with $$16$$ which gives us $$16+10+12=38$$ and $$14+6+18=38$$, with the final number, $$8$$, we have $$18+8+12=38$$ and the diagram is complete. The numbers adjacent to $$2$$ and $$18$$, with their sum, are $$16+20+8+6=\color{blue}{50}$$
• Nice approach. +1 Oct 15, 2019 at 4:00
• Placing 2 and 4 next to 20 is a bit arbitrary. Why not 2 and 6? Couldn't it also work? You could have started more logically by noting that 2 can only be in a line with 20 and 16 and 10 only with 12 and 16. Apr 10, 2022 at 10:13
• @FlorianF Placing 2 and 6 next to 20 puts 16 and 12 on their respective vertices, leaving 14 and 18 on the remaining side needing the 6 again for 38. Your logic leading to 20-2-16-10-12 leaves only one place for 18, proving the solution unique. Apr 10, 2022 at 12:19<|endoftext|>
| 4.625 |
4,211 |
Keratoconus (keh-rah-toe-cone-us) affects your cornea, the clear dome-shaped window at the front of your eye. Your cornea focuses light into your eye to help produce a clear image.
In keratoconus, your cornea becomes weaker and thinner at its centre. This thinning causes it to bulge outwards in an irregular cone shape. This can make your vision blurry and distorted, as light being focused by your cornea forms an unclear image on your retina, at the back of your eye.
Keratoconus usually develops in your teens or 20s and can worsen over time. It cannot be treated with eye drops or other medications. In the early stages, glasses may help correct vision. As keratoconus develops glasses may no longer help, but most people can still get a good level of vision by wearing contact lenses. Depending on your keratoconus sometimes contact lens can be uncomfortable to wear for long periods.
Some treatments are becoming available which can prevent sight being more affected or try to improve the shape of the cornea to give a better level of vision. A treatment called collagen cross-linking can help to slow down the development of keratoconus, to prevent further changes to the cornea. This can mean that someone can continue to get a good level of sight with glasses or contact lenses. Corneal implants can also be used to try and improve the shape of the cornea to give better vision with contact lenses.
For some people keratoconus can continue to worsen to a stage where contact lenses no longer give clear vision. If this happens, it is possible to have a corneal transplant, which can improve sight to a good level again.
Keratoconus causes your cornea to develop an irregular, uneven shape. This affects how it focuses light onto your retina, at the back of your eye. It can make your vision blurry and distorted. This type of focusing problem is known as "irregular astigmatism".
Keratoconus can also cause your eye to become more short-sighted (myopic). This makes distant objects appear blurred, while nearer objects are clearer.
You might find bright light uncomfortable and have trouble seeing things in very bright light. This is because the uneven cornea scatters light more as it enters your eye. You may also see halos around lights, which can affect your vision at night, for example, when you are driving. Visit our page on light sensitivity for more information about the condition and tips on coping.
Your cornea is made up of a number of layers and is normally smooth and clear. Your cornea is also very strong. This is because the largest, middle section of the cornea, the stroma, is made up of many regular-shaped bundles of the connective tissue, collagen, which are firmly joined together.
The surface of your cornea is very sensitive. It contains many nerve endings and can detect even the smallest piece of dirt or fluff. It also has a transparent skin layer that acts as a barrier between your eye and the outside world, helping to protect it from injury and infection.
Your cornea is important for sight. It bends and focuses light into the eye. Light is then focused further by the eye's lens onto the retina, at the back of the eye. Your retina converts light into electrical signals. These are then sent to the brain where they are interpreted as the images we understand as sight.
The main treatment for keratoconus is to try and correct the vision problems caused by the irregular, cone-shaped cornea. At first you might be able to wear glasses, but if keratoconus worsens, you may only be able correct your sight with hard (rigid gas permeable) contact lenses.
If you wear glasses and your cornea is becoming steeper and cone-like, you might need more powerful lenses to correct your sight. Stronger, more powerful lenses can be thicker and heavier. They can also cause your vision to be distorted when you are looking through the edges of the lens.
Because keratoconus makes your cornea thinner and more flexible, this can also mean that your prescription might change more rapidly and you may need to change your glasses more often.
If glasses are no longer improving your vision, usually the next type of lens you will try are rigid gas permeable (RGP) or "hard" contact lenses.
Hard lenses give a more even shape to your cornea, improving focusing. RGP lenses are made of strong, breathable plastic. There are lots of different RGP lenses available. Some are specifically designed to fit corneas affected by keratoconus. The brand of lens that is most comfortable can be quite individual and you may have to try out a few different ones to find the best for you.
When you blink you leave a thin layer, called the tear film, over the front of your eye. The tear film keeps the front of your eye healthy and also helps with the focusing of light into your eye.
