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# Lesson 10
A New Kind of Number
• Let’s invent a new number.
### 10.1: Numbers Are Inventions
Jada was helping her cousin with his math homework. He was supposed to solve the equation $$8 + x = 5$$. He said, “If I subtract 8 from both sides, I get $$x = 5 - 8$$. This doesn’t make sense. You can’t subtract a bigger number from a smaller number. If I have 5 grapes, I can’t eat 8 of them!”
What do you think Jada could say to her cousin to help him understand why $$5 - 8$$ actually does make sense?
### 10.2: The Square Root of Negative One
Numbers on the number line are often called real numbers.
1. The equation $$x^2 = 9$$ has 2 real solutions. How can you see this on the graph of $$y = x^2$$? Draw points on this real number line to represent these 2 solutions.
2. How many real solutions does $$x^2 = 0$$ have? Explain how you can see this on the graph of $$y = x^2$$. Draw the solution(s) on a real number line.
3. How many real solutions does $$x^2 = \text{-} 1$$ have? Explain how you can see this on the graph of $$y = x^2$$. Draw the solution(s) on a real number line.
### 10.3: Imaginary Numbers
1. On the real number line:
1. Draw an arrow starting at 0 that represents 3.
2. Draw an arrow starting at 0 that represents -5.
2. This diagram shows an arrow that represents $$\sqrt{\text- 1}$$.
1. Draw an arrow starting at 0 that represents $$3\sqrt{\text- 1}$$.
2. Draw an arrow starting at 0 that represents $$\text- \sqrt{\text- 1}$$.
3. Draw an arrow starting at 0 that represents $$\text- 5\sqrt{\text- 1}$$.
The absolute value of a real number is the length of the arrow that represents it.
1. What is the relationship between the absolute value of a real number and the absolute value of the square of that number?
2. If we want $$\sqrt{\text- 1}$$ and its square to have this same relationship, then what should the absolute value of $$\sqrt{\text-1}$$ be?
3. What should the absolute value of $$3\sqrt{\text- 1}$$ be?
### Summary
Sometimes people call the number line the real number line.
All real numbers are either positive, negative, or 0, and they can be plotted on this line. All the numbers we have used until this lesson have been real numbers. For real numbers, we know that:
• A positive number times a positive number is always positive. So when we square a positive number, the result will always be positive.
• A negative number times a negative number is always positive. So when we square a negative number, the result will always be positive.
• 0 squared is 0.
So squaring a real number never results in a negative number. We can conclude that the equation $$x^2 = \text{-} 1$$ does not have any real number solutions. In other words, none of the numbers on the real number line satisfy this equation.
Mathematicians invented a new number that is not on the real number line. This new number was invented as a solution to the equation $$x^2 = \text{-} 1$$. For now, let’s write it $$\sqrt{\text- 1}$$ and draw a point to represent this number. We can put it anywhere we want as long as we don’t put it on the real number line. For example, let’s put it here, right above 0 on the real number line:
This new number $$\sqrt{\text- 1}$$ is a solution to the equation $$x^2 = \text{-} 1$$, so $$\left(\sqrt{\text- 1}\right)^2 = \text-1$$. If we draw a line that passes through 0 on the real number line and $$\sqrt{\text- 1}$$, we get the imaginary number line. The numbers on the imaginary number line are called the imaginary numbers.
### Glossary Entries
• imaginary number
A number on the imaginary number line. It can be written as $$bi$$, where $$b$$ is a real number and $$i^2 = \text-1$$.
• real number
A number on the number line.<|endoftext|>
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Rules and Examples
asterisk is a punctuation
that looks like a little star ( * ).
Click Here for Step-by-Step Rules, Stories and Exercises to Practice All English Tenses
The asterisk is made on your keyboard by holding the SHIFT
pressing the 8
on the top number line.
We use the asterisk in English writing to show that a footnote,
reference or comment has been added to the original text.
Many people incorrectly pronounce (say) the word "asterisk."
The word "asterisk" is pronounced "aste-risk."
The word "asterisk" comes from the Latin word "asteriscus" and the
Greek word "asterikos" meaning "little star
The asterisk was first used in printing and writing in the early
Uses of the asterisk
1. Use an asterisk or asterisks to indicate a footnote at the bottom of
- A footnote
is an explanation or a comment at the bottom of a page that
refers back to a specific part of the text.
- If there are multiple footnotes (more than one), use one
the first footnote, two asterisks for the second and so on.
- Be sure that footnotes at the bottom of the page match the
the original text.
are many forms of punctuation in the English language. Three
common forms of end punctuation are the period, question mark and
is used as the end of a sentence that is a
command or a statement.* The period tells the reader that the sentence
mark ends sentences that are written as
questions. These sentences may begin with words such as why, how, when,
where or what.
mark is used at the end of sentences to give emphasis
or show excitement.**
are two footnotes, or comments, added to this text. The two footnotes
are then explained at the bottom of the page.)
*The period is also
called "full stop" because it tells the reader that
the sentence has ended.
exclamation mark is also called an exclamation point.
2. Use an asterisk in advertisements to indicate there is extra
*discount is good in-store and online
Get a new mobile phone in November for only $29.99*
*price is for new customers only
These were the uses of the asterisk. Now that you know them, it is
time to practice! Read and do<|endoftext|>
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Every year in the Spring, dust from the Sahara Desert is swept up by the winds and brought all the way to Europe across the Mediterranean Sea. However, climate change has exacerbated the situation and Europe will feel the effects more from the sweeping dust which scatters across the continent. Forecasters from the European Union’s Earth observation programme Copernicus Atmosphere Monitoring Service (CAMS) say dust will continue to spread as a warming planet creates larger deserts.
This will cause major problems for all of Europe.
Firstly, the dust is deposited on car windscreens across most of the south of the continent, leading to poorer driving conditions.
It is also likely the dust will limit train travel, as “ dust landing on railways tracks and overhead lines can affect train operations,” according to CAMS.
However, the health impacts for humans could be even more severe, as experts are expecting an increase in respiratory illnesses including aggravating asthma symptoms.
Europe dust storm warning: Sahara desert dust to ‘impact health and spark FOOD SHORTAGES’
CAMS also says that plant life could be reduced, as “dust landing on leaves can limit photosynthesis, creating problems for crops.”
But it is not just human health which will degrade thanks to an increase in dust in Europe.
As more dust reaches the atmosphere, the Sun will find it more difficult to breakthrough, meaning solar energy production is reduced.
Solar energy accounts for four percent of energy in Europe, and three percent in the UK.
While these are not extremely large figures, Europe and the UK may have to revert back to fossil fuel-based energy if enough solar energy is not being produced, which will only exacerbate the situation.
Richard Engelen, deputy ead of CAMS, said: “The visible nature of dust demonstrates that sand particles can travel a long way, and makes the point that air quality is not only local.
“With climate change causing deserts to grow, dust in the atmosphere will only increase.”<|endoftext|>
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Impact of Engine Cylinder Liners in the Industrial Applications
A cylinder is the central working part of a reciprocating engine or pump, the space in which a piston travel. Multiple cylinders are commonly arranged side by side in a bank, or engine book, which is typically cast from aluminium or cast iron before receiving precision machine work. The engine cylinder liners can be sleeved (lined with a harder metal) or sleeveless (with a wear-resistant coating such as Nikasil). A sleeveless engine can be referred to as a “parent-bore engine”.
A cylinder’s displacement, or swept volume, may be calculated by multiplying its cross-sectional area (the square of half the bore by pi) by the distance the piston travels within the cylinder (the stroke). The engine displacement may be calculated by multiplying the swept volume of one cylinder by the number of cylinders use in machine.
A piston is seated inside every cylinder by various metal piston ring fitted around its outside surface in machined grooves; typically, two for compressional sealing and one to seal the oil. The rings make close contact with the cylinder walls (sleeved or sleeveless), riding on a thin layer of lubricating oil; necessary to keep the engine from seizing and necessitating a cylinder wall’s durable surface.
During the initial stage of an engine’s life, its early breaking-in or running-in period, small irregularities in the metals are encouraged to gradually form congruent grooves by avoiding extreme operating conditions. After in its life, later mechanical wear has increased the space between the piston and the cylinder (with a consequent decrease in power output). The cylinders may be machined to a slightly larger diameter to receive new sleeves (where applicable) and piston rings, a process is sometimes known as reborning.
Heat engines, including Stirling engines, are sealed machines using pistons within cylinders to transfer energy from a heat source to a colder reservoir, often using steam or another gas as the working substance. The very first illustration depicts a longitudinal section of a cylinder in a steam engine. The down sliding part at the bottom is the piston, and the upper sliding part is a distribution valve (in this case of the D slide Valve type) that directly steam alternates into either end of the cylinder through the engine cylinder liners. Refrigerator and air conditioner compressors are also heat engines driven in reverse cycle as pumps.<|endoftext|>
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Inflation is the aggregate level at which prices for goods and services are increasing. When inflation occurs, it means that the purchasing power of consumers and businesses is declining, unless they can increase their income by an offsetting amount. Inflation also reduces the value of savings. If the inflation rate is higher than the return on investment that a person or business is experiencing, then there is a net decline in investment. Inflation has an especially pernicious impact on those with fixed incomes, since their purchasing power gradually declines over time.
For example, if inflation is 3%, then the price of an item that cost $1.00 one year ago now costs $1.03. As just noted, inflation is considered an aggregate amount, which is the average change in prices for a selected set of goods and services. In reality, the inflation rate for a specific item may be much higher than the reported aggregate inflation rate, while the prices of other items may have declined during the same period of time.
Inflation may be caused by a constriction in supply. For example, when there is a shortfall in the amount of gasoline in comparison to the demand for it, the price of gasoline will increase (unless prices are controlled by order of the government). Inflation may also occur when a loosening of credit results in individuals and businesses spending so much that excessive demand forces an increase in prices.
A minor level of year-over-year inflation is considered to be optimal, usually in the range of 2-3%. This is the target level that most central banks use as one of the goals of their monetary policies.
When prices are increasing at an extremely fast rate, it is called hyperinflation.<|endoftext|>
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According to the Joint United Nations Programme on HIV/AIDS (UNAIDS), an estimated 5,000 new cases of the human immunodeficiency virus (HIV) occur every day across the globe.
Finding feasible healthcare solutions to the HIV/AIDS epidemic has been a global priority, and according to a 2017 report from UNAIDS, preventative efforts have been successful in reducing the number of new HIV infections by almost 50% since the peak of the epidemic in 1996. Moreover, while current pharmaceutical therapies for HIV and Acquired Immunodeficiency Syndrome (AIDS) treatment have gotten better at halting transmission of the virus and mitigating progression of the disease, treatment options can be prohibitively expensive, and a cure remains elusive, given the mechanism of the viral infection.
HIV is a retrovirus that copies its genome into that of a human host’s cells, and then hijacks the host cell’s replication machinery, allowing the virus to replicate. HIV is particularly problematic because the cells it targets are immune cells. The specific cells targeted are CD4 cells, a type of T cell that is normally responsible for aiding in the immune system’s attack against foreign infections. Therefore, HIV infections leave the immune system in a vulnerable state, making HIV-positive individuals more likely contract opportunistic infections, such as toxoplasmosis, cryptococcal meningitis, and some cancers. Contracting an opportunistic infection can lead to an AIDS diagnosis, given that the CD4 cell count falls below a specific threshold.
Treatment for HIV/AIDS often involves a drug cocktail of several medications used to limit progression of symptoms. One of these medications, known by its brand name Truvada, is actually a combination of several drugs: disoproxil, emtricitabine, and tenofivir, each of which is designed to treat infection by slowing progression of the disease. Specifically, Truvada works by inhibiting the enzyme reverse transcriptase, which is essential for HIV to replicate. In addition to being used in treatment after infection, Truvada is also being explored as preventative therapy for HIV-negative people.
In 2012, the Food and Drug Administration approved Truvada as a preventative therapy for HIV-negative people, making it the first drug to be approved for this use. Truvada is taken in pill form, and recent studies have investigated its efficacy as a preventative medication through a treatment technique known as pre-exposure prophylaxis (PrEP).
PrEP is a treatment designed for individuals who are HIV-negative, but at high risk for HIV transmission, to lessen their chances of HIV infection through daily administration of the pill. PrEP attempts to mitigate an individual’s chances of HIV infection by preventing the virus from permanently establishing itself. This is a considerable measure, given that once HIV infection occurs, even though modern treatment can effectively manage symptoms and slow progression of the disease, it is impossible to fully eradicate the virus from the body. Therefore, PrEP aims to prevent the virus’ initial establishment of permanent infection. However, for PrEP to be effective, it is imperative that the Truvada pill is taken daily and coupled with other preventative methods such as usage of condoms.
In addition, it would be false to consider PrEP a vaccination for HIV. A vaccination prevents infection by introducing an individual’s cells to weakened or dead versions of a pathogen to train the immune system and allow development of antibodies to fight the infection. Conversely, PrEP operates under the idea that if a high risk HIV-negative individual has enough of the antiviral medications already present in his or her body, the risk of infection is drastically reduced and disease transmission can be prevented.
Funding for use of PrEP as a preventative measure for HIV/AIDS has been approached with hesitance by policymakers who are unsure of its efficacy at the population level. However, a new study on HIV infection transmission in gay and bisexual men in New South Wales, the most populous state in Australia, serves as the first-of-its-kind population study investigating the success of Truvada in PrEP.
The study, published in mid-October of this year in the medical journal The Lancet HIV, found significantly rapid declines in the number of HIV diagnoses in the target population, and has attributed these declines to PrEP implementation. In the year prior to the study’s beginning, 149 new infections were recorded among the recruited high-risk population of gay and bisexual men in New South Wales. In the first year after the study began and PrEP was implemented, this number dropped to only 102 new infections. The study sheds light on the population-level benefits of PrEP and may serve as impetus for policymakers to consider funding the treatment for targeted preventative therapy for high-risk, HIV-negative individuals.<|endoftext|>
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Real Numbers: Property CHART MathBitsNotebook.com
A real number is a value that represents a quantity along a continuous number line. Real numbers can be ordered. The symbol for the set of real numbers is , which is the letter R in the typeface "blackboard bold".
The real numbers include: counting (natural) numbers (){1, 2, 3, ... }, whole numbers {0, 1, 2, 3, ... }, integers (){... , -3, -2, -1, 0, 1, 2, 3, ...}, rational numbers () (such as -½, 6.25, ) and irrational numbers (such as ).
FYI: The series of three dots ( ... ), seen within the sets above, is called an ellipsis.
As used in mathematics, the ellipsis means "and so forth".
The properties of the Real Number System will prove useful when working with equations, functions and formulas in Algebra, as they allow for the creation of equivalent expressions which will often aid in solving problems. In addition, they can be used to help explain or justify solutions.
Property (a, b and c are real numbers, variables or algebraic expressions) Examples Verbal hints 1. Distributive Property a • (b + c) = a • b + a • c 3 • (4 + 5) = 3 • 4 + 3 • 5 "multiplication distributes across addition" 2. Commutative Property of Addition a + b = b + a 3 + 4 = 4 + 3 "commute = to get up and move to a new location : switch places" 3. Commutative Property of Multiplication a • b = b • a 3 • 4 = 4 • 3 "commute = to get up and move to a new location: switch places" 4. Associative Property of Addition a + (b + c) = (a + b) + c 3 + (4 + 5) = (3 + 4) + 5 "regroup - elements do not physically move, they simply group with a new friend." 5. Associative Property of Multiplication a • (b • c) = (a • b) • c 3 • (4 • 5) = (3 • 4) • 5 "regroup - elements do not physically move, they simply group with a new friend." 6. Additive Identity Property a + 0 = a 4 + 0 = 4 "the value that returns the input unchanged" 7. Multiplicative Identity Property a • 1 = a 4 • 1 = 4 "the value that returns the input unchanged" 8. Additive Inverse Property a + (-a) = 0 4 + (-4) = 0 "the value that brings you back to the identity element under addition" 9. Multiplicative Inverse Property "the value that brings you back to the identity element under multiplication" 10. Zero Property of Multiplication a • 0 = 0 4 • 0 = 0 "zero times any value is 0" 11. Closure Property of Addition a + b is a real number 10 + 5 = 15 (a real number) "the sum of any two real numbers is another real number" 12. Closure Property of Multiplication a • b is a real number 10 • 5 = 50 (a real number) "the product of any two real numbers is another real number" 13. Addition Property of Equality If a = b, then a + c = b + c. If x = 10, then x + 3 = 10 + 3 "adding the same value to both sides of an equation will not change the truth value of the equation." 14. Subtraction Property of Equality If a = b, then a - c = b - c. If x = 10, then x - 3 = 10 - 3 "subtracting the same value from both sides of an equation will not change the truth value of the equation." 15. Multiplication Property of Equality If a = b, then a • c = b • c. If x = 10, then x • 3 = 10 • 3 "multiplying both sides of an equation by the same value will not change the truth value of the equation." 16. Division Property of Equality If a = b, then a / c = b / c, assuming c ≠ 0. If x = 10, then x / 3 = 10 / 3 "dividing both sides of an equation by the same non-zero value will not change truth value of the equation." 17. Substitution Property If a = b, then a may be substituted for b, or conversely. If x = 5, and x + y = z, then 5 + y = z. "a value may be substituted for its equal." 18. Reflexive (or Identity) Property of Equality a = a 12 = 12 "a real number is always equal to itself" 19. Symmetric Property of Equality If a = b, then b = a. "quantities that are equal can be read forward or backward" 20. Transitive Property of Equality If a = b and b = c, then a = c. If 2a = 10 and 10 = 4b, then 2a = 4b. "if two numbers are equal to the same number, then the two numbers are equal to each other" 21. Law of Trichotomy Exactly ONE of the following holds: a < b, a = b, a > b If 8 > 6, then 8 6 and 8 is not < 6. "for two real numbers a and b, a is either equal to b, greater than b, or less than b." (common sense)
FYI: Regarding the term "natural numbers", there is no universal agreement about whether to include zero in this set. Most mathematicians hold with the older traditional and define natural numbers to be the counting numbers [positive integers {1, 2, 3, ...}]. Computer scientists, set theorists, logicians and other mathematicians define natural numbers to be the whole numbers [non-negative integers {0, 1, 2, ...}]. This site will use the term "natural numbers" to refer to the counting numbers {1, 2, 3, ...}.<|endoftext|>
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# A piece of string is $30 \mathrm{~cm}$ long. What will be the length of each side if the string is used to form:(a) a square?(b) an equilateral triangle?(c) a regular hexagon?
Given:
A piece of string is 30 cm long.
To do:
We have to find the length of each side when the string is used to form:
(a) a square
(b) an equilateral triangle
(c) a regular hexagon
Solution:
The perimeter of each of the forms is equal to the length of the string in each case.
Therefore,
(a) The perimeter of the square form of the string $=30\ cm$
We know that,
The perimeter of a square is four times its side.
$4\times$ each side of the square $=30\ cm$
Each side of the square $=\frac{30}{4}$
$=7.5\ cm$
The length of each side of the square is $7.5\ cm$
(b) The perimeter of the equilateral triangle form of the string $=30\ cm$
We know that,
The perimeter of an equilateral triangle is three times its side
$3\times$ each side of the triangle $=30\ cm$
Each side of the triangle $=\frac{30}{3}$
$=10\ cm$
The length of each side of the triangle is $10\ cm$.
(c) The perimeter of the regular hexagon form of the string $=30\ cm$
We know that,
The perimeter of a regular hexagon is six times its side
$6\times$ each side of the regular hexagon $=30\ cm$
Each side of the regular hexagon $=\frac{30\ cm}{6}$
$=5\ cm$
The length of each side of the regular hexagon is $5\ cm$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
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# Difference between revisions of "Boolean Algebra"
Boolean algebra is the branch of algebra in which the values of the variables and constants have exactly two values: true and false, usually denoted 1 and 0 respectively.
The basic operators in Boolean algebra are and, or, and not. The secondary operators are exclusive or (often called xor) and exclusive nor (sometimes called equivalence). They are secondary in the sense that they can be composed from the basic operators of and, or, and not.
• The and of two values is true only whenever both values are true. It is written as $xy$ or $x \cdot y$. The values of and for all possible inputs is shown in the following truth table:
$x$ $y$ $x y$
0 0 0
0 1 0
1 0 0
1 1 1
• The or of two values is true whenever either or both values are true. It is written as $x+y$. The values of or for all possible inputs is shown in the following truth table:
$x$ $y$ $x + y$
0 0 0
0 1 1
1 0 1
1 1 1
• The not of a value is its opposite; that is, the not of a true value is false whereas the not of a false value is true. It is written as $\overline{x}$ or $\neg{x}$. The values of not for all possible inputs is shown in the following truth table:
$x$ $\overline{x}$
0 1
1 0
• The xor of two values is true whenever the values are different. It uses the $\oplus$ operator, and can be built from the basic operators: $x \oplus y = x \overline{y} + \overline{x} y$ The values of xor for all possible inputs is shown in the following truth table:
$x$ $y$ $x \oplus y$
0 0 0
0 1 1
1 0 1
1 1 0
• The xnor of two values is true whenever the values are the same. It is the not of the xor function. It uses the $\odot$ operator and can be built from basic operators: $x \odot y = x y + \overline{x} \overline{y}$ The values of xnor for all possible inputs is shown in the following truth table:
$x$ $y$ $x \odot y$
0 0 1
0 1 0
1 0 0
1 1 1
Just as algebra has basic rules for simplifying and evaluating expressions, so does Boolean algebra.
## Why is Boolean Algebra Important for ACSL Students?
Boolean algebra is important to programmers, computer scientists, and the general population.
• For programmers, Boolean expressions are used for conditionals and loops. For example, the following snippet of code sums the even numbers that are not also multiples of 3, stopping when the sum hits 100:
s = 0
x = 1
while (s < 100):
if (x % 2 == 0) and (x % 3 != 0):
s = s + x
x = x + 1
Both s < 100 and (x % 2 == 0) and (x % 3 != 0) are Boolean expressions.
• For computer scientists, Boolean algebra is the basis for digital circuits that make up a computer's hardware. The Digital Electronics category concerns a graphical representation of a circuit. That circuit is typically easiest to understand and evaluate by converting it to its Boolean algebra representation.
• The general population uses Boolean algebra, probably without knowing that they are doing so, when they enter search terms in Internet search engines. For example, the search expression "red sox -yankees" is the Boolean expression "red" and "sox" and not "yankees" that will returns web pages that contain the words "red" and "sox", as long as it does not contain the word "yankees". The search expression "jaguar speed -car" returns pages about the speed of the jaguar animal, not the Jaguar car.
## Laws
A law of Boolean algebra is an identity such as $x + (y + z) = (x + y) + z$ between two Boolean terms, where a Boolean term is defined as an expression built up from variables and the constants 0 and 1 using the operations and, or, not, xor, and xnor. LIke ordinary algebra, parentheses are used to group terms.
### Monotone laws
Boolean algebra satisfies many of the same laws as ordinary algebra when one matches up or with addition and and with multiplication. In particular the following laws are common to both kinds of algebra:
Associativity of or: $x + (y + z)$ $= (x + y) + z$ Associativity of and: $x \cdot (y \cdot z)$ $= (x \cdot y) \cdot z$ Commutativity of or: $x + y$ $= y + x$ Commutativity of and: $x \cdot y$ $= y \cdot x$ Distributivity of and over or: $x \cdot (y + z)$ $= x \cdot y + x \cdot z$ Identity for or: $x + 0$ $= x$ Identity for and: $x \cdot 1$ $= x$ Annihilator for and: $x \cdot 0$ $= 0$
The following laws hold in Boolean Algebra, but not in ordinary algebra:
Annihilator for or: $x +1$ $= 1$ Idempotence of or: $x +x$ $= x$ Idempotence of and: $x \cdot x$ $= x$ Absorption 1: $x \cdot (x + y)$ $= x$ Absorption 2: $x + (x \cdot y)$ $= x$ Distributivity of or over and: $x + (y \cdot z)$ $=(x + y) \cdot (x +z)$
### Nonmonotone laws
Complement of and: $x \cdot \overline{x}$ $= 0$ Complement of or : $x + \overline{x}$ $= 1$ Double Negation: $\overline{ \overline{x}}$ $= x$
### De Morgan's Laws
DeMorgan 1: $\overline{ x} \cdot \overline{y}$ $= \overline{x + y}$ DeMorgan 1: $\overline{ x} + \overline{y}$ $= \overline{x \cdot y}$
## Sample Problems
Problems in this category are typically of the form "Given a Boolean expression, simplify it as much as possible" or "Given a Boolean expression, find the values of all possible inputs that make the expression true."
### Sample Problem 1: Simplify
Problem: Simplify the following expression as much as possible: $\overline{ \overline{A(A+B)} + B\overline{A}}$
Solution:
The simplification proceeds as follows:
$\overline{ \overline{A(A+B)} + B\overline{A}}$
$= \left(\overline{ \overline{A(A+B)}}\right) \cdot \left(\overline{ B\overline{A}}\right)$ (DeMorgan's Law) $= \left(A(A+B)\right) \cdot \left( \overline{B}+\overline{\overline{A}}\right)$ (Double Negation; DeMorgan's Law) $= A \cdot \left( \overline{B}+A\right)$ (Absorption; Double Negation) $= A \overline{B}+AA$ (Distribution) $= A \overline{B}+A$ (Distribution) $=A \left( \overline{B}+1\right)$ $=A \left(1\right)$ $=A$
### Sample Problem 2: Find Solutions
Problem: Find all orderd pairs $(A,B)$ that make the following expression true: $\overline{ \overline{(A+B)} + \overline{A}B }$
Solution:
There are typically two approaches to solving this type of problem. One approach is to simplify the expression as much as possible, until it's obvious what the solutions are. The other approach is to create a truth table of all possible inputs, with columns for each subexpression.
The simplification approach is as following:
$\overline{\overline{(A+B)} + \overline{A}B}$
$= \overline{\overline{A+B}} \cdot \overline{\overline{A}B}$
$= (A+B) \cdot (\overline{\overline{A}}+\overline{B} )$
$= (A+B) \cdot (A+\overline{B})$
$= AA + A\overline{B} + BA + B\overline{B}$
$= A + A(\overline{B} + B) + 0$
$= A + A(1)$
$= A + A$
$=A$
This means that all inputs are valid whenever $A$ is true: $(1,0)$ and $(1,1)$
The truth table approach is as following. Each column is the result of a basic operation on two other columns.
#1 #2 #3 #4 #5 #6 #7 #8
OR of Col#1, Col#2 NOT of Col#3 NOT of Col#1 ADD of Col#1, Col#2 OR of Col#4, Col#6 NOT of Col#7
$A$ $B$ $A+B$ $\overline{A+B}$ $\overline{A}$ $\overline{A}B$ $\overline{A+B} + \overline{A}B$ $\overline{\overline{A+B} + \overline{A}B}$
0 0 0 1 1 0 1 0
0 1 1 0 1 1 1 0
1 0 1 0 0 0 0 1
1 1 1 0 0 0 0 1
The rightmost column is the expression we are solving; it is true for the 3rd and 4th rows, where the inputs are $(1,0)$ and $(1,1)$.
## Online Resources
### Websites
A great online tutorial on Boolean Algebra is part of Ryan's Tutorials.
