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ZIMSEC O Level Geography Notes: Surface Water Flow and the Origin of rivers
Surface water flow and origin of rivers
- Rain falling down on land flows down the slope as sheet flow, rill flow and gully flow all of which contribute to stream discharge.
- Underground water oozes at certain points called springs and also contributes stream discharge.
- It is a type of overland flow or downslope movement of water which takes the form of a thin, continuous film over relatively smooth soil or rock surfaces
- is generated when rain falling onto the earth’s surface flows over the whole surface as a thin layer of water.
- It commonly occurs at the head of the watershed where the slope is gentle and the surface flat e.g. artificial surfaces, rocks etc.
- Rills are shallow channels (no more than a few tens of centimetres deep) cut into soil by the erosive action of flowing water.
- As the slope steepens,the amount of water increases and sheet flow encounters surface irregularities sheet flow turns into small shallow channels or rivulets known as rills.
- Rills in turn join up with other rills and form gullies.
- A gully is a landform created by running water, eroding sharply into soil, typically on a hillside.
- Gullies resemble large ditches or small valleys, but are metres to tens of metres in depth and width.
- The process by which gullies are formed is called gullying.
- A gully may grow in length by means of headward erosion at a knickpoint.
- Gullies are sometimes known as dongas.
- Gullies empty into streams which are perennial rivers.
The results of water erosion
- Sheet flow results in sheet erosion
- This results in the washing away of fertile top soils and shallow soils.
- Rock surfaces and plant roots are also exposed by sheet wash.
- Rill flow results in rill erosion.
- Gully flow results in gullies also known as dongas.
- Both Rill and gully erosions results in the formation of dongas and ravines.
The problems of dongas.
- Can lead to some areas becoming inaccessible as they are difficult to cross especially when it comes to carts and motor vehicles.
- Disrupts communication lines such as roads.
- Reduces the area available for crops pastures and settlements.
- Can lead to the uprooting of trees.
- Contribute to siltation.
- Humans and animals can fall into these ravines leading to injuries.
To access more topics go to the Geography Notes page.<|endoftext|>
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The Romans were well established across England by the end of the first century and it became necessary to establish vineyards to supply their armies with wine. They couldn’t go thirsty after all! It is generally agreed that Romans brought the vines with them to Britain, leaving behind evidence of consumption and even vineyards on a commercial scale. Although Roman reign eventually came to an end after 400 years, their gift of wine had already begun to entangle its roots into British culture. Rarely drunk by the wider population, the Christian Church began to incorporate wine into their ceremonies and by the time the Romans left, the only vineyards that remained were those tied to the monasteries.
In the Christmas of 1085, William the Conquerer commissioned the writing of the Domesday Book recording a total of 46 vineyards across the country, 12 of which being tied to a monastery. Although the number of vineyards had grown considerably to 139 sizeable plots, that number started to decrease shortly after the reign of Henry VIII. Some put it down to the change in climate while others put it down to the dissolution of the monasteries, but the true reason remains under debate.
1662 marks the date of Christopher Merrett’s revolutionary work. As an English scientist and physician, Merrett discusses in his paper the first deliberate case of secondary fermentation. By adding additional sugar and molasses after the primary fermentation, Merrett discovered a secondary phase of fermentation was induced, allowing for the essential fizz of the sparkling wine to develop. Although accidental secondary fermentation was known to occur prior to Merrett’s paper, the weak wood-fired glass could not withstand the enormous pressure built up in the process (indeed, the pressure can build to 3 times that of a car tyre). The development of tougher, coal fired glass in 17th century England was crucial in the development of sparkling wines. The higher combustion rate in the coal furnaces strengthened the glass, allowing for the secondary fermentation process without the hazard of exploding. This adds an interesting dimension to the traditional assumption that Dom Pérignon invented sparkling wine in Champagne around 1697.
By the end of the First World War no vineyard existed in England at all. It fell to Ray Barrington Brock to dispel this myth creating a research station, developing a wealth of data that went on to become the backbone of wine growing knowledge. Brock’s work demonstrated the grape varietals and techniques that flourished in the mild British climate. This essential revelation revitalised England’s love for winemaking, leading to the planting of the first commercial English vineyards of the modern time in Hampshire in 1951, the first vineyard to be planted in Britain for the production of wine for sale since 1875.
From the 1950s onwards, Britain has not just been at the forefront of the production of sparkling wine, but also at the forefront of the consumption being the largest consumer of Champagne, Prosecco and more recently Cava. Perhaps it was Winston Churchill, perhaps the heady days of the 60s, or perhaps it is simply down to the English mood of celebration. Whatever the cause, our love of sparkling wines is well established. Nowadays, English sparkling wine is setting new global standards while remaining truly English.
Since the 1970s, there have been a huge number of plantings in Britain, and in 2016 there were over 500 commercial vineyards in the UK, totalling over 2,000ha and producing around 4 million bottles a year. In the last 10 years, hectarage of planted vines has more than doubled, and is set to grow by another 50% by 2020. Some of the larger producers have multiple vineyards, Nyetimber being the largest with seven separate sites covering 171-ha. Other large producers include Gusbourne 93-ha, Denbies 90-ha, and Chapel Down 78-ha. Many of the larger producers such as Chapel Down, Ridgeview, and Camel Valley also buy grapes from vineyards which are usually under long-term contract to them or with which they have grape supply agreements. Together, the largest 100 producers control around 75% of the UK’s wine production.
In general, the current climate is suited to growing grapes for sparkling wine production and approximately 66% of all wine made in the UK is sparkling, 24% still whites, and 10% red or rose. Chardonnay, Pinot Noir and Pinot Meunier account for over 50% of varietal plantings in the UK. Bacchus, Seyval Blanc, Reichensteiner, Rondo, Ortega and Muller Thurgau are the other most planted varieties.
Order your bottle of Gusbourne Brut Rose now and get it delivered free!
Blog posted by Laura
(Look for the continuation of the British sparkling wine and discover more about the award winning estate Gusbourne)<|endoftext|>
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This complex terrain on Jupiter's moon, Europa, shows an area centered at 8 degrees north latitude, 275.4 degrees west longitude, in the trailing hemisphere. As Europa moves in its orbit around Jupiter, the trailing hemisphere is the portion which is always on the moon's backside opposite to its direction of motion. The area shown is about 62 miles by 87 miles (100 kilometers by 140 kilometers). The complex ridge crossing the picture in the upper left corner is part of a feature that can be traced hundreds of miles across the surface of Europa, extending beyond the edge of the picture. The upper right part of the picture shows terrain that has been disrupted by an unknown process, superficially resembling blocks of sea ice during a springtime thaw. Also visible are semicircular mounds surrounded by shallow depressions. These might represent the intrusion of material punching through the surface from below and partial melting of Europa's icy crust. The resolution of this image is about 200 yards (180 meters); this means that the smallest visible object is about a quarter of a mile across.
This picture of Europa was taken by Galileo's Solid State Imaging system from a distance of 11,100 miles (17,900 kilometers) on the spacecraft's sixth orbit around Jupiter, on Feb. 20, 1997.<|endoftext|>
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More than 70% of all bird species – over 5,000 species altogether – are perching birds, or Passerines. They have feet with three toes pointing forwards and one backwards, to help them cling to a perch.
Perching birds sing – this means that their call is not a single sound, but a sequence of musical notes.
Songbirds, such as thrushes, warblers and nightingales, are perching birds with especially attractive songs.
Usually only male songbirds sing – and mainly in the mating season, to warn off rivals and attract females.
Sparrows are small perching birds found in many parts of the world. Sparrows are seed-eaters with the house sparrow specializing in grain. Changes in farming practices are thought to account for this bird’s dramatic decline in numbers in Britain.
Starlings often gather on overhead cables ready to migrate.
Sparrows are small, plump birds, whose chirruping song is familiar almost everywhere.
Starlings are very common perching birds which often gather in huge flocks, either to feed or to roost.
All the millions of European starlings in North America are descended from 100 set free in New York’s Central Park in the 1890s.
Many perching birds, including mynahs, are talented mimics. The lyre bird of southeastern Australia can imitate car sirens and chainsaws, as well as other birds.
The red-billed quelea of Africa is the world’s most abundant bird. There are over 1.5 billion of them.<|endoftext|>
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Compound word game
£ 9.99 + P&P (Free UK P&P!)
Compound words can be made by joining two or more existing words together. The combination of these words creates a new word.
The words when joined up to form a compound word do not always keep their original meaning, for example invoice or breakfast.
Most children initially struggle with compound words as they read the words as two separate words. We have devised a word matching game to assist children in blending the words together. This helps children understand language structure.
The cards are colour coded to assist children with recognition as to which word begins the word and which word ends the word, i.e. red cards signify the beginning of the word and yellow cards signify the ending of the word.
The word on the cards is in bold letters so that it is clearly displayed to the child. The cards are numbered 1-3 in the top right hand corner of the card, which denotes the level of difficulty of the word, 3 being the most difficult.<|endoftext|>
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In most of the programming languages (C/C++, Java, etc), the use of else statement has been restricted with the if conditional statements. But Python also allows us to use the else condition with for loops.
The else block just after for/while is executed only when the loop is NOT terminated by a break statement.
Else block is executed in below Python 3.x program:
1 2 3 No Break
Else block is NOT executed in below Python 3.x program:
Such type of else is useful only if there is an if condition present inside the loop which somehow depends on the loop variable.
In the following example, the else statement will only be executed if no element of the array is even, i.e. if statement has not been executed for any iteration. Therefore for the array [1, 9, 8] the if is executed in third iteration of the loop and hence the else present after the for loop is ignored. In case of array [1, 3, 5] the if is not executed for any iteration and hence the else after the loop is executed.
For List 1: list contains an even number For List 2: list does not contain an even number
As an exercise, predict the output of following program.
This article is contributed by Harshit Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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Improved By : tarun2207<|endoftext|>
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These resources are suitable for lecturers and tutors to use in seminar groups, but many are also suitable for students engaged in independent study.
Useful Glossary of Economics terms.
The video-clips provide:
- Real world examples illustrating principles of economics.
- Animated graphics boards demonstrating how to answer questions raised in the introduction.
- Mathematical concepts are developed by an on-camera presenter, and illustrated with simple animations.
- Provide links to suggested Question Bank materials (from the film download page).
The teaching and learning guides are prepared with the lecturer or tutor in mind, and are designed to:
- Present innovative and interactive approaches to teaching mathematical concepts to economics students.
- Suggest top tips, and ways to incorporate the question bank and video clips into teaching activities.
- Provide links to external resources.
This resource can be used by students and lecturers for:
- Diagnostic testing
- Practice exercises
- Problem sheets
- Video support
These real-world case studies put in context mathematical concepts. They:
- Demonstrate the relevance of mathematical concepts.
- Can be used in seminars/tutorials.
- Aid revision.
Similar resources for statistics in social science are available from the DeSTRESS project.
More resources for teaching Maths for Economists are listed on the Economics Network site.
If you have any comments please contact The Economics Network.<|endoftext|>
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Related Articles
Factorization
• Last Updated : 03 Jan, 2021
The factor is a number or algebraic expression that divides another number or expression evenly i.e its remainder is 0. (or) factors are small numbers when multiplied gives other numbers. For Example 1, 2, 4, 7, 14, 28 are factors of number 28.
• Prime factor form: If we write a number in form of the product of prime factors. Then it is called prime factors form.
Example: 70 = 2 * 5 * 7; (here 2, 5, 7 are factors of 70; speacially these are also called prime factors as these (2, 5, 7 are prime numbers)
• Algebraic expression factors: Similarly, we can express algebraic expressions as the product of their factors. If an algebraic expression cannot be reduced further then it is its factor.
Example: 8xy = 8 * x * y (here 8xy is formed by multiplication of numbers(8, x, y) are factors of that number)
## Factorization
Factorization is nothing but writing a number as the product of smaller numbers. It is the decomposition of a number (or) mathematical objects into smaller or simpler numbers/objects. The process includes in factorization are:
### 1. Method of Common Factors
• Step 1: First, split every term of algebraic expression into irreducible factors
• Step 2: Then find the common terms among them.
• Step 3: Now the product of common terms and the remaining terms give the required factor form.
Example: Factorise 3x + 18?
Solution:
Step 1: First splitting every term into irreducible factors.
3x = 3 * x;
18 = 2 * 3 * 3;
Step 2: Next step to find the common term
3 is the only common term
Step 3: Now the roduct of common terms and remaining terms is 3(x + 6)
So 3(x + 6) is the required form.
### 2. Factorization by Regrouping
Sometimes the terms of the given expression should be arranged in suitable groups in such a way. So that all the groups have a common factor.
Example 1: Factorise x2 + yz + xy + xz?
Solution:
Here we don’t have a common term for all. So we are taking (x2 + xy) as one group and (yz + xz) as another group.
Factor form of (x2 + xy) = (x * x) + (x * y)
= x(x + y)
Factor form of (yz + xz) = (y * z) + (x * z)
= z(x + y)
After combining them,
x2 + yz + xy + xz = x(x + y) + z(x + y)
Taking (x + y) as common we get,
x2 + yz + xy + xz = (x + y) (x + z)
Example 2: Factorise 2xy + 3 + 2y + 3x?
Solution:
2xy + 2y + 3x + 3 [here we are rearranging terms to check if we get common terms or not]
2y (x + 1) + 3(x + 1)
(2y + 3) (x + 1)
### 3. Factorization using Identities
There are many standard identities. Some of them are given below:
i. (a + b)2 = a2 + 2ab + b2
ii. (a – b)2 = a2 – 2ab + b2
iii. a2 – b2 = (a + b) (a – b)
Example 1: Factorise x2 + 8x + 16?
Solution:
This is in the form of (a + b)2 = a2 + 2ab + b2
x2 + 8x + 16 = x2 + 2 * x * 4 + 42
= (x + 4)2
= (x + 4) (x + 4)
Example 2: Factorise a2 – 20a + 100?
Solution:
This in the form of (a – b)2 = a2 – 2ab + b2
a2 – 20a + 100 = a2 – 2 * a * 10 + 102
= (a – 10)2
= (a – 10) (a – 10)
Example 3: Factorise 25x2 – 49?
Solution:
This in the form of a2 – b2 = (a + b) (a – b)
25x2 – 49 = (5x)2 – 72
= (5x + 7) (5x – 7)
### 4. Factors of the form (x + a)(x + b)
In this method we need to factorise given expression such that (x + a) (x + b) = x2 + (a + b)x + ab.
Example: Factorise m2 + 10m + 21?
Solution:
This is in the form of (x + a) (x + b) = x2 + (a + b)x + ab
Where x = m; (a + b) = 10; ab = 21;
On solving we get a = 3; b = 7;
On substitution we get
m2 + 10m + 21 = m2 + (3 + 7)m + 3 * 7
= (m + 3)(m + 7)
## Division Of Algebraic Expressions
### 1. Division of monomial by a monomial
Example 1: Divide 35abc by 5ab?
Solution:
Convert each term into irreducable form
35abc = 5 * 7 * a * b * c
5ab = 5 * a * b
Normal division,
35abc / 5ab = 5 * 7 * a * b * c / 5 * a * b
= 7c
Example 2: Divide 14x5 by 2x3?
Solution:
14x5 = 2 * 7 * x * x * x * x * x
2x3 = 2 * x * x * x
14x5 / 2x3 = 2 * 7 * x * x * x * x * x / 2 * x * x * x
= 7x2
### 2. Division of polynomial by a monomial
Example: Divide 8x4 – 16x3 + 12x2 + 4x by 4x?
Solution:
8x4 – 16x3 + 12x2 + 4x / 4x
=> 4x(2x3 – 4x2 – 3x – 1) / 4x (Taking 4x as common and dividing it with denominator)
=> 2x3 – 4x2 – 3x – 1
### 3. Division of polynomial by a polynomial
Example: Divide 16a2 + 8 by 4a + 2?
Solution:
16a2 + 8 / 4a + 2
=> 4a(4a + 2) / (4a + 2) (here we took 4a common)
=> 4a
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes arrow_drop_up<|endoftext|>
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Academy of Literacy converts the memory challenged into logic based learners.
Teaches the ALPHabetic Principle in a meaningful & predictable format.
The skill of critical analysis is key to a learner's success in the information age.
Critical analytical thinking skills are the fundamental building blocks of PhonKnowLedgy.
It bridges the gap between the traditional back-to-the-basics - phonics
and the call for a more scientific system such as - constructivism
PhonKnowLedgy, serves as a key to the information age by
spanning the gap between traditional phonology and the need
for a more knowledge based system.
A scientific system of constructing learning that bridges
Abstract Phonology AND Knowledge children bring to learning
refers to the sounds letters make...
...and to the units of sound that make up words
means to understand
is a vein of ore--
here, veins of knowledge so fundamental that they run through the youngest learners.
The letter "g" is known as the "thirsty letter" in the PhonKnowLedgy process. It makes the same sound as does a thirsty child gulping down a drink. By drawing this kind of analogy between a child's behaviour and the abstract sound of the letter "g", the student is developing analogic thinking that makes 90% of written English predictable and logical. Making the letter sounds meaningful helps students to string them together to create the blocks of sound that make up whole words.
Phonology --the accurate pronunciation of letters-- is once again taking its place in education as a way of fostering literacy. It is time to move forward beyond the sounds of the letters to the units of sound that become logical to learners based on their own knowledge --PhonKnowLedgy.
Rules for dividing words into syllables are arbitrary and confusing for even the most astute of learners.
PhonKnowledgy refers to the natural rhythm of the words in songs by dividing words into bytes of sound...
...that are readily identified. Bytes of sound in words become meaningful knowledge when they are compared to members of a family.
• Just as individual members form a family, so bytes of sound form a word.
• Each individual needs a heart to be alive and for a byte-of-sound the heart is a Vowel.<|endoftext|>
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Jump to a New ChapterIntroducing the New ACT (and Ending World Hunger)General Strategies for Taking the ACTThe ACT English TestStrategies for the English TestUsage/Mechanics Questions on the English TestRhetorical Skills Questions on the English TestThe New ACT Writing TestThe ACT Math TestStrategies for the Math TestACT Math SubjectsThe ACT Reading TestStrategies for the Reading TestPassages and Questions on the Reading TestThe ACT Science Reasoning TestStrategies for the Science Reasoning TestPassages and Questions on the Science Reasoning TestPractice Tests Are Your Best Friends
10.1 Pre-Review Review 10.2 Pre-Algebra 10.3 Elementary Algebra 10.4 Intermediate Algebra
10.5 Coordinate Geometry 10.6 Plane Geometry 10.7 Trigonometry
Coordinate Geometry
Coordinate geometry is geometry dealing primarily with the line graphs and the (x, y) coordinate plane. The ACT Math Test includes nine questions on coordinate geometry. The topics you need to know are:
1. Number Lines and Inequalities
2. The (x,y) Coordinate Plane
3. Distance and Midpoints
4. Slope
5. Parallel and Perpendicular Lines
6. The Equation of a Line
7. Graphing Equations
8. Conic Sections
Most of the questions on coordinate geometry focus on slope. About two questions on each test will cover number lines and inequalities. The other topics are usually covered by just one question, if they are covered at all.
Number Lines and Inequalities
Number line questions generally ask you to graph inequalities. A typical number line graph question will ask you:
What is the graph of the solution set for 2(x + 5) > 4?
To answer this question, you first must solve for x.
1. Divide both sides by 2 to get: x + 5 > 2
2. Subtract 5 from both sides to get: x > –3
3. Now you match x > –3 to its line graph:
The circles at the endpoints of a line indicating an inequality are very important when trying to match an inequality to a line graph. An open circle at –3 denotes that x is greater than but not equal to –3. A closed circle would have indicated that x is greater than or equal to –3.
For the solution set –3 < x < 3, where x must be greater than –3 and less than 3, the line graph looks like this:
The (x,y) Coordinate Plane
The (x,y) coordinate plane is described by two perpendicular lines, the x-axis and the y-axis. The intersection of these axes is called the origin. The location of any other point on the plane (which extends in all directions without limit) can be described by a pair of coordinates. Here is a figure of the coordinate plane with a few points drawn in and labeled with their coordinates:
As you can see from the figure, each of the points on the coordinate plane receives a pair of coordinates: (x,y). The first coordinate in a coordinate pair is called the x-coordinate. The x-coordinate of a point is its location along the x-axis and can be determined by the point’s distance from the y-axis (x = 0 at the y-axis). If the point is to the right of the y-axis, its x-coordinate is positive, and if the point is to the left of the y-axis, its x-coordinate is negative.
The second coordinate in a coordinate pair is the y-coordinate. The y-coordinate of a point is its location along the y-axis and can be calculated as the distance from that point to the x-axis. If the point is above the x-axis, its y-coordinate is positive; if the point is below the x-axis, its y-coordinate is negative.
The ACT often tests your understanding of the coordinate plane and coordinates by telling you the coordinates of the vertices of a defined geometric shape like a square, and asking you to pick the coordinates of the last vertex. For example:
In the standard (x,y) coordinate plane, 3 corners of a square are (2,–2), (–2,–2), and (–2,2). What are the coordinates of the square’s fourth corner?
The best way to solve this sort of problem is to draw a quick sketch of the coordinate plane and the coordinates given. You’ll then be able to see the shape described and pick out the coordinates of the final vertex from the image. In this case, the sketch would look like this:
A square is the easiest geometric shape which a question might concern. It is possible that you will be asked to deal with rectangles or right triangles. The method for any geometric shape is the same, though. Sketch it out so you can see it.
Distance
The ACT occasionally asks test takers to measure the distance between two points on the coordinate plane. Luckily, measuring distance in the coordinate plane is made easy thanks to the Pythagorean theorem. If you are given two points, and their distance will always be given by the following formula:
The distance between two points can be represented by the hypotenuse of a right triangle whose legs are of lengths and The following diagram shows how the formula is based on the Pythagorean theorem (see p. ).
Here’s a sample problem:
Calculate the distance between (4,–3) and (–3,8).
To solve this problem, just plug the proper numbers into the distance formula:
The distance between the points is which equals approximately 13.04.
Finding Midpoints
Like finding the distance between two points, the midpoint between two points in the coordinate plane can be calculated using a formula. If the endpoints of a line segment are and then the midpoint of the line segment is:
In other words, the x- and y-coordinates of the midpoint are the averages of the x- and y-coordinates of the endpoints.
Here is a practice question:
What is the midpoint of the line segment whose endpoints are (6,0) and (3,7)?
All you have to do is plug the end points into the midpoint formula. According to the question, and
Slope
The slope of a line is a measurement of how steeply the line climbs or falls as it moves from left to right. More technically, the slope is a line’s vertical change divided by its horizontal change, also known as “rise over run.” Given two points on a line, and the slope of that line can be calculated using the following formula:
The variable most often used to represent slope is m.
So, for example, the slope of a line that contains the points (–2,–4) and (6,1) is:
Positive and Negative Slopes
You can easily determine whether the slope of a line is positive or negative just by looking at the line. If a line slopes uphill as you trace it from left to right, the slope is positive. If a line slopes downhill as you trace it from left to right, the slope is negative.
You can determine the relative magnitude of the slope by the steepness of the line. The steeper the line, the more the “rise” will exceed the “run,” and the larger and, consequently, the slope will be. Conversely, the flatter the line, the smaller the slope will be.
For practice, look at the lines in the figure below and try to determine whether their slopes are positive or negative and which have greater relative slopes:
Lines l and m have positive slopes, and lines n and o have negative slope. In terms of slope magnitude, line l > m > n > o.
Special Slopes
It can be helpful to recognize a few slopes by sight.
• A line that is horizontal has a slope of 0. Since there is no “rise,” and thus
• A line that is vertical has an undefined slope. In this case, there is no “run,” and Thus and any fraction with 0 in its denominator is, by definition, undefined.
• A line that makes a angle with a horizontal has a slope of 1 or –1. This makes sense because the “rise” equals the “run,” and or
Line a has slope 0 because it is horizontal. Line b has slope –1 because it makes a angle with the horizontal and slopes downward as you move from left to right. Line c has slope 1 because it makes a angle with the horizontal and slopes upward as you move from left to right. Line d has undefined slope because it is vertical.
Parallel and Perpendicular Lines
Parallel lines are lines that don’t intersect. In other words, parallel lines are lines that share the exact same slope.
Perpendicular lines are lines that intersect at a right angle (or 90%). In coordinate geometry, perpendicular lines have negative reciprocal slopes. That is, a line with slope m is perpendicular to a line with a slope of –1/m.
In the figure below are three lines. Lines q and r both have a slope of 2, so they are parallel. Line s is perpendicular to both lines q and r, and thus has a slope of –1/2.
On the ACT, never assume that two lines are parallel or perpendicular just because they look that way in a diagram. If the lines are parallel or perpendicular, the ACT will tell you so. (Perpendicular lines can be indicated by a little square located at the place of intersection, as in the diagram above.)
Equation of a Line
We’ve already shown you how to find the slope of a line using two points on the line. It is also possible to find the slope of a line using the equation of the line. In addition, the equation of a line can help you find the x- and y-intercepts of the line, which are the locations where the line intersects with the x- and y-axes. This equation for a line is called the slope-intercept form:
where m is the slope of the line, and b is the y-intercept of the line.
Finding the Slope Using the Slope-Intercept Form
If you are given the equation of a line that matches the slope-intercept form, you immediately know that the slope is equal to the value of m. However, it is more likely that the ACT will give you an equation for a line that doesn’t exactly match the slope-intercept form and ask you to calculate the slope. In this case, you will have to manipulate the given equation until it resembles the slope-intercept form. For example,
What is the slope of the line defined by the equation 5x + 3y = 6?
To answer this question, isolate the y so that the equation fits the slope-intercept form.
The slope of the line is –5/3.
Finding the Intercepts Using the Slope-Intercept Form
The y-intercept of a line is the y-coordinate of the point at which the line intersects the y-axis. Likewise, the x-intercept of a line is the x-coordinate of the point at which the line intersects the x-axis. In order to find the y-intercept, simply set x = 0 and solve for the value of y. To find the x-intercept, set y = 0 and solve for x.
To sketch a line given in slope-intercept form, first plot the y-intercept, and then use the slope of the line to plot another point. Connect the two points to form your line. In the figure below, the line y = –2x + 3 is graphed.
Since the slope is equal to –2, the line descends two units for every one unit it moves in the positive x direction. The y-intercept is at 3, so the line crosses the y-axis at (0,3).
Graphing Equations
For the ACT Math Test, you should know how the graphs of certain equations look. The two equations that are most important in terms of graphing are and
If you add lesser-degree terms to the equations, these graphs will shift around the origin but retain their basic shape. You should also keep in mind what the negatives of these equations look like:
Conic Sections
Occasionally, the ACT will test your knowledge of parabolas, circles, or ellipses. These topics do not regularly appear on the ACT, but it still pays to prepare: if these topics do appear, getting them right can separate you from the crowd.
Parabolas
A parabola is a “U”-shaped curve that can open either upward or downward.
A parabola is the graph of a quadratic function, which, you may recall, follows the form The equation of a parabola gives you quite a bit of information about the parabola.
1. The vertex of the parabola is
2. The axis of symmetry of the parabola is the line
3. The parabola opens upward if a > 0, and downward if a < 0.
4. The y-intercept is the point (0, c).
Circles
A circle is the collection of points equidistant from a given point, called the center of the circle. Circles are defined by the formula:
where (h,k) is the center of the circle, and r is the radius. Note that when the circle is centered at the origin, h = k = 0, so the equation simplifies to:
That’s it. That’s all you need to know about circles in coordinate geometry. Once you know and understand this equation, you should be able to sketch a circle in its proper place on the coordinate system if given its equation. You should also be able to figure out the equation of a circle given a picture of its graph with coordinates labeled.
Ellipses
An ellipse is a figure shaped like an oval. It looks like a circle somebody sat on, but it is actually a good deal more complicated than a circle, as you can see from all the jargon on the diagram below.
The two foci are crucial to the definition of an ellipse. The sum of the distances from the foci to any point on the ellipse is constant. To understand this visually, look at the figure below. The quantity is constant for each point on the ellipse.
The line segment containing the foci of an ellipse with both endpoints on the ellipse is called the major axis. The endpoints of the major axis are called the vertices. The line segment perpendicularly bisecting the major axis with both endpoints on the ellipse is the minor axis. The point midway between the foci is the center of the ellipse. When you see an ellipse, you should be able to identify where each of these components would be.
The equation of an ellipse is:
where a, b, h, and k are constants. With respect to this formula, remember that:
1. The center of the ellipse is (h,k).
2. The length of the horizontal axis is 2a.
3. The length of the vertical axis is 2b.
4. If a > b, the major axis is horizontal and the minor axis is vertical; if b > a, the major axis is vertical and the minor axis is horizontal.
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Before you attach the 9V battery, all points in the circuit are at ground (0V). To understand how the voltage regulator circuit works, first consider how it operates without the capacitor.
With the switch in the off position, a voltage is put on the Zener diode through R14.
(A Zener diode is a unique type of diode that allows the current to flow forward the same way as an ideal diode, but can also allow it to flow in reverse when the voltage is higher than a certain value known as the breakdown voltage. It was named after the physicist who discovered this phenomenon, Clarence Zener. The device is commonly used to provide a reference point for voltage regulators, or to protect other semiconductors from momentary voltage pulses).
R14 is used so the Zener diode will be in the reverse breakdown mode. As a result, the voltage will be held at a constant 5.6V all across the diode. The 5.6V in this example is also maintained along the transistor and load reistor base emitter junction in the series. Because the voltage is greater than 0.7V, the diode at the base emitter is biased towards the front and the current flows into the base of the transistor.
This means that as long as the Zener diode is in reverse mode and the base emitter is biased to the front, the voltage across the resistor will be fixed at 5.6 – 0.7 = 4.9V.
To understand how the circuit is able to always hold the voltage around a constant 5V, consider the flow of the current in the circuit. As the current flows through R14, it splits between the base emitter diode and the Zener diode.
If Ib is the current flowing into the the base, a current equal to ßIb will flow from the battery into the collector. Because the transistor is set to function in the forward mode, a current equal to (ß + 1)Ib will flow out of the emitter and pass through the load resistor.
If there is a change in the load resistor, then the current flowing through the Zener diode will change as well, so that the base current and the emitter current can maintain the correct value to put out the required 5 V all across the load resistor.
Since the capacitor is the component in the circuit capable of storing the electrical charge, it is used in this example to help maintain the output of the voltage regulator at a constant rate over time.
The voltage change rate across a capacitor is proportionate to the amount of current flowing from it, divided by the amount of electricity it is able to store. Therefore, the bigger the capacitor, the less of a voltage output change there will be over time for a fixed current drain.<|endoftext|>
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by Siddharthan Surveswaran
When you pass through any wasteland in your locality you may notice a pale green plant with thick leaves and small purplish-white flowers. Some even grow this plant in front of their houses for religious purposes. This plant is calledCalotropis, and it is a widespread kind of milkweed, originally from Africa. When you pluck a leaf of the Calotropis plant a gush of milky sap or latex oozes out from the wound and therefore these plants are aptly named as milkweeds.
Milkweeds flowers are perhaps the most interesting of all the many kinds of flowers in the world. In general, all flowers have male and female parts. The male part is called the stamen, which consists of a long stalk upon which a small capsule is attached. This capsule is called the anther and it holds pollen grains, which are the male cells. The female part of the flower is called the pistil which consists of a tube, called the style, and a head-like structure, called the stigma. In other flowers the male and female parts are separate, but milkweed flowers are remarkable because the male and female structures are united into an elegant circular structure that looks like a crown. Isn’t that wonderful? Like a king and queen ruling from the same throne.
Pack of pollens
Now, the point of flowers is reproduction – getting the pollen grains of one flower to land on the stigma of another so that fertilization can happen and the next generation (that is, the seeds) can be formed. Many plants use wind to disperse their pollen grains to neighbouring flowers, while other plants attract insects, which carry the pollen grains to other flowers. The clever milkweeds manage this more efficiently by packing their pollen into little bags called pollinia. (The only other plants that do this are orchids.) These pollen bags are embedded into the crown of the flower.
Why so complicated? When a bee comes to visit the milkweed flower for nectar it tramples on the crown and releases the pollen bags which then readily stick to the bee’s legs. The pollinia now hitch a ride with the bee, which visits the next flower for its next dose of nectar. The pollinia then land on the stigma of another flower and the milkweed fruits are formed.
The awesomeness of milkweeds doesn’t end there! Their seeds have a whisk of silky hairs attached to their head. These hairs catch the wind and carry the seeds like a parachute, dropping them off in distant places. This mechanism is responsible for the spread of these plants over long distances thereby colonizing newer and newer areas. Isn’t it amazing that these plants are so clever? Next time you see a milkweed plant, take a closer look. But be sure to wash your hands carefully after you touch the plant: its milky latex contains toxic substances to stop animals from eating it. The wonders of the humble milkweed go on and on!
Picture: Flowers of the Calotropis plant. Can you see the crown in the centre of the flowers, with their five anthers?
Credit: Mayur Y. Kamble<|endoftext|>
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The gas industry is already feeding gas from biogas into the gas network. The plan for the future is to use the network of gas pipelines as a composite system in which natural gas, biogas as well as hydrogen and synthetic methane produced from renewable electricity are combined to form one huge energy source. The natural gas pipelines in Germany have a total length of more than 505,000 km and are already transporting double the amount of energy than the entire electricity grid.
Much more promising is the option of converting renewable electricity into gas and storing it in the gas pipelines. This would make it possible to use demand side management (adjusting the demand for electricity to the amounts being generated at that time), for instance at times when the wind is blowing so hard that more electricity is being produced than can be taken up by the grids, or to render the production side more flexible. It would also be possible for excess electricity to be converted into other forms of energy. One option here is power-to-gas technology. This technology uses electricity to split water into hydrogen and oxygen (electrolysis). The hydrogen is then fed into the gas network. In a second step, additional chemical reactions can be triggered to produce methane from the hydrogen. The gas that is produced in this way can then be used for industrial purposes, for heating, and for transport. It can also be used to power turbines, which convert the energy back into electricity.
Multiple conversions do, however, result in high levels of energy loss. This is why this solution is not yet economically viable. In the medium-term, however, power-to-gas technology could provide an affordable solution for storing large amounts of electricity in way that is economically viable. It would also forge a strong link between the gas network and electricity from renewables.<|endoftext|>
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Relations and Mapping are important topics in Algebra. Relations & Mapping are two different words and have different meanings mathematically. Let’s get deep into the article to know all about the Relations, Mapping, or Functions like Definitions, Types of Relations, Solved Examples, etc.
## What is a Relation?
A relation is a collection of ordered pairs. Relation in general defines the relationship between two different sets of information. An ordered pair is a point that has both x and y coordinates. Let us consider two sets x and y and set x has a relation with y. The values of set x are called Domain and the values of set y are called Range.
Relations can be represented using three different notations i.e. in the form of a table, graph, mapping diagram.
Example: (2, -5) is an ordered pair.
### What is Mapping?
Mapping denotes the relation from Set A to Set B. Relation from A to B is the Subset of AxB. Mapping the oval on the left-hand side denotes the values of Domain and the oval on the right-hand side denotes the values of Range.
From the above diagram, we can say the ordered pairs are (1,c) (2, n) (5, a) (7, n).
Set{ 1, 2, 5, 7} represents the domain.
{a, c, n} is the range.
A function is a relation that derives the output for a given input.
Remember that all functions are relations but not all relations are functions.
### Types of Relations
There are 8 different types of Relations and we have mentioned each of them in the following modules along with Examples.
• Empty Relation
• Universal Relation
• Identity Relation
• Inverse Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Equivalence Relation
Empty Relation:
Empty Relation is the one in which there will be no relation between elements of a set. It is also called a Void Relation. For instance, Set A = {3, 4, 5} then void relation can be R = {x, y} where | x- y| = 7.
For an Empty Relation R = φ ⊂ A × A
Universal Relation:
In Universal Relation every element of a set is related to each other. Consider a set A = {a, b, c}. Universal Relations will be R = {x, y} where, |x – y| ≥ 0.
For Universal Relation R = A × A
Identity Relation:
Every element of a set is related to itself in an Identity Relation. Consider a Set A = {a, b, c} then Identity Relation is given by I = { a,a}, {b,b}, {c, c}
I = {(a, a), a ∈ A}
Inverse Relation:
In Inverse Relation when a set has elements that are inverse pairs of another set. If Set A = {(a,b), (c,d)} then inverse relation will be R-1 = {(b, a), (d, c)}
R-1 = {(b, a): (a, b) ∈ R}
Reflexive Relation:
Every element maps to itself in Reflexive Relation. Consider Set A = { 3, 4} then Reflexive Relation R = {(3, 3), (4, 4), (3, 4), (4, 3)}. Reflexive Relation is given by (a, a) ∈ R
Symmetric Relation:
In the case of Symmetric Relation if a = b, then b = a is also true. Relation R is symmetric only if (b, a) ∈ R is true when (a,b) ∈ R.
aRb ⇒ bRa, ∀ a, b ∈ A
Transitive Relation:
In case of a transitive relation if (x, y) ∈ R, (y, z) ∈ R then (x, z) ∈ R
aRb and bRc ⇒ aRc ∀ a, b, c ∈ A
Equivalence Relation:
A Relation symmetric, reflexive, transitive at the same time is called an Equivalence Relation.
### How to Convert a Relation to Function?
A relation that follows the rule that every X- Value associated with only one Y-Value is called a Function.
Example
Is A = {(1, 4), (2, 5), (3, -8)}?
Solution:
Since the set has no duplicates or repetitions in the X- Value, the relation is a function.
### Mapping Diagrams
Mapping Diagram consists of two columns in which one denotes the domain of a function f whereas the other column denotes the Range. Usually, Arrows or Lines are drawn between domain and range to denote the relation between two elements.
One-to-One Mapping
Each element of the range is paired with exactly one element of the domain. The function represented below denotes the One-to-One Mapping.
Many to One Mapping
If one element in the range is associated with more than one element in the domain is called many to one mapping. In the below diagram you can see the second number II is associated with more than one element in the domain.
One to Many Mapping
If one element in the domain is mapped with more than elements in the range then it is called One to Many Mapping. In the below diagram the first element in the domain is mapped to many elements in the range therefore it is called One to Many Mapping.
Hope you got a complete idea on Relations and Mapping Concept. If you need any other help you can ask us through the comment section and we will get back to you at the earliest possibility.<|endoftext|>
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Preparation: Obtain a number of pictures of events during Jesus’ life. Post these around the Primary room. Gather crayons, markers, and/or pencils and a sheet of paper for each child.
Presentation: Give each child paper and crayons and have them create a picture of an event from Jesus’ life. Point out the pictures of Jesus posted around the room. If needed, have their teachers help explain the posted pictures to the children, as they work. Give the children a set amount of time to work on their picture.
Ask a number of children to show their pictures and briefly tell the story the pictures depict. After each picture, discuss what Jesus did. Ask the children what they can learn from that story and how they can be more like Jesus.
Bear your testimony of Jesus Christ. Sing songs about Jesus Christ, such as:
“When Jesus Christ Was Baptized” (Children’s Songbook, p. 102)
“Jesus Loved the Little Children” (Children’s Songbook, p. 59)
“Jesus Once Was a Little Child” (Children’s Songbook, p. 55)
“He Sent His Son” (Children’s Songbook, p. 34)<|endoftext|>
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# Geometry Set A 2014-2015 SSC (English Medium) 10th Standard Board Exam Question Paper Solution
Geometry [Set A]
Date: March 2015
Duration: 2h
[5] 1 | Solve any five sub-questions :
[1] 1.1
In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find (A(triangleABC))/(A(triangleDCB))
Concept: Properties of Ratios of Areas of Two Triangles
Chapter: [0.01] Similarity
[1] 1.2
Find the slope and y-intercept of the line y = -2x + 3.
Concept: Intercepts Made by a Line
Chapter: [0.04] Co-ordinate Geometry
[1] 1.3
In the following figure, in ΔABC, BC = 1, AC = 2, ∠B = 90°. Find the value of sin θ.
Concept: Heights and Distances
Chapter: [0.06] Trigonometry
[1] 1.4
Find the diagonal of a square whose side is 10 cm.
Concept: Circumference of a Circle
Chapter: [0.07] Mensuration [0.07] Mensuration
[1] 1.5
The volume of a cube is 1000 cm3. Find the side of a cube.
Concept: Surface Area of a Combination of Solids
Chapter: [0.07] Mensuration
[1] 1.6
If two circles with radii 5 cm and 3 cm respectively touch internally, find the distance between their centres.
Concept: Touching Circles
Chapter: [0.03] Circle
[8] 2 | Solve any four sub-questions :
[2] 2.1
If sin θ =7/25, where θ is an acute angle, find the value of cos θ.
Concept: Trigonometric Ratios of Complementary Angles
Chapter: [0.06] Trigonometry
[2] 2.2
Draw ∠ABC of measure 120° and bisect it.
Concept: Basic Geometric Constructions
Chapter: [0.05] Geometric Constructions
[2] 2.3
Find the slope of the line passing through the points M(4,0) and N(-2,-3).
Concept: Slope of a Line
Chapter: [0.04] Co-ordinate Geometry
[2] 2.4
Find the area of the sector whose arc length and radius are 14 cm and 6 cm respectively.
Concept: Areas of Sector and Segment of a Circle
Chapter: [0.07] Mensuration
[2] 2.5
In the following figure, in Δ PQR, seg RS is the bisector of ∠PRQ.
PS = 11, SQ = 12, PR = 22. Find QR.
Concept: Similarity of Triangles
Chapter: [0.01] Similarity
[2] 2.6
In the following figure, if m(arc DXE) = 120° and m(arc AYC) = 60°. Find ∠DBE.
Concept: Areas of Sector and Segment of a Circle
Chapter: [0.07] Mensuration
[9] 3 | Solve any three sub-questions :
[3] 3.1
In the following figure, Q is the centre of a circle and PM, PN are tangent segments to the circle. If ∠MPN = 60°, find ∠MQN.
Concept: Number of Tangents from a Point on a Circle
Chapter: [0.03] Circle [0.03] Circle
[3] 3.2
Draw the tangents to the circle from the point L with radius 3 cm. Point ‘L’ is at a distance 8 cm from the centre ‘M’.
Concept: Construction of Tangents to a Circle
Chapter: [0.05] Geometric Constructions
[3] 3.3
The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.
Concept: Properties of Ratios of Areas of Two Triangles
Chapter: [0.01] Similarity
[3] 3.4
Two buildings are in front of each other on either side of a road of width 10 metres. From the top of the first building which is 20 metres high, the angle of elevation to the top of the second is 45°. What is the height of the second building?
Concept: Heights and Distances
Chapter: [0.06] Trigonometry
[3] 3.5
Find the volume and surface area of a sphere of radius 8.4 cm.
(pi=22/7)
Concept: Surface Area of a Combination of Solids
Chapter: [0.07] Mensuration
[8] 4
[4] 4.1
Prove that ‘the opposite angles of a cyclic quadrilateral are supplementary’.
Chapter: [0.03] Circle
[4] 4.2
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
Concept: Trigonometric Identities
Chapter: [0.06] Trigonometry
[4] 4.3
A test tube has diameter 20 mm and height is 15 cm. The lower portion is a hemisphere. Find the capacity of the test tube. (π = 3.14)
Concept: Surface Area of a Combination of Solids
Chapter: [0.07] Mensuration
[10] 5 | Solve any two sub-questions :
[5] 5.1
Prove that the angle bisector of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
Concept: Similarity of Triangles
Chapter: [0.01] Similarity
[5] 5.2
Write down the equation of a line whose slope is 3/2 and which passes through point P, where P divides the line segment AB joining A(-2, 6) and B(3, -4) in the ratio 2 : 3.
Concept: Division of a Line Segment
Chapter: [0.04] Co-ordinate Geometry [0.05] Geometric Constructions
[5] 5.3
ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.
Concept: Division of a Line Segment
Chapter: [0.04] Co-ordinate Geometry [0.05] Geometric Constructions
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## Maharashtra State Board previous year question papers 10th Standard Board Exam Geometry with solutions 2014 - 2015
Maharashtra State Board 10th Standard Board Exam Geometry question paper solution is key to score more marks in final exams. Students who have used our past year paper solution have significantly improved in speed and boosted their confidence to solve any question in the examination. Our Maharashtra State Board 10th Standard Board Exam Geometry question paper 2015 serve as a catalyst to prepare for your Geometry board examination.
Previous year Question paper for Maharashtra State Board 10th Standard Board Exam Geometry-2015 is solved by experts. Solved question papers gives you the chance to check yourself after your mock test.
By referring the question paper Solutions for Geometry, you can scale your preparation level and work on your weak areas. It will also help the candidates in developing the time-management skills. Practice makes perfect, and there is no better way to practice than to attempt previous year question paper solutions of Maharashtra State Board 10th Standard Board Exam.
How Maharashtra State Board 10th Standard Board Exam Question Paper solutions Help Students ?
• Question paper solutions for Geometry will helps students to prepare for exam.
• Question paper with answer will boost students confidence in exam time and also give you an idea About the important questions and topics to be prepared for the board exam.
• For finding solution of question papers no need to refer so multiple sources like textbook or guides.<|endoftext|>
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Why does the thyroid gland need special protection after a release of radioactive material?
The thyroid gland needs iodine to produce hormones that regulate the body’s energy and metabolism. The thyroid absorbs available iodine from the bloodstream. The thyroid gland cannot distinguish between stable (regular) iodine and radioactive iodine and will absorb whatever it can. In babies and children, the thyroid gland is one of the most radiation-sensitive parts of the body. Most nuclear accidents release radioactive iodine into the atmosphere which can be absorbed into the body. When thyroid cells absorb too much radioactive iodine, it can cause thyroid cancer to develop several years after the exposure. Babies and young children are at highest risk. The risk is much lower for people over age 40. Thyroid cancer seems to be the only cancer whose incidence rises after a radioactive iodine release. Potassium iodide protects only the thyroid, but it is the organ at greatest risk from radioactive iodine.
What is Potassium Iodide (KI)?
Potassium iodide (KI) is the same form of iodine used to iodize table salt. KI floods the thyroid with iodine, thus preventing radioactive iodine from being absorbed. If taken at the proper time, KI protects the thyroid from radioactive iodine from all sources – air, food, milk, and water. KI is a non-prescription drug that can be bought over the internet and at some pharmacies. KI is made in pill and liquid forms. KI products approved are: Iosat Tablets (130 mg), ThyroSafe Tablets (65 mg) and ThyroShield Solution (65 mg/ml). Properly packaged, KI’s shelf life is at least 5 years and possibly as long as 11 years. If you take a very old pill, it may not work fully but it won’t hurt you.
What is the proof that KI works?
After the 1986 Chornobyl (formerly called “Chernobyl”) nuclear accident, shifting winds blew a radioactive cloud over Europe. As many as 3,000 people exposed to that radiation developed thyroid cancer over the next 10 years. Most victims had been babies or young children living in Ukraine, Belarus, or Russia at the time of the accident. The region of excess risk extended up to a 200 mi radius from Chornobyl. Poland, immediately adjacent to Belarus and Ukraine, distributed KI to >95% of their children within 3 days of the accident and does not appear to have had an increase in thyroid cancer.
Who should take KI?
Since children are at the highest risk to exposure to radioactive iodine, KI should be available to all children. Also, because of the risk to the developing fetus, pregnant women should also take KI in the event of a nuclear accident. Adults are at a lower risk but still may benefit from KI. In addition to KI, priority should be given to evacuation, sheltering (staying in an unventilated room with the doors and windows closed) and avoiding contaminated food, milk, and water. KI should not take the place of any other protective measure.
When should KI be taken?
KI fills the thyroid cells and prevents the gland from absorbing radioactive iodine for approximately 24 hours. People should take one dose a day while they are being exposed to radioactive iodine until the risk no longer exists. KI should be used only under instruction from local health authorities. Not every radioactive release includes the radioactive iodine that can cause thyroid cancer. For example, a “dirty bomb” is not likely to contain radioactive iodine because it has a short halflife. (A “dirty bomb” is a conventional bomb mixed with radioactive material, and designed to explode spewing out the radioactive isotopes and contaminating a wide area.) Health authorities can determine which radioactive isotopes are released during a nuclear event. If radioactive iodine is released, then health authorities will advise on when and how long to take KI.
What are the recommended KI doses?
The FDA recommends the following doses:
0 – 1 months 15 mg
1 months- 3 years 30 – 35 mg
3-12 years 65 mg
>12 years 130 mg
The easiest way to prepare a 16-mg dose for a newborn 0-1 month is to dissolve a 130-mg pill in 8 oz of a clear liquid and feed the newborn 1 oz of the liquid.
Who should not take KI?
Millions of people have taken KI but few serious sideeffects have been reported. The only people who should not take KI are those who have had a major allergic reaction to iodine. During a nuclear emergency, KI’s benefit far outweighs any potential risk. Adults over age 40 do not need KI at all unless they are exposed to extremely high levels of radioactive iodine. Patients with thyroid disease can safely take the pills in the FDA recommended doses. If taken long enough, KI can cause temporary hypothyroidism (underactive thyroid gland). “Long enough” is different for every person. Prolonged treatment can become a serious problem for very young children. Such children should be seen afterward by a health professional. Patients with Graves’ hyperthyroidism or with autonomous functioning thyroid nodules should also be seen.