Dry eye is caused by a problem with your tear film. It can make your eyes feel dry, scratchy, irritated and uncomfortable and can also make your eyes watery. Having keratoconus can make it more likely for you to develop dry eye. This is because it can be harder for your eyelids to spread your tears over the uneven cornea.
Wearing contact lenses can also make it more likely for you to develop dry eye. Dry eye can make contact lenses uncomfortable, making it harder for you to wear them for long periods.
There is no cure for dry eye, but treating it can make contact lens wear more comfortable. An optician or ophthalmologist would be able to diagnose dry eye and discuss what can be done to help. They can recommend treatments, such as artificial tear eye drops, which can help to make your eyes feel more comfortable. Your optician would also be able to try out different lenses with you to help you to find the best ones for you.
If you find it too uncomfortable to wear RGP lenses for large amounts of time or, if keratoconus has worsened and RGPs no longer improve your sight, other types of lenses may help. Your optician should be able to discuss other contact lens options, which could include:
If glasses no longer correct your vision, your ophthalmologist might say that it is "clinically necessary" or that there is a "medical need" for you to have contact lenses.
There is an NHS patient charge, which is set nationally for clinically necessary contact lenses. This charge is reviewed by the Department of Health every year and usually covers you for a period of six months if you need further changes in your prescription.
You may also be entitled to help with the cost of glasses and contact lenses if you qualify for help with NHS costs under the NHS voucher scheme arrangements. Your optician would be able to tell you if you are entitled to help with these costs.
If you are under 16 or under 19 and in full time education there is no charge for replacing your contact lenses.
If you are over 16 and not in full time education it is likely that you will have to pay for replacement lenses, even if you are entitled to help with the cost of glasses or contact lenses. Your hospital department would be able to confirm if you need to pay for replacements if yours are lost or damaged.
Although your contact lenses are clinically necessary, your hospital is not responsible for providing you with cleaning and sterilising solutions. This means that you will normally have to buy them..
Keratoconus is usually diagnosed when someone is in their late teens or early 20s. When it is diagnosed, keratoconus can normally been seen in both eyes, but it is usually worse in one eye, than the other.
About 1 in 2,000 people has keratoconus. It affects men and women equally. Keratoconus may also be more common in certain ethnic groups, for example, keratoconus affects about 1 in 450 South Asian people.
The cause of keratoconus is currently unknown. For a small number of people, about 10-14 per cent, keratoconus can run in the family, but this means that most people have no family history of keratoconus.
It is thought that people who have allergies might be more likely to develop keratoconus. Allergies can cause your eyes to become itchy and uncomfortable making you more likely to rub them.
If someone rubs their eyes a lot over a long time, this could make the cornea weaker, causing keratoconus, although this is not true for everyone. If you do experience allergies, it is important to seek treatment for these.
Some research also suggests that corneas affected by keratoconus may not be as good at healing from everyday wear and tear, or have less of the important fibres that give the cornea its strength.
Keratoconus can appear or get worse at any age. Keratoconus tends to worsen more quickly the younger someone is, and it can continue to develop until someone’s mid-thirties. At this point it may slow down or stop developing.
If you have another medical condition you might be more likely to develop keratoconus. These can include the eye conditions Leber's congenital amaurosis and retinitis pigmentosa, as well as the genetic conditions osteogenesis imperfecta, Ehlers-Danlos, Marfan, Turner and Down’s syndromes.
Children and adults with these conditions are generally more at risk of some eye conditions, so it is important for them to have regular eye tests. Not everyone who has these conditions will develop keratoconus. Most people with keratoconus do not have any other eye or medical conditions.
Your optician or ophthalmologist can carry out a number of tests to diagnose keratoconus and measure any changes to the shape of your cornea over time. These include:
Collagen cross-linking, also known as CXL or C3R, is a treatment used to treat progressive (worsening) keratoconus. The aim of this treatment is to stop things getting worse, although, for some people cross-linking can also cause an improvement.
The middle layer of your cornea, the stroma, is made up of a web of collagen fibres held together with links. This gives your cornea its shape and strength. In keratoconus these links may be less strong causing your cornea to bulge outwards in an irregular cone shape.
It is thought that CXL works by increasing the number of naturally occurring collagen cross-links in your cornea, making it stronger. Treatment can be carried out by an ophthalmologist or sometimes nurse specialist and involves removing a small area of the surface of your cornea, known as the epithelium, and then applying vitamin B2 (riboflavin) drops. Your cornea is then exposed to ultraviolet-A (UVA) light.
Cross-linking treatments take around 30-60 minutes and you do not need to stay in hospital overnight. You would normally be awake for the procedure.