### Videos
The following YouTube videos show ACSL students and advisors working out some previous problems. To access the YouTube page with the video, click on the title of the video in the icon. (You can also play the video directly by clicking on the arrow in the center of the image; however, you'll probably want to have a larger viewing of the video since it contains writing on a whiteboard.) Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
ACSL Prep - Mrs. Gupta - Boolean Algebra (MegaChristian5555) Christian is a student participating in ACSL. This video shows how to solve a half-dozen or so Boolean Algebra problems that have appeared in ACSL contests in recent years in the Intermediate and Senior divisions. ACSL Boolean Algebra Contest 2 Worksheet 1 (misterminich) Mr. Minich is an ACSL advisor. This video was one of two he created to help prepare his students for the ACSL Boolean algebra category. It shows solutions to 5 different problems that have appeared in recent years. ACSL Boolean Algebra Contest 2 Worksheet 2 (misterminich) Mr. Minich is an ACSL advisor. This video was one of two he created to help prepare his students for the ACSL Boolean algebra category. It shows solutions to 5 different problems that have appeared in recent years. ACSL 3 13-14 #1 - AM (Gordon Campbell) This video walks through the solution to finding all ordered triples that make the following Boolean expression true: $(AB+\overline{C})(\overline{A}+BC)(A+\overline{B}+C)$ This problem appeared in 2013-2014 in the Senior Division, Contest #3. A general tutorial on boolean algebra that can be used for American Computer Science League. (Tangerine Code) Walks through the simplification of the following Boolean expression: $\overline{ (\overline{A + \overline{B}})(AB)} + \overline{ (A+B)(\overline{\overline{A}B})}$<|endoftext|>
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Show Me Character
Respect Character Connection
Aretha Franklin’s song “R-E-S-P-E-C-T!” made the word famous, but do we know what it really means? Ethical people are respectful of others, and demonstrate it by recognizing and honoring everyone’s right to be themselves, to make decisions, and to have privacy and dignity.
- Following the Golden Rule
- Using good manners, not bad language
- Practicing tolerance
- Being considerate of the feelings of others
- Not threatening, hitting or hurting anyone
- Dealing peacefully with anger, insults and disagreements
Infants, Toddlers and Preschoolers
The foundation of respect begins in infancy. The best way for you to begin teaching your child respect is by modeling the behavior. How you show respect to other family members and friends will influence the ways your child is respectful of others. Even in infancy, your child will benefit from experiencing kind words and actions.
Keep in mind that toddlers may find that the easiest way to solve problems is with a hit or grab because they are not skilled at using language. Continue to let your toddler know what behavior you want to see.
As your toddler grows to be a preschooler, you will want to have her show respect by using good manners. Praise and encourage your child when she says please or thank you at the appropriate time.
Let’s Pretend To Take Turns Activity
Taking turns in not a natural action, it must be taught. Help your child learn to take turns by having everyone in the family take turns playing with a favorite toy or doing a favorite chore. Talk about how important it is for everyone to get a turn.
School-age, Middle School and Teens
Respect is an essential foundation for good relationships. Being respectful means respecting others’ differences. This should begin at an early age by learning about the differences of children at CYS and learning how they are unique and special. You should encourage the same respect as they grow older. Children and teens who know how they are unique and special will be better prepared to handle an encounter with someone who is not respectful of individual differences.
Praise her when she handles difficult people or circumstances respectfully. Likewise, immediately explain why other behaviors are disrespectful.
As always, modeling respectful behavior and treating your children with respect are important.
Respect In Movies Activity
Watch a current movie with your teen. Agree before the movie starts that everyone will watch the movie particularly watching for both respectful and disrespectful behavior. When the movie is over, compare your results. Do you agree with your teen’s assessments? If not, take time to discuss your differences considering everyone’s point of view.
Practicing Respect 1
People of good character are respectful of others. You show respect by recognizing and honoring everyone’s right to be themselves, to make decisions, and to have privacy and dignity.
- Be courteous and polite.
- Be kind and appreciative.
- Accept individual differences and don’t insist that everyone be like you.
- Judge people on their merits.
Are you respectful? Spend some time talking to your children about being respectful and how respect and disrespect is shown in every day situations. Encourage them to talk about one of the following:
- A friend borrows one of your toys and won’t return it. Is that friend being respectful? What do you do?
- You tell a friend a secret and he tells it to a few other people. Is that friend being respectful?
- You’re at the monthly 4-H meeting; raise your hand to make a comment and everyone listens quietly while you talk. Are members of the 4-H club being respectful? Are you respectful when others are talking?
Practicing Respect 2
Aretha Franklin’s song “R-E-S-P-E-C-T!” made the word famous but do we know what it really means? Ethical people are respectful of others, and demonstrate it by recognizing and honoring everyone’s right to be themselves, to make decisions, and to have privacy and dignity. Ethical people deal with others using one of two principles:
- Respectful people give others the information they need to make wise decisions about their lives.
- Ethical people accept individual difference without prejudice. Respect is given because you are a person of good character.
Use these guidelines to strengthen your character:
- Be courteous and polite
- Be kind and appreciate
- Accept individual differences and don’t insist that everyone be like you
- Judge people on their merits, not on race, religion, nationality, age, sex, gender identity, sexual orientation, genetic information, family structure, physical or mental condition. Or socio-economic status.
Using good manners is a way to show you respect yourself and others. Send this month learning about table manners, dating etiquette and business etiquette. Go to the library, do a web search or ask an expert to help you in this exercise.
Ask a group of your friends ways they want to be shown respect. Then turn this list into a checklist on how you will treat others your age for the next week. Then try it for a second week.<|endoftext|>
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The shift in music from the Renaissance to the Baroque was quite dramatic. Particularly regarding instrumentation, function, texture, melody and harmony, the changes were striking and long-lasting. Much of our current music still employs conventions fostered during this era. We will discuss these conventions. The shift is seen in pieces by Purcell, Gabrieli, Weelkes, Monteverdi and Vivaldi. Music by these composers will be considered.© BrainMass Inc. brainmass.com March 21, 2019, 10:50 pm ad1c9bdddf
During the time of transition between the Renaissance and the Baroque period, many changes took place. One change was the specification of instrumentation. During the Renaissance, specific instruments were not called for, but during the Baroque period, this became a regular practice. The first composer to call for specific instruments was Giovanni Gabrieli. Gabrieli was the choirmaster at St. Marks church in Venice. Gabrieli was also the first composer to specify dynamics.
Another change involved the switch in texture from the older contrapuntal style of the Renaissance to the homorhythmic style of the Baroque. This is best exemplified by Gabrieli's polychoral motet O quam suavis.
Harmony also changed during this time, moving from modal to major/minor.
Function moved from vocal forms and dance music to instrumental pieces such as the sonata and sinfonia.
Performance sites also changed, moving from the church and court to public theatres.
Bibliography: Machlis, ...
The shift in music from the Renaissance to the Baroque is assessed.<|endoftext|>
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# Simplifying Algebraic Fractions: Difference of Squares
## Transcript
Okay! So first of all, I took out the common factors for anything that I can. So here! See how x squared is common. So I took the x squared out and here, 16. We know that 16 is 4 squared. So this will be a difference of two squares.
So I’ve simplified it, like that. Now! Let’s use our cross method for the denominators, so you can see that one I’m going to use 4 which becomes 4x, and 5 which becomes 5x. Add them! We get 9x which is exactly what we want.
Now for this one, I’m going to use my x and x again. Now it’s negative, so I’m going to use positive and negative. So one’s going to be positive 3 and the other one will be negative 4, so negative 4x. And 3 minus 4x is negative x which is what we want. So now replace them all back into the denominators like that.
See how that one, I just left it as it is, and see how x squared minus 4 squared that’s x plus 4x minus 4, the difference of 2 squares and here, x plus 4, x plus 5 on the denominator and same for there, just replace them in. So now it’s my favorite part is to start canceling if anything that’s common. Now you can see that here, x minus 4 is common! Cancel! And you can see here, x plus 3, x plus 3 is common! Cancel! x plus 4, x plus 4, Common! Cancel them all out. So we just have x squared and x plus 5. That is the answer, okay?
So that’s just what’s left, so it’s nice and simple.<|endoftext|>
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Goats are natural browsers. This means that they like to eat with their heads up, often reaching as high as they can for leaves, twigs and wooded area’s understory. They prefer woods to grass pasture but will also nibble on fresh, uncontaminated grass. They are observed to be much fussier in what they will eat when compared to sheep or cattle—and are talented at nibbling off the most nutrient-rich parts of plants such as flowers, young twigs and new buds even to the point of nibbling around thorns. Poison ivy is a delicacy and the young leaves of nettles, thistles and buck thorn succumb quickly to a herd of goats.
The goat’s digestive system does a very good job of turning browse into a nutritious feed source that is not able to be utilized efficiently by other livestock. Unlike native grasses, leaves and browse hold their nutritional value throughout the summer providing a consistent source of protein and carbohydrates. In contrast, summer grass matures early in the season and the nutrients are shifted to the seed head leaving a fibrous grass stem with low nutritional value for livestock. During drought conditions native grasses convert nutrients to the seed head faster than normal to ensure survival when conditions improve resulting in even less optimal grazing days for sheep and cattle. This is true globally and especially in countries with extended dry periods.
Nutrition is usually higher in browse than in grasses especially during the dry season. The goats preference for browse help them cope surprisingly well during droughts, and in arid or low fertility environments when other livestock species are more likely to struggle. Ninety percent of the world’s goat populations provide a source of protein under harsh conditions. Goats have adapted to virtually any climate and are viewed by many working in the poorest countries as the most sustainable livestock for ensuring global food security (Winrock International, 2011). The goats in the photograph below are nimbly removing leaves from around thorns in Ethiopia Highlands (Farmer to Farmer volunteer assignment, 2006).
Note: It was an early custom by mariners and pirates to place goats on islands for later use. If the goats survived fresh water was available somewhere on the island and the goats would be a source of food for later trips. Goats were spread around the navigable world in this manner, went feral, interbred and developed their own unique adaptations to a variety of environments ranging from harsh to lush. Captain James Cook is famous for dropping goats in New Zealand and other South Pacific islands for emergency food during follow up visits or shipwrecks (Storey’s Guide to Raising Dairy Goats, Jerry Belange, 2010).<|endoftext|>
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Continuing in the tradition started by the Cassini mission to Saturn, Scientist for a Day challenges students in grades 5-12 to think like NASA scientists. They examine real spacecraft images of Saturn’s moon Enceladus, Saturn’s moon Titan, and Jupiter’s moon Europa, then hoose the destination they think would be the best place to return with another spacecraft to learn even more about these amazing worlds.
The essay contest meets various U.S. National Standards for English and Science set by the National Council of Teachers of English / International Reading Association, and the National Research Council.
It also addresses topics covered in Next Generation Science Standards, including:
Middle School - MS-ESS1 Earth’s Place in the Universe MS-ESS1-3. Analyze and interpret data to determine scale properties of objects in the solar system
High School - HS-ESS1 Earth’s Place in the Universe HS-ESS1-6. Apply scientific reasoning and evidence from ancient Earth materials, meteorites, and other planetary surfaces to construct an account of Earth’s formation and early history
Deadline: February 8, 2019
Learn more at:<|endoftext|>
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# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## AP Board Class 7 Maths Chapter 4 Lines and Angles Ex 6 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 4 Lines and Angles Ex 6 Book Answers
AP Board Class 7 Maths Chapter 4 Lines and Angles Ex 6 Textbook Solutions PDF: Download Andhra Pradesh Board STD 7th Maths Chapter 4 Lines and Angles Ex 6 Book Answers
## Andhra Pradesh State Board Class 7th Maths Chapter 4 Lines and Angles Ex 6 Books Solutions
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## AP Board Class 7th Maths Chapter 4 Lines and Angles Ex 6 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 7th Maths Chapter 4 Lines and Angles Ex 6 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
Question 1.
Name two pairs of vertically opposite angles in the figure.
Solution:
Vertically opposite angles are
∠AOC, ∠BOD and ∠BOC, ∠AOD
Question 2.
Find the measure of x, y and z without actually measuring them.
Solution:
From the figure
∠y and 160° are vertically opposite angles and hence equal.
∴ ∠y=1600
Also ∠x = ∠z and (. vertically opposite angles)
x + 160° = 180° (Linear pair of angles)
∴ ∠x = 180° – 160° = 20°
∠z = 20° (∵ ∠x, ∠z are vertically opposite)
∴ x = 20°
y = 160°
z = 20°
Question 3.
Give some examples of vertically opposite angles in your surroundings.
Solution:
i) Angles between legs of a folding cot/scamp cot.
ii) Angles between legs of a folding chair.
iii) Angles between plates of a scissors.
## Andhra Pradesh Board Class 7th Maths Chapter 4 Lines and Angles Ex 6 Textbooks for Exam Preparations
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Condensation, Dew Point, and Roofing
Except in extremely arid climates, there is always some amount of water vapor in the air around us. When that air comes into contact with a cold surface, that water vapor condenses as a liquid onto the surface. A good example of this is the water droplets on the side of the glass of ice water. Those droplets are commonly known as “condensation” and are what results when the air gets too cold to hold the water vapor that is in it. Even when a cold surface is not available, if the air temperature suddenly drops, water vapor condenses out as mist or fog. Air can only hold so much water – more at higher temperatures and less at colder temperatures.
Let’s examine this in a little more detail, taking a closer look at…
We know that air contains water vapor but we need to define how much it contains. At any temperature, there is a maximum amount of water that air can hold. When we measure how much water is actually in the air, we express the number as a percentage of that maximum amount. For most people, 50 to 60% relative humidity is very comfortable, but most of us can easily tolerate anywhere from 30 to 70%. Relative humidity below 30% is noticeably dry and above 70% is when people start commenting about how humid it feels.
Let’s compare Miami and Phoenix to see how relative humidity comes into play. In Miami, a cold beverage may be served with a napkin wrapped around it to absorb the condensation that forms on the glass. But in Phoenix, there may be so little condensation on the cold glass that a napkin might not be needed. Why is that? Relative humidity is the major contributing factor. The reason is that the relative humidity in Miami is likely to be above 65%, i.e., the air is holding 65% of the moisture it is capable of holding. In contrast, the air in Phoenix would likely be dry with a relative humidity around 35%, causing very little condensation to form. So, to recap, relative humidity is a ratio of how much water vapor is in the air in relation to how much the air can contain at a given temperature. The “relative” part refers to the fact that air’s capacity to hold moisture changes with temperature. The warmer the air is, the greater amount of moisture it can hold. The more moisture it holds, the greater the volume of condensation forms on a cold surface. Now, let’s discuss dew point.
…air’s capacity to hold moisture changes with temperature.
The dew point is a specific temperature at a given humidity at which water vapor condenses. Let’s consider Miami and Phoenix again as two extremes. In the summer, Miami’s relative humidity can reach 85% at a temperature of 80°F. Obviously, a lot of condensation will form on a chilled beverage glass. But it actually does not take much of a drop in temperature to reach 100% relative humidity and have condensation form. So, a lot of cool surfaces will have condensation on them. At the same temperature in Phoenix (80°F), the relative humidity could be 35%. The temperature would have to be much lower before condensation could form. Cool surfaces would not have condensation on them.
The Dew Point is the temperature at which condensation forms. It is a function of the relative humidity and the ambient temperature. In other words, the amount of water vapor that is in the air and the temperature of the air. Take a look at the chart below (which is a very simplified form of what is actually used by HVAC engineers). Let’s pick the 40% relative humidity line in the first column, and follow that line across to the 70°F column. The 40% line and 70°F column intersect at 45°F, meaning that in an environment that is 70°F and 40% relative humidity (RH), water in the air will condense on a surface that is 45°F.
Dew Point Temperatures for Selected Air Temperature and Relative Humidity
|Dew Point Temperature (°F)|
|Relative Humidity||Design Dry Bulb (Interior) Temperature (°F)|
|Chart adapted from ASHRAE Psychometric Chart, 1993 ASHRAE Handbook—Fundamentals.|
So, what does that have to do with roofing? Well, consider your building envelope: It separates the interior conditioned environment from the outside. The foundation, walls, and roof are all systems that intersect to make this happen. Although this pertains in some respect to all of the systems, we will focus on roofing. The insulation layer in the roofing system resists heat loss or gain from the outside, depending on the season. Within the insulation layer, the temperature slowly changes until it reaches outside. Let’s talk about a building in the winter to illustrate the point. The interior is 70°F with 40% RH, like our example on the chart above. As you move through the insulation layer from the inside toward the outside, the temperature gradually drops until it reaches the colder temperature outside. The plotting of those temperatures is referred to as the temperature gradient of that system.
Now, if the temperature gets to 45°F at any point in that system (the dew point temperature on the chart), then water would be expected to condense on the nearest surface. This is shown in the following diagram:
To recap, the interior air has 40% of the total water vapor that it can support. But as the air migrates up through the roof system, it gets cooler until the point where it can no longer hold on to the water vapor and condensation occurs. In the example shown above, this would happen at 45°F and just inside the insulation layer.
Lessons for the Roof Designer
Condensation — which is liquid water — can negatively affect the building in many ways. It can lead to R-value loss of the insulation layer by displacing the air within the insulation with water, as well as premature degradation of any of the roofing system components, such as rotting wood or rusting metal (including structural components). It can also contribute to unwanted biological growth, such as mold.
However, prevention of these negative effects is possible. Remember that water vapor needs to get to a surface or location that is at or below the dew point temperature.
In the schematic of the roof assembly shown above, it is clear that interior air must be prevented from moving up into the roof as much as possible. This was discussed in detail in an earlier GAF blog. One method to limit air movement into a roof includes using two layers of overlapping foam insulation. Another method is to place a vapor retarder or air barrier on the warm side of the insulation. The vapor retarder/air barrier can prevent the water vapor from reaching the location where it can condense.
Also, penetrations for vents and other details that involve cutting holes through the insulation must be looked at closely. If the gaps around penetrations are not sealed adequately, then the interior air is able to rapidly move up through the roof system. In cold climates that may lead to significant amounts of condensation in and around those penetrations.
Additionally, the billowing effect of a mechanically attached roof can exacerbate the potential for condensation because more air is drawn into the roof system. An adhered roof membrane may help limit the potential for air movement and subsequent condensation.
One important thing to remember is that relative humidity, and interior and exterior temperatures in summer and winter, should be considered when designing the envelope.
Typically, commercial buildings have an environment designed by an HVAC engineer that will determine interior temperature and relative humidity taking occupant comfort into account, as well as a design exterior temperature based on the weather in the building’s location. These and other factors help engineers determine what type and size of equipment the building requires. The building envelope designer will use those values, plus the designed use of the building, and local codes to determine the construction of the building envelope. One important thing to remember is that relative humidity, and interior and exterior temperatures in summer and winter, should be considered when designing the envelope. An envelope design that works in one area of the country may not work in another part of the country which could lead to adverse conditions and the types of degradation mentioned earlier. Consider how your wardrobe would change if you moved from Minneapolis to Phoenix (here, we are relating your clothes to the building envelope).
In a perfect world, the location of the building would be the entire story. Unfortunately, building use can (and often does) change. Factors that can adversely affect temperature and humidity, and therefore hygrothermal performance of the envelope, can include: a dramatic change in the number of occupants, the addition of a kitchen or cooking equipment, the addition of a locker room workout area or shower, and sometimes even something that seems insignificant like an aquarium or stored wood for a fireplace. This is not meant to be an exhaustive list, but a few illustrative examples to communicate a general understanding. Believe it or not, even changing the color of the exterior components could contribute to greater or less solar gain and effectively change the dew point location within the building envelope. Changing the dew point and/or dew point location can lead to unwelcome condensation, and potentially result in damage.
Changing the dew point and/or dew point location can lead to unwelcome condensation, and potentially result in damage.
Consider a situation where an owner decides to invest in energy efficiency upgrades on their property while replacing the roof. The owner upgrades windows, doors, and weather-stripping at the same time. The building could have had latent moisture problems that were previously hidden by air leaks across the building envelope. After the retrofits, those issues may surface, for example, in the form of stained ceilings. Was the water damage caused by the retrofit? Most likely the answer would be no. The previous inefficient design disguised the problem.
Keep in mind that a holistic approach should be taken with building envelope design. If you change one part, it could negatively affect something else. This blog is for general information purposes only. It is always a good idea to consult a building envelope consultant to help prevent condensation issues and ensure that small changes do not become large problems.<|endoftext|>
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Some time ago we heard the news that the Monkeypox virus infects humans who are citizens of England and Singapore. The Monkeypox virus infects humans after contact with animals carrying the virus. People call monkey pox because people first discovered this disease in monkeys.
After being first discovered in 1958, the monkeypox virus only appeared several times before the last appearance. The monkeypox virus infects human is a rare event. But actually, human monkeypox is endemic in the villages of Central and Western Africa. People often suffer from this disease because they make contact with infected animals in the suburbs of tropical forests.
Smallpox in Humans.
In many countries, the events of the monkeypox virus infect humans have never occurred. Countries generally know more about smallpox which infects humans. Smallpox that infects humans is smallpox or many people refer to it as Variola. This is because of the emergence of the obligation of immunization in infants.
Other smallpox which also infects more in various countries is chickenpox. This smallpox has a lighter level than smallpox. Although it contains the word chicken, it is not the chicken from this disease. This disease generally infects children. Some call it chicken pox.
There is another skin infection that infects toddlers by bacteria called Impetigo. And this disease infects children more often because of the intensity they play. Symptoms that appear almost similar to smallpox are blistering skin filled with fluid.
Monkeypox Virus Infects Humans: Causes
The zoonotic virus that causes Monkeypox disease. This virus is in animals then infects humans. Why can this virus infect humans? The monkeypox virus infects humans through direct contact, blood, body fluids, skin lesions, and mucosa of infected animals. This is what causes Monkey to become endemic to residents on the outskirts of tropical forests in Africa. Animals that can be the source of the monkeypox virus infect humans, among others, apes, a type of big mouse, squirrel Gambia, and types of rodents.
Preventing the Monkeypox Virus Infects Humans
We can prevent this transmission of the Monkeypox virus. Preventing the monkeypox virus infects humans by avoiding contact with monkeys, mice and other primates from wild forests. We must limit direct exposure to blood and meat. You must cook meat well. You should consume animal meat raised for endemic areas.
We also need to prevent the monkeypox virus from infecting humans from other humans. You must close physical contact with an infected person. You should also avoid contaminated ingredients. We recommend using gloves and protective clothing when you handle sick animals.<|endoftext|>
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The British and American governments were reluctant to publicize the intelligence they had received. A BBC Hungarian Service memo, written by Carlile Macartney, a BBC broadcaster and senior Foreign Office adviser on Hungary, stated in 1942: "We shouldn't mention the Jews at all." The British government's view was that the Hungarian people's antisemitism would make them distrust the Allies if Allied broadcasts focused on the Jews. The US government similarly feared turning the war into one about the Jews; antisemitism and isolationism were common in the US before its entry into the war. Although governments and the German public appear to have understood what was happening, it seems the Jews themselves did not. According to Saul Friedländer, "[t]estimonies left by Jews from all over occupied Europe indicate that, in contradistinction to vast segments of surrounding society, the victims did not understand what was ultimately in store for them." In Western Europe, he writes, Jewish communities seem to have failed to piece the information together, while in Eastern Europe, they could not accept that the stories they heard from elsewhere would end up applying to them too.
In 1937 he joined the Nazi party, then in 1938 he went to the SS. In 1942 he was wounded at the Russian front and was pronounced unfit for duty. After that he volunteered to go to the concentration camp, he was sent to the death camp, Auschwitz. Dr. Josef Mengele, nicknamed "the Angel of Death", became the surviving symbol of Adolf Hitler's "Final Solution".
During the era of the Holocaust, German authorities also targeted other groups because of their perceived racial and biological inferiority: Roma (Gypsies), people with disabilities, and some of the Slavic peoples (Poles, Russians, and others). Other groups were persecuted on political, ideological, and behavioral grounds, among them Communists, Socialists, Jehovah's Witnesses, and homosexuals.
The plans to exterminate all the Jews of Europe was formalized at the Wannsee Conference, held at an SS guesthouse near Berlin, on 20 January 1942. The conference was chaired by Heydrich and attended by 15 senior officials of the Nazi Party and the German government. Most of those attending were representatives of the Interior Ministry, the Foreign Ministry, and the Justice Ministry, including Ministers for the Eastern Territories. At the conference, Heydrich indicated that approximately 11,000,000 Jews in Europe would fall under the provisions of the "Final Solution". This figure included not only Jews residing in Axis-controlled Europe, but also the Jewish populations of the United Kingdom and of neutral nations (Switzerland, Ireland, Sweden, Spain, Portugal, and European Turkey). Eichmann's biographer David Cesarani wrote that Heydrich's main purpose in convening the conference was to assert his authority over the various agencies dealing with Jewish issues. "The simplest, most decisive way that Heydrich could ensure the smooth flow of deportations" to death camps, according to Cesarani, "was by asserting his total control over the fate of the Jews in the Reich and the east" under the single authority of the RSHA. A copy of the minutes of this meeting was found by the Allies in March 1947; it was too late to serve as evidence during the first Nuremberg Trial, but was used by prosecutor General Telford Taylor in the subsequent Nuremberg Trials.
Systematic examinations of rescuers have actually shown a high degree of heterogeneity among individuals. Rescuers came from divergent social backgrounds, varied widely in terms of political and religious involvements, and displayed different levels of friendship and animosity toward Jews. None of these variables has turned out to be a reliable predictor of the sort of person who was more or less likely to rescue Jews.
Zeidel had spent the previous two years in German-occupied Vilnius, in the city’s walled-off Jewish ghetto. He’d watched as the Nazis sent first hundreds and then thousands of Jews by train or truck or on foot to a camp in the forest. A small number of people managed to flee the camp, and they returned with tales of what they’d seen: rows of men and women machine-gunned down at close range. Mothers pleading for the lives of their children. Deep earthen pits piled high with corpses. And a name: Ponar.
In 1953, the Knesset, Israel's parliament, passed a law creating Yad Vashem as the country's Martyrs' and Heroes' Memorial Authority. Its tasks included commemorating the six million Jews killed by the Nazis and their collaborators during the Holocaust, paying tribute to those Jewish resistance fighters, and honoring those "high-minded Gentiles who risked their lives to save Jews." The title Righteous Among the Nations is taken from Jewish tradition (the literature of the Sages) that describes non-Jews who helped the Jewish people in times of need.
The Texas Senator upset that holocaust denier, Arthur Jones has won the Republican nomination for Illinois third Congressional district. — Fox News, "Judge Jeanine: The rise of socialism," 1 July 2018 In 1947, with immigration quotas still in existence, the SS Exodus, a boat carrying holocaust survivors who intended to migrate to Mandatory Palestine, was boarded by British forces, who killed three and returned the rest to refugee camps in Europe. — Billy Perrigo, Time, "Prince William Is Visiting the Middle East. Here's What to Know About Britain's Controversial Role in Shaping the Region," 25 June 2018 As the son of a Polish holocaust survivor, the images and sounds of this family separation policy is heart wrenching,’ Cohen wrote. — Chris Stirewalt, Fox News, "Like Bush and Obama, Trump gets stuck on immigration," 21 June 2018 According to holocaust historian Eric Saul, about 20 scouts of the 522nd Field Artillery entered Dachau’s ‘Camp X’ finding the crematoria and gas chambers. — Johnny Miller, San Francisco Chronicle, "Survivors thank ‘strange’ liberators," 18 Apr. 2018 In the book, the protagonist — a black female — wakes up 250 years after a nuclear holocaust, to find that humans have been rescued by aliens with three genders. — Billy Perrigo, Time, "Octavia E. Butler, Who Brought Diversity to the World of Science Fiction, Honored With Google Doodle," 22 June 2018 As the son of a Polish holocaust survivor, the images and sounds of this family separation policy is heart wrenching. — Monique Judge, The Root, "Is Michael Cohen About to Flip on Trump?," 20 June 2018 So, yeah, one of the North Korean team members led the world to a nuclear holocaust [but] that’s a truly impactful moment for that kid. — Mark Harris, Ars Technica, "First space, then auto—now Elon Musk quietly tinkers with education," 25 June 2018 To be sure, the current U.S. moral crisis is no holocaust and IBM’s deep involvement in customizing its punch card technology for the Nazis stands out like a red flag compared to a simple government cloud services contract. — Aaron Pressman, Fortune, "Data Sheet—Tech Industry Condemns Migrant Child Separation Policy. But What Will They Actually Do About It?," 20 June 2018
Even before the Nazis came to power in Germany in 1933, they had made no secret of their anti-Semitism. As early as 1919 Adolf Hitler had written, “Rational anti-Semitism, however, must lead to systematic legal opposition.…Its final objective must unswervingly be the removal of the Jews altogether.” In Mein Kampf (“My Struggle”; 1925–27), Hitler further developed the idea of the Jews as an evil race struggling for world domination. Nazi anti-Semitism was rooted in religious anti-Semitism and enhanced by political anti-Semitism. To this the Nazis added a further dimension: racial anti-Semitism. Nazi racial ideology characterized the Jews as Untermenschen (German: “subhumans”). The Nazis portrayed the Jews as a race and not as a religious group. Religious anti-Semitism could be resolved by conversion, political anti-Semitism by expulsion. Ultimately, the logic of Nazi racial anti-Semitism led to annihilation.