Why worry so much about thyroid cancer if most people survive it?
In general, 90% of patients survive thyroid cancer. The post-Chornobyl cancers have been aggressive and have been unusual in affecting children younger than 10 years of age. Thyroid cancer survivors always remain at risk for recurrence and require lifelong medical care. Likewise, the people who were exposed to radioactive iodine from the Chornobyl accident but have not developed thyroid cancer remain at risk for life and must continue to be tested. The demands of regular testing and care for this large population are putting a heavy burden on both patients and health care systems.
How should KI be incorporated into an overall emergency plan?
KI is an adjunct to evacuation, sheltering (staying in an unventilated room with the doors and windows closed), and avoiding contaminated food, milk, and water. KI should not take the place of any other protective measures
Won’t having KI pills lull people into a false sense of security?
Not likely. Local authorities recommend that people leave the vicinity of a nuclear emergency as quickly as possible. People are being taught that KI is just a supplement to evacuation.
Why waste time taking a pill if you’re being told to evacuate?
Nuclear releases are unpredictable and traffic jams are likely to delay speedy evacuation. People should take their KI before they evacuate, following instructions from local health officials.
Why offer KI to people just within 10 or 20 miles of a plant? Can’t radiation be harmful farther away?
No one can predict how far a radioactive iodine cloud might spread. After Chornobyl, higher than expected rates of thyroid cancer were found more than 200 miles away from the nuclear plant. Thus, no one can predict how far from a nuclear plant, one should distribute KI if it is to protect every person who might be exposed to radioactive iodine.
Because there is no right answer, it has been recommended three levels of coverage, determined by distance from the nuclear plant:
- 0 – 50 mi Pre-distribute KI to households, keep stockpile near
- 50 – 200 mi Stockpile KI in local public facilities (hospitals, schools, police and fire stations)
- > 200 mi Make KI available from HHS National stockpile
What are other countries doing?
The World Health Organization endorses KI distribution. France, Ireland, Sweden, and Switzerland not only stockpile KI but predistribute KI to their populations.<|endoftext|>
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When it comes to dealing with the cosmos, we humans like to couch things in familiar terms. When examining exoplanets, we classify them based on their similarities to the planets in our own Solar System – i.e. terrestrial, gas giant, Earth-size, Jupiter-sized, Neptune-sized, etc. And when measuring astronomical distances, we do much the same.
For instance, one of the most commonly used means of measuring distances across space is known as an Astronomical Unit (AU). Based on the distance between the Earth and the Sun, this unit allows astronomers to characterize the vast distances between the Solar planets and the Sun, and between extra-solar planets and their stars.
According to the current astronomical convention, a single Astronomical Unit is equivalent to 149,597,870.7 kilometers (or 92,955,807 miles). However, this is the average distance between the Earth and the Sun, as that distance is subject to variation during Earth’s orbital period. In other words, the distance between the Earth and the Sun varies in the course of a single year.
During the course of a year, the Earth goes from distance of 147,095,000 km (91,401,000 mi) from the Sun at perihelion (its closest point) to 152,100,000 km (94,500,000 mi) at aphelion (its farthest point) – or from a distance of 0.983 AUs to 1.016 AUs.
History of Development:
The earliest recorded example of astronomers estimating the distance between the Earth and the Sun dates back to Classical Antiquity. In the 3rd century BCE work, On the Sizes and Distances of the Sun and Moon – which is attributed to Greek mathematician Aristarchus of Samos – the distance was estimated to be between 18 and 20 times the distance between the Earth and the Moon.
However, his contemporary Archimedes, in his 3rd century BCE work Sandreckoner, also claimed that Aristarchus of Samos placed the distance of 10,000 times the Earth’s radius. Depending on the values for either set of estimates, Aristarchus was off by a factor of about 2 (in the case of Earth’s radius) to 20 (the distance between the Earth and the Moon).
The oldest Chinese mathematical text – the 1st century BCE treatise known as Zhoubi Suanjing – also contains an estimate of the distance between the Earth and Sun. According to the anonymous treatise, the distance could be calculated by conducting geometric measurements of the length of noontime shadows created by objects spaced at specific distances. However, the calculations were based on the idea that the Earth was flat.
Famed 2nd century CE mathematician and astronomer Ptolemy relied on trigonometric calculations to come up with a distance estimate that was equivalent to 1210 times the radius of the Earth. Using records of lunar eclipses, he estimated the Moon’s apparent diameter, as well as the apparent diameter of the shadow cone of Earth traversed by the Moon during a lunar eclipse.
Using the Moon’s parallax, he also calculated the apparent sizes of the Sun and the Moon and concluded that the diameter of the Sun was equal to the diameter of the Moon when the latter was at it’s greatest distance from Earth. From this, Ptolemy arrived at a ratio of solar to lunar distance of approximately 19 to 1, the same figure derived by Aristarchus.
For the next thousand years, Ptolemy’s estimates of the Earth-Sun distance (much like most of his astronomical teachings) would remain canon among Medieval European and Islamic astronomers. It was not until the 17th century that astronomers began to reconsider and revise his calculations.
This was made possible thanks to the invention of the telescope, as well as Kepler’s Three Laws of Planetary Motion, which helped astronomers calculate the relative distances between the planets and the Sun with greater accuracy. By measuring the distance between Earth and the other Solar planets, astronomers were able to conduct parallax measurements to obtain more accurate values.
By the 19th century, determinations of about the speed of light and the constant of the aberration of light resulted in the first direct measurement of the Earth-Sun distance in kilometers. By 1903, the term “astronomical unit” came to be used for the first time. And throughout the 20th century, measurements became increasingly precise and sophisticated, thanks in part to accurate observations of the effects of Einstein’s Theory of Relativity.
By the 1960s, the development of direct radar measurements, telemetry, and the exploration of the Solar System with space probes led to precise measurements of the positions of the inner planets and other objects. In 1976, the International Astronomical Union (IAU) adopted a new definition during their 16th General Assembly. As part of their System of Astronomical Constants, the new definition stated:
“The astronomical unit of length is that length (A) for which the Gaussian gravitational constant (k) takes the value 0.01720209895 when the units of measurement are the astronomical units of length, mass and time. The dimensions of k² are those of the constant of gravitation (G), i.e., L³M-1T–2. The term “unit distance” is also used for the length A.”
In response to the development of hyper-precise measurements, the International Committee for Weights and Measures (CIPM) decided to modify the the International System of Units (SI) in 1983. Consistent with this, they redefined the meter to be measured in terms of the speed of light in vacuum.
However, by 2012, the IAU determined that the equalization of relativity made the measurement of AUs too complex, and redefined the astronomical unit in terms of meters. In accordance with this, a single AU is equal to 149597870.7 km exactly (92.955807 million miles), 499 light-seconds, 4.8481368×10-6 of a parsec, or 15.812507×10-6 of a light-year.
Today, the AU is used commonly to measure distances and create numerical models for the Solar System. It is also used when measuring extra-solar systems, calculating the extent of protoplanetary clouds or the distance between extra-solar planets and their parent star. When measuring interstellar distances, AUs are too small to offer convenient measurements. As such, other units – such as the parsec and the light year – are relied upon.
The Universe is a huge place, and measuring even our small corner of it producing some staggering results. But as always, we prefer to express them in ways that are as relatable and familiar.
We’ve written many interesting articles about distances in the Solar System here at Universe Today. Here’s How Far are the Planets from the Sun?, How Far is Mercury from the Sun?, How Far is Venus from the Sun?, How Far is Earth from the Sun?, How Far is Mars from the Sun?, How Far is Jupiter from the Sun?, How Far is Saturn from the Sun?, How Far is Uranus from the Sun?, How Far is Neptune from the Sun?, How Far is Pluto from the Sun?
If you’d like more information about the Earth’s orbit, check out NASA’s Solar System Exploration page.
We’ve also recorded an episode of Astronomy Cast dedicated to the measurement of distances in astronomy. Listen here, Episode 10: Measuring Distance in the Universe.
- NASA: Near Earth Object Program – Astronomical Unit (AU)
- EarthSky – What is an astronomical unit?
- Cool Cosmos – Astronomical Unit
- Wikipedia – Astronomical Unit<|endoftext|>
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Lesson 1001: Representing Boolean Expressions with Truth Tables
A helpful way to compare and visualize boolean expressions is with truth tables. Truth tables allow you to evaluate the expression quickly yourself. To demonstrate this, lets create a truth table for the expression:
First, we must make our ordered input columns for the two inputs in this expression, A and B:
Next, we must evaluate the first term. The order of operations tells us that the operation we will perform last is the OR. So, lets move from left to right with our terms. First, let's place AB on the chart, since both of the operands required, A and B are already on the chart for us.
Now, we must evaluate the second term, which is NOT(A XOR B). There are two reasons why we will perfor the A XOR B first: 1) the order of operations tells us so, and 2) we do not yet know the result of the NOTs operand.
We can now place NOT(A XOR B) on the truth table, and simply perform the NOT of the previous column.
Finally, we can perform the OR using its two operands on the chart, by simply glancing at the two columns. We get as the final evaluation:
This example brings up an interesting point. The two final operands are quite distant from each other on the truth table. Had we performed the first, AB, at the end, they would have been next to each other because AB requires no prior evaluation: its operands are inputs. Things like this make your truth table a little easier to follow, but do not make a significant increase of ease.
Construct a truth table for the Boolean expression ~(A+B)(BC) (that's NOT(A OR B) AND (B AND C) ). A basic skeleton is constructed for you, just fill in the letters and numbers in the appropriate boxes.
``` | | | A+B | | BC |
---+---+---+-----+--------+----+------------
| | | | | |
| | | | | |
| | | | | |
| | | | | |
| | | | | |
| | | | | |
| | | | | |
| | | | | |
```
DIGital
An Online Digital Circuitry Course<|endoftext|>
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Have Fun Learning English
British Culture, British Customs and British Traditions
With dictionary look up - Double click on any word for its definition.
This section is in advanced English and is only intended to be a guide, not to be taken too seriously!
The Summer Solstice is also known as Midsummer's Day. It is the time of the year when the distance of the Sun from the equator is at its greatest. Pagans believe it is a time to celebrate the achievement of man under the guidance and protection of 'Mother Earth' and 'Bel' (hence Beltane) resulting from their mutual veneration. In ancient Celtic and Wiccan beliefs such a time and event is symbolized by the 'Cauldron' and the 'Spear'. The druids name for this solstice was Alban Heruin (literally the light of the shore).
The main focus of the Solstice is Stonehenge, the most famous prehistoric site in Europe. Over 5,000 years old this prehistoric monolith is one of the great mysteries of Britain. How were these enormous stones transported from South Wales, erected and aligned to the movements of the heavens?
We have written an article on Stonehenge here.
If you want to read more about access to Stonehenge you can visit the English Heritage website here.
If you want to know more about the mystery of the stones at Stonehenge then visit the longest running Stonehenge website.<|endoftext|>
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noun of place
- (grammar) In Arabic and other Semitic languages, a noun formed from a root and expressing the sense of "place of X".
The Arabic noun of place is most often formed from the root of a verb and means “the place for doing x”, where x is the meaning of the verb. It is typically formed by placing the three consonants of a root in the pattern مَفْعَل (mafʿal): for instance, the tool noun from the root ك ت ب (k-t-b) is مَكْتَب (maktab, “desk”). In the case of مَكْتَب (maktab), the basic (Form I) verb from the root is كَتَبَ (kataba, “to write”), so مَكْتَب (maktab) literally means “place for writing”.<|endoftext|>
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# Rules how to round numbers to the nearest hundredth
In mathematics called rounding operation, which reduces the number of digits in the number using a replacement, given certain rules.If you are interested in the question of how to round numbers to the hundredths, then the first step is to deal with all the existing rules of rounding.There are several ways of how to round the number:
1. Statistical - used in the refinement of the number of residents.Speaking about the number of citizens called only an approximate value and not an exact figure.
2. Half - half going round to the nearest even number.
3. Round to a smaller number (rounded to zero) - this is the easiest round, in which discards all the "extra" digits.
4. Rounding up to a larger number - if the signs that want to round is not equal to zero, the number is rounded.This method of using providers or mobile operators.
5. nonzero round - the number is rounded according to the rules, but when the result should be 0, then rounding is done "from scratch."
6. Alternating rounding - when the N + 1 is equal to 5
, the number of turns in less rounded, in a big way.
For example, you need to round a number 21.837 to the hundredths.After rounding your correct answer should be 21.84.Here's why.The number 8 is included in the category of tenths, therefore the category of 3 hundredths, and 7000th.7 more than 5, so we increase the 3-ku 1, that is up to 4.It's quite easy if you know a few rules:
1. The latter retains the figure is increased by one in the event that the first cast by the front of her - more than 5. If this figure is equal to 5, and behind it there are stillany other digit, as previous increases by 1.
For example, we need to round up to the tenth: 54.69 = 54.7, or 7.357 = 7.4.
If you asked a question about how to round numbers to the nearest hundredth, act similarly to the above embodiment.
2. The latter retains the figure remains unchanged if the first of the cast, which stands in front of her less than 5.
Example: 96.71 = 96.7.
3. The last of the stored numbers remain unchanged, provided that it is even, and if the first of the cast - the number 5, and behind it there is no longer any figures.If the reserves figure - odd, it increases by 1.
Examples: 84.45 = 63.75 = 84.4 or 63.8.
Note.In many schools, students give a simplified version of the rules of rounding, so it pays to keep this in mind.In them all the numbers remain the same if they go after numbers from 0 to 4 and is increased by 1, provided that the following is a number from 5 to 9. competently solve problems with rounding according to strict rules, but if the school wound easy option,to avoid confusion is to stick to it.We hope you understand how to round numbers to the nearest hundredth.Rounding
in life is necessary for the convenience of working with numbers and directions of measurement accuracy.Currently definition appeared such as anti-rounding.For example, when counting of votes of any round of studies is considered bad manners.Stores also use anti-round to create the impression among customers more favorable prices (for example, write 199 instead of 200).We hope that the question of how to round numbers to the hundredths or tenths, now you can answer yourself.
Most Popular<|endoftext|>
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A man’s reproductive system is specifically designed to produce, store, and transport sperm. Unlike the female genitalia, the male reproductive organs are on both the interior and the exterior of the pelvic cavity. They include:
Sperm production occurs in the testicles. Upon reaching puberty, a man will produce millions of sperm cells every day, each measuring about 0.002 inches (0.05 millimeters) long.
There is a system of tiny tubes in the testicles. These tubes, called the seminiferous tubules, house the germ cells that hormones — including testosterone, the male sex hormone — cause to turn into sperm. The germ cells divide and change until they resemble tadpoles with a head and short tail.
The tails push the sperm into a tube behind the testes called the epididymis. For about five weeks, the sperm travel through the epididymis, completing their development. Once out of the epididymis, the sperm move to the vas deferens.
When a man is stimulated for sexual activity, the sperm are mixed with seminal fluid — a whitish liquid produced by the seminal vesicles and the prostate gland — to form semen. As a result of the stimulation, the semen, which contains up to 500 million sperm, is pushed out of the penis (ejaculated) through the urethra.
The process of going from a germ cell to a mature sperm cell capable of egg fertilization takes around 2.5 months.<|endoftext|>
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Note: This document is a work in progress. You can help improve it.
Block versus inline elements
There are two important categories of elements in HTML which you should know about. They are block-level elements and inline elements.
- Block-level elements form a visible block on a page — they will appear on a new line from whatever content went before it, and any content that goes after it will also appear on a new line. Block-level elements tend to be structural elements on the page that represent, for example, paragraphs, lists, navigation menus, footers, etc. A block-level element wouldn't be nested inside an inline element, but it might be nested inside another block-level element.
- Inline elements are those that are contained within block-level elements and surround only small parts of the document’s content, not entire paragraphs and groupings of content. An inline element will not cause a new line to appear in the document; they would normally appear inside a paragraph of text, for example an
<a>element (hyperlink) or emphasis elements such as
<em> is an inline element, so as you can see below, the first three elements sit on the same line as one another with no space in between. On the other hand,
<p> is a block-level element, so each element appears on a new line, with space above and below each (the spacing is due to default CSS styling that the browser applies to paragraphs).
<em>first</em><em>second</em><em>third</em> <p>fourth</p> <p>fifth</p> <p>sixth</p><|endoftext|>
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Terminologies in Vibration:
Let me just start with the basic terms before we get to the really tough ones.
- Amplitude: This is the term that specifies the movement of the body. Well just think how we measure movement. It can be
- Displacement – amount of movement from one point to another. E.g. I just walked 100 meters.
- Velocity – the rate of movement, E.g. I moved the 100 meters in 10 seconds
- Acceleration – The rate of change of velocity. E.g. The car has the capability to go from 0 mph to 100 mph in 8 Seconds.
These terms form the basis for the amplitude of vibration. Any measurement of vibration is normally denoted in the above three terms only. This also is incidentally the Y – Axis in any vibration related “graphs” as you call them. A small illustration below is given herewith to help you in understanding the terms better.
- Frequency: While this is a common term in English, denoting how often something occurs, the same thing applies in vibration too. This denotes how frequently something occurs. For example, the full moon appears once in a month, certain features appear at regular intervals or they are made to appear at regular intervals based on their relative motion. We do have a small mathematical formula for rotating members to know their frequency, it is
Frequency,(Hz) = speed in rpm/60
The Hz denotes Hertz, the unit for frequency.
These two terms are the primary ones that everyone needs to know. But we need to know more about two important terminologies before we move on to look at the other terms. Vibration is frequently related with time. To be more specific we can say that vibration is more of a time-dependent data. This is why we need to know more about the two important terms before we proceed to the other terms. They are:
Time Domain: To say in a very simple way we can think of this as a graph with Time in the X – Axis and Amplitude in the Y – Axis. You can assume the amplitude to be for example the amount of height a body jumps due to vibration. You can also keep the unit as "mm". A good example of the visualizing Time Domain data would be the Sine Wave. This is the form in which we visualize or look at the vibration data with respect to time. But this tends to give lesser information on vibration. Let us discuss more on this in my later articles.
Frequency or Spectrum Domain: This is another dimension to the Time Domain data you see. Again visualize this as a graph with Frequency in the X – Axis and Amplitude in the Y – Axis. Just see the adjoining figure to clearly understand what Time Domain and Frequency Domain are.
Understanding these two domains is the real foundation for anyone who needs to pursue a career in the field of Vibration Engineering. We will look more into other terminologies in the second chapter of this article.<|endoftext|>
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On this day in 1909, the United States government purchased its first military aircraft, designed by the Wright brothers and costing $30,000.
Earlier, in 1908, the U.S. Army Signal Corps requested bids for a two-seat observation aircraft. Orville Wright went to Fort Myer, Virginia, with the Wright airplane to demonstrate its capability. Midway through the trials, however, the Wright airplane malfunctioned and crashed, severely injuring Orville and killing his passenger, Lt. Thomas Selfridge—the first fatality in a powered airplane.
With a new airplane, the brothers returned to Fort Myer in 1909 and successfully completed their demonstration. On August 2, 1909, the Signal Corps officially accepted the Wright airplane. The 1909 Wright Military Flyer was used to train Army pilots.<|endoftext|>
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Courses
# Sample Solution Paper 1 - Term- 1 Math, Class 8 Class 8 Notes | EduRev
## Class 8 : Sample Solution Paper 1 - Term- 1 Math, Class 8 Class 8 Notes | EduRev
``` Page 1
CBSE VIII | Mathematics
Sample Paper 1 - Solution
CBSE Board
Class VIII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
2.5 is greater than 2.4 and less than 2.6 and 2.7.
Hence, on a number line, 2.5 will lie between 2.4 and 2.7.
= 5 + 5 + 3 = 13
1 is a square as well as a triangular number.
Cube root of (-8) x (-343) x (125) = (-2) x (-7) x 5 = 70
On cross multiplying, we get 4x + 28 = 3x - 3
4x - 3x = - 3 - 28
x = -31
x + 31 = 0
Page 2
CBSE VIII | Mathematics
Sample Paper 1 - Solution
CBSE Board
Class VIII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
2.5 is greater than 2.4 and less than 2.6 and 2.7.
Hence, on a number line, 2.5 will lie between 2.4 and 2.7.
= 5 + 5 + 3 = 13
1 is a square as well as a triangular number.
Cube root of (-8) x (-343) x (125) = (-2) x (-7) x 5 = 70
On cross multiplying, we get 4x + 28 = 3x - 3
4x - 3x = - 3 - 28
x = -31
x + 31 = 0
CBSE VIII | Mathematics
Sample Paper 1 - Solution
In a rhombus, if one angle is 70
o
, then its opposite angle will be 70
o
.
(Since, opposite angles of a rhombus are equal)
o o o
Value of component 1
360 360 90
Total va
Central
lue 4
angle ? ? ? ? ?
Square root = 2 × 3 × 3 × 5 = 90
Actual C.P. = C.P. + Overhead expenses = Rs. (900 + 200) = Rs. 1100
Profit = S.P. – C.P. = Rs. 1200 – Rs. 1100 = Rs. 100
100 1
Profit percent= 100 9 %
1100 11
??
Section B
13.
(by commutativity)
(by distributivity)
14. Let x be the greater part.
Then, 64 - x is the smaller part.
Then 3x = 5(64 - x)
3x = 320 - 5x
3x + 5x = 320
8x = 320
x = 40
So, the two parts are 40 and 24.
Page 3
CBSE VIII | Mathematics
Sample Paper 1 - Solution
CBSE Board
Class VIII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
2.5 is greater than 2.4 and less than 2.6 and 2.7.
Hence, on a number line, 2.5 will lie between 2.4 and 2.7.
= 5 + 5 + 3 = 13
1 is a square as well as a triangular number.
Cube root of (-8) x (-343) x (125) = (-2) x (-7) x 5 = 70
On cross multiplying, we get 4x + 28 = 3x - 3
4x - 3x = - 3 - 28
x = -31
x + 31 = 0
CBSE VIII | Mathematics
Sample Paper 1 - Solution
In a rhombus, if one angle is 70
o
, then its opposite angle will be 70
o
.
(Since, opposite angles of a rhombus are equal)
o o o
Value of component 1
360 360 90
Total va
Central
lue 4
angle ? ? ? ? ?
Square root = 2 × 3 × 3 × 5 = 90
Actual C.P. = C.P. + Overhead expenses = Rs. (900 + 200) = Rs. 1100
Profit = S.P. – C.P. = Rs. 1200 – Rs. 1100 = Rs. 100
100 1
Profit percent= 100 9 %
1100 11
??
Section B
13.
(by commutativity)
(by distributivity)
14. Let x be the greater part.
Then, 64 - x is the smaller part.
Then 3x = 5(64 - x)
3x = 320 - 5x
3x + 5x = 320
8x = 320
x = 40
So, the two parts are 40 and 24.
CBSE VIII | Mathematics
Sample Paper 1 - Solution
15. Opposite sides are equal in a parallelogram
Therefore, 3y + 1 = 19
Or, 3y = 19 - 1
Or, 3y = 18
Or, y = 6
Also, 4x+3 = 23
Or, 4x = 23-3
Or, 4x = 20
Or, x = 5
Hence x = 5 and y = 6.
16. Let the number of chairs in each row be x.
Then, the number of rows = x.
Total number of chairs in the auditorium
? ?
2
x x x ? ? ?
But the number of chairs that the auditorium can accommodate = 1764 (given)
? ?
??
? ? ? ? ? ?
? ? ? ?
?
2
x 1764
2 2 3 3 7 7
x 2 3 7
42
Hence, the number of chairs in each row is 42.
17. Here,
Total number of outcomes = 10 + 25 = 35
Let E be the event of getting a prize.
Number of outcomes favourable to event E = 10
? ?
Number of favourable outcomes 10 2
Total number of outcome
P E
s7
35
? ? ?
2
Probability of getting a prize
7
?
Page 4
CBSE VIII | Mathematics
Sample Paper 1 - Solution
CBSE Board
Class VIII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
2.5 is greater than 2.4 and less than 2.6 and 2.7.
Hence, on a number line, 2.5 will lie between 2.4 and 2.7.
= 5 + 5 + 3 = 13
1 is a square as well as a triangular number.
Cube root of (-8) x (-343) x (125) = (-2) x (-7) x 5 = 70
On cross multiplying, we get 4x + 28 = 3x - 3
4x - 3x = - 3 - 28
x = -31
x + 31 = 0
CBSE VIII | Mathematics
Sample Paper 1 - Solution
In a rhombus, if one angle is 70
o
, then its opposite angle will be 70
o
.
(Since, opposite angles of a rhombus are equal)
o o o
Value of component 1
360 360 90
Total va
Central
lue 4
angle ? ? ? ? ?
Square root = 2 × 3 × 3 × 5 = 90
Actual C.P. = C.P. + Overhead expenses = Rs. (900 + 200) = Rs. 1100
Profit = S.P. – C.P. = Rs. 1200 – Rs. 1100 = Rs. 100
100 1
Profit percent= 100 9 %
1100 11
??
Section B
13.
(by commutativity)
(by distributivity)
14. Let x be the greater part.
Then, 64 - x is the smaller part.
Then 3x = 5(64 - x)
3x = 320 - 5x
3x + 5x = 320
8x = 320
x = 40
So, the two parts are 40 and 24.
CBSE VIII | Mathematics
Sample Paper 1 - Solution
15. Opposite sides are equal in a parallelogram
Therefore, 3y + 1 = 19
Or, 3y = 19 - 1
Or, 3y = 18
Or, y = 6
Also, 4x+3 = 23
Or, 4x = 23-3
Or, 4x = 20
Or, x = 5
Hence x = 5 and y = 6.
16. Let the number of chairs in each row be x.
Then, the number of rows = x.
Total number of chairs in the auditorium
? ?
2
x x x ? ? ?
But the number of chairs that the auditorium can accommodate = 1764 (given)
? ?
??
? ? ? ? ? ?
? ? ? ?
?
2
x 1764
2 2 3 3 7 7
x 2 3 7
42
Hence, the number of chairs in each row is 42.
17. Here,
Total number of outcomes = 10 + 25 = 35
Let E be the event of getting a prize.
Number of outcomes favourable to event E = 10
? ?
Number of favourable outcomes 10 2
Total number of outcome
P E
s7
35
? ? ?
2
Probability of getting a prize
7
?
CBSE VIII | Mathematics
Sample Paper 1 - Solution
18. Since ABCD is an isosceles trapezoidal, we have AD = BC
Therefore, AD = BC = 4 cm.
Now the perimeter of given trapezium
= AB + BC + CD + DA
= 12 + 4 + 8 + 4
= 28
Hence, the perimeter of the given trapezium is 28 cm.
19. Total number of votes = (1136 + 7636 + 11628) = 20400.
11628
Therefore, required percentage 100 % 57%
20400
??
? ? ?
??
??
20. Since, the measure of each angle of rectangle is 90
o
.
Therefore, ?ADC = y = 90
o
o
This gives, x + 45
o
= 90
o
x = 90
o
- 45
o
= 45
o
Hence, x = 45
o
and y = 90
o
.
21. It can be observed that the squares of the given numbers can be found by first
writing the counting number up to the number of 1's and then writing the reverse
counting till 1.
Thus, 1111
2
= 1234321
11111
2
= 123454321
22. By cross multiplication, we get,
3(3x-1) = 4(2x+5)
9x-3 = 8x +20
9x-8x-3 = 20 (transposing 8x to LHS)
9x-8x = 20+3 (transposing -3 to RHS)
x = 23
23.
Page 5
CBSE VIII | Mathematics
Sample Paper 1 - Solution
CBSE Board
Class VIII Mathematics
Term I
Sample Paper 1 - Solution
Time: 2 ½ hours Total Marks: 80
Section A
2.5 is greater than 2.4 and less than 2.6 and 2.7.
Hence, on a number line, 2.5 will lie between 2.4 and 2.7.
= 5 + 5 + 3 = 13
1 is a square as well as a triangular number.
Cube root of (-8) x (-343) x (125) = (-2) x (-7) x 5 = 70
On cross multiplying, we get 4x + 28 = 3x - 3
4x - 3x = - 3 - 28
x = -31
x + 31 = 0
CBSE VIII | Mathematics
Sample Paper 1 - Solution
In a rhombus, if one angle is 70
o
, then its opposite angle will be 70
o
.
(Since, opposite angles of a rhombus are equal)
o o o
Value of component 1
360 360 90
Total va
Central
lue 4
angle ? ? ? ? ?
Square root = 2 × 3 × 3 × 5 = 90
Actual C.P. = C.P. + Overhead expenses = Rs. (900 + 200) = Rs. 1100
Profit = S.P. – C.P. = Rs. 1200 – Rs. 1100 = Rs. 100
100 1
Profit percent= 100 9 %
1100 11
??
Section B
13.
(by commutativity)
(by distributivity)
14. Let x be the greater part.
Then, 64 - x is the smaller part.
Then 3x = 5(64 - x)
3x = 320 - 5x
3x + 5x = 320
8x = 320
x = 40
So, the two parts are 40 and 24.
CBSE VIII | Mathematics
Sample Paper 1 - Solution
15. Opposite sides are equal in a parallelogram
Therefore, 3y + 1 = 19
Or, 3y = 19 - 1
Or, 3y = 18
Or, y = 6
Also, 4x+3 = 23
Or, 4x = 23-3
Or, 4x = 20
Or, x = 5
Hence x = 5 and y = 6.
16. Let the number of chairs in each row be x.
Then, the number of rows = x.
Total number of chairs in the auditorium
? ?
2
x x x ? ? ?
But the number of chairs that the auditorium can accommodate = 1764 (given)
? ?
??
? ? ? ? ? ?
? ? ? ?
?
2
x 1764
2 2 3 3 7 7
x 2 3 7
42
Hence, the number of chairs in each row is 42.
17. Here,
Total number of outcomes = 10 + 25 = 35
Let E be the event of getting a prize.
Number of outcomes favourable to event E = 10
? ?
Number of favourable outcomes 10 2
Total number of outcome
P E
s7
35
? ? ?
2
Probability of getting a prize
7
?
CBSE VIII | Mathematics
Sample Paper 1 - Solution
18. Since ABCD is an isosceles trapezoidal, we have AD = BC
Therefore, AD = BC = 4 cm.
Now the perimeter of given trapezium
= AB + BC + CD + DA
= 12 + 4 + 8 + 4
= 28
Hence, the perimeter of the given trapezium is 28 cm.
19. Total number of votes = (1136 + 7636 + 11628) = 20400.
11628
Therefore, required percentage 100 % 57%
20400
??
? ? ?
??
??
20. Since, the measure of each angle of rectangle is 90
o
.
Therefore, ?ADC = y = 90
o
o
This gives, x + 45
o
= 90
o
x = 90
o
- 45
o
= 45
o
Hence, x = 45
o
and y = 90
o
.
21. It can be observed that the squares of the given numbers can be found by first
writing the counting number up to the number of 1's and then writing the reverse
counting till 1.
Thus, 1111
2
= 1234321
11111
2
= 123454321
22. By cross multiplication, we get,
3(3x-1) = 4(2x+5)
9x-3 = 8x +20
9x-8x-3 = 20 (transposing 8x to LHS)
9x-8x = 20+3 (transposing -3 to RHS)
x = 23
23.
CBSE VIII | Mathematics
Sample Paper 1 - Solution
24.
Section C
25. Represent the months on the x-axis and the number of visitors on the y-axis.
The double bar graph is as follows:
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
## Mathematics (Maths) Class 8
144 videos|328 docs|48 tests
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;<|endoftext|>
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# Functional Relationships
## functional_relationship.jpg
The equation for the natural frequency of a spring-mas system is deduced using the Rayleigh method. This method seeks to find exponents that make the basic dimensions match.
By User:Lzyvzl at en.wikipedia [Public domain], from Wikimedia Commons
The form of equations can often be deduced from an analysis of the dimensions of the variables or parameters that affect the solution to a problem. The process can best be illustrated by an example.
The form of equations can often be deduced from an analysis of the dimensions of the variables or parameters that affect the solution to a problem. The process can best be illustrated by an example.
The natural frequency of a spring-mass system is observed to be related to the stiffness, k, of the spring and the mass, m. What is the functional relationship between the spring stiffness and mass?
In equation form, the frequency is some function , F, of k and m
A functional relationship that uses exponents for each variable can be written as
where the coefficient C and exponents a and b are unknowns. The use of this functional relationship is known as Rayleigh’s Method. The exponents a and b can be determined through dimensional analysis while the coefficient C must be determined through experiment or some other method.
The basic dimensions of the various terms are substituted into the equation
The powers of each basic dimension must be equal for the equation to be dimensionally consistent. Therefore,
Solving the two equations, gives a = ½ and b=-1/2. The functional relationship can thus be written as
The true relationship can be determined analytically to be
A comparison shows that the coefficient C is equal to 1/2π. The correct functional relationship was determined purely from making sure that the dimensions were correct.
The number of basic dimensions and number of exponents must be equal for this method to work. In the example, there are two basic dimension, M and t, and two exponents, a and b. This results in two equations with two unknowns, which can be solved to find the unknown exponents.
## Learning Objectives:
Learn how to determine the functional relationship between parameters using Rayleigh’s method.
Learn how to determine when the Rayleigh method will or will not work.<|endoftext|>
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# Factoring Cubic Polynomials – Steps, Definition With Examples
Welcome, Brighterly learners! It’s time to embark on another fascinating mathematical adventure. Today, our journey is going to take us deep into the world of algebra, where we’ll unravel the mystery behind cubic polynomials.
Now, you might be wondering, “What are cubic polynomials?” or “Why do we even need to learn about them?” Here at Brighterly, we understand these questions, and we’re committed to making sure you not only get the answers but also enjoy the process of learning. Because we believe that every new concept you learn is a key to unlock a new door in the vast mansion of mathematics!
Factoring cubic polynomials may sound a bit complex, but we promise you, it’s just like solving an engaging puzzle. And we all love puzzles, don’t we? They challenge us, make us think, and when we finally solve them, they give us an incredible sense of achievement. Learning about cubic polynomials and how to factor them is no different!
## What Is Factoring Cubic Polynomials?
Factoring cubic polynomials is an essential concept in algebra that serves as a gateway to more complex mathematical ideas. It might sound intimidating, but don’t worry! We’ll break it down into bite-sized chunks to make it fun and digestible.
At its heart, factoring cubic polynomials is the process of breaking down a complex polynomial (a mathematical expression involving many terms) into simpler factors that, when multiplied together, give the original polynomial. It’s a lot like breaking down a chocolate bar into individual squares or breaking down a sentence into individual words. Factoring makes the polynomial easier to understand and work with.
## Definition of Polynomials
In simple terms, a polynomial is an algebraic expression made up of ‘terms’ that are separated by ‘+’ or ‘-‘ signs. These terms consist of variables (like x or y) and coefficients (the numbers in front of the variables) raised to a non-negative integer exponent. For example, 2x² – 5x + 3 is a polynomial with three terms.
## Definition of Cubic Polynomials
A cubic polynomial is a specific type of polynomial with a degree of three. The degree of a polynomial refers to the highest exponent in the polynomial. So, a cubic polynomial will always have one term where the variable is raised to the power of three. For example, x³ – 4x² + 3x – 2 is a cubic polynomial.
## Properties of Polynomials and Cubic Polynomials
### Properties of Polynomials
Polynomials have several exciting properties. For instance, they’re closed under addition, subtraction, and multiplication, which means when you add, subtract, or multiply two polynomials, you always get another polynomial. Also, the degree of a polynomial gives us a lot of information about its shape and the number of solutions it has.
### Properties of Cubic Polynomials
Cubic polynomials, being a type of polynomial, share these properties, but also have a few unique ones. Most notably, a cubic polynomial will always have at least one real root (solution), and it can have up to three. Also, the graph of a cubic polynomial is always a continuous curve without any breaks or sharp turns.
## Difference Between Polynomials and Cubic Polynomials
The primary difference between polynomials and cubic polynomials lies in their degrees. While a polynomial can have any non-negative integer degree, a cubic polynomial always has a degree of three. This difference affects their properties, such as the number of solutions they can have and the shape of their graphs.
## Steps to Factor Cubic Polynomials
### Writing Steps for Factoring Polynomials
Factoring polynomials generally involves identifying common factors among the terms, grouping similar terms, and using different factoring techniques such as difference of squares or the distributive property.
### Writing Steps for Factoring Cubic Polynomials
Factoring cubic polynomials can be a bit trickier. You’ll often need to use a method called the factor theorem or synthetic division to identify one factor first and then break down the remaining quadratic polynomial. But don’t worry! We’ll be delving into these steps in detail in future articles.
## Practice Problems on Factoring Cubic Polynomials
Sharpen your skills with these practice problems. Remember, the more you practice, the more comfortable you’ll become with these concepts!
1. Factor the cubic polynomial x³ – 6x² + 11x – 6.
2. Factor the cubic polynomial 2x³ – 9x² + 12x – 4.
3. Factor the cubic polynomial 3x³ – x² – 4x + 4.
## Conclusion
We hope this journey into the world of cubic polynomials has been enlightening and fun-filled! At Brighterly, our aim is not just to help you solve problems but to foster your love and curiosity for mathematics.
We’ve seen what cubic polynomials are, explored their unique properties, and learned how to factor them. It might have been a bit challenging, but remember, challenges are what make learning exciting and meaningful. They are the stepping stones towards growth and mastery.
Factoring cubic polynomials is a powerful tool in your mathematical toolkit. As you continue your adventures in mathematics, you’ll encounter this concept again and again. But worry not, because with practice and the Brighterly spirit of curiosity and persistence, you’ll be factoring cubic polynomials like a pro in no time!
As always, remember, math is not a subject to be feared, but a fascinating world to be explored. So keep exploring, keep learning, and let Brighterly guide you on your mathematical journey. Until our next adventure, happy learning!
## Frequently Asked Questions on Factoring Cubic Polynomials
### What is a cubic polynomial?
A cubic polynomial is an algebraic expression with a degree of three. This means it will always have one term where the variable (like x or y) is raised to the power of three. For example, an expression like x³ – 4x² + 3x – 2 is a cubic polynomial.
### How do you factor cubic polynomials?
Factoring cubic polynomials usually involves a multi-step process that might include using the factor theorem or synthetic division. The goal is to break down the cubic polynomial into simpler factors. We’ll detail this process in future posts, but if you’re eager to start practicing, check out resources like Khan Academy or Math is Fun for more immediate guidance.
### What’s the difference between polynomials and cubic polynomials?
While a polynomial is an algebraic expression that can have any non-negative integer degree, a cubic polynomial is a specific type of polynomial that always has a degree of three. This means a cubic polynomial will always have a term with the variable raised to the power of three.
Information Sources
Math Catch Up Program
• Learn Math Simple - Fast - Effective
• Overcome math obstacles and reach new heights with Brighterly.
Simple - Fast - Effective
Overcome math obstacles and reach new heights with Brighterly.<|endoftext|>
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Mixed Age Year 4 and 5 Decimals Step 17 Resource Pack – Classroom Secrets | Classroom Secrets
All › Mixed Age Year 4 and 5 Decimals Step 17 Resource Pack
# Mixed Age Year 4 and 5 Decimals Step 17 Resource Pack
## Step 17: Mixed Age Year 4 and 5 Decimals Resource Pack
Mixed Age Year 4 and 5 Decimals Step 17 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 4 Halves and Quarters for Spring Block 4.
### What's included in the Pack?
This Mixed Age Year 4 and 5 Decimals Step 17 pack includes:
• Mixed Age Year 4 and 5 Decimals Step 17 Teaching PowerPoint with examples.
• Year 5 Subtract Mixed Numbers 2 Varied Fluency with answers.
• Year 5 Subtract Mixed Numbers 2 Reasoning and Problem Solving with answers.
#### National Curriculum Objectives
Mathematics Year 4: (4F6a) Recognise and write decimal equivalents to 1/4 , 1/2 , 3/4
Differentiation for Year 5 Subtract Mixed Numbers 2:
Varied Fluency
Developing Questions to support writing half, quarter and three quarters as decimals.
Expected Questions to support writing fractions equivalent to half, quarter and three quarters as decimals.
Greater Depth Questions to support writing fractions equivalent to half, quarter and three quarters as decimals. Multiple answers possible.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Problem Solving)
Developing Find the odd one out where two pairs of simplified fractions and decimals are given with one odd answer.
Expected Find the odd one out where two pairs of simplified or equivalent fractions and decimals are given with one odd answer.
Greater Depth Find the odd one out where three pairs of equivalent fractions and decimals are given with one odd answer.
Questions 2, 5 and 8 (Reasoning)
Developing Explain who travels the furthest using simplified fractions and decimals. Two statements to compare.
Expected Explain who travels the furthest using equivalent fractions and decimals. Two statements to compare.
Greater Depth Explain who travels the furthest using equivalent fractions and decimals. Three statements to compare.
Questions 3, 6 and 9 (Problem Solving)
Developing Use the digit clues to find the missing decimal or simplified fractions.
Expected Use the digit clues to find the missing decimal or equivalent fractions.
Greater Depth Use the digit clues to find the missing decimal or equivalent fractions. There may be multiple answers.
This resource is available to download with a Premium subscription.<|endoftext|>
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# Density
All your Questions answered in one place
In this chapter we’re going to be exploring. You shall understand arrangement and permutation problems, calculating permutations when there is a repeated object. Ordering r objects from n. You should have some prior knowledge of basic probability.
## What is Density?
Density is the measurement or comparison of how heavy different types of materials are. Instead of measuring how heavy a material is, we measure how heavy a cube of each material would be. Consider this; What weighs more, a tonne of gold or a tonne of feathers. You would think that a tonne of gold weighs more but you would be wrong. Both materials weigh the same 1 tonne. For example 10 cm3 of gold weighs 79.2 grams. What would 1 cm3 of gold weigh? 1 cm3 of gold would weigh 7.92 grams. That means gold has 7.92 grams for every centimetre cube or 7.92g per cm3 We say that it has a density of 7.92 g/cm3
## Difference between weight and mass
The two terms weight and mass are similar, they describe how heavy an object is. Although they are similar they are not the same. Mass is the measure of how much substance there is in an object, and weight is the measure of how heavy an object is. Weight depends on gravity and would vary depending on where you’re in the universe. An object that feels very heavy on planet earth will weigh differently on the moon or Mars. Meaning that it might be impossible to lift a car on earth but it will be easy on the moon. Mass is different in that, it stays the same no matter what planet or where you’re in the universe.
## Finding Density
454 grams of cheese make up a block of 4cm by 6cm by 10cm. What is the density of cheese in g/cm3. The information that we require here is grams and cm3. We know that cm3 refers to volume. We need to find volume first. Now we know that 454g is in the 240cm3 cubes. The density is the number of grams per cm3, so the density of the cheese is; Some times we know how dense different materials are per cm3, for example the density of Aluminium is 2.79g/cm3 and the density of diamond is 3.52g/cm3. We can use these values to find the mass when the volume is more than one or for different volumes. In some cases we have to find the volume first. ;
//Comments In this chapter we’re going to be exploring. You shall understand arrangement and permutation problems, calculating permutations when there is a repeated object. Ordering r objects from n. You should have some prior knowledge of basic probability.
## What is Density?
Density is the measurement or comparison of how heavy different types of materials are. Instead of measuring how heavy a material is, we measure how heavy a cube of each material would be. Consider this; What weighs more, a tonne of gold or a tonne of feathers. You would think that a tonne of gold weighs more but you would be wrong. Both materials weigh the same 1 tonne. For example 10 cm3 of gold weighs 79.2 grams. What would 1 cm3 of gold weigh? 1 cm3 of gold would weigh 7.92 grams. That means gold has 7.92 grams for every centimetre cube or 7.92g per cm3 We say that it has a density of 7.92 g/cm3
## Difference between weight and mass
The two terms weight and mass are similar, they describe how heavy an object is. Although they are similar they are not the same. Mass is the measure of how much substance there is in an object, and weight is the measure of how heavy an object is. Weight depends on gravity and would vary depending on where you’re in the universe. An object that feels very heavy on planet earth will weigh differently on the moon or Mars. Meaning that it might be impossible to lift a car on earth but it will be easy on the moon. Mass is different in that, it stays the same no matter what planet or where you’re in the universe.
## Finding Density
454 grams of cheese make up a block of 4cm by 6cm by 10cm. What is the density of cheese in g/cm3. The information that we require here is grams and cm3. We know that cm3 refers to volume. We need to find volume first. Now we know that 454g is in the 240cm3 cubes. The density is the number of grams per cm3, so the density of the cheese is; Some times we know how dense different materials are per cm3, for example the density of Aluminium is 2.79g/cm3 and the density of diamond is 3.52g/cm3. We can use these values to find the mass when the volume is more than one or for different volumes. In some cases we have to find the volume first. ;<|endoftext|>
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## Word Problems Involving Multiplicative Comparisons
written by: Donna Ventura • edited by: Tricia Goss • updated: 3/30/2014
Students will use multiplication or division to solve word problems involving multiplicative comparisons.
• slide 1 of 5
### Lesson Objective
The lesson is aligned to the Common Core State Standards for Mathematics – 4.OA.2 Operations and Algebraic Thinking - Multiply or divide to solve word problems involving multiplicative comparison.
• slide 2 of 5
Calculator
• slide 3 of 5
### Using Multiplication to Solve Word Problems Involving Multiplicative Comparisons
Part A
Look at the word problem below.
Jacob’s dog weighs 3 times as much as Aaron’s dog. Aaron’s dog weighs 14 pounds. How much does Jacob’s dog weigh?
This word problem can be represented by a multiplication equation. The weight of Jacob’s dog is unknown. Let w represent the weight of Jacob’s dog.
w = 3 x 14
w = 42
Jacob’s dog weighs 42 pounds.
Part B
Look at the word problems below. Write a multiplication equation for each word problem. Solve the equations and label your answers.
1. Ashley has a kitten that weighs 2 times as much as Carly’s kitten. Carly’s kitten weighs 4 pounds. How much does Ashley’s kitten weigh?
2. Jasmine has 2 times as many stickers as Chloe. Chloe has 12 stickers. How many stickers does Jasmine have?
1. Let k = the weight of Ashley’s kitten. k = 2 x 4, k = 8, Ashley’s kitten weighs 8 pounds.
2. Let s = the number of stickers Jasmine has. s = 2 x 12, s = 24, Jasmine has 24 stickers.
• slide 4 of 5
### Using Division to Solve Word Problems involving Multiplicative Comparisons
Part A
Look at the word problem below.
Darren’s dog weighs 36 pounds. Darren’s dog weighs 3 times as much as Corey’s dog. How much does Corey’s dog weigh?
This word problem can be represented by a division equation. The weight of Corey’s dog is unknown. Let c represent the weight of Corey’s dog.
c = 36 / 3
c = 12
Cory’s dog weighs 12 pounds.
Part B
Look at the word problems below. Write a multiplication equation for each word problem. Solve the equations and label your answers.
1. Andy’s kitten weighs 9 pounds. Andy’s kitten weighs 3 times as much as Dave’s kitten. How much does Dave’s kitten weigh?
2. James has 51 stickers. James has 3 times as many stickers as Chuck. How many stickers does Chuck have?
1. Let d = the weight of Dave’s kitten. d = 9/3, d = 3, Dave’s kitten weighs 3 pounds.
2. Let c = the number of stickers Chuck has. c = 51/3, c = 17, Chuck has 17 stickers.
• slide 5 of 5
### Individual or Group Work
Use multiplication or division to solve the word problems about a video game. Write the equation for each word problem. Solve the equations and label your answers.
1. Ken’s score on a video game is 540 points. Ken’s score is 4 times as many points as Alex’s score. How many points is Alex’s score?
2. Tanya scored 3 times as many points as Stacy on a video game. Stacy scored 57 points. How many points did Tanya score?
3. Peter scored 4 times as many points as Bob on a video game. Bob scored 84 points. How many points did Peter score?
4. Ben’s score on a video game is 990 points. Ben’s score is 3 times as many points as John’s score. How many points is John’s score?
5. Emily scored 2 times as many points as Sarah on a video game. Sarah scored 63 points. How many points did Emily score?<|endoftext|>
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In honor of Youth Sports Safety Month, let’s visit a topic that has many parents and coaches concerned when it comes to young athletes: concussions.
A concussion is a traumatic brain injury (TBI) in which an impact to the body – usually a blow to the head or any hit resulting in whiplash – causes the brain to move within the skull.
When it comes to high school players, the risk is even more serious. According to the Sports Concussion Institute,
Recent research demonstrates that high school athletes not only take longer to recover after a concussion when compared to collegiate or professional athletes, but they also may experience greater severity of symptoms and more neurological disturbances as measured by neuropsychological and postural stability tests… Because the frontal lobes of the human brain continue to develop until age 25, it is vital to manage youth concussions very conservatively to ensure optimal neurological development and outcomes.