The main aim of CXL is to stop keratoconus getting worse, preventing your vision from deteriorating in the future. For some people this treatment can also help to flatten and regulate the shape of your cornea, improving vision, although this is not the case for everyone. Cross-linking is usually carried out on one eye at a time, and may need to be repeated.
There is some promising evidence that CXL is very successful in stopping the development of keratoconus and more evidence is being collected all the time, to help CXL become a regular treatment.
Corneal implants are flexible C-shaped plastic rings which are inserted into the edges of your cornea. These aim to flatten the cornea, to correct its irregular shape. For some people this can allow for a better contact lens fit or better correctable vision with glasses.
Usually two C-shaped rings are inserted and this procedure can be done under local or general anaesthetic (while you are awake or asleep). A small cut is made in the cornea and channels are created into which the implants are placed.
At the moment there is no good evidence that shows that corneal implants prevent worsening of, or reverse keratoconus.
Corneal implants and epithelium‑off collagen cross-linking aren’t routinely available on the NHS.
Although the National Institute for Health and Care Excellence (NICE), the body that approves treatments for use on the NHS, has said that corneal implants and epithelium‑off cross-linking work well enough and are safe enough to use on the NHS, at present the is no guidance that these treatments should receive routine NHS funding. This means that these treatments are not available everywhere.
Some hospitals offer CXL as routine but others may not. If your ophthalmologist feels that these treatments would be suitable for you, they may need to apply for specific funding from your local health provider for you to receive them.
Corneal hydrops is a rare complication of keratoconus. It occurs when fluid from inside your eye enters your cornea through breaks in the membrane at the back of the cornea. This fluid causes the cornea to become swollen.
This swelling can affect your vision, making it blurry, even with your contact lenses in. It can also give your cornea a cloudy or milky appearance. Hydrops can cause irritation or pain, light sensitivity and make your eye watery (teary) and red.
The breaks in your cornea will usually take at least three months to heal. You will usually be advised not to wear contact lenses while your eye is recovering. If you feel any discomfort or pain, your ophthalmologist can give you eye drops to make your eye feel more comfortable.
Once the breaks have healed most people find that their vision improves again. You may need a new contact lens prescription after hydrops as your cornea may have changed shape.
Some people can also find contact lenses are more comfortable and stay in better after having hydrops. This is because hydrops can cause scarring which flattens the cornea. Less commonly, you might be left with scarring that could make your vision in that eye worse.
Most people with keratoconus are able to get good vision with contact lenses. But, if wearing contact lenses becomes very uncomfortable and you can only wear them for very short amounts of time, or your corneas are scarred and you can’t get much improvement of your sight with contact lenses, your ophthalmologist may suggest that you have a corneal transplant.
A corneal transplant is surgery to remove all or part of a damaged cornea and replace it with healthy, clear cornea tissue from the eye of a donor who has died. It is possible to carry out transplants which replace all, or only some layers of your cornea, with healthy donor tissue.
Only about 10-25 per cent of people with keratoconus need to have a corneal transplant. In keratoconus two types of transplant are commonly used:
In both types of transplant your ophthalmologist uses tiny stitches to hold the new donor cornea in place. If you have a DALK transplant these stitches are usually removed six months after the surgery. If you have a PK transplant stitches are not removed before 12 months. It can take up to 18 months for your vision to improve to the best possible level.
Both DALK and PK transplants work well in people with keratoconus. Studies have shown that for people with keratoconus 95 per cent of PK transplants are still healthy five years after surgery. Newer DALK transplants also seem to have an even lower risk of failure, to last for longer and to have a shorter recovery times, compared to PK ones.
Although PK transplants are an effective treatment, 50 per cent are no longer working at 20 years. This means that if you are younger you might be advised to wait longer before having a corneal transplant, as you are more likely to require a number of transplants in your lifetime. Although it is possible to replace a failed or rejected transplant, known as a "re-graft", the risk of rejection and failure goes up each time a transplant is done.
The risks of both PK and DALK are low, but after both types of surgery it can take a long time for vision to recover. You will also still usually need to wear contact lenses after the transplant to get the best possible vision.
Following a transplant you will also need to use steroid eye drops for at least six months and in some cases indefinitely, to prevent rejection of the new transplanted tissue.
Transplants also carry the risk of cataracts, the clouding of the lens in your eye, and glaucoma, a condition where pressure at the front of your eye damages the optic nerve at the back of your eye. These risks are partly due to the need for steroids following the surgery.
Your ophthalmologist will discuss when it is right for you to have a corneal transplant and which transplant may be the best one for you.