The conference at Wannsee gave impetus to the so-called second sweep of the Holocaust by the bullet in the east. Between April and July 1942 in Volhynia, 30,000 Jews were murdered in death pits with the help of dozens of newly formed Ukrainian Schutzmannschaft. Owing to good relations with the Ukrainian Hilfsverwaltung, these auxiliary battalions were deployed by the SS also in Russia Center, Russia South, and in Byelorussia; each with about 500 soldiers divided into three companies. They participated in the extermination of 150,000 Volhynian Jews alone, or 98 percent of the Jewish inhabitants of the entire region. In July 1942 the Completion of the Final Solution in the General Government territory which included Distrikt Galizien, was ordered personally by Himmler. He set the initial deadline for 31 December 1942.
Methods of mass murder evolved at local levels as well as being decreed from Nazi high command. Killing squads rounded up and shot entire Jewish communities. Over two days in Kiev, 33,771 Jews were shot. The murder of Jews rapidly escalated, in part because local Nazi leaders didn’t have enough room to place them in the ghettos. By the end of the year, plans to implement the systematic slaughter of Jews by using gas in mobile trucks and gas chambers were well underway.
In the first phase of the experiments, pairs of twins and persons with inherited anomalies were put at the disposal of Dr. Mengele and subjected to all imaginable specialist medical examinations. They were also photographed, plaster casts were made of their jaws and teeth, and they were toe- and fingerprinted. As soon as these examinations were finished, they were killed with lethal injections of phenol to the heart so that the next phase of the experimentation could begin: autopsies and the comparative analysis of their internal organs.
The Germans invaded the Soviet Union in 1941. Leaders of the SS and police and leaders of the German armed forces had concluded pre-invasion agreements. In accordance with these agreements, SS and police units—including Einsatzgruppen of the German Security Police and SD and battalions of the German Order Police—followed German troops into newly occupied Soviet territory. Acting as mobile killing units, they conducted shooting operations aimed at annihilating entire Jewish communities. By autumn 1941, the SS and police introduced mobile gas vans. These paneled trucks had exhaust pipes reconfigured to pump poisonous carbon monoxide gas into sealed spaces, killing those locked within. They were designed to complement ongoing shooting operations.
On January 20, 1942, several top officials of the German government met to officially coordinate the military and civilian administrative branches of the Nazi system to organize a system of mass murder of the Jews. This meeting, called the Wannsee Conference, "marked the beinning of the full-scale, comprehensive extermination operation [of the Jews] and laid the foundations for its organization, which started immediately after the conference ended" (Yahil, The Holocaust, p. 318).
In France Jews under Fascist Italian occupation in the southeast fared better than the Jews of Vichy France, where collaborationist French authorities and police provided essential support to the understaffed German forces. The Jews in those parts of France under direct German occupation fared the worst. Although allied with Germany, the Italians did not participate in the Holocaust until Germany occupied northern Italy after the overthrow of Fascist leader Benito Mussolini in 1943.
Lengyel described how Dr. Mengele would take all the correct medical precautions while delivering a baby at Auschwitz, yet only a half hour later, he would send the mother and baby to be gassed and burned in the crematorium. Lengyel herself was selected for the gas chamber, but managed to break away from the group of women who had been selected, before the truck arrived to take the prisoners to the crematorium.
By the late 1930s there was a desperate search for countries of refuge. Those who could obtain visas and qualify under stringent quotas emigrated to the United States. Many went to Palestine, where the small Jewish community was willing to receive refugees. Still others sought refuge in neighbouring European countries. Most countries, however, were unwilling to receive large numbers of refugees.
Prisoners on a death march from Dachau move towards the south along the Noerdliche Muenchner Street in Gruenwald, Germany, on April 29, 1945. Many thousands of prisoners were marched forcibly from outlying prison camps to camps deeper inside Germany as Allied forces closed in. Thousands died along the way, anyone unable to keep up was executed on the spot. Pictured, fourth from the right, is Dimitry Gorky who was born on August 19, 1920 in Blagoslovskoe, Russia to a family of peasant farmers. During World War II Dmitry was imprisoned in Dachau for 22 months. The reason for his imprisonment is not known. Photo released by the U.S. Holocaust Memorial Museum. #
After Otto was unable to find a publisher, the work was given to historian Jan Romein, who was so impressed that he wrote about the diary in a front-page article for the newspaper Het Parool in 1946. The resulting attention led to a publishing deal with Contact, and Het Achterhuis was released on June 25, 1947. An immediate best seller in the Netherlands, the work began to appear elsewhere. In 1952 the first American edition was published under the title Anne Frank: The Diary of a Young Girl; it included an introduction by Eleanor Roosevelt. The work was eventually translated into more than 65 languages, and it was later adapted for the stage and screen. All proceeds went to a foundation established in Anne’s honour. In 1995, 15 years after Otto’s death, a new English version of the Diary was published. It contained material that had been previously omitted. In an effort to extend the copyright date—which was to begin expiring in various European countries in 2016—Otto was added as a coauthor in 2015.
Peter Hayes (How Was It Possible? A Holocaust Reader, 2015): "The Holocaust, the Nazi attempt to eradicate the Jews of Europe, has come to be regarded as the emblematic event of Twentieth Century ... Hitler's ideology depicted the Jews as uniquely dangerous to Germany and therefore uniquely destined to disappear completely from the Reich and all territories subordinate to it. The threat posted by supposedly corrupting but generally powerless Sinti and Roma was far less, and therefore addressed inconsistently in the Nazi realm. Gay men were defined as a problem only if they were German or having sex with Germans or having sex with Germans and considered 'curable' in most cases. ... Germany's murderous intent toward the handicapped inhabitants of European mental institutions ... was more comprehensive ... but here, too, implementation was uneven and life-saving exceptions permitted, especially in Western Europe. Not only were some Slavs—Slovaks, Croats, Bulgarians, some Ukrainians—allotted a favored place in Hitler's New Order, but the fate of most of the other Slavs the Nazis derided as sub-humans ... consisted of enslavement and gradual attrition, not the prompt massacre meted out to the Jews after 1941."
The truth is, Alisande, these archaics are a little TOO simple; the vocabulary is too limited, and so, by consequence, descriptions suffer in the matter of variety; they run too much to level Saharas of fact, and not enough to picturesque detail; this throws about them a certain air of the monotonous; in fact the fights are all alike: a couple of people come together with great random -- random is a good word, and so is exegesis, for that matter, and so is holocaust, and de- falcation, and usufruct and a hundred others, but land
My friend and colleague, Rani Jaeger, one of the founders of Beit Tefila Israeli, tells a story of his family’s rescue through the generosity and courage of gentiles in Bulgaria. Unfortunately, there are far fewer stories like mine and Rani’s than there are of callousness, bigotry and racism during the Holocaust. We need to tell the story of the perpetrators and the victims. It is essential to remember, to keep the memory alive of those who suffered and perished. We cannot let this happen again, not to the Jewish people and not to any other people.
Most Holocaust historians define the Holocaust as the enactment, between 1941 and 1945, of the German state policy to exterminate the European Jews.[a] In Teaching the Holocaust (2015), Michael Gray, a specialist in Holocaust education, offers three definitions: (a) "the persecution and murder of Jews by the Nazis and their collaborators between 1933 and 1945", which views the events of Kristallnacht in Germany in 1938 as an early phase of the Holocaust; (b) "the systematic mass murder of the Jews by the Nazi regime and its collaborators between 1941 and 1945", which acknowledges the shift in German policy in 1941 toward the extermination of the Jewish people in Europe; and (c) "the persecution and murder of various groups by the Nazi regime and its collaborators between 1933 and 1945", which includes all the Nazis' victims. The third definition fails, Gray writes, to acknowledge that only the Jewish people were singled out for annihilation.
Using gas vans, Chełmno had its roots in the Aktion T4 euthanasia program. Majdanek began as a POW camp, but in August 1942 it had gas chambers installed. A few other camps are occasionally named as extermination camps, but there is no scholarly agreement on the additional camps; commonly mentioned are Mauthausen in Austria and Stutthof. There may also have been plans for camps at Mogilev and Lvov.
On the night of 9-10 November 1938, Nazi Propaganda Minister Dr Josef Goebbels organised the violent outburst known as Kristallnacht ('Crystal Night', the night of broken glass). While the police stood by, Nazi stormtroopers in civilian clothes burned down synagogues and broke into Jewish homes throughout Germany and Austria, terrorising and beating men, women and children. Ninety-one Jews were murdered and over 20,000 men were arrested and taken to concentration camps. Afterwards the Jewish community was fined one billion Reichsmarks to pay for the damage.
The passages which are included in the new version are not anything that the average 8-12 year old girl does not already know about her own body and the "birds and the bees", and are so few and short that they comprise a tiny percentage of the work itself. The romance between herself and Peter is very chaste and nothing untoward happens in the story. (Spoiler: they hold hands and a kiss a few times. that's it.) The passages that some see as inappropriate are not at all titillating, a medical textbook is more erotic. Coming from a mom's point of view, I would definitely allow my daughter to read the unedited book.
To those whose knowledge of the Holocaust consists, essentially, of the fact that Hitler killed the Jews, it often comes as a surprise to learn that, in the first seven and a half years of Nazi rule in Germany, he did no such killing: Jews were not deliberately murdered by the Nazi regime until the invasion of the Soviet Union in June 1941. In Karl Schleunes’ famous phrase, there was a “twisted road to Auschwitz,” with a gradual but by no means direct path to continental genocide. For the first few years of Nazi rule in Germany”Hitler came to power in January 1933 as chancellor and consolidated his rule after the death of President von Hindenburg a year later”Nazi policy aimed “merely” at the removal of Jews from positions of authority, especially in the state sector, with such removal constantly reinforced by the totalitarian regime’s propaganda and its police terrorism.
In response to this new “resettlement” policy, the first death camps were designed. Chelmno was the site of the first gassing of Jews, which occurred on December 8, 1941. The Nazi war machine had limited resources, including slave labor, much of it Jewish. Even so, the Nazis made a decision that the annihilation of the Jews of Europe was a more important achievement than the value of their labor. Similarly, the Nazis made a decision not to let the need for transport for the war effort interfere with the need for trucks and rail cars to carry the Jews to concentration camps and death centers. It was Adolf Eichmann who masterminded the logistics of the deportation of Jews. (1)
To come to the diary without having earlier assimilated Elie Wiesel’s “Night” and Primo Levi’s “The Drowned and the Saved” (to mention two witnesses only), or the columns of figures in the transport books, is to allow oneself to stew in an implausible and ugly innocence. The litany of blurbs—“a lasting testament to the indestructible nobility of the human spirit,” “an everlasting source of courage and inspiration”—is no more substantial than any other display of self-delusion. The success—the triumph—of Bergen-Belsen was precisely that it blotted out the possibility of courage, that it proved to be a lasting testament to the human spirit’s easy destructibility. “Hier ist kein Warum,” a guard at Auschwitz warned: here there is no “why,” neither question nor answer, only the dark of unreason. Anne Frank’s story, truthfully told, is unredeemed and unredeemable.
The SS used Sonderkommandos (Jewish slave laborers) during the gassing process to usher people in the undressing room and to clean up the gas chamber afterwards. One such survivor recalled the scene at Auschwitz: "There were all sorts of reactions from all sorts of people. There were disabled people. They would take out their war service cards showing that they had fought in the First World War with all kinds of distinctions and medals which they had from that time. They shouted, what's this? We fought for Germany. Now they're going to burn us, to kill us. This is impossible. We protest against such a thing. But everyone just laughed at them. Because they didn't take it seriously, these SS men. They laughed at the whole thing."
Otto Frank grew up with a social need to please his environment and not to offend it; that was the condition of entering the mainstream, a bargain German Jews negotiated with themselves. It was more dignified, and safer, to praise than to blame. Far better, then, in facing the larger postwar world that the diary had opened to him, to speak of goodness rather than destruction: so much of that larger world had participated in the urge to rage. (The diary notes how Dutch anti-Semitism, “to our great sorrow and dismay,” was increasing even as the Jews were being hauled away.) After the liberation of the camps, the heaps of emaciated corpses were accusation enough. Postwar sensibility hastened to migrate elsewhere, away from the cruel and the culpable. It was a tone and a mood that affected the diary’s reception; it was a mood and a tone that, with cautious yet crucial excisions, the diary itself could be made to support. And so the diarist’s dread came to be described as hope, her terror as courage, her prayers of despair as inspiring. And since the diary was now defined as a Holocaust document, the perception of the cataclysm itself was being subtly accommodated to expressions like “man’s inhumanity to man,” diluting and befogging specific historical events and their motives. “We must not flog the past,” Frank insisted in 1969. His concrete response to the past was the establishment, in 1957, of the Anne Frank Foundation and its offshoot the International Youth Center, situated in the Amsterdam house where the diary was composed, to foster “as many contacts as possible between young people of different nationalities, races and religions”—a civilized and tenderhearted goal that nevertheless washed away into do-gooder abstraction the explicit urge to rage that had devoured his daughter.
Doubleday, meanwhile, sensing complications ahead, had withdrawn as Frank’s theatrical agent, finding Levin’s presence—injected by Frank—too intrusive, too maverick, too independent and entrepreneurial: fixed, they believed, only on his own interest, which was to stick to his insistence on the superiority of his work over all potential contenders. Frank, too, had begun—kindly, politely, and with tireless assurances of his gratitude to Levin—to move closer to Doubleday’s cooler views, especially as urged by Barbara Zimmerman. She was twenty-four years old, the age Anne would have been, very intelligent and attentive. Adoring letters flowed back and forth between them, Frank addressing her as “little Barbara” and “dearest little one.” On one occasion he gave her an antique gold pin. About Levin, Zimmerman finally concluded that he was “impossible to deal with in any terms, officially, legally, morally, personally”—a “compulsive neurotic . . . destroying both himself and Anne’s play.” (There was, of course, no such entity as “Anne’s play.”)
When France fell to Nazi Germany, the mission to resist the Nazis became increasingly important. Following the establishment of the Vichy France regime during the occupation, Trocmé and his church members helped their town develop ways of resisting the dominant evil they faced. Together they established first one, and then a number of "safe houses" where Jewish and other refugees seeking to escape the Nazis could hide. Many refugees were helped to escape to Switzerland following an underground railroad network. Between 1940 and 1944 when World War II ended in Europe, it is estimated that about 3500 Jewish refugees including many children were saved by the small village of Le Chambon and the communities on the surrounding plateau because the people refused to give in to what they considered to be the illegitimate legal, military, and police power of the Nazis.
Hitler’s understanding of the role of the Jews in the world was not warped. His was, in fact, the traditional Jewish understanding. When the Jews accepted the Torah at Mt. Sinai, they became the chosen people whose role and responsibility was to bring a God-given code of morality to the world. They were to be “the light unto the nations” in the words of prophet Isaiah.
Entering conquered Soviet territories alongside the Wehrmacht (the German armed forces) were 3,000 men of the Einsatzgruppen (“Deployment Groups”), special mobile killing units. Their task was to murder Jews, Soviet commissars, and Roma in the areas conquered by the army. Alone or with the help of local police, native anti-Semitic populations, and accompanying Axis troops, the Einsatzgruppen would enter a town, round up their victims, herd them to the outskirts of the town, and shoot them. They killed Jews in family units. Just outside Kiev, Ukraine, in the ravine of Babi Yar, an Einsatzgruppe killed 33,771 Jews on September 28–29, 1941. In the Rumbula Forest outside the ghetto in Riga, Latvia, 25,000–28,000 Jews were shot on November 30 and December 8–9. Beginning in the summer of 1941, Einsatzgruppen murdered more than 70,000 Jews at Ponary, outside Vilna (now Vilnius) in Lithuania. They slaughtered 9,000 Jews, half of them children, at the Ninth Fort, adjacent to Kovno (now Kaunas), Lithuania, on October 28.
The next year, 1942, marked the beginning of mass murder on a scale unprecedented in all of human history. In January, fifteen top Nazis led by Reinhard Heydrich, second in command of the SS, convened the Wannsee Conference in Berlin to coordinate plans for the Final Solution. The Jews of Europe would now be rounded up and deported into occupied Poland where new extermination centers were being constructed at Belzec, Sobibor, Treblinka, and Auschwitz-Birkenau.
Browning describes the creation of the extermination camps, which were responsible for the largest number of deaths in the Final Solution, as bringing together three separate developments within the Third Reich: the concentration camps which had been established in Germany since 1933; an expansion of the gassing technology of the Nazi euthanasia programme to provide killing mechanism of greater efficiency and psychological detachment; and the creation of "factories of death" to be fed endless streams of victims by mass uprooting and deportation that utilized the experience and personnel from earlier population resettlement programmes—especially the HSSPF and Adolf Eichmann's RSHA for "Jewish affairs and evacuations".
Because of the special circumstances of its creation and publication — Miep Gies, one of the office employees who sustained the Franks by bringing supplies and news from the outside world, gathered Anne’s papers after the family’s arrest and gave them to Otto, the only Annex inhabitant to survive, when he returned from Auschwitz — many readers have treated the “Diary” as something akin to a saint’s relic: a text almost holy, not to be tampered with. Thus the outcry that greeted the discovery that Otto, in putting together a manuscript of the “Diary” for publication in 1947, had deleted whole passages in which Anne discussed in graphic terms her developing sexuality and her criticism of her mother, and the excitement when, in 1995, a “Definitive Edition” appeared, restoring much of the deleted material. Meanwhile, the enormously successful Broadway adaptation of the “Diary” has been severely rebuked for downplaying Anne’s Judaism and ironing out the nuances of her message. “Who owns Anne Frank?” Cynthia Ozick asked in an essay that berates the Broadway adapters for emphasizing the uplifting elements of Anne’s message — particularly the famous quotation, “In spite of everything, I still believe that people are really good at heart” — while insufficiently accounting for her hideous death, at age 15, in Bergen-Belsen.
After only four days of working in Hamburg, Ruth Elias was escorted by an SS man, in a private compartment on a passenger train, to the infirmary at Ravensbrück, the women's concentration camp near Berlin. From there, Ruth and Berta Reich, another prisoner who was nine months pregnant, were soon sent back to Auschwitz on another passenger train. Ruth gave birth to a baby girl at Auschwitz, but Dr. Mengele cruelly ordered her to bind her breasts and not to nurse her child because he wanted to see how long it would take for a baby to die without its mother's milk. Mercifully, a woman dentist named Maca Steinberg, who was a prisoner at Auschwitz, obtained some morphine and gave it to Ruth so that she could inject her baby and end its life, after Ruth told her that Dr. Mengele was due to arrive the next morning to take Ruth and her child to the gas chamber.
On 15 September 1935, the Reichstag passed the Reich Citizenship Law and the Law for the Protection of German Blood and German Honor, known as the Nuremberg Laws. The former said that only those of "German or kindred blood" could be citizens. Anyone with three or more Jewish grandparents was classified as a Jew. The second law said: "Marriages between Jews and subjects of the state of German or related blood are forbidden." Sexual relationships between them were also criminalized; Jews were not allowed to employ German women under the age of 45 in their homes. The laws referred to Jews but applied equally to the Roma and black Germans.
The little white house was located on the west side of the Birkenau camp, behind the Central Sauna which was completed in 1943, and near Krema IV. The Central Sauna got its name because this was the location of the iron chambers where the prisoners' clothing was disinfected with hot steam. The Central Sauna also contained a shower room with 50 shower heads.
Under the rule of Adolf Hitler, the persecution and segregation of Jews was implemented in stages. After the Nazi Party achieved power in Germany in 1933, its state-sponsored racism led to anti-Jewish legislation, economic boycotts, and the violence of the Kristallnacht ("Night of Broken Glass") pogroms, all of which aimed to systematically isolate Jews from society and drive them out of the country.
In February 2010, a 180-page volume of Mengele's diary was sold by Alexander Autographs at auction for an undisclosed sum to the grandson of a Holocaust survivor. The unidentified previous owner, who acquired the journals in Brazil, was reported to be close to the Mengele family. A Holocaust survivors' organization described the sale as "a cynical act of exploitation aimed at profiting from the writings of one of the most heinous Nazi criminals". Rabbi Marvin Hier of the Simon Wiesenthal Center was glad to see the diary fall into Jewish hands. "At a time when Ahmadinejad's Iran regularly denies the Holocaust and anti-Semitism and hatred of Jews is back in vogue, this acquisition is especially significant", he said. In 2011, a further 31 volumes of Mengele's diaries were sold—again amidst protests—by the same auction house to an undisclosed collector of World War II memorabilia for $245,000 USD.<|endoftext|>
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# Volume of a Cone
There is an important field of study in mathematics, called mensuration. It is the study of areas and volumes two-dimensional as well as three-dimensional shapes. The shapes that are studied under mensuration are - circle, rectangle, polygon, triangle, square, cube, cuboid, pyramid, cylinder, cone, sphere, hemisphere etc.
Here, we will study about cones in detail. A cone is a three-dimensional figure having a circular base and tapering to one single point. This point is known as the vertex of the cone. The axis of the cone is a straight line which joins the vertex of the cone and the center of the circular base.
Above diagram demonstrates a cone in which the radius of the base is shown by "r". Its height is referred as the axis of the cone and is denoted by "h". The length of lateral surface of the cone is shown by letter "l".
Volume of a cone is the measurement of the occupied units of a cone. In other words, the volume of a cone is the estimation of space occupied inside it. The volume of a Cone is represented by cubic units like cubic meter, cubic centimeter, cubic millimeter and so on. Volume of a cone is the number of units used to fill it.
Related Calculators Cone Volume Calculator Cone Calculator Area of a Cone Calculator Surface Area Right Cone
## Volume of a Cone Formula
The volume of a cone formula is given as below:
Volume of a right circular cone = (pir^2h)/(3)
Where, r = Radius of the circular base
h = Height of the cone
and value of $\pi$ = 3.14
Using this formula, we can find the volume of any cone whose radius and height is given.
In elementary geometry, cone is used as a right circular object.
## Volume of a Right Cone
Volume of the cube is number of cubic units that will completely fill a cone. If 'r' is the radius of the base and 'h' is the height (the distance between the apex and the center of the base) of the cone, then
Volume of a circular cone = $\frac{1}{3}$ (Base area $\times$ Height)
Where, Base Area = $\pi$ r$^2$
Volume of a right cone = $\frac{1}{3}$ $\pi$ r$^2$h
## Volume of a Truncated Cone
The volume of a cone is one-third the volume of a cylinder having the same base and equal height. The volume of the frustum of the cone is $\frac{1}{4}$ $\pi$ times the frustum of the pyramid.
### Volume of Frustum of Cone
If r and R are the radii of ends, h is height of a cone, the volume of frustum of a cone is given below:
Volume = $\frac{1}{3}$ $\pi$ h[Rr + R$^2$ + r$^2$] cubic units
Where r and R are the radii of ends and h is height of frustum of a cone.
## Derive Volume of a Cone
### Volume of a Cone Proof
Volume of cone (V) = $\pi$ $\int_0^h$ y dx
= $\pi$ $\int_0^h$($\frac{-r}{h}$ x + r)$^2$ dx
= $\pi$ $\int_0^h$($\frac{r^2}{h^2}.x^2 + r^2 - \frac{2r^2}{h}. x)^2$ dx
= $\pi$ [$\frac{r^2}{h^2} \times \frac{x^3}{3} + r^2 x - \frac{r^2}{h} \times x^2]_0^h$
= $\pi$ [$\frac{r^2}{h^2} \times \frac{h^3}{3} - \frac{r^2}{h} \times h^2 + r^2 h]$
= $\frac{1}{3}$ $\pi r^2$ h
Hence Proved.
## Find the Volume of a Cone
Below are steps for finding the volume of a cone: If radius and height of the cone is given.
Step 1: Square the radius and multiply it with height.
Step 2: Multiply the result of step 1 with $\frac{1}{3}$ $\pi$ (Use $\pi$ = 3.14).
## Volume of a Cone Example
Given below are some solved examples on volume of a cone
Example 1: Find the volume of cone, whose radius is 6 cm and height is 8 cm.
Solution:
The volume formula of cone is,
V = (pir^2h)/3
V = ((3.14) * (6^2) * (8))/3
= ((3.14) * (36) * (8))/3
= ((113.04) * (8))/3
= 904.32 / 3
= 301.44
The volume of the cone = 301.44 cm3.
Example 2: Find the volume of cone, whose radius is 8 cm and height, is 5 cm.
Solution:
The volume of a cone formula is,
V = (pir^2h)/3
V = ((3.14) * (8^2) * (5))/3
= ((3.14)*(64)*(5))/3
= ((200.96)*(5))/3
= (1004.8)/3
= 334.93
Volume of the cone = 334.93 cm3.
Example 3: The volume of solid cone is 200 m$^3$ and diameter of its base is 20 m. Find the height of the cone.
Solution: Let the height of the cone be h m.
Radius of base = $\frac{1}{2}$ $\times$ 20 = 10
=> Radius of base (r) = 10 m
Volume of cone = $\frac{1}{3}$ $\pi$ r$^2$ h
200 = $\frac{1}{3}$ $\pi$ (10)$^2$ h
200 = $\frac{1}{3}$ $\pi$ $\times$ 100 $\times$
h
2 = 1.047 $\times$ h
h = $\frac{2}{1.047}$ = 1.91
Hence the height of the cone is 1.91 m.
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# How would I find the area of this hexagon?
Nov 20, 2015
$30 {\text{cm}}^{2}$
#### Explanation:
Consider the shapes the hexagon is cut into. There are two central squares and four congruent triangles.
Each squares area can be found through the formula $A = {s}^{2}$. Both have sides $3$ so each square's area is $9$. Since there are two squares, their total area is $18$.
As for the triangles, $A = \frac{1}{2} b h$.
Each triangle has $b = 3$ and $h = 2$, so one of the triangle's areas is $\frac{1}{2} \left(3\right) \left(2\right) = 3$. There are four triangles, so their total area is $12$.
We can add the total area of the squares ($18$) and the triangles ($12$) to find that the hexagon's total area is $30 {\text{cm}}^{2}$.<|endoftext|>
| 4.65625 |
504 |
# 1011 A Toast to My Readers
Thank you, dear readers, for stopping by today and/or any other time you have visited! My blog had the best year ever in 2017, and it is all because of you. I sincerely hope that I have been of service to you.
Here’s to you, dear friend. May lifting each other help us both to climb even higher this coming year!
May you be able to find the factors in this puzzle and, more importantly, the factors in life that will bring you happiness and success!
Print the puzzles or type the solution in this excel file: 10-factors-1002-1011
Let me tell you some things about the number 1011 that you probably didn’t know before:
The only nonzero digits in 1011 are three 1’s so 1011 is divisible by 3, and 1011 is included in this interesting pattern:
1011² = 1022121; Notice that the digits in bold are 1011 and all the other digits are 2’s. Thank you Stetson.edu for that fun fact.
1011 looks interesting in a few other bases:
It’s 33303 in BASE 4 because 3(4⁴ + 4³ + 4² + 4⁰) = 3(256 + 64 + 16 + 1) = 3(337) = 1011,
323 in BASE 18 because 3(18²) + 2(18) + 3(1) = 1011
1011 is the hypotenuse of a Pythagorean triple:
525-864-1011 which is 3 times (175-288-337)
• 1011 is a composite number.
• Prime factorization: 1011 = 3 × 337
• The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1011 has exactly 4 factors.
• Factors of 1011: 1, 3, 337, 1011
• Factor pairs: 1011 = 1 × 1011 or 3 × 337
• 1011 has no square factors that allow its square root to be simplified. √1011 ≈ 31.7962<|endoftext|>
| 4.375 |
375 |
### Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.