With the severity of concussion after-effects becoming more apparent down the line for athletes, waiting to return to play is vital. According to the Centers for Disease Control (CDC), “Those who survive a TBI can face effects lasting a few days to disabilities which may last the rest of their lives. Effects of TBI can include impaired thinking or memory, movement, sensation (e.g., vision or hearing), or emotional functioning (e.g., personality changes, depression).”
Headache and dizziness are the most common symptoms of a concussion, but almost half of people experience no symptoms and up to 90 percent of concussions occur when there is not loss of consciousness. Of those that do experience symptoms, it can take up to two weeks to recover.
Only neurological testing can determine if a concussion has occurred and when it’s safe to get back into the game. If you suspect a player has sustained a concussion, discontinue play and consult with a doctor immediately.
More concussion stats*:
- ER visits for sports-related concussions in high school athletes have risen about 60% between 2001 and 2009.
- Estimated 47% of athletes do not report feeling any symptoms after a concussive blow
- 5-10% of athletes will experience a concussion in any given sport season
- Football is the most common sport with concussion risk for males (75% chance for concussion)
- Soccer is the most common sport with concussion risk for females (50% chance for concussion)
- 78% of concussions occur during games
- Concussion sufferers can take up to 2 weeks to recover
*According to the CDC and the Sports Concussion Institute.<|endoftext|>
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The resource has been added to your collection
In this brief pre-lecture quiz, students are asked to consider concepts that they have learned in previous lectures, and to answer the questions in a short period of time; hopefully without the aid of lecture notes. The quiz allows students to settle into the lecture, arrive late (many are coming from work), and refresh their geochemical knowledge from previous weeks. In this worksheet students balance chemical equations, work out numbers of moles and masses of compounds, and determine delta H and delta G for a common Earth-surface reaction. This is appropriate for first-year Geology/Earth Sciences students.
This resource has not yet been reviewed.
Not Rated Yet.<|endoftext|>
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# Multiples of 19
Created by: Team Maths - Examples.com, Last Updated: May 23, 2024
## Multiples of 19
Multiples of 19 are numbers that can be obtained by multiplying 19 with any integer. These numbers are part of the broader category of integers and follow the basic principles of multiplication. In mathematics, multiples are closely related to factors and divisors, as they represent the product of a given number and an integer. Understanding multiples of 19 helps in various mathematical operations, including finding common multiples and solving problems related to divisibility.
## What are Multiples of 19?
Multiples of 19 are numbers that result from multiplying 19 by any integer (e.g., 19, 38, 57, 76, etc.). These multiples are part of the set of integers and follow the principle of multiplication. They are used in various mathematical contexts, such as finding common multiples and understanding divisibility.
Prime Factorization of 19: 19 = 1 × 19 First 10 multiples of 19: 19, 38, 57, 76, 95, 114, 133, 152, 171, 190.
## For example, 38, 76, 114 and 171 are all multiples of 19, 183 is not a multiple of 19 for the following reasons:
! Here’s the updated table including the remainder:
Table of 19
## Important Notes
• Definition: Multiples of 19 are the products obtained by multiplying 19 with any integer.
• Sequence: The sequence of multiples of 19 starts as 19, 38, 57, 76, 95, 114, and continues infinitely.
• Formula: A multiple of 19 can be represented as 19×n, where n is any integer.
• Common Multiples: Multiples of 19 are used to find common multiples with other numbers, aiding in solving least common multiple (LCM) problems.
• Divisibility: A number is a multiple of 19 if it is divisible by 19 without leaving any remainder.
• Applications: Understanding multiples of 19 is essential in arithmetic operations, algebra, and number theory.
• Examples: Some examples of multiples of 19 include 38 (19×2), 95 (19×5), and 152 (19×8).
## Examples on Multiples of 19
19: Calculation: 19×1=19
38: Calculation: 19×2=38
57: Calculation: 19×3=57
76: Calculation: 19×4=76
95: Calculation: 19×5 = 95
114: Calculation: 19×6 = 114
133: Calculation: 19×7 = 133
152: Calculation: 19×8 = 152
171: Calculation: 19×9 = 171
190: Calculation: 19×10 = 19
### Example: 200
• Calculation: 200÷19 = 10 remainder 10
• Reason: 200 is not a multiple of 19 because 200÷19 does not result in an integer (remainder 10).
### Example: 217
• Calculation: 217÷19 = 11 remainder 8
• Reason: 217 is not a multiple of 19 because 217÷19 does not result in an integer (remainder 8).
## What is a multiple of 19?
A multiple of 19 is any number that can be expressed as 19×n, where n is an integer. Examples include 19, 38, 57, 76, and so on.
## How do you find the 5th multiple of 19?
To find the 5th multiple of 19, multiply 19 by 5: 19×5 = 95
So, the 5th multiple of 19 is 95.
## Is 114 a multiple of 19?
No, 114 is not a multiple of 19. When you divide 114 by 19, the result is not an integer:
114÷19 = 6 (remainder 0, not an exact multiple)
## What are the first five multiples of 19?
The first five multiples of 19 are 19, 38, 57, 76, and 95.
## How can you verify if a number is a multiple of 19?
To verify if a number is a multiple of 19, divide the number by 19. If the result is an integer with no remainder, it is a multiple. For example, 95 divided by 19 equals 5, so 95 is a multiple of 19.
## Is 190 a multiple of 19?
Yes, 190 is a multiple of 19 because: 19×10 = 190
So, the 10th multiple of 19 is 190.
## How are multiples of 19 used in real-life scenarios?
Multiples of 19 can be used in scenarios like scheduling, where events might occur every 19 days, or in financial contexts, where payments might be structured around multiples of 19 units.
## Is 209 a multiple of 19?
No, 209 is not a multiple of 19. When you divide 209 by 19, the result is not an integer: 209÷19 = 11
## What is the relationship between multiples of 19 and factors of 19?
Multiples of 19 are numbers that can be divided by 19 without a remainder, while factors of 19 are numbers that can multiply together to give 19. Since 19 is a prime number, its only factors are 1 and 19 itself.
## What is the 10th multiple of 19?
The 10th multiple of 19 is found by multiplying 19 by 10: 19×10 = 190
So, the 10th multiple of 19 is 190.
Text prompt<|endoftext|>
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Tularemia is caused by the bacteria Francisella tularensis and is typically found in animals, especially rodents, rabbits, and hares.
People become infected with tularemia through the bite of infected insects, most commonly ticks and deer flies, or through skin contact with infected animal tissue. The bacteria can also be inhaled when infected animal tissue is broken up into small particles and spread in the air, such as when an infected carcass is mowed over.
Symptoms of Tularemia Include
- Skin ulcers
- Swollen and painful lymph glands
- Inflamed eyes
- Sore throat
- Mouth sores
Symptoms can also include
- Abrupt onset of fever
- Muscle aches
- Joint pain
- Dry cough
- Difficulty breathing
- Bloody sputum
- Respiratory failure
Tularemia is treatable when detected in early stages.
Public health officials recommend the following precautions:
- Stay out of areas inhabited by wild rodents. If you must enter areas frequented by wild rodents, always wear insect repellent that is effective against ticks, biting flies, and mosquitoes and contains DEET or oil of lemon eucalyptus.
- Do not go barefoot in an area where rabbits have died. The tularemia bacteria can persist in the environment for several months after it is detected.
- Consider wearing a dust mask when mowing or blowing vegetation in areas where animal die-offs have occurred.
- Prevent your pets from hunting or eating wild rodents or rabbits. Infected pets, such as cats, may in turn transmit the disease to people.
- Avoid all contact with wild rodents, including squirrels and rabbits; do not feed or handle them.
- Avoid ticks. The best protection for pets, especially cats, is to keep them indoors. If outdoors with pets, keep them out of heavily wooded areas, which are ideal habitats for ticks.
- Never touch sick or dead animals with your bare hands. If an animal must be moved, place it in a garbage bag using a long-handled shovel, and place the bag in an outdoor garbage can.
- Avoid drinking unpurified water from streams or lakes; keep your pets from doing the same.
- Don’t mow over animal carcasses, and consider using a dust mask when doing landscape work.
- See a health care provider if you become ill with a high fever and/or swollen lymph nodes. Tularemia is a treatable illness when diagnosed early.
- Contact a veterinarian if your pet becomes ill with a high fever and/or swollen lymph nodes.
In the United States, human cases of tularemia have been reported from every state except Hawaii, with the majority occurring in south-central and western states.
- Tularemia (Centers for Disease Control & Prevention)
- Ticks (Centers for Disease Control & Prevention)
- Boulder County Health Information Line at 303-441-1460
To report an animal die-off in Boulder County, call 303.441.1564.<|endoftext|>
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A hyperlink is a word, phrase, or image that you can click on to jump to a new document or a new section within the current document. Hyperlinks are found in nearly all Web pages, allowing users to click their way from page to page. Text hyperlinks are often blue and underlined, but don't have to be. When you move the cursor over a hyperlink, whether it is text or an image, the arrow should change to a small hand pointing at the link. When you click it, a new page or place in the current page will open.
Hyperlinks, often referred to as just "links," are common in Web pages, but can be found in other hypertext documents. These include certain encyclopedias, glossaries, dictionaries, and other references that use hyperlinks. The links act the same way as they do on the Web, allowing the user to jump from page to page. Basically, hyperlinks allow people to browse information at hyperspeed.<|endoftext|>
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Respond to the questions below either as a direct post or by linking a Google Doc or Word Doc. If it’s a Google Doc, make sure it’s in your Humanities folder so that I can see it. 1. The book will talk about six families who rules China for two millennia. How long is two millennia? Two millennia is equal to 2,000 years. 2. Using the Table of Contents, list the six dynasties. Han Dynasty, Tang Dynasty, Song Dynasty, Yuan Dynasty, Ming Dynasty, and Qing Dynasty. 3.
According to Chinese legend, how is the fish related to the dragon? The fist dragon was a fish that swam up to the top of a mountain along the way the fish encountered many difficulties including a locked gate at the top but was successful. After the fish reached the top, it transformed into a dragon. 4. List three reasons that the number nine is important in Chinese culture. Number nine is associated with dragon (the emperor) and the word jiu sounds like the Chinese word for long life. 5.
According to the Chinese Zodiac, 2012 is the year of the dragon. What is the animal for 2013 (You may have to look this up online. )? Snake is the Zodiac Sign for 2013. 6. If you had a phone in China, why would you not want the number 1414-4444? The number “four” in Chinese sounds like the word for death and therefore considered to be a bad omen to use this number. 7. List three reasons that the color red is important in China. In China, the color red symbolizes success, happiness, and good luck. 8.
List three reasons that the color yellow is important in China. In China, the color yellow symbolizes the center. Yellow was also the color that was reserved for the emperor and members of the imperial family. 9. Find two pieces of evidences on pages 12–13 that silk was extremely valuable to China over a thousand years ago. Silk was an extremely valuable item because it was very complicated to make. Silk was even used as money and most important and rich people around the world wore clothing made from Chinese silk.<|endoftext|>
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Title: Building Bridges
Author: Tammy Enz
Through simple text and engaging photographs, readers learn about bridges and the science behind them. Experiment with building paper trusses and testing them by following the step-by-step instructions illustrated in the book. What type of bridge will span across long distances? Find out, and build one using cardboard, textbooks, straws, and string! The author is a civil engineer and has written dozens of engineering and science books for young people. Other books in the Young Engineers series are: Building Structures and Towers, Building Vehicles that Fly, and Building Vehicles that Roll.<|endoftext|>
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Creates a graphical chart (as opposed to textplots of the XYPLOT command) of any number of dependent variables (yVectors) versus one independent variable (xVector). The first vector contains all the "X" values and each of the remaining vectors contains a set of "Y" values that correspond to the "X" values. An example of this form of the command is:
XYGRAPH xVec y1Vec y2Vec y3Vec y4Vec
Note that there is only one X vector and that it applies to all the Y vectors. All the vectors must be the same length.
If the pairs keyword is present, then the command will plot multiple curves defined by pairs of vectors, the first of a pair containing the X values, the second containing the Y values. Both members of each pair must be the same length, but each pair may be of a different length. An example of this form of the command is:
XYGRAPH PAIRS x1Vec y1Vec x2Vec y2Vec x3Vec y3Vec
Note that there is one X vector for each Y vector.
If the scatter keyword is present it must be followed by an argument indicating how many of the Y vectors immediately following the argument are to be shown as scattergraphs. Any remaining vectors or vector pairs will be graphed as lines. The argument may be a literal number or a vector variable. If it's a vector, only its first element will be used. Here are examples of the usage of this keyword:
XYGRAPH SCATTER 1 xVec y1Vec y2Vec y3Vec
The above command says that y1Vec will be shown as a scattergraph and that y2Vec and y3Vec will be displayed as line graphs. All the Y vectors will share the same X vector, xVec.
XYGRAPH PAIRS SCATTER 1 x1Vec y1Vec x2Vec y2Vec x3Vec y3Vec
The above command, as before, says that y1Vec will be shown as a scattergraph and that y2Vec and y3Vec will be displayed as line graphs. But this time, each of the Y vectors has its own X vector preceding it because of the pairs keyword.
The chart will by default have linear-scaled axes. If the log keyword is present, the graph's Y-axis will have a logarithmic scale. If the loglog keyword is present, both axes will have logarithmic scales. An exception is that if the X or Y data contains any values that are illegal for logs (I.e, they are negative or zero) then the keyword will be ignored and the scales will be linear.
The chart is displayed in the Statistics101 Output Window in a tab of its own. There are three titles or labels associated with the graph: the tab title, the X-axis label, and the Y-axis label. The optional keyword title establishes the title to be displayed in the graph's tab. It takes a literal string (enclosed in double quotes) or a string variable argument that becomes the tab's title. If the title keyword and its argument are omitted, then the tab's title will display the Y-axis label. If there is no Y-axis label, then the tab's title will be the name of the first input yVector.
The axis labels are not associated with any keyword. If they are omitted, then the name of the first yVector becomes the Y-axis label and the name of the xVector becomes the X-axis label. If they are present, they are assigned based on their order, the first being the Y-axis label and the second being the X-axis label. If you want to specify an x-axis label but you want to keep the default Y-axis label (the name of the first yVector), you can use an empty string ("") as the Y-axis label, followed by the X-axis label. This maintains the required order.
If you don't want a label for either or both axes, use a blank string (" ") for that axis label or for both axis labels. Note the difference between a blank string and an empty string: the blank string has at least one blank in it and only blanks. The empty string has nothing between the two quote marks.
The following program illustrates the operations of the XYGRAPH command with one independent variable (degrees) and three dependent variables (sines, cosines and offsetSines). Cut and paste it into Statistics101's edit window and run it.
'Plot an xygraph of sines and cosines: INCLUDE "lib/mathConstants.txt" MULTIPLY 1,180 5 degrees degrees = 5 * 1,180 radians = degrees * degToRad sines = SIN(radians) cosines = COS(radians) offsetRads = radians + pi / 4 SIN offsetrads offsetSines XYGRAPH "Circular Functions" degrees sines cosines offsetSines
Here is the output of the above program:
The next program shows the XYGRAPH command graphing two independent graphs. Note the use of the pairs keyword:
INCLUDE "lib/complexCommands.txt" starAngles = (234 90 306 162 18 234) * degToRad TORECT 1 starAngles starX starY circleAngles = 0,361,1 * degToRad TORECT 1 circleAngles circleX circleY XYGRAPH pairs starX starY circleX circleY
Here is the result:
The next program generates a combined scatter and line graph:
INCLUDE "lib/predictCommand.txt" INCLUDE "lib/mathConstants.txt" COPY 1,19 trend COPY 1976,1994 year COPY (5.04 5.54 7.93 11.19 13.36 16.38 12.26 9.09 10.23 \ 8.10 6.81 6.66 7.57 9.21 8.10 5.69 3.52 3.02 4.21) interestRate COPY (7.7 7.1 6.1 5.8 7.1 7.6 9.7 9.6 7.5 7.2 7.0 \ 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1) unemploymentRate COPY (165.4 193.1 230.2 259.8 259.7 272.0 260.6 294.9 \ 348.8 377.4 407.7 419.4 432.3 443.7 442.2 403.4 435.0 \ 464.5 506.9) newConstruction REGRESS newConstruction trend interestRate unemploymentRate coefficients PRINT coefficients PREDICT coefficients yvec trend interestRate unemploymentRate PRINT table year yvec scatterCount = 1 XYGRAPH scatter 1 trend newConstruction yVec
Here is the graph it produces:
If you want to display a line graph with its data points identified you can duplicate the Y vector with the first Y instance being a scatter graph and the second being a line graph, like this:
SEED 123456 LET x = 1,10 'Make some fake data SAMPLE 10 10,19 y1 SAMPLE 10 10,19 y2 XYGRAPH scatter 2 x y1 y2 y1 y2<|endoftext|>
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## Bayes Theory and Its Facts
Bayes Theory and Its Facts
##### January 6, 2021
Boutros El-Gamil
Bayes Theory and Its Facts
January 6, 2021
Boutros El-Gamil
# 1. Example and Bayesian Treatment
John is software engineer with 15+ years of technical experiences in IT. He applied to the CTO position in a startup company. In the final interview, and among a short list of 5 applicants , John was double qualified than the other four candidates. Do you think that John got the job?
If I am willing to guess, I will assume that John definitely got the job, says with 90% chance. I think also that you agree with me, do you? Based on the available information, John has over-performed all other applicants in the last round of interviews. Accordingly, he is the most likely to get the job. But does Bayes’ Theory shares us the same opinion? Let us see how Bayes’ Theorem guesses the answer!
The first step in Bayes’ Theory is to estimate the probability that John got the job. This assumption is usually called the hypothesis (H). As all applicants in the short list have equal opportunity to get the job, John – as well as the other four – has equal probability of 1/5 (or 20%) to get the job. This estimation is called the prior and can be written as:
P(H) = \frac{1}{5} = 20\%
There is another piece of information; that is, John is double qualified than the other four applicants after the final evaluation. This piece of information is called the evidence (E), and it is the information piece that should update our prior believe (John got the job). The goal now is to calculate the probability that John got the job (i.e. P(H)), given the fact that John was double-qualified for the job than all other applicants (i.e. E). This probability is called the posterior and it is mathematically written as P(H|E).
But how can we update our hypothesis (H) using the evidence (E)? Here comes Bayes’ Theorem into the picture. Bayes’ Theory binds the prior and the posterior by a third component called the likelihood. The likelihood is nothing but the opposite conditional probability to the posterior (i.e. P(E|H)) or the probability that John was double-qualified than all other applicants, given that John got the Job.
In order to calculate the likelihood of our example, we need to quantify our evidence with numbers. Suppose the last interview round includes an evaluation sheet of 10 points. In this evaluation sheet John got 8/10 while each other applicant got only 4/10. We can translate these ratios into probabilities as follows:
\begin{equation*}
P(\textit{John was double-qualified $|$ John got the job}) = P(E|H) = 0.8
\end{equation*}
\begin{equation*}
P(\textit{John was double-qualified $|$ John did not get the job}) = P(E|\neg H) = 0.4
\end{equation*}
Now we can write Bayes’ Theorem as:
\begin{equation*}
Posterior \propto Likelihood \times Prior
\end{equation*}
or:
\begin{equation*}
P(H|E) \propto P(E|H) \times P(H)
\end{equation*}
Note that both sides are proportional to each other! in order to equalize them, we need to normalize the right side by using the marginal probability. Marginal probability is the total probability of seeing the evidence E, whether the hypothesis H holds or not. It is calculated as follows:
\begin{equation*}
\begin{split}
\textit{marginal prob.} & = P(E \cap H) + P(E \cap\neg H)\\
& = P(E|H) \times P(H) + P(E|\neg H) \times P(\neg H)
\end{split}
\end{equation*}
Now we can re-write Bayes’ Theorem as:
\begin{equation*}
P(H|E) =\dfrac{P(E|H) \times P(H)}{P(E|H) \times P(H) + P(E|\neg H) \times P(\neg H)}
\end{equation*}
For simplicity and symmetry, the denominator is usually written as P(E):
\begin{equation*}
{\boxed{P(H|E) =\dfrac{P(E|H) \times P(H)}{P(E)} } }
\end{equation*}
Let’s plug in our numbers and calculate the posterior given the above data:
\begin{split}
P(H|E) & =\dfrac{P(E|H) \times P(H)}{P(E|H) \times P(H) + P(E|\neg H) \times P(\neg H)}\\\\
& = \frac{0.8 \times 0.2}{0.8 \times 0.2 + 0.4 \times 0.8} = \frac{0.16}{0.16+0.32} = \frac{0.16}{0.48} \\\\
&= \frac{1}{3} \approx 33\%
\end{split}
The above result says that according to Bayes’ Theorem, John has 33% chance to be the CTO, given that he double-qualified all other applicants! Does this result make sense to you?
# 2. Interpretation
The easiest way to understand Bayes’ Rule is to think of it as a function of time $$f(T)$$. At the first moment (say $$T_0$$), we build our hypothesis (prior). At each next moment, we update our prior using Bayes’ equation. In our example, we can consider the first moment $$T_0$$ is when the startup creates the short list of the five candidates. At this point, we build our prior hypothesis as follows:
\begin{equation*}
P(\textit{John got the job},T_0) = \frac{1}{5} = P(\textit{another candidate got the job},T_0)
\end{equation*}
At time point $$T_1$$, the startup has interviewed the five candidates, and it turns out that John is double-performed than the other 4 candidates. Now we update our prior as follows:
\begin{equation*}
P(\textit{John got the job},T_1) = 2 \times P(\textit{another candidate got the job},T_1)
\end{equation*}
Now the probabilities must change to meet the new evidence. To calculate the new probabilities, we assign each candidate (other than John) a probability of $$\frac{1}{6}$$, such that:
\begin{equation*}
P(\textit{another candidate got the job},T_1) = \frac{1}{6}\approx 17\%
\end{equation*}
\begin{equation*}
P(\textit{John didn’t get the job},T_1) = 4\times\frac{1}{6} = \frac{2}{3}\approx 76\%
\end{equation*}
\begin{equation*}
\begin{split}
P(\textit{John got the job},T_1) & = 1 – \frac{2}{3}\\
& = \frac{1}{3}\approx 33\%\\
& = 2 \times P(\textit{another candidate got the job},T_1)
\end{split}
\end{equation*}
This is the same result we calculated using Bayes’ Theorem. After updating our prior with the new evidence, the probability that John got the job increased from 20% to 33%, and doubled the probability that another candidate got the job (around 17%). The result – however – did not neglect the fact that John still has four qualified competitors, and the sum of probabilities of those competitors should be considered in the final evaluation. This leads us to the first fact of Bayes’ Theorem.
# 3. Fact #1: New evidences do not neglect or overwrite the prior, but update It
Bayes’ Theorem does not working like human minds. While human minds usually overwrite information and highlight the most significant pieces of it, Bayes’ Theory has a strong memory that cumulate information over time and update its outputs afterwards. In our example, even if John is 3 times more qualified than other candidates in the final interview, his chances to get the job according to Bayes’ Rule is around 43%. If he is 4 times more qualified than others, his chance shall not exceed 50%. As you notice, Bayes’ Rule increases John chances according to his achievements in the interview (the evidence) without ignoring the fact that he is one of 5 qualified candidates for the job.
# 4. Fact #2: Each new evidence can drastically change the final estimation
This fact seems to contradict the first one, but it holds in all cases. In order to illustrate this fact, we add an update to the above example:
… while the startup founders were deciding about the CTO position, they received a feedback from a big investor who was ready to fund their startup. The investor asked however, that the company CTO must be 5 days a week in the company’ office. The founders asked John to relocate to their city, but John refused their demand for social reasons. He offered instead to be in the office only 1 day/week, and the rest of the week working from home. The other 4 candidates were ready to work full time from office. After this update, what is the chance that John got the job?
In this update, a new evidence (say $$E_2$$) has been added to the problem. To plug it into the Bayes’ Rule, we need to quantify the new information in both likelihood and marginal probability. Let’s start with the likelihood:
P(\textit{CTO work from office $|$ John got the job}) = \frac{1}{5} = P(E_2|H)\\\\
P(\textit{CTO work from office $|$ John didn’t get the job}) = 1 = P(E_2|\neg H)
And the marginal probability:
\begin{equation*}
\begin{split}
& P(E_1,E_2|H)) \times P(H) + P(E_1,E_2|\neg H)) \times P(\neg H) =\\
& (0.8 \times 0.2 \times 0.2) + (0.4 \times 1 \times 0.8) = 0.352
\end{split}
\end{equation*}
Now we update the posterior as follows:
\begin{equation*}
\begin{split}
P(H|E_1,E_2) & = \frac{P(E_1,E_2|H) \times P(H)}{P(E_1,E_2|H) \times P(H) + P(E_1,E_2|\neg H) \times P(\neg H)}\\\\
& = \frac{0.8 \times 0.2 \times 0.2}{0.352} = 0.09 = 9\%
\end{split}
\end{equation*}
As the result shows, the new evidence has dropped John chance to get the job from 33% down to 9%. This gives us an example of how much a new evidence can dramatically change a Bayesian estimation.
# 5. Fact #3 Likelihood is NOT a probability of events, it’s a probability of hypotheses!
The Likelihood is the basic – yet the most ambiguous part – of Bayes’ Rule. Many people treat likelihoods as probability values, and they get confused when discover that it does not meet probability postulations. The fact is that likelihood is a special type of probability called probability of hypotheses or probability of probabilities.
This means the likelihood calculations are build upon hypotheses, not upon data or measurements (at least within the focus of this article). Because the impeded hypotheses are not disjoint and exhaustive probabilities, the generated likelihoods do not follow the postulations of common (i.e. event) probabilities. For instance, the likelihood and its complement(s) must not sum up to 1. In addition, likelihoods can not be interpreted as informative ratios.
Back to John’s case, we saw that after the last interview it turns out that John was double qualified than all other candidates. We express this likelihood as follows:
P(\textit{John was double-qualified $|$ John got the job}) = P(E|H) = 0.8\\ \\
P(\textit{John was double-qualified $|$ John did not get the job}) = P(E|\neg H) = 0.4
As you see, the likelihood $$P(E|H)$$ and its complement $$P(E|\neg H)$$ do not sum up to 1. This makes sense; like for example if a number of students took an exam, it is very unlikely that the sum of their marks equals exactly the full mark of the exam. Moreover, the likelihood values mentioned above can not be exclusively interpreted as percentages, as these values are just proportional to each other. I assume here that the ratio is 0.8:0.4. You may assume it’s 1:0.5 or 0.6:0.3. As all these ratios are equivalent to 2:1, plugging any of them leads to the same posterior estimation.
# 6. Fact #4 Likelihoods are meaningless, unless compared to each other!
There are endless ways of estimating the probability of a hypothesis. Accordingly, no body can be 100% sure of his, or others’ estimations of a given question. For example, if you give a golden ring to 12 jewelers, and ask each of them to estimate the ring weight by hand. Each one shall give an approximate value according to her/his experience and hand sensitivity. Let’s say their estimations (in grams) are $$\{4,3.9,3.85,3.8,4,3.9,3.9,3.85,3.9,3.85,3.95,3.9\}$$.
If we want to evaluate these estimations, we can’t do this selectively, as all estimations have the same degree of uncertainty. The solution is to measure the likelihood of each weight, and the weight with maximum likelihood estimation gains the lowest degree of uncertainty (or highest reliability).
\begin{equation*}
\begin{split}
\textit{Likelihood}(3.8)= \frac{1}{12}\\
\textit{Likelihood}(3.85)= \frac{3}{12}\\
\textit{Likelihood}(3.9)= \frac{5}{12}\\
\textit{Likelihood}(3.95)= \frac{1}{12}\\
\textit{Likelihood}(4)= \frac{2}{12}
\end{split}
\end{equation*}
From this simple evaluation, we find that 3.9 is the weight with maximum likelihood, and it gains a probability of 42%. This technique is know as the Maximum Likelihood Estimation (MLE), which is widely used in data analysis.
In addition to MLE, we can assign a range of reliability to the MLE analysis. In our example, the following graph shows that 90% of data lies between 3.87 and 3.93.
This range is called the Credible Interval CI (or Highest Posterior Density HPD), and it indicates that 90% of the hypotheses are more likely to be between 3.87 and 3.93. You can expand or shrink the credible interval to meet your analysis requirements. Following are credible intervals of 80% (left) and 95% (right).
# 7. Fact #5 Marginal likelihood is constant over all hypotheses!
I mentioned earlier that marginal likelihood is just a normalizer that ensures the generated posterior is a valid ratio. Because marginal likelihood is data-independent, it has no influence on the posterior estimations. Let’s have a look at the posterior estimation of a hypothesis $$H$$ given an evidence $$E$$:
\begin{equation*}
P(H|E) =\dfrac{P(E|H) \times P(H)}{P(E|H) \times P(H) + P(E|\neg H) \times P(\neg H)}
\end{equation*}
By replacing hypothesis $$H$$ with it’s complement $$\neg H$$, the posterior is calculated as follows:
\begin{equation*}
P(\neg H|E) =\dfrac{P(E|\neg H) \times P(\neg H)}{P(E|H) \times P(H) + P(E|\neg H) \times P(\neg H)}
\end{equation*}
Note here that the denominator on the right side has not changed, which emphasizes the neutrality of marginal likelihood over different hypotheses. This fact has a lot of consequences in real life applications. For instance, if you decide to apply Bayes Rule in its simplest form (i.e. MLE), you don’t need (and you can’t) calculate the marginal likelihood, due to the absence of prior probability $$P(H)$$. If you build a Naive Bayes Classifier, you also don’t need to compute marginal likelihood, because it has no influence on the classification results.
# 8. Proof of Bayes Theorem
As an appendix to this article, I want to give a simple proof of Bayes Theorem, which is very simple and straightforward. Bayes Rule could be easily inferred using joint probability axiom of 2 independent events $$A$$ and $$B$$:
\begin{equation*}
P(A \cap B) = P(A|B) \times P(B)
\end{equation*}
In the same way, we can define the joint probability of $$B$$ and $$A$$:
\begin{equation*}
P(B \cap A) = P(B|A) \times P(A)
\end{equation*}
As joint probabilities are interchangeable (i.e. $$P(A \cap B)=P(B \cap A)$$), we can also write that:
\begin{equation*}
P(A|B) \times P(B) = P(B|A) \times P(A)
\end{equation*}
which gives us the Bayes Theorem:
\begin{equation*}
P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}
\end{equation*}<|endoftext|>
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A timer is a software facility that allows functions to be invoked at some future moment, after a given time interval has elapsed; a time-out denotes a moment at which the time interval associated with a timer has elapsed.
Timers are widely used both by the kernel and by processes. Most device drivers use timers to detect anomalous conditions — floppy disk drivers, for instance, use timers to switch off the device motor after the floppy has not been accessed for a while, and parallel printer drivers use them to detect erroneous printer conditions.
Timers are also used quite often by programmers to force the execution of specific functions at some future time (see the later section "The setitimer( ) and alarm( ) System Calls").
Implementing a timer is relatively easy. Each timer contains a
field that indicates how far in the future the timer should expire. This
field is initially calculated by adding the right number of ticks to the
current value of
jiffies. The field
does not change. Every time the kernel checks a timer, it compares the
expiration field to the value of
jiffies at the current moment, and the timer
jiffies is greater than
or equal to the stored value.
Linux considers two types of timers called dynamic timers and interval timers . The first type is used by the kernel, while interval timers may be created by processes in User Mode.
One word of caution about Linux timers: since checking for timer functions is always done ...<|endoftext|>
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# How do you solve 2x^2 - 250x + 5000 = 0?
Mar 31, 2018
25,100
#### Explanation:
$2 {x}^{2} - 250 x + 5000 = 0$
$\Rightarrow 2 \left({x}^{2} - 125 x + 2500\right) = 0$
$\Rightarrow {x}^{2} - 125 x + 2500 = 0$
$\Rightarrow {x}^{2} - 25 x - 100 x + 2500 = 0$
$\Rightarrow x \left(x - 25\right) - 100 \left(x - 25\right) = 0$
$\Rightarrow \left(x - 25\right) \left(x - 100\right) = 0$
$\Rightarrow \left(x - 25\right) = 0 , \left(x - 100\right) = 0$
rArr x = 25 & 100
Mar 31, 2018
$x = 25 \text{ or } x = 100$
#### Explanation:
$\text{take out a "color(blue)"common factor } 2$
$\Rightarrow 2 \left({x}^{2} - 125 x + 2500\right) = 0$
$\text{the factors of + 2500 which sum to - 125 are - 25 and - 100}$
$\Rightarrow 2 \left(x - 25\right) \left(x - 100\right) = 0$
$\text{equate each factor to zero and solve for x}$
$x - 25 = 0 \Rightarrow x = 25$
$x - 100 = 0 \Rightarrow x = 100$<|endoftext|>
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## Hit the Target
### Posted by misshampson | Posted in Addition, Four Operations, Number, Place Value | Posted on September 2, 2016
This is a Maths warm up game that can be used to practise both Place Value and Addition skills.
Teachers set a target on the board and in like ability pairs students use a calculator to try and add numbers together to hit the target. This can be used to practise the other operations as well.
For example, the target is 25
Students 1 puts 4 into the calculator
Student 2 needs to figure out what they need to add to 4 to get 25. In this case, they add 21.
Variation: students can play it where they have to add multiple numbers together to hit the target.
For example, if the target is 71.
Student 1 puts 12 into the calculator
Student 2 adds 9 to 12 which equals 21 and so on and so forth.
## Figure Me Out Project
### Posted by Miss Gardiner | Posted in Addition, Division, Four Operations, Multiplication, Subtraction | Posted on August 17, 2016
Students create questions based around themselves with numerical answers, e.g. age, house number, number of siblings, number of letters in their first name, shoe size.. (the list is endless). Once students have the answers, they have to create equations to be worked out to get to their answer, e.g ‘My Age’ Answer=12, Equation could be 2×6= 10+2= or even more advanced using BODMAS, decimals, fractions etc.
Once students have created the equations and answers, they present these on a poster with the equation displayed on a piece of paper or a post it note that can be opened up to display the answer underneath.
Figure Me Out Project
## The Array game.
### Posted by Miss Gardiner | Posted in Addition, Multiplication | Posted on August 17, 2016
Students will need:
– Two coloured pencils
– Two grey leads
– Graphing paper
– Dice (this can be modified according to levels)
– Dice mats
Students work in pairs.
Students roll two dice.
They colour in the squares on the graph paper that the dice multiply to, an array. For example if they roll a 2 and a 3. So it would be 2×3=6 they would then colour a box 2 by 3 squares to equal 6 squares in their chosen colour on the page and write the multiplication sum in grey lead on the squares.
It would then be the next child’s turn. They would then roll the dice and colour squares in the other coloured pencil.
Once all the squares are coloured in the children will need to add the total of their squares together, they can use their chosen addition strategy.
The student with the most squares coloured wins.
## Addition Go Fish
### Posted by Miss Gardiner | Posted in Addition | Posted on August 17, 2016
2 or 3 players
Use a standard deck of cards with tens and face cards removed. Aces are worth one. Deal five cards to each player then take out one card and set it aside without looking at it. If a player has any two cards that add to 10 (eg:
3 + 7), s/he lays the pair on the table, face up. Once all players have laid down all their “10” pairs, the first player asks any other player for a card what would complete a “10” pair in his/her hand. If the other player has the requested card, he/she must hand it over and the first player may continue asking for cards, from the same person or anyone else. If the player doesn’t have the requested card, s/he says, “Go fish!” and the first player takes the top card from the stack of undealt cards. If a player runs out of cards, s/he draws a new one at the beginning of a new turn and continues play. When all the cards are matched up, there will be one card without a pair (the one removed from the deck at the beginning of the game). The person who winds up with the most cards wins.
EXTENSION; Use subtraction and the numbers to equal zero.
## Bike Ride using Cartesian Plane
### Posted by Miss Lawrence | Posted in Addition, Angles, Cartesian Plane, Coordinates, Location, Mapping | Posted on June 29, 2016
Have students map out a bike ride that goes for a specific distance (e.g. – 5km).
• Listing coordinates of intersections and landmarks
• Writing directions
• Calculating the time it would take to complete the bike ride, allowing for walking across roads
• Investigating height of terrain to see vertical meters covered
• Measure the angle of turns made
Challenges could also be made as to how much distance is covered in a particular quadrant.
Map attached with Cartesian Plane.
Cartesian Plane – Cranbourne East
## Apple tree
### Posted by lisar | Posted in Addition | Posted on September 18, 2015
Students will be drawing an apple tree in their books and selecting a handful of green and red counters/apples to put on their tree. Children will have to draw these apples onto their tree and create an equation, e.g. 5 green apples + 7 red apples = 12 apples altogether.
Click on the link below for lesson plan
apple tree activity
## Sweets Shop
### Posted by lisar | Posted in Addition, Money | Posted on September 17, 2015
This activity can carry on from the chocolate bar & lolly shop activity (featured in the Fractions section) by giving each “lolly” a price. Alternatively, you could have the students design a sweets shop (with cakes, slices, etc) or an ice-cream shop (with a range of flavours, cones and toppings!). Students may either write the individual prices onto the lolly shop sheet or may design their own price list. These can be adapted to ability levels. It is up to teachers how far that they want to take the next stage. Ideas are:
• Ask other students to choose a number of items from their shop, which they would like to “purchase”. Owner of the shop has to add up the total of the items. You may even have them work out the change.
• Set up a “shop” where the shop items are drawn or constructed out of materials. Using play money (and calculators if needed), students are given the opportunity to go shopping at each other’s shop and “buy” items using their play money. When an item is purchased, shop owners can cut out the item from their sheet and physically give them the item. They must also work out the total price of the items purchased, which the customer is to give them and the shop owner works out the change to give back.
* Please note: if you are allowing students to cut out their items from their poster, you may want to photocopy them just in case!
## Cover Up (Addition Game)
### Posted by lisar | Posted in Addition | Posted on September 17, 2015
Students will work in pairs and each receive a game board with numbers 1 to 12 along the side and two 6 sided dice. Each player will take turns rolling the dice and add the numbers together, they will then place a counter on that number. The first person to place all their counters on their numbers wins the round.
### Posted by lisar | Posted in Addition | Posted on September 17, 2015
• Teacher explicitly teaches addition strategies relevant to the students level of understanding. Around the room are a variety of tools that students may use to help them work out addition equations such as rulers, number lines, dice, counters, toys. Children roam the room and choose any of the materials to help them complete the equations. All children are using the materials together/next to each other regardless of ability.
• Students can generate their own addition equations through rolling dice (a variety of dice can be used to suit a students ability such as six sided dot dice, 6 sided numeral dice, 10 or 20 sided dice). Students could also take a small handful/scoop of two different counter colours.
• Teachers can choose a resource and fishbowl a group to demonstrate.
## Four Operations Board Game
### Posted by lisar | Posted in Addition, Division, Four Operations, Multiplication, Subtraction | Posted on September 17, 2015
Students design and create a game board in the style of their choice.
Students create question cards using the four operations. (i.e 8 addition equations, 8 subtraction, 8 multiplication and 8 division) There can be a mixture of number sentences and worded problems.
Each process is written on a different colour card.
Game boards include sections where players pick up cards as well as general game instructions. (i.e move back 2 spaces. Roll a 5 to move again)
The game board can also be used for:
• Measurement (to measure how long your journey is)
• Money (collecting and taking away money amounts along the way)<|endoftext|>
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Vegetation around the world is on the move, and climate change is the culprit, according to a new analysis of global vegetation shifts led by a University of California, Berkeley, ecologist in collaboration with researchers from the U.S. Department of Agriculture Forest Service.
In a paper published today in the journal Global Ecology and Biogeography, researchers present evidence that over the past century, vegetation has been gradually moving toward the poles and up mountain slopes, where temperatures are cooler, as well as toward the equator, where rainfall is greater.
Moreover, an estimated one-tenth to one-half of the land mass on Earth will be highly vulnerable to climate-related vegetation shifts by the end of this century, depending upon how effectively humans are able to curb greenhouse gas emissions, according to the study.
The results came from a meta-analysis of hundreds of field studies and a spatial analysis of observed 20th century climate and projected 21st century vegetation.
The meta-analysis identified field studies that examined long-term vegetation shifts in which climate, rather than impacts from local human activity such as deforestation, was the dominant influence. The researchers found 15 cases of biome shifts since the 18th century that are attributable to changes in temperature and precipitation.
“This is the first global view of observed biome shifts due to climate change,” said the study’s lead author Patrick Gonzalez, a visiting scholar at the Center for Forestry at UC Berkeley’s College of Natural Resources. “It’s not just a case of one or two plant species moving to another area. To change the biome of an ecosystem, a whole suite of plants must change.”
The researchers calculated that from 1901 to 2002, mean temperatures significantly increased on 76 percent of global land, with the greatest warming in boreal, or subarctic, regions. The most substantial biome shifts occurred where temperature or precipitation changed by one-half to two standard deviations from 20th century mean values.
Some examples of biome shifts that occurred include woodlands giving way to grasslands in the African Sahel, and shrublands encroaching onto tundra in the Arctic.
“The dieback of trees and shrubs in the Sahel leaves less wood for houses and cooking, while the contraction of Arctic tundra reduces habitat for caribou and other wildlife,” said Gonzalez, who has served as a lead author on reports of the Intergovernmental Panel on Climate Change (IPCC). “Globally, vegetation shifts are disrupting ecosystems, reducing habitat for endangered species, and altering the forests that supply water and other services to many people.”
To identify the areas most vulnerable to future vegetation shifts, the researchers combined statistical analyses of observed climate data from the 20th century with models of vegetation change in the 21st century.
Based upon nine different combinations of IPCC greenhouse gas emissions scenarios and climate models, the researchers divided the world’s land into five classes – from very high to very low – of vulnerability to biome shifts.
“Scientists had not quantified this risk before,” said Gonzalez. “We developed a simple classification system that natural resource management agencies can use to identify regions in greatest need of attention and planning. We have worked with the U.S.D.A. Forest Service and the U.S. Fish and Wildlife Service to explore the application of our results to adaptation of natural resource management.”
Gonzalez said that because of limited resources, it may be prudent to focus on protecting areas of greater resilience to ecological changes so that they can serve as refuges for plants and animals. “It is also useful to identify places of higher vulnerability, because agencies will need to consider adaptation measures for vulnerable ecosystems,” he said. “Some shifts in vegetation could increase fuel for wildfires, for example, so prescribed burning may be necessary to reduce the risk of catastrophic fires.”
“Approximately one billion people now live in areas that are highly to very highly vulnerable to future vegetation shifts,” said Gonzalez. “Ecosystems provide important services to people, so we must reduce the emissions that cause climate change, then adapt to major changes that might occur.”<|endoftext|>
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## Search This Blog
An expression containing the fourth power of x, that is x^4, can be factored just like a quadratic expression. For example:
x^4 + 3x^2 + 2
The above expression contains x^4, but it can be considered a quadratic expression and factored just like quadratic expressions. For a moment, forget that there is an x^4 in it. Just look at the the numbers in the expression. Comparing with standard quadratic expression ax^2 + bx + c,
• a = 1
• b = 3
• c = 2
Product of a and c is 2.
So you need to find two numbers whose product is 2 and sum is 3. Try to find them first without looking at the solution below. Let me leave some blank space between this and the solution so that you don't see the answer before trying to solve it :~)
If you were able to solve it, good! The two numbers are 1 and 2, because 1 times 2 is 2 and 1 plus 2 is 3. Now replace 3x in the equation (which is the middle term) with these numbers 2x + x to get
x^4 + 2x^2 + x^2 + 2
Rearrange the terms so that the two two's are together,
x^4 + x^2 + 2x^2 + 2
Factor out x^2 from the first two terms and 2 from the next two terms
x^2(x^2 + 1) + 2(x^2 + 1)
Factor out (x^2 + 1) from the expression,
(x^2 + 1)(x^2 + 2x)
### ~ * ~
Thus we factored an expression having a degree of four (that is x^4) as if it was a quadratic expression because it had
• three terms in it
• did not contain an x^3
So we notice that we can factor all three termed expressions, called trinomials, as long as the degree is even, there are three terms, the second term has a degree half that of the first term, and the third term is either a constant or has a degree that is half that of the second term. So if you have x^6 + 3x^3 + 2, you can factor it just like you factored the x^4 one above, as follows:
x^6 + 3x^3 + 2 = x^6 + x^3 + 2x^3 + 2 = x^3(x^3 + 1) + 2(x^3 + 1) = (x^3 + 1)(x^3 + 2)
Similarly if you have an expression with x^8 you can factor it out as follows:
x^8 + 3x^4 + 2 = x^8 + x^4 + 2x^4 + 2 = x^4(x^4 + 1) + 2(x^4 + 1) = (x^4 + 1)(x^4 + 2)
### Summary
The above method is called splitting the middle term. You can factor any trinomial by the method of splitting the middle term as long as its highest power (degree) is even, degree of its second term is half that of the first one and the third term is either a constant or has a degree that is half of the second term's degree.<|endoftext|>
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Today’s infographic is simple and to the point. A big part of grade school and even college and onward, is writing papers. Some professions write more papers than others, but it is still an important skill in order to get your point across. This infographic uses venn diagrams to convey the importance of different parts of papers, and to show how they interact with one another. It also shows how much of your paper should include each part.
Of course every paper should begin with an introduction and end with a conclusion. It should also include several point in the middle, that are introduced and concluded in the introduction and conclusion. But how should the middle be laid out? That is up to the author, but it should there is a bit of a formula.
This infographic does a great job of showing that there should be pros and cons. You should always share how your paper may be argued against, and go ahead and prove some of these points wrong. In addition, a good paper should show why the information is important. Why should someone read your paper?
Show this to your students whenever a paper is assigned. Make sure your students are ready to write a good paper, and know what is involved in writing such a paper.
WARNING added later by David Warlick: This is one of those really useful and appealing infographics with embed codes for posting on your own blog or english class web site – that, when you go to the source, you find a quite well-implemented paper mill. We should all be aware that infographics have become the new BAIT to get you to the web sites that make money for people who are often providing unscrupulous and fraudulent services.<|endoftext|>
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# 11.6: Scores on Job Application Tests
Our data come from three groups of 10 people each, all of whom applied for a single job opening: those with no college degree, those with a college degree that is not related to the job opening, and those with a college degree from a relevant field. We want to know if we can use this group membership to account for our observed variability and, by doing so, test if there is a difference between our three group means. We will start, as always, with our hypotheses.
### Step 1: State the Hypotheses
Our hypotheses are concerned with the means of groups based on education level, so:
$\begin{array}{c}{\mathrm{H}_{0}: \text { There is no difference between the means of the education groups }} \\ {\mathrm{H}_{0}: \mu_{1}=\mu_{2}=\mu_{3}}\end{array} \nonumber$
$\mathrm{H}_{A}: \text { At least one mean is different } \nonumber$
Again, we phrase our null hypothesis in terms of what we are actually looking for, and we use a number of population parameters equal to our number of groups. Our alternative hypothesis is always exactly the same.
### Step 2: Find the Critical Values
Our test statistic for ANOVA, as we saw above, is $$F$$. Because we are using a new test statistic, we will get a new table: the $$F$$ distribution table, the top of which is shown in Figure $$\PageIndex{1}$$:
The $$F$$ table only displays critical values for $$α$$ = 0.05. This is because other significance levels are uncommon and so it is not worth it to use up the space to present them. There are now two degrees of freedom we must use to find our critical value: Numerator and Denominator. These correspond to the numerator and denominator of our test statistic, which, if you look at the ANOVA table presented earlier, are our Between Groups and Within Groups rows, respectively. The $$df_B$$ is the “Degrees of Freedom: Numerator” because it is the degrees of freedom value used to calculate the Mean Square Between, which in turn was the numerator of our $$F$$ statistic. Likewise, the $$df_W$$ is the “$$df$$ denom.” (short for denominator) because it is the degrees of freedom value used to calculate the Mean Square Within, which was our denominator for $$F$$.
The formula for $$df_B$$ is $$k – 1$$, and remember that k is the number of groups we are assessing. In this example, $$k = 3$$ so our $$df_B$$ = 2. This tells us that we will use the second column, the one labeled 2, to find our critical value. To find the proper row, we simply calculate the $$df_W$$, which was $$N – k$$. The original prompt told us that we have “three groups of 10 people each,” so our total sample size is 30. This makes our value for $$df_W$$ = 27. If we follow the second column down to the row for 27, we find that our critical value is 3.35. We use this critical value the same way as we did before: it is our criterion against which we will compare our obtained test statistic to determine statistical significance.
### Step 3: Calculate the Test Statistic
Now that we have our hypotheses and the criterion we will use to test them, we can calculate our test statistic. To do this, we will fill in the ANOVA table. When we do so, we will work our way from left to right, filling in each cell to get our final answer. We will assume that we are given the $$SS$$ values as shown below:
Table $$\PageIndex{1}$$: ANOVA Table
Source $$SS$$ $$df$$ $$MS$$ $$F$$
Between 8246
Within 3020
Total
These may seem like random numbers, but remember that they are based on the distances between the groups themselves and within each group. Figure $$\PageIndex{2}$$ shows the plot of the data with the group means and grand mean included. If we wanted to, we could use this information, combined with our earlier information that each group has 10 people, to calculate the Between Groups Sum of Squares by hand. However, doing so would take some time, and without the specific values of the data points, we would not be able to calculate our Within Groups Sum of Squares, so we will trust that these values are the correct ones.