Keratoconus affects the shape of your cornea, causing it to bulge outwards in an irregular cone shape. This can make your vision blurry and distorted, as light being focused by your uneven cornea forms an unclear image on your retina, the back of your eye.
Keratoconus can worsen over time, but most people are able to have good vision wearing contact lenses. Corneal collagen cross-linking can for most people prevent sight being more affected. Your ophthalmologist would be able to give you more information about whether or not this treatment would be helpful for you.
Much less commonly, keratoconus can cause changes in the cornea that can mean you may no longer be able to have good vision with contact lenses. If this happens, it is possible to have a corneal transplant, which can improve your sight to a good level. Only about 10-25 per cent of people with keratoconus ever reach this stage.
Being diagnosed with an eye condition can be very upsetting. You may find that you are worried about the future and how you will manage with a change in your vision. All these feelings are natural.
Some people may want to talk over some of these feelings with someone outside their circle of friends or family. RNIB can help, with our Helpline and our Sight Loss Counselling team. Your GP or social worker may also be able to help you find a counsellor if you think this would help you.
Ask your ophthalmologist, optometrist or GP about low vision aids, like a magnifier, and ask for a referral to your local low vision service. You should also ask whether you are eligible to register as sight impaired (partially sighted) or severely sight impaired (blind). Registration can act as your passport to expert help and sometimes to financial concessions. Even if you aren’t registered a lot of this support is still available to you.
Local social services should also be able to offer you information on staying safe in your home and getting out and about safely. They should also be able to offer you some practical mobility training to give you more confidence when you are out.
Our Helpline can also give you information about the low vision services available, and our website offers lots of practical information about adapting to changes in your vision and products that make everyday tasks easier.
The RNIB Helpline is your direct line to the support, advice and products you need. We'll help you to find out what's available in your area and beyond, both from RNIB and other organisations.
Low vision services can help people make the most of their sight. They:
UK Keratoconus Self Help and Support Association is a registered charity that works to raise awareness of keratoconus, as well as providing information and support for to those affected by the condition, living in the UK.<|endoftext|>
| 3.6875 |
389 |
# One half of a number increased by 16 is four less than two thirds of the number. What are the numbers?
Oct 24, 2016
The number is $\textcolor{g r e e n}{72}$
#### Explanation:
Let the number be $n$
the number increased by $16$
$\textcolor{w h i t e}{\text{XXX}} n + 16$
one half of the number increased by $16$
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} \left(n + 16\right)$
two thirds of the number
$\textcolor{w h i t e}{\text{XXX}} \frac{2}{3} n$
one half of the number increased by $16$
is $4$ less than
two thirds of the number
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} \left(n + 16\right) = \frac{2}{3} n - 4$
Multiply both sides by $6$ to get rid of the fractions
$\textcolor{w h i t e}{\text{XXX}} 3 \left(n + 16\right) = 4 n - 24$
Simplify
$\textcolor{w h i t e}{\text{XXX}} 3 n + 48 = 4 n - 24$
Subtract $3 n$ from both sides
$\textcolor{w h i t e}{\text{XXX}} 48 = n - 24$
Add $24$ to both sides
$\textcolor{w h i t e}{\text{XXX}} 72 = n$
or
$\textcolor{w h i t e}{\text{XXX}} n = 72$<|endoftext|>
| 4.71875 |
406 |
# Mr.Kapur withdrew rs.25000 from an ATM.If he received 150 notes in denominations of rs.500 and rs.100,find the no. of notes of each denomination?although i know how to do it, but it is a 2 variable method...can u suggest a 1 variable method?ans-25,125thanx
Suppose the number of Rs.500 notes be x.
Total number of notes = 150
Then, number of Rs.100 notes = $\left(150-x\right)$
Amount of 150 notes = $500x+100\left(150-x\right)$
Total amount withdrew from an ATM = Rs.25000
So we have;
$500x+100\left(150-x\right)=25000\phantom{\rule{0ex}{0ex}}⇒500x+15000-100x=25000\phantom{\rule{0ex}{0ex}}⇒400x=25000-15000\phantom{\rule{0ex}{0ex}}⇒x=\frac{10000}{400}=25$
Therefore number of Rs.500 notes = 25 and number of Rs.100 notes = $150-25=125$
• 7
let the number 500 rs notes be x
and number of 100rs notes be (150-x)
according to the condition
25000 = x(500) + (150 - x)(100)
25000 = 500x + 15000 - 100x
25000 = 500x - 100x +15000
25000 -15000 = 400x
10000 = 400x
x=10000/400
x=25
number of 500 rs notes = x=25 notes
number of 100rs notes = 150 - x=150-25 = 125 notes
• 3
What are you looking for?<|endoftext|>
| 4.40625 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.