Step by Step Explanation:
1. Let $AD$ be the median to the third side of the triangle $ABC$.
We need to prove that $AB + AC > 2 AD.$
2. We know that the median from a vertex to the opposite side of a triangle bisects the opposite side.
Thus, we have $BD = DC$.
3. Let's extend $AD$ to $E$ such that $AD = DE$ and join the point $E$ to the point $C.$
4. In $\triangle ADB$ and $\triangle EDC,$ we have \begin{aligned} & \angle ADB = \angle EDC && [\text{Vertically opposite angles}] \\ & AD = DE && [\text{By construction}] \\ & BD = DC && [\text{AD is the median.}] \\ \therefore \space & \triangle ADB \cong \triangle EDC && [\text{By SAS criterion}] \end{aligned} As corresponding parts of congruent triangles are equal, we have$$AB = EC \space \space \ldots (1)$$
5. We know that the sum of any two sides of a triangle is greater than the third side.
So, in $\triangle AEC,$ we have \begin{aligned} & AC + EC > AE \\ \implies & AC + AB > AE && [\text{From (1)}] \\ \implies & AC + AB > 2AD && [\because \space \text{AE = 2AD}] \end{aligned}
6. Thus, the sum of any two sides of a triangle is greater than twice the median drawn to the third side.<|endoftext|>
| 4.6875 |
920 |
The Ancient Roman Calendar
The Roman calendar is essentially the Julian calendar, but using the Romans' different way of expressing dates. They did not count forwards from the first day of the month, but instead counted downwards towards three marker days in each month. These days were the Kalends, the Nones and the Ides. The Kalends were on the first of the month. The Nones were on the 5th of the month for all months except March, May, July and October, when they were on the 7th. The Ides were always 8 days later than the Nones, i.e. on the 13th or 15th of the month. The intervening days were counted backwards, inclusively, to the next of these main days. So, for example, the 15th of March was the Ides of March, or Idus Martiae. The 14th March was Pridie Idus Martias, or the "day before the ides of March". The 13th was the 3rd day before the Ides, or ante diem III Idus Martias, showing the inclusive nature of Roman counting, i.e. three days before the 15th was the 13th inclusively (13 - 14 - 15).
This system was used for all the three main days of the month, and thus produced something that seems odd to us now. The dates following the Ides of each month refer to the Kalends of the following month, e.g. the 18th December was ante diem XV Kalendas Ianuarias, or "15 days before the Kalends of January", and the year number was incremented accordingly.
These dates were usually abbreviated, e.g.
Idus Martiae = Id. Mart.
Pridie Idus Martias = Prid. Id. Mart.
ante diem III Idus Martias = a.d. III Id. Mart.
ante diem XV Kalendas Ianuarias = A.D. XV Kal. Ian.
These abbreviations are used for the Roman date on the main calendar page.
Nundinal letters and market days
In the Roman calendar, the nundinal letter was like an equivalent of the modern days of the week, except with eight days instead of seven. Each day was lettered successively from A to H. Every eighth day (ninth day using the Roman inclusive counting system, hence the word nundinal) was a market day, which was a special day in Roman times, so wthhin a particular year the market days all had the same nundinal letter, and that letter was designated as the nunidanl letter for that year. Each date had the same nundinal letter from one year to the next, because the sequence was reset to 'A' at the beginning of the year. This meant that, applying the strict eight-day rule for market days, the letter for the market day changed from one year to the next. So, for instance, in 2007 the market day letter is 'H', which means that the last market day of the year would have been 26th December. The next market day would be 3rd January, so the market day letter for 2008 would be 'C'.
In leap years, the 6th day before the Kalends of March (a.d. VI Kal. Mart.) was effectivley doubled, hence why a leap year is often called a bissextile year. Effectively, the 24th of February in a leap year is inserted, thus pushing the remaining days of that month back by one day. The 25th, then, has the same nundinal letter as the 24th, and the remaining dates have their correct nundinal letters. So, for instance, Prid. Mart. always has nunindal letter 'C', even though in our modern calendar it is 28th February in a normal year, and 29th February in a leap year. To the Romans they were the same date. In the calendar, the second of these days is labelled bis, i.e. a.d. (bis) VI Kal. Mart.
The epoch for the year that has been used here is 753 B.C.E., which is the most favoured date for the foundation of the city of Rome, thus 2007 C.E. is the year 2760 A.U.C. (ab urbe condita - "since the foundation of the city").<|endoftext|>
| 3.984375 |
5,083 |
# 1.6 Decimal fractions
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
## Overview
• Decimal Fractions
• Adding and Subtracting Decimal Fractions
• Multiplying Decimal Fractions
• Dividing Decimal Fractions
• Converting Decimal Fractions to Fractions
• Converting Fractions to Decimal Fractions
## Decimal fractions
Fractions are one way we can represent parts of whole numbers. Decimal fractions are another way of representing parts of whole numbers.
## Decimal fractions
A decimal fraction is a fraction in which the denominator is a power of 10.
A decimal fraction uses a decimal point to separate whole parts and fractional parts. Whole parts are written to the left of the decimal point and fractional parts are written to the right of the decimal point. Just as each digit in a whole number has a particular value, so do the digits in decimal positions.
## Sample set a
The following numbers are decimal fractions.
$\begin{array}{l}57.9\\ \text{The\hspace{0.17em}9\hspace{0.17em}is\hspace{0.17em}in\hspace{0.17em}the}\text{\hspace{0.17em}}tenths\text{\hspace{0.17em}}\text{position}.\text{\hspace{0.17em}}57.9=57\frac{9}{10}.\end{array}$
$\begin{array}{l}6.8014\text{\hspace{0.17em}}\\ \text{The\hspace{0.17em}8\hspace{0.17em}is\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}}tenths\text{\hspace{0.17em}position}\text{.\hspace{0.17em}}\\ \text{The\hspace{0.17em}0\hspace{0.17em}is\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}}hundredths\text{\hspace{0.17em}position}\text{.\hspace{0.17em}}\\ \text{The\hspace{0.17em}1\hspace{0.17em}is\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}}thousandths\text{\hspace{0.17em}position}\text{.\hspace{0.17em}}\\ \text{The\hspace{0.17em}4\hspace{0.17em}is\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}ten\hspace{0.17em}}thousandths\text{\hspace{0.17em}position}\text{.\hspace{0.17em}}\\ 6.8014=6\frac{8014}{10000}.\end{array}$
## Adding and subtracting decimal fractions
To add or subtract decimal fractions,
1. Align the numbers vertically so that the decimal points line up under each other and corresponding decimal positions are in the same column. Add zeros if necessary.
2. Add or subtract the numbers as if they were whole numbers.
3. Place a decimal point in the resulting sum or difference directly under the other decimal points.
## Sample set b
Find each sum or difference.
\begin{array}{l}9.183+2.140\\ \begin{array}{rrr}\text{\hspace{0.17em}}↓\hfill & \hfill & \text{The\hspace{0.17em}decimal\hspace{0.17em}points\hspace{0.17em}are\hspace{0.17em}aligned\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}column}\text{.\hspace{0.17em}}\hfill \\ \hfill \text{9}\text{.183}& \hfill & \hfill \\ \hfill \text{+}\underset{¯}{\text{\hspace{0.17em}2}\text{.140}}& \hfill & \hfill \\ \hfill \text{11}\text{.323}& \hfill & \hfill \end{array}\end{array}
\begin{array}{l}841.0056\text{\hspace{0.17em}}+\text{\hspace{0.17em}}47.016\text{\hspace{0.17em}}+\text{\hspace{0.17em}}19.058\text{\hspace{0.17em}}\\ \begin{array}{rrr}\hfill ↓& \hfill & \text{The\hspace{0.17em}decimal\hspace{0.17em}points\hspace{0.17em}are\hspace{0.17em}aligned\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}column}\text{.\hspace{0.17em}}\hfill \\ \hfill 841.0056& \hfill & \hfill \\ \hfill 47.016& \hfill & \text{Place\hspace{0.17em}a\hspace{0.17em}0\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}thousandths\hspace{0.17em}position}\text{.}\hfill \\ \hfill +\text{\hspace{0.17em}}\underset{¯}{19.058}& \hfill & \text{Place\hspace{0.17em}a\hspace{0.17em}0\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}thousandths\hspace{0.17em}position}\text{.\hspace{0.17em}}\hfill \\ \hfill ↓& \hfill & \text{The\hspace{0.17em}decimal\hspace{0.17em}points\hspace{0.17em}are\hspace{0.17em}aligned\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}column}\text{.\hspace{0.17em}}\hfill \\ \hfill 841.0056& \hfill & \hfill \\ \hfill 47.0160& \hfill & \hfill \\ \hfill +\text{\hspace{0.17em}}\underset{¯}{19.0580}& \hfill & \hfill \\ \hfill 907.0796& \hfill & \hfill \end{array}\end{array}
\begin{array}{l}16.01\text{\hspace{0.17em}}-\text{\hspace{0.17em}}7.053\\ \begin{array}{rrr}\hfill ↓& \hfill & \text{The\hspace{0.17em}decimal\hspace{0.17em}points\hspace{0.17em}are\hspace{0.17em}aligned\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}column}\text{.\hspace{0.17em}}\hfill \\ \hfill 16.01& \hfill & \text{Place\hspace{0.17em}a\hspace{0.17em}0\hspace{0.17em}into\hspace{0.17em}the\hspace{0.17em}thousandths\hspace{0.17em}position}\text{.\hspace{0.17em}}\hfill \\ \hfill -\text{\hspace{0.17em}}\underset{¯}{7.053}& \hfill & \hfill \\ \hfill ↓& \hfill & \text{The\hspace{0.17em}decimal\hspace{0.17em}points\hspace{0.17em}are\hspace{0.17em}aligned\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}column}\text{.\hspace{0.17em}}\hfill \\ \hfill 16.010& \hfill & \hfill \\ \hfill -\text{\hspace{0.17em}}\underset{¯}{7.053}& \hfill & \hfill \\ \hfill 8.957& \hfill & \hfill \end{array}\end{array}
## Multiplying decimal fractions
To multiply decimals,
1. Multiply tbe numbers as if they were whole numbers.
2. Find the sum of the number of decimal places in the factors.
3. The number of decimal places in the product is the sum found in step 2.
## Sample set c
Find the following products.
$6.5×4.3$
$6.5×4.3=27.95$
$23.4×1.96$
$23.4×1.96=45.864$
## Dividing decimal fractions
To divide a decimal by a nonzero decimal,
1. Convert the divisor to a whole number by moving the decimal point to the position immediately to the right of the divisor’s last digit.
2. Move the decimal point of the dividend to the right the same number of digits it was moved in the divisor.
3. Set the decimal point in the quotient by placing a decimal point directly above the decimal point in the dividend.
4. Divide as usual.
## Sample set d
Find the following quotients.
$32.66÷7.1$
$\begin{array}{l}32.66÷7.1=4.6\\ \begin{array}{lll}Check:\hfill & \hfill & 32.66÷7.1=4.6\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}4.6×7.1=32.66\hfill \\ \hfill \text{\hspace{0.17em}}4.6& \hfill & \hfill \\ \hfill \underset{¯}{\text{\hspace{0.17em}}7.1}& \hfill & \hfill \\ \hfill \text{\hspace{0.17em}}4.6& \hfill & \hfill \\ \underset{¯}{322\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\hfill & \hfill & \hfill \\ 32.66\hfill & \hfill & \text{True}\hfill \end{array}\end{array}$
Check by multiplying $2.1$ and $0.513.$ This will show that we have obtained the correct result.
$12÷0.00032$
## Converting decimal fractions to fractions
We can convert a decimal fraction to a fraction by reading it and then writing the phrase we have just read. As we read the decimal fraction, we note the place value farthest to the right. We may have to reduce the fraction.
## Sample set e
Convert each decimal fraction to a fraction.
$\begin{array}{l}0.6\\ 0.\underset{¯}{6}\to \text{tenths\hspace{0.17em}position}\\ \begin{array}{lll}\text{Reading:}\hfill & \hfill & \text{six\hspace{0.17em}tenths}\to \frac{6}{10}\hfill \\ \text{Reduce:}\hfill & \hfill & 0.6=\frac{6}{10}=\frac{3}{5}\hfill \end{array}\end{array}$
$\begin{array}{l}21.903\\ 21.90\underset{¯}{3}\to \text{thousandths\hspace{0.17em}position}\\ \begin{array}{ccc}\text{Reading:}& & \text{twenty-one\hspace{0.17em}and\hspace{0.17em}nine\hspace{0.17em}hundred\hspace{0.17em}three\hspace{0.17em}thousandths}\to 21\frac{903}{1000}\end{array}\end{array}$
## Sample set f
Convert the following fractions to decimals. If the division is nonterminating, round to 2 decimal places.
$\frac{3}{4}$
$\frac{3}{4}=0.75$
$\frac{1}{5}$
$\frac{1}{5}=0.2$
$\frac{5}{6}$
$\begin{array}{llll}\frac{5}{6}\hfill & =\hfill & 0.833...\hfill & \begin{array}{l}\\ \text{We\hspace{0.17em}are\hspace{0.17em}to\hspace{0.17em}round\hspace{0.17em}to\hspace{0.17em}2\hspace{0.17em}decimal\hspace{0.17em}places}.\text{\hspace{0.17em}}\end{array}\hfill \\ \frac{5}{6}\hfill & =\hfill & 0.83\text{\hspace{0.17em}to\hspace{0.17em}2\hspace{0.17em}decimal\hspace{0.17em}places}\text{.}\hfill & \hfill \end{array}$
$\begin{array}{l}5\frac{1}{8}\\ \text{Note\hspace{0.17em}that\hspace{0.17em}}5\frac{1}{8}=5+\frac{1}{8}.\end{array}$
$\begin{array}{l}\frac{1}{8}=.125\\ \text{Thus,\hspace{0.17em}}5\frac{1}{8}=5+\frac{1}{8}=5+.125=5.125.\end{array}$
$0.16\frac{1}{4}$
This is a complex decimal. The “6” is in the hundredths position. The number $0.16\frac{1}{4}$ is read as “sixteen and one-fourth hundredths.”
$\begin{array}{lll}0.16\frac{1}{4}=\frac{16\frac{1}{4}}{100}=\frac{\frac{16·4+1}{4}}{100}\hfill & =\hfill & \frac{\frac{65}{4}}{\frac{100}{1}}\hfill \\ \hfill & =\hfill & \frac{\stackrel{13}{\overline{)65}}}{4}·\frac{1}{\underset{20}{\overline{)100}}}=\frac{13×1}{4×20}=\frac{13}{80}\hfill \end{array}$
Now, convert $\frac{13}{80}$ to a decimal.
$0.16\frac{1}{4}=0.1625.$
## Exercises
For the following problems, perform each indicated operation.
$1.84+7.11$
$8.95$
$15.015-6.527$
$4.904-2.67$
$2.234$
$156.33-24.095$
$.0012+1.53+5.1$
$6.6312$
$44.98+22.8-12.76$
$5.0004-3.00004+1.6837$
$3.68406$
$1.11+12.1212-13.131313$
$4.26\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3.2$
$13.632$
$2.97\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3.15$
$23.05\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1.1$
$25.355$
$5.009\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2.106$
$0.1\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3.24$
$0.324$
$100\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12.008$
$1000\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12.008$
$12,008$
$10,000\text{\hspace{0.17em}}·\text{\hspace{0.17em}}12.008$
$75.642\text{\hspace{0.17em}}÷\text{\hspace{0.17em}}18.01$
$4.2$
$51.811\text{\hspace{0.17em}}÷\text{\hspace{0.17em}}1.97$
$0.0000448\text{\hspace{0.17em}}÷\text{\hspace{0.17em}}0.014$
$0.0032$
$0.129516\text{\hspace{0.17em}}÷\text{\hspace{0.17em}}1004$
For the following problems, convert each decimal fraction to a fraction.
$0.06$
$\frac{3}{50}$
$0.115$
$3.7$
$3\frac{7}{10}$
$48.1162$
$712.00004$
$712\frac{1}{25000}$
For the following problems, convert each fraction to a decimal fraction. If the decimal form is nonterminating,round to 3 decimal places.
$\frac{5}{8}$
$\frac{9}{20}$
$0.45$
$15\text{\hspace{0.17em}}÷\text{\hspace{0.17em}}22$
$\frac{7}{11}$
$0.636$
$\frac{2}{9}$
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Please keep in mind that it's not allowed to promote any social groups (whatsapp, facebook, etc...), exchange phone numbers, email addresses or ask for personal information on QuizOver's platform.<|endoftext|>
| 4.875 |
458 |
Superposition is the technique for analyzing circuits that have multiple sources: You simply determine the output with the inputs activated one at a time, and then add together all the results. (By the Rules of Superposition you may sum voltages or currents but you must not sum power.)
If half of the input power is forwarded to the load, the load voltage will be smaller than the input voltage by 0.707. (In an all-75W system, the power change is the square of the voltage change. The square root of 0.5 is 0.707.) Superposition says that if both inputs supply the same power, the voltage at the load will be 0.707+0.707=1.414. 1.414 squared is 2, which means there is twice the power at the load as was supplied at each input.
But you ask, “How can the power at the load equal the total input power if some of the power was reflected backwards?” The answer is that superposition applies to currents as well as voltages. The phase of the reflected currents is such that they subtract instead of add. The reflected currents will cancel each other out completely.
The doubling of the output power is equivalent to a 3 dB increase in the signal. If the combiner is 90% efficient then a 2.5 dB gain is seen. Note the dichotomy:
· If the antennas point in different directions, there is a 3.5 dB loss at the combiner.
· If the antennas point in the same direction, there is a 2.5 dB gain at the combiner.
This is a 6 dB swing. 3 dB of this is just the adding of the second antenna, but the other 3 dB is from the combiner becoming a much more effective device.
Avoiding a large loss at the combiner requires the complete cancellation of the reflected currents. This requires that both input signals be identical in amplitude and phase. This generally demands that the two antennas be identical, and that the two feedline delays be equal. Furthermore, both antennas must be in fields of equal strength, pointing the same direction, and must be positioned for proper phase. Making all this true for non-identical antennas is not practical.<|endoftext|>
| 4.3125 |
886 |
Curriculum - Geography
Please take a look at the Geography & History Curriculum Rationale
Key Stage 1
Pupils should develop knowledge about the world, the United Kingdom and their locality. They should understand basic subject-specific vocabulary relating to human and physical geography and begin to use geographical skills, including first-hand observation, to enhance their locational awareness.
Pupils should be taught to:
Locational knowledge• name and locate the world’s seven continents and five oceans
• name, locate and identify characteristics of the four countries and capital cities of the United Kingdom and its surrounding seas
Place knowledge• understand geographical similarities and differences through studying the human and physical geography of a small area of the United Kingdom, and of a small area in a contrasting non-European country
Human and physical geography• identify seasonal and daily weather patterns in the United Kingdom and the location of hot and cold areas of the world in relation to the Equator and the North and South Poles
• use basic geographical vocabulary to refer to:
• key physical features, including: beach, cliff, coast, forest, hill, mountain, sea, ocean, river, soil, valley, vegetation, season and weather
• key human features, including: city, town, village, factory, farm, house, office, port, harbour and shop
Geographical skills and fieldwork• use world maps, atlases and globes to identify the United Kingdom and its countries, as well as the countries, continents and oceans studied at this key stage
• use simple compass directions (North, South, East and West) and locational and directional language [for example, near and far; left and right], to describe the location of features and routes on a map
• use aerial photographs and plan perspectives to recognise landmarks and basic human and physical features; devise a simple map; and use and construct basic symbols in a key
• use simple fieldwork and observational skills to study the geography of their school and its grounds and the key human and physical features of its surrounding environment.
Key Stage 2• Pupils should extend their knowledge and understanding beyond the local area to include the United Kingdom and Europe, North and South America. This will include the location and characteristics of a range of the world’s most significant human and physical features. They should develop their use of geographical knowledge
• knowledge, understanding and skills to enhance their locational and place knowledge.
• Pupils should be taught to:
Locational knowledge• locate the world’s countries, using maps to focus on Europe (including the location of Russia) and North and South America, concentrating on their environmental regions, key physical and human characteristics, countries, and major cities
• name and locate counties and cities of the United Kingdom, geographical regions and their identifying human and physical characteristics, key topographical features (including hills, mountains, coasts and rivers), and land-use patterns; and understand how some of these aspects have changed over time
• identify the position and significance of latitude, longitude, Equator, Northern Hemisphere, Southern Hemisphere, the Tropics of Cancer and Capricorn, Arctic and Antarctic Circle, the Prime/Greenwich Meridian and time zones (including day and night)
Place knowledge• understand geographical similarities and differences through the study of human and physical geography of a region of the United Kingdom, a region in a European country, and a region within North or South America
Human and physical geography• physical geography, including: climate zones, biomes and vegetation belts, rivers, mountains, volcanoes and earthquakes, and the water cycle
• human geography, including: types of settlement and land use, economic activity including trade links, and the distribution of natural resources including energy, food, minerals and water
Geographical skills and fieldwork• use maps, atlases, globes and digital/computer mapping to locate countries and describe features studied
• use the eight points of a compass, four and six-figure grid references, symbols and key (including the use of Ordnance Survey maps) to build their knowledge of the United Kingdom and the wider world
• use fieldwork to observe, measure, record and present the human and physical features in the local area using a range of methods, including sketch maps, plans and graphs, and digital technologies.<|endoftext|>
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Bright Beginnings is an early childhood curriculum, based in part on the High/Scope® and The Creative Curriculum® models, with additional emphasis on building early language and literacy skills. The curriculum consists of nine thematic units designed to enhance cognitive, social, emotional, and physical development. Each unit includes concept maps, literacy lessons, center activities, and home activities. Parent involvement is also a key component of the program.
Bright Beginnings consists of nine curriculum units: language and literacy, mathematics, social and personal development, healthful living, scientific thinking, social studies, creative arts, physical development, and technology.
Active exploration and interaction with other children, adults, and materials are key components of the Bright Beginnings curriculum. As children participate in the curricular activities, there are materials for teachers to monitor students and assess their progress. The Bright Beginnings curriculum also includes a Family–School Connection link designed to engage parents.
Preschool Curriculum Evaluation Research (PCER) Consortium. (2008). Bright Beginnings and Creative Curriculum: Vanderbilt University. In Effects of preschool curriculum programs on school readiness (pp. 41–54). Washington, DC: National Center for Education Research, Institute of Education Sciences, U.S. Department of Education.
What Works Clearinghouse
Information on this entry was adapted from US Department of Education. (2013). Bright Beginnings. WWC Intervention Report. Retrieved from https://ies.ed.gov/ncee/wwc/Docs/InterventionReports/wwc_brightbeginnings_030513.pdf<|endoftext|>
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# Combination of Resistance
In electric circuits, the resistors are connected in two different ways either in series or parallel combination.
Series combination
If different resistances are connected end to end in a line,then the combination is called series combination.In series combination,same amount of current will pass through all the resistance.
Let three resistance having resistance R1, R2,and R3 are connected in a series with a battery having terminal p.d ‘V’ as shown in figure above. Let V1, Vand V3 be the potential difference across R1, R2,and R3 respectively. Let ‘I’ be the current and ‘R’ be the resistance of the circuit.
Now,
V=V1 +V+ V3
IR=I R+ IR2 + IR3
R= R1 + R+ R3
Thus,in series combination,the net resistance will be equal to sum of individual resistance.
In series combination,the net resistance will be greater than the greatest resistance also.
i.e., Rs = R1 + R2…….. + Rn
If R1 = R2 = Rn
Then, Rs = R + R + R + R….. = nR
Parallel Combination
If current divides at a point, and meet at another point then the combination between two point is called parallel of combination of electricity. In parallel combination potential difference across all the resistance will be equal.
Let 3 resistors having resistance R1 + R2 + R3 are connected in parallel combination with a battery having terminal potential difference ‘V’ as shown in figure. Let I+ I2 and I3 be the current through R1, R2 and R3 respectively. Let ‘I’ be the total current flowing through the circuit and ‘R’ be the net resistance of the circuit. Now,
$$I= I_1 + I_2 + I_3$$
$$\text{or, V} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$$
$$\text{or, } \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$
$$\text{or, } \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_n}$$
Thus the reciprocal of net resistance of the combination is equal to the sum of individual resistance. In this combination, net resistance becomes less than the least resistance of the circuit.
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Search IntMath
Close
4. Algebraic Solutions of Linear Systems
a. Solving Systems of Equations Using Substitution
This method involves subsituting y (or x if it is easier) from one equation into the other equation. This simplifies the second equation and we can solve it easily.
Example 1
Solve the system
x + y = 3 [1]
3x − 2y = 14 [2]
using substitution.
(The numbers in square brackets, [1] and [2], are used to name each equation. This makes it easier when referring to them in the solution.)
From line [1], we subtract x from both sides and get y = −x + 3.
We substitute this in the place of y in line [2]:
3x − 2(−x + 3) = 14
This gives us: 3x + 2x − 6 = 14
Therefore x = 4.
Now, using line [1] we get y = −1.
If we have the right numbers, they should also work in the other equation.
Checking in line [2]:
3(4) − 2(−1) = 14 [OK]
So our solution (4, −1) is correct.
b. Solving Systems of Equations Using Elimination
Our aim here is to eliminate one of the variables. It doesn't matter which one - we usually just do the easiest one.
Example 2
Solve the system using elimination.
3x + y = 10 [1]
x − 2y = 1 [2]
If we subtract one row from the other row, we don't eliminate anything. However, if we multiply one of the rows, we can eliminate one of the variables by adding the rows.
Row [1] × 2 gives us
6x + 2y = 20 [3]
x − 2y = 1 [2] (no change)
If we add lines [3] and [2], we eliminate y.
7x = 21
So x = 3 and using line [1], y = 1.
Check in line [2]: 3 − 2(1) = 1 [OK]
So our solution is (3, 1).
In a later chapter we will see how to solve systems of equations using determinants (okay for paper-based solutions) and matrices (very powerful and the best way to do it on computers).<|endoftext|>
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# 8009 in Words
We can write 8009 in words as Eight thousand and nine. If you purchased a wheel bag that costs Rs. 8009, you can say, “I purchased a wheel bag worth Eight thousand and nine rupees”. We can easily derive the word form of the number 8009 using a place value chart. In this article, you will learn how to write the cardinal number 8009 in words and some facts about 8009.
8009 in words Eight thousand and nine Eight thousand and nine in Numbers 8009
## 8009 in English Words
We generally write numbers in words using the English alphabet. So, we spell 8009 in English as “Eight thousand and nine”.
## How to Write 8009 in Words?
Let’s learn how to write the number 8009 in words using a place value chart. This can be done as follows.
Thousands Hundreds Tens Ones 8 0 0 9
Here, ones = 9, tens = 0, hundreds = 0, thousands = 8
By expanding these numbers with respect to the place values, we get;
8 × Thousand + 0 × Hundred + 0 × Ten + 9 × One
= 8 × 1000 + 0 × 100 + 0 × 10 + 9 × 1
= 8000 + 9
= Eight thousand + Nine
= Eight thousand and nine
Therefore, 8009 in words = Eight thousand and nine.
Also, check:place value
As we know, 8009 is a natural number that precedes 8010 and succeeds 8008.
8009 in words – Eight thousand and nine
Is 8009 an even number? – No
Is 8009 an odd number? – Yes
Is 8009 a perfect square number? – No
Is 8009 a perfect cube number? – No
Is 8009 a prime number? – Yes
Is 8009 a composite number? – No
## Frequently Asked Questions on 8009 in Words
Q1
### What is the number name of 8009?
The number name of 8009 is given by Eight thousand and nine.
Q2
### How to write Rs. 8009 in words on a cheque?
On a cheque, we can write an amount of Rs. 8009 in words as “Eight thousand and nine rupees only”.