We were given the sums of squares values for our first two rows, so we can use those to calculate the Total Sum of Squares.
Table $$\PageIndex{2}$$: Total Sum of Squares
Source $$SS$$ $$df$$ $$MS$$ $$F$$
Between 8246
Within 3020
Total 11266
We also calculated our degrees of freedom earlier, so we can fill in those values. Additionally, we know that the total degrees of freedom is $$N – 1$$, which is 29. This value of 29 is also the sum of the other two degrees of freedom, so everything checks out.
Table $$\PageIndex{3}$$: Total Sum of Squares
Source $$SS$$ $$df$$ $$MS$$ $$F$$
Between 8246 2
Within 3020 27
Total 11266 29
Now we have everything we need to calculate our mean squares. Our $$MS$$ values for each row are just the $$SS$$ divided by the $$df$$ for that row, giving us:
Table $$\PageIndex{4}$$: Total Sum of Squares
Source $$SS$$ $$df$$ $$MS$$ $$F$$
Between 8246 2 4123
Within 3020 27 111.85
Total 11266 29
Remember that we do not calculate a Total Mean Square, so we leave that cell blank. Finally, we have the information we need to calculate our test statistic. $$F$$ is our $$MS_B$$ divided by $$MS_W$$.
Table $$\PageIndex{5}$$: Total Sum of Squares
Source $$SS$$ $$df$$ $$MS$$ $$F$$
Between 8246 2 4123 36.86
Within 3020 27 111.85
Total 11266 29
So, working our way through the table given only two $$SS$$ values and the sample size and group size given before, we calculate our test statistic to be $$F_{obt} = 36.86$$, which we will compare to the critical value in step 4.
### Step 4: Make the Decision
Our obtained test statistic was calculated to be $$F_{obt} = 36.86$$ and our critical value was found to be $$F^* = 3.35$$. Our obtained statistic is larger than our critical value, so we can reject the null hypothesis.
Reject $$H_0$$. Based on our 3 groups of 10 people, we can conclude that job test scores are statistically significantly different based on education level, $$F(2,27) = 36.86, p < .05$$.
Notice that when we report $$F$$, we include both degrees of freedom. We always report the numerator then the denominator, separated by a comma. We must also note that, because we were only testing for any difference, we cannot yet conclude which groups are different from the others. We will do so shortly, but first, because we found a statistically significant result, we need to calculate an effect size to see how big of an effect we found.
## Contributors
• Foster et al. (University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus)<|endoftext|>
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What is Art?
Interactions between the elements and principles of art help artists to organize sensorially pleasing works of art while also giving viewers a framework within which to analyze and discuss aesthetic ideas.
Evaluate the frameworks we can use to analyze and discuss works of art
- The interplay between the principles and elements of art provide a language with which to discuss and analyze works of art.
- The principles of art include: movement, unity, harmony, variety, balance, contrast , proportion and pattern.
- The elements of art include: texture , form , space , shape, color, value and line .
- How best to define the term art is a subject of constant contention.
- Since conceptual art and postmodern theory came into prominence, it has been proven that anything can be termed art.
- Formalism:The study of art by analyzing and comparing form and style—the way objects are made and their purely visual aspects.
What is Art?
Art is a highly diverse range of human activities engaged in creating visual, auditory, or performed artifacts— artworks—that express the author’s imaginative or technical skill, and are intended to be appreciated for their beauty or emotional power.
The oldest documented forms of art are visual arts, which include images or objects in fields like painting, sculpture, printmaking , photography, and other visual media . Architecture is often included as one of the visual arts; however, like the decorative arts, it involves the creation of objects where the practical considerations of use are essential, in a way that they usually are not in another visual art, like a painting.
Art may be characterized in terms of mimesis (its representation of reality), expression, communication of emotion, or other qualities. Though the definition of what constitutes art is disputed and has changed over time, general descriptions center on the idea of imaginative or technical skill stemming from human agency and creation. When it comes to visually identifying a work of art, there is no single set of values or aesthetic traits. A Baroque painting will not necessarily share much with a contemporary performance piece, but they are both considered art.
Despite the seemingly indefinable nature of art, there have always existed certain formal guidelines for its aesthetic judgment and analysis. Formalism is a concept in art theory in which an artwork’s artistic value is determined solely by its form, or how it is made. Formalism evaluates works on a purely visual level, considering medium and compositional elements as opposed to any reference to realism , context, or content.
Art is often examined through the interaction of the principles and elements of art. The principles of art include movement, unity, harmony, variety, balance, contrast, proportion and pattern. The elements include texture, form, space, shape, color, value and line. The various interactions between the elements and principles of art help artists to organize sensorially pleasing works of art while also giving viewers a framework within which to analyze and discuss aesthetic ideas.
What Does Art Do?
A fundamental purpose inherent to most artistic disciplines is the underlying intention to appeal to, and connect with, human emotion.
Examine the communication, utilitarian, aesthetic, therapeutic, and intellectual purposes of art
- The decorative arts add aesthetic and design values to the objects we use every day, such as a glass or a chair.
- Art therapy is a relatively young type of therapy that focuses on the therapeutic benefits of art-making, using different methods and theories.
- Since the introduction of conceptual art and postmodern theory, it has been proven that anything can, in fact, be termed art.
- It can be said that the fine arts represent an exploration of the human condition and the attempt at a deeper understanding of life.
- human condition:The characteristics, key events, and situations which compose the essentials of human existence, such as birth, growth, emotionality, aspiration, conflict, and mortality.
- fine arts:Visual art created principally for its aesthetic value.
- aesthetic:Concerned with artistic impact or appearance.
A fundamental purpose common to most art forms is the underlying intention to appeal to, and connect with, human emotion. However, the term is incredibly broad and is broken up into numerous sub-categories that lead to utilitarian , decorative, therapeutic, communicative, and intellectual ends. In its broadest form, art may be considered an exploration of the human condition, or a product of the human experience.
The decorative arts add aesthetic and design values to everyday objects, such as a glass or a chair, transforming them from a mere utilitarian object to something aesthetically beautiful. Entire schools of thought exist based on the concepts of design theory intended for the physical world.
Art can function therapeutically as well, an idea that is explored in art therapy. While definitions and practices vary, art therapy is generally understood as a form of therapy that uses art media as its primary mode of communication. It is a relatively young discipline, first introduced around the mid-20th century.
Historically, the fine arts were meant to appeal to the human intellect, though currently there are no true boundaries. Typically, fine art movements have reacted to each other both intellectually and aesthetically throughout the ages. With the introduction of conceptual art and postmodern theory, practically anything can be termed art. In general terms, the fine arts represent an exploration of the human condition and the attempt to experience a deeper understanding of life.
What Does Art Mean?
The meaning of art is shaped by the intentions of the artist as well as the feelings and ideas it engenders in the viewer.
Evaluate the perspectives behind the meaning of art
- The meaning of art is often shared among the members of a given society and dependent upon cultural context.
- The nature of art has been described by philosopher Richard Wollheim as “one of the most elusive of the traditional problems of human culture.”
- Some purposes of art may be to express or communicate emotions and ideas, to explore and appreciate formal elements for their own sake, or to serve as representation.
- Art, at its simplest, is a form of communication and means whatever it is intended to mean by the artist.
- mimesis:The representation of aspects of the real world, especially human actions, in literature and art.
The meaning of art is often culturally specific, shared among the members of a given society and dependent upon cultural context. The purpose of works of art may be to communicate political, spiritual or philosophical ideas, to create a sense of beauty (see aesthetics), to explore the nature of perception, for pleasure, or to generate strong emotions. Its purpose may also be seemingly nonexistent.
The nature of art has been described by philosopher Richard Wollheim as “one of the most elusive of the traditional problems of human culture.” It has been defined as a vehicle for the expression or communication of emotions and ideas, a means for exploring and appreciating formal elements for their own sake, and as mimesis or representation. More recently, thinkers influenced by Martin Heidegger have interpreted art as the means by which a community develops for itself a medium for self-expression and interpretation.
Art, in its broadest sense, is a form of communication. It means whatever the artist intends it to mean, and this meaning is shaped by the materials, techniques, and forms it makes use of, as well as the ideas and feelings it creates in its viewers . Art is an act of expressing feelings, thoughts, and observations.
What Makes Art Beautiful?
Beauty in terms of art refers to an interaction between line, color, texture, sound, shape, motion, and size that is pleasing to the senses.
Define “aesthetics” and “beauty” as they relate to art
- Beauty in art can be difficult to put into words due to a seeming lack of accurate language.
- An aesthetic judgment cannot be an empirical judgment but must instead be processed on a more intuitive level.
- Aesthetics is the branch of philosophy that deals with the nature and appreciation of art, beauty, and taste. Aesthetics is central to any exploration of art.
- For Immanuel Kant, the aesthetic experience of beauty is a judgment of a subjective, but common, human truth.
- For Arthur Schopenhauer, aesthetic contemplation of beauty is the freest and most pure and truthful that intellect can be, and is therefore beautiful.
- Art is often intended to appeal to, and connect with, human emotion.
- aesthetics:The branch of philosophy dealing with the nature of art, taste, and the creation and appreciation of beauty.
- intuitive:Spontaneous, without requiring conscious thought; easily understood or grasped by instinct.
What makes art beautiful is a complicated concept, since beauty is subjective and can change based on context. However, there is a basic human instinct, or internal appreciation, for harmony, balance, and rhythm which can be defined as beauty. Beauty in terms of art usually refers to an interaction between line, color, texture , sound, shape, motion, and size that is pleasing to the senses.
Aesthetics is the branch of philosophy that deals with the nature and appreciation of art, beauty, and taste. Aesthetics is central to any exploration of art. The word “aesthetic” is derived from the Greek “aisthetikos,” meaning “esthetic, sensitive, or sentient. ” In practice, aesthetic judgment refers to the sensory contemplation or appreciation of an object (not necessarily a work of art), while artistic judgment refers to the recognition, appreciation, or criticism of a work of art.
Numerous philosophers have attempted to tackle the concept of beauty and art. For Immanuel Kant, the aesthetic experience of beauty is a judgment of a subjective, but common, human truth. He argued that all people should agree that a rose is beautiful if it indeed is. There are many common conceptions of beauty; for example, Michelangelo’s paintings in the Sistine Chapel are widely recognized as beautiful works of art. However, Kant believes beauty cannot be reduced to any basic set of characteristics or features.
For Arthur Schopenhauer, aesthetic contemplation of beauty is the freest and most pure that intellect can be. He believes that only in terms of aesthetics do we contemplate perfection of form without any kind of worldly agenda.
Beauty in art can be difficult to put into words due to a seeming lack of accurate language. An aesthetic judgment cannot be an empirical judgment but must instead be processed on a more intuitive level.
Art and Human Emotion
Sometimes beauty is not the artist’s ultimate goal. Art is often intended to appeal to, and connect with, human emotion. Artists may express something so that their audience is stimulated in some way—creating feelings, religious faith, curiosity, interest, identification with a group, memories, thoughts, or creativity. For example, performance art often does not aim to please the audience but instead evokes feelings, reactions, conversations, or questions from the viewer . In these cases, aesthetics may be an irrelevant measure of “beautiful” art.
Who Is an Artist?
An artist is a person who is involved in the wide range of activities that are related to creating art.
Summarize the evolution of the term “artist” and its predecessors
- In ancient Greece and Rome there was no word for “artist,” but there were nine muses who oversaw a different field of human creation related to music and poetry, with no muse for visual arts.
- During the Middle Ages , the word “artista” referred to something resembling “craftsman.”
- The first division into major and minor arts dates back to the 1400s with the work of Leon Battista Alberti.
- The European Academies of the 16th century formally solidified the gap between the fine and the applied arts which exists in varying degrees to this day.
- Currently an artist can be defined as anyone who calls him/herself an artist.
- muses:Goddesses of the inspiration of literature, science, and the arts in Greek mythology.
- Pop art:An art movement that emerged in the 1950s that presented a challenge to traditions of fine art by including imagery from popular culture such as advertising and news.
- fine arts:The purely aesthetic arts, such as music, painting, and poetry, as opposed to industrial or functional arts such as engineering or carpentry.
An artist is a person who is involved in the wide range of activities that are related to creating art. The word has transformed over time and context, but the modern understanding of the term denotes that, ultimately, an artist is anyone who calls him/herself an artist.
In ancient Greece and Rome, there was no word for “artist.” The Greek word “techne” is the closest that exists to “art” and means “mastery of any art or craft.” From the Latin “tecnicus” derives the English words “technique,” “technology,” and “technical.” From these words we can denote the ancient standard of equating art with manual labor or craft.
Each of the nine muses of ancient Greece oversaw a different field of human creation. The creation of poetry and music was considered to be divinely inspired and was therefore held in high esteem. However, there was no muse identified with the painting and sculpture; ancient Greek culture held these art forms in low social regard, considering work of this sort to be more along the lines of manual labor.
During the Middle Ages, the word “artista” referred to something resembling “craftsman,” or student of the arts. The first division into “major” and “minor” arts dates back to the 1400s with the work of Leon Battista Alberti, which focused on the importance of the intellectual skills of the artist rather than the manual skills of a craftsman. The European academies of the 16th century formally solidified the gap between the fine and the applied arts, which exists in varying degrees to this day. Generally speaking, the applied arts apply design and aesthetics to objects of everyday use, while the fine arts serve as intellectual stimulation.
Currently, the term “artist” typically refers to anyone who is engaged in an activity that is deemed to be an art form. However, the questions of what is art and who is an artist are not easily answered. The idea of defining art today is far more difficult than it has ever been. After the exhibition during the Pop Art movement of Andy Warhol’s Brillo Box and Campbell’s Soup Cans, the questions of “what is art?” and “who is an artist?” entered a more conceptual realm. Anything can, in fact, be art, and the term remains constantly evolving.<|endoftext|>
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Given two integers L and R. Find the number of perfect powers in the given range [L, R]. A number x is said to be perfect power if there exists some integers a > 0, p > 1 such that x = ap.
Input : 1 4 Output : 2 Explanation : Suitable numbers are 1 and 4 where 1 can be expressed as 1 = 12 and 4 can be expressed as 4 = 22 Input : 12 29 Output : 3 Explanation : Suitable numbers are 16, 25 and 27.
Approach : Let’s fix some power p. It’s obvious that there are no more than 1018/p numbers x such that xp doesn’t exceed 1018 for a particular p. At the same time, only for p = 2 this amount is relatively huge, for all other p ≥ 3 the total amount of such numbers will be of the order of 106. There are 109 squares in the range [1, 1018], so can’t store them to answer our query.
Either, generate all of powers for p ≥ 2 and dispose of all perfect squares among them or generate only odd powers of numbers like 3, 5, 7, etc. Then answer to query (L, R) is equal to the amount of generated numbers between L and R plus some perfect squares in range.
- The number of perfect squares in the range is the difference of floor value of square root of R and floor value of square root of (L – 1), i.e. (floor(sqrt(R)) – floor(sqrt(L – 1)). Note that due to precision issues the standard sqrt might produce incorrect values, so either use binary search or sqrtl inbuilt function defined in cmath (Check here for more description of sqrtl).
- To generate those odd powers of numbers. First of all, do precomputation of finding such numbers that can be expressed as power of some number upto 1018 so that we can answer many queries and no need to process them again and again for each query. Start by iterating a loop from 2 to 106 (since we are calculating for powers p ≥ 3 and 106 is the maximum number whose power raised to 3 cannot exceed 1018), for each value we insert its square into a set and check further if that value is already a perfect square (already present in the set), we do not find any other powers of that number (since any power of a perfect square is also a perfect square). Otherwise, run an inside loop to find odd powers of the number until it exceeds 1018 and insert into another set say ‘s’. By this approach, we haven’t pushed any perfect square in the set ‘s’.
Hence the final answer would be sum of number of perfect squares in the range and difference of upper value of R and lower value of L (using binary search).
Below is the implementation of above approach in C++.
Number of powers between 12 and 29 = 3 Number of powers between 1 and 100000000000 = 320990
- N expressed as sum of 4 prime numbers
- Check if a number can be expressed as power | Set 2 (Using Log)
- Check if a number can be expressed as x^y (x raised to power y)
- Modulo power for large numbers represented as strings
- Sum of all even numbers in range L and R
- Count Odd and Even numbers in a range from L to R
- Sum of all the prime numbers in a given range
- Count of numbers from range [L, R] whose sum of digits is Y
- Sum of range in a series of first odd then even natural numbers
- No of pairs (a[j] >= a[i]) with k numbers in range (a[i], a[j]) that are divisible by x
- Sum of all odd length palindromic numbers within the range [L, R]
- Count all the numbers in a range with smallest factor as K
- K-Primes (Numbers with k prime factors) in a range
- XOR of numbers that appeared even number of times in given Range
- Count numbers in range such that digits in it and it's product with q are unequal
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.<|endoftext|>
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# Where permutations and combinations are used
## Combinatorics
### 1. Permutations
• \ (k = n \)
(i.e. all elements \ (k \) of the basic set \ (n \) are considered)
• The order of the elements is taken into account
### 1.1 Permutation without repetition
A Permutation without repetition is an arrangement of \ (n \) objects that are all distinguishable.
We have \ (n \) distinguishable objects that we want to arrange in a row next to each other in \ (n \) places.
There are \ (n \) placement options for the first object. For the second object there are \ ((n-1) \) possibilities, for the third object \ ((n-2) \) .... and for the last object there is only one possibility.
In mathematical notation, it looks like this:
\ (n \ cdot (n-1) \ cdot (n-2) \ cdot \ text {...} \ cdot 1 = n! \)
In short, you can write \ (n! \) For a permutation without repetition. This term is called "faculty".
example
There are five different colored balls in an urn. How many ways are there to arrange the balls in a row?
solution
\ (5! = 5 \ times 4 \ times 3 \ times 2 \ times 1 = 120 \)
Answer: There are 120 possibilities to arrange five different colored balls in a row.
>> You can find more on this topic in the chapter on permutation without repetition. <<
### 1.2 permutation with repetition
A Permutation with repetition is an arrangement of \ (n \) objects, some of which are indistinguishable.
We have already learned that there are \ (n \) ways to distribute \ (n \) distinguishable (!) Objects to \ (n \) places.
However, if exactly \ (k \) objects are identical, then these can be exchanged in their places without resulting in a new order. In this way, exactly \ (k! \) Arrangements are the same.
The number of permutations of \ (n \) objects, of which \ (k \) are identical, is calculated as follows
\ [\ frac {n!} {k!} \]
If there is not just one, but \ (s \) groups, each with \ (k_1, k_2 \ cdot \ text {...} \ cdot k_s \) identical objects, the formula reads
\ [\ frac {n!} {k_1! \ cdot k_2! \ cdot \ text {...} \ cdot k_s!} \]
example
There are three blue and two red balls in an urn. How many ways are there to arrange the balls in a row?
solution
\ [\ frac {5!} {3! \ cdot 2!} = \ frac {5 \ times 4 \ times 3 \ times 2 \ times 1} {(3 \ times 2 \ times 1) \ times (2 \ times 1)} = 10 \]
Answer: There are 10 ways to arrange three blue and two red balls in a row.
>> You can find more on this topic in the chapter on permutation with repetition. <<
### 2. Variations
• \ (k (i.e. only one sample - i.e. \ (k \) elements of the basic set \ (n \) - is considered)
• The order of the elements is taken into account
### 2.1 Variation without repetition
At a Variation without repetition \ (k \) are selected from \ (n \) objects taking into account the order, whereby each object can only be selected once.
There are \ (n \) placement options for the first object. For the second object there are (n-1) possibilities, for the third object (n-2) .... and for the last object there are still (n-k + 1) possibilities.
\ [n \ cdot (n-1) \ cdot (n-2) \ cdot \ text {...} \ cdot (n-k + 1) = \ frac {n!} {(n-k)!} \]
example
There are five different colored balls in an urn. Three balls should be drawn without replacing (= without repetition) and observing the order. How many options are there?
solution
\ [\ frac {5!} {(5-3)!} = \ frac {5!} {2!} = \ frac {5 \ times 4 \ times 3 \ times 2 \ times 1} {2 \ times 1 } = 5 \ times 4 \ times 3 = 60 \]
Answer: There are 60 ways to draw 3 out of 5 balls without replacing them, observing the order.
>> You can find more on this topic in the chapter on variation without repetition. <<
### 2.2 Variation with repetition
At a Variation with repetition \ (k \) are selected from \ (n \) objects taking into account the order, whereby objects can also be selected multiple times.
There are \ (n \) options for the first object. Since objects can be selected several times, there are also options for the second, third and kth object. The following applies accordingly:
\ [n \ cdot n \ cdot \ text {...} \ cdot n = n ^ k \]
example
There are five different colored balls in an urn. Three balls should be drawn with replacement (= with repetition) and observing the order. How many options are there?
solution
\ [5 ^ 3 = 5 \ times 5 \ times 5 = 125 \]
Answer: There are 125 possibilities to move 3 out of 5 balls with replacement, observing the order.
>> You can find more on this topic in the chapter on variation with repetition. <<
### 3. Combinations
• \ (k (i.e. only one sample - i.e. \ (k \) elements of the basic set \ (n \) - is considered)
• The order of the elements is not taken into account
### 3.1 Combination without repetition
At a Combination without repetition \ (k \) are selected from \ (n \) objects regardless of the order, whereby each object can only be selected once.
The only difference between a variation without repetition and a combination without repetition is the fact that in the combination - in contrast to the variation - the order of the objects does not matter.
We already know the formula for the variation without repetition
\ [\ frac {n!} {(n-k)!} \]
The \ (k \) selected objects can be arranged in \ (k! \) Different ways. However, since the order in the combination is irrelevant, the formula reads accordingly
\ [\ frac {n!} {(n-k)! \ cdot k!} = {n \ choose k} \]
\ ({N \ choose k} \) is also called the binomial coefficient.
example
There are five different colored balls in an urn. Three balls should be drawn without replacing (= without repetition) and without considering the order. How many options are there?
solution
\ [{5 \ choose 3} = 10 \]
Answer: There are 10 ways to move 3 out of 5 balls without replacing them, regardless of the order.
>> You can find more on this topic in the chapter on combining without repetition. <<
### 3.2 Combination with repetition
At a Combination with repetition \ (k \) are selected from \ (n \) objects regardless of the order, whereby objects can also be selected multiple times.
The only difference between a combination without repetition and a combination with repetition is the fact that the combination with repetition allows the objects to be selected more than once.
We already know the formula for the combination without repetition
\ [\ frac {n!} {(n-k)! \ cdot k!} = {n \ choose k} \]
By modifying the numerator and the denominator, we finally arrive at the formula for a combination with repetition
\ [\ frac {(n + k-1)!} {(n-1)! \ cdot k!} = {n + k-1 \ choose k} \]
example
There are five different colored balls in an urn. Three balls should be drawn with replacement (= with repetition) and regardless of the order. How many options are there?
solution
\ [{5 + 3-1 \ choose 3} = {7 \ choose 3} = 35 \]
Answer: There are 35 possibilities to move 3 out of 5 balls with replacement, regardless of the order.
>> You can find more on this topic in the chapter on combination with repetition. <<
If you are dealing with combinatorics for the first time, you must first deal with the above formula and practice it. In an exam, however, you not only have to master the formula, but also know when to use which formula. (Very few teachers will write which case is next to the assignment.)
In combinatorial tasks, two questions arise:
1. What formula do I have to use in this exercise?
2. What's the formula
When it comes to the second question, only one thing helps: PRACTICE! (... and memorize!).
However, very few pupils and students have problems memorizing the formulas. It is more difficult to determine which formula to use for a specific task.
By systematic approach you get to the right formula quickly ...
Are all elements of the basic set relevant to the task?
(yes) -> permutation
Are all elements distinguishable from one another?
(yes) -> Permutation without repetition
(no) -> Permutation with repetition
(no) -> variation or combination
Does the order matter?
(yes) -> variation
Are all elements distinguishable from one another?
(in the urn model: without replacing?)
(yes) -> Variation without repetition
(no) -> Variation with repetition
(no) -> combination
Are all elements distinguishable from one another?
(in the urn model: without replacing?)
(yes) -> Combination without repetition
(no) -> Combination with repetition<|endoftext|>
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Practice the problems of Math in Focus Grade 4 Workbook Answer Key Chapter 8 Practice 2 Adding Decimals to score better marks in the exam.
Fill in the blanks. Write each sum as a decimal.
Example
0.02 + 0.04 = 2 hundredths + 4 hundredths
= 6 hundredths
= 0.06
Question 1.
0.03 + 0.07 = ___ hundredths + ____ hundredths
= ____ hundreths
= _____
0.03 + 0.07 = 0.10.
Explanation:
0.03 + 0.07 = 3 hundredths + 7 hundredths
= 10 hundreths
= 0.10.
Question 2.
0.06 + 0.08 = ___ hundredths + ____ hundredths
= ____ hundreths
= _____
0.06 + 0.08 = 0.14.
Explanation:
0.06 + 0.08 = 6 hundredths + 8 hundredths
= 14 hundreths.
= 0.14.
Question 3.
0.09 + 0.05 = ___ hundredths + ____ hundredths
= ____ hundreths
= _____
0.09 + 0.05 = 0.14.
Explanation:
0.09 + 0.05 = 9 hundredths + 5 hundredths
= 14 hundreths.
= 0.14.
Fill in the blanks.
Question 4.
4 hundredths + 7 hundredths
= ___ hundredths
Regroup the hundredths.
___ hundredths = ___ tenth ___ hundredth
Step 2
3 tenths + 8 tenths + ___ tenth = ___ tenths
Regroup the tenths.
___ tenths = ___ one and ___ tenths
Step 3
2 ones + 0 ones + ___ one = ___ ones
So, 2.34 + 0.87 = ___.
2.34 + 0.87 = 3.21.
Explanation:
Step 1:
4 hundredths + 7 hundredths
= 11 hundredths.
Regroup the hundredths.
11 hundredths = 1 tenth 1 hundredth.
Step 2:
3 tenths + 8 tenths + 1 tenth = 12 tenths.
Regroup the tenths.
12 tenths = 1 one and 2 tenths.
Step 3:
2 ones + 0 ones + 1 one = 3 ones.
Question 5.
0.02 + 0.35 = 0.37.
Explanation:
Step 1:
2 hundredths + 5 hundredths
= 7 hundredths.
Step 2:
0 tenths + 3 tenths = 3 tenths.
Step 3:
0 ones + 0 ones = 0 ones.
Question 6.
0.06 + 0.46 = 0.52.
Explanation:
Step 1:
6 hundredths + 6 hundredths
= 12 hundredths.
Regroup the hundredths.
12 hundredths = 1 tenth 2 hundredth.
Step 2:
0 tenths + 4 tenths + 1 tenth = 5 tenths.
Step 3:
0 ones + 0 onesĀ = 0 ones
Write in vertical form. Then add.
Question 7.
$0.57 +$0.29 = $_____ Answer:$0.57 + $0.29 =$0.86.
Explanation:
Question 8.
3.6 + 0.54 = ____
3.6 + 0.54 = 4.14.
Explanation:
Question 9.
$0.78 +$0.88 = $____ Answer:$0.78 + $0.88 =$1.66.
Explanation:
Question 10.
7.25 + 1.78 = ___
7.25 + 1.78 = 9.03.
Explanation:
Derek hops two steps on each number line. Which decimal does he land on? Write the correct decimal in each box.
Example
Question 11.
Explanation:
10.0 + 1.26 + 2.57
=11.26 + 2.57
= 13.83.
Question 12.
Explanation:
50.0 + 2.69 + 1.83
= 52.69 + 1.83
= 54.52.
Question 13.<|endoftext|>
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0
What is the greatest common factor of 8 and 11?
Updated: 4/28/2022
Wiki User
8y ago
The GCF of 8 and 11 is 1.
The factors of 8 are 1, 2, 4, and 8.
The factors of 11, which is a Prime number, are 1 and 11.
The only common factor is 1, so it is the greatest common factor, which means the 8 and 11 are relatively prime.
You can also determine this by looking at the prime factors.
The prime factors of 8 are 2, 2, and 2.
The prime factors of 11 are 11.
There are no prime factors in common, so the numbers are relatively prime, which means their only common factor is 1.
Wiki User
8y ago
Wiki User
5y ago
The GCF is 1.
The GCF is 1.
Earn +20 pts
Q: What is the greatest common factor of 8 and 11?
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Related questions
What is the greatest common factor of 4 8 13 and 11?
The greatest common factor (GCF) is: 1
What is the greatest common factor of 33 and 88?
The greatest common factor of 88 and 33 is 11. 11 x 3= 33, 11 x 8= 88
What are the greatest common factor of 8 and 32?
The greatest common factor of 8 and 32 is 8.
What are the common factors and greatest common factor of 11?
There is neither a greatest common factor nor common factors of a single number, such as 11, because there cannot be any form of common factor without two or more numbers to compare. Common factors are factors that the numbers being compared have in common. The greatest common factor is the largest factor that all the numbers being compared have in common. Thus, since there are not two or more numbers to compare, there are neither common factors nor a greatest common factor.The factors of 11 are 1 and 11.The prime factor of 11 is 11; it is a prime number.Examples:The common factors of 11 and 44 are 1 and 11; the greatest common factor is 11.The common factors of 11 and 56 are only 1; the greatest common factor is 1.The common factors of 11 and 71 are only 1; the greatest common factor is 1.The common factors of 11 and 132 are 1 and 11; the greatest common factor is 11.The common factors of 8, 11, and 33 are only 1; the greatest common factor is 1.* because 11 is a prime number there is no common factor just it self !
The GCF is 1.
The greatest common factor of 32 and 24?
The Greatest Common Factor (GCF) is: 8
What is the greatest common factor of 8 and 400?
The greatest common factor of 8 and 400 is 8.
What is the greatest common factor of 1624 and 8?
The greatest common factor of 1624 , 8 = 8
What is the Greatest common factor of 8 and 1000?
The Greatest Common Factor (GCF) is: 8.
What is the greatest common factor for 8 and 11?
It is: 1 because 11 is a prime number
What is the greatest common factor of 8 and 68?
The Greatest Common Factor (GCF) of 8 and 68 is: 4<|endoftext|>
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# Number Representation and Arithmetic in Various Numeral Systems
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1 1 Number Representation and Arithmetic in Various Numeral Systems Computer Organization and Assembly Language Programming Adapted by Yousef Shajrawi, licensed by Huong Nguyen under the Creative Commons License A numeral system is a collection of symbols used to represent small numbers, together with a system of rules for representing larger numbers. Each numeral system uses a set of digits. The number of various unique digits, including zero, that a numeral system uses to represent numbers is called base or radix. Base - b numeral system b basic symbols (or digits) corresponding to natural numbers between 0 and b 1 are used in the representation of numbers. To generate the rest of the numerals, the position of the symbol in the figure is used. The symbol in the last position has its own value, and as it moves to the left its value is multiplied by b. We write a number in the numeral system of base b by expressing it in the form N(b), with n+1 digit for integer and m digits for fractional part, represents the sum: in the decimal system. Decimal, Binary, Octal and Hexadecimal are common used numeral system. The decimal system has ten as its base. It is the most widely used numeral system, because humans have four fingers and a thumb on each hand, giving total of ten digit over both hand. Modern computers use transistors that represent two states with either high or low voltages. Binary digits are arranged in groups to aid in processing, and to make the binary numbers shorter and more manageable for humans. Thus base 16 (hexadecimal) is commonly used as shorthand. Base 8 (octal) has also been used for this purpose. 1
2 2 Decimal System Decimal notation is the writing of numbers in the base-ten numeral system, which uses various symbols (called digits) for no more than ten distinct values (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) to represent any number, no matter how large. These digits are often used with a decimal separator which indicates the start of a fractional part, and with one of the sign symbols + (positive) or (negative) in front of the numerals to indicate sign. Decimal system is a place-value system. This means that the place or location where you put a numeral determines its corresponding numerical value. A two in the one's place means two times one or two. A two in the one-thousand's place means two times one thousand or two thousand. The place values increase from right to left. The first place just before the decimal point is the one's place, the second place or next place to the left is the ten's place, the third place is the hundred's place, and so on. The place-value of the place immediately to the left of the "decimal" point is one in all place-value number systems. The place-value of any place to the left of the one's place is a whole number computed from a product (multiplication) in which the base of the number system is repeated as a factor one less number of times than the position of the place. For example, 5246 can be expressed like in the following expressions 2
3 3 Binary System The binary number system is base 2 and therefore requires only two digits, 0 and 1. The binary system is useful for computer programmers, because it can be used to represent the digital on/off method in which computer chips and memory work. A binary number can be represented by any sequence of bits (binary digits), which in turn may be represented by any mechanism capable of being in two mutually exclusive states. Counting in binary is similar to counting in any other number system. Beginning with a single digit, counting proceeds through each symbol, in increasing order. Decimal counting uses the symbols 0 through 9, while binary only uses the symbols 0 and 1. When the symbols for the first digit are exhausted, the next-higher digit (to the left) is incremented, and counting starts over at 0. A single bit can represent one of two values, 0 or 1.Binary numbers are convertible to decimal numbers. Example: 3
4 4 Hexadecimal System The hexadecimal system is base 16. Therefore, it requires 16 digits. The digits 0 through 9 are used, along with the letters A through F, which represent the decimal values 10 through 15. Here is an example of a hexadecimal number and its decimal equivalent: The hexadecimal system (often called the hex system) is useful in computer work because it is based on powers of 2. Each digit in the hex system is equivalent to a four-digit binary number. Table below shows some hex/decimal/binary equivalents. Hexadecimal Digit Decimal Equivalent Binary Equivalent A B C D E F F FF
5 5 Octal System Binary is also easily converted to the octal numeral system, since octal uses a radix of 8, which is a power of two (namely, 2 3, so it takes exactly three binary digits to represent an octal digit). The correspondence between octal and binary numerals is the same as for the first eight digits of hexadecimal in the table above. Binary 000 is equivalent to the octal digit 0, binary 111 is equivalent to octal 7, and so forth. Converting from octal to binary proceeds in the same fashion as it does for hexadecimal: And from octal to decimal: 5
6 6 Converting from decimal to base b To convert a decimal fraction to another base, say base b, you split it into an integer and a fractional part. Then divide the integer by b repeatedly to get each digit as a remainder. Namely, with value of integer part = d n 1 d n 2...d 2 d 1 d 0(10), first divide value by b the remainder is the least significant digit a 0. Divide the result by b, the remainder is a 1.Continue this process until the result is zero, giving the most significant digit, a n 1. Let's convert (10) to hexadecimal: 6
7 7 Unsigned Integers Unsigned integers are represented by a fixed number of bits (typically 8, 16, 32, and/or 64) With 8 bits, can be represented; With 16 bits, can be represented; In general, an unsigned integer containing n bits can have a value between 0 and 2 n 1 If an operation on bytes has a result outside this range, it will cause an overflow Signed Integers The binary representation discussed above is a standard code for storing unsigned integer numbers. However, most computer applications use signed integers as well; i.e. the integers that may be either positive or negative. There are three common methods to represent signed numbers. In all three methods, positive numbers are represented in the same way, whereas negative numbers are represented differently. Moreover, in all three methods, the most significant bit indicates whether the number is positive (msb=0) or negative (msb=1). Sign-Magnitude The most significant bit is the sign (0 positive, 1 negative). The magnitude (which is the absolute value of the number) is stored in the remaining bits, as an unsigned value. For example, an 8-bit signed magnitude representation of decimal 13 is while 13 is Note that using 8-bit signed magnitude, one can represent integers in the range 127 ( ) to +127 ( ). Sign-magnitude has one peculiarity, however. The integer 0 can be represented in two ways: = +0 and = 0. One s Complement In this method, positive numbers are represented the same way as in sign-magnitude. Given a positive number (msb=0), we get the corresponding negative number by complementing each bit (1 becomes 0, and 0 becomes 1). This works also in the other direction (from negative to positive). For example, an 8-bit one s complement representation of decimal 13 is while 13 is Using 8-bit one s complement, we can represent integers in the range 127 ( ) to +127 ( ). Like sign-magnitude, the number 0 has two representations: = +0 and = 0. 7
8 8 Two's Complement This is the most common representation scheme. Positive numbers are represented the same way as in the other methods. Given a positive number (msb=0), the corresponding negative number is obtained by the following steps: Complement each bit (getting in fact a 1 s complement number) Add one to the complemented number. This works also in the other direction (from negative to positive). Example: = , = Example : Performing two's complement on the decimal 42 to get -42 using 8-bit representation: 42= Convert to binary Complement the bits Add 1 to the complement Result is -42 in two's complement Notice that in two s complement, there is a single representation for 0. The range of numbers that can be represented in two s complement is asymmetric. For example, using 8-bit two s complement, we can represent integers in the range 128 ( ) to +127 ( ). If you try to convert -128 to the corresponding positive number, you end up getting -128 again. In other words, the lowest negative number has no complement. 8
10 10 Like addition, the advantage of using two's complement is the elimination of examining the signs of the operands to determine if addition or subtraction is needed. For example, subtracting 5 from 15 is really adding 5 to 15, but this is hidden by the two's-complement representation: (borrow) (15) ( 5) =========== (20) Another example is a subtraction operation where the result is negative: = 20: (borrow) (15) (35) =========== ( 20) Overflow is detected the same way as for addition, by examining the two leftmost (most significant) bits of the borrows; overflow has occurred if they are different. 10
11 11 Multiplication and Division of Integers Binary Multiplication Multiplication in the binary system works the same way as in the decimal system: 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1, and no carry or borrow bits Example: = = 23(base 10) = 3(base 10) carries = 69(base 10) Another Example: Binary division Follow the same rules as in decimal division. 11
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# Math in Focus Grade 7 Chapter 10 Lesson 10.2 Answer Key Finding Probability of Events
This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Lesson 10.2 Finding Probability of Events detailed solutions for the textbook questions.
## Math in Focus Grade 7 Course 2 B Chapter 10 Lesson 10.2 Answer Key Finding Probability of Events
### Math in Focus Grade 7 Chapter 10 Lesson 10.2 Guided Practice Answer Key
Copy and complete. Solve.
Question 1.
You roll a fair number die.
a) Find the probability of getting a four.
There are outcomes when you roll a number die. All the outcomes are equally likely.
So, the probability of getting a four is .
Answer:6 outcomes; 1/6.
Explanation:
There are 6 possible outcomes: {1, 2, 3, 4, 5, 6}
For each number, we throw a dice once then the probability of getting each number is 1/6.
The probability of number 1 is 1/6
The probability of number 2 is 1/6
The probability of number 3 is 1/6
The probability of number 4 is 1/6
The probability of number 5 is 1/6
The probability of number 6 is 1/6.
b) Find the probability of getting a seven.
It is impossible to get a seven when you roll a standard number die.
So, the probability of getting a seven is .
Answer: 0
Explanation:
Yes, it is impossible to get seven when you roll a standard number die. In a rolling dice there will be no 7. Therefore, the probability of getting seven is 0.
Solve.
Question 2.
When you spin the spinner, what is the probability that the arrow will point to a number?
Answer: 0
Explanation:
Observe the spinner carefully,
In the spinner, all the sections are having alphabets only.
There are no numbers in the spinner. The arrow cannot show the number when we spin.
Complete.
Question 3.
Max has 4 short-sleeved shirts and 5 long-sleeved shirts in his closet. X is the event of Max randomly choosing a long-sleeved shirt. Find P(X). Express the probability as a fraction.
Event X has favorable outcomes.
P(X) = $$\frac{\text { Number of outcomes favorable to event } X}{\text { Total number of equally likely outcomes }}$$. Use the formula.
=
There are long-sleeved shirts out of 9 shirts.
Answer:
The number of short-sleeved shirts=4
The number of long-sleeved shirts=5
The total number of shirts=4+5=9
Event X be choosing randomly long-sleeved shirts.
Event X=5
There are 5 long-sleeved shirts.
Here we need to find out the probability.
Formula:
The most common formula used to determine the likelihood of an event is given below:
probability=number of favourable events/total number of outcomes
P(X)=n(X)/n(S)
Where P(X)=probability
n(X)=number of favourable events
n(S)=total number of outcomes (shirts).
P(X)=5/9.
Therefore, the probability is 5/9.
Solve.
Question 4.
A box contains 28 pink ribbons and 12 green ribbons. You randomly take a ribbon from the box without looking. Find the probability of picking a pink ribbon. Express the probability as a percent.
Answer: 70%
Explanation:
The number of pink ribbons is there in a box=28
The number of green ribbons is there in a box=12
The total number of ribbons=28+12=40
The probability of finding of picking the pink ribbon=X
P(X)=n(X)/n(R)
P(X)=28/40
p(X)=0.7
Here asked that percentage. So we need to write the answer in percentage.
To find the probability of percentage:
Finally, take the answer you got and move the decimal point to the right two places or multiply the decimal by 100. Your answer will be the per cent probability that the desired outcome will take place.
Probability of percentage=0.7 x 100 = 70%
Therefore, the probability of percentage is 70%.
Question 5.
Ten cards have the following numbers printed on them: 3, 6, 9, 11, 19, 27, 35, 39, 40, and 45. A card is randomly drawn from the ten cards. Let W be the event of getting a number that is an odd number greater than 20. Let V be the event of getting a prime number.
a) List all the outcomes favourable to events W and V.
Answer:
Odd numbers: Odd numbers are the numbers that cannot be divided by 2 evenly. It cannot be divided into two separate integers evenly. If we divide an odd number by 2, then it will leave a remainder. The examples of odd numbers are 1, 3, 5, 7, etc.
Event W is the getting a number that is an odd number greater than 20
W={27, 35, 39, 45}
Prime numbers: A prime number is an integer, or whole number, greater than 1 that is only divisible by 1 and itself. In other words, a prime number only has two factors, 1 and itself.
Event V is the getting a prime number.
V= {3, 11, 9}
b) Draw a Venn diagram for the sample space and the two events. Place all possible outcomes in the Venn diagram.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
Event W is getting a number that is an odd number greater than 20
W={27, 35, 39, 45}
Event V is getting a prime number.
V= {3, 11, 9}
representation of Venn diagram:
c) Are events Wand V mutually exclusive? Explain.
Answer: Yes, W and V events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).
d) Find P(W) and P(V).
Answer:2/5; 3/10
The above-given numbers: 3, 6, 9, 11, 19, 27, 35, 39, 40, and 45.
Event W is getting a number that is an odd number greater than 20
W={27, 35, 39, 45}
P(W)=N(W)/N(T)
P(W)=4/10
P9W)=2/5
Event V is getting a prime number.
V= {3, 11, 9}
P(V)=N(V)/N(T)
P(V)=3/10
Question 6.
A letter is selected at random from the state name RHODE ISLAND. Let C be the event of getting a consonant. Let H be the event of getting a letter that comes after H in the alphabet.
a) Draw a Venn diagram for the sample space and the two events. Place all possible outcomes in the Venn diagram.
Answer:
Explanation:
Event C is getting a consonant.
Consonants are R, H, D, S, L, N
Vowels are A, E, I, O
Event H be the getting a letter that comes after H in the alphabet.
H after the alphabets are I, O
Note: A ∩ B = B ∩ A
b) Are events C and H mutually exclusive? Explain.
Answer: No, C and H are not mutually exclusive because they overlapped each other.
Two sets are said to be joint sets when they have at least one common element.
c) Find P(C) and P(H).
Answer:7/11;6/11
Explanation:
The above-given word: RHODE ISLAND
The total number of letters in a word=11
The number of consonants= 7
P(C)=N(C)/N(T)
P(C)=7/11
Event H be the getting a letter that comes after H in the alphabet.
P(H)=N(H)/N(T)
P(H)=6/11.
Question 7.
You randomly choose a month from the twelve months in a year. Let A be the event of randomly choosing a month that has the letter a in its name.
a) Draw a Venn diagram to represent events A and A’. Give the meaning of event A’, the complement of event A.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
A={January, February, March, April, May, August]
A’={June, July, September, October, November, December}
A’ is the complement of A. This represents elements that are neither in set A.
A’ represents the months are having without letter ‘a’.
b) What outcomes are favourable to event A’?
Answer:
A’={June, July, September, October, November, December}
A’ is the complement of A. This represents elements that are neither in set A.
c) Find P(A) and P(A’).
Answer:
Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen. The meaning of probability is basically the extent to which something is likely to happen. This is the basic probability theory, which is also used in the probability distribution, where you will learn the possibility of outcomes for a random experiment. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
The formula for probability: The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
probability of an event to happen P(E)=number of favourable events/total number of outcomes.
A be the event of randomly choosing a month that has the letter ‘a’ in its name.
P(A)=N(A)/N(T)
P(A)=6/12
P(A)=1/2
A’ is the event having no letter in the month that is choosen.
P(A’)=N(A’)/N(T)
P(A’)=6/12
P(A’)=1/2.
Complete.
Question 8.
40% of the apples in an orchard are green and the rest of the apples are red. 5% of the red apples are rotten.
a) Copy the Venn diagram to represent the information.
Answer:
The above-given question:
Assume the total apples percentage=100%
The percentage of green apples=40%
The percentage of red apples=100-40=60%
The percentage of red apples are rotten=5%
According to the Venn diagram:
Therefore, the red apples are 60%
the green apples are 40%
b) If you pick an apple at random in the orchard, what is the probability the apple you pick is red that is not rotten? Give your answer as a decimal and as a percent.
Of all the red apples, of them are not rotten.
of 60% = • 0.60 Write percents as decimals and multiply.
= Simplify.
= % Write as a percent.
The probability of picking a red apple that is not rotten is .
Answer: 57%
Of all the red apples 95 of them are not rotten.
In simplification:
95 of 60%
=95*60/100
=57%
Solve.
Question 9.
Among the 200 jellybeans in a bag, 3 out of every 5 are blue jellybeans. The blue jellybeans consist of light blue ones and dark blue ones in the ratio 2:1.
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
The total number of jellybeans=200
In that every 5 jellybeans we have 3 blue jellybeans. from this, we can write the number of jellybeans.
Divide 200/5=40
And multiply 3 by 40.
40*3=120.
Therefore, the number of blue jellybeans=120.
The rest of the bags means which are not blue=200-120=80.
b) What fraction of the jellybeans are light blue?
Answer:2/5
Explanation:
The ratio is already given=2:1
The light blue jellybeans ratio is 2
The dark blue jellybeans ratio is 1
We are taking 5 bags every time in that every 3 bags is blue colour only.
According to the question, the light blue was asked.
So the light blue ratio is 2 and the total is 5.
Therefore, the fraction of the light blue jellybeans is 2/5.
c) If you pick a jellybean randomly from the bag, what is the probability that a jellybean that is light blue is selected? Give your answer as a decimal.
Answer:0.4
Explanation:
The ratio is already given=2:1
The light blue jellybeans ratio is 2
The dark blue jellybeans ratio is 1
We are taking 5 bags every time in that every 3 bags is blue colour only.
According to the question, the light blue was asked.
So the light blue ratio is 2 and the total is 5.
Therefore, the fraction of the light blue jellybeans is 2/5.
P(A)=N(A)/N(T)
P(A)=2/5
P(A)=0.4
Therefore, the probability is 0.4
### Math in Focus Course 2B Practice 10.2 Answer Key
Solve.
Question 1.
You roll a fair number die with faces labeled 1 to 6.
a) What is the probability of rolling an odd number?
Answer:
Probability when a dice is rolled:
total number of possible outcomes=6 (1, 2, 3, 4, 5, 6)
Probability of having 1 on the roll-up of dice:
Occurrence of 1 on top of dice/ Total number of possible outcomes of dice
= 1/6
Each of the outcomes of the dice has an equal probability of 1/6. Each outcome is equally likely to come on the top when the dice is rolled. Hence, the rolling of dice can be considered to be a fair probability scenario.
The odd numbers present on the die is 1, 3, 5
Therefore, out of the 6 total occurrences, the possible outcomes are 3.
Probability of obtaining an odd number = 3/6 = 1/2.
b) What is the probability of rolling a number less than 3?
Answer: 1/3
The probability of rolling a number less than 3 (so, 1 or 2) is 2 out of 6, or 1/3
Therefore, out of the 6 total occurrences, the possible outcomes are 2
Probability of obtaining a number less than 3=2/6=1/3.
c) What is the probability of rolling a prime number?
Answer:
total numbers that can occur=1, 2, 3, 4, 5, 6
Prime numbers that can occur in a roll=2, 3, 5
Total sample space n=6
Total favourable cases= m=3
P(prime number)=m/n=3/6
P(prime number)=1/2
therefore, the probability of rolling a prime number=1/2.
d) What is the probability of rolling a number greater than 1?
Answer:5/6
The total numbers that can occur=1, 2, 3, 4, 5, 6
The numbers that can occur in a roll that are greater than 1=2, 3, 4, 5, 6
The total sample space n =6
Total favourable cases= m = 5
P(greater than 1)=m/n=5/6
Therefore, the probability of rolling a dice greater than 1 is 5/6.
Question 2.
Abigail randomly chooses a disk from 6 green, 4 black, 2 red, and 2 white disks of the same size and shape.
a) What is the probability of getting a red disk?
Answer:
The total number of disks= n =6+4+2+2=14
The favourable cases=m=2
P(red disk)=m/n=2/14
P(red disk)=1/7
Therefore, the probability of getting a red disk=1/7.
b) What is the probability of not getting a green or white disk?
Answer:
The total number of disks= n =6+4+2+2=14
The number of disks that are green=6
The number of disks that are white=2
The disks that are not getting=m =6+2=8
P(not getting)=m/n
P(not getting)=8/14
P(not getting)=4/7
therefore, the probability of not getting green and white disks are 4/7.
c) What is the probability of getting a black disk?
Answer:
The total number of disks= n =6+4+2+2=14
The favourable cases=m=4
P(black disk)=m/n=4/14
P(black disk)=2/7
Therefore, the probability of getting a black disk=2/7.
Question 3.
A letter is randomly chosen from the word MATHEMATICS. What is the probability of choosing letter M?
Answer:
The above-given word: MATHEMATICS
The total number of letters = n = 11
The letter we need to select is ‘M’.
The favourable cases = m = 2
because we can select M two times.
P(letter M)=m/n
P(letter M)=2/11
Therefore, the probability of choosing letter M is 2/11.
Question 4.
There are 6 red marbles and 10 white marbles in a bag. What is probability of randomly choosing a white marble from the bag?
Answer:5/8
Explanation:
The number of red marbles=6
The number of white marbles=10
The total number of marbles=n=16
The favourable cases=m=10
The probability of choosing white marbles=P
P(white marbles)=m/n
P(white marbles)=10/16
P(white marbles)=5/8
Therefore, the probability of choosing white marbles are 5/8.
Question 5.
Numbers made up of two digits are formed using the digits 2, 3, and 4 with no repeating digits.
a) List all possible outcomes.
Answer:
The above-given numbers:2, 3, and 4
The possible outcomes are 23, 24, 32, 34, 42, 43
b) Find the probability of randomly forming a number greater than 32.
Answer:1/2
Explanation:
sample space={3, 4}
The total number= n= 2
According to the given numbers above, numbers obtained greater than 32 randomly=m=1 (34)
P(greater than 32)=m/n
P(greater than 32)=1/2.
Therefore, the probability of randomly forming a number greater than 32 is 1/2.
c) Find the probability of randomly forming a number divisible by 4.
Answer:1/3
The two-digit numbers formed using the given numbers 2, 3, and 4
The possible outcomes are 23, 24, 32, 34, 42, 43
Sample space={23, 24, 32, 34, 42, 43}
The total number of outcomes= n = 6
The total favourable cases= m = 2
Those cases are 24, 32 which are divisible by 4.
P(divisible by 4)=m/n
P(divisible by 4)=2/3
P(divisible by 4)=1/3.
Therefore, 1/3 is the probability of randomly forming a number divisible by 4.
Question 6.
Jack picks a letter randomly from the following list: h, i, m, o, p, q, r, t, u, and x.
The event V occurs when Jack picks a vowel.
a) Draw a Venn diagram to represent the information. Explain the meaning of the complement of event V.
Answer:
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done. Complementary events occur when there are only two outcomes, for example clearing an exam or not clearing an exam. The complement means the exact opposite of an event.
Definition: For any event A, there exists another event A‘ which shows the remaining elements of the sample space S. A’ denotes complementary event of A.
: A’ = S – A.
Event A and A’ are mutually exclusive and exhaustive.
Consider the example of tossing a coin. Let P(E) denote the probability of getting a tail when a coin is tossed.
Then the probability of getting ahead is denoted by P(E’)
According to the definition, the complement of event V is the exact opposite of event V.
In complement of V (V’) all the letters are consonants, not vowels.
b) Find P(V) and P(V’).
Answer:
The number of consonants is nothing but the complement of V
The total number of letters=10
The number of vowels= V =3 (i, o, u)
P(V)=m/n
P(V)=3/10
Therefore, the probability of P(V) is 3/10
The number of consonants (V’)= 7 (h, m, p, q, r, t, x)
P(V’)=7/10
Thus the probability of the complement of V is 7/10.
Question 7.
A number is randomly selected from 1 to 20. X is the event of selecting a number divisible by 4. Y is the event of getting a prime number.
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
Event X is selecting a number divisible by 4.
Event Y is getting a prime number.
N(X)={4, 8, 12, 16, 20}
N(Y)={2, 3, 5, 7, 11, 13, 17, 19}
b) Are events X and Y mutually exclusive? Explain.
Answer: Yes, X and Y events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).
c) Find P(X) and P(Y).
Answer:
sample space X={4, 8, 12, 16, 20}
sample space Y={2, 3, 5, 7, 11, 13, 17, 19}
The total numbers=n=20
The number of numbers divisible by 4=m=5
P(divisible by 4)=m/n
P(divisible by 4)=5/20
P(divisible by 4)=1/4
Therefore the probability of the numbers divisible by 4 is 1/4.
The number of prime numbers=8
P(prime numbers)=8/20
P(prime numbers)=2/5
Therefore, the probability of getting prime numbers is 2/5.
Question 8.
A dodecahedron number die has 12 faces. Each face is printed with one of the numbers from 1 to 12. Suppose you roll a fair dodecahedron number die and record the value on the top face. Let A be the event of rolling a number that is a multiple of 3. Let B be the event of rolling a number that is a multiple of 4.
a) Draw a Venn diagram to represent the information.
Answer:
The numbers will get when we roll a dice=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
Probability of having 1 on the roll-up of dice:
Occurrence of 1 on top of dice/ Total number of possible outcomes of dice
= 1/12
Each of the outcomes of the dice has an equal probability of 1/12. Each outcome is equally likely to come on the top when the dice is rolled. Hence, the rolling of dice can be considered to be a fair probability scenario.
The possible outcome for 1 is 1/12
The possible outcome for 2 is 1/12
E(A) is the rolling number that is a multiple of 3.
E(B) in the event of rolling a number that is a multiple of 4.
The numbers multiple of 3 is 3, 6, 9, 12
The numbers multiple of 4 is 4, 8, 12
b) From the Venn diagram, tell whether events A and B are mutually exclusive. Explain your answer.
Answer: Yes, A and B events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).
c) Find P(A) and P(B).
Answer:
E(A) is the rolling number that is a multiple of 3.
E(B) in the event of rolling a number that is a multiple of 4.
The numbers multiple of 3 is 3, 6, 9, 12
The numbers multiple of 4 is 4, 8, 12
P(A)=E(A)/N(T)
P(A) is the probability of numbers that are multiples of 3
E(A)= the number of multiples that occurred in an event.
N(T) is the total number that is in the dice.
P(A)=4/12
P(A)=1/3
Therefore, the probability is 1/3.
P(B)=E(B)/N(T)
P(B) is the probability of numbers that are multiples of 4
E(B)= the number of multiples that occurred in an event.
N(T) is the total number that is in the dice.
P(B)=3/12
P(B)=1/4
Therefore, the probability is 1/4.
Question 9.
This year, some students in the Drama Club have the same first names. Name tags for the students are shown below.
One of the name tags is selected at random. Event E occurs when the name has the letter e. Event J occurs when the name tag is .
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagram: Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operation performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
E(E) is the name having with letter ‘e’.
E(J) is when the name tag occurred ‘John’.
b) List all the types of outcomes of event E’, the complement of event E.
Answer: John, Mary
Explanation:
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done. Complementary events occur when there are only two outcomes, for example clearing an exam or not clearing an exam. The complement means the exact opposite of an event.
Definition: For any event A, there exists another event A‘ which shows the remaining elements of the sample space S. A’ denotes complementary event of A.
: A’ = S – A.
Event A and A’ are mutually exclusive and exhaustive.
Consider the example of tossing a coin. Let P(E) denote the probability of getting a tail when a coin is tossed.
Then the probability of getting ahead is denoted by P(E’)
According to the definition, the complement of an event E is the exact opposite of event E.
In complement of E (E’) in a word not having letter ‘e’.
c) Are events E and J mutually exclusive? Explain your answer.
Answer: Yes, E and J events are mutually exclusive. Because they are disjoint sets.
Explanation:
Mutually exclusive: In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. In other words, mutually exclusive events are called disjoint events. If two events are considered disjoint events, then the probability of both events occurring at the same time will be zero.
If A and B are the two events, then the probability of disjoint of event A and B is written by:
Probability of Disjoint (or) Mutually Exclusive Event = P ( A and B) = 0.
If A and B are mutually exclusive events then its probability is given by P(A Or B) or P (A U B).
d) Find P(E), P(E’), and P(J).
Answer:
sample space={Peter, Peter, Peter, James, James}
The number of event E occurs= N(E) = 5
The n\total number of names= N(T)= 9
P(E)=N(E)/N(T)
P(E)=5/9
Therefore, the probability of P(E) is 5/9.
P(E’)=N(E’)/N(T)
E’=John, Mary
The number of event E’ occurs=2
P(E’)=2/9
Therefore, the probability of E’ is 2/9.
P(J)=N(J)/N(T)
The number of event J occurs=3
P(J)=3/9
P(J)=1/3
therefore, the probability of J is 1/3.
Question 10.
Math Journal Explain why mutually exclusive events are not necessarily complementary.
Answer:
mutually exclusive events are not necessarily complementary because there will be two independent events.
Two events associated with a random experiment are said to be mutually exclusive if both cannot occur together in the same trial. Mutually-exclusive events are also known as disjoint events.
It is also important to distinguish between independent and mutually exclusive events. Independent events are those which do not depend on one another; while mutually exclusive events cannot occur together at one time.
For example: In the experiment of throwing a die, the events A = {1, 4} and B = {2, 5, 6} are mutually exclusive events.
In the same experiment, the events A = {1, 4} and C = {2, 4, 5, 6} are not mutually exclusive because, if 4 appears on the die, then it is favourable to both events A and C.
If A and B are two events, then A or B or (A ⋃ B) denotes the event of the occurrence of at least one of the events A or B.
A and B or (A ⋂ B) is the event of the occurrence of both events A and B.
If A and B happen to be mutually exclusive events, then P(A ⋂ B) = 0.
Question 11.
At a middle school, 39% of the student’s jog and 35% of the students do aerobic exercise. One out of every five students who do aerobic exercise also jogs.
a) Draw a Venn diagram to represent the information.
Answer:
Venn diagrams are the diagrams that are used to represent the sets, relation between the sets and operations performed on them, in a pictorial way. Venn diagram, introduced by John Venn (1834-1883), uses circles (overlapping, intersecting and non-intersecting), to denote the relationship between sets. A Venn diagram is also called a set diagram or a logic diagram showing different set operations such as the intersection of sets, the union of sets and the difference of sets. It is also used to depict subsets of a set.
Jog=39%; aerobic=35%
One out of every five students who do aerobic exercise also jogs
– so if you divide students into set of 5 :
Jog=5*7=35; Aerobic=5*7=35
so 7 sets we have.
And subtract one student from every set of 7; so 39-7=32 and 35-7=28
Common students who do both are 7%
b) What per cent of the students do both activities?
Answer: 7%
Jog=39%; aerobic=35%
One out of every five students who do aerobic exercise also jogs
– so if you divide students into sets of 5 :
Jog=5*7=35; Aerobic=5*7=35
so 7 sets we have.
And subtract one student from every set of 7; so 39-7=32 and 35-7=28
Common students who do both are 7%
c) What fraction of the students only jog?
Answer:8/25
Only jog students in fractions=32/100
Therefore, the fraction of students only jog is 8/25
d) What is the probability that a randomly selected student at the middle school does neither activity? Give your answer as a decimal.
Answer:0.33
Neither do any activity=100-(32+7+28)
=100-67
=33
In decimals,
=33/100
=0.33
Question 12.
A teacher chooses a student at random from a class with 20 boys and 36 girls. 25% of the students wear glasses. 15 boys in the class do not wear glasses.
a) Draw a Venn diagram to represent the information.
Answer:
the total number of students in a class=56
The number of boys=20
The number of girls=36
The number of boys does not wear glasses=15
The remaining boys who do not wear glasses=20-15=5
Now there is a chance for 5 boys and 36 girls wearing glasses.(36+5=41)
The number of students wearing glasses=25/100*41=10.56 (nearest to 11)
b) What fraction of the students in the class are girls who do not wear glasses?
Answer:
the total number of students in a class=56
The number of boys=20
The number of girls=36
The fraction of students in the class are girls who do not wear glasses=X
25% are wearing glasses remaining 75% may not wear
X=75*36/100
X=27/56
X=1/2
c) What is the probability that a randomly selected student is a boy who wears glasses?
Answer:
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done.
P(A)=N(A)/N(T)
P(A)=5/56
P(A)=11.2
therefore, 11.2 is the probability that a randomly selected student is a boy who wears glasses
Question 13.
Alex has a pair of red socks, a pair of white socks, and a pair of black socks in his drawer. Unfortunately, the socks are not matched up with each other. Alex reaches into the drawer in the dark and pulls out two socks.
a) What ¡s the probability that the two socks are the same color?
Answer:1/5
Explanation:
Alex has pair of red socks, white socks, black socks.
He can take out the possible outcomes={RR, WW, BB}
Probability of drawing socks:
Picking the first sock of the two with the same colour has the probability of 2/6, and therefore picking the second sock with the same colour has a probability of (2−1)/(6−1)=1/5 respectively.
Occurrence of 1 pair of socks/ Total number of possible outcomes
= 1/5
Each of the outcomes of the same socks has an equal probability of 1/5. Each outcome is equally likely to come on the top when the same colour of socks is drawn. Hence, it can be considered to be a fair probability scenario.
b) What is the probability that the two socks are different colors?
Answer: 4/5
Alex has pair of red socks, white socks, black socks.
He can take out the possible outcomes of two different socks={RW, WB, RB}
Picking the first sock of the two with the different colours has the probability of 2/6, and therefore picking the second sock with the different colour has a probability of (6−2)/(6−1)=4/5 respectively.
c) Are the two events described in a) and b) complementary? Explain.
Answer: yes, the vents are complementary because two socks either match or do not match.
The probability of an event is a measure of the chance of occurrence of an event when an experiment is done. Complementary events occur when there are only two outcomes, for example clearing an exam or not clearing an exam. The complement means the exact opposite of an event.
Definition: For any event A, there exists another event A‘ which shows the remaining elements of the sample space S. A’ denotes complementary event of A.
: A’ = S – A.
Event A and A’ are mutually exclusive and exhaustive.
Consider the example of tossing a coin. Let P(E) denote the probability of getting a tail when a coin is tossed.
Then the probability of getting ahead is denoted by P(E’)
Question 14.
A small town has a population of 3,200. 30% of the townspeople speak Italian, 20% speak French, and the rest do not speak either of these languages 360 people speak both Italian and French.
a) Draw a Venn diagram to represent the information.
Answer:
The total population of small town=3,200
The number of people who speaks italian=3200*30/100=960
The number of people who speaks French=3200*20/100=640
The number of people who speak both Italian and French=360
The rest of the people do not speak both languages=X
X=960+640+360-3200
X=1960-3200
X=1240
Therefore, 1240 people do not speak both languages.
b) What per cent of the townspeople speak only Italian?
Answer:
We already know that 30% of the people speak only Italian.
The above-given the percentage of the people who speaks only Italian.
c) If you randomly pick a person in the town and speak to the person in Italian, what is the probability that the person does not understand you?
Answer:
The total population of small town=3,200
The number of people who speaks italian=3200*30/100=960
The number of people who speaks French=3200*20/100=640
The number of people who speak both Italian and French=360
The rest of the people do not speak both languages=X
X=960+640+360-3200
X=1960-3200
X=1240
Randomly picking a person and speaking Italian to that person and asking the probability that he does not understand the Italian.
It means there is a chance of picking the person who speaks French and cannot speak both Italian and French.
So here 2 possible cases.
The total people for these 2 possible cases=640+1240=1880
P(A)=2/1880
P(A)=1/940.
Question 15.
The 6,000 oranges harvested at an orange grove are a combination of Valencia and Navel oranges. The ratio of Valencia oranges to Navel oranges is 7 : 5. The owner of the orange grove finds that 1 in every 20 Valencia oranges and 1 in every 25 Navel oranges are rotten.
a) What fraction of the oranges is not rotten?
Answer:229/240
the total number of oranges=6000
The ratio of valencia oranges=7
The ratio of Navel oranges=5
The total number of valencia oranges=X
X=6000*7/12
X=3500
1 in every 20 valencia oranges are rotten:
Valencia oranges rotten=3500/20
Valencia oranges rotten=175
Valencia oranges are not rotten=3500-175
valencia oranges not rotten=3325
Total navel oranges=6000*5/(7+5)
=6000*5/12
=2500
Navel oranges rotten=2500/25
Navel oranges rotten=100
navel oranges, not rotten=2500-100
navel oranges, not rotten=2400
Fraction of oranges not rotten:
=6000-(175+100)/6000
=6000-275/6000
=5725/6000
=229/240
b) What is the probability that a randomly selected orange is a good orange?
Answer:
This formula is the number of favourable outcomes to the total number of all the possible outcomes that we have already decided in the Sample Space.
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
the probability that a randomly selected orange is a good orange=P(A)
oranges that are not rotten=229/240.
P(A)=229/240
c) What is the probability that a randomly selected orange is a rotten Valencia?
Answer:7/240
The probability of an Event = (Number of favourable outcomes) / (Total number of possible outcomes)
P(A) = n(E) / n(S)
P(A) < 1
Here, P(A) means finding the probability of an event A, n(E) means the number of favourable outcomes of an event and n(S) means the set of all possible outcomes of an event.
Probability that a randomly selected orange is rotten valencia
valencia oranges rotten=175/2000
P(Valencia oranges rotten)=7/240<|endoftext|>
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# How do you write 0.000000235 in scientific notation?
Nov 8, 2015
$2.35 \cdot {10}^{-} 7$
#### Explanation:
To get our answer, first know that when you write a number in scientific notation, the number that is being multiplied by ${10}^{x}$ must be between 1 and 10. Next, let's move the decimal point to a point where if will agree with the rule. If you move over the decimal point 7 spots to the right, you will get 2.35. Since we moved 7 spots to the RIGHT, this means that the exponent will be negative (which is completely normal in context of scientific notation). Using the number we got (2.35) and the exponent of -7, we can safely say that the answer is $2.35 \cdot {10}^{-} 7$.
Hope this helps!
-Arbegla
Nov 8, 2015
$0.000000235 = 2.35 \times {10}^{- 7}$
#### Explanation:
$0.000000235$
Move the decimal to the right until it is after the $2$, so that there is one nonzero digit in front of the decimal. Since the decimal was moved to the right seven places, the exponent will be $- 7$.
$2.35 \times {10}^{- 7}$<|endoftext|>
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# Machine Kinematics Questions and Answers – Belt, Rope and Chain Drives
This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Belt, Rope and Chain Drives”.
1. The velocity ratio of two pulleys connected by an open belt or crossed belt is
a) directly proportional to their diameters
b) inversely proportional to their diameters
c) directly proportional to the square of their diameters
d) inversely proportional to the square of their diameters
Explanation: It is the ratio between the velocities of the driver and the follower or driven.
Let d1 = Diameter of the driver,
d2 = Diameter of the follower,
N1 = Speed of the driver in r.p.m., and
N2 = Speed of the follower in r.p.m.
∴ Length of the belt that passes over the driver, in one minute
= π d1.N1
Similarly, length of the belt that passes over the follower, in one minute
= π d2 . N2
Since the length of belt that passes over the driver in one minute is equal to the length of belt that passes over the follower in one minute, therefore
π d1.N1 = π d2 . N2
∴ Velocity ratio, N2/N1 = d1/d2.
2. Two pulleys of diameters d1 and d2 and at distance x apart are connected by means of an open belt drive. The length of the belt is
a) π /2 (d1 + d2) 2x + (d1 + d2)2/4x
b) π /2 (d1 – d2) 2x + (d1 – d2)2/4x
c) π /2 (d1 + d2) 2x + (d1 – d2)2/4x
d) π /2 (d1 – d2) 2x + (d1 + d2)2/4x
Explanation: None.
3. In a cone pulley, if the sum of radii of the pulleys on the driving and driven shafts is constant, then
a) open belt drive is recommended
b) cross belt drive is recommended
c) both open belt drive and cross belt drive are recommended
d) the drive is recommended depending upon the torque transmitted
Explanation: In a cross belt drive, both the pulleys rotate in opposite directions. If sum of the radii of the two pulleys be constant, then length of the belt required will also remain constant, provided the distance between centres of the pulleys remain unchanged.
4. Due to slip of the belt, the velocity ratio of the belt drive
a) decreases
b) increases
c) does not change
d) none of the mentioned
Explanation: The result of the belt slipping is to reduce the velocity ratio of the system. As the slipping of the belt is a common phenomenon, thus the belt should never be used where a definite velocity ratio is of importance.
5. When two pulleys of different diameters are connected by means of an open belt drive, then the angle of contact taken into consideration should be of the
a) larger pulley
b) smaller pulley
c) average of two pulleys
d) none of the mentioned
Explanation: None.
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6. The power transmitted by a belt is maximum when the maximum tension in the belt (T) is equal to
a) TC
b) 2TC
c) 3TC
d) 4TC
Explanation: When the power transmitted is maximum, 1/3rd of the maximum tension is absorbed as centrifugal tension.
T = 3TC
where TC = Centrifugal tension.
7. The velocity of the belt for maximum power is
a) √T/3m
b) √T/4m
c) √T/5m
d) √T/6m
Explanation: We know that T1 = T– TC and for maximum power TC = T/3
T1 = T – T/3 = 2T/3
the velocity of the belt for the maximum power, v = √T/3m
where m = Mass of the belt in kg per metre length.
8. The centrifugal tension in belts
a) increases power transmitted
b) decreases power transmitted
c) have no effect on the power transmitted
d) increases power transmitted upto a certain speed and then decreases
Explanation: None.
9. When the belt is stationary, it is subjected to some tension, known as initial tension. The value of this tension is equal to the
a) tension in the tight side of the belt
b) tension in the slack side of the belt
c) sum of the tensions in the tight side and slack side of the belt
d) average tension of the tight side and slack side of the belt
Explanation: When the driver starts rotating, it pulls the belt from one side (increasing tension in the belt on this side) and delivers it to the other side (decreasing the tension in the belt on that side). The increased tension in one side of the belt is called tension in tight side and the decreased tension in the other side of the belt is called tension in the slack side.
10. The relation between the pitch of the chain ( p) and pitch circle diameter of the sprocket (d) is given by
a) p = d sin (600/T)
b) p = d sin (900/T)
c) p = d sin (1200/T)
d) p = d sin (1800/T)
Explanation: It is given by p = d sin (1800/T).
Sanfoundry Global Education & Learning Series – Machine Kinematics.
To practice all areas of Machine Kinematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]<|endoftext|>
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Assignment #8
For EMAT 6680
Authored By
INTRODUCTION:
In this write-up we will take a look at a triangle ABC. First we will find the orthocenter, H, of this triangle. Then we will construct the triangles, HBC, HAB, and HAC along with their circumcenters and circumcircles. We will also construct the circumcenter and circumcircle for triangle ABC. The diagram of these constructions is pictured below. After constructing this it appears that the four circumcircles are all the same size. By same size, I mean that they appear to have the same area. Therefore, the hope of this write up is to prove this conjecture.
Before we being the proof we can use GSP to give us a measurement of the circles to give us an indication of whether or not the conjecture is true. Here are the results:
GREEN-Area(Circle C(hbc)H(ab)) = 53.258 square cm
BLACK-Area(Circle C(abc)H(ab)) = 53.258 square cm
BLUE-Area(Circle C(hac)H(bc)) = 53.258 square cm
RED-Area(Circle C(hab)H(ac)) = 53.258 square cm
PROOF:
To prove our conjecture it is sufficient to prove that the area of the black circle is equal to the area of any of the other three circles. Let us choose the red circle and prove that it's area is equal to the area of the black circle. It should be clear that the area of the green and blue circles could be proven equal in a similar fashion.
First note that the segment AB is a chord for both the black and red circles. Also, recall that the point Chab is the center of the red circle and the point Cabc is the center of the black circle. To prove that the two circles have equal area it is sufficient to show that the radius of each circle is equal. At this point we must recognize the fact that the centers Chab and Cabc are reflections of each other across the chord AB. Therefore, segment JChab is congruent to segment JCabc. Also, we know that segment ChabCabc is perpendicular to chord AB. Therefore angles AJChab and AJCabc are congruent. Also, AJ is congruent to itself. Therefore we have congruent triangles by SAS. Hence, AChab is congruent to ACabc. Since these two segments are also the radii of the chosen circles it remains that the two circles have equal areas and are thus congruent. Therefore, similarly, all four of the original triangles are congruent.<|endoftext|>
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# In a cube, given ${{E}_{x}}=5Ax+2B$, $A=10N/C.m$, $B=5N/C$ and $L=10cm$ .Find out: (A) Total electric flux through the cube(B) Net charge enclosed by the cube(C) What if the frame of reference is at the centre of the cube?
Last updated date: 01st Mar 2024
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Hint: We have been given the variation of the electric field along the x-axis. This means that only the faces along the x-axis are of our concern because the flux along all the other faces would be zero. Once we have found the total electric flux through the cube, we can use Gauss’ law to find the charge enclosed by the cube.
Formula Used:
$\phi =E.ds.\cos \theta$, $\phi =\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$
Complete step by step solution:
Since we know that electric flux passing through a surface due to electric field in the region is given as $\phi =E.ds.\cos \theta$ where $E$ is the electric field in the region, $ds$ is the area of cross-section of the surface and $\theta$ is the angle between the electric field and the surface through which the flux passes
Look at the figure below
Only the faces P and Q are along the x-axis, the remaining faces are perpendicular to the electric field and the cosine of $\text{90 }\!\!{}^\circ\!\!\text{ }$ is zero.
Now, to calculate the electric flux through the face P, we have the electric field ${{E}_{P}}=5Ax+2B$
Substituting the values, we get ${{E}_{P}}=5(10)(0)+2(5)=10N/C$ since at the face P $x=0$
Area of cross-section \begin{align} & (ds)={{L}^{2}}={{(10cm)}^{2}} \\ & \Rightarrow ds={{(0.1m)}^{2}}=0.01{{m}^{2}}\left[ \because 1cm=0.01m \right] \\ \end{align}
The electric flux through the face P,
\begin{align} & {{\phi }_{P}}=E.ds.\cos \theta \\ & \Rightarrow {{\phi }_{P}}=\left( 10N/C \right).\left( 0.01{{m}^{2}} \right).\cos 180{}^\circ \\ & \Rightarrow {{\phi }_{P}}=-0.1N.{{m}^{2}}/C\left[ \because \cos 180{}^\circ =-1 \right] \\ \end{align}
Similarly, the electric field at the face Q, ${{E}_{Q}}=5(10)(0.1)+2(5)=15N/C$ since at the face Q $x=L=0.1m$
Hence, electric flux through the face Q,
\begin{align} & {{\phi }_{Q}}=E.ds.\cos \theta \\ & \Rightarrow {{\phi }_{Q}}=\left( 15N/C \right).\left( 0.01{{m}^{2}} \right).\cos 0{}^\circ \\ & \Rightarrow {{\phi }_{Q}}=-0.15N.{{m}^{2}}/C\left[ \because \cos 0{}^\circ =1 \right] \\ \end{align}
Now, total electric flux through the cube
\begin{align} & \phi ={{\phi }_{P}}+{{\phi }_{Q}} \\ & \Rightarrow \phi =(0.15-0.01)=0.05N.{{m}^{2}}/C \\ \end{align}
Now, from Gauss’ law, we know that $\phi =\dfrac{{{q}_{in}}}{{{\varepsilon }_{0}}}$ where ${{q}_{in}}$ is the enclosed charge and ${{\varepsilon }_{0}}$ is the permittivity of free space
Hence ${{q}_{in}}=\phi \times {{\varepsilon }_{0}}$ where ${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}F/m$
Substituting the values, we get
\begin{align} & {{q}_{in}}=(0.05)\times (8.85\times {{10}^{-12}}) \\ & \Rightarrow {{q}_{in}}=4.42\times {{10}^{-13}}C \\ \end{align}
If the frame of reference is at the centre of the cube, the total flux passing through the cube would be zero (since both the faces P and Q would have equal distances from the reference and would cancel the electric flux) and hence the charge enclosed by the cube would be zero.
Note: The permittivity of free space is a physical constant used often in electromagnetism. It represents the capability of a vacuum to permit electric fields. Gauss's law has a close mathematical similarity with several laws in other areas of physics, such as Gauss's law for magnetism and Gauss's law for gravity. Any inverse-square law can be formulated in a way similar to Gauss's law.<|endoftext|>
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Blue Green Algae
What is Blue Green Algae?
Blue-green algae, or cyanobacteria, are a group of photosynthetic bacteria that can release harmful toxins and skin irritants. Naturally present in very low concentrations in lakes and streams, a combination of key environmental factors and weather conditions can favor blue-green algae growth and form blooms and surface scums where toxins may become concentrated enough to become a health threat. Blue green algae is also referred to as “Harmful Algal Blooms” or “HABs.”
Exposure to blue-green has the potential to cause illness in humans and animals. A person can become exposed to blue-green algae through swimming, inhaling water spray (e.g., when boating or water skiing), or by swallowing contaminated water. Blooms can also have a negative impact on the Lake by discoloring water and producing foul odors that result from the decomposition of blue-green algae cells.
- Do not swim in or swallow water containing algae scums/blooms; Do not swallow lake water even if algae is not present.
- Do not boat, water ski, jet ski, etc. over affected water.
- Do not let children play with scum layers, even from the shore.
- Wash and dry any clothing and equipment that has come in contact with algae scums.
- Do not let pets drink or swim in waters experiencing blooms.
- Immediately rinse off pets with clean water after contact with surface water before they get a chance to groom and ingest toxins that may have collected in their fur.
- Never treat surface waters with chemicals, including herbicides or algaecides, without a permit from the Department of Environmental Conservation: www.dec.ny.gov.
- Always take a shower after coming into contact with any surface water (whether or not a blue-green algae bloom is present).
- Consider medical attention of you experience sumptoms of neausea, vomiting, or diarrhea; skin, eye or throat irriation; allergic reactions or breathing difficulties. Report Symptoms to the Livingston County Department of Health
Blue-green algae cells can form into blooms and surface scums. Cyanobacteria are most often blue-green in color, but can also be green, blue, reddish-purple, or even brown. Blooms can take on the appearance of “pea soup,” green or blue paint, or form puffy clumps that float on the surface. Blue-green algae is often confused with filamentous algae which can also collect on the surface. If the bloom in question has a hair-like or stringy appearance, it is likely filamentous algae, which is not a health concern.
To Report a Suspect Algae Bloom of for Current Lake Status Please Contact:
Livingston County Department of Health - Center for Environmental Health
Phone: (585) 243-7280
Current Health Alerts
What is Livingston County Doing to Address this Threat?
In response to widespread blue green algae blooms on nearby lakes in 2011, the Conesus Lake Watershed Council developed a Blue Green Algae Detection and Response Plan overseen by the Livingston County Department of Health. Updated in 2015, the plan outlines testing, response and public notification procedures in the event of a blue green algae bloom. The implementation of the Conesus Lake Watershed Management Plan helps to reduce nutrients flowing off the landscape and into the Lake, which feeds algae blooms of all varieties.<|endoftext|>
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# Motion with constant velocity – problems and solutions
1. A car travels at a constant 10 m/s. Determine distance after 10 seconds and 60 seconds.
Solution
Constant speed 10 meters/second means car travels 10 meters every 1 second.
After 2 seconds, the car travels 20 meters,
After 5 seconds, the car travels 50 meters,
After 10 seconds, the car travels 100 meters,
After 60 seconds, the car travels 600 meters.
Read : Dimensional analysis – problems and solutions
2. A car travels along a straight road at constant 72 km/h. Determine the car’s distance after 2 minutes and 5 minutes.
Solution
72 km/h = (72)(1000 meters) / 3600 seconds = 72,000 / 3600 seconds = 20 meters/second.
The constant speed at 20 meters/second means car travels 20 meters every 1 second.
After 120 seconds or 2 minutes, car travels 20 meters x 120 = 2400 meters,
After 300 seconds or 5 minutes, car travels 20 meters x 300 = 6000 meters.
Read : Atomic theory of Rutherford, Thomson, Bohr – problems and solutions
3. A body travels along a straight road for 100 meters in 50 seconds. Determine the speed of the body.
Solution
100 meters / 50 seconds = 10 meters / 5 seconds = 2 meters/second.
4. Determine speed according to the diagram below….
Solution
Speed = Distance / time elapsed
Speed = 2 meters / 1 second = 4 meters / 2 seconds = 6 meters / 3 seconds = 8 meters / 4 seconds = 2 meters/second.
5. Cars A and B approach each other on parallel tracks. When the distance between the two cars is 100 meters, car A moves at a constant speed of 10 m/s, car B moves at a constant speed of 40 m/s. Determine (a) the distance of car A before passing car B (b) time interval before car B passing car A.
Solution
Car A moving with a constant speed at 10 meters/second, means car A moves as far as 10 meters every 1 second. After 2 seconds, A car move as far as 20 meters.
Car B moves with a constant speed at 40 meters/second, means car B moves as far as 40 meters every 1 second. After 2 seconds, car B moves as far as 80 meters.
20 meters + 80 meters = 100 meters.
(a) The distance of car A before passing car B is 20 meters. The distance of car B before passing car A is 80 meters.
(b) Time interval of car B before passing car A is 2 seconds. Time interval of car A before passing car B is 2 seconds
5. If the speedometer of a car shows 108 km/h, determine the distance traveled by car in one minute.
Solution :
The speedometer is an instrument to measure speed. The speed of a car is 108 km/hour.
108 km / h = (108) (1000 meters) / 3600 seconds = 30 meters/second.
1 minute = 60 seconds
The speed of the car 30 meters/second means the car moves as far as 30 meters in 1 second.
After 1 second, the car moves as far as 1 x 30 meters = 30 meters.
After 2 seconds, the car moves as far as 2 x 30 meters = 60 meters.
After 60 seconds, the car moves as far as 60 x 30 meters = 1800 meters.
6. Tom throws a ball straight to Andrew. Tom and Andrew are separated as far as 10.08 meters. The ball is thrown horizontally and moves at 20 m/s (ignore gravity). Andrew hits the ball 4.00 x 10-3 seconds after the ball was thrown. If the hitter moves at a constant speed of 5.00 m/s, the ball is hit by the hitter after the hitter moves as far as…
Known :
The distance between Tom and Andrew = 10.08 meters
Ball’s speed (v) = 20 m/s
The time interval (t) = 4 x 10-3 seconds = 0.004 seconds
Hitter’s speed (v) = 5 m / s
Wanted: The ball is hit by hitter after the ball moves as far as…
Solution :
Ball’s distance :
s1 = v t = (20) (0.004) = 0.08 meters
Hitter’s distance :
s2 = v t = 5 t
Ball’s distance + hitter’s distance = distance between Tom and Andrew.
0.08 + 5 t = 10.08
5 t = 10.08 – 0.08
5 t = 10
t = 10/5
t = 2 seconds
Hitter’s distance :
s2 = v t = 5 t = (5) (2) = 10 meters
7. A hunter with his car is chasing a deer. The car moves at 72 km/h and the deer run at speeds of 64.8 km/h. When the distance between the car and the deer is 2012 meters, the hunter fired his shotgun. Bullets out of the gun at 200 m/s. Determine the time interval of the deer getting shot.
A. 0.5 s
B. 1 s
C. 1.25 s
D. 1.5 s
Known :
Speed of car (vb) = 72 km/h = (72)(1000 m) / 3600 s = 20 m/s
Speed of deer (vr) = 64.8 km/h = (64.8)(1000 m) / 3600 s = 64800 m / 3600 s = 18 m/s
When the bullet is fired, the distance between the car and the deer (s) = 202 meters
Speed of fire (vp) = 20 m/s + 200 m/s = 220 m/s
Weapons held by hunters who are in a car that moves with a speed of 20 m/s so that the speed of car is also added to the speed of the bullet.
Wanted: Determine the time interval of the deer getting shot
Solution :
Think of cars and deer moving at a constant velocity.
Equation : v = s / t or s = v t
v = speed, s = distance, t = time interval
Distance = 202 + Xr = 202 + vr t = 202 + 18 t
Distance = Yp = vp t = 220 t
Distance traveled by deer = distance traveled by bullet
202 + 18 t = 220 t
202 = 220 t – 18 t
202 = 202 t
t = 202/202
t = 1 second
The correct answer is B.
Read : Isobaric thermodynamics processes - problems and solutions
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Bohr’s Model of Atom
Neils Bohr founded the Institute of Theoretical Physics at the , now known as the Niels Bohr Institute, in 1920. He predicted the existence of a new zirconium-like element, which was named hafnium,after the Latin name for Copenhagen, where it was discovered. Later, the element Bohrium was named after him.
In September 1943, word reached Bohr that he was about to be arrested by the Germans, and he fled to Sweden. From there, he was flown to Britain, and later became part of the British mission to the Manhattan Project.
Bohr proposed the atomic model in 1913 according to which the electrons travel in circular orbits around the nucleus due to electrostatic force. It was actually the modification of Rutherford’s atomic model who predicted electronic cloud around a positively charged nucleus. Rutherford’s model however faced some technical difficulties when viewed in terms of classical mechanics. This model did not use orbits of fixed positions. The model was defective because according to classical mechanics the orbiting electron will lose energy and ultimately fall into nucleus. Also as the orbit will get smaller and smaller the frequency of the emitted radiation will increase, which is not seen in emission spectra. Such a model of an atom was not acceptable since it shows that all atoms are unstable which is not applicable.
Bohrs atomic model was also based on the classical mechanics but a quantum concept was introduced into it by stating that the electrons can revolve around the nucleus only in certain allowed circular paths. He postulated that:
1. Electrons revolve around the nucleus in circular orbits
2. The electronic orbits are fixed i.e. they can revolve around the nucleus at certain discrete set of distances (energy levels) having particular energies. Thus while orbiting the acceleration of electron does not result in the emission of radiation as by the classical electromagnets (quantum concept).
3. Electrons can only lose or gain energy by jumping from one energy level to other and the emission or absorbance of radiation is determined by the band gap between two levels.
Bohrs model was based upon Plancks quantum theory of radiation introduced by Max Planck in 1900 which stated that energy can only be emitted or absorbed in discrete amounts.
The energy difference between the electronic orbits can be found by using Planks Law equation which is:
∆E = E2 – E1
Where h is Plancks’s constant and v is the frequency of radiation emitted. According to the postulate 3 the laws of classical mechanics were valid for the motion of electrons only when viewed in terms of quantum rule i.e. the electrons can only radiate a certain discrete amount of energy by jumping from one orbit to another. Thus the energy of the electron in an orbit is fixed and so is angular momentum. The angular momentum is an integral multiple of fixed unit given by:
L = n h/2π = nħ
Where ħ = h/2π and n = 1, 2, 3…is a principal quantum number and its lowest value is 1. Bohr model was successfully applied to calculate the orbital energies of hydrogen and hydrogen like atoms.
Electrons energy level of hydrogen atom
Bohr model provided the expression for calculating the energy level of hydrogen atom (and hydrogen like atoms) by considering two concepts, the classical and quantum concept. The first is the classical mechanical concept that electrons revolve in circular orbit due to electrostatic force and centripetal force is equal to the Coulomb force:
Where me is the mass of electron, assumed to be much less than the nuclear mass. Rearranging the above equation provided the speed of electrons:
The total energy of electron at any radius is given by:
The negative total energy suggests that it require energy to take electron out of the orbit away from proton. Now according to the quantum rule, the angular momentum is an integral multiple of nħ:
Putting the value of velocity and rearranging this equation would give the value of radius of nth orbit of an atom:
For n=1 the smallest radius of hydrogen atom can be found (Z=1 for hydrogen) and its value comes out to be 5.29 × 10-11 m. Similarly the Bohrs model enabled the calculation of energy of nth orbit by putting the expression of rn in the energy expression.
Bohrs model was an important step to understand the absorption and emission spectra of atoms which contained discrete lines. Bohr model related these discrete lines in the atomic spectra to the difference in the electronic orbits in atoms. However this was not achieved by Bohr himself because the idea that the electrons can behave as material waves was suggested eleven years later but still the Bohrs concept of discrete energy levels was a significant step in understanding the electrons behavior in atom and towards the development of quantum mechanics.<|endoftext|>
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# How to find a length of a "radius" not centered in a circle?
• B
• hm_tested
In summary, the conversation discusses how the radius of a circle is typically uniform around the circle when it is at the center. However, if the radius is shifted away from the center, called r', there is a way to determine the minimum length, L, between the point and the side of the circle using the known original radius, r, and the angle, θ, between r' and L. The attached image provides clarification and drawing connections between points can help with the trigonometry needed to find L.
hm_tested
Generally in a circle, the radius of the circle is uniform around the circle due to it being at the center, this is the obvious part. However, let's say the the radius was shifted away from the center so that it is somewhere in the circle, in this case called r'. Given that the original radius, r, is known, r' is the minimum length between the point and the side of the circle, and the angle, θ, is known such that it is the angle between r' and the desired length, L, is it possible to determine L? See the attached image for clarification.
#### Attachments
6.1 KB · Views: 542
Sure. Drawing connections from the left point to the center and to the upper end of r' should help. A bit of trigonometry with the formed triangles will lead to L.
Try joining r' and r.It's seems that r is centre here.You need to work with theta and r' to get answer L
## 1. What is the definition of a radius not centered in a circle?
A radius not centered in a circle is the distance between the center of the circle and any point on the circumference that is not the center. In other words, it is the length of a line segment that connects the center of the circle to a point on the edge of the circle.
## 2. How do you find the length of a radius not centered in a circle?
To find the length of a radius not centered in a circle, you can use the Pythagorean theorem. First, measure the distance from the center of the circle to the point on the circumference. Then, measure the distance from the center to the edge of the circle along a line perpendicular to the first distance. Finally, use the Pythagorean theorem (a² + b² = c²) to find the length of the radius.
## 3. Can a radius not centered in a circle be negative?
No, a radius cannot be negative. The length of a radius is always a positive value, as it represents a distance.
## 4. How does the position of the radius affect its length?
The position of the radius does not affect its length. The length of a radius is solely determined by the distance from the center of the circle to a point on its circumference.
## 5. Can you find the length of a radius not centered in a circle without knowing the circle's center?
Yes, you can find the length of a radius not centered in a circle without knowing its center. As long as you have the distance from the center to the point on the circumference, you can use the Pythagorean theorem to calculate the length of the radius.
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Skip to 0 minutes and 5 secondsJEREMY: Let's talk about data types. A type is a set of values that are related, a family of values that belong together. So far we've seen the Bool data type, which consists of values True and False. We've also used the Int data type, which consists of whole number values from some minimum to some maximum. These bounds may differ based on whether your OS is 32-bit or 64-bit. What else have we looked at? Characters, lists. Oh, yes, and pairs as well. We've also looked at function types that describe argument and return values. Now, we want to think about user-defined types, creating custom data types for our own program.
Skip to 0 minutes and 57 secondsFirst, we'll consider a simple type that consists of a finite set of alternative values. Apparently, in the Amazon jungle, there is a tribe who count one, two, many. That is, they have no distinct words for larger integers than two. We could represent this type in Haskell as follows-- data SimpleNum equals One or Two or Many. If you have GHCi open, why not type this in your interactive session right away? The key word "data" indicates we are defining a new type. The name of the type and the names of the values should start with capital letters. The alternative values are separated with a vertical bar character. Let's have a look at some of these values. One-- ah, we can't see it.
Skip to 2 minutes and 1 secondWe need to be able to print it out. Add deriving Show to the type definition.
Skip to 2 minutes and 10 secondsShow is a type class. We'll talk in more detail about this later. For now, let's just understand that any type must derive Show if we are to print out its values.
Skip to 2 minutes and 26 secondsData SimpleNum equals One or Two or Many deriving Show. OK.
Skip to 2 minutes and 40 secondsGreat. Now, let's write a function to convert from int to SimpleNum values. First, I'm going to turn on multi-line support in GHCi so I can format my function definition over several lines. I use the set command to do this. OK. Let's go for it. Let convert 1 equal One, convert 2 equal Two, convert underscore equal Many. So convert takes an int input-- well, really a value that has a type belonging to the Num data class-- again, more later-- and return a SimpleNum output. Let's try this out.
Skip to 3 minutes and 32 secondsConvert 1.
Skip to 3 minutes and 36 secondsConvert 300. OK. Map convert 1 to 5. Perfect.
Skip to 3 minutes and 51 secondsRight. That's a custom data type with alternative values, otherwise known as a Sum data type. Now let's think about a record data type that stores a portfolio of values. Hm. How about cricket scores? When a team bats in cricket, you need to know two integer values. The first is the number of runs scored by the team. The second is the number of players who are out, i.e. have been eliminated. So a good score for the New Zealand cricket team might be 350 for 4. That's 350 runs scored for the loss of four players. We can represent this as a Product data type. Data CricketScore equals Score Char list int, int, deriving Show.
Skip to 4 minutes and 58 secondsScore is a type constructor that takes a string and two int arguments and returns a cricket score value.
Skip to 5 minutes and 12 secondsIn general, these kinds of custom data types are called algebraic data types. The alternative values relate to algebraic sums, and the record values relate to algebraic products. I'll spare you the hairy Type Theory for now. But at least you know to Google for algebraic data types if you are keen to discover more about type theory. Let's conclude with what we've learned in this session.
Skip to 5 minutes and 45 secondsFirst, we use the "data" keyword to define a new custom type. Second, types must derive the Show type class if we want to see their values printed out. Third, we used vertical bars to specify alternatives, or sum data types. And we used type constructors to build record types like score values in cricket, which are product data types. Thanks. Goodbye.
Define Your Own Data Types
So far we have looked at the basic built-in data types like Int and Boolean. We can combine these basic types to generate custom data types, using algebraic sums and products.
© University of Glasgow<|endoftext|>
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In the late Roman Empire and early Byzantine Empire, an exarch was a governor of a particular territory. From the end of 3rd century or early 4th, every Roman diocese was governed by vicarius who was titled "exarch" in eastern parts of the Empire where the Greek language and the use of Greek terminology dominated. even though Latin was the language of the imperial administration from the provincial level up until the 440s (Greek translations were sent out with the official Latin text). In Greek texts the Latin title is spelled 'Bicarios.' The office of exarch as a governor with extended political and military authority was later created in Byzantine Empire, with jurisdiction over a particular territory, usually a frontier region at some distance from the capital Constantinople.
In the Eastern Christian Churches (Eastern Orthodox, Oriental Orthodox and Eastern Catholic), the term exarch has three distinct uses: metropolitan who holds the office of exarch is the deputy of a patriarch and holds authority over bishops of the designated ecclesiastical region (thus, a position between that of patriarch and regular metropolitan); or, an auxiliary or titular bishop appointed to be exarch over a group of the faithful not yet large enough or organized enough to be constituted an eparchy or diocese (thus the equivalent of a vicar apostolic); and, a priest or deacon who is appointed by a bishop as his executive representative in various fields of diocesan administration (in the Byzantine Empire, executive exarchs were usually collecting diocesan revenues for local bishops).
- 1 Political exarchs
- 2 Ecclesiastical exarchs
- 2.1 Early tradition
- 2.2 Eastern Orthodox Churches
- 2.3 Oriental Orthodox Churches
- 2.4 Latin Church
- 2.5 Modern Eastern Catholic Churches
- 2.6 List of Eastern Catholic exarchates
- 3 References
- 4 Sources
- 5 External links
In the civil administration of the Byzantine Empire the exarch was, as stated above, the imperial governor of a large and important region of the Empire. The Exarchates were a response to weakening imperial authority in the provinces and were part of the overall process of unification of civil and military offices, initiated in early form by Justinian I, which would lead eventually to the creation of the Thematic system by either the Emperor Heraclius or Constans II.
After the dissolution of the Western Empire in the late fifth century, the Eastern Roman Empire remained stable through the beginning of the Middle Ages and retained the ability for future expansion. Justinian I reconquered North Africa, Italy, Dalmatia and finally parts of Spain for the Eastern Roman Empire. However, this put an incredible strain on the Empire's limited resources. Subsequent emperors would not surrender the re-conquered land to remedy the situation. Thus the stage was set for Emperor Maurice to establish the Exarchates to deal with the constantly evolving situation of the provinces.
In Italy the Lombards were the main opposition to Byzantine power. In North Africa the Amazigh or Berber princes were ascendant due to Roman weakness outside the coastal cities. The problems associated with many enemies on various fronts (the Visigoths in Spain, the Slavs and Avars in the Balkans, the Sassanid Persians in the Middle East, and the Amazigh in North Africa) forced the imperial government to decentralize and devolve power to the former provinces.