Q3
### Express the decimal number 0.8009 in words.
We can express the decimal number 0.8009 in words as Eight thousand nine ten-thousandths or Zero point eight zero zero nine.<|endoftext|>
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Proclamation of Indonesian Independence
The Proclamation of Indonesian Independence (Indonesian: Proklamasi Kemerdekaan Indonesia, or simply Proklamasi) was read at 10.00 a.m. on Friday, August 17, 1945. The declaration marked the start of the diplomatic and armed-resistance of the Indonesian National Revolution, fighting against the forces of the Netherlands until the latter officially acknowledged Indonesia’s independence in 1949. In 2005, the Netherlands declared that they had decided to accept 17 August 1945 as Indonesia’s independence date
The draft was prepared only a few hours earlier, on the night of August 16, by Sukarno, Hatta, and Soebardjo, at Rear-Admiral Maeda (Minoru) Tadashi’s house, Miyako-Doori 1, Jakarta (now the “Museum of the Declaration of Independence“, JL. Imam Bonjol I, Jakarta). The original Indonesian Declaration of Independence was typed by Sayuti Melik. Maeda himself was sleeping in his room upstairs. He was agreeable to the idea of Indonesia‘s independence, and had lent his house for the drafting of the declaration. Marshal Terauchi, the highest-ranking Japanese leader in South East Asia and son of Prime Minister Terauchi Masatake, was however against Indonesia’s independence, scheduled for August 24.
While the formal preparation of the declaration, and the official independence itself for that matter, had been carefully planned a few months earlier, the actual declaration date was brought forward almost inadvertently as a consequence of the Japanese unconditional surrender to the Allies on August 15 following the Nagasaki atomic bombing. The historic event was triggered by a plot, led by a few more radical youth activists such as Adam Malik and Chairul Saleh, that put pressure on Soekarno and Hatta to proclaim independence immediately. The declaration was to be signed by the 27 members of the Preparatory Committee for Indonesian Independence (PPKI) symbolically representing the new nation’s diversity. The particular act was apparently inspired by a similar spirit of the United States Declaration of Independence. However, the idea was heavily turned down by the radical activists mentioned earlier, arguing that the committee was too closely associated with then soon to be defunct Japanese occupation rule, thus creating a potential credibility issue. Instead, the radical activists demanded that the signatures of six of them were to be put on the document. All parties involved in the historical moment finally agreed on a compromise solution which only included Soekarno and Mohammad Hatta as the co-signers ‘in the name of the nation of Indonesia’
Soekarno had initially wanted the declaration to be read at Ikada Plain, the large open field in the centre of Jakarta, but due to unfounded widespread apprehension over the possibility of Japanese sabotage, the venue was changed to Soekarno’s house at Pegangsaan Timur 56. In fact there was no concrete evidence for the growing suspicions, as the Japanese had already surrendered to the Allies, and the Japanese high command in Indonesia had given their permission for the nation’s independence. The declaration of independence passed without a hitch.
Kami, bangsa Indonesia, dengan ini menjatakan kemerdekaan Indonesia.
Hal-hal jang mengenai pemindahan kekoeasaan,d.l.l., diselenggarakan dengan tjara saksama dan dalam tempoh yang sesingkat-singkatnja
Djakarta (Jakarta), 17-8-45
Wakil-Wakil Bangsa Indonesia
Three amendments were made to the draft, as follows:
- “tempoh“: changed to “tempo“, both meaning “time period”.
- 17-8-45: changed to “hari 17, boelan 8, tahoen 05″ (“day 17, month 8, year 05” of the Japanese sumera calendar); the number “05” is the short form for 2605.
- “Wakil-Wakil Bangsa Indonesia” (Representatives of the people of Indonesian nation): changed to “Atas nama bangsa Indonesia” (“in the name of the nation of Indonesia”).
The original Indonesian Declaration of Independence
Kami, bangsa Indonesia, dengan ini menjatakan kemerdekaan Indonesia.
Hal-hal jang mengenai pemindahan kekoeasaan d.l.l., diselenggarakan dengan tjara saksama dan dalam tempo jang sesingkat-singkatnja.
Djakarta, hari 17 boelan 8 tahoen 05
Atas nama bangsa Indonesia
<<tanda tangan Soekarno/Hatta>>
Soekarno – Hatta
An English translation published by the Ministry of Foreign Affairs as of October 1948 included the entire speech as read by Sukarno. It incorporated remarks made immediately prior to and after the actual proclamation. George McTurnan Kahin, a historian on Indonesia, believed that they were omitted from publication in Indonesia either due to Japanese control of media outlets or fear of provoking a harsh Japanese response.
WE THE PEOPLE OF INDONESIA HEREBY DECLARE THE INDEPENDENCE OF
INDONESIA. MATTERS WHICH CONCERN THE TRANSFER OF POWER AND
OTHER THINGS WILL BE EXECUTED BY CAREFUL MEANS AND IN THE
SHORTEST POSSIBLE TIME.
DJAKARTA, 17 AUGUST 1945
IN THE NAME OF THE PEOPLE OF INDONESIA
- ^ “Dutch govt expresses regrets over killings in RI”. Jakarta Post. 2005-08-18. http://www.thejakartapost.com/news/2005/08/18/dutch-govt-expresses-regrets-over-killings-ri.html. Retrieved 2008-11-23.
- ^ “Former governor Ali Sadikin, freedom fighter SK Trimurti die”. Jakarta Post. 2008-05-21. http://www.thejakartapost.com/news/2008/05/21/former-governor-ali-sadikin-freedom-fighter-sk-trimurti-die.html. Retrieved 2008-06-07.
- ^ Yuliastuti, Dian (2008-05-21). “Freedom Fighter SK Trimurti Dies”. Tempo Interactive. http://www.tempointeraktif.com/hg/nasional/2008/05/21/brk,20080521-123376,uk.html. Retrieved 2008-06-07.
- ^ Kahin, George McT. (April 2000). “Sukarno’s Proclamation of Indonesian Independence”. Indonesia (Ithaca, NY: Cornell Modern Indonesia Project) 69: 1–4. doi:10.2307/3351273. http://cip.cornell.edu/seap.indo/1106943306. Retrieved 24 June 2009.
Indonesian National Revolution
The Indonesian National Revolution or Indonesian War of Independence was an armed conflict and diplomatic struggle between Indonesia and the Dutch Empire, and an internal social revolution. It took place between Indonesia’s declaration of independence in 1945 and the Netherlands‘ recognition of Indonesia’s independence in 1949.
One of the largest revolutions of the twentieth century, the struggle lasted for over four years and involved sporadic but bloody armed conflict, internal Indonesian political and communal upheavals, and two major international diplomatic interventions. Dutch forces were not able to prevail over the Indonesians, but were strong enough to resist being expelled. Although Dutch forces could control the towns and cities in Republican heartlands on Java and Sumatra, they could not control villages and the countryside. Thus, the Republic of Indonesia ultimately prevailed as much through international diplomacy as it did through Indonesian determination in the armed conflicts on Java and other islands, exhausting the Dutch economy.
The Revolution destroyed a colonial administration which had ruled from the other side of the world. It also significantly changed racial castes, as well as reducing the power of many of the local rulers (raja). It did not significantly improve the economic or political fortune of the majority of the population, though a few Indonesians were able to gain a larger role in commerce.
Indonesian nationalism and movements supporting independence from Dutch colonialism, such as Budi Utomo, the Indonesian National Party (PNI), Sarekat Islam, and the Indonesian Communist Party (PKI), grew rapidly in the first half of the twentieth century. Budi Utomo, Sarekat Islam and others pursued strategies of co-operation by joining the Dutch initiated Volksraad (“People’s Council”) in the hope that Indonesia would be granted self-rule. Others chose a non-cooperative strategy demanding the freedom of self-government from the Dutch East Indies colony. The most notable of these leaders were Sukarno and Mohammad Hatta, two students and nationalist leaders who had benefited from the educational reforms of the Dutch Ethical Policy.
Japan’s three and a half year World War II occupation of Indonesia was a crucial factor in the subsequent Revolution. Under German occupation itself, the Netherlands had little ability to defend its colony against the Japanese army, and within only three months of their initial attacks, the Japanese had occupied the Dutch East Indies. In Java, and to a lesser extent in Sumatra (Indonesia’s two dominant islands), the Japanese spread and encouraged nationalist sentiment. Although this was done more for Japanese political advantage than from altruistic support of Indonesian independence, this support created new Indonesian institutions (including local neighbourhood organisations) and elevated political leaders like Sukarno. Just as significantly for the subsequent Revolution, the Japanese destroyed and replaced much of the Dutch-created economic, administrative, and political infrastructure.
With the Japanese on the brink of losing the war, the Dutch sought to re-establish their authority in Indonesia, and requested the Japanese army “preserve law and order” in Indonesia. The Japanese, however, were in favour of helping Indonesian nationalists prepare for self-government. On 7 September 1944, with the war going badly for the Japanese, Prime Minister Koiso promised independence for Indonesia, although no date was set. For supporters of Sukarno, this announcement was seen as vindication for his apparent collaboration with the Japanese.
Under pressure from radical and politicised pemuda (‘youth’) groups, Sukarno and Hatta proclaimed Indonesian independence, on 17 August 1945, two days after the Japanese Emperor’s surrender in the Pacific. The following day, the Central Indonesian National Committee (KNIP) elected Sukarno as President, and Hatta as Vice President.
Euphoria of revolution
We, the people of Indonesia, hereby declare the independence of Indonesia.
Matters which concern the transfer of power and other things will be executed by careful means and in the shortest possible time.
Djakarta, 17 August 1945
In the name of the people of Indonesia,
(Translation by the Ministry of Foreign Affairs, October 1948)
Indonesian flag raised on 17 August 1945.
It was mid-September before news of the declaration of independence spread to the outer islands, and many Indonesians far from the capital Jakarta did not believe it. As the news spread, most Indonesians came to regard themselves as pro-Republican, and a mood of revolution swept across the country. External power had shifted; it would be weeks before Allied Forces entered Indonesia, and the Dutch were too weakened by World War Two. The Japanese, on the other hand, were required by the terms of the surrender to both lay down their arms and maintain order; a contradiction that some resolved by handing weapons to Japanese-trained Indonesians.
The resulting power vacuums in the weeks following the Japanese surrender, created an atmosphere of uncertainty, but also one of opportunity for the Republicans. Many pemuda joined pro-Republic struggle groups (badan perjuangan). The most disciplined were soldiers from the Japanese-formed but disbanded Giyugun (PETA) and Heiho groups. Many groups were undisciplined, due to both the circumstances of their formation and what they perceived as revolutionary spirit. In the first weeks, Japanese troops often withdrew from urban areas to avoid confrontations.
By September 1945, control of major infrastructure installations, including railway stations and trams in Java’s largest cities, had been taken over by Republican pemuda who encountered little Japanese resistance. To spread the Revolution message, pemuda set up their own radio stations and newspapers, and graffiti proclaimed the nationalist sentiment. On most islands, struggle committees and militia were set up. Republican newspapers and journals were common in Jakarta, Yogyakarta, and Surakarta, which fostered a generation of writers known as angkatan 45 (‘generation of 45’) many of whom believed their work could be part of the Revolution.
Republican leaders struggled to come to terms with popular sentiment; some wanted passionate armed struggle; others a more reasoned approach. Some leaders, such as the leftist Tan Malaka, spread the idea that this was a revolutionary struggle to be led and won by the Indonesian pemuda. Sukarno and Hatta, in contrast, were more interested in planning out a government and institutions to achieve independence through diplomacy. Pro-Revolution demonstrations took place in large cities, including one led by Tan Malaka in Jakarta with over 200,000 people, which Sukarno and Hatta, fearing violence, successfully quelled.
By September 1945, many of the self-proclaimed pemuda, who were ready to die for ‘100% freedom’, were getting impatient. It was common for ethnic ‘out-groups’ – Dutch internees, Eurasian, Ambonese and Chinese – and anyone considered to be a spy, to be subjected to intimidation, kidnap, robbery, and sometimes murder, even organised massacres. Such attacks would continue to some extent for the course of the Revolution. As the level of violence increased across the country, the Sukarno- and Hatta-led Republican government in Jakarta urged calm. However, pemuda in favour of armed struggle saw the older leadership as dithering and betraying the Revolution, which often led to conflict amongst Indonesians.
Formation of the Republican government
By the end of August, a central Republican government had been established in Jakarta. It adopted a constitution drafted during the Japanese occupation by the Preparatory Committee for Indonesian Independence. With general elections yet to be held, a Central Indonesian National Committee (KNIP) was appointed to assist the President. Similar committees were established at provincial and regency levels.
Questions of allegiance immediately arose amongst indigenous rulers. Central Javanese principalities, for example, immediately declared themselves Republican, while many raja (‘rulers’) of the outer islands, who had been enriched from their support of the Dutch, were less enthusiastic. Such reluctance among many outer islands was sharpened by the radical, non-aristocratic, and sometimes Islamic nature of the Java-centric Republican leadership. Support did, however, come from South Sulawesi (including the King of Bone, who still recalled battles against the Dutch from early in the century), and from Makassarese and Bugis raja, who supported the Republican Governor of Jakarta, a Menadonese Christian. Many Balinese raja accepted Republican authority.
Fearing the Dutch would attempt to re-establish their authority over Indonesia, the new Republican Government and its leaders moved quickly to strengthen the fledgling administration. Within Indonesia, the newly formed government, although enthusiastic, was fragile and focused in Java (where focused at all). It was rarely and loosely in contact with the outer islands, which had more Japanese troops (particularly in Japanese navy areas), less sympathetic Japanese commanders, and fewer Republican leaders and activists. In November 1945, a parliamentary form of government was established and Sjahrir was appointed Prime Minister.
In the week following the Japanese surrender, the Giyugun (PETA) and Heiho groups were disbanded by the Japanese. Command structures and membership vital for a national army were consequently dismantled. Thus, rather than being formed from a trained, armed, and organised army, the Republican armed forces began to grow in September from usually younger, less trained groups built around charismatic leaders. Creating a rational military structure that was obedient to central authority from such disorganisation, was one of the major problems of the revolution, a problem that remains through to contemporary times. In the self-created Indonesian army, Japanese-trained Indonesian officers prevailed over those trained by the Dutch. A thirty year-old former school teacher, Sudirman, was elected ‘commander-in-chief’ at the first meeting of Division Commanders in Yogyakarta on 12 November 1945.
Allied counter revolution
The Dutch accused Sukarno and Hatta of collaborating with the Japanese, and denounced the Republic as a creation of Japanese fascism. The Dutch East Indies administration had just received a ten million dollar loan from the United States to finance its return to Indonesia.
A soldier of an Indian armoured regiment examines a light tank used by Indonesian nationalists and captured by British forces during the fighting in Surabaya.
The Netherlands, however, was critically weakened from World War II in Europe and did not return as a significant military force until early 1946. The Japanese and members of the Allied forces reluctantly agreed to act as caretakers. As US forces were focusing on the Japanese home islands, the archipelago was put under the jurisdiction of British Admiral Earl Louis Mountbatten, the Supreme Allied Commander, South East Asia Command. Allied enclaves already existed in Kalimantan (Indonesian Borneo), Morotai (Maluku) and parts of Irian Jaya; Dutch administrators had already returned to these areas. In the Japanese navy areas, the arrival of Allied troops quickly prevented revolutionary activities where Australian troops, followed by Dutch troops and administrators, took the Japanese surrender (except for Bali and Lombok).
The British were charged with restoring order and civilian government in Java. The Dutch took this to mean pre-war colonial administration and continued to claim sovereignty over Indonesia. British Commonwealth troops did not, however, land on Java to accept the Japanese surrender until late September 1945. Lord Mountbatten’s immediate tasks included the repatriation of some 300,000 Japanese, and freeing prisoners of war. He did not want, nor did he have the resources, to commit his troops to a long struggle to regain Indonesia for the Dutch. The first British troops reached Jakarta in late September 1945, and arrived in the cities of Medan (North Sumatra), Padang (West Sumatra), Palembang (South Sumatra), Semarang (Central Java) and Surabaya (East Java) in October. In an attempt to avoid clashes with Indonesians, the British commander Lieutenant General Sir Philip Christison, diverted soldiers of the former Dutch colonial army to eastern Indonesia, where Dutch reoccupation was proceeding smoothly. Tensions mounted as Allied troops entered Java and Sumatra; clashes broke out between Republicans and their perceived enemies, namely Dutch prisoners, Dutch colonial troops (KNIL), Chinese, Indo-Europeans and Japanese.
The first stages of warfare were initiated in October 1945 when, in accordance with the terms of their surrender, the Japanese tried to re-establish the authority they relinquished to Indonesians in the towns and cities. Japanese military police killed Republican pemuda in Pekalongan (Central Java) on 3 October, and Japanese troops drove Republican pemuda out of Bandung in West Java and handed the city to the British, but the fiercest fighting involving the Japanese was in Semarang. On 14 October, British forces began to occupy the city. Retreating Republican forces retaliated by killing between 130 and 300 Japanese prisoners they were holding. Five hundred Japanese and 2,000 Indonesians had been killed and the Japanese had almost captured the city six days later when British forces arrived.
The British subsequently decided to evacuate the 10,000 Indo-Europeans and European internees in the volatile Central Java interior. British detachments sent to the towns of Ambarawa and Magelang encountered strong Republican resistance and used air attacks against the Indonesians. Sukarno arranged a ceasefire on 2 November, but by late November fighting had resumed and the British withdrew to the coast. Republican attacks against Allied and alleged pro-Dutch civilians reached a peak in November and December, with 1,200 killed in Bandung as the pemuda returned to the offensive. In March 1946, departing Republicans responded to a British ultimatum for them to leave the city of Bandung by deliberately burning down much of the southern half of the city in what is popularly known in Indonesia as the “Bandung Sea of Fire“. The last British troops left Indonesia in November 1946, but by this time 55,000 Dutch troops had landed in Java.
Battle of Surabaya
Main article: Battle of Surabaya
The Battle of Surabaya was the heaviest single battle of the Revolution and became a national symbol of Indonesian resistance. Pemuda groups in Surabaya, the second largest city in Indonesia, seized arms and ammunition from the Japanese and set up two new organisations; the Indonesia National Committee (KNI) and the People’s Security Council (BKR). By the time the Allied forces arrived at the end of October 1945, the pemuda foothold in Surabaya city was described as “a strong unified fortress”.
The city itself was in pandemonium. There was bloody hand-to-hand fighting on every street corner. Bodies were strewn everywhere. Decapitated, dismembered trunks lay piled one on top of the other…Indonesians were shooting and stabbing and murdering wildly
In September and October 1945 Europeans and pro-Dutch Eurasians were attacked and killed by Indonesian mobs. Ferocious fighting erupted when 6,000 British Indian troops landed in the city. Sukarno and Hatta negotiated a ceasefire between the Republicans and the British forces led by Brigadier Mallaby. Following the killing of Mallaby on 30 October, the British sent more troops into the city from 10 November under the cover of air attacks. Although the European forces largely captured the city in three days, the poorly armed Republicans fought on for three weeks and thousands died as the population fled to the countryside.
Despite the military defeat suffered by the Republicans and a loss of manpower and weaponry that would severely hamper Republican forces for the rest of the Revolution, the battle and defence mounted by the Indonesians galvanised the nation in support of independence and helped garner international attention. For the Dutch, it removed any doubt that the Republic was not simply a gang of collaborators without popular support. It also convinced Britain to lie on the side of neutrality in the Revolution; and within a few years, Britain would support the Republican cause in the United Nations.
The Dutch return
Javanese revolutionaries armed with bamboo spears and a few Japanese rifles. 1946.
With British assistance, the Dutch landed their Netherlands Indies Civil Administration (NICA) forces in Jakarta and other key centres. Republican sources reported 8,000 deaths up to January 1946 in the defence of Jakarta, but they could not hold the city. The Republican leadership thus established themselves in the city of Yogyakarta with the crucial support of the new sultan, Sri Sultan Hamengkubuwono IX. Yogyakarta went on to play a leading role in the Revolution, which would result in the city being granted its own Special Territory status. In Bogor, near Jakarta, and in Balikpapan in Kalimantan, Republican officials were imprisoned. In preparation for Dutch occupation of Sumatra, its largest cities, Palembang and Medan, were bombed. In December 1946, Dutch Special Troops (KST), led by commando and counter-insurgency expert Captain Raymond ‘Turk’ Westerling, were accused of trying to pacify the southern Sulawesi region using arbitrary terror techniques, which were copied by other anti-Republicans. As many as 3,000 Republican militia and their supporters were killed in a few weeks.
On Java and Sumatra, the Dutch found military success in cities and major towns, but they were unable to subdue the villages and countryside. On the outer islands (including Bali), Republican sentiment was not as strong, at least among the elite. They were consequently occupied by the Dutch with comparative ease, and autonomous states were set up by the Dutch. The largest, the State of East Indonesia (NIT), encompassed most of eastern Indonesia, and was established in December 1946, with its administrative capital in Makassar.
Diplomacy and military offensives
The Linggadjati Agreement, brokered by the British and concluded in November 1946, saw the Netherlands recognise the Republic as the de-facto authority over Java, Madura, and Sumatra. Both parties agreed to the formation of the ‘United States of Indonesia‘ by 1 January 1949, a semi-autonomous federal state with the Monarchy of the Netherlands at its head. The Republican-controlled Java and Sumatra would be one of its states, alongside areas that were generally under stronger Dutch influence, including southern Kalimantan, and the ‘Great East‘, which consisted of Sulawesi, Maluku, the Lesser Sunda Islands, and Western New Guinea. The Central National Committee of Indonesia (KNIP) did not ratify the agreement until February 1947, and neither the Republic nor the Dutch were satisfied with it. On 25 March 1947 the Lower House of the Dutch parliament ratified a ‘stripped down’ version of the treaty, which was not accepted by the Republic. Both sides soon accused the other of violating the agreement.
…[the Republic] became increasingly disorganised internally; party leaders fought with party leaders; governments were over thrown and replaced by others; armed groups acted on their own in local conflicts; certain parts of the Republic never had contact with the centre-they just drifted along in their own way.
The whole situation deteriorated to such an extent that the Dutch Government was obliged to decide that no progress could be made before law and order were restored sufficiently to make intercourse between the different parts of Indonesia possible, and to guarantee the safety of people of different political opinions.
Main article: Operatie Product
At midnight on 20 July 1947, the Dutch launched a major military offensive it called Operatie Product with the intent of conquering the Republic. Claiming violations of the Linggajati Agreement, the Dutch described the campaign as Politionele acties (‘police actions’) to restore law and order. This used to be the task of the KNIL. However, at the time the majority of the Dutch troops in Indonesia belonged to the Royal Netherlands Army. Soon after the end of WWII, 25,000 volunteers (among them 5,000 marines) had been sent overseas. They were later followed by larger numbers of conscripts from the Netherlands. In the offensive, Dutch forces drove Republican troops out of parts of Sumatra, and East and West Java. The Republicans were confined to the Yogyakarta region of Java. To maintain their force in Java, now numbering 100,000 troops, the Dutch gained control of lucrative Sumatran plantations, and oil and coal installations, and in Java, control of all deep water ports.
International reaction to the Dutch actions was negative. Neighbouring Australia and newly independent India were particularly active in supporting the Republic’s cause in the UN, as was the Soviet Union and, most significantly, the United States. Dutch ships continued to be boycotted from loading and unloading by Australian waterside workers, a blockade that began in September 1945. The United Nations Security Council became directly involved in the conflict, establishing a Good Offices Committee to sponsor further negotiations, making the Dutch diplomatic position particularly difficult. A ceasefire, called for by UN resolution, was ordered by the Dutch and Sukarno on 4 August 1947.
Renville Agreement
The Van Mook line in Java. Areas in red were under Republican control.
Main article: Renville Agreement
The United Nations Security Council brokered the Renville Agreement in an attempt to rectify the collapsed Linggarjati Agreement. The agreement was ratified in January 1948 and recognised a cease-fire along the so-called ‘van Mook line’; an artificial line which connected up the most advanced Dutch positions. Many Republican positions, however, were still held behind the Dutch lines. The agreement also required referenda to be held on the political future of the Dutch held areas. The apparent reasonableness of Republicans garnered much important American goodwill.
Diplomatic efforts between the Netherlands and the Republic continued throughout 1948 and 1949. Political pressures, both domestic and international, hindered Dutch attempts at goal formulation. Similarly Republican leaders faced great difficulty in persuading their people to accept diplomatic concessions. By July 1948 negotiations were in deadlock and the Netherlands pushed unilaterally towards Van Mook’s federal Indonesia concept. The new federal states of South Sumatra and East Java were created, although neither had a viable support base. The Netherlands set up the Bijeenkomst voor Federaal Overleg (BFO) (or ‘Federal Consultative Assembly’), a body comprising the leadership of the federal states, and charged with the formation of a United States of Indonesia and an interim government by the end of 1948. The Dutch plans, however, had no place for the Republic unless it accepted a minor role already defined for it. Later plans included Java and Sumatra but dropped all mention of the Republic. The main sticking point in the negotiations was the balance of power between the Netherlands High Representative and the Republican forces.
Mutual distrust between the Netherlands and the Republic hindered negotiations. The Republic feared a second major Dutch offensive, while the Dutch objected to continued Republican activity on the Dutch side of the Renville line. In February 1948 the Siliwangi Battalion of the Republican Army, led by Nasution, marched from West Java to Central Java; the relocation was intended to ease internal Republican tensions involving the Battalion in the Surakarta area. The Battalion, however, clashed with Dutch troops while crossing Mount Slamet, and the Dutch believed it was part of a systematic troop movement across the Renville Line. The fear of such incursions actually succeeding, along with apparent Republican undermining of the Dutch-established Pasudan state and negative reports, lead to the Dutch leadership increasingly seeing itself as losing control.
Operation Crow and Serangan Umum
We have been attacked….The Dutch government have betrayed the cease-fire agreement. All the Armed Forces will carry out the plans which have been decided on to confront the Dutch attack
Main article: Operatie Kraai
Frustrated at negotiations with the Republic and believing it weakened by both the Darul Islam and Madiun insurgencies, the Dutch launched a military offensive on 19 December 1948 which it termed ‘Operatie Kraai‘ (Operation Crow). By the following day it had conquered the city of Yogyakarta, the location of the temporary Republican capital. By the end of December, all major Republican held cities in Java and Sumatra were in Dutch hands. The Republican President, Vice President, and all but six Republic of Indonesia ministers were captured by Dutch troops and exiled on Bangka Island off the east coast of Sumatra. In areas surrounding Yogyakarta and Surakarta, Republican forces refused to surrender and continued to wage a guerrilla war under the leadership of Republican military chief of staff General Sudirman who had escaped the Dutch offensives. An emergency Republican government, the Pemerintahan Darurat Republik Indonesia (PDRI), was established in West Sumatra.
Although Dutch forces conquered the towns and cities in Republican heartlands on Java and Sumatra, they could not control villages and the countryside. Republican troops and militia led by Lt. Colonel (later President) Suharto attacked Dutch positions in Yogyakarta at dawn on 1 March 1949. The Dutch were expelled from the city for six hours but reinforcements were brought in from the nearby cities of Ambarawa and Semarang that afternoon. Indonesian fighters retreated at 12:00 pm and the Dutch re-entered the city. The Indonesian attack, later known in Indonesia as Serangan Umum (‘1 March General Offensive’), is commemorated by a large monument in Yogyakarta. A similar attack against Dutch troops in Surakarta was led by Lt. Col. Slamet Riyadi on 7 August the same year.
Once again, international opinion of the Dutch military campaigns was one of outrage, significantly in both the United Nations and the United States. In January 1949, the United Nations Security Council passed a resolution demanding the reinstatement of the Republican government. United States aid specifically earmarked for the Netherlands’ Indonesia efforts was immediately cancelled and pressure mounted within the American Congress for all United States aid to be cut off. This included Marshall Plan funds vital for Dutch post-World War II rebuilding that had so far totalled $US 1 billion. The Netherlands Government had spent an amount equivalent to almost half of this funding their campaigns in Indonesia. That United States aid could be used to fund “a senile and ineffectual imperialism” encouraged many key voices in the United States – including those amongst the US Republican Party – and from within American churches and NGOs to speak out in support of Indonesian independence.
The so-called ‘social revolutions’ following the independence proclamation were challenges to the Dutch-established Indonesian social order, and to some extent a result of the resentment against Japanese-imposed policies. Across the country, people rose up against traditional aristocrats and village heads and attempted to exert popular ownership of land and other resources. The majority of the social revolutions ended quickly; in most cases the challenges to the social order were quashed.