The term Exarch most commonly refers to the Exarch of Italy, who governed the area of Italy and Dalmatia, still remaining under Byzantine control after the Lombard invasion of 568. The exarchate's seat was at Ravenna, whence it is known as the "Exarchate of Ravenna". Ravenna remained the seat of the Exarch until the revolt of 727 over Iconoclasm. Thereafter, the growing menace of the Lombards and the split between eastern and western Christendom that Iconoclasm caused made the position of the Exarch more and more untenable. The last Exarch was killed by the Lombards in 751.
A second exarchate was created by Maurice to administer northern Africa, formerly a separate praetorian prefecture, the islands of the western Mediterranean and the Byzantine possessions in Spain. The capital of the Exarchate of Africa was Carthage. The exarchate proved both financially and militarily strong, and survived until the Arab Muslim conquest of Carthage in 698.
The term exarch entered ecclesiastical language at first for a metropolitan (an archbishop) with jurisdiction not only for the area that was his as a metropolitan, but also over other metropolitans within local political dioceses. Since imperial vicarius (governor of a political diocese) was often called "exarch" in eastern, Greek speaking parts of the Empire, it became customary for the metropolitans of the diocesan capitals (Ephesus in Diocese of Asia, Heraclea in Thrace and Caesarea in Cappadocia) also to use the title "exarch" in order to emphasize their precedence and primatial status over other metropolitans within local political dioceses.
The Council of Chalcedon (451), which gave special authority to the see of Constantinople as being "the residence of the emperor and the Senate," in its canons spoke of diocesan "exarchs", placing all metropolitans in dioceses of Asia, Thrace an Pontus (including metropolitans-exarchs of Ephesus, Heraclea and Caesarea) under the jurisdiction of the Archbishop of Constantinople. Metropolitans-exarchs of Ephesus tried to resist the supreme jurisdiction of Constantinople, but eventually failed since imperial government supported the creation of a centralized Patriarchate.
When the proposed government of universal Christendom by five patriarchal sees (Rome, Constantinople, Alexandria, Antioch and Jerusalem, known as the pentarchy), under the auspices of a single universal empire, was formulated in the legislation of Emperor Justinian I (527-565), especially in his Novella 131, and received formal ecclesiastical sanction at the Council in Trullo (692), the name "patriarch" became the official one for the heads of major autocephalous churches, and the title of "exarch" was further demoted by naming all metropolitans as "patriarchal exarchs" in their ecclesiastical provinces. The advance of Constantinople put an end to privileges of three older, original exarchates, which fell back to the state of ordinary metropolitan sees.
Local ecclesiastical development in some regions also included the title of exarch. Since the Church of Cyprus was declared autocephalous (431), its Primate received the title of Exarch of Cyprus. On the similar principle the Archbishop of Mount Sinai and Raithu is an exarch, though in this case, as in that of Cyprus, modern Eastern Orthodox usage generally prefers the title "Archbishop".
Eastern Orthodox Churches
In modern ecclesiastical practice of the Eastern Orthodox Church, the title of exarch was often used to designate the highest hierarchical office under the rank of patriarch. When Russian Patriarch Adrian of Moscow died in 1700, Emperor Peter the Great abolished the patriarchal office and appointed Metropolitan Stefan Yavorsky as "exarch" and head of the governing council of the Russian Orthodox Church.
After imperial Russia annexed Georgia (eastern part in 1801, and western part in 1810), the ancient Georgian Orthodox Church (autocephalous since 750, whose head was since 1008 styled Catholicos-Patriarchs) was reorganized into Georgian Exarchate, and the newly appointed Exarch of Georgia (since 1817 always an ethnic Russian) sat in the Russian Holy Synod at St. Petersburg. Since the entire region of Caucasus fell under the Russian rule, jurisdiction of Georgian Exarchate was expanded, encompassing territories of modern-day Georgia, Armenia and Azerbaijan. On 7 April 1917, the Georgian Patriarchate was restored for the Archbishops of Mtsheta and Tbilisi, with the style Catholicos-Patriarch of All Georgia, and the title Exarch of Georgia was extinguished, but only for the Georgian part of the Exarchate. The Russian Orthodox Church and its exarch Platon (Rozhdestvensky) kept their jurisdiction over non-Georgian parts of the Caucasian region, and for those territories Caucasian Exarchate of the Russian Orthodox Church was created in the summer of 1917, with metropolitan Platon as Exarch of Caucasus. In the spring of 1918, he was succeeded by metropolitan Cyril (Smirnov) as new Exarch of Caucasus, but after his transfer to another post in the spring of 1920 no new exarch was appointed.
On 28 February 1870 the twenty-year-old struggle between Greeks and Bulgarians for the control of the Orthodox Church in Bulgaria culminated when the Ottoman Sultan Abd-ul-Aziz created an independent Bulgarian ecclesiastical organization, known as the Bulgarian Exarchate. The Orthodox Church in Bulgaria had now become independent of the Greek-dominated Ecumenical Patriarchate of Constantinople. The Bulgarian Exarch, who resided in Constantinople, became the most famous bearer of the title of exarch; his adherents throughout region were called exarchists, as opposed to the Greek patriarchists.The ensuing struggle, waged especially in Macedonia, was not only religious but had a conspicuous political dimension of a contention between competing Greek and Bulgarian national aims. For more information see Bulgarian Exarchate and Bulgarian Orthodox Church.
In 1921, eparchies of the Russian Orthodox Church in Ukraine were reorganized as Ukrainian Exarchate of the Russian Orthodox Church, headed by patriarchal exarch with seat in Kiev. The Ukrainian Exarchate existed until 1990 when it was granted a higher degree of ecclesiastical autonomy within the Moscow Patriarchate. In 1989, an autonomous Belarusian Exarchate of the Russian Orthodox Church was formed, with jurisdiction over eparchies in Belarus.
During the 20th century, the pentarchy-number principle, already abandoned in the case of Bulgaria (10th century), Serbia (14th century) and Russia (16th century), gave way to the desire of the now politically independent orthodox nations to see their sovereignty reflected in ecclesiastical autonomy – autocephaly – and the symbolic title to crown it: a 'national' Patriarch. For example, Bulgarian Exarchate was raised to the rank of Patriarchate in 1953.
In the Eastern Orthodox Church, the office of exarch can be also given to a special deputy of a Patriarch, with jurisdiction over a community outside the home territory of the Patriarchate. Thus, in the United States there are Exarchs representing, among others, the Serbian, Romanian, Bulgarian and Jerusalem Patriarchs. The style of the Exarchs of the Patriarchate of Jerusalem is "Exarch of the Holy Sepulcher".
The Mexican Orthodox parishes in five deaneries (Mexico City, D.F., State of Mexico, State of Jalisco, State of Veracruz and State of Chiapas) of the Orthodox Church in America are governed as the "Exarchate of Mexico", currently under the leadership of Bishop Alejo of Mexico City.
Oriental Orthodox Churches
The Oriental Orthodox Patriarch of Antioch currently has under his authority an Exarch in India, known by the ancient title Maphrian, although he is popularly referred to as Catholicos. This is not to be confused with the autocephalous Catholicate of the East, which is also located in India.
Historically, there have been a very few cases of the civil title of Exarch granted by the civil authority to prelates of the Latin Church, as when Emperor Frederick I named the Archbishop of Lyon Exarch of Burgundy in 1157.
However, the ecclesiastical title of Exarch has disappeared in the Western Catholic Church, being replaced by the terms "Primate" (ranking above Metropolitan Archbishop) and "Apostolic Vicar" (ranking below Suffragan Bishop).
Modern Eastern Catholic Churches
In Eastern Catholic Churches (of Eastern tradition but in full communion with the Bishop of Rome, the Pope), the ecclesiastical title of Exarch is in common use, just as with its Orthodox counterparts.
These Churches are, in general, not identified with a particular liturgical rite. Thus, no less than fourteen of them use the one same Byzantine Rite, mostly in one or other of only two languages, Greek and Church Slavonic, but they maintain their distinct identities. Because of population shifts, half or so of these Churches have not just exarchates but full-scale eparchies (bishoprics) or even archeparchies (archdioceses) outside their original territory.
Apostolic exarch is usually a consecrated bishop of a titular see to whom the Pope, as Bishop of the Roman See of the Apostle Peter, has entrusted the pastoral care of the faithful of an autonomous Eastern Catholic particular Church sui iuris in an area, not raised to the rank of eparchy (diocese), that is situated outside the home territory of an Eastern Catholic Church. The office of apostolic exarch thus corresponds to what in the Latin Church is called an Apostolic vicar. Apostolic exarchates are generally exempt (immediately subject to the Holy See), with limited oversight by the Patriarch, Major Archbishop or Metropolitan in chief of the particular Eastern Church. It there is no metropolitan in a particular Eastern Catholic church, apostolic exarchates in their territories are directly subjected to Rome. For example, Byzantine Catholic Apostolic Exarchate of Serbia belongs to the Byzantine Catholic Church of Croatia and Serbia, but since there is no metropolitan in that church, Apostolic Exarch of Serbia is directly subjected to the Holy See.
Patriarchal exarch is appointed in those Eastern Catholic churches whose head is styled as patriarch. Office of patriarchal exarch is often (not always) given to a consecrated bishop of a titular see. Their appointments are limited to the traditional territory of their church, with main task of governing the region not yet raised to the rank of eparchy (diocese). They may be suffragan to an archdiocese or archeparchy of the Eastern Catholic Church, or be immediately subject to the Patriarch.
Archiepiscopal exarch is appointed in those Eastern Catholic churches whose head is styled as Major Archbishop. Office of archiepiscopal exarch is also usually given to a consecrated bishop of a titular see. Appointment of archiepiscopal exarchs is limited to the traditional territory of their particular church. They also may be suffragans to an archdiocese or archeparchy of their Eastern Catholic Church, or be immediately subject to the Major Archbishop.
In particular cases, usually because of illness or some other problem, an exarch of any rank can be assisted by the appointment of a colleague who is called Coadjutor exarch. The position of coadjutor exarch towards his superior exarch is similar to the position of Latin coadjutor bishop towards his superior diocesan bishop. Coadjutor exarchs are appointed with rights of succession. For example, in 1993 titular Bishop Christo Proykov of Briula was appointed Coadjutor to Apostolic Exarch of Sofia, Methodius Stratiev, and when the latter died in 1995 coadjutor exarch succeeded him as the new Apostolic Exarch.
In practice, exarch of any rank can be additionally assisted by an auxiliary exarch, who is appointed in order to help the exarch in administration of his exarchate. Position of auxiliary exarch towards his superior exarch is similar to position of Latin auxiliary bishop towards his superior diocesan bishop. Auxiliary exarchs are appointed without the rights of succession.
List of Eastern Catholic exarchates
The following Eastern Catholic exarchates can be found in the 2006 Annuario Pontificio and newer sources. The Apostolic Exarchates are exempt, i.e. immediately subject to the Holy See, rather than to their Patriarch or other head of the particular Church
- Bulgarian Greek Catholic Church: comprises only the
- Greek Byzantine Catholic Church:
- Melkite (Greek) Catholic Church:
- Russian Greek Catholic Church:
- Ruthenian Greek Catholic Church:
- Ukrainian Greek Catholic Church:
- Apostolic Exarchate in Germany and Scandinavia for the Ukrainians for Germany and Scandinavia (Finland, Norway, Sweden, and Denmark)
- Maronite Church:
- Syriac (Syrian) Catholic Church :
- Armenian Catholic Church:
- Armenian Catholic Church:
- Armenian Catholic Patriarchal Exarchate of Damascus (Syria)
- Armenian Catholic Patriarchal Exarchate of Jerusalem and Amman (Palestine, Israel and Jordan)
- Melkite (Greek) Catholic Church:
- Maronite Church:
- Syriac (Syrian) Catholic Church:
- Syriac Catholic Patriarchal Exarchate of Bassorah and the Gulf (Iraq, Kuwait etc.)
- Syriac Catholic Patriarchal Exarchate of Jerusalem (Palestine, Israel and Jordan)
- Syriac Catholic Patriarchal Exarchate of Turkey
- Ukrainian (Greek) Catholic Church, in Ukraine:
- Ukrainian Catholic Archiepiscopal Exarchate of Donetsk
- Ukrainian Catholic Archiepiscopal Exarchate of Kharkiv
- Ukrainian Catholic Archiepiscopal Exarchate of Lutsk
- Ukrainian Catholic Archiepiscopal Exarchate of Odessa
- Ukrainian Catholic Archiepiscopal Exarchate of Krym (Crimea), on the Russian-annexed Crimea, with cathedral see at Simferopol
Former Eastern Catholic Exarchates
(probably still incomplete)
Former Eastern Catholic Exarchates in the Old World
- in Europe - Byzantine Rite
- Greek Catholic Apostolic Exarchate of Turkey of Europe (now of Istanbul)
- Hungarian Catholic Apostolic Exarchate of Miskolc (Hungary; promoted to eparchy)
- Apostolic Exarchate of Łemkowszczyzna
- Ukrainian Catholic Archiepiscopal Exarchate of Donetsk-Kharkiv (Ukraine; split in both named cities)
- Ukrainian Catholic Archiepiscopal Exarchate of Lutsk-Volyn (Ukraine; split in ? )
- Ukrainian Catholic Archiepiscopal Exarchate of Odessa-Crimea (Ukraine; split in both named parts)
- Apostolic Exarchate of Serbia and Montenegro (2003–2013)
- Byzantine Catholic Apostolic Exarchate of Serbia (2013–2018, elevated as Eparchy in 2018)
- Apostolic Exarchate of Macedonia (former, elevated as an Eparchy in 2018)
- in Asia - Armenian Rite
- Armenian Catholic Patriarchal Exarchate of Jerusalem (Palestine, Israel and Jordan, now 'Jerusalem and Amman')
- Armenian Catholic Patriarchal Exarchate of Syria (suppressed)
- in Asia - Antiochian Rite
- Syrian Catholic Patriarchal Exarchate of Lebanon (national; suppressed)
- Syro-Malankara Catholic Exarchate in the United States (USA; promoted Eparchy of St. Mary, Queen of Peace, of the United States of America and Canada)
- in Asia - Syro-Oriental Rite
- Syro-Malabar Apostolic Exarchate of Chanda (India; promoted eparchy)
- in Africa - Alexandrian Rite
- Apostolic Exarchate of Addis Abbeba (Ethiopic Catholic; promoted Metropolitanate sui iuris)
- Apostolic Exarchate of Asmara (Eritrean Catholic)
- in Africa - Antiochian Rite
Former Eastern Catholic Exarchates in the New World
- in the Americas - Antiochian Rite
- in the Americas - Armenian Rite
- Armenian Catholic Apostolic Exarchate of Latin America and Mexico
- Armenian Catholic Apostolic Exarchate of the USA and Canada
- in the Americas - Byzantine Rite
- Romanian Catholic Apostolic Exarchate of the USA (US and Canada)
- Ruthenian Catholic Apostolic Exarchate of the USA
- Melkite Catholic Apostolic Exarchate of the USA
- Ukrainian Catholic Apostolic Exarchate of Canada
- Ukrainian Catholic Apostolic Exarchate of Central Canada
- Ukrainian Catholic Apostolic Exarchate of Eastern Canada
- Ukrainian Catholic Apostolic Exarchate of Edmonton (Canada)
- Ukrainian Catholic Apostolic Exarchate of Manitoba (Canada)
- Ukrainian Catholic Apostolic Exarchate of Saskatoon (Canada)
- Ukrainian Catholic Apostolic Exarchate of Stamford (US)
- Ukrainian Catholic Apostolic Exarchate of Toronto (Canada)
- Ukrainian Catholic Apostolic Exarchate of the USA
- Ukrainian Catholic Apostolic Exarchate of Western Canada
- in the Americas - Syro-Oriental Rite
- Chaldean Catholic Apostolic Exarchate of the USA
- Syro-Malabar Catholic Church:
- Ostrogorsky 1956.
- Meyendorff 1989.
- A. Fortescue, Orthodox Eastern Church, 21-25.
- Fortescue 1908, p. 295, 304-305, 351.
- "Vladimir Moss, The Orthodox Church in the Twentieth Century". romanitas.ru.
- "Dioceses - Diocese of Mexico". www.oca.org.
- Cheney, David M. "Serbia (Apostolic Exarchate) [Catholic-Hierarchy]". www.catholic-hierarchy.org.
- Cheney, David M. "Bishop Christo Proykov [Catholic-Hierarchy]". www.catholic-hierarchy.org.
- "Catholic Dioceses in the World (Apostolic Exarchates)". www.gcatholic.org. Retrieved 2018-12-28.
- Fortescue, Adrian (1908). The Orthodox Eastern Church. London: Catholic Truth Society.
- Ostrogorsky, George (1956). History of the Byzantine State. Oxford: Basil Blackwell.
- Meyendorff, John (1989). Imperial unity and Christian divisions: The Church 450-680 A.D. The Church in history. 2. Crestwood, NY: St. Vladimir's Seminary Press.
- Nedungatt, George, ed. (2002). A Guide to the Eastern Code: A Commentary on the Code of Canons of the Eastern Churches. Rome: Oriental Institute Press.
- This article incorporates text from a publication now in the public domain: Adrian Fortescue (1913). . In Herbermann, Charles (ed.). Catholic Encyclopedia. New York: Robert Appleton.
- Encyclopædia Britannica (11th ed.). 1911. .
- Giga-Catholic Information: Rites - Apostolic Exarchates - Patriarchal Exarchates - Archiepiscopal Exarchates
- WorldStatesmen- Religious Organisations<|endoftext|>
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Dengue fever is a disease caused by a family of viruses transmitted by mosquitoes. It is an acute illness of sudden onset that usually follows a benign course with symptoms such as headache, fever, exhaustion, severe muscle and joint pain, swollen lymph nodes (lymphadenopathy), and rash. The presence of fever, itchy rash, and headache (the “dengue triad”) is characteristic of dengue. Other signs of dengue fever include bleeding gums, severe pain behind the eyes, and red palms and soles.
Dengue (pronounced DENG-gay) can affect anyone but tends to be more severe in people with compromised immune systems. Because it is caused by one of five serotypes of the dengue virus, it is possible to get dengue fever multiple times. However, an attack of dengue produces immunity for a lifetime to that particular viral serotype to which the patient was exposed.
Dengue goes by other names, including “breakbone fever” or “dandy fever.” Victims of dengue often have contortions due to the intense pain in the joints, muscles, and bones, hence the name breakbone fever. Slaves in the West Indies who contracted dengue were said to have dandy fever because of their postures and gait.
Dengue hemorrhagic fever is a more severe form of the viral illness. Symptoms include headache, fever, rash, and evidence of bleeding (hemorrhage) in the body. Petechiae (small red spots or purple splotches or blisters under the skin), bleeding in the nose or gums, black stools, or easy bruising are all possible signs of hemorrhage. This form of dengue fever can be life-threatening and can progress to the most severe form of the illness, dengue shock syndrome.
How is dengue fever contracted? Is dengue fever contagious?
The virus is contracted from the bite of a striped Aedes aegypti mosquito that has previously bitten an infected person. The mosquito flourishes during rainy seasons but can breed in water-filled flower pots, plastic bags, and cans year-round. One mosquito bite can cause the disease.
The virus is not contagious and cannot be spread directly from person to person. It is mosquito-borne, so there must be a person-to-mosquito-to-another-person pathway. The full life cycle of the virus involves the mosquito as the vector (transmitter) and the human as the source of infection.
What is the incubation period for dengue fever?
After being bitten by a mosquito carrying the virus, the incubation period for dengue fever ranges from three to 15 (usually five to eight) days before the signs and symptoms of dengue appear in stages.
What are dengue fever symptoms and signs?
Dengue fever starts with symptoms of chills, headache, pain in the back of the eyes that may worsen upon moving the eyes, appetite loss, feeling unwell (malaise), and low backache. Painful aching in the legs and joints occurs during the first hours of illness. The temperature rises quickly as high as 104 F (40 C), with relatively low heart rate (bradycardia) and low blood pressure (hypotension). The eyes become reddened. A flushing or pale pink rash comes over the face and then disappears. The lymph nodes in the neck and groin are often swollen.
High fever and other signs of dengue last for two to four days, followed by a rapid drop in body temperature (defervescence) with profuse sweating. This precedes a period with normal temperature and a sense of well-being that lasts about a day. A second rapid rise in temperature follows. A characteristic itchy rash (small red spots, called petechiae) appears along with the fever and spreads from the extremities to cover the entire body except the face. The palms and soles may be bright red and swollen.
What tests do health care providers use to diagnose dengue fever?
The diagnosis of dengue fever is usually made when a patient exhibits the typical clinical symptoms of headache, high fever, eye pain, severe muscle aches, and petechial rash and has a history of being in an area where dengue fever is endemic. Dengue fever can be difficult to diagnose because its symptoms overlap with those of many other viral illnesses, such as West Nile virus and chikungunya fever.
Health care professionals may use a blood test called the DENV Detect IgM Capture ELISA to diagnose people with dengue fever. The FDA notes that the test may also give a positive result when a person has a closely related virus, such West Nile disease.
What is the treatment for dengue fever?
Because dengue fever is caused by a virus, there are no specific antibiotics to treat it. Antiviral medications are also not indicated for dengue fever. For typical dengue, the treatment is concerned with relief of the symptoms and signs. Home remedies such as rest and fluid intake (oral rehydration) are important. Pain relievers such as aspirin and nonsteroidal anti-inflammatory drugs (NSAIDs) should only be taken under a doctor’s supervision because of the possibility of worsening bleeding complications. Acetaminophen (Tylenol) and codeine may be given for severe headache and for joint and muscle pain (myalgia).
Patients hospitalized for dengue may receive IV fluids.
Carica papaya leaf extract (papaya leaf) has been shown in several clinical studies to be an effective treatment for dengue fever.
How long does dengue fever last?
The acute phase of dengue with fever and muscle pain (myalgia) lasts about one to two weeks. Convalescence is accompanied by a feeling of weakness (asthenia) and fatigue, and full recovery often takes several weeks.
What is the prognosis for typical dengue fever?
The prognosis for dengue is usually good. The worst symptoms of the illness typically last one to two weeks, and most patients will fully recover within several additional weeks.
Typical dengue is fatal in less than 1% of cases; however, dengue hemorrhagic fever is fatal in 2.5% of cases. If dengue hemorrhagic fever is not treated, mortality (death) rates can be as high as 20%-50%.
What is dengue hemorrhagic fever?
Dengue hemorrhagic fever (DHF) is a specific syndrome that tends to affect children under 10 years of age. This complication of dengue fever causes abdominal pain, hemorrhage (bleeding), and circulatory collapse (shock). DHF is also called Philippine, Thai, or Southeast Asian hemorrhagic fever or dengue shock syndrome.
DHF starts abruptly with continuous high fever and headache. There are respiratory and intestinal symptoms with sore throat, cough, nausea, vomiting, and abdominal pain. Shock occurs two to six days after the start of symptoms with sudden collapse, cool, clammy extremities (the trunk is often warm), weak pulse, and blueness around the mouth (circumoral cyanosis).
In DHF, there is bleeding with easy bruising, red or purple blood spots in the skin (petechiae), spitting up blood (hematemesis), blood in the stool (melena), bleeding gums, and nosebleeds (epistaxis). Pneumonia is common, and inflammation of the heart (myocarditis) may be present.
Patients with DHF must be monitored closely for the first few days since shock may occur or recur precipitously (dengue shock syndrome). Cyanotic (having a bluish coloration to the skin and mucus membranes) patients are given oxygen. Vascular collapse (shock) requires immediate fluid replacement. Blood transfusions may be needed to control bleeding.
The mortality (death) rate with DHF is significant. With proper treatment, the World Health Organization estimates a 2.5% mortality rate. However, without proper treatment, the mortality rate rises to 20%. Most deaths occur in children. Infants under 1 year of age are especially at risk of dying from DHF.
What are potential complications of dengue fever?
If dengue fever is severe, complications include leakage of fluid from the bloodstream causing fluid accumulation in the extremities, respiratory distress, severe bleeding, or organ impairment. Without proper treatment, these symptoms can be fatal.
Dengue hemorrhagic fever (DHF; see above) is a complication of dengue that usually affects children under 10 years of age when it occurs. This complication of dengue fever starts abruptly with continuous high fever and headache. DHF causes abdominal pain, sore throat, cough, nausea, vomiting, hemorrhage (bleeding), and circulatory collapse (shock). It can be fatal.
Another complication is postinfectious fatigue syndrome, which can occur in about one-quarter of hospitalized dengue patients.
Is it possible to prevent dengue fever?
The transmission of the virus to mosquitoes must be interrupted to prevent the illness. To this end, patients are kept under mosquito netting until the second bout of fever is over and they are no longer able to transmit the virus to a biting mosquito.
The prevention of dengue fever requires control or eradication of the mosquitoes carrying the virus that causes dengue. In nations plagued by dengue fever, people are urged to empty stagnant water from old tires, trash cans, and flower pots. Governmental initiatives to decrease mosquitoes also help to keep the disease in check but have been poorly effective.
To prevent mosquito bites, wear long pants and long sleeves. For personal protection, use mosquito-repellant sprays that contain DEET when visiting places where dengue is endemic. There are no specific risk factors for contracting dengue fever except living in or traveling to an area where the mosquitoes and virus are endemic. Limiting exposure to mosquitoes by avoiding standing water and staying indoors for two hours after sunrise and before sunset will help, as the Aedes aegypti mosquito is a daytime biter with peak periods of biting around sunrise and sunset. It may bite at any time of the day and is often hidden inside homes or other dwellings, especially in urban areas.
Is there a dengue fever vaccine?
In April 2016, the WHO approved Sanofi Pasteur’s Dengvaxia (CYD-TDV), a live recombinant tetravalent vaccine for dengue fever. Dengvaxia can be administered as a three-dose series in people 9-45 years of age who live in areas where dengue is endemic.
Dengvaxia was initially approved in 2015 for use only in Mexico, the Philippines, Brazil, and El Salvador.
Several other vaccines for dengue are undergoing clinical trials, but none have yet been approved for use.
Original Article: Medicinenet
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Newbie
# A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
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In this Question A two-digit number is such that the product of its digit is given. if a number is substracted to the number, the digits interchange their places.
You have to find the number.
This is the Important question based on Linear Equations in two Variables Chapter of R.S Aggarwal book for ICSE & CBSE Board.
This is the Question Number 17 Of Exercise 3E of RS Aggarwal Solution.
Share
1. Solution :-
Let the tens place digit be x.
And the units place digit be y.
According to the Question,
⇒ xy = 18
⇒ y = 18/x …..(i)
And, (10x + y) – 63 = 10y + x
⇒ 9x – 9y = 63
⇒ x – y = 7 …. (ii)
Putting y’s value in Eq (ii), we get
⇒ x – 18/x = 7
⇒ x² – 18 = 7x
⇒ x² – 7x – 18 = 0
⇒ x² – 9x + 2x – 18 = 0
⇒x(x – 9) + 2(x – 9) = 0
⇒ (x – 9) (x + 2) = 0
⇒ x – 9 = 0 or x + 2 = 0
⇒ x = 9, – 2 (As x can’t be negative)
⇒ x = 9
Putting x’s value in Eq (i), we get
⇒ xy = 18
⇒ 9y = 18
⇒ y = 18/9
⇒ y = 2
Number = 92
Hence, the required number is 92
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Algeria Table of Contents
Figure 2. Roman North Africa, Fourth Century B.C. to Third Century A.D.
As Carthaginian power grew, its impact on the indigenous population increased dramatically. Berber civilization was already at a stage in which agriculture, manufacturing, trade, and political organization supported several states. Trade links between Carthage and the Berbers in the interior grew, but territorial expansion also resulted in the enslavement or military recruitment of some Berbers and in the extraction of tribute from others. By the early fourth century B.C., Berbers formed the single largest element of the Carthaginian army. In the Revolt of the Mercenaries, Berber soldiers rebelled from 241 to 238 B.C. after being unpaid following the defeat of Carthage in the First Punic War. They succeeded in obtaining control of much of Carthage's North African territory, and they minted coins bearing the name Libyan, used in Greek to describe natives of North Africa. The Carthaginian state declined because of successive defeats by the Romans in the Punic Wars; in 146 B.C. the city of Carthage was destroyed. As Carthaginian power waned, the influence of Berber leaders in the hinterland grew. By the second century B.C., several large but loosely administered Berber kingdoms had emerged. Two of them were established in Numidia, behind the coastal areas controlled by Carthage (see fig. 2). West of Numidia lay Mauretania, which extended across the Moulouya River in Morocco to the Atlantic Ocean. The high point of Berber civilization, unequaled until the coming of the Almohads and Almoravids more than a millennium later, was reached during the reign of Masinissa in the second century B.C. After Masinissa's death in 148 B.C., the Berber kingdoms were divided and reunited several times. Masinissa's line survived until A.D. 24, when the remaining Berber territory was annexed to the Roman Empire.
Data as of December 1993<|endoftext|>
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What is the course about?
Starting with the basics of cell biology you will learn about how genetic information is carried in our DNA on genes and chromosomes. You will learn how genes influence evolution and how we inherit our physical characteristics. You will also learn how our health may be determined by our genes, and how genes contribute to cancer and other diseases. The course will deliberate the uses of genetic manipulation and its uses in our modern lives – from ‘GM’ foodstuffs, medical research to DNA testing for paternity and forensic analysis.
What will we cover?
- Basics of cell biology.
- Where do we find DNA?
- DNA structure and function
- How genes work and how do they carry information?
- Inheritance and evolution.
- Genetic diseases and cancer development.
- DNA testing for genetic diseases, paternity tests, forensic analysis.
- Uses of DNA technology and genetic manipulation for medicine, health and food.
What will I achieve?
By the end of this course you should be able to...
- Describe how the structure of DNA is directly linked to its function.
- State what is meant by the terms genes, chromosomes and genetic traits.
- Explain how genetic information is carried from one generation to the next.
- Outline some of the ways in which our genes and the environment govern our health – nature vs nurture.
- Recognise and evaluate genetic manipulation – and its uses.
What level is the course and do I need any particular skills?
This is an introductory course and is open to all. No prior knowledge of the subject is required. However, you will
need a good grasp of English to keep up with the course and to participate fully in discussions.
How will I be taught, and will there be any work outside the class?
Classes will be taught through a mixture of illustrated talks and discussions, along with short problem-solving exercises. Materials will include projected images and paper handouts, as well as internet based resources/media.
Are there any other costs? Is there anything I need to bring?
Bring a notepad and pen. There are no additional costs for the course.
When I've finished, what course can I do next?
HS094 The history of science.
General information and advice on courses at Build is available from the Student Centre and Library on Monday to Friday from 12:00 – 19:00.
See the course guide for term dates and further details<|endoftext|>
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# Lesson 6: Polygons and Angles
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1 Lesson 6: Polygons and Angles Selected Content Standards Benchmark Assessed: G.4 Using inductive reasoning to predict, discover, and apply geometric properties and relationships (e.g., patty paper constructions, sum of the angles in a polygon) Lesson Focus In this lesson, students will discover relationships between polygons and angles. It includes the following aspects: Determine the sum of the measures of the interior angles of a triangle Recognize the relationship between interior and exterior angles Relate every convex polygon to the triangle Develop a formula to determine the sum of the interior angles of a polygon GEE 21 Connection The skills that will be addressed in this lesson include the following: Understand the relationship between triangles and all polygons Demonstrate the ability to use the interior angle theorem to determine angle measure Identify interior and exterior angles of a polygon Translating content Standards into Instruction Whether done inductively or deductively, it is important that all students understand how to find the measure of an interior angle in any polygon. This lesson is intended to help students understand the foundation of the interior and exterior angle theorem. A. The first thing students should understand is why there are 180º in a triangle. 1. Have each student draw a triangle. Do not be specific; the more types, the better. Label the angles inside the triangle with numbers. 2. On a separate sheet of paper have them draw a segment with a point near the center. They can measure the segment with a protractor to determine its measure to be 180º. 3. Have students tear off all three angles of the triangle and re-arrange them with each vertex on the point on the line segment. 4. Discuss with the students to determine that everyone made the same discovery. B. Next we will relate the triangle to every convex polygon. 36
2 1. Have students draw convex polygons with 4, 5, 6, 7, 8, 9, and 10 sides. You can provide this to save a little time, but it is better to have all different sizes and shapes. 2. Have each student choose one vertex to highlight in each polygon. 3. Have students use a straight edge to connect the chosen vertex to every other vertex in the polygon. This is a good time to define diagonals. 4. Help students see that each polygon is now full of triangles. C. Begin developing a formula. 1. Students need to make a chart of the results. Name of Figure #Sides # Triangles Interior Angle Sum Quadrilateral Pentagon Hexagon Heptagon Octagon Nonagon Decagon Students will see the pattern emerging from the chart. Be sure to ask them why there are 2 less triangles than sides. 3. Develop the formula (n-2)180 from the discovery lesson. 4. Now, using the interior angle theorem, challenge students to find the unknown angle measures in Student Worksheet 1. D. Students should use their original diagrams to discover the exterior angle theorem. 1. Using a color different from the one use in the original diagram, have students extend one side of each angle. 2. Now have students use the supplement theorem to determine the measure of each exterior angle. 3. Extend the chart previously used to include sum of exterior angles. Name of Figure #Sides # Triangles Interior Angle Sum Exterior Angle Sum Quadrilateral Pentagon Hexagon Heptagon Octagon Nonagon Decagon
3 Sources of Evidence about Student Learning A. Have students work through the discovery lesson. B. Have students complete the student worksheet. C. Have students use a protractor to measure angles to verify their results using the angle sum theorems. GEE 21 Connection On the GEE 21 test, students may be required to A. Determine interior or exterior angle measure given partial information B. Identify formulas related to interior and exterior sums Attributes of Student Work at the Got-It level A. Students can successfully complete the student worksheet. B. Students can explain why (n-2) 180 = sum of interior angles. C. Students can identify the measure of interior and exterior angles in polygons. 38
4 Lesson 6: Relationships of Polygons and Angles Student Worksheet 39
5 40
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### Inscribed Angle Theorem and Its Applications
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### Geometry of Minerals
Geometry of Minerals Objectives Students will connect geometry and science Students will study 2 and 3 dimensional shapes Students will recognize numerical relationships and write algebraic expressions
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Lesson 2: Pythagorean Theorem Selected Content Standards Benchmarks Addressed: G-5-M Making and testing conjectures about geometric shapes and their properties G-7-M Demonstrating the connection of geometry
### CHAPTER 6. Polygons, Quadrilaterals, and Special Parallelograms
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### Unit 6 Grade 7 Geometry
Unit 6 Grade 7 Geometry Lesson Outline BIG PICTURE Students will: investigate geometric properties of triangles, quadrilaterals, and prisms; develop an understanding of similarity and congruence. Day Lesson
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## Gambling Mathematics: Why Will The Casino Win in the Long Run
Gambling in a casino and winning once or twice might hook you, but if you play often, there is a very small chance that you would win. Why? Because the games are mathematically designed for the casino to win.
To make it simple, let us use a simple die game for our discussion. Two dice are rolled. If the sums of the number of dots are 2, 3, 4, 10, 11, and 12, the player wins. If the sums are 5, 6, 7, 8, and 9, the house (the casino) wins. » Read more
## The Experimental and Theoretical Probability Series
Last week, we have completed the 5-part Experimental and Theoretical Probability Series. To those who have not read it, below is the list of posts.
1. Experimental and Theoretical Probability Part 1. This post discusses about the sum of the number of dots of two rolled standard cubical dice. A spreadsheet is used to simulate the rolling 1000 times and sums are recorded and tallied. A step-by-step instruction in doing the simulation is provided.
2. Experimental and Theoretical Probability Part 2. This post confirms the findings in Part 1. Two more experiments are conducted — the first one is rolling the dice 2000 times, and the other is rolling it again 3000 times.
3. Experimental and Theoretical Probability Part 3. The third part is a discussion on why the findings in Part 1 and Part 2 are such. The ways of getting a particular sum is discussed in this post.
4. Experimental and Theoretical Probability Part 4. The fourth part discusses the relationships between the experiments and the findings in Part 3. The formal definitions of Experimental and Theoretical probabilities are also discussed here.
5. Experimental and Theoretical Probability Part 5. The fifth part summarizes the series and give real-life examples that use experimental and theoretical probability.
I hope you have enjoyed reading this series. Watch out for more Math and Multimedia Tutorial Series.
## Experimental and Theoretical Probability 5
This is the fifth and the final part of the Experimental and Theoretical Probability Series. In this post, we are going to summarize what we have discussed in the previous four posts, and we are going to talk about some real-life applications of experimental and theoretical probability.
Standard Cubical Dice
Experimental Probability, as we have discussed in the fourth part of this series, may be obtained by conducting experiments and recording the results. It is the ratio of the number of times an event occurs to the total number of trials. In the first part of this series, we experimented rolling to dice 1000 times (via a spreadsheet) and we tallied the sums. We recorded the that sum 2 occurred 29 times out of 1000 trials. We can say that the experimental probability of getting a 2 from that particular experiment is 29/1000. » Read more
1 2 3<|endoftext|>
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Hong Kong
Stage 1 - Stage 3
# Complementary results
Lesson
There are connections between the trigonometric functions that can make simplifications possible.
In a right-angled triangle, the two acute angles together make a right-angle. We say the acute angles are complementary (to one another). If the two acute angles have measures $\alpha$α and $\beta$β, then $\alpha+\beta=90^\circ$α+β=90° and so, $\beta=90^\circ-\alpha$β=90°α.
The diagram below illustrates the following relationships.
$\cos\alpha=\frac{b}{h}=\sin\beta=\sin\left(90^\circ-\alpha\right)$cosα=bh=sinβ=sin(90°α)
$\sin\alpha=\frac{a}{h}=\cos\beta=\cos\left(90^\circ-\alpha\right)$sinα=ah=cosβ=cos(90°α)
$\cot\alpha=\frac{b}{a}=\tan\beta=\tan\left(90^\circ-\alpha\right)$cotα=ba=tanβ=tan(90°α)
Thus, the 'co' in complementary explains the meaning of cosine in relation to sine, and to cotangent in relation to tangent.
The statements
$\cos\alpha\equiv\sin\left(90^\circ-\alpha\right)$cosαsin(90°α)
$\sin\alpha\equiv\cos\left(90^\circ-\alpha\right)$sinαcos(90°α) and
$\cot\alpha\equiv\tan\left(90^\circ-\alpha\right)$cotαtan(90°α)
are called identities because they are true whatever the value of the angle $\alpha$α.
These identities are true not only in right-angled triangle trigonometry, but they also hold for angles of any size. This can be confirmed by thinking about the geometry in the unit circle diagram that is used for defining the trigonometric functions of angles of any magnitude.
#### Example
Simplify the relation $\sin\left(90^\circ-\theta\right)=\sqrt{3}\sin\theta$sin(90°θ)=3sinθ
it will be a good plan to try to rearrange the equation so that the trigonometric functions are on one side and the coefficients are on the other. We divide both sides by $\sin\left(90^\circ-\theta\right)$sin(90°θ) and also by $\sqrt{3}$3 to obtain $\frac{1}{\sqrt{3}}=\frac{\sin\theta}{\sin\left(90^\circ-\theta\right)}$13=sinθsin(90°θ). But, $\sin\left(90^\circ-\theta\right)$sin(90°θ) is just $\cos\theta$cosθ. So, the simplification we seek is
$\tan\theta=\frac{1}{\sqrt{3}}$tanθ=13.
We recognise an exact value for $\tan$tan and conclude that $\theta=30^\circ$θ=30° if $\theta$θ is acute. You should check that there is also a third quadrant solution, $\theta=210^\circ$θ=210°.
#### Worked Examples
##### Question 1
By finding the ratio represented by $\sin\theta$sinθ, $\cos\theta$cosθ and $\tan\theta$tanθ in the given figure, we want to prove that $\frac{\sin\theta}{\cos\theta}=\tan\theta$sinθcosθ=tanθ.
1. Write down the expression for $\sin\theta$sinθ.
2. Write down the expression for $\cos\theta$cosθ.
3. Hence, form an expression for $\frac{\sin\theta}{\cos\theta}$sinθcosθ.
4. Write down the expression for $\tan\theta$tanθ.
5. Does $\frac{\sin\theta}{\cos\theta}=\tan\theta$sinθcosθ=tanθ?
Yes
A
No
B
##### Question 2
Prove that $\frac{\tan x\cos x}{\sin x}=1$tanxcosxsinx=1.
##### Question 3
Simplify the following expression using complementary angles:
$\frac{\sin51^\circ}{\cos39^\circ}$sin51°cos39°<|endoftext|>
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“The Rockets’ Red Glare”:
Francis Scott Key and the Bombardment of Fort McHenry
In this lesson, students have examined the bombardment of Fort McHenry and the creation and history of “The Star-Spangled Banner.” Those interested in learning more, will find much useful information on the web.
Fort McHenry National Monument and Historic Shrine
Fort McHenry’s website contains a wealth of material on the fort, on the battle, on the defenders, and on the composing of the national anthem. It also includes a Teacher's Guide with lesson plans, bibliography and teacher evaluation, and curriculum-based lesson plans for Grades 4 and 8.
Star-Spangled Banner National Historic Trail, National Park Service
This website highlights a number of places associated with the Chesapeake Campaign of the War of 1812, with special emphasis on the events leading up to the writing of the national anthem.
Heritage Education Services, National Park Service
The Teaching with Historic Places website contains more than 135 curriculum-based lesson plans on places listed in the National Register of Historic Places. It includes lessons on other important American icons, including Independence Hall: International Symbol of Freedom; The Liberty Bell: From Obscurity to Icon; and The Washington Monument: Tribute in Stone.
National Register of Historic Places, National Park Service
The “Fort McHenry” National Register registration form is available online. Students interested in more information on the fortifications themselves and how they changed over time can find a detailed history of the fort in this form. It also includes a useful glossary of the old, fascinating terms used for various elements of forts and fortifications.
National Park Service Travel Itineraries
The Discover Our Shared Heritage travel itinerary for Baltimore includes, among other historic sites in the city, Fort McHenry and the Flag House, the home of Mary Pickersgill, who designed and fabricated the “Star-Spangled Banner.”
The National Register of Historic Places online itinerary Places Reflecting America’s Diverse Cultures highlights the historic places and stories of America’s diverse cultural heritage. This itinerary seeks to share the contributions various peoples have made in creating American culture and history.
U. S. Army Center of Military History
This website contains the chapter on the War of 1812 in The United States Army and the Forging of a Nation, 1775-1917, Volume 1 of American Military History, published by the U.S. Army Center of Military History in 2004. Clearly and engagingly written, it includes a detailed account of the origin and conduct of the war, and the forces affecting its outcome.
Smithsonian Institution, The Star-Spangled Banner
This website is tied to the original Star-Spangled Banner, on display at the National Museum of American History in Washington. It includes information on the War of 1812, the national anthem, and the restored flag that flew over Fort McHenry. It also provides a variety of lesson plans for primary through middle school students and bibliographies for both general and young readers.
Maryland Historical Society, Star-Spangled Banner Sheet Music Collection
This online exhibit includes the earliest extant manuscript version of Key’s poem, one of only two known copies of its earliest printing, and a great deal of other printed and manuscript material relating to the song.
The Star-Spangled Banner Flag House
This website highlights the Baltimore home and place of business of Mary Pickersgill, maker of the flag that inspired Francis Scott Key’s poem. The house is a National Historic Landmark and offers living history, special events, and a variety of educational programs.
Library of Congress, A Guide to the War of 1812
This website identifies and links to digital materials related to the War of 1812 that are available elsewhere on the Library of Congress website. In addition, it provides links to external websites focusing on the War of 1812 and a bibliography containing selections for both general and younger readers.
Library of Congress, American Memory
This website contains one of only five copies of the first setting of Key’s words to music as “The Star-Spangled Banner.”
Library of Congress, Historic American Buildings Survey
This website contains extensive documentation for Fort McHenry, including maps and photographs.
James Madison Center, War of 1812
This website contains information on the war at sea and in the northwest.
For further reading
The Star-Spangled Banner: The Making of an American Icon was written by Lonn Taylor, Kathleen M. Kendrick, and Jeffrey L. Brodie (New York: HarperCollins, for the Smithsonian Institution, 2008) on the occasion of the re-opening of the Star-Spangled Banner exhibit at the Smithsonian Institution’s National Museum of American History in 2008. It brings together a wealth of historical information on the War of 1812, on the Battle of Baltimore, and on Francis Scott Key and the writing of “The Star-Spangled Banner” in one beautifully illustrated volume. It also discusses the later history of the anthem and the flag, including the recent restoration of the original flag that flew over Fort McHenry on September 14, 1814.<|endoftext|>
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# Geometry Constructions
Related Topics:
More Lessons for High School Geometry
More Lessons for Geometry
Math Worksheets
A series of free, online High School Geometry Video Lessons and solutions.
Videos, worksheets, and activities to help Geometry students.
In these lessons, we will learn
• how to construct and duplicate a line segment
• how to construct and duplicate an angle
• how to construct a perpendicular bisector
• how to construct a perpendicular line through a point on the line
• how to construct a perpendicular line through a point not on the line.
### Duplicating a Line Segment
When constructing a line segment, we use a compass and straightedge to first draw a ray or line and then a point that will serve as an endpoint of the new segment. Next, we measure the given segment with a compass and make a mark with the pencil end. Without changing the spacing of the compass, place the sharp end of the compass on the point drawn on the new line/ray, and make a mark on the line/ray.
How to duplicate a line segment using a compass and straightedge.
How to use a compass and straightedge to construct a congruent segment.
### Duplicating an Angle
When constructing an angle, first swing an arc from the vertex of your angle. Then, swing a congruent arc from the new vertex. Return to the original angle; the drawn arc intersected the sides of the angle - measure this distance. On the new angle, place the sharp end of the compass on the intersection of the arc and ray and draw another arc. Draw a ray connecting the new vertex with the point of intersection.
How to duplicate an angle using a compass and straightedge.
How to use a compass and straightedge to construct a congruent angle.
### Constructing the Perpendicular Bisector
When looking at a line segment, there is only one line that will pass through the midpoint that will be a constant distance between the two endpoints. This line is called the perpendicular bisector. To construct the perpendicular bisector, we first find the midpoint of the line segment and then use a compass and straightedge to draw the perpendicular line.
How to define a perpendicular bisector
How to use a straight edge and a compass to bisect a segment.
### Constructing a Perpendicular at a Point on a Line
When constructing a perpendicular bisector, we are specifically being asked to construct a line perpendicular to a line through the midpoint. To construct a perpendicular to a line through a point like the midpoint, we use a process similar to constructing a perpendicular to a line through a point not on the line. To construct a perpendicular, we use a compass and straightedge to determine a point equidistant from two equidistant points on the line.
Perpendicular Line(1) Construction
How to use a compass and straightedge to construct a perpendicular line through a point on the line.
### Constructing a Perpendicular to a Line
The shortest distance between a point not on a line and a line is along the perpendicular to the line. Constructing a perpendicular to a line uses the same process as constructing the perpendicular bisector of a line segment, but with one additional step. The first step is to swing an arc from the point and intersect the line in two places, which creates a segment that can be bisected.
Perpendicular Line(2) Construction
How to use a compass and straightedge to construct a perpendicular line through a point not on the line.<|endoftext|>
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Last Updated on August 3, 2023 by user
Tedious math word problems are the bane of every 4th grader’s existence, but when a child reads a creative or exciting word problem, it changes everything. Engage a child’s mind just once, and you can spark a lifelong love of math as well as a passion for learning. Problems relating to concepts that your fourth grader already understands make it easier for them to understand fundamental concepts while polishing their problem-solving skills. Your 4th grader might not care about calculating the meeting point of two trains traveling at different speeds from opposite directions. Give them a word problem they can solve by playing with shapes or counting handshakes. By giving your child a problem concerning things he or she already finds interesting, you can help them find an interest.
### 1. The Bottle Water Project
Challenging 4th grade math problems involves multi-step questions. Consider this one: “Your class is collecting bottled water to donate to a needy community after a tornado. The teacher wants you to collect 300 bottles of water. You brought in three packages that have six bottles in each package. Your friend brings in six packages with six bottles in each package. How many do you still need to collect?”
This problem is challenging because it involves two steps. First, the student must determine how many bottles of water have been collected. Then, they must figure out how many are left.
The students have collected 54 bottles in all. This leaves 246 water bottles to collect.
The above is a perfect example of the teacher going beyond just 4th grade math worksheets, instead, focusing on something that the students will likely participate in sometime in their lives.
2. Finding the Perimeter or Area of Unusual Shapes
Finding the perimeter and area of standard shapes is not difficult for advanced fourth graders, as this is a foundational concept in many 4th grade math problems. To make it more challenging, create shapes that are unusual or unexpected.
For example, draw an octagon made from right angles. This will create several different rectangles. To find the perimeter, your child will first need to determine the length of any unmarked sides. Then, he will need to add all eight sides together. To find the area, the student will need to break the shape into the smaller rectangles, figure the area of each of these, and then add the areas together.