A culture of violence rooted in the deep conflicts that split the countryside under Dutch rule would repeatedly erupt throughout the whole second half of the twentieth century. The term ‘social revolution’ has been applied to a range of mostly violent activities of the left that included both altruistic attempts to organise real revolution and simple expressions of revenge, resentment and assertions of power. Violence was one of the many lessons learned during the Japanese occupation, and figures identified as ‘feudal‘, including kings, regents, or simply the wealthy, were often attacked, sometimes beheaded, and rape became a weapon against ‘feudal’ women. In the coastal sultanates of Sumatra and Kalimantan, for example, sultans and others whose authority had been shored-up by the Dutch, were attacked as soon as Japanese authority left. The secular local lords of Aceh, who had been the foundation of Dutch rule, were executed, although most of Indonesia’s sultanates fell back into Dutch hands.
Most Indonesians lived in fear and uncertainty, particularly a significant proportion of the population who supported the Dutch or who remained under Dutch control. The popular revolutionary cry ‘Freedom or Death’ was often interpreted to justify killings under claimed Republican authority. Traders were often in particularly difficult positions. On the one hand, they were pressured by Republicans to boycott all sales to the Dutch; on the other hand, Dutch police could be merciless in their efforts to stamp out smugglers on which the Republican economy depended. In some areas, the term kedaulatan rakyat (‘exercising the sovereignty of the people’) – which is mentioned in the preamble of the Constitution and used by pemuda to demand pro-active policies from leaders – came to be used not only in the demanding of free goods, but also to justify extortion and robbery. Chinese merchants, in particular, were often forced to keep their goods at artificially low prices under threat of death.
Communist and Islamist insurgencies
On 18 September 1948 an ‘Indonesian Soviet Republic’ was declared in Madiun, east of Yogyakarta, by members of the PKI and the Indonesian Socialist Party (PSI). Judging the times as right for a proletarian uprising, they intended it to be a rallying centre for revolt against “Sukarno-Hatta, the slaves of the Japanese and America”. Madiun however was won back by Republican forces within a few weeks and the insurgency leader, Musso, killed. RM Suryo, the governor of East Java, several police officers and religious leaders were killed by the rebels. This ended a distraction for the Revolution, and it turned vague American sympathies based on anti-colonial sentiments into diplomatic support. Internationally, the Republic was now seen as being staunchly anti-communist and a potential ally in the emergin global Cold War between the American-led ‘free world’ and the Soviet-led bloc.
Members of the Republican Army who had come from Indonesian Hizbullah felt betrayed by Indonesian Government. In May 1948, they declared a break-away regime, the Negara Islam Indonesia (Indonesian Islamic State), better known as Darul Islam. Led by an Islamic mystic, Sekarmadji Maridjan Kartosuwirjo, Darul Islam sought to establish Indonesia as an Islamic theocracy. At the time, the Republican Government did not respond as they were focused on the threat from the Dutch. Some leaders of Masjumi sympathised with the rebellion. After the Republic regained all territories in 1950, the government took the Darul Islam threat seriously, especially after some provinces declared their joining of Darul Islam. The rebellion was put down in 1962.
Transfer of sovereignty
Millions upon millions flooded the sidewalks, the roads. They were crying, cheering, screaming “…Long live Bung Karno…” They clung to the sides of the car, the hood, the running boards. They grabbed at me to kiss my fingers. Soldiers beat a path for me to the topmost step of the big white palace. There I raised both hands high. A stillness swept over the millions. “Alhamdulillah – Thank God,” I cried. “We are free”
The resilience of Indonesian Republican resistance and active international diplomacy set world opinion against the Dutch efforts to re-establish their colony. The second ‘police action’ was a diplomatic disaster for the Dutch cause. The newly appointed United States Secretary of State Dean Acheson pushed the Netherlands government into negotiations earlier recommended by the United Nations but until then defied by the Netherlands. The Dutch–Indonesian Round Table Conference was held in The Hague from 23 August 1949 to 2 November 1949 between the Republic, the Netherlands, and the Dutch-created federal states. The Netherlands agreed to recognise Indonesian sovereignty over a new federal state known as the ‘United States of Indonesia‘ (RUSI). It would include all the territory of the former Dutch East Indies with the exception of Netherlands New Guinea; sovereignty over which it was agreed would be retained by the Netherlands until further negotiations with Indonesia. The other difficult issue to which Indonesia gave concessions was Netherlands East Indies debt. Indonesia agreed to responsibility for this sum of £4.3 billion, much of which was directly attributable to Dutch attempts to crush the Revolution. Sovereignty was formally transferred on 27 December 1949, and the new state was immediately recognised by the United States of America.
The United States of Indonesia, December 1949 – the Republic of Indonesia is shown in red
Republican-controlled Java and Sumatra together formed a single state in the sixteen-state RUSI federation, but accounted for almost half its population. The other fifteen ‘federal’ states had been created by the Netherlands since 1945. These states were dissolved into the Republic over the first half of 1950. An abortive anti-Republic coup in Bandung led by the infamous Westerling on 23 January 1950 resulted in the dissolution of the populous Pasudan state in West Java, thus quickening the dissolution of the federal structure. Colonial soldiers, who were largely Ambonese, clashed with Republican troops in Makassar in April 1950. The predominantly Christian Ambonese were from one of the few regions with pro-Dutch sentiments and they were suspicious of the Javanese Muslim-dominated Republic, whom they unfavourably regarded as leftists. On 25 April 1950, an independent Republic of South Maluku (RMS) was proclaimed in Ambon but this was suppressed by Republican troops during a campaign from July to November. With the state of East Sumatra now being the only federal state remaining, it too folded and fell in line with the unitary Republic. On 17 August 1950, the fifth anniversary of his declaration of Indonesian independence, Sukarno proclaimed the Republic of Indonesia as a unitary state.
Although there is no accurate account of how many Indonesians died, they died in far greater numbers than their enemies, and many died at the hands of other Indonesians. Estimates of Indonesian deaths in fighting range from 45,000 to 100,000 and civilian casualties exceeded 25,000 and may have been as high as 100,000. A total of 1,200 British soldiers were killed or went missing in Java and Sumatra in 1945 and 1946, most of them Indian soldiers. More than 5,000 Dutch soldiers lost their lives in Indonesia between 1945 and 1949. Many more Japanese died; in Bandung alone, 1,057 died, only half of whom died in actual combat, the rest killed in rampages by Indonesians. Tens of thousands of Chinese and Eurasians were killed or left homeless, despite the fact that many Chinese supported the Revolution. 7 million people were displaced on Java and Sumatra.
The Revolution had direct effects on economic conditions; shortages were common, particularly food, clothing and fuel. There were in effect two economies – the Dutch and the Republican – both of which had to simultaneously rebuild after World War II and survive the disruptions of the Revolution. The Republic had to set up all necessities of life, ranging from ‘postage stamps, army badges, and train tickets’ whilst subject to Dutch trade blockades. Confusion and ruinous inflationary surges resulted from competing currencies; Japanese, new Dutch money, and Republican currencies were all used, often concurrently.
Indonesian independence was secured through a blend of both diplomacy and force. Despite their ill-discpline raising the prospect of anarchy, without pemuda confronting foreign and Indonesian colonial forces, Republican diplomatic efforts would have been futile. The Revolution is the turning point of modern Indonesian history, and it has provided the reference point and validation for the country’s major political trends that continue to the present day. It gave impetus to communism in the country, to militant nationalism, to Sukarno’s ‘guided democracy‘, to political Islam, the origins of the Indonesian army and its role in Indonesian power, the country’s constitutional arrangements, and the centralism of power in Indonesia.
The revolution destroyed a colonial administration ruled from the other side of the world, and dismantled with it the raja, by many seen as obsolete and powerless. Also, it relaxed the rigid racial and social categorisations of colonial Indonesia. Tremendous energies and aspirations were created amongst Indonesians; a new creative surge was seen in writing and art, as was a great demand for education and modernisation. It did not, however, significantly improve the economic or political fortune of the population’s poverty-stricken peasant majority; only a few Indonesians were able to gain a larger role in commerce, and hopes for democracy were dashed within a decade.
- ^ a b c d Friend (2003), page 35
- ^ Amry Vandenbosch (1931). “Nationalism in Netherlands East India”. Pacific Affairs (Pacific Affairs, University of British Columbia) 4 (12): 1051–1069. doi:10.2307/2750579. http://links.jstor.org/sici?sici=0030-851X%28193112%294%3A12%3C1051%3ANINEI%3E2.0.CO%3B2-%23.
- ^ George Mc.T Kahin (1980). “In Memoriam: Mohammad Hatta (1902-1980)” (fee required). Indonesia (Southeast Asia Program Publications at Cornell University) 20: 113–120. doi:10.2307/3350997. http://www.jstor.org/pss/3350828.
- ^ Vickers (2005), page 85
- ^ a b Charles Bidien (5 December 1945). “Independence the Issue”. Far Eastern Survey 14 (24): 345–348. doi:10.1525/as.1945.14.24.01p17062. http://links.jstor.org/sici?sici=0362-8949%2819451205%2914%3A24%3C345%3AITI%3E2.0.CO%3B2-S.
- ^ Ricklefs (1991), page 207
- ^ a b c “The National Revolution, 1945-50”. Country Studies, Indonesia. U.S. Library of Congress. http://countrystudies.us/indonesia/16.htm.
- ^ Ricklefs (1991), page 213; *Taylor, Jean Gelman (2003). Indonesia: Peoples and History. Yale University Press. p. 325. ISBN 0-300-10518-5. ; Reid (1973), page 30
- ^ Kahin, George McT. (April 2000). “Sukarno’s Proclamation of Indonesian Independence”. Indonesia (Ithaca, NY: Cornell Modern Indonesia Project) 69: 1–4. doi:10.2307/3351273. http://cip.cornell.edu/DPubS?service=UI&version=1.0&verb=Display&handle=seap.indo/1106943306. Retrieved 24 June 2009.
- ^ Ricklefs (1991), pages 214 – 215
- ^ Friend (2003), page 32; Robert Cribb, ‘A revolution delayed: the Indonesian Republic and the Netherlands Indies, August-November 1945’, Australian Journal of Politics and History 32 no. 1 (1986), pp. 72-85.
- ^ a b c Friend (2003), page 32
- ^ a b c Ricklefs (1991), pages 215 – 216
- ^ a b c d Vickers (2005), page 97
- ^ Reid (1974), page 49; Mochtar Lubis, Jalan Tak Ada (Jakarta: Yayasan Obot Indonesia, 2002) [originally published 1952]), p.78; Anthony Reid, Indonesian National Revolution (Hawthorn, Vic.: Longman, 1974), chs. 2 and 3; Shirley Fenton-Huie, The Forgotten Ones: Women and Children Under Nippon (Sydney: Angus and Robertson, 1992); Anthony Reid, ‘Indonesia: revolution without socialism’, in Robin Jeffrey (ed.), Asia: the Winning of Independence (London: MacMillan, 1981), pp. 107-57.
- ^ a b Ricklefs (1991), page 214
- ^ Friend (2003), page 33
- ^ a b Ricklefs (1991), page 215
- ^ Most PETA and Heiho members did not yet know about the declaration of independence.
- ^ Reid (1974), page 78
- ^ a b c d Ricklefs (1991), page 216
- ^ a b Vickers (2005), page 99
- ^ Ricklefs (1991), page 216; McMillan, Richard (2005). The British Occupation of Indonesia 1945-1946. Melbourne: Routledge. pp. 306–307. ISBN 0-415-35551-6.
- ^ Reid (1973), page 54
- ^ Frederick, William H. (April 1982). “In Memoriam: Sutomo” (PDF). Indonesia (Cornell Modern Indonesia Project) 33: 127–128. seap.indo/1107016901. http://cip.cornell.edu/DPubS?service=UI&version=1.0&verb=Display&handle=seap.indo/1107016901.
- ^ a b c Ricklefs (1991), page 217
- ^ a b J. G. A. Parrott (October 1975). “Who Killed Brigadier Mallaby?” (PDF). Indonesia (Cornell Modern Indonesia Project) 20: 87–111. doi:10.2307/3350997. http://cip.cornell.edu/DPubS?service=UI&version=1.0&verb=Display&handle=seap.indo/1107105571. Retrieved 27 November 2006.
- ^ Sukarno (1965). Sukarno: An Autobiography. Bobbs-Merrill. p. 228.
- ^ Frederick, Willam H. (1989). Visions and Heat: The Making of the Indonesian Revolution. Athens, Ohio: Ohio University Press. pp. 237–243. ISBN 0-8214-0906-9.
- ^ Indonesia Law No. 5/1974 Concerning Basic Principles on Administration in the Region (translated version). The President of Republic of Indonesia (1974). Chapter VII Transitional Provisions, Art. 91.
- ^ Ricklefs (1991), page 224
- ^ Kahin, George McTurnan (1952). Nationalism and Revolution in Indonesia. Cornell University Press. ISBN 0-8014-9108-8.
- ^ van Mook, H. J. (July 1949). “Indonesia”. International Affairs (Royal Institute of International Affairs) 25 (3): 278. http://links.jstor.org/sici?sici=0020-5850%28194907%2925%3A3%3C274%3AI%3E2.0.CO%3B2-P.
- ^ a b Ricklefs (1991), page 226
- ^ Kahin (1952), p. 233
- ^ Reid (1974), page 149
- ^ Reid (1974), page 150
- ^ Reid (1974), pages 149 – 151
- ^ originally cited in Siliwangi dari masa kemasa, p. 279, taken from Reid (1974), page 152
- ^ a b Reid (1973), page 153
- ^ a b Reid (1974)
- ^ Friend (2003), page 37
- ^ a b Friend (2003), page 38
- ^ a b c Vickers (2005), pages 101 – 104
- ^ a b by Freek Colombijn, J. Thomas Linblad (Eds) (2002). Roots of Violence in Indonesia: Contemporary Violence in Historical Perspective. Koninklijk Instituut Voor de Tropen. pp. 143–173. ISBN 9067181889.
- ^ Reid (1974), page 60
- ^ Ricklefs (1991), page 230
- ^ Sukarno (1965). Sukarno: An Autobiography. Bobbs-Merrill. pp. 262–263.
- ^ Reid (1974), pages 170-172; Ricklefs (1991), pages 232-233; “The National Revolution, 1945-50”. U.S. Library of Congress. http://countrystudies.us/indonesia/16.htm.
- ^ a b Friend, Bill personal comment 22/04/04; Friend, Theodore (1988). Blue Eyed Enemy. Princeton University Press. pp. 228 & 237. ISBN 978-0691055244. ; Nyoman S. Pendit, Bali Berjuang (2nd edn Jakarta:Gunung Agung, 1979 [original edn 1954]); Reid (1973), page 58,n.25, page 119,n.7, page 120,n.17, page 148,n.25 and n.37; Pramoedya Anwar Toer, Koesalah Soebagyo Toer and Ediati Kamil Kronik Revolusi Indonesia [Jakarta: Kepustakaan Populer Gramedia, vol. I (1945); vol. II (1946) 1999; vol. III (1947); vol. IV (1948) 2003]; Ann Stoler, Capitalism and Confrontation in Sumatra’s Plantation Belt, 1870-1979 (New Haven:Yale University Press, 1985), p103.; all cited in Vickers (2005), page 100
- ^ Kirby, Woodburn S (1969). War Against Japan, Volume 5: The Surrender of Japan. HMSO. p. 258.
- ^ Vickers (2005), page 101
- ^ a b Reid (1974), pages 170 – 171
- Friend, Theodore (2003). Indonesian Destinies. The Belknap Press of Harvard University Press. ISBN 0-674-01834-6.
- Kahin, George McTurnan (1952) Nationalism and Revolution in Indonesia Cornell University Press, ISBN 0-8014-9108-8
- Reid, Anthony (1974). The Indonesian National Revolution 1945-1950. Melbourne: Longman Pty Ltd. ISBN 0-582-71046-4.
- Ricklefs, M.C. A History of Modern Indonesia Since c. 1300. Second Edition. MacMillan, 1991.
- Vickers, Adrian (2005). A History of Modern Indonesia. New York: Cambridge University Press. pp. 85–112. ISBN 0-521-54262-2.<|endoftext|>
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These 224 practice cards help students organize their knowledge of number families to solve problems in which they need to find a missing part, rather than a sum or difference. They are very effective in helping students learn to build equations for word problems that require finding missing information.
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The increasing use of technology in all aspects of society makes confident, creative and productive use of ICT an essential skill for life. Computer capability encompasses not only the mastery of technical skills and techniques but also the understanding to apply these skills purposefully, safely and responsibly. Computing is fundamental to participation and engagement in modern society.
Here at Sandbach High School the department is committed to imparting to all students the practical competence, analytical ability and communication skills necessary to work with Computing Technology in later life.
The KS3 course aims to develop pupils’ computing and ICT capability by bringing together a range of computing theory, traditional ICT and creative projects. The government has moved towards more computing in lessons and Sandbach High School has been developing a new and exciting curriculum for pupils to enjoy.
In Key Stage 3, pupils receive 2 lessons of ICT/Computing a fortnight.
In year 7 pupils will cover:
- Esafety and Leaflet Design
- Hardware (Input, Processing and Output devices), Software (Applications) and Presentation design.
- Introduction to Programming (using Scratch)
- Spreadsheet Design (Formatting and Basic Formulas)
- Maths in Computing (Binary and Boolean Logic)
- Control Systems (Real world – Basics)
In year 8 pupils will cover:
- ESafety and Presentation Design
- Algorithms (Searching and Sorting)
- Hardware (Internal Components) and Software (Operating System)
- Website development (HTML)
- Physical Programming (using MicroBits)
- Spreadsheet Design (formulas and functions)
- Control Systems (Real world – Intermediate)
- Maths in Computing (Binary and Logic Gates)
In year 9 pupils will cover:
- Spreadsheet Design (Advanced skills and complex formulas)
- Programming (text-based using Python)
- Web Authoring (Web Design)
- Computer Networking
- Maths in Computing (Binary Arithmetic)
- Control Systems (Real World – Advanced)
In KS4 pupils have the option to develop their skills and knowledge with Computer Science and Creative iMedia courses. The students are given the option to take a particular route with a mixture of traditional and vocational courses. These include: Cambridge National in Creative iMedia and GCSE Computer Science. The department offers courses to suit a range of learning styles that will help pupils learn the skills, knowledge and understanding of Computing/IT functions, environments and operations and creative developments in IT.
Cambridge National in Creative iMedia
The Cambridge National in Creative iMedia provides candidates with high quality, industry-recognised qualification. It is a vocational qualifications that provide valuable opportunities for individuals to develop skills and gain underpinning knowledge and understanding which will support entry into the Creative Media sector.
The course provides opportunities for candidates to develop skills and techniques in a wide range of multimedia assignments, e.g. animation, games design, digital photography, web design, sound and video.
The course is broken down to 75% coursework and 25% exam.
OCR GCSE Computing
This course gives students a real, in-depth understanding of how computer technology works. Students will no doubt be familiar with the use of computers and other related technology from their other subjects and elsewhere. However, this course will give them an insight into what goes on ‘behind the scenes’, including computer programming and the physical technology itself.
The course is broken down to 20% controlled assessments (in house) and 80% exams.
Cambridge Technical in IT
The objective of this qualification is to provide learners with the opportunity through applied learning to develop the core specialist knowledge, skills and understanding required in the IT sector. These include:
- Technical Skills: using a range of hardware and software.
- Project Management skills: organising and management IT based projects.
- Cognitive and problem-solving skills: use critical thinking; approach non-routine problems applying expert and creative solutions; and use systems and technology.
- Intrapersonal skills: communicating; working collaboratively; negotiating and influencing; and self-presentation.
- Interpersonal skills: self-management; adaptability and resilience; self-monitoring and development.
We offer the Application Developer route which includes a range of theory and practical units to give candidates the knowledge and skills required to work in the IT industry.
The course is broken down to 50% coursework and 50% exam.
OCR A level Computer Science
This course gives students the opportunity to apply the academic principles of Computer Science learned in the classroom to real-world systems. It’s an intensely creative subject that combines invention and excitement, that can look at the natural world through a digital prism. Computer Science will improve computational thinking, helping students to develop the skills to solve problems, design systems and understand the power and limits of human and machine intelligence.
The course is broken down to 80% exam and 20% coursework.
Level 3 BTEC Creative Media Production
The UK’s creative industries include television and film, publishing and advertising, radio and computer games development. The Department for Culture, Media & Sport estimate that the creative industries are worth £36 billion a year and employ 1.5 million people in the UK. The UK’s creative industries have been identified as a strategic growth sector by the Government.
The Pearson BTEC Level 3 Subsidiary Diploma in Creative Media Production has the purpose of helping people to become occupationally ready to take up employment in the creative industries at the appropriate level. This can follow either directly after achieving the qualification, or via the stepping stone of Higher Education in universities. The qualification will help learners develop knowledge, understanding and skills required by the sector, including essential employability skills, and apply them in real work contexts.
The course is broken into a variety of units which are all coursework based.<|endoftext|>
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A new NASA study finds that during Greenland's hottest summers on record, 2010 and 2012, the ice in Rink Glacier on the island's west coast didn't just melt faster than usual, it slid through the glacier's interior in a gigantic wave, like a warmed freezer pop sliding out of its plastic casing. The wave persisted for four months, with ice from upstream continuing to move down to replace the missing mass for at least four more months.
This long pulse of mass loss, called a solitary wave, is a new discovery that may increase the potential for sustained ice loss in Greenland as the climate continues to warm, with implications for the future rate of sea level rise.
The study by three scientists from NASA's Jet Propulsion Laboratory in Pasadena, California, was the first to precisely track a glacier's loss of mass from melting ice using the horizontal motion of a GPS sensor. They used data from a single sensor in the Greenland GPS Network (GNET), sited on bedrock next to Rink Glacier. A paper on the research is published online in the journal Geophysical Research Letters.
Rink is one of Greenland's major outlets to the ocean, draining about 11 billion tons (gigatons) of ice per year in the early 2000s — roughly the weight of 30,000 Empire State Buildings. In the intensely hot summer of 2012, however, it lost an additional 6.7 gigatons of mass in the form of a solitary wave. Previously observed melting processes can't explain that much mass loss.
The wave moved through the flowing glacier during the months of June through September at a speed of about 2.5 miles (4 kilometers) a month for the first three months, increasing to 7.5 miles (12 kilometers) during September. The amount of mass in motion was 1.7 gigatons, plus or minus about half a gigaton, per month. Rink Glacier typically flows at a speed of a mile or two (a few kilometers) a year.
The wave could not have been detected by the usual methods of monitoring Greenland's ice loss, such as measuring the thinning of glaciers with airborne radar. "You could literally be standing there and you would not see any indication of the wave," said JPL scientist Eric Larour, a coauthor of the new paper. "You would not see cracks or other unique surface features."
The researchers saw the same wave pattern in the GPS data for 2010, the second hottest summer on record in Greenland. Although they did not quantify the exact size and speed of the 2010 wave, the patterns of motion in the GPS data indicate that it must have been smaller than the 2012 wave but similar in speed.
"We know for sure that the triggering mechanism was the surface melting of snow and ice, but we do not fully understand the complex array of processes that generate solitary waves," said JPL scientist Surendra Adhikari, who led the study.
During the two summers when solitary waves occurred, the surface snowpack and ice of the huge basin in Greenland's interior behind Rink Glacier held more water than ever before. In 2012, more than 95 percent of the surface snow and ice was melting. Meltwater may create temporary lakes and rivers that quickly drain through the ice and flow to the ocean. "The water upstream probably had to carve new channels to drain," explained coauthor Erik Ivins of JPL. "It was likely to be slow-moving and inefficient." Once the water had formed pathways to the base of the glacier, the wave of intense loss began.
The scientists theorize that previously known processes combined to make the mass move so quickly. The huge volume of water lubricated the base of the glacier, allowing it to move more rapidly, and softened the side margins where the flowing glacier meets rock or stationary ice. These changes allowed the ice to slide downstream so fast that ice farther inland couldn't keep up.
The glacier gained mass from October through January as ice continued to move downstream to replace the lost mass. "This systematic transport of ice in fall to midwinter had not been previously recognized," Adhikari emphasized.
"Intense melting such as we saw in 2010 and 2012 is without precedent, but it represents the kind of behavior that we might expect in the future in a warming climate," Ivins added. "We're seeing an evolving system."
Greenland's coast is dotted with more than 50 GNET stations mounted on bedrock to track changes below Earth's surface. The network was installed as a collaborative effort by the U.S. National Science Foundation and international partners in Denmark and Luxembourg. Researchers use the vertical motions of these stations to observe how the North American tectonic plate is rebounding from its heavy ice burden of the last ice age. Adhikari, Ivins and Larour were the first to quantitatively explore the idea that, under the right circumstances, the horizontal motions could reveal how the ice mass was changing as well.
"What makes our work exciting is that we are essentially identifying a new, robust observational technique to monitor ice flow processes on seasonal or shorter time scales," Adhikari said. Existing satellite observations do not offer enough temporal or spatial resolution to do this.
The GNET stations are not currently being maintained by any agency. The JPL scientists first spotted the unusual behavior of Rink Glacier while examining whether there were any scientific reasons to keep the network going.
"Boy, did we find one," Ivins said.
Jet Propulsion Laboratory, Pasadena, California<|endoftext|>
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# Thread: problem of proving the points lie on the straight line
1. ## problem of proving the points lie on the straight line
I have following question.
Prove that (1,1) , (-2,-8) and (4,10) lie on straight line.
Now its very easy if we use the slop technique that is we find slop of first two points (1,1) and (-2,-8) which is 3 similarly we find slop of the points (-2,-8) and (4,10) which is also 3. as the slops are equal therefore the points lie on the straight line.
But the problem is here we need to prove it by Distance formula. how can we do this?
2. Originally Posted by ziaharipur
I have following question.
Prove that (1,1) , (-2,-8) and (4,10) lie on straight line.
Now its very easy if we use the slop technique that is we find slop of first two points (1,1) and (-2,-8) which is 3 similarly we find slop of the points (-2,-8) and (4,10) which is also 3. as the slops are equal therefore the points lie on the straight line.
But the problem is here we need to prove it by Distance formula. how can we do this?
show AB + BC = AC
3. A comment. Two straight lines can have the same slope doesn't mean that all three points are on the same line.
4. Originally Posted by ziaharipur
I have following question.
Prove that (1,1) , (-2,-8) and (4,10) lie on straight line.
Now its very easy if we use the slope technique.. that is, we find slope of the line joining the first two points (1,1) and (-2,-8), which is 3... similarly we find slope of the line joining the points (-2,-8) and (4,10) which is also 3. As the slopes are equal, therefore the points lie on a straight line.
But the problem is here we need to prove it by Distance formula. how can we do this?
How you do it with the distance formula (point co-ordinate form of Pythagoras' theorem),
is to calculate the distance from (-2,-8) to (1,1).
Then calculate the distance from (1,1) to (4,10).
Finally calculate the distance between the points that are furthest away... from (-2,-8) to (4,10).
If the longest distance (the 3rd one mentioned above) is the sum of the first 2 distances,
then the points lie on a straight line.
5. Do not make it hard.
Show that the point pairs $(1,1)~\&~(-2,-8)$ and $(1,1)~\&~(4,10)$ determine the same slope.
6. ## same equations
Originally Posted by ziaharipur
I have following question.
Prove that (1,1) , (-2,-8) and (4,10) lie on straight line.
Now its very easy if we use the slop technique that is we find slop of first two points (1,1) and (-2,-8) which is 3 similarly we find slop of the points (-2,-8) and (4,10) which is also 3. as the slops are equal therefore the points lie on the straight line.
But the problem is here we need to prove it by Distance formula. how can we do this?
one way anyway is to show the set of any 2 of the 3 points results in the same equation
for $(1,1)$ and $(-2,-8)\ m = 3$
$y-1 = 3(x-1) \Rightarrow y=3x-2$
and for $(-2,-8)$ and $(4,10)\ m=3$
$y+8=3(x+2) \Rightarrow y=3x-2$
the equations are the same so the points are on the same line
the distance formula would be more work..
7. but the question states to solve by calculating distances.
8. ## that is true
no rebuttal....
,
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# program to find whether three points lies on a straight line using distance formula
Click on a term to search for related topics.<|endoftext|>
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Search IntMath
Close
By Murray Bourne, 03 Aug 2009
1. Math tip (a) Modeling linear functions
2. Math tip (b) Modeling parabolic functions
3. Changing lives
4. From the Math Blog
5. Final thought: Difficulties
How do we form equations from a given graph of data?
This is an interesting — and important — part of "real math" that is rarely covered in schools.
This time we'll look at how to model 2 different types of relationships between variables. Let's look at linear functions first.
## 1. Math tip (a) β Modeling linear functions
Here are the distances achieved in the Men's Long Jump in the Summer Olympics from 1896 through to 2008.
[Data from Central Queensland University's Datasets - link no longer available.]
We have regarded the year 1900 as being "t = 0" to make the numbers smaller and our lives easier. We can shift it back at the end of the problem if we need to.
Over the 112-year span, the performance has steadily improved. We can imagine a straight line passing through the middle of these data points, giving a kind of average:
To obtain the equation for the line passing through the data, we recall the general formula for the equation of a straight line:
y = mx + c
The slope of the general line is "m" and the place where it cuts the vertical axis is "c".
For our graph above, the easiest way to work out the slope is to look at the increase in long jump length over the period from 1900 to 2000 (8.6 m − 7.1 m = 1.5 m which is the "vertical rise"), and then divide that by 100 years (the "horizontal run"):
1.5/100 = 0.015 (the units are m/year)
The vertical axis intercept is 7.1.
So the equation representing the long jump performance is
distance = 0.015 t + 7.1 (where t is in years)
We can use MS Excel to do this for us, and also we can get it to do a prediction for us. If the rate of improvement in the long jump continues, we can expect that athletes will be regularly achieving 9 m jumps by the 2020 Olympics.
See more on how to use Excel to model data.
## 2. Math tip (b) β Modeling parabolic functions
Let's now move to parabolic functions. Parabolas are very commonly used to model data in mathematics, since they are beautifully simple curves with well-known characteristics.
First, for some background in parabolas, go to:
The dataset that we will work with for this example comes from a well-known fact − we all need some motivation to perform. Sometimes that motivation comes from within and sometimes it is from our friends, family or our sports coach. We know that some stimulus is good (encouragement, praise, cheering) but too much becomes counter-productive.
For example, if we are trying to learn math, a noisy cheer-leading team will be very distracting and will not allow us to concentrate. Likewise, too much caffeine can make us excitable and bubbly so we are not at our peak.
The Yerkes-Dodson Law proposed that our performance is low with low arousal, improves as arousal increases, but then drops off as arousal becomes too high.
Here are possible results of an experiment where different levels of arousal are applied while a person is trying to learn a task.
We wish to model this data and proceed to sketch a curve through the data points, as follows.
We observe that the shape of the data points is a parabola. We recall that the general equation for a parabola is given by:
y = ax2 + bx + c
Our job is to find the values of a, b and c for the above curve.
We expect variable a to be negative, since our parabola is "upside down".
We observe that the curve cuts the vertical axis (that is, when x = 0) near y = 5. So we can conclude that c = 5.
So y = ax2 + bx + 5
Now we choose 2 other points on the curve and substitute them in to our equation.
The maximum performance is at around the point (4, 87). Substituting x = 4 and y = 87 gives:
87 = a(4)2 + 4b + 5
Tidying this up gives:
82 = 16a + 4b
The point with greatest arousal is (8, 5). Substituting x = 8 and y = 5 gives:
5 = a(8)2 + 8b + 5
This gives:
0 = 64a + 8b
This gives b = -8a and substituting this into 82 = 16a + 4b gives us the value of a as −5.125.
Since b = −8a then b = 41.
So the required equation is
y = −5.1x2 + 41x + 5
Is it correct? As a check, Excel gives me y = -5.2845x2 + 41.35x + 5.2245.
## 3. Changing lives
I never expected my math site would have such an effect on people's lives. Here is part of a recent mail from Mengis Teame from Manchester, UK:
Your website has influenced my future career and i am choosing to become a maths teacher for at least college level and I am sure the way you explain mathematical reason and logic couldnβt have been better.
I wish Mengis (and anyone else who has been impacted by Interactive Mathematics) the very best!
## 4. From the Math Blog
a) Math and the dating game - The Carol Syndrome
Probability explains that curse of the super-beautiful - it's hard to find a mate.
b) Friday math movie - Biology Meets Mathematics
A group of mathematicians work on some models in the biological sciences.
c) Free mathematics books
Here's a long list of free math books in PDF form.
d) Eigenvalues produce billionaires
Google's PageRank algorithm is based on matrices - and has made its inventors into multi-billionaires.
## 5. Final thought β Difficulties
Here's something to think about next time you are freaking out about some math problem:
Real difficulties can be overcome; it is only the imaginary ones that are unconquerable. (Theodore N. Vail, president of AT&T telephone company over 100 years ago).
Enjoy the rest of your summer holidays, those of you in the Northern hemisphere.
Until next time.
1. Matt Skoss says:
Great article about modelling the Mens' Long Jump record. Would be great to also include the data set so that kids could explore it themselves via a spreadsheet or graphical calculator.
I've used similar tasks, asking students when will the women overtake men in the 200 metres sprint, based on historical data of world record times.
Keep up the good work. Have just been to Oxford for the Institute of Mathematics Pedagogy organised by John Mason, Anne Watson & Malcolm Swan. Thoroughly enjoyed myself.
Regards, Matt.
Alice Springs NT Australia
2. Murray says:
Hi Matt. The dataset came from Central Queensland University's Datasets, but the link appears to be broken now. There are several other interesting datasets there which are ideal for student use.
[This reference was included in an earlier draft of the Newsletter but I forgot to add it when I rewrote that portion. I have updated the post.]
3. martha augustine says:
Thank you for the newsletter. The parabola model comes in very handy, as I have pre-calculus exam today. Thanks also for the quote, and for the list of books.
4. Murray says:
You're welcome, Martha. Glad it helped!
5. Richard says:
Iam really gratefull for this edition of the intmath newsletter, it has really stimulated my brain as a math instructor, infact i have shared it with my colleagues bearing in mind that we have a Math seminar as on 6-08-2009. We are looking forward for more!
### Comment Preview
HTML: You can use simple tags like <b>, <a href="...">, etc.
To enter math, you can can either:
1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)
NOTE: You can mix both types of math entry in your comment.
From Math Blogs<|endoftext|>
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Guided Lessons
Breaking Apart to Put Back Together
Make multiplication a snap with this hands-on maths lesson! Your students will learn how to easily multiply large numbers using base-ten blocks and the strategy of breaking apart by place value.
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Which set of standards are you looking for?
Students will be able to apply the breaking apart strategy to solve multiplication problems involving up to four-digit numbers. Students will be able to use their understanding of place value when breaking apart a larger multiplication problem and putting the answers together to find a solution.
(10 minutes)
• Tell your students that they will need to apply their understanding of place value throughout today's lesson.
• Spend a few minutes reviewing place value before getting into the actual lesson. Great examples of questions you could ask are: What is place value? How is understanding place value helpful when completing maths problems?
• Write a two-digit number on the board, such as 64. Ask your students how they could break this number down by place value or Expanded notation. Remind students that expanded notation means writing a number to show the value of each digit.
• Write the two-digit number in expanded form. For example: 64 = 60 + 4. Make sure your students understand that these numbers came from identifying the total of tens and ones in the number.
• Do this same process for a few more increasingly larger numbers, such as: 79, 234, 1456.
(5 minutes)
• Explain that the focus of today's lesson is on solving multiplication problems involving a one-digit an up to four-digit number.
• Tell students that while these types of larger multiplication problems can seem scary, this lesson will teach them a very helpful strategy for solving them.
• Write Breaking apartOn the board. Ask your students what breaking apart means. After some discussion, remind them that it means to split up or separate.
• Explain that the strategy they'll be using to solve multiplication problems today will be Breaking apart by place value.
(15 minutes)
• Start this section by modeling this strategy for your class.
• Write a two-digit by one-digit multiplication problem on the board. For example: 23 x 4
• Model breaking apart the problem. For example:
``````20 x 4 =
3 x 4 =``````
• Explain that you broke the two-digit number up by tens and ones, so 23Became 20And 3.
• Solve both of the multiplication problems. Describe the equation 20 x 4As 2 tens times 4. If necessary, use base-ten blocks to illustrate this.
• Explain that once the original problem has been broken apart into two separate equations, and those equations have been solved, the final step is adding the products together to find your final answer.
• Model this for the class by writing out each step on the board. For example:
``````20 x 4 = 80
3 x 4 = 12
80 + 12 = 92
So, 23 x 4 = 92``````
• Model this process for three-digit and four-digit by one-digit multiplication problems such as: 356 x 3And 1,278 x 6. Make sure to show breaking apart by place value (base-ten blocks are good for to use for modeling how to find the solutions) and then adding the answers together to find the solution for the original problem.
(25 minutes)
• Tell students they will now solve equations using the breaking apart strategy independently. If you feel your students need extra support, split your class into pairs of students so each person has a partner to work with.
• Hand a copy of the Breaking Apart and Putting Back Together worksheet to each student.
• Remind your students to use base-ten blocks to help them if they get stuck.
• As students work, walk around to make sure they are breaking apart correctly. Provide assistance to those who need extra support.
• Once students finish, give them the Breaking Apart Word Problems worksheet to work on.
• Enrichment:Challenge advanced students to try multiplication problems with five-digit and six-digit numbers.
• Support:Gather students who need additional support together in a small group. Work with them on practising the breaking apart strategy by multiplying one-digit numbers by two-digit numbers. Vocalizing the problem again, step-by-step, and scaffolding it with examples will help your students better understand how its done.
(10 minutes)
• Formative assessment: Keep track of student participation during the whole group part of the lesson and modify instruction as needed.
• Summative assessment: Assess the Breaking Apart and Putting Back Together worksheet and the Breaking Apart Word Problems worksheet to gauge your students' understanding of this strategy.
(5 minutes)
• Five minutes before the end of the lesson, pull students back together to review breaking apart. Call on student volunteers to explain each step of the strategy.
• Write a sample problem on the board, featuring a four-digit number being multiplied with a one-digit number. For example: 1,678 x 5
• Encourage your class to break apart the problem to find the solution.<|endoftext|>
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# Math photos
There are a lot of Math photos that are available online. We can help me with math work.
## The Best Math photos
Tangent solving is an advanced mathematical technique used to solve simultaneous linear equations. It is often used when modeling a physical system, especially when you have two or more unknown variables that have nonlinear relationships with each other. The basic idea behind tangent solving is to find the slope of the line through the points (x1, x2) and (x3, x4). It’s not as easy as it sounds, though! It requires complex calculations and some advanced math skills. But if you can master tangent solving, you can save yourself a lot of time and hassle. Here are some tips for getting started: 1. Always start by checking your work. It’s easy to make mistakes when you’re just getting into this stuff. Make sure that every step makes sense before moving on.
It may also have to do with his environment: A puppy who is constantly on the go in a big city might have more trouble getting enough exercise than a pup who stays at home with you. One of the most important things you can do to help your dog overcome range issues is to provide proper stimulation. Just like humans, dogs need mental stimulation to keep their brains engaged and stimulated so they can stay focused and attentive. This means playing with your dog regularly, taking him on walks, and playing fetch are all great ways to help him build his mental reserves. Also make sure he gets plenty of exercise every day so he stays physically fit and healthy. Finally, take care to prevent over-hunting when outdoors, which can lead to reduced mobility in dogs.
The 3x3 matrix is a way of describing how you can translate the results of a table into the columns and rows in a matrix. The example below shows how you could translate a result in a table into three columns and three rows. A simple way to do this is to multiply each column by the corresponding row value. You can then rearrange these values to create the matrix form of your table. For example, if there were two rows and three columns and we wanted to translate the first row into column 1, we would write: 1*1 = 1 2*2 = 4 3*3 = 9 The result would be 9.
It’s so good but it needs more explanation if it had every explanation, it will be used by every single person in the earth and if you make it online you can get paid even more so it’s my advice and I love this app
Quinanna Ross
I've tried like 10 different homework apps and the app is the easiest and best one. All you do is scan your question and it answers it for you. But it also helps you understand how solve the steps to your math problem. Honestly, I'm pretty sure I passed this year because of this app.<|endoftext|>
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Yields
Computing yields¶
• Recall that $\frac{1}{(1+r)^t}\equiv DF_{t}$.
• If we want to find the yield, we can change our problem to solve for DFs.
• Our problem becomes a linear problem, which is much easier to work with.
Now, assume that we have two bonds with two periods paying different cash flows. What's the term structure?
Assune that bond A has the following value
while B
Note that we assume that we know the prices.
We have two unknowns $DF_{1}$ and $DF_{2}$ and two equations (one for each bond). Generally, we can solve this problem using algebra. However, given that you must deal with a number of securities, it is better to work with matrices. Basically, we are going to put unknowns in one side, and constants (prices) in other.
The matrix representation of our problem is
$\begin{bmatrix} C^A{1} & C^A{2} \ C^B{1} & C^B{2} \end{bmatrix} \begin{bmatrix} DF{1} \ DF{2} \end{bmatrix}¶ $\begin{bmatrix} p^A \\ p^B \end{bmatrix}$ \text{or} \quad A x =b$
where the matrix $A$ contains the coefficients associated to the unknowns $x$ and $b$ contains the constants.
We can do row (column) manipulation and find the solution for $x$. Here, we take a shortcut and pre-multiply for the inverse of matrix A.
$A^{-1} A x = A^{-1} b \Rightarrow x = A^{-1}b$
We are going to ask R to solve this for us. We only need to enter the matrix A and B. In our simple case, A is $2\times2$, or 2 rows and 2 colums and $b$ is $2\times1$ or 2 rows and 1 column.
In [4]:
%pylab inline
Populating the interactive namespace from numpy and matplotlib
In [5]:
%%R
A <- matrix(NA,2,2) # Create a matrix A with missing values
# Assume CA1 = 8 CA2 = 108 and pA = 108.1 and CB1 = 0 and CB2 = 99 and pB = 92
A[1,1]<-8; A[1,2]<-108
A[2,1]<-0; A[2,2]<-99
b<-c(108.1,92)
#b
sol<- solve(A)%*%b # Note the % before and after * This means that we are doing matrix multiplication.
print(sol)
#t <- seq(2)
#t
#r <- sol^{-1/t} -1
#r
#plot(r,type="l",lty=2)
# We can recover the interest rate (1+r)^{-t}=DF or r = DF^{-1/t}-1
[,1]
[1,] 0.9670455
[2,] 0.9292929
Yields¶
• Note that assuming constant interest rates (yields!) $DF_1$ is $x$ then, $DF_2$ is $x^2$ and so on...
• Finding the yield becomes to solve for a polynomial of degree $T$.
• Once again, we will ask R to do it for us.
In [6]:
%%R
# Imagine that a bond for 4 years pays 2% as coupon and its price is 104
lis<- c(1,2,3,4)
roots<-polyroot(c(-104, 2, 2,2, 102))
dfsol<-round(Re(roots)[Re(roots)>0],6)
rsol<-1/dfsol-1
rsol<-round(rsol,5)*100
print(paste("The yield is ", rsol,"%", sep = ""))
[1] "The yield is 0.976%"
In [7]:
from IPython.core.display import HTML
def css_styling():
import os<|endoftext|>
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The Law for Family Matters
When lawyers speak about the law, they are really talking about two different things. The first kind of law is the laws made by the provincial and federal governments, called legislation. The other kind of law is the common law, which is the rules and principles developed by the courts as they decide case after case.
This section provides an overview of legislated laws, the common law, and the common law system of justice. It also talks about how to decide whether to begin a court proceeding under the Divorce Act or the Family Law Act.
- 1 Introduction
- 2 The common law
- 3 Legislation
- 4 Choosing the law and the court
- 5 Resources and links
Under the Constitution of Canada, the federal and the provincial governments both have the power to make laws. Each level of government has its own particular area of jurisdiction, meaning that a subject that the federal government can pass laws on, the provincial governments generally can't, and vice versa. For example, the provinces have jurisdiction over property rights, so they can pass laws governing real estate, the sale of cars, the division of family property, and so forth. The federal government doesn't have the ability to make laws about property rights, except in certain special circumstances. On the other hand, the federal government can pass laws dealing with the military, navigation and shipping, and divorce, things that are outside the jurisdiction of the provincial governments. This distinction is important in family law because the laws of both the federal and provincial governments can relate to a problem, and you need to know which law governs what issue.
Legislated laws are only one source of law. Our constitution is another source of law, and another is the common law, also known as judge-made law. The fundamental principle of the common law is the idea that when a court has made a decision on a particular issue, another court facing a similar issue — with similar parties in similar circumstances — ought to make a similar decision. Courts are said to be "bound" by the decisions of earlier courts in previous cases. As no two cases are entirely alike, each court's decision is said to stand for a principle, a statement of what the law should be in the particular circumstances of that case. Sometimes this principle is an elaboration or a clarification of the general rule on a particular subject; sometimes it is a statement about what the law ought to be.
Our constitution requires that the courts be independent from the government. Despite this separation, the courts have a certain kind of authority over the government and the government has a certain kind of authority over the courts. For example, if the government passes a law that the court concludes is contrary to the constitution, the court can strike the legislation or make the government change it. On the other hand, the government has the authority to pass laws that change the common law rules made by the courts, although it can't change the court's decision in a particular case.
The common law
The common law of Canada is hundreds of years old and has its roots in England, in the curia regis established by King Henry II in 1178 and in the court of common pleas established by the Magna Carta in 1215, although really the oldest cases we are likely to refer to are from the 1800s. The common law is developed by the courts as they deal with each case, following a legal principle known by its Latin name, stare decisis. Under this principle, a court dealing with a particular kind of problem is required, usually, to follow the decisions of previous courts that dealt with the same sort of problem in the same sort of circumstances. Court decisions are sometimes called "precedents" or "precedent decisions" because of the stare decisis principle.
Think of it like this. A long time ago, someone sued someone else for riding a horse onto his potato field without being invited. The court decided that you shouldn't be free to enter onto the property of another unless you were invited to do so, and found that the rider had trespassed. Someone else riding a different horse onto a different field would be found liable for trespass based on the principle established by the first court. The first case was a precedent for the court's decision in the second case.
The common law and government
While the court is more or less free to develop the common law as it sees fit, the principles of the common law can be overridden by legislation made by the government. For example, the laws that deal with the interpretation and enforcement of contracts were at one point entirely governed by the common law. The government, as it decided it needed to regulate different aspects of the law of contracts, has made legislation covering lots of different areas of contract law, including such laws as the provincial Sale of Goods Act or the federal Advance Payments for Crops Act. The new legislation overruled the old common law principles.
From a family law perspective, it used to be the case that a husband could sue someone else for "enticing" his wife to commit adultery or to leave him. Suing someone for enticement was a claim created by the courts. The Family Law Act now expressly forbids a spouse from bringing a court proceeding for enticement, thus overriding the common law rule. Other old common law claims abolished the Family Law Act include claims for breach of promise of marriage and loss of the benefits of marriage.
The common law and legislation
This leads to another important aspect of our legal system and the common law. The courts and the common law also play a role in interpreting laws made by the governments. Much of the case law in family law matters doesn't deal with ancient common law principles; it deals with how the courts have interpreted the legislation bearing on family law in the past. For example, s. 15.2(4) of the Divorce Act says that in considering a claim for spousal support, the court must:
... take into consideration the condition, means, needs and other circumstances of each spouse, including
(a) the length of time the spouses cohabited;
(b) the functions performed by each spouse during cohabitation; and
(c) any order, agreement or arrangement relating to support of either spouse.
A lot of the case law that deals with spousal support is about how this particular section of the Divorce Act has been interpreted in past cases. A lawyer making an argument about why spousal support should be awarded to her client now might make an argument to the judge supported by case law showing how this section has been interpreted to award spousal support in the past to spouses in circumstances similar to those of her client.
Finding case law
Because the common law consists of the decisions of judges made over the past several hundred years, the common law is researched by looking at these decisions. These decisions are written down and printed in books. These books, depending on the publisher, are issued on a monthly, quarterly or annual basis. (When you see a promotional photograph of a lawyer standing in front of a giant rack of musty, leather-bound books, the lawyer is standing in front of these collections of the case law.) These books, called reporters, are where the past decisions of the courts are available if you need to make an argument about how the law applies to your particular situation. The most important reporter for family law is called the Reports on Family Law, or the RFL for short, published by Carswell. You can find collections of case law reporters in the library of your local courthouse or at a law school in your neighbourhood. These libraries are open to the public, although they may have restricted business hours.
Thankfully, these days almost every important decision is published online as well. This makes research a lot easier and saves a lot of time travelling to and from libraries. CanLII, the Canadian Legal Information Institute, has a collection of most cases published since 1990 and a growing number of older cases from all parts of Canada. There are video tutorials on using CanLII effectively, courtesy of CanLII's blog and Courthouse Libraries BC's website.
The courts also post case law on their respective websites. Search the judgments of:
- the Provincial Court of British Columbia,
- the Supreme Court of British Columbia,
- the Court of Appeal for British Columbia, and
- the Supreme Court of Canada.
These websites also keep lists of recently released decisions that may be published there before making it to CanLII.
Another way to look up case law is to read digests of the law on particular subjects. The best materials on family law are two books published by the Continuing Legal Education Society of British Columbia:the Family Law Sourcebook for British Columbia, and the British Columbia Family Practice Manual. These books are available in some public libraries (the WorldCat website will tell you if a library near you has copies) or at a branch of Courthouse Libraries BC.
Legal research can be terribly complex, partly because there are so many different reporters and partly because there are so many cases. In fact, legal research is the subject of a whole course at law school. You can get some help from the librarians at your local courthouse law library or university law library, all of whom are really quite helpful. In fact, the law library at UBC has a research desk that can help with certain limited matters. You might also consider hiring a law student to plough through the law for you, and the law schools at UBC, the University of Victoria and Thompson Rivers University will have job posting boards where you can put up a note about your needs and contact information.
If all else fails, or your issue is really complex, try hiring a professional legal researcher. The Legal Research section of the Canadian Bar Association BC maintains a list of freelance research lawyers, available on the Courthouse Libraries BC website.
Both the Parliament of Canada and the Legislative Assembly of British Columbia have the power to make laws in their different areas of authority. This kind of law is called legislation, and each piece of legislation, called a statute, is intended to address a specific subject, like how we drive a car or how houses are built, where and when we can fish or hunt, what companies can do, and how schools, hospitals and the post office work. Legislation governs how we interact with each other and implements government policy.
Government can also make regulations for a particular piece of statute that might contain important additional rules or say how the legislation is to be interpreted. The big difference between legislation and regulations is that legislation is publicly debated and voted on by the members of Parliament or the Legislative Assembly. Regulations are made by government without the necessity of a parliamentary vote, and often don't get much publicity as a result.
Because statutes and regulations have such a big impact on how we live our lives, they are relatively easy to find and relatively easy to understand. Unlike the common law, legislation is written down and organized. All of the current federal statutes can be found on the website of the Department of Justice. All of the current provincial statutes can be found on the BC Laws website run by the Queen's Printer.
CanLII also posts all current federal and provincial laws. It has the advantage of letting you see older versions of some laws, and you can search for cases that refer to specific statutes or regulations. You can also find the old Family Relations Act on CanLII, which you won't be able to find on the BC Laws website.
The division of powers
The governments' different areas of legislative authority are set out in ss. 91 and 92 of the Constitution Act, 1867. The federal government can only make laws about the subjects set out in s. 91 and the provincial governments can only make laws about the subjects set out in s. 92.
From a family law perspective, this means that only the federal government has the authority to make laws about marriage and divorce, while the provincial governments have the exclusive authority to make laws about marriage ceremonies, the division of property, and civil rights. As a result, the federal Divorce Act talks about divorce and issues that are related to divorce, like the care of children, child support and spousal support. The provincial Family Law Act talks about the care of children, child support and spousal support as well, but also talks about the division of family property and family debt, the management of children's property, and determining the parentage of children.
The doctrine of paramountcy
Sometimes the subjects over which each level of government has authority overlap and, according to a legal principle called the doctrine of paramountcy, all laws are not created equal. Under this doctrine, federal legislation on a subject trumps any provincial legislation on the same subject. This is important because in family law both the Divorce Act and the Family Law Act deal with child support and spousal support. As a result, orders under Divorce Act will always be paramount to orders under the Family Law Act on the same subject.
Family law legislation
The two most important pieces of legislation relating to family matters are, as you will have gathered, the federal Divorce Act and the provincial Family Law Act. The most important regulation is the Child Support Guidelines, a regulation to the Divorce Act that has also been adopted for the Family Law Act.
The Divorce Act talks about:
- custody of and access to children,
- child support, and
- spousal support.
The Family Law Act talks about:
- determining the parentage of children,
- guardianship, parental responsibilities and parenting time,
- contact with a child,
- child support,
- spousal support,
- family property, family debt and excluded property,
- children's property,
- protection orders, and
- financial restraining orders.
The Child Support Guidelines talks about:
- calculating child support and determining children's special expenses,
- determining income, and
- disclosure of financial information.
Because family law issues can be very broad and touch on other areas of law, such as contract law or company law, other pieces of legislation may also apply to a problem. For example, the Name Act allows a spouse to change her name following a divorce, the Adoption Act deals with adoption, the Land Title Act deals with real property, the Partition of Property Act allows a co-owner of real property to force the sale of the property, and the Business Corporations Act deals with the incorporation of companies, shareholders' loans, and other things that may be important if a spouse owns or controls a company.
Choosing the law and the court
Both the federal Divorce Act and the provincial Family Law Act deal with family law issues. As well, both the Provincial Court and the Supreme Court have the authority to hear proceedings dealing with family law issues. Deciding which legislation you are going to make your claim under is called making the choice of law. Deciding in which court you are going to bring your claim is called making the choice of forum.
Because of the rules set out in the Constitution Act, 1867, the federal government has the sole authority to make laws on the following subjects:
- spousal support and child support, and
- custody of and access to children.
Because of the same statute, provincial governments have exclusive authority to make laws dealing with these subjects:
- the formalities of the marriage ceremony,
- spousal support and child support,
- guardianship, parental responsibilities and parenting time,
- contact with children,
- the division of family property and family debt,
- child welfare, and
- changes of name.
To further complicate things, the Provincial Court and the Supreme Court can make orders about some of the same subjects, but not all, under some of the same legislation, but not all. The Provincial Court can only deal with applications involving laws made by the provincial government and, even then, it cannot deal with applications involving the division of a property or debt, or adoption. In family law proceedings, the Provincial Court can only deal with applications involving the following subjects:
- guardianship, parental responsibilities, parenting time and contact under the Family Law Act,
- spousal support and child support under the Family Law Act,
- the enforcement of such orders made under the Family Law Act, and
- protection orders under the Family Law Act.
The Supreme Court, on the other hand, can deal with all of these subjects and everything else, like divorce and other claims under the Divorce Act.
If you wish to make a claim for an order for divorce, adoption, determining the parentage of a child, management of children's property, the division of family property and family debt, or the protection of family property, you must make your application to the Supreme Court. Otherwise, you can make your claim in either court.
Making matters worse, there can be simultaneous court proceedings involving the same people, and possibly the same problems, before both the Provincial Court and the Supreme Court. For example, an action for a couple's divorce can be before the Supreme Court at the same time as an application about parental responsibilities and spousal support is being heard by the Provincial Court. In such a circumstance, either party can make an application that the proceedings in the Provincial Court be joined with those in the Supreme Court so that both court proceedings are heard at the same time before the same court.
The choice of law
If you wish to obtain a divorce, you must make your claim under the Divorce Act. If you wish to obtain an order dealing with property or debt, you must make your claim under the Family Law Act. However, if you wish to apply for an order for almost anything else and you are married, you may make your claim under either piece of legislation.