This incorporates a few math concepts and theories that your child will be able to master.
### 3. Number of Handshakes
This word problem reads like this: “There are five people at a party. If everyone shakes hands one time with all of the other partygoers, how many handshakes occurred?”
This problem requires the student to realize that the first person shakes hands with all of the other people, creating four handshakes. However, the second person only has to shake hands with the remaining three, creating three new handshakes, followed by two for the third person and one for the fourth person. By the time the handshakes reach the final person, he will have already shaken hands with everyone in the line. So, the answer is 4+3+2+1=10. This is a challenging 4th grade math problem because it does not follow a specific formula, but requires some additional thinking. Presenting multi-step word problems like this are important for your child’s development.
Sometimes, coming up with challenging 4th grade math problems for a student who needs the extra enrichment is not easy. Thinkster Math can help. This teacher-designed program offers the exact level of challenge that advanced students need to stay engaged and excited about math. If you have a student who loves fourth grade math, the game-like play and personalized teaching offered by Thinkster can offer an additional math challenge. If you have a student who struggles to engage in fourth grade math, game-like play, math worksheets, and personalized teaching offered by Thinkster can offer an additional math challenge.
We not only have the expertise of carefully vetted teachers but our tutors also have the experience of working with many other fourth-graders who have all kinds of learning styles. There will be no one standard unit that goes out to all students. That’s because every fourth grade student comes to us with different math skills. Some may understand divisibility rules perfectly while that same child may struggle with place value understanding.
Our tutors get it. Instead of applying rigorous rules about ‘how’ to learn they are busy coming up with ways that your individual child will learn – whether that’s a new math game or a smaller unit of study. If your child better retains what he or she writes down, maybe a math worksheet will be in order – or not. We do not pigeonhole kids in this program. It is highly individualized and tailored specifically for your child.
And the best part? It’s all online using tools your child is likely already familiar with. You can access your materials any time, anywhere. Is your child having a problem with tonight’s homework on multi-digit multiplication? You don’t have to surf the internet or look for a helpful video – just go to Thinkster.
One problem we see often is the child thinks they understand as the teacher explains the problem in class, an algebra problem for example, but once they get home and it’s time to execute the same concept on his or her homework, the concept is fuzzy.
Or does your child have to remember how to find the area model? We all know what it’s like trying to remember something that went on in a meeting without taking proper notes – and we’re adults. It’s just as difficult if not more so for a child struggling with learning not only mathematics but how to take notes to begin with.
Get your child the help they need by getting Thinkster. Our very experienced online tutors are available on your time schedule, not ours. Our lessons are individualized to your student, we don’t expect your student to fit a preconceived role.
And the benefits of Thinkster will surely multiply beyond just mathematics. A student that formerly struggled with math can become more confident in their studies. That confidence can translate into the rest of his or her academic life as well as all facets of their life.
We all want our children to succeed, using Thinkster is easy for the whole family. No getting to a tutoring session ‘on time’ because it’s on your time.
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Has your child been getting bored with 4th grade math problems? try out these 3 challenging problems to excite and motivate them.
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## Recommended Articles
### Mastering Math: How to Crack Challenging Math Problems in Under a Minute!
Math, often regarded as the king of all sciences, can sometimes pose challenges that leave even the brighte...<|endoftext|>
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# Column Method Multiplication – Definition, Examples | How to do Column Method Multiplication?
Are you confused about the column method multiplication? you landed on the correct page where you will get full information about column method multiplication. This includes the definition of the column method multiplication, steps to follow the column method multiplication. You can also find examples of column method multiplication by completely going through this article.
Do Refer: Expansion Method of Multiplication
## Column Method Multiplication | Long Multiplication
In column method multiplication, One number is written underneath another number and the numbers are multiplied together. It is also called long method multiplication.
For example 3 5
*2 5
——–——–——–
8 7 5
——–——–——–
### Column Method Multiplication 3 Digit by 2 Digit
The steps to be followed for the column method multiplication of a three-digit number by a two-digit number is as follows:
1. Multiply the one’s digit of the number by one’s digit of the multiplier.
2. Multiply the tens digit of the number by one’s digit of the number.
3. Multiply the Hundred’s digit of the number by one’s digit of the multiplier.
4. Multiply the one’s digit of the number by the tens digit of the multiplier.
5. Multiply the tens digit of the number by the tens digit of the number.
6. Multiply the Hundred’s digit of the number by the tens digit of the multiplier.
The same procedure has to be followed for the multiplication of the two-digit number by the two-digit number.
### Examples of Column Method Multiplication
Example 1:
Multiply 53, 23 by column method multiplication.
Solution:
1. Multiply the multiplicand by one’s digit of the multiplier.
5 3
*3ones
——–——–—
1 5 9
——–——–—
2. Multiply the multiplicand by tens digit of the multiplier.
5 3
*2tens
——–——–——–
1 0 6
——–——–——–
159 ones+106 tens
=159*1+106*10
=159+1060
=1,219.
The product of 53,23 is 1,219.
It means 5 3
* 2 3
——–——–——–——–
1 5 9 —-> 53*3=159
1 0 6 0 —-> 53*20=1060
1
——–——–——–——–——–
1 2 1 9 ——>53*23=1219
——–——–——–——–——–
Example 2:
Find Multiplication of 455,32 by column method multiplication?
Solution:
1. Multiply the multiplicand by one’s digit of the multiplier.
4 5 5
* 2 ones
——–——–——–——–
9 1 0
——–——–——–——–
2. Multiply the multiplicand by tens digit of the multiplier.
4 5 5
*3tens
——–——–——–——–
1 3 6 5
——–——–——–——–
910 ones+1365tens
=910*1+1365*10
=910+13650
=14560.
Multiplication of 455,32 is 14560.
It means
4 5 5
*3 2
——–——–——–
9 1 0 —->455*2=910
1 3 6 5 0 —–>455*30=13650
1
——–——–——–——–
1 4 5 6 0 —–>455*32=14560
——–——–——–——–——–
Example 3:
Find the multiplication of 1234,34 by column method Multiplication.
Solution:
1. Multiply the multiplicand by one’s digit of the multiplier.
1 2 3 4
*4ones
——–——–——–——–
4 9 3 6
——–——–——–——–
2. Multiply the multiplicand by tens digit of the multiplier.
1 2 3 4
*3 tens
——–——–——–——–
3 7 0 2
——–——–——–——–
1234*4ones+1234*3tens
=1234*4+1234*30
=4936+37020
=41,956.
Similarly, we can do multiplication of the four-digit number, multiplication of the five-digit number by two-digit numbers.
### Column Method Multiplication of 4 Digit by 2 Digit Examples
Example 1:
Find the multiplication of 34251,62 by column method multiplication?
Solution:
1.Multiply the multiplicand by one’s digit of the multiplier.
3 4 2 5 1
*2
——–——–——–——–
6 8 5 0 2
——–——–——–——–
2.Multiply the multiplicand by tens digit of the multiplier.
3 4 2 5 1
*6 tens
——–——–——–——–
2 , 0 5, 5 0 6
——–——–——–——–——–<|endoftext|>
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Divergent thinking is the ability to elaborate and cultivate diverse and original ideas with fluency and speed. It’s the snowball effect that occurs when mulling over one idea and instantly coming up with a variety of “what ifs,” connecting possibilities to find a broad range of solutions. This type of thinking is essential in business since it is necessary for innovation and can influence the overall success of an organization.
How does one measure the divergent thinking abilities of his or her staff?
There are seven elements associated with creative thinking abilities used to help identify divergent thinkers. They are:
- Fluency – The ability to come up with many ideas that, in turn, influence the number of possible solutions to a problem.
- Flexibility – Having different perceptions of certain ideas that can change the nature of a problem.
- Elaboration – Building on an idea to exhaust all avenues when problem solving.
- Originality – Developing fresh or daring ideas for processes that are considered to be “outside the box” of the normal ways of doing things.
- Complexity – Being able to peel back the layers of a seemingly difficult concept to understand it better.
- Imagination – Inventing ingenious ideas or products.
- Curiosity – Unabashed and eager to ask questions, probe further into concepts, and learn more about ideas or concepts to increase fluency.
One test used to gauge one’s creativity is Guilford’s Alternative Uses exam. It asks students to name all alternative uses for a common object and “grades” them on specific factors such as originality, fluency, flexibility and elaboration. These factors can be very insightful for leaders looking for certain qualities in their team dynamic.
Another test, The Candle Problem, challenges an individual to think outside the box and come up with unconventional ways to use a common item to solve a problem.
Many methods can be applied to measure the divergent thinking prowess of your team. By understanding where their strengths lie, you can develop a strong team of thinkers who will inevitably lead your organization toward innovation and future success.
To discover more on how you can implement divergent and critical thinking, Stir It Up with a workshop from one of our Ascendis Experts.<|endoftext|>
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Stars known as red dwarfs might be far more common than previously thought, enough to triple the total number of stars known in the universe, a new study suggests.
These new findings could also boost the number of planets that could harbor life, astronomers announced today (Dec. 1).
Red dwarfs are dim compared to stars such as our sun and just 10 to 20 percent as massive. As such, astronomers could not detect them in galaxies other than our Milky Way and its closest neighbors until now.
Now astronomers have used powerful scientific instruments on the W.M. Keck Observatory in Hawaii to detect the faint signature of red dwarfs in eight massive so-called elliptical galaxies located between about 50 million and 300 million light-years away.
"It is remarkable that we can measure and quantify the light of these incredibly feeble stars in galaxies outside of the Milky Way," researcher Pieter van Dokkum, an astronomer at Yale University, told SPACE.com.
Elliptical galaxies are some of the largest galaxies in the universe. The largest of these galaxies were thought to hold more than 1 trillion stars (compared with the 400 billion stars in our Milky Way). [Top 10 Star Mysteries]
The new finding suggests there may be five to 10 times as many stars inside elliptical galaxies than previously thought, which would triple the total number of known stars in the universe, researchers said.
The final, total tally for the number of stars in the universe still remains murky, the scientists cautioned. Determining the total star population of the cosmos requires an accurate number for galaxies as well, they added.
"It's this latter quantity that is quite uncertain. Nevertheless, best estimates are around 100 sextillion – a 1 with 23 zeros," van Dokkum said. "Our results would then triple that number, but again, there's a large additional uncertainty associated with this estimate."
The researchers' computer models based on these findings suggest that red dwarfs are far more common than expected, with these galaxies each possessing roughly 20 times more red dwarfs on average than the Milky Way, said researcher Charlie Conroy, an astronomer at the Harvard-Smithsonian Center for Astrophysics.
"The abundance of these stars is very surprising — there are many more than we had expected," van Dokkum said.
All in all, red dwarfs could make up at least 80 percent of the total number of stars and at least 60 percent of all the mass found in stars.
"No one knew how many of these stars there were," van Dokkum said. "Different theoretical models predicted a wide range of possibilities, so this answers a longstanding question about just how abundant these stars are."
These findings suggest that galaxies might contain less of the mysterious substance labeled "dark matter" than before thought. Instead, red dwarfs could contribute more mass than realized.
"We need to revise current estimates of the masses and star formation activity of nearby and distant galaxies," van Dokkum said.
Moreover, the increased number of red dwarfs means there could increase the number of planets orbiting stars, which in turn boosts the number of potentially habitable planets.
To date, astronomers have discovered just over 500 alien planets around other stars.
"There are possibly trillions of Earths orbiting these stars," van Dokkum said. He added that the red dwarfs they discovered, which are typically more than 10 billion years old, have been around long enough for complex life to evolve. "It's one reason why people are interested in this type of star."
Van Dokkum and Conroy detail their findings in the Dec. 2 issue of the journal Nature.
This article was provided by SPACE.com, a sister site of Live Science.<|endoftext|>
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2021-11-18
Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.
Othrom
f(x) is continous at x=a if and only if
$\underset{x\to a}{lim}f\left(x\right)=f\left(a\right)$
$\underset{x\to a}{lim}f\left(x\right)=\underset{x\to a}{lim}\frac{2x+3}{x-2}$
Divisionlaw for lomits
$\underset{x\to a}{lim}\frac{2x+3}{x-2}=\frac{\underset{x\to a}{lim}2x+3}{\underset{x\to a}{lim}x-2}$
$\frac{\underset{x\to a}{lim}2x+3}{\underset{x\to a}{lim}x-2}=\frac{\underset{x\to a}{lim}2x+\underset{x\to a}{lim}3}{\underset{x\to a}{lim}x-2}$
$\frac{\underset{x\to a}{lim}2x+\underset{x\to a}{lim}3}{\underset{x\to a}{lim}x-\underset{x\to a}{lim}2}=\frac{2\cdot \underset{x\to a}{lim}x+\underset{x\to a}{lim}3}{\underset{x\to a}{lim}x-\underset{x\to a}{lim}2}$
$\frac{2\cdot \underset{x\to a}{lim}x+\underset{x\to a}{lim}3}{\underset{x\to a}{lim}x-\underset{x\to a}{lim}2}=\frac{2a+3}{a-2}$
$\frac{2a+3}{a-2}=f\left(a\right)$
Hence proved that f(x) is continuous for all $a\ne 2$
Do you have a similar question?<|endoftext|>
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The resource has been added to your collection
Students prepare for this activity by working with a unidirectional flume with a sand bed. We adjust water depth, flow velocity, and channel slope to achieve a range of bed states, in an effort for them to understand the controls on bedforms. This portion of the activity could be done in lecture or via another exercise that makes use of digital video of actual experiments. The activity itself is a jigsaw: students form groups of three, each group responsible for plotting depth vs. velocity plots of bedform state for a single sand grain size range (0.10-0.14 mm, 0.5-0.64 mm, and 1.3-1.8 mm). These data are provided to them as Excel files and the data were directly 'stolen' from the original depth vs. velocity plots in Middleton and Southard (1984), Mechanics of Sediment Movement, SEPM Short Course Number 3. Datathief software (available free on the web) was used to steal the data. The data are arranged in columns: depth, velocity, and bedform type. Students must plot each of the different bedform types with a different symbol, then they have to define field boundaries. It is critical that they have never seen the original plots in their textbook. The goal is for them to derive them on their own, not to regurgitate what is in their textbook or elsewhere. After they complete their plots for each grain size range, the groups re-arrange themselves into groups of three with one representative from each of the grain size groups. They then must try to evaluate the effects of changing grain size on bedform state. Finally, after completing the exercise, the bedform analysis is linked to the cross stratification that is produced under conditions of high sediment fallout rates and the given bed state. The activity gives students practice working with realistic datasets, exposure to the role of physical modeling in sedimentary geology, and a chance to plot and interpret real data. Furthermore, it really solidifies the link between cross stratification and its dynamic interpretation from the rock record.
This resource has not yet been reviewed.
Not Rated Yet.<|endoftext|>
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## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 7 Decimals Check Your Progress
Question 1.
Convert the following decimal numbers into fractions (in lowest terms).
(i) 6.015
(ii) 0.876
(iii) 23.375
Solution:
(i) 6.015
Question 2.
Write the following fractions as decimals numbers:
Solution:
Question 3.
Arrange the following decimal numbers in ascending order:
(i) 123.8, 74.205, 74.209, 7.4209
(ii) 85.01, 85.1, 85.001, 85.103
Solution:
(i) 7.4209, 74.205, 74.209, 123.8
(ii) 85.001, 85.01, 85.1, 85.103
Question 4.
Arrange the following decimal numbers in desending order:
(i) 6.45, 4.65, 6.405, 64.5, 6.54
(ii) 73.5, 35.7, 7.35, 7.53, 7.035
Solution:
(i) 64.5 > 6.54 > 6.45 > 6.405 > 4.65
(ii) 73.5 > 35.7 > 7.53 > 7.35 > 7.035
Question 5.
If the school bags of Garima and Nakul weigh 5.2 kg and 4.832 kg respectively, find
(i) the total weight.
(ii) the difference in weight of the bags.
Solution:
Weight of Garima’s bag = 5.2 kg.
Weight of Nakul’s bag = 4.832 kg.
Question 6.
Evaluate the following:
(i) 31.42 – 17.853 -6.43
(ii) 13.01 – 5.428 – 3.703 + 2.99.
Solution:
(i) 31.42 – 17.853 – 6.43
Question 7.
By how much does the sum of 15.453 and 31.647 exceed the sum of 18.47 and 19.506?
Solution:
Question 8.
Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Solution:
Distance travelled everyday
= 20 km 50 m = 20 km + 50 m
= 20 km + $$\frac{50}{1000}$$ km
= 20 km + 0.050 km [∵ $$\frac{1}{1000}$$km = 0.001 km]
= (20 + 0.050) km = 20.050 km
Distance travelled by bus
= 10 km 200 m = 10 km + 200 m
= 10 km + $$\frac{200}{1000}$$ km
= 10 km + 0.200 km [∵ $$\frac{1}{1000}$$ km = 0.001 km]
= (10 + 0.200) km = 10.200 km
∴ Distance travelled by auto
= 20.050 km – 10.200 km = 9.850 km
Question 9.
Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour (aata). Find the total weight of his purchases.
Solution:
Weight of rice purchased
= 5 kg 400 g = 5 kg + 400 g
= 5 kg + $$\frac{400}{1000}$$ kg [∵ 1 g = $$\frac{1}{1000}$$ kg]
= 5 kg + 0.400 kg
= (5 + 0.400) kg = 5.400 kg
Weight of sugar purchased
= 2 kg 20 g = 2 kg + 20 g
= 2 kg = $$\frac{20}{1000}$$ kg [∵ 1 g = $$\frac{1}{1000}$$ kg]
= 2 kg + 0.020 kg = 2.020 kg
Weight of flour purchased
= 10 kg 850 g = 10 kg + 850 g
= 10 kg + $$\frac{850}{1000}$$ kg [∵ 1 g = $$\frac{1}{1000}$$ kg]
= 10 kg + 0.850 kg
= (10 + 0.850) kg = 10.850 kg
∴ Total weight of his purchases is
Question 10.
1 kg of pure milk contains 0.263 kg of fat. How much fat is there 15.5 kg of milk?
Solution:
1 kg of pure milk contains fat = 0.263 kg
∴ 15.5 kg of milk contain fat
Question 11.
The product of two numbers is 15.275. If one number is 4.7, find the other.
Solution:
The product of two number = 15.275
One number = 4.7
∴ Then other is = $$\frac{15.275}{4.7}=3.25$$<|endoftext|>
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5000 years BC
Cotton was first cultivated around 7000 years ago in the Indus Valley in what is now Pakistan. The important point to remember here is that the cotton wasn't just wild but was cultivated. It an industry which was to spread globally. Cloth weaving and fabric manufacture spread throughout India and across to the Mediterranean countries over the centuries. This state of affaires remained through to the Middle Ages however cotton farming did spread to the Americas where it was also widely grown and processed.
Right up to the late medieval period cotton was an imported product in northern Europe with little knowledge of how it was actually produced. But things were changing. The East India Company had a hold over India to the extent that although the cotton was grown in India it was against the law to process the raw cotton into cloth. This processing would be moved to the industrial mills of England and specifically Manchester (which was nicknamed 'cottonopolis'). With the industrial Revolution in full flow England became the worlds leading exporter of cotton fabric without actually growing any cotton. With all the machinery to speed the processes the only thing needed was a ready supply of cotton.
Fortunes won and lost
From then on the fortunes of people, businesses, banks and even countries were decided on where cotton came from. During the American civil war blockades denied the British the raw cotton from the southern states needed to keep the factory wheels turning. Egyptian cotton plantations were expanded with huge loans from European banks being invested only for the market to return to to the America once the civil war was over; Egypt was left bankrupt.
For years and years before the American civil war much of the cotton was picked by slave labour in the south but the end of the war saw share croppers both black and white farming and providing much of the raw cotton for the waiting mills.
The 21st century
Globally the early 20th century saw this highly labour intensive industry change yet again. Machines were taking the place of people. The delicate bolls of cotton once picked by hand were mechanically harvested and processed. The invention in the early 19th century and development of the cotton 'gin' was central to this. Its the machine used to gently separate the cotton from the unwanted seeds.
These days the top cotton producers are China, USA, India, Pakistan and Brazil with China producing most. The main uses for cotton range from textiles including clothes and bedding to unusual applications like fishing nets to coffee filters. Cotton is also mixed together with synthetics like rayon and polyester to create hard wearing fabrics.
No alternative to luxury cotton
Over the past 7000 years cotton has been central to creating economies, supporting civilisations and cultures. It's started wars and ended wars. Lost fortunes and made fortunes. Bankrupted countries and built economies. It's a staple crop for many countries and up to yet no man-made alternative has come close to replacing it, it remains a unique commodity... Sleep tight.
To find out more about our luxury bedding that is available up to 1000 thread count, simply request your free brochure below or call 01777 869 669.<|endoftext|>
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Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 14 Polygons are provided here with simple step-by-step explanations. These solutions for Polygons are extremely popular among Class 8 students for Math Polygons Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 8 Math Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
#### Question 1:
Find the measure of each exterior angle of a regular
(i) pentagon
(ii) hexagon
(iii) heptagon
(iv) decagon
(v) polygon of 15 sides.
Exterior angle of an n-sided polygon = ${\left(\frac{360}{n}\right)}^{o}$
(i) For a pentagon:
(ii) For a hexagon:
(iii) For a heptagon:
(iv) For a decagon:
(v) For a polygon of 15 sides:
#### Question 2:
Is it possible to have a regular polygon each of whose exterior angles is 50°?
Each exterior angle of an n-sided polygon = ${\left(\frac{360}{n}\right)}^{o}$
If the exterior angle is 50°, then:
Since n is not an integer, we cannot have a polygon with each exterior angle equal to 50°.
#### Question 3:
Find the measure of each interior angle of a regular polygon having
(i) 10 sides
(ii) 15 sides.
For a regular polygon with n sides:
(i) For a polygon with 10 sides:
(ii) For a polygon with 15 sides:
#### Question 4:
Is it possible to have a regular polygon each of whose interior angles is 100°?
Each interior angle of a regular polygon having n sides =
If each interior angle of the polygon is 100°, then:
Since n is not an integer, it is not possible to have a regular polygon with each interior angle equal to 100°.
#### Question 5:
What is the sum of all interior angles of a regular
(i) pentagon
(ii) hexagon
(iii) nonagon
(iv) polygon of 12 sides?
Sum of the interior angles of an n-sided polygon = $\left(n-2\right)×180°$
(i) For a pentagon:
(ii) For a hexagon:
(iii) For a nonagon:
(iv) For a polygon of 12 sides:
#### Question 6:
What is the number of diagonals in a
(i) heptagon
(ii) octagon
(iii) polygon of 12 sides?
Number of diagonal in an n-sided polygon = $\frac{n\left(n-3\right)}{2}$
(i) For a heptagon:
$n=7⇒\frac{n\left(n-3\right)}{2}=\frac{7\left(7-3\right)}{2}=\frac{28}{2}=14$
(ii) For an octagon:
$n=8⇒\frac{n\left(n-3\right)}{2}=\frac{8\left(8-3\right)}{2}=\frac{40}{2}=20$
(iii) For a 12-sided polygon:
$n=12⇒\frac{n\left(n-3\right)}{2}=\frac{12\left(12-3\right)}{2}=\frac{108}{2}=54$
#### Question 7:
Find the number of sides of a regular polygon whose each exterior angle measures:
(i) 40°
(ii) 36°
(iii) 72°
(iv) 30°
Sum of all the exterior angles of a regular polygon is ${360}^{o}$.
(i)
(ii)
(iii)
(iv)
#### Question 8:
In the given figure, find the angle measure x.
Sum of all the interior angles of an n-sided polygon = $\left(n-2\right)×180°$
∴ x = 105
#### Question 9:
Find the angle measure x in the given figure.
For a regular n-sided polygon:
Each interior angle = $180-\left(\frac{360}{n}\right)$
In the given figure:
∴ x = 108
#### Question 1:
How many diagonals are there in a pentagon?
(a) 5
(b) 7
(c) 6
(d) 10
(a) 5
For a pentagon:
$n=5$
#### Question 2:
How many diagonals are there in a hexagon?
(a) 6
(b) 8
(c) 9
(d) 10
(c) 9
Number of diagonals in an n-sided polygon = $\frac{n\left(n-3\right)}{2}$
For a hexagon:
#### Question 3:
How many diagonals are there in an octagon?
(a) 8
(b) 16
(c) 18
(d) 20
(d) 20
For a regular n-sided polygon:
Number of diagonals =: $\frac{n\left(n-3\right)}{2}$
For an octagon:
$n=8\phantom{\rule{0ex}{0ex}}\frac{8\left(8-3\right)}{2}=\frac{40}{2}=20$
#### Question 4:
How many diagonals are there in a polygon having 12 sides?
(a) 12
(b) 24
(c) 36
(d) 54
(d) 54
For an n-sided polygon:
Number of diagonals = $\frac{n\left(n-3\right)}{2}$
#### Question 5:
A polygon has 27 diagonals. How many sides does it have?
(a) 7
(b) 8
(c) 9
(d) 12
(c) 9
#### Question 6:
The angles of a pentagon are x°, (x + 20)°, (x + 40)°, (x + 60)° and (x + 80)°. The smallest angle of the pentagon is
(a) 75°
(b) 68°
(c) 78°
(d) 85°
(b) 68°
Sum of all the interior angles of a polygon with n sides = $\left(n-2\right)×180°$
#### Question 7:
The measure of each exterior angle of a regular polygon is 40°. How many sides does it have?
(a) 8
(b) 9
(c) 6
(d) 10
(b) 9
#### Question 8:
Each interior angle of a polygon is 108°. How many sides does it have?
(a) 8
(b) 6
(c) 5
(d) 7
(c) 5
Each interior angle for a regular n-sided polygon = $180-\left(\frac{360}{n}\right)$
#### Question 9:
Each interior angle of a polygon is 135°. How many sides does it have?
(a) 8
(b) 7
(c) 6
(d) 10
(a) 8
#### Question 10:
In a regular polygon, each interior angle is thrice the exterior angle. The number os sides of the polygon is
(a) 6
(b) 8
(c) 10
(d) 12
(b) 8
For a regular polygon with n sides:
Each exterior angle = $\frac{360}{n}$
Each interior angle = $180-\frac{360}{n}$
#### Question 11:
Each interior angle of a regular decagon is
(a) 60°
(b) 120°
(c) 144°
(d) 180°
(c) 144°
Each interior angle of a regular decagon = $180-\frac{360}{10}=180-36={144}^{o}$
#### Question 12:
The sum of all interior angles of a hexagon is
(a) 6 right ∠s
(b) 8 right ∠s
(c) 9 right ∠s
(d) 12 right ∠s
(b)
Sum of all the interior angles of a hexagon is $\left(2n-4\right)$ right angles.
For a hexagon:
#### Question 13:
The sum of all interior angles of a regular polygon is 1080°. What is the measure of each of its interior angles?
(a) 135°
(b) 120°
(c) 156°
(d) 144°
(a) 135°<|endoftext|>
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# Solve Recurrence Relation Using Iteration/Substitution Method
Iteration/Substitution Method
The Iteration Method, is also known as the Iterative Method, Backwards Substitution, Substitution Method, and Iterative Substitution. It is a technique or procedure in computational mathematics used to solve a recurrence relation that uses an initial guess to generate a sequence of improving approximate solutions for a class of problems, in which the n-th approximation is derived from the previous ones.
Let’s take a look at a recurrence relation problem and try to solve it. By solve it, here I mean get it in a “closed form” . A Closed-Form Solution is an equation that solves a given problem in terms of functions and mathematical operations from a given generally-accepted set. For example, an infinite sum would generally not be considered closed-form.
Let’s start with the recurrence relation, T(n) = 2 * T(n/2) + 2, and try to get it in a closed form. Note that ‘T’ stands for time, and therefore T(n) is a function of time that takes in input of size ‘n’.
T(n) = 2T(n/2) + 2
Now we need to figure out what T(n/2) is. We can do that by taking “n/2” , and putting it into our original function T(n), to get the following:
Note: We are just replacing n with n/2.
T(n/2) = 2T( (n/2) / 2) + 2
= 2T( n/4 ) + 2
So our original equation looks like the following when k= 2
T(n) = 2T(n/2) + 2
= 2( 2T( n/4 ) + 2) + 2
= 4T(n/4) + 4 + 2
That was our second iteration, so this means k=2 .Now we must figure out what T(n/4) is. We can do that by taking “n/4” , and putting it into our original function T(n), to get the following:
T(n/4) = 2T( (n/4) / 2 ) + 2
= 2T(n/8) + 2
So our original equation looks like the following when k=3
T(n) = 4T(n/4) + 4 + 2
= 4(2T(n/8) + 2) + 4 + 2
= 8T(n/8) + 8+ 4 + 2
That was our third iteration, so this means k=3 . Now we must figure out what T(n/8) is. We can do that by taking “n/8” , and putting it into our original function T(n), to get the following:
T(n/8) = 2T( (n/8) / 2 ) + 2
= 2T(n/16) + 2
So our original equation looks like the following when k=4
T(n) = 8T(n/8) + 8+ 4 + 2
=8(2T(n/16) + 2) + 8+ 4 + 2
= 16T(n/16)+16 + 8+ 4 + 2
That was our fourth iteration, so this means k=4 . Now we must figure out what T(n/16) is. We can do that by taking “n/8” , and putting it into our original function T(n), to get the following:
T(n/16) = 2T( (n/16) / 2 ) + 2
= 2T(n/32) + 2
So our original equation looks like the following when k=5
T(n) = 16T(n/16)+16 + 8+ 4 + 2
=16(2T(n/32) + 2)+16 + 8+ 4 + 2
= 32T(n/32)+ 32 +16 + 8+ 4 + 2
Okay let’s stop here and see if we can see a pattern, if not then you should continue this method or process until you do.
When k=1, we had T(n) = 2T(n/2) + 2
When k=2, we had T(n) = 4T(n/4) + 4 + 2
When k=3, we had T(n) = 8T(n/8) + 8+ 4 + 2
When k=4, we had T(n) = 16T(n/16)+16 + 8+ 4 + 2
When k=5, we had T(n) = 32T(n/32)+ 32 +16 + 8+ 4 + 2
When k=1, we had T(n) = 2T(n/2) + 2
= T(n) = 2T(n/2) + 2 →T(n) = 2T(n/2) + 2(1)
When k=2, we had T(n) = 4T(n/4) + 4 + 2
= T(n) = 4T(n/4) + 4 + 2 → T(n) = 4T(n/4) + 2(2 + 1)
When k=3, we had T(n) = 8T(n/8) + 8+ 4 + 2
= T(n) = 8T(n/8) + 8+ 4 + 2 → T(n) = 8T(n/8) + 2( 4 + 2 +1)
When k=4, we had T(n) = 16T(n/16)+16 + 8+ 4 + 2
= T(n) = 16T(n/16)+16 + 8+ 4 + 2 → T(n) = 16T(n/16)+2(8+ 4 + 2 +1)
When k=5, we had T(n) = 32T(n/32)+ 32 +16 + 8+ 4 + 2
=T(n) = 32T(n/32)+ 32 +16 + 8+ 4 + 2
= T(n) = 32T(n/32)+ 2 (16 + 8+ 4 + 2 + 1)
Notice the general form in terms of ‘k’ looks like:
T(n) = 2^k * T(n/ (2^k) + 2 *Summation from i=0 to k-1 of 2^i
The summation from i=0 to k-1 of 2^i is 2^(k) — 1.
So our general form looks like below:
T(n) = 2^k * T(n/ (2^k) )+ 2 *(2^k — 1)
T(n) = 2^k * T(n/ (2^k) )+ 2^(k+1) — 2
When does our recurrence relation stop ? Well it stops when T(n/(2^k)) hits the base case. You may say, but we were never given a base case. That’s okay because every recurrence relation has a base case, and it is always some constant value. So if a base case isn’t provided, we can make one up. So our base case will be T(1) = C, which means when our input is of size 1, the time it takes to execute the function ‘T’ is ‘C’ unit(s) of time, where ‘C’ is some constant unit of time. This means when n = 1, T(n=1) = C.
We must now get our T(n) function from our general form to stop, and therefore need T(n/(2^k))=C, this implies n/(2^k) must equal 1, since T(n/(2^k)=1) = C. Let’s do some algebra below.
= n/(2^k) = 1
= n = (2^k)
= log base 2 of n = k
We now not only have our stopping case, but can also put our value ‘k’ in terms of n, for the entire equation, by plugging in ‘log base 2 of n’ for ‘k’.
T(n) = 2^(log base 2 of n) * T(n/ (2^(log base 2 of n)) )+ 2 *(2^(log base 2 of n) — 1)
T(n) = n * T(n/n) + 2( n — 1)
T(n) = n T(1) + 2n — 2
T(n) = n*(C) + 2n — 2
T(n) = Cn + 2n — 2
T(n) = n ( C+2) — 2
So our guess is that the closed form of our recurrence relation is:
T(n) = n ( C+2) — 2, where C is some constant
NOTE: n ( C+2) — 2 is O(n)
If C = 0 then the equation is
T(n) = n(0 + 2) — 2
T(n) = 2n — 2 = 2( n-1)
Because the closed form is a guess we need to show that our guess is correct and to do that we can use Mathematical Induction.
There are many places to learn more about recurrence relation online, and at school. I have a course on Udemy.com called “Recurrence Relation Made Easy” where I help students to understand how to solve recurrence relations and asymptotic terms such as Big-O, Big Omega, and Theta, be sure to check it out !
Thanks for reading this article I hope its helpful to you all ! Keep up the learning, and if you would like more computer science, programming and algorithm analysis videos please visit and subscribe to my YouTube channels (randerson112358 & compsci112358 )
Written by<|endoftext|>
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Saturday , July 24 2021
Calculators
# How to Calculate Center of Gravity
In physics, center of gravity is the point where the entire weight of a body can be assumed to be concentrated. There is another term used in science, called center of mass, which is the point where the entire mass of the body can be assumed to be concentrated. We need to calculate the mean position of the mass of the object to find its center of mass. These two points are same only in the case of uniformly distributed gravitational force on the object.
## Definition of center of gravity
In short definition, the center of gravity is the geometric mean position of the weight of the object. Whereas center of mass is the mean position of total mass of the object.
A simple definition for kids is that, It is such a point on an object from which if it is allowed to suspended through a string, then the body will be stable and do not topple.Using this point of view, a complex object of weight W can be considered as a particle of weight W at its center of gravity. For many scientific calculations, we can describe the motion of any complex object in terms of the motion of its center of gravity; also the rotation of the object can be considered around its center of gravity.
## Calculating center of gravity
Calculating the center of gravity of a simple object like ball is very easy, if it is uniform, its center of gravity will be at its center. But for complex shaped objects, generally calculus is required to calculate the center of gravity. If the object has a line of symmetry, the center of gravity will lie on that line.For a system of objects, we can use the following formula,
Center of gravity = (sum of product of masses and their positions)/ sum of masses of the objects
where,
is the position of each mass in the system, and
is the nos of mass in the system
In 2D problems of physics the center of gravity of all the particles will lie in the same plane, making the calculation easier, but in 3D all the three coordinates are involved and it also represents a more practical situation.
Let us assume a system of particles of mass and position of their center of gravity as respectively.
Then the center of gravity of the system can be found by following method.
## Examples of center of gravity
Center of gravity has a huge relevance in our everyday life.
Taller things topple easily than shorter ones, because the center of gravity of a tall object is farther from the surface of earth than the shorter. Thus, racing cars are designed with their center of gravity closer to the ground, since it reduces their chances to tip over while they are at a fast speed.
Also, airplanes and rockets rotate around their center of gravity.
Therefore, center of gravity plays a vital role in various aspects of the real life.
Let us see a calculation example below
Consider a lamppost of height 20 m, equally distributed mass 10 kg. A bird of mass 2 kg is sitting at the top of the lamppost. What would be the center of gravity of the system consisting of the lamppost and the bird?
Since the lamppost is vertically standing and the bird is sitting at the top of it. The axis of the center of gravity of the bird and the post will be same. We can consider the mass of the bird to be concentrated at the top of the lamppost, that is, at 20 m from the ground. The center of gravity of the lamppost will be at its geometric center, that is, 10 m.
Now using,
Thus the center of gravity of the system will be situated at a height of 11.66 m from ground.
### Authored by Kiran Mandia
I am a computer science student at Indian Institute of Technology, Kharagpur; and a writer who loves to read about various topics.I also like making short films.
## 10 Interesting Facts To Know About Photosynthesis
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## Calculate Power Consumption and Average Monthly Electricity Bills of Your Home
Every household that uses electricity needs to pay monthly electricity bills, which is important part ...
### One comment
1. This helped me impress my present girlfriend!! Thanks a lot.<|endoftext|>
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Of the many fictional and theoretical Faster Than Light (FTL) travel methods, Warp differs in that it does not instantaneously take you to your destination. In order to travel faster than light, you either have to bend space itself to create a wormhole or change your own mass to be lighter than that of a photon (which is close to massless). Warp Drives utilizes the latter.
In Star Trek it is described as a bubble around the ship that warps the space around the ship, rather than the ship itself, thus allowing the ship to be virtually massless. Thus the so-called time barrier is broken, and no time dilation occurs for the ship (time dilation is when time moves slower for things objects move closer to the speed of light relative to objects that move slower).
This fictional method of propulsion was first introduced in the novel Islands In Space by John W. Campbell in 1931. It was most famously adopted in Star Trek: The Original Series and has been used in every Star Trek series and film since then. But is such a thing possible?
In 1994, physicist Miguel Alcubierre theorized that a field of energy-density lower than vacuum could be created and this would create negative mass, allowing for warp speeds. Using this method, called the Alcubierre Drive, one would not move through space in a local frame of reference, but instead contract space in front of the craft and expanding it again behind the craft. Later calculations proposed this would require exotic matter to fuel it. Exotic matter is matter that is not constructed with protons and electrons, but some other exotic subatomic particles.
If such matter were to be discovered, the drive would be theoretically possible, as it does not conflict with Einstein’s field equations.
In 2012, NASA researcher Harold White theorized that it would be possible to build a Alcubierre Drive, if the shape of the drive and the field were changed from the originally proposed one. NASA is currently researching the technology and the preliminary research has been promising. We might have Warp Drives in the near future.<|endoftext|>
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When thinking about cost behavior, think about how the cost behaves in total. A variable cost is a cost that varies in total. The cost increases or decreases based on a related activity.
The formula for total variable cost is:
Total Variable Cost = Variable Rate X Activity
### Assume a constant rate
For planning and decision making purposes, we assume that the variable rate is constant. This allows for a single variable in the calculations. Only the activity will change. Now, that is not always the case, but as long as we are within the relevant range for our decision, we can assume that the rate will stay the same.
### But isn’t it fixed if the rate stays the same?
Remember that a variable cost varies in total. The rate might stay the same but once you multiply the rate by varying levels of activity, the total variable cost will change.
Imagine that you are selling candy bars as a fundraiser for a club to which you belong. Your cost is 50 cents per candy bar and the club sells the candy bars for \$1 each. If the club sells 200 candy bars, what is the total variable cost? Is it 50 cents? No, that is the cost of a single candy bar. If you sell 200, you would need to multiply that by 50 cents for each of the candy bars sold.
200 candy bars X 50 cents per candy bar = \$100
What if the club sold 500 candy bars? The total variable cost would be \$250.
Here is a graph of the total variable cost of candy bars for the fundraiser:
Notice that if no candy bars are sold, there is no cost. The more candy bars that are sold, the higher the cost. The cost line is a straight line. The slope of the line is equal to the variable rate. For each additional unit sold, the line increases at a rate of 50 cents. Think of the formula of a line: y=mx + b, where y is your y coordinate, x is your x coordinate, m is the slope and b is the y-intercept (the point where the line hits the y-axis).
The formula for total variable cost is: y=mx. The y-intercept for a variable cost is always zero because if there is no activity, there is no cost. Therefore, the line will always start at 0,0. The slope of the line, m, is your variable rate. The activity is x. See your math teacher was right when he or she told you you would use this stuff someday!
Frequently, you will see textbooks show the formula for the slope of a line as the formula for cost equations.
#### Related Videos
Cost Behavior: Fixed, Variable, Step and Mixed
Fixed and Variable costs as per unit and total costs
We have previously discussed cost objects and assigning costs to cost objects. One object that is used frequently by job costing.
Job costing is a cost allocation method used by companies that make custom products. Imagine a cabinet maker who makes custom cabinets for homes. Each cabinet is custom built based on measurements made inside the customer’s home. All jobs require different amounts of material, labor, and overhead. Therefore, each job has a unique total cost.
Direct costs are easy to assign to jobs. It is easy to calculate the total cost of direct materials based on the materials used in the job. Companies use job cost sheets to record the cost of materials used on the job.
Direct labor is easy to calculate as we know how much each of the employees earns per hour and how many hours the employees worked on the job. Employees track which jobs they are working on throughout the week so that direct labor costs can be added to the job cost sheets.
Overhead can cause problems for organizations that use job costing. Without allocating overhead to jobs, we do not have an accurate idea of the cost of the job. Since overhead is such a large part of the cost of operating a business, we cannot ignore it. So how do we allocate it?
We could wait until the end of the year, when we know actual numbers but that does not help us determine if our jobs are profitable when they are completed. If we waited until we had all the overhead spending calculated for the year, it would be January of 2017 before we know if jobs completed in 2016 were profitable or not. We can’t wait that long if we want to determine if we are making profit on our jobs.
We could use overhead costs for the current or previous month and apply those costs to the job, but what if our overhead costs vary greatly from month to month. For example, what if we get the annual renewal for our insurance policies in March. Those insurance policies cost \$100,000. Is it ethical to allocate that \$100,000 only to the jobs that are completed in March or April? No, those costs should be allocated to all jobs completed throughout the year, not just those unlucky customers who happened to place in order with our company in March or April.
If we don’t know the actual costs, how can we allocate overhead fairly to all of our jobs? We are going to use estimates.
Most businesses complete budgets prior to the beginning of each year. These budgets include estimated overhead and estimated activity. We can use those estimates to calculated a rate that we can apply to our jobs. We call this rate the predetermined overhead rate.
Predetermined overhead rate = Estimated Overhead / Estimated Activity
The company can choose any activity it believes will most accurately apply the overhead costs. Most companies use direct labor hours or machine hours to allocate overhead costs.
Example #1
K’s Premier Cabinets uses job costing to calculate the cost of jobs as they are completed. The company estimates that it will have \$1,250,000 in overhead costs in 2015. The company believes that employees will work 200,000 hours and that 150,000 machine hours will be used during 2015. Calculate the predetermined overhead rate assuming that the company uses direct labor hours to allocate overhead to jobs.
Predetermined overhead rate is estimated overhead divided by estimated activity. The example states that estimated activity is 200,000 direct labor hours.
\$1,250,000 / 200,000 direct labor hours = \$6.25 per direct labor hour (don’t forget to label your numbers)
### Applying the predetermined overhead rate to jobs
Alright, so we have a rate. Now what?
We must apply the rate to our jobs. The rate from our example above is \$6.25 per direct labor hour. How do we apply this to our jobs? Based on the direct labor hours worked on the job. For each direct labor hour worked, we will add \$6.25 of overhead to the job.
Applied overhead is overhead added to a job by taking the predetermined overhead rate multiplied by the actual activity. Applied overhead is added to direct materials and direct labor to calculate total job cost.
Total Job Cost = Direct Materials + Direct Labor + Applied Overhead
Every time a job is completed, overhead is applied to the job. The total cost of all the jobs completed over the course of the year is cost of goods sold.
Example #2
K’s Premier Cabinets completes job #322 on July 7. The job used 45 direct labor hours and 30 machine hours. The job consumed \$1,800 worth of materials. The average direct labor rate is \$18.00 per hour and the company uses the predetermined overhead rate calculated in Example #1. Calculate the total cost of job #322.
There are three components of job cost: direct materials, direct labor and applied overhead.
We are told that direct materials are \$1,800.
We must calculate total direct labor cost. Direct labor is a variable cost. Our rate is \$18.00 per direct labor hour. Our driver is 45 hours. If we calculate rate x activity, our total direct labor cost for job #322 is:
\$18.00 per direct labor hour X 45 direct labor hours = \$810
Lastly, we just apply overhead to the job. Overhead is applied by taking our predetermined overhead rate and multiplying it by activity. We are given activity for direct labor hours and for machine hours. Which one do we use? Remember when I said to label your answer? This is precisely why it is so important to label your rate! So what was our rate?
\$6.25 per direct labor hour
Therefore, we are going to apply the rate based on direct labor hours.
\$6.25 per direct labor hour X 45 direct labor hours = \$281.25
We now have the components of job cost. Now to add them up and see what the total cost of the job is.
Direct materials = \$1,800
Direct labor = \$810
Applied overhead = \$281.25
Total job cost = \$2,891.25
### Final Thoughts
One of the most important pieces of advice I can give you about allocating overhead using a predetermined overhead rate is keeping the terminology straight in your mind.
Estimated overhead is calculated at the beginning of the year before any work begins. Applied overhead is calculated throughout the year as jobs are completed and added to the cost of the jobs.
First calculate your predetermined overhead rate using estimates and LABEL YOUR ANSWER! Then apply overhead to jobs by using the predetermined overhead rate and actual activity.
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### 8.NS.1
Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
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### 8.NS.2
Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., ?2). For example, by truncating the decimal expansion of ?2, show that ?2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
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### 8.EE.1
Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 × 3–5 = 3–3 = 1/33 = 1/27.
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### 8.EE.2
Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that ?2 is irrational.
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Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3 × 108 and the population of the world as 7 × 109, and determine that the world population is more than 20 times larger.
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Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology.
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Butterflies fluttering around the garden are a beautiful symbol of summer. Unfortunately, there have been sharp declines in different butterfly species, due to habitat loss, pollution and invasive plants. That matters because these flying insects aren’t just pretty — they are important plant pollinators and valuable food for birds and other insects.
Fortunately, you can help. By following a few simple gardening tips and by growing plants that attract butterflies, you can support these beautiful critters for future generations to enjoy.
Photo via Bill Gracey/Flickr Creative Commons
What to plant?
Butterflies have very specific needs for food, based on their life stage. Larvae depend on certain plants, for instance, while the adults may later feed on different parts of various plants.
Some advice on how to attract butterflies…
- Go native: “Many butterflies and native flowering plants depend on each other for survival and reproduction,” reports the National Wildlife Federation. That’s why planting native plants is always a good idea. Look for native plants specific to your region for best results.
Photo via e3000/Flickr Creative Commons
- Flower types: Butterflies like flat or clustered blossoms in purple, pink, red, orange or yellow colors, according to the National Wildlife Federation. Select flowers that bloom at different types, so there are continuous blossoms for butterflies to enjoy. Some flower examples are asters, bee balm, calendula, lupine, verbena and phlox.
- Butterfly bush: It’s true that butterfly bush (Buddleia spp.) attracts plenty of the winged insects. But keep in mind that in some areas, such as the Pacific Northwest, butterfly bush has been declared a noxious weed and should be avoided.
- Feed caterpillars: It’s important to grow host plants that caterpillars eat, so they will stay around in your garden when they eventually become butterflies. For instance, to attract viceroys, try planting willows, cottonwoods and aspens. For silver-spotted skipper, grow wisteria.
Swallowtail photo by SFAJane/Flickr Creative Commons
Other butterfly gardening tips
- Think sun: Butterflies only drink nectar from plants grown in full sun, or at least 6 hours of sun daily. The insects use the sun to navigate and warm their wings for flight. For best results, give butterflies a sunny garden that’s sheltered from the wind.
- Provide support: Create butterfly-friendly areas in your garden by leaving flat stones for them to rest in the sun. A shallow pan with moist wet sand can be buried in the garden soil to provide a place for butterflies to drink water and extract minerals.
- Forget chemicals: Accept a little imperfection in the garden, and avoid using pesticides, herbicides or fungicides, which have chemicals that kill caterpillars and butterflies.
Monarch photo via Puzzler4879/Flickr Creative Commons
No article about attracting butterflies to the garden would be complete without mentioning the monarch butterfly, which needs special attention right now.
According to Xerces, the monarch butterfly has declined by 90 percent from its 20-year average since the mid-1990s. “If monarchs were people, that would be like losing every living person in the United States except those in Florida and Ohio,” according to this non-profit organization studying this issue.
To help save monarchs, start growing milkweed in your garden. Monarchs need milkweed to survive. It’s the only plant that the caterpillars eat, and the butterflies need milkweed to lay eggs. Make sure you grow a milkweed that’s specific for your growing area.
Learn more about monarchs, milkweed and how you can help by visiting Monarch Joint Venture.
For other helpful advice about monarchs and other butterflies, visit National Wildlife Federation.
You might also enjoy our post on attracting pollinators to your garden.
How do you attract butterflies to your garden?
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Description of trematodes
Trematodes are representatives of worms from the type of flatworms. If you translate their name from the Greek language, it will look like "having a sucker." Another name for the trematode class is flukes. Representatives are quite numerous and only about forty species can develop in their human body.Representatives of trematodes cause severe pathological processes in the host, in which they parasitize. These diseases are called trematodoses and can even be fatal. Such types of diseases as opisthorchiasis, schistosomiasis, fascioliasis, clonorchosis and dicroceliosis are distinguished.
Who are the trematodes
If we consider what the flukes look like, then it is correct to note that the shape of their body looks like a leaf. But different species have their own characteristics and differences. Photo trematodes shows that their outlines may differ so much that they take on an almost round shape or pear-shaped shape.
The size of the worms can vary too. If we talk about all possible representatives, it is from a few millimeters to a meter and a half. Those that are parasitic in humans, can reach eight centimeters.
Characteristics of the external and internal structure
The structure of trematodes is quite primitive. Their body is deprived of a cavity like that of nematodes or segments like cestodes. Trematodes consist of a special type of epithelium called the syncytial tegument. The whole worm is as if placed in a bag consisting of muscles and skin.At one end of the trematode is a suction cup, which simultaneously serves as an oral and anus. It is necessary for feeding the parasite. Another sucker is located below and is used by trematodes as a fastening tool.
On a note!
The parasite lacks a sensory system, but otherwise the trematodes are full-fledged microscopic organisms.
Trematode digestive system
Oral sucker has a bottom on which a mouth is located. This hole continues into the pharynx and esophagus. The next section of the digestive tract are two blindly ending intestinal canals. Leftovers that have not been digested go the same way as they came.
Trematode excretory system
From each cell surrounded by cilia, near the trematodes, very thin tubules are drawn, which group together and form larger reservoirs, which are called canaliculi, and by means of them leave the opening at the rear end of the parasite.
Trematode nervous system
The trematodes have some semblance of the brain, represented by a ganglion.From this node along the body stretch the nerves, 2 each in the ventral, dorsal and lateral parts. Those trunks that are located on the sides, have connections with each other.
The larval stage of the flukes has certain sensory organs, represented by the ocelli and epithelial receptors.
Genital system of trematodes
Trematodes have a hermaphroditic reproductive system, but their fertilization is of a cross-type. This does not apply to those parasites that belong to the inhabitants of the blood. These species are dioecious and all life are in pairs. Their female is smaller and the male places it in a specialized fold. There the female exists in the state attached to it all her life.
The morphological features of trematode eggs are an oval shape, an impenetrable shell and a special cap by means of which the larva leaves the egg and enters the environment. In some species, eggs have an irregular shape and have thorns and processes. The color of eggs in trematodes ranges from light yellow to dark brown.
Features of the life cycle of trematodes
Before considering the general characteristics of the life cycle of trematodes, it is necessary to understand some of the concepts.
- Marita. An individual worm that has reached puberty.
- Miracidia.This is the first stage of the larval stage.
- Sporocista. Second larval stage. During this stage, the larva develops the possibility of parthenogenetic reproduction.
- Redia and cercarium. This is the third and fourth larval stage, respectively.
- Adolescarius is a reincarnation in an adult, which, once in the body, may be a provocateur of diseases.
- The ultimate owner. This is the name of an organism in which a mature parasite lives and multiplies.
- The owner of the intermediate type. It is an organism in which worms develop, being in the larval stage.
The scheme of the life course of trematodes has significant differences from other pathogenic pathogens. They have in addition to the main and intermediate hosts.
On a note!
The development cycle of trematodes is more complex in comparison with other worms. In addition to the final and intermediate host, there are also additional ones. As the last various animals act.
Mature helminth lays eggs, which, with secretions of the main host, fall into the water or soil. The embryo formed in the egg for the next stage of development must enter the mollusk.Otherwise, he just dies. The development of larvae occurs in the mollusk, and the formation of cercariae, which have all the signs of an adult. Having formed, the cercariae leave the mollusk and exit into the aquatic environment, where they swim, awaiting a new host. This may be optional or final. In the first case, trematodes turn into metacercariae and overgrow with a capsule. The larvae of some species are attached to plants growing near the coast and turn into adolescaria. These are the main stages of development of trematodes.
On a note!
Many of the formed helminth larvae are doomed to death due to such a complex development process, but nature has compensated for this with the ability of the larvae to reproduce, as well as a large number of eggs that produce mature individuals.
What are the flukes
There are many different types of flukes (members of class Trematoda) and the table below shows their distribution and the possibility of human infection.
|Worm type||Infection, habitat|
|Fasciolopis||Plant food. Eastern Eurasia|
|Heterophis||Mullet or tilapia.European East, Asian and African countries|
|Metagonimus||Meat carp or trout. Siberian latitudes|
|Gastrodiscoides||Plant food. Indian and Philippine Islands|
|Clonorchis or Chinese fluke||Any fish. East Asia and North American countries|
|Hepatic liver||Plant food. Everywhere|
|Cat litter||Crabs and crustaceans from freshwater. Asian countries|
|Pulmonary flukes||Carp family of fish. European and Middle Eastern regions|
The most famous flukes, which parasitize in man, are divided into schistosome and non-chistomea. The first live in the blood, and the rest in the lungs or liver. Blood flukes are parasitic in the blood of their host and live in hot countries. Infection with such a parasite is easy when bathing in dirty waters. This is due to the fact that cercariae of this species emerge from mollusks and move freely in the water column until they stumble upon a person into whom they enter, penetrating through the skin.
Other species of trematodes enter the human body in the form of metacercariae, which a person swallows along with fish, crustaceans, or plants that have not been carefully processed.They parasitize in the gastrointestinal system.<|endoftext|>
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### Archive
Archive for the ‘Other stuff’ Category
## How to calculate your horizon distance
While on a recent trip to the remote South Atlantic island of St Helena (exile place of Napoleon, and location of Edmond Halley’s observatory) [blog post to follow!] I ascended the highest mountain on the island, Diana’s Peak.
At 823m above sea level it commanded splendid views of the island, but the most striking thing was the unbroken 360° view of the horizon. I did a quick calculation in my head of how far I could see, and that forms the basis of this blog post: how do you calculate your horizon distance?
It turns out it’s pretty straight forward if you know a little simple maths. It helps to start by drawing a picture, so I did:
Definitely not to scale
For an observer of height h above sea level, the horizon distance is D. The Rs in this diagram are the radius of the planet you’re standing on, in this case the Earth. The only real assumption here is that you’re seeing a sea level horizon.As you can see you can draw a right-angled triangle where one side is D, the other is R, and the hypotenuse (the side opposite the right angle) is R + h.
Using Pythagoras’s Theorem, discovered around 2500 years ago, the square of the hypotenuse is equal to the sum of the squares of the other two sides. So we can say that:
(R + h)2 = R2 + D2
If you expand the part to the left of the bracket you get (R + h)2 = R2 + 2Rh + h2 so that:
R2 + 2Rh + h= R2 + D2
There’s an R2 term on both sides of the calculation so you can cancel them out, leaving:
2Rh + h= D2
Therefore the horizon distance, D, is:
D = √(2Rh+h2)
Here’s where you can make life much simpler for yourself. In almost every case R is much, much larger than h, which means that 2Rh is much, much larger than h2 so you can just ignore h2 and your equation simplifies to:
D ≈ √2Rh
(the ≈ sign here means “almost equals”. Honestly.)
So if you know R and h you can calculate D. To make this calculation easily you can carry round the value of √2R in your head meaning you only have to calculate √h and multiply those two numbers together.
So for the Earth, R is 6371000m, so √2R is 3569.6. Multiplying this by √h in metres would give you D in metres, so lets convert that into km to make things easier. This means dividing this number by 1000, giving an answer of 3.5696 which is ≈ 3.5.
So as a rough rule of thumb, your horizon distance on Earth,
D = 3.5 x √h
where D is measured in km and h in metres.
On Diana’s Peak, at 823m high, √h = 28.687… which multiplied by 3.5 gives a horizon distance of almost exactly 100km!
This is pretty cool, and is true of anywhere you can see the sea from a heigh of 823m.
One final calculation which sprung to mind on the mountain top was the area of sea I could see, which is easy to work out using the fact that the area of a circle is πr2, where r in this case is D, or 100km.
π is 3.14159 which means that the area of sea I could see was 31415.9 km2. Just a tad larger than Belgium, at 30528 km2.
And in that Belgium-sized circle of ocean was only one ship, the RMS St Helena that was taking me home the following day.
If you’re on Mars your horizon distance is shorter, at 2.6√h. On Mercury it’s smaller still at 2.2√h. This is due to Mars and Mercury being much smaller than the Earth, and so their surfaces curve away from you quicker. Venus is almost exactly the same size as the Earth (only a fraction smaller) so there you’d have to use the same calculation as here on Earth, 3.5√h.
Hovering above the surface of Jupiter your horizon would stretch to 11.8√h and on Saturn to 10.8√h. Uranus and Neptune are about the same size, giving a horizon distance of 7.1√h.
Mercury 2.2√h
Venus 3.5√h
Earth 3.5√h
Mars 2.6√h
Jupiter 11.8√h
Saturn 10.8√h
Uranus 7.1√h
Neptune 7.1√h
What about the dwarf planets? Being so small their surfaces will curve away from you very quickly, shortening your horizon distance. One of the smallest spherical objects in the solar system is the dwarf planet Ceres (as in cereal), which is the largest object amongst the fragments of rock in the asteroid belt. Your horizon distance on Ceres is almost exactly √h, making that a pretty simple horizon calculation!
Categories: Other stuff
## John Dobson, “The Sidewalk Astronomer”, 1915-2014
John Dobson, known as “the sidewalk astronomer“, and the inventor of the Dobsonian telescope mount, died yesterday 15 January 2014.
John Dobson, 1915-2014
I had the pleasure of meeting John back in 2006 when he visited Glasgow Science Centre to give a talk in the planetarium. As planetarium manager at the time I was in John’s company for most of the two days he was in Glasgow, and became well used to his unorthodox – and mischievous – teaching style.
“How many stars are there in our solar system?” he asked, with a twinkle in his eye.
“One…?” I ventured, smelling a trap.
“Actually there are three. Jupiter and Saturn are stars too. They define a star as something that emits visible light. But Jupiter shines in infra-red; it gives off more than twice what it takes from the sun, but astronomers don’t see that, because they’re too retarded. You see, that’s the problem with talking to me – I throw you a curve ball.”
His most controversial views were regarding the Big Bang Model. It’s fair to say that John wasn’t a believer* – he told me he was “allergic” to it! He instead promoted a steady state model, where the receding galaxies “fell off the edge” and got recycled back in… To be honest I didn’t really follow his arguments too closely, but there’s no doubting the entertainment of his delivery.
Certainly the audience who came to Glasgow Science Centre’s planetarium to see John talk were thoroughly entertained. Perhaps not in a very orthodox way, nor with theories that were widely accepted in the astronomy community, nor indeed about what they thought they’d hear John speak about. The talk had been billed as “hear John Dobson, inventor of the Dobsonian telescope mount, talk about his revolutionary design and his passion for sidewalk astronomy”. He gave that five minutes at the start of the talk; the rest was non-standard cosmology and poking fun at the consensus.
He didn’t credit himself as the inventor of the Dobsonian mount. After all, astronomers had been using elements of it for a long time before John put them all together and began popularising this low-cost mount.
A Dobsonian Telescope Mount
A Dobsonian mount is effectively a spinning plate with a cradle on it for holding the telescope. The plate spins allowing you to move the telescope from side to side (the azimuth co-ordinate, in astronomy speak) and the telescope can tilt in the cradle allowing you to move it up and down (the altitude co-ordinate). This is a far simpler mount than the alternatives, and can be built out of everyday items at low cost, meaning that more of your budget can go on the telescope tube itself, building larger tubes to collect more light (we call these large telescopes “light buckets”) and so get clearer, sharper images.
John himself never called them Dobsonians, instead referring to them as “sidewalk telescopes”. “For hundreds of years, wars were fought using cannon on ‘Dobsonian’ mounts; it’s nothing new,” he would say. But his design was innovative, and it brought the universe a little bit closer to us.
My primary telescope – the Skywatcher 250 PX – uses a Dobsonian mount. It’s an ideal scope for public astronomy events as it’s very quick to set up, and is really easy to operate. Want to move it? Just nudge or pull it.
John’s mount, and his passion for showing people the universe through a telescope, led to his popularising of “sidewalk astronomy”, which involves standing with a telescope out in a busy street in your town and showing passers-by views of the cosmos. Of course due to light pollution in towns and cities you’re limited as to what you can show, but the Moon and planets are easily visible from wherever you are. If your unsuspecting passer-by has never seen the rings of Saturn, or the moons on Jupiter, or mountain ranges and craters on our own Moon, you can be sure that their few minutes with your telescope will amaze them.
John Dobson will be remembered as the grandfather of sidewalk astronomy, but I’ll remember him most fondly as the very eccentric and enthusiastic man that I spent a couple of days with in Glasgow in 2006. The most vivid memory I have of John is taking him out for dinner the night he arrived in Glasgow, as I did with all visiting speakers. He didn’t eat meat, he informed me. And he didn’t eat processed food in restaurants. He saw that I was beginning to look worried. I suggested he might like a salad. With a grin he said “Why pay for a salad when there’s perfectly edible stuff just laying around?”, at which point he began rummaging in the flowerbeds for edible plants and weeds…
* Not that I’m a believer. I don’t believe in the Big Bang, rather I accept it as a model for the universe which fits all of the observations that we make. It’s true, to the limits of our current observations.
Categories: Other stuff, Stargazing
## Light Pollution and Birds: Early Bird Survey
The negative effect of light pollution on wildlife has long been known, specifically – but not exclusively – its effect on bats, bugs, and sea turtles. Now the British Trust for Ornithology (BTO) are running an Early Bird Survey, asking people in the UK to monitor the pre-dawn feeding times of garden birds to see what – if any – effect light pollution is having.
To take part you need to get up before dawn* on 9** January 2014 (tomorrow, as I write this), watch your garden bird feeders, and record the times that the first ten species arrive to feed. You can download the full instructions here (pdf), and submit your observations here.
* dawn occurs at different times around the UK, so you should find your sunrise time and get up half an hour earlier than that, during civil twilight.
** observations on 10, 11, and 12 January are welcome too.
As the BTO website says:
Winter is not an easy time for birds. They need extra energy to keep warm, especially during long winter nights. To cope with this, they lay down extra fat reserves, though small birds quite often only lay down enough for a single night. Longer nights not only affect the amount of energy a bird uses, they also reduce the amount of time that birds can feed in. Birds, therefore, have to make the most of the daylight hours to replenish their energy reserves before it gets dark.
The 2004 BTO Shortest Day Survey, run in association with BBC Radio 4, investigated the patterns behind birds arriving at garden bird feeders first thing on a winter’s morning. Building on observations from the Shortest Day Survey, the Early Bird Survey will investigate what effect, if any, light and heat pollution have on the feeding patterns of birds during a cold winter’s morning.
## Scots in Space
I was delighted to hear that two groups from Glasgow were winners in last night’s UK Space Conference‘s Arthur Clarke Awards 2011.
Clyde Space, a “leading supplier of small and micro spacecraft systems”, was given the Arthur Clarke Award 2011 for Achievement in Space Commerce, while the University of Strathclyde’s Advanced Space Concepts Laboratory, which “undertakes frontier research on visionary space systems”, was given the Arthur Clarke Award 2011 for Achievement in Space Research.
Congratulations to both, and it’s exciting to me as a Scot and a resident of Glasgow that these two groups, located within 5 miles of one another, are leading the UK in space research and commerce.
## The Circle of Little Animals
This is an auspicious time of the year.
The Sun, on its yearly circuit of the sky*, moves gradually along the ecliptic, a line which is the projection of our solar system’s disk onto our night sky. This line, the ecliptic, is also known as the zodiac, a term originating from the late 14th Century, and deriving from the Greek literally translating as “circle of little animals”. (Incidentally our word for zoo derives from the same origin).
The Circle of Little Animals
To the ancient Greeks this was indeed a circle of little animals, featuring: a ram, Ares; a bull, Taurus; Pisces the Fish; and many other well known (for all the wrong reasons) constellations. It also features three humans: Aquarius the Water Carrier, said to represent Ganymede, beloved of Zeus; and Gemini the Twins, Castor and Pollux.
The only zodiac constellation which is inanimate is Libra the Scales, taken from Babylonian astrology. The Greeks however didn’t recognise Libra; instead they thought that the stars here marked out Scorpius’ claws, which they considered to be a separate sign.
So the twelve constellations that lie along the ecliptic are most well-known due to astrology, a pseudo-science that suggests there is some significance to which constellation the Sun was in when you were born. This is, of course, bullshit.
There are so many reasons why astrology should be laughed off as pre-scientific magical thinking (no evidence, no mechanism by which it might work, inconsistent etc) but next time you meet an astrologer ask them what star sign you would be if you were born between 30 November and 18 December. If they tell you Sagittarius (if they’re a Hindu astrologer they might also say Scorpius) then tell them they are plain wrong.
On these 18.4 days of the year the Sun is wonderfully absent from the usual twelve zodiac signs, It is still there, however, gracefully moving along the ecliptic, but between 30 November and 18 December it is in the constellation of Ophiuchus the Serpent-bearer.
Ophiuchus does not appear in any astrologer’s zodiac. Back in the early days of astrology, when it was first dreamed up several thousand years ago, there were indeed only twelve constellations lying along the zodiac. Over the past few millennia however the Earth’s axis has wobbled slightly (an effect called precession) with the result that the line of the ecliptic has moved with respect to the constellations, and so an interloper, Ophiuchus, has crept in.
So celebrate all those of you born between 30 November and 18 December (one person in twenty share this star sign); you’re not a Sagittarian at all; you’re an Ophiuchan.
It’s still all bullshit though.
* the Sun, of course, does not orbit the Earth, it’s the other way around. It just looks like it does from down here…
Categories: Other stuff, Skepticism
## Awards [Blush]
This year I was fortunate enough to be the recipient if two fantastic astronomy awards, and a nominee for another.
In April 2010 I was nominated for the UK Space Conference’s Arthur Clarke Award for Public Promotion of Space (it was won by EADS Astrium).
Then last month, while speaking at the Federation of Astronomical Societies’ 2010 Convention, I was presented with two awards:
The 2010 Eric Zucker Award for Outstanding Contribution to Astronomy, awarded by the Federation of Astronomical Societies
and
The 2010 Joy Grifiths Award for Meritorious Efforts in the Cause of Darker Skies, awarded by the British Astronomical Association’s Campaign for Dark Skies
Needless to say I was chuffed to win these awards, and they have pride of place on my workdesk:
My Awards!
## Summer Hiatus
It’s been exactly five months since my last post, due to a combination of factors: Sam went back to work after maternity leave, and I dropped my workload to 2.5 days a week to help with childcare; I went freelance on top of the end of my last contract; and the lack of dark skies over the Summer meant a drop in my astronomy output!
Normal blogging will hopefully resume as of now!
Categories: Other stuff
## Fly the Cloudy Skies
I haven’t posted anything in the last couple of weeks due to my being “stranded” in the delightful island of Sark by the Eyjafjallajökul volcano eruption in Iceland which grounded most flights into and out of the UK for a week.
The view of the Icelandic Volcano from our cottage in Sark
There was no real hardship being on Sark for an extra week, especially given that every single night we were there it was clear (and dark). Indeed it seemed clearer on the nights during the air flight ban, and the daytime skies certainly contained fewer, if any, clouds. And no contrails.
Contrails over Europe
These facts are related, as contrails from planes quickly disipate in the atmosphere, spreading out to become indistinguishable from thin cirrus or cirrostratus clouds.
Contrails, short for “condensation trails”, are the visible trails of condensed water vapour in the sky left behind by exhausts of aircraft. They are, essentially, man-made clouds.
This got me thinking; just how much of the “cloud cover” that affects astronomers so badly is down to contrails?
It turns out I’m not the only one worrying about this. In fact it’s been thought a problem for quite some time.
Gerry Gilmore, of Cambridge University (and of Max Alexander‘s excellent Explorers of the Universe portrait exhibition) was saying back in 2006 that the problem of contrails might make ground based telescopes worthless by the year 2050. One small reason, amongst many, to limit our use of planes.
The last time there was such a significant drop in the number of planes in the sky was after the terrorist attacks of 11 September 2001, and back then astronauts on board the International Space Station noticed a decrease in the number of contrails over the US.
The same was true between 16 and 22 April 2010 when, for a few days, flights over Europe stopped, stranding thousands overseas, but giving a brief and welcome respite to astronomers who were able, for once, to enjoy truly clear skies.
For more info than you could ever want on contrails and their impact on ground-based astronomy, see Holger Pederson’s website with links to many articles and reports.
Also worth a look is the apparently now-defunct National Contrail Network, which formed a part of a masters research project at the University of Lancashire. The official homepage link doesn’t work, but there’s plenty of data and info that is accessible.
National Contrail Network
Finally, have a look at the Cloud Appreciation Society‘s gallery of contrail images: they’re actually rather beautiful.
Trails and Halo, Denmark © Jesper Grønne.
Categories: Other stuff Tags: ,
## Five Useful iPhone Apps for Astronomers
March 26, 2010 1 comment
Here’s a brief overview of the five iPhone apps that I, as an astronomer, find indispensible (in no particular order):
Weather Pro (£2.39) Let’s face it, no weather forecast is 100% right, but this app gives you more info than most, allowing you to figure out in advance whether there’ll be clear skies. Magic Hour – formerly VelaClock (£2.39) So, it’s going to be a clear night. Now you need to know: when is sunset? when does astronomical twilight end? Magic hour is your app for that. Also displays moonrise and moonset times, this is the perfect app for figuring out exactly when your skies will be dark. Starwalk (£1.79) The very best of all iPhone astronomy apps, Starwalk has an exquisite interface, and is packed full of features. Great for beginners wanting to find their way around the sky, and for experts who want to dig a little deeper. Satellite Visibility (£1.79) Showing iridium flares, ISS & Hubble passes, and many other sats too. Impress your friends by knowing where and when (to the exact second) satellites are due to pass overhead. Reeder (£1.79) OK, so not immediately anything to do with astronomy, but with this RSS aggregator and subscriptions to some astronomy blog feeds (Bad Astronomy, Universe Today…), you’ll be able to keep abreast of all breaking astronomy news.
So only two of these are directly related to astronomy, but I’d be lost without any of these, and for a total cost of just over a tenner these apps are great value for money.
## Communicating Astronomy with the Public (CAP)
I have just returned to the UK after attending the amazing CAP2010 conference in Cape Town. This five day meeting of science communicators and astronomers from all over the world was an incredible opportunity to swap ideas and discuss what went well during IYA2009, and what we need to build on.
The conference was blogged by many, but nowhere more thoroughly than the psychohistorian’s blog
For me highlights of the conference include:
Dr Chris Engelbrecht discussing his SkyRanger training programme for African park rangers and safari guides
Amelia Ortiz-Gil describing planetarium activities for the blind an partially sighted, “The Sky is in Your Hands”
Sze-leung Cheung covering the Dark Skies Awareness switch-off programme “Dim It!” in Hong Kong
All in all an amazing time.
The true star of the show though was Cape Town itself. It’s a stunning city, towered over by Table Mountain, which I spent a day climbing on the last Saturday of the conference, with Marek Kukula and Carolina Odman.
Carolina Odman in the Hely-Hutchison Reservoir on Table Mountain
Categories: Other stuff<|endoftext|>
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# Mathematical Journeys: Solve Only for What You Need
Standardized test math doesn't behave like normal math. On a normal math test, your knowledge of the concepts and material is being tested, using (hopefully) fair test questions. On a standardized test, though, they're looking for you to think outside the box, to apply math concepts and algorithms to unusual situations, and to really understand what they're looking for and find the quickest way to go about it. Let's take a question from a recent GRE student's lesson:
If 4x – 5y = 10 and 6y – 3x = 22, then what is x + y?
Now, this is a set of two equations with two variables each, so it looks to me like a perfect candidate for solving as a system. If I were solving this one on a regular math test, I'd start off trying the substitution method, since I'm more comfortable with that one. So let's explore that one first:
I'll start by solving the first equation for y:
4x – 5y = 10
- 5y = 10 – 4x
y = (-10/5) – (4/-5)x
y = -2 + (4/5)x
Then I'll plug that in for y in the second equation:
6(-2 + [4/5]x) – 3x = 22
-12 + (24/5)x – 3x = 22 Now we have to convert the 3x into a fraction
-12 + (24/5)x – (15/5)x = 22
-12 + (9/5)x = 22
(9/5)x = 34
x = 34 (5/9)
x = 170/9
Then plug that back in for x in the first equation:
y = -2 + (4/5)(170/9)
y = 136/9
And, FINALLY, find the quantity asked for in the problem by adding x and y together:
x + y = (170/9) + (136/9)
x + y = 306/9
x + y = 34
Well, that's one way to find the answer, but that took a long time, with lots of large numbers, and lots of potential for mistakes. This is a standardized test, remember, so time is a factor here. Take a look at the question again. It's asking for x + y. Why wouldn't it be asking simply for x, or y, or even x and y, for that matter? Is it because x + y is a much cleaner number? Is it to be ornery? To make you waste time?
Well, to be honest, the answer to that last question is yes, but not in the way you might think. In our math classes, we're hardwired to try to solve for x – we want to end up with a nice clean number to equal one of our variables. It's the way most math classes work; manipulate the equation until it tells you the missing piece of information. The test builders know that, and they know that everyone's first instinct in a math problem is to try to solve for x. But in this case, they're not asking for the value of x; they're asking for the value of an expression containing x. And they're doing that very deliberately – because finding x + y is much easier than finding x.
Take a look at our system again – this time I'll re-arrange it slightly in preparation for using the addition method to solve it:
4x – 5y = 10
– 3x + 6y = 22
See it yet? Use the addition method – don't even modify anything – and add straight down the columns:
4x – 5y = 10
– 3x + 6y = 22
x + y = 32
Well, would you look at that? That's the answer they're looking for – and you'll notice it's not the same answer as our previous attempt. Not only would you have wasted a bunch of time going through all those hoops to solve with substitution, but you would have gotten the question wrong to boot!
It's an odd way of looking at a math problem, but one of the biggest strategies I tell my students is to not think about the test as a math test. It's a logic test that happens to involve numbers. Here, the test is remembering to only solve for what you need. Don't bother getting all the way down to x if x won't help you in the end. Sometimes they're asking for a quantity because going any further past that quantity will only cause you grief. Solve for the quantity they ask for, and no more.
\$45p/h
Ellen S.
Math and Writing Geek
500+ hours
if (isMyPost) { }<|endoftext|>
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### Algebra Intro
Home >> Science >> Math (Algebra)
## Algebra 100 Intro
Algebra is a scary word to some people, myself included, but it doesn't have to be. Really, it's just a fancy way of complicating the simple math we normally do anyway everyday. The equations that represent these daily transactions are often unknown or perhaps unrealized in our day-to-day activities, but they can provide useful new perspectives on how we interact in society.
### Uses of Algebra
Algebra is useful in many others ways too though. All the functions we perform on the daily can, in one way or another, be represented using Algebraic equations.
When we shop for groceries on a limited budget, for example, we might have to account for the taxes on certain goods that aren't exempt from taxes. Perhaps multiple each item by the number of mouths to feed, and then maybe replace one of each to account for someone with allergies.
Maybe you have three kids, and each eats two pieces of fruit each school day, three vegetables each day, and a pint of ice cream each weekend day, or some other blend of purchases that must be repeated frequently in order to survive.
That problem can be represented with an algebraic math equation.
When we want to purchase anything that is on sale or discounted a certain percent, we must employ simple algebra equations to fulfill our needs. When we learn any other science subject, we must first have a basic understanding of the principles of Algebra and Algebraic equations in order to understand those topics.
To understand algebra at its simplest form, certain concepts must first be assimilated and understood. In this quick math course, we are going to review the basics of algebra and most of the concepts that will be needed in later courses.
Let's begin with math in its simplest form, and evolve our course from that elementary example. Algebra has equations that are read from left to right. That will be important later because when two operators are equal in an equation, then the equation begins being read from the left.
Next is the order of operators: 1) Parentheses, 2) Multiplication and Division are equal, 3) Exponents, 4) then addition and subtraction come last.
So, for example, (4-(2-5*2)^2-3)/(-3) = -21 is obviously incorrect, since it's positive twenty one (+21). Let's find out why.
Don't want to wait? Learn it now
For example, we know that:
1 + 1 = 2. So therefore, if x = 1, then x + x = 2
because x is 1 and 1 + 1 = 2.
That's easy.
Let's try that again.
2 + 1 = 3
If a variable named y is equal to 1,
then 2 + y = 3
because we defined the variable y to equal 1.
Now what if we don't know one of the numbers, perhaps the 2. In that case, when we write the above equation, instead of a 2, we place a variable[D: a character representing a numeric value or number] in its place.
Let's try it again.
The variable p = 3
And the variable q is unknown
But we know the following equation is true:
p + p = q
Then q must equal 6, right?
Because p = 3,
Then p + p = 3 + 3 = 6
Therefore, q = 6.
### Variables
A variable is anything representing something else in mathematics or science. Computers, physics, engineering, and many more science topics all rely on variables. A variable acts as a symbol of a number we do not yet know for sure. So now, with a variable of X in place of the 2, the equation would be written as follows:
1 + 1 = X Or we could say:
1 + x = 2
You can decide what character you use as a variable, but keep in mind the following:
If you draw or write a symbol not available on a keyboard, then when you want to type the equation and your brain is working already on making sure you copy it right from paper, then remembering that cool symbol you drew before is now another thing to keep track of and could result in errors down the line. Meaning, keep it simple or as simple as possible for the sake of ease of use or understanding.
### Tangent
As equations get more complicated, small mistakes in the beginning could grow into big mistakes later. Avoiding complications will reduce errors.
### Variable Symbols
Also, using symbols such as a dollar sign ('\$') or something you found on your new mobile device, perhaps whatever this symbol is: ¥, could already be used as a reserved symbol representing something else; for example, π = 3.14 approximately.
Let me explain. While you are free to choose any character or symbol at your discretion[LR: discrete, discretionary, lat: cogitare, sp: crer], you are not the first person to learn Algebra. Many others have learned the hard way that if everyone uses different variables to represent the same thing, then when someone else looks at your version of the same equation, precious time will be wasted translating your variable choices to the ones they will have understood.
So, what people did with their fancy public school systems served a la suburbua was agree on certain variables being standards for certain things.
One day, I'll link to a comprehensive[D: All inclusive but not necessarily complete] spreadsheet of all industry standard variables so we can all enjoy that saved time on better moments, but for now, here's some guidelines.
### Variable Guidelines
Typical math numerals are italicized letters, usually lower case, but sometimes upper case, such as if you choose X, because a lower case X could be confused with a multiplication symbol. So of course, to add to the confusion, lower case x is in fact the standard for multiplying any two numbers but also for teaching variables in Algebra classes. I don't specifically remember ever running across the equation: X x X = X^2, but you can see how using a variable that appears identical to the symbols that modify it can be confusing. The asterisk ('*') is also used as the standard for multiplying but specifically with computer programs and programmers, and has in some cases replaced the letter X on graphic user interfaces ('GUI').
The upper carat symbol ('^') is a standard for symbolizing a power or exponent of a number. For example, X^2 is X to the second power or X squared.
Another example is the lower case i, which is reserved in advanced mathematical theorems for the imaginary number. Rest assured, any mathematics dependent upon the imaginary number or required to prove another equation true follows these three important rules:
1) The imaginary number will never be useful in anybody's future,
2) The imaginary number will never have any positive affect on your life or anybody else's life that we know, and
3) The imaginary number-based equations are still theories and even if they result in equations that do not include the imaginary number in them and seem to be true by all accounts, they were still proven using a fictitious algorithm based upon an imagined number that doesn't exist, which means the equations are not reverse engineer verifiable.
Back to math that can help our daily lives. Other standard symbols include:
t for time,
v for velocity,
d for distance (also diameter),
n for number,
g for gravity,
f for force,
h for height,
l for length,
p for price,
two numbers next to each other, each surrounded by parentheses symbolizes the multiplication symbol even though it is omitted, so (4)(3) = 12.
That same equation can be written (4)*(3)=12, or (4)x(3)=12, or 4*3 = 12, or 4 x 3 = 12.
All of those equations are the same.
x, y, and z together are used to symbolize each axis of a three-dimensional plane,
a for acceleration (also alpha)
/ forward slash for dividing the first number into the second,
< and > are the less than and greater than symbols, respectively. These symbols should not be confused with the same symbols in the context of programming HTML tags or when used to symbolize flow of text.
There's plenty more, but as you can see, most of the variables we thought we would get to invent ourselves have already been standardized.
Let's continue learning the math.
The equation we were working on is simple:
If:
1 + 1 = X And because we know that 1 and another 1 gives us 2, then we can logically deduce that in order for the equation to be true, X must equal 2;
Otherwise, the equation would not be true and the equal sign would have to be replaced with a greater than or less than symbol for any other value of X other than 2.
1 + 1 < 3
1 + 1 > 1.5
So that is an introduction to Algebra. Next, let's solve: 1 + y = 2z
## Summary
Algebra is intended to simplify our lives, but in order to do so, we must first understand algebraic equations, learn basic math concepts, and feel comfortable working with variables instead of numbers.<|endoftext|>
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# Fixed-Point Representation
Positive integers, including zero, can be represented as unsigned numbers. However, to represent negative integers, we need a notation for negative values. In ordinary arithmetic, a negative number is indicated by a minus sign and a positive number by a plus sign. Because of hardware limitations, computers must represent everything with 1's and D's, including the sign of a number. As a consequence, it is customary to represent the sign with a bit placed in the leftmost position of the number. The convention is to make the sign bit equal to 0 for positive and to 1 for negative.
binary point :
In addition to the sign, a number may have a binary (or decimal) point. The position of the binary point is needed to represent fractions, integers, or mixed integer-fraction numbers. The representation of the binary point in a register is complicated by the fact that it is characterized by a position in the register. There are two ways of specifying the position of the binary point in a register: by giving it a fixed position or by employing a floating-point representation. The fixed-point method assumes that the binary point is always fixed in one position. The two positions most widely used are (1) a binary point in the extreme left of the register to make the stored number a fraction, and (2) a binary point in the extreme right of the register to make the stored number an integer. In either case, the binary point is not actually present, but its presence is assumed from the fact that the number stored in the register is treated as a fraction or as an integer. The floating-point representation uses a second register to store a number that designates the position of the decimal point in the first register. Floating-point r.epresentation is discussed further in the next section
## Frequently Asked Questions
+
Ans: Since we are dealing with unsigned numbers, there is really no way to get an unsigned result for the second example. view more..
+
Ans: The direct method of subtraction taught in elementary schools uses the borrow concept. In this method we borrow a 1 from a higher significant position when the minuend digit is smaller than the corresponding subtrahend digit. view more..
+
Ans: The r's complement of an n-digit number N in base r is defined as r' - N for N * D and D for N = D. Comparing with the (r - I)'s complement, we note that the r's complement is obtained by adding I to the (r - I)'s complement since r' - N = [(r' - I) - N] + I. view more..
+
Ans: Positive integers, including zero, can be represented as unsigned numbers. However, to represent negative integers, we need a notation for negative values. In ordinary arithmetic, a negative number is indicated by a minus sign and a positive number by a plus sign. view more..
+
Ans: When an integer binary number is positive, the sign is represented by 0 and the magnitude by a positive binary number. When the number is negative, the sign is represented by 1 but the rest of the number may be represented in one of three possible ways: view more..
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Ans: The addition of two numbers in the signed-magnitude system follows the rules of ordinary arithmetic. If the signs are the same, we add the two magnitudes and give the sum the common sign. If the signs are different, we subtract the smaller magnitude from the larger and give the result the sign of the larger magnitude. view more..
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Ans: Subtraction of two signed binary numbers when negative numbers are in 2' s complement form is very simple and can be stated as follows: Take the 2's complement of the subtrahend (including the sign bit) and add it to the minuend (including the sign bit). A carry out of the sign bit position is discarded. view more..
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Ans: When two numbers of n digits each are added and the sum occupies n + 1 digits, we say that an overflow occurred. When the addition is performed with paper and pencil, an overflow is not a problem since there is no limit to the width of the page to write down the sum. view more..
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Ans: An overflow condition can be detected by observing the carry into the sign bit position and the carry out of the sign bit position. If these two carries are not equal, an overflow condition is produced. view more..
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Ans: The representation of decimal numbers in registers is a function of the binary code used to represent a decimal digit. A 4-bit decimal code requires four flip-flops for each decimal digit. view more..
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Ans: The floating-point representation of a number has two parts. The first part represents a signed, fixed-point number called the mantissa. The second part designates the position of the decimal (or binary) point and is called the exponent. The fixed-point mantissa may be a fraction or an integer. For exam ple, the decimal number +6132.789 is represented in floating-point with a fraction and an exponent as follows: view more..
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Ans: A floating-point number is said to be normalized if the most significant digit of the mantissa is nonzero. For example, the decimal number 350 is normalized but 00035 is not. Regardless of where the position of the radix point is assumed to be in the mantissa, the number is normalized only if its leftmost digit is nonzero. view more..
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Ans: In previous sections we introduced the most common types of binary-coded data found in digital computers. Other binary codes for decimal numbers and alphanumeric characters are sometimes used. Digital computers also employ other binary codes for special applications. A few additional binary codes encountered in digital computers are presented in this section. view more..
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Ans: Binary codes for decimal digits require a minimum of four bits. Numerous different codes can be formulated by arranging four or more bits in 10 distinct possible combinations. A few possibilities are shown in Table 3-6. view more..
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Ans: The ASCII code (Table 3-4) is the standard code commonly used for the transmission of binary information. Each character is represented by a 7-bit code and usually an eighth bit is inserted for parity (see Sec. 3-6). The code consists of 128 characters. Ninety-five characters represent graphic symbols that include upper- and lowercase letters, numerals zero to nine, punctuation marks, and special symbols view more..
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Ans: Binary information transmitted through some form of communication medium is subject to external noise that could change bits from 1 to 0, and vice versa. An error detection code is a binary code that detects digital errors during transmission. The detected errors cannot be corrected but their presence is indicated. The usual procedure is to observe the frequency of errors. If errors occur infrequently at random, the particular erroneous information is transmitted again. If the error occurs too often, the system is checked for malfunction view more..
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Ans: Parity generator and checker networl<s are logic circuits constructed with exclusive-OR functions. This is because, as mentioned in Sec. 1·2, the exclusive-OR function of three or more varia.bles is by definition an odd function. An odd function is a logic function whose value is binary 1 if, and only if, an odd function number of variables are equal to 1. According to this definition, the P( even) is the exclusive-OR of x, y, and l because it is equal to 1 when either one or all three of the variables are equal to I (Table 3-7). The P(odd) function is the complement of the P(even) function. view more..
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Ans: A digital system Is an interconnection of digital hardware module. that at'ClOinpl.lsh a specific Wormation-proceaslna taslc. Digital systems vary in size and complexi.ty interacting digital &om a few integrated circuits to a complex of interconnected and computers. Digital system design invariably UBeS a modular approach. The modules are constructed &om such digital components as ules registet&, are in decoders, terconnected arithmetic with common elements data and control paths , and control logic. The to fonn various moda digital computer system. view more..
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Video: GCSE Mathematics Foundation Tier Pack 3 • Paper 1 • Question 24
GCSE Mathematics Foundation Tier Pack 3 • Paper 1 • Question 24
03:43
Video Transcript
Five bars of soap have a total weight of 750 grams. Four bottles of soap have a total weight of 600 grams. Calculate the total weight of four bars of soap and six bottles of soap. Give your answer in kilograms.
In order to be able to answer this question, we first need to work out the weight of a single bar of soap and a single bottle of soap. We’ll start with the bars. We’re told in the question remember that five bars of soap have a total weight of 750 grams.
To find the weight of one bar, we need to divide 750 by five, which we can do using a short division or bus stop method. There is one five in seven with a remainder of two, there are five fives in 25 with no remainder, and there are no fives in zero. So 750 divided by five is 150.
We’ve worked out then that one bar of soap has a weight of 150 grams. But we need to find the weight of four bars of soap. To find this then, we need to multiply 150 by four, which is the same as doubling it and then doubling it again. Doubling 150 gives 300 and that’s because doubling 15 gives 30 and then doubling 300 gives 600. So we found that the weight of four bars of soap is 600 grams.
Next, let’s consider the bottles. We’re told in the question that four bottles of soap have a total weight of 600 grams. To find the weight of one bottle then, we need to divide 600 by four. We could do this by halving and then halving again, but there’s actually an easier way.
Earlier in the question, we saw that 150 multiplied by four was 600. So the reverse of this means that 600 divided by four is 150. And this gives us the answer to the calculation that we’re looking for. So the weight of one bottle of soap is 150 grams. But we need to work out the weight of six bottles of soap.
To find this, we need to multiply 150 by six. We could do this using a column multiplication method or we can be a little bit smart about this. We already know that four multiplied by 150 is 600. And we can work out that two multiplied by 150 is 300. So to find six multiplied by 150, we just need to add these two values together. And it gives 900. This tells us then that the weight of six bottles of soap is 900 grams.
We then just need to add the total weight of the four bars and the six bottles together, which we can do using a column addition method. 600 plus 900 is 1500. Remember though that this is 1500 grams. And the question asks to give our answer in kilograms.
There are 1000 grams in a kilogram. So to convert this answer in grams into a value in kilograms, we need to divide by 1000. To do so, we can keep the decimal point, which in this case would be at the end of the number, fixed and move all of the digits three places to the right. So 1500 divided by 1000 is equal to 1.5.
The total weight of four bars and six bottles of soap is 1.5 kilograms.<|endoftext|>
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When you describe a photo it is important to say where the people are and what the people are doing. These activities help you to practise using the present continuous to say what someone is doing in a photo. You will also learn new words about school. Listen to a description of the photograph above.
▶Task 1 - children's things
▶Task 2 - common verbs and nouns
▶Task 3 - compound nouns
▶Task 4 - gap fill typing
▶Task 5 - your turn
Use the new words and present continuous to describe this picture of school.
Think about these questions to help you:
- How many people are in the photo?
- Who are they?
- What are they wearing?
- Where are they going?
- Where is the mother going after the school run?
In this photo we can see a mother. She is taking her son to school. It is the morning. It is the time when parents take their children to school. We call this time – the school run. Parents do the school run in the morning and the afternoon. Some people go by car. Some go by bus and many go on foot. In this photo, the boy’s mother is taking her son to school on foot. She is pushing a red pram with the boy’s baby brother or sister inside. The little boy is walking next to her. He is holding onto the pram. He is wearing a school uniform. He has a white polo shirt, grey shorts, black shoes and grey socks. He is carrying a blue book bag. Behind the boy we can see three girls in grey skirts. They are walking to school. They are wearing backpacks to carry their books. I think his mother is going to go home after she does the school run. She is not wearing work clothes. Or maybe she is going to take her baby to a play group.
More like this
- Do you take your children to school?
- How do you go?
- Where do you go after the school run?<|endoftext|>
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A melody (from Greek μελῳδία, melōidía, "singing, chanting"), also tune, voice, or line, is a linear succession of musical tones that the listener perceives as a single entity. In its most literal sense, a melody is a combination of pitch and rhythm, while more figuratively, the term can include successions of other musical elements such as tonal color. It may be considered the foreground to the background accompaniment. A line or part need not be a foreground melody. Melodies often consist of one or more musical phrases or motifs, and are usually repeated throughout a composition in various forms. Melodies may also be described by their melodic motion or the pitches or the intervals between pitches (predominantly conjunct or disjunct or with further restrictions), pitch range, tension and release, continuity and coherence, cadence, and shape. The true goal of music—its proper enterprise—is melody. All the parts of harmony have as their ultimate purpose only beautiful melody. Therefore, the question of which is the more significant, melody or harmony, is futile. Beyond doubt, the means is subordinate to the end.<|endoftext|>
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# Roots and Powers of Algebraic Expressions
### Subtracting Integers: Rules & Examples
In this lesson, we will cover definitions, rules, and examples for subtracting integers. We'll also cover some helpful tips when dealing with tricky math problems.
### What Are Integers?
Integers are whole numbers (numbers that aren't a fraction or a decimal). They include both positive and negative numbers. Positive numbers either have no sign or a plus (+) sign in front of the number. Negative numbers always have a negative (-) sign in front of the number. You can see negative numbers and positive numbers represented on a number line like this:
Many students find subtracting integers confusing, and why not? There are too many signs, right? Why can't everything be simple? It can be! When subtracting integers you have to keep a few basic things in mind.
Subtraction is the opposite of addition. Wait, what? Yes! It's true! Subtraction is really just the opposite of addition. So when looking at any subtraction problem, think to yourself: add the opposite. For example, say you have the problem 5 - 2 = 3. You can get the same answer by adding (instead of subtracting) if you write the problem as 5 + (-2) = 3
Still a little confused? Try to think of integers in terms of money. Take the above problem, for example. You have \$5 in your pocket. You owe your friend \$2. Now ask yourself: are you going to have any money left over? Yes, because \$5 - \$2 = \$3. The answer is a positive number.
### Subtracting Double Negatives
Try this problem 5 - (-3)=?
Many people do a double take at problems like this. There are two negatives. What to do now? In math, double negatives cancel one another out - just like in English. For example, a double negative statement in English such as 'I cannot not go to this class' really means the same thing as this positive statement 'I have to go to this class'. It works the same way with mathematics. A double negative cancels out, and you are left with a positive number.
Remember, subtracting integers is like adding the opposite. So, we would start with the problem like this: 5+
What do we do with (-3)? What's the opposite? Right, positive three is the opposite of a negative three. So now my problem looks like this: 5 + 3 = 8.
### Subtraction Rules and Examples
If you're still having some issues with subtracting integers, here are some basic rules to follow:
• Negative - Positive = Negative Why? Remember, if you are adding the opposite, that positive number will change to a negative number.
o Example: -5 - 10 = -15
• Positive - Negative = Positive Why? Same as earlier, except this time your negative number will be the opposite, making it into a positive number.
o Example: 14 - (-3) = 17
Helpful hint: What you can take away from these two rules: if you have a positive and a negative number you are subtracting, take the sign from the first number and then add the two numbers together.
Negative - Negative = Negative + Positive Why? Change that double negative into a positive. The answer is a bit trickier here. Subtract the two numbers. Your sign will be the sign of the greatest number. See the example for this one!
o
Example: -6 - (-10) = ? To solve, change the double negative so that -6 + 10 = ? Take the sign from greatest number (10) and subtract. The answer is 4.
Positive - Positive = Positive + Negative Why? Remember, add the opposite here. Then, just as before, you will take the sign of the greatest number and subtract the two numbers for your answer.
o
Example: 8 - 9 = ? To solve, add the opposite number so that you have 8 + (-9) = ? Then, take the sign from the greatest number (9) and subtract. The answer is -1.
### Lesson Summary:
Integers are positive and negative numbers. In mathematical terms, subtracting means to add the opposite. Pay attention to your signs when subtracting integers, and when you get stuck, think of the numbers in terms of having or owing money! When subtracting double negatives, remember that double negatives cancel one another out, leaving a positive number.<|endoftext|>
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Longitude stress of $1{\text{ }}kg/m{m^2}$ is applied on a wire. The percentage increase in length is ($Y= 10^{11} N/m^2$)A) 0.002B) 0.001C) 0.003D) 0.01
Last updated date: 03rd Mar 2024
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Hint: Stress is defined as the force experienced by the object which causes a change in the object. The strain is defined as the change in the shape of an object when stress is applied. Stress is measurable and has a unit. The strain is a dimensionless quantity and has no unit.
Complete step by step solution:
Given data:
Stress $=$ $1{\text{ }}kg/m{m^2}$
Young’s Modulus, $Y = 10^{11} N/m^2$
Percentage increase in length $=$$?$
We know that
$Y$ $=$ $\dfrac{{Stress}}{{Strain}}$
Substituting the values of Stress and Young’s Modulus, we get
$\Rightarrow$ $10^{11} = \dfrac{1}{{strain}}$
$\therefore$ $Strain = \dfrac{{1{\text{ }}kg/m{m^2}}}{{{{10}^{11}}}} = \dfrac{1}{{{{10}^{ - 6}}}} \times \dfrac{1}{{{{10}^{11}}}}$ $($On conversion of $mm^2$ to $m^2$ $)$
$\Rightarrow$ Strain $=$ $\dfrac{1}{{{{10}^5}}}$
$\Rightarrow $$\dfrac{{\Delta l}}{l}$$ =$ $\dfrac{1}{{{{10}^5}}}$
$\therefore$ Percentage increase in length
$\Rightarrow$$\dfrac{{\Delta l}}{l}$$\times$100 $=$ $\dfrac{1}{{{{10}^5}}}$$\times$100
$\Rightarrow$ Percentage increase in the length of wire$=$0.001
Hence the correct option for the problem is B.
Note: 1) Stress can also be defined as the restoring force per unit area of the material.
2) Strain can also be considered as a fractional change in either length (when tensile stress is considered) or volume (when bulk stress is considered).
3) Young’s modulus is also termed as the modulus of elasticity.<|endoftext|>
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