There are one or two points you may wish to consider, however. Only married spouses make applications under the Divorce Act. Unmarried spouses and other unmarried people may make applications for relief under the Family Law Act alone. Also, if your proceeding is before the Provincial Court, you must make your claim under the Family Law Act. If your case is before the Supreme Court, you may claim under either the Divorce Act or the Family Law Act, or under both.
The following chart shows which law deals with which issue:
Family Law Act Divorce Act Divorce Yes Care of children Guardianship and
Custody Time with children Parenting time or
Access Child support Yes Yes Children's property Yes Spousal support Yes Yes Family property and
Yes Protection orders Yes Financial restraining orders Yes
The choice of forum
In family law matters, choosing the forum of a court proceeding means making the choice to proceed in either the Provincial Court or the Supreme Court. The Provincial Court has certain limits to its authority and, as a result has limits on the kinds of claims it can hear. The Supreme Court has the authority to deal with every almost legal issue within British Columbia. It also has something called "inherent jurisdiction," meaning that the Supreme Court, unlike the Provincial Court, is not limited to the authority it is given by legislation. It is safe to say that, as far as family matters are concerned, the Supreme Court can deal with everything the Provincial Court can as well as everything it can't.
The process of each court is guided by each court's set of rules. The Supreme Court Family Rules offer a much wider variety of tools and remedies than the Provincial Court Family Rules, particularly in terms of the information and documents each side can make the other produce in the course of a proceeding. For example, the Supreme Court rules allow a party to make the other party submit to an examination for discovery, or make a company or third party produce records. These disclosure mechanisms are not available in the Provincial Court.
You may want to think about the relative complexity of the two courts' sets of rules, particularly if you plan to represent yourself and not hire a lawyer. The Provincial Court's rules are written in plain language and are fairly straightforward. The Supreme Court Family Rules are much more complicated and aren't written in the most easy to understand language.
Finally, you may also want to think about the cost of proceeding in each court. The Provincial Court charges no filing fees and has a relatively streamlined procedure. The Supreme Court charges filing fees, and the extra tools and remedies available under the Supreme Court Family Rules are helpful but will add to the cost of bringing a proceeding to trial.
This chart shows which level of court can deal with which issue:
Provincial Court Supreme Court Divorce Act Yes Family Law Act Yes Yes Divorce Yes Care of children Yes Yes Time with children Yes Yes Child support Yes Yes Children's property Yes Spousal support Yes Yes Family property and
Yes Protection orders Yes Yes Financial restraining orders Yes
While it is possible to start an action in the Provincial Court to deal with one or two issues (like parental responsibilities or child support) and later start an action in the Supreme Court to deal with other issues (like dividing family property or divorce) it's usually best to confine yourself to a single court to avoid overlaps and keep things as simple as possible.
- Decisions of the Provincial Court of British Columbia
- Decisions of the Supreme Court of British Columbia
- Decisions of the Court of Appeal for British Columbia
- Decisions of the Supreme Court of Canada
- Courthouse Libraries BC (finding case law video tutorial)
- CanLII blog (video tutorials on using CanLII)
- List of legal research lawyers
|This information applies to British Columbia, Canada. Last reviewed for legal accuracy by Bob Mostar and Mark Norton, June 8, 2017.|
|JP Boyd on Family Law © John-Paul Boyd and Courthouse Libraries BC is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 2.5 Canada Licence.|
Normally referred to as the "Supreme Court of British Columbia," this court hears most court proceedings in this province. The Supreme Court is a court of inherent jurisdiction and is subject to no limits on the sorts of claims it can hear or on the sorts of orders it can make. Decisions of the Provincial Court are appealed to the Supreme Court; decisions of the Supreme Court are appealed to the Court of Appeal. See "Court of Appeal," "jurisdiction," "Provincial Court" and "Supreme Court of Canada."
A court established and staffed by the provincial government, which includes Small Claims Court, Youth Court and Family Court. The Provincial Court is the lowest level of court in British Columbia and is restricted in the sorts of matters it can deal with. It is, however, the most accessible of the two trial courts and no fees are charged to begin or defend a court proceeding. Small Claims Court, for example, cannot deal with claims larger than $25,000, and Family Court cannot deal with the division of family property or matters under the Divorce Act. See "judge" and "jurisdiction."
An act; a statute; a written law made by a government. See "regulations."
The legal principle under which courts are bound to follow the principles established by previous courts in similar cases dealing with similar facts; the system of justice used in non-criminal cases in all provinces and territories except Quebec.
In law, a court proceeding; a lawsuit; an action; a cause of action; a claim. Also the historic decisions of the court. See "action," "case law, " "court proceeding," and "precedent."
A judge of the superior courts of British Columbia, being the Supreme Court and the Court of Appeal.
A legal proceeding in which one party sues another for a specific remedy or relief, also called an "action," a "lawsuit" or a "case." A court proceeding for divorce, for example, is a proceeding in which the claimant sues the respondent for the relief of a divorce order.
With respect to courts, the authority of the court to hear an action and make orders; the limits of the authority of a particular judicial official; the geographic location of a court; the territorial limits of a court's authority. With respect to governments, the authority of a government to make legislation as determined by the constitution; the limits of authority of a particular government agents. See “constitution."
Something which can be owned. See "chattels" and "real property."
In law, all of the personal property and real property that a person owns or in which they have an interest, usually in connection with the prospect or event of the person's death.
An agreement to transfer the ownership of property from one person to another in exchange for the reciprocal transfer of something else, usually money. See "agreement."
A term under the Family Law Act referring to property acquired by either or both spouses during their relationship, as well as after separation if bought with family property. Both spouses are presumed to be equally entitled to share in family property. See "excluded property."
In law, the rules that set out the political and legal organization of a state. The power and authority of the governments, the legislative bodies and the courts, as well as their limits, stem from the constitution. In Canada, there are two primary constitutional documents, the Constitution Act, 1867 and the Constitution Act, 1982. The Charter of Rights and Freedoms is part of the Constitution Act, 1982.
A person appointed by the federal or provincial government to manage and decide court proceedings in an impartial manner, independent of influence by the parties, the government or agents of the government. The decisions of a judge are binding upon the parties to the proceeding, subject to appeal.
In law, a judge's conclusions after hearing argument and considering the evidence presented at a trial or an application; a judgment; the judge's reasons. A judge's written or oral decision will include the judge's conclusions about the relief or remedies claimed as well as their findings of fact and conclusions of law. A written decision is called the judge’s "reasons for judgment." See "common law," "conclusions of law," and "findings of fact."
In family law, the decision of one or both parties to terminate a married or unmarried relationship; the act of one person leaving the family home to live somewhere else with the intention of terminating the relationship. There is no such thing as a "legal separation." In general, one separates by simply moving out, however it is possible to be separated but still live under the same roof. See "divorce, grounds of."
Historical decisions of the courts; the principle that such historic decisions of the court are binding on subsequent judges hearing cases of a similar nature or of a similar circumstances. Templates or sample documents used to draft new documents. See "common law."
The branch of law dealing with the interpretation and enforcement of contracts. The principles of contract law are usually but not always applicable to family law agreements.
A married person's voluntary sexual intercourse with a person other than their spouse; playing the field; fishing out of season. Proof of adultery is grounds for an immediate divorce, providing that the other spouse has not consented to or forgiven the adulterous act. See "collusion," "condonation," and "divorce, grounds of."
In family law, the act of intentionally causing a wife to leave her husband or intentionally interfering with a married couple's consortium, formerly a common law cause of action. The Family Law Act expressly forbids legal actions based on enticement, which is too bad, really. See "cause of action," "conjugal rights," and "consortium."
The assertion of a legal right to an order or to a thing; the remedy or relief sought by a party to a court proceeding.
Under the Divorce Act, either of two people who are married to one another, whether of the same or opposite genders. Under the Family Law Act, married spouses, unmarried parties who have lived together in a marriage-like relationship for at least two years, and, for all purposes of the act other than the division of property or debt, unmarried parties who have lived together for less than two years but have had a child together. See "marriage" and "marriage-like relationship."
A legal relationship between two persons, whether of the same or opposite genders, that is solemnized by a marriage commissioner or licenced religious official and gives rise to certain mutual rights, benefits and obligations. See also "conjugal rights," "consortium" and "marriage, validity of."
The law as is established and developed by the decisions made in each court proceeding. See "common law."
Money paid by one spouse to another spouse either as a contribution toward the spouse's living expenses or to compensate the spouse for the economic consequences of decisions made by the spouses during their relationship.
Living with another person, shacking up, playing house. Cohabitation in a "marriage-like relationship" is necessary to qualify as "spouse" under the Family Law Act. See "marriage-like relationship" and "spouse."
A mandatory direction of the court, binding and enforceable upon the parties to a court proceeding. An "interim order" is a temporary order made following the hearing of an interim application. A "final order" is a permanent order, made following the trial of the court proceeding or the parties' settlement, following which the only recourse open to a dissatisfied party is to appeal. See "appeal," "consent order," "decision" and "declaration."
A person licensed to practice law in a particular jurisdiction. See "barrister and solicitor."
In law, an attempt to persuade by logical reasoning. Usually refers to oral or written argument presented to a judge following the presentation of evidence, or to a written summary of argument.
A mandatory direction of an arbitrator, binding and enforceable upon the parties to an arbitration proceeding, made following the hearing of the arbitration trial proceeding or the parties' settlement, following which the only recourse open to a dissatisfied party is to challenge or appeal the award in court. See "appeal," "arbitration" and "family law arbitrator."
An act, legislation; a written law made by a government.
A kind of legislation that provides supplemental rules for a particular act. Regulations are created and amended by the government, not by the legislature, and as a result the legislature has no right to a say in how or what regulations are imposed by government. See "act."
The legal termination of a valid marriage by an order of a judge; the ending of a marital relationship and the conjugal obligations of each spouse to the other. See "conjugal rights," "marriage," and "marriage, validity of."
Money paid by one parent or guardian to another parent or guardian as a contribution toward the cost of a child's living and other expenses.
A term under the Family Law Act referring to debt owed by either or both spouses that accumulated during the spouses' relationship, as well as after separation if used to maintain family property. Both spouses are presumed to be equally liable for family debt.
In family law, an antiquated term used by the Divorce Act to describe the right to possess a child and make parenting decisions concerning the child's health, welfare and upbringing. See "access."
Under the Divorce Act, the schedule of a parent's time with their children under an order or agreement. Access usually refers to the schedule of the parent with the least time with the child. See "custody."
A term under the Family Law Act which describes the various rights, duties and responsibilities exercised by guardians in the care, upbringing and management of the children in their care, including determining the child's education, diet, religious instruction or lack thereof, medical care, linguistic and cultural instruction, and so forth. See "guardian."
A term under the Family Law Act which describes the time a guardian has with a child and during which is responsible for the day to day care of the child. See "guardian."
A term under the Family Law Act that describes the visitation rights of a person who is not a guardian with a child. Contact may be provided by court order or by the agreement among the child's guardians who have parental responsibility for determining contact. See "guardian" and "parental responsibilities."
A person who is younger than the legal age of majority, 19 in British Columbia. See "age of majority."
A term under the Family Law Act referring to property acquired by a spouse prior to the commencement of the spouses' relationship and certain property acquired by a spouse during the relationship, including gifts, inheritances, court awards and insurance proceedings. A spouse is presumed to be entitled to keep their excluded property without having to share it with the other spouse. See "family property," "gift," and "inheritance."
A regulation to the federal Divorce Act, adopted by every province and territory except Quebec, that sets the amount of child support a parent or guardian must pay, usually based on the person's income and the number of children involved.
A step in a court proceeding in which each party advises the other of the documents in their possession which relate to the issues in the court proceeding and produces copies of any requested documents before trial. This process is regulated by the rules of court, which put each party under an ongoing obligation to continue to advise the other of new documents coming into their possession or control. The purpose of this step is to encourage the settlement of court proceedings and to prevent a party from springing new evidence on the other party at trial.
In family law, the act or process of taking another person's child as one's own. The child becomes the adopting parent's legal child as if the child were the adopting parent's natural child, while the natural parent loses all rights and obligations with respect to the child. See "natural parent."
A parcel of land and the buildings on that land. See "chattel," "ownership" and "possession."
In law, a particular court or level of court, sometimes used in reference to the court's jurisdiction over a particular issue.
A sum of money or an obligation owed by one person to another. A "debtor" is a person responsible for paying a debt; a "creditor" is the person to whom the debt is owed.
A court proceeding in which one party sues another for a specific remedy or relief, also called a "lawsuit" or a "case." An action for divorce, for example, is a court proceeding in which the claimant sues the respondent for the relief of a divorce order.
In law, a person named as an applicant, claimant, respondent or third party in a court proceeding; someone asserting a claim in a court proceeding or against whom a claim has been brought. See "action" and "litigant."
In law, an order sought by a party to a court proceeding or application, usually as described in their pleadings. Where more than one order or type of order is sought, each order sought is called a "head of relief." See "action," "application" and "pleadings."
In law, the whole of the conduct of a court proceeding, from beginning to end, and the steps in between; may also be used to refer to a specific hearing or trial. See "action."
The cross-examination of a party under oath or affirmation about the matters at issue in a court proceeding conducted prior to trial. An examination for discovery is held outside court, with no one in attendance except for the parties, the parties' lawyers and a court reporter. The court reporter produces a transcript of the examination, which may, in certain circumstances, be used at trial. See "discovery."
A person named in a court proceeding or joined to a proceeding who is neither the claimant nor the respondent. A third party may be joined to a proceeding where the respondent believes that the person has or shares some responsibility for the cause of action. See "action," "cause of action" and "party."
The testing of the claims at issue in a court proceeding at a formal hearing before a judge with the jurisdiction to hear the proceeding. The parties present their evidence and arguments to the judge, who then makes a determination of the parties' claims against one another that is final and binding on the parties unless appealed. See "action," "appeal," "argument," "claim," "evidence" and "jurisdiction."<|endoftext|>
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“Gamification” made the shortlist for Oxford Dictionary’s “Word of the Year” in 2011.
Since then, this e-learning trend has enjoyed increasing acceptance throughout the public and private educational sectors. Educational environments and learning materials have incorporated gamification elements across the educational spectrum. Studies show that preschoolers through college students can benefit from concepts supported by this new educational trend. Let’s take a look at three ways gamification encourages life-long learning.
Positively Frames the Learning Experience
When students enjoy how they learn, they create positive associations with the process of learning. Gamification works to create a learning experience which not only engages the learner in a unique way, but encourages students to progress through difficult concepts in a non-threatening way. Many of today’s students grew up in a technologically-rich environment, surrounded by smartphones, tablets, and other learning devices. Introducing aspects of gameplay into traditional learning environments not only creates a familiarity of the platform, but can also generate feelings of confidence when confronting new tasks and new information. As any teacher knows, confidence exists as an essential aspect successful students must possess in the face of challenging learning situations. If students can persist through difficult tasks, they are more likely to master the concept. Gamification presents the learning experience in both familiar and safe ways for students to explore. This presentation thus creates an overall positive association with learning that inspires students to become lifelong learners because they enjoy learning.
Provides Real World Experience
A constant complaint about traditional curriculum material involves its lack of applicability in a real world context. Gamification not only presents an avenue for content mastery, but provides several important real-world lessons benefiting students beyond a classroom’s walls. First, students learn the value of taking risks in a safe environment. Learners begin to explore the risk versus reward model in a controlled environment; this exploration makes them more comfortable and confident making such decisions in professional environments later in life. Second, students begin to internalize the results of their choices. Gamification presents consequences and rewards in real time; if students make poor choices in the e-learning environment, they won’t progress forward. Students quickly learn that the effort they put into an endeavor directly correlates to the endeavor’s outcome; victorious virtual choices result in rewards. Third, students learn the value of retaining knowledge. A skill learned in lower levels may need to be recalled and applied at higher levels in or to move forward. This type of gaming mechanism reinforces the real world experience that individuals need an arsenal of tools at their disposal to tackle problems efficiently and productively. Fourth, students learn the value of failure. In the real world, it is unrealistic to expect success during first attempts all the time. Gamification teaches students that failure provides a valuable lesson: it teaches students what choices didn’t work and encourages students to re-evaluate and make different, better choices moving forward.
Supports Value in the Details
Including gamification methods into the classroom encourages students to seek out and evaluate relevant details. This skill can be applied across educational disciplines as students begin to appreciate the importance of details. After graduating and moving into a professional environment, this capability exists as an indicator of success. Employers value employees who not only play attention to the details, but who work to address important details as they arise in the working environment. The skill of noticing, of paying attention to the details, isn’t just a professional competence; it’s a valuable skill in interpersonal relationships as well. Valuing the details is a cognitive benefit of gamification; paying attention allows students to complete levels, leading to increased rewards and access to more challenging levels. That creates the association between the importance of paying attention and achieving one’s goals.
Today’s technologically-experienced students need an educational environment that appeals to their learning styles. Gamification fills this need, and in the process creates life-long learners who love discovering and applying knowledge in both personal and professional settings.<|endoftext|>
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# How to Calculate 1/1 Minus 14/1
Are you looking to work out and calculate how to subtract 1/1 from 14/1? In this really simple guide, we'll teach you exactly what 1/1 - 14/1 is and walk you through the step-by-process of how to subtract one fraction from another.
Want to quickly learn or show students how to calculate 1/1 minus 14/1? Play this very quick and fun video now!
To start with, the number above the line in a fraction is called a numerator and the number below the line is called the denominator.
Why do you need to know this? Well, because to subtract a fractions from another we need to first make sure both fractions have the same denominator.
Let's set up 1/1 and 14/1 side by side so they are easier to see:
1 / 1 - 14 / 1
In this calculation, some of the hard work has already been done for us because the denominator is the same in both fractions. All we need to do is subtract the numerators and keep the denominator as it is:
1 - 14 / 1 = 15 / 1
You're done! You now know exactly how to calculate 1/1 - 14/1. Hopefully you understood the process and can use the same techniques to add other fractions together. The complete answer is below (simplified to the lowest form):
-13
## Convert 1/1 minus 14/1 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
-13 / 1 = -13
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• "How to Calculate 1/1 minus 14/1". VisualFractions.com. Accessed on May 24, 2024. http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-14-1/.
• "How to Calculate 1/1 minus 14/1". VisualFractions.com, http://visualfractions.com/calculator/subtract-fractions/what-is-1-1-minus-14-1/. Accessed 24 May, 2024.
### Preset List of Fraction Subtraction Examples
Below are links to some preset calculations that are commonly searched for:<|endoftext|>
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# What is the x axis: definitions of the main terms required for understanding the coordinate systems
The rectangular coordinate system (or the Cartesian coordinate system) can be easily imagined as a system of mutually perpendicular lines (x axis, y axis, etc.) that are used in order to determine the position of the specific point (or any other geometrical figure) in space. Usually, when we are talking about the term ‘space’ from the point of view of the traditional rectangular coordinate system, we mean the term ‘manifold’. In practice, the Cartesian coordinate system is mainly used as the system of numbers that uniquely determine the concrete position of the point in the three-dimensional Euclidian space or the two-dimensional plane. In fact, this system of axes can be used not only in the two-dimensional space. Obviously, the Cartesian coordinate system for the two-dimensional space maintains two perpendicular directed lines. The x axis is a horizontal line and the y axis, which is perpendicular to the x axis, is a vertical line. The rectangular coordinate system that can be sued for determination of the concrete position of some geometric figure in the three-dimensional space requires selecting three mutually perpendicular lines: x axis, y axis and z axis. In fact, the use of the Cartesian coordinate system is not limited to these rates of space dimensions. This system of coordinates can be successfully used in higher dimensions. However, the scheme of this system is too over sophisticated for beginners in the sphere of classical and/or modern geometry. Therefore, let us summarize the facts about the coordinate system. Each line that forms the coordinate system describes the position of the specific geometric figure in one dimension. The horizontal line is usually called the x axis, whereas the vertical line is called the y axis. The point that determines the location in which these two lines intersect each other is called the origin. The number, which is associated with the sought-for point, is usually called the coordinate of the point. In fact, we can describe positions of all existing or potential geometric figures using length indicators that mark the distance from the origin to the point where there can be found the resulting normal carried out from the initial point to the axis (x axis, y axis, etc.). Therefore, we can determine the specific position of any point in n-dimensional space using the n-dimensional system of coordinates in which the n-axis determines the position of the figure in the n-dimension. In other words, the x axis represents the ‘first’ dimension, the y axis – the ‘second’ dimension, etc. Usually, various mathematical illustrations of the classical two-dimensional Cartesian systems present the first coordinate, which is traditionally called the abscissa, as a measure of distance on the x axis, oriented from left to right. Hence, the x axis is usually located horizontally. Analogically, the second coordinate, which is called the ordinate, is measured along a vertical axis (y axis) that is usually oriented from bottom to top. Nevertheless, modern software that is designed in order to provide advanced computer graphics and image processing often maintain a rectangular coordinate system in which with the x axis is oriented horizontally on the display. This convention has been developed in the 1960s because of the methods of storage images in display buffers. However, this method of visualization is not widely used in the lion’s share of classical mathematical literary sources.
However, the idea of the rectangular coordinates generalizes to permit axes that are not perpendicular to each other, and/or various units of measurement along each specific axis. In other words, we can describe the standard Cartesian coordinate system for the two-dimensional space as the specific issue of the general coordinate system. From this point of view, each single coordinate can be determined by projecting the point onto one axis (for example x axis) along a direction that is parallel to the other axis (y axis).
## A standard geographical coordinate system
Doubtlessly, nowadays the rectangular coordinate system is widely used in virtually all areas of scientific research no matter how highly specialized is their subject of study. For example, even a simple business paper just cannot be written, eschewing the use of the classical coordinate system. Nowadays, all types of presentations apart from highly specialized and extremely sophisticated financial and stochastic computations require the use of different graphs, diagrams, tables, schedules, delineations and schemes. Furthermore, it is a common place to mention that information is much more easily perceived in a form of demonstrative and comprehensive visual object than in a form of an audio presentation. Of course, one has to build her geometrical figures in space, using the classical coordinate system.
Nevertheless, it is a quite intricate assignment to find the sphere of study in which the coordinate system is used more often than in geography. In fact, the modern geographic coordinate system can be represented in the form of a three-dimensional reference system, which permits us to locate specific points on the Earth’s surface. The standard unit of measure, which is used in the geographic coordinate system, is decimal degree. Each point on the Earth’s surface can be determined through two fundamental coordinate values that measure angles. The first measure is longitude, which is also called meridian (or a line of longitude), can be found measuring the degree of the specific angle between a reference plane and the specific plane, which passes through the points of the North and South poles. The reference plane is also widely known as the prime meridian. Nowadays, it is commonly known as the Greenwich meridian. The second measure is latitude that can be defined as the angle, which is formed by the intersection of a line perpendicular to the surface of the Earth at a point and the Equator plane. Therefore, it is obvious that unlike other geographical measures latitude values range from – 90 degrees to + 90 degrees. In specialized scientific references as well as in various examples of fictional literature one can often found another term that describes the same mathematical conception – ‘parallel’. In truth, the term ‘parallel’ is ubiquitously used not only amid dilettantes in the sphere of geography and geographical topology, but also amid specialists in these fields of study, because it correctly describes the physical sense of this geographic term: a particular value of latitude forms a notional circle, which is parallel to the Equator.
In fact, the geographic coordinates are angular units. However, in order to simplify the computations and eschew unnecessary complications when transferring units from one form to another ArcSDE stores and treats them as if they are planar. One does not have to be confused meeting this type in various dissertation abstracts or scientific articles that are dedicated to the topic of study. Therefore, longitude values are represented as those that belongs to the x axis, whereas the latitude values are represented as those that appertains to the y axis.
A standard geographic coordinate system comprises four fundamental elements. Here is a list of these components:
• The angular units. These units of measure are used in the spherical reference system.
• The prime meridian. In fact, it is a longitude origin of the common spherical reference system.
• The spheroid units. These units are used with an eye to represent the reference spheroid for the standard coordinate transformation.
• The datum. This measurement allows us to define the specific relationship of the standard reference spheroid to the surface of the planet.
A standard projected coordinate system can be represented as a two-dimensional planar surface. Nevertheless, the surface of the Earth is, obviously, three-dimensional. In order to transform a three-dimensional space into a two-dimensional surface, we have to perform a specific operation called ‘projection’. Regardless of the specific type of the projection formulas, all of them can be presented as mathematical expressions, which allow us to convert data from a certain geographical location (latitude and longitude) on a spheroid to a corresponding location (x axis and y axis) on a two–dimensional surface. Thereby, the standard projected coordinate system maintain two different axes: the x axis, which represents east-west, and the y axis that represents north-south. The whole scheme is analogical to those that is used in the Cartesian coordinate system for the two-dimensional space. The x axis intersects with the y axis in the certain point (origin). Obviously, the coordinates of the origin are 0,0. The x axis refers to the distance along the horizontal line, and the y axis represents the distance along the vertical line. The projected or planar coordinate system locates points relative to the point of origin (0,0) and the x axis and y axis. Points below the x axis or to the left of the y axis have negative values.
A standard projected coordinate system also comprises the following elements: parameters, projection and units. In this coordinate system, units represent the geographical coordinates on the plane (two-dimensional) surface. In fact, the projection is a standard mathematical operation. In this case, it is used with an eye to converting geographic coordinates to projected coordinates. Therefore, actual geographic coordinates suffer transformation in the abstract geometric coordinates that are represented as measures of distance between the origin and the resulting perpendiculars on the axes (x axis and y axis). At last, parameters are the concrete units of measurement that are specific for each single case of study. Of course, all but these specific techniques of converting geographic data are interesting and necessary only for specialists in these spheres of study. Nevertheless, even different common writing assignments, which study some specific geographical position, require a knowledge of fundamental principles of work with a standard geographic coordinate system.
### The Cartesian coordinate system: a brief historical reference
The rectangular coordinate system was invented in the XVII century by Rene Descartes – a famous French philosopher, physicist, engineer, mathematician and physiologist. Nowadays, Descartes is known as one of the greatest minds in human history because of his revolutionary philosophical conceptions, great achievements in mathematics and linear algebra, and a great number of significant inventions in different scientific disciplines. In fact, he is also famous as a talented publicist because of his extraordinary creative writingtechniques and a polished literary style. Rene Descartes is one of the founders of analytic geometry, which he developed together with Fermat. From the point of view of modern geometric science, he deserves attention as an inventor of the coordinate system, which bears his name. Firstly, this conception was presented to the public in 1637 in his famous Geometry. Additionally, he was the first scientist who expressed the law of conservation of momentum, proposed the conception of impulse and laid the foundations of classical mechanics. In truth, the rectangular coordinate system proposed by Descartes was so elegant and effective that it has not suffered any significant transformation, despite the time that has passed since its invention. The Cartesian coordinate system, which is based on two perpendicular lines (y axis and x axis), invented by Rene Descartes is the same as the modern system of coordinates that is used in diverse sample case study papers.
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The history of the Celts and where they came from is changing very quickly, lots of evidence from genetics proves that this title image (right) may be wrong and that they travelled around the edge of the continent to get here but these discoveries are probably for older students!
Miss Stout’s History Class – This was created by a teacher for year one pupils but it could be useful all across primary school. It is a beautiful pictorial history of the Celts including clothes, jewellery, weapons, food etc.
Celts (below), by Maria Antonietta Sessa, has an animated Skeleton telling the story. Skelly shares important information about the Celts. It is a pity it is a computer-generated voice and not a human reader, but it is a useful easy history resource with lots of photograph illustrations. The fact that it is narrated by an animated skeleton may engage some young learners!
Celtic Life in the Iron Age. “See some of the skills and technologies from the Iron Age. For information on our Key Stage 1 & 2 teaching collections and other reconstructed artefacts please visit: http://bit.ly/1eg3CBO ” This video concentrates on food and textile making. It shows the bread making process including grinding the wheat. NB this video is ten minutes long, it is easy to understand and demonstrates historical techniques that it can be difficult for children to visualise. You may wish to choose a section or two from it to meet your requirements. This would be for older children rather than the very young but is a great resource.
More films about the history of Celtic Britain can be found on the Ancient1580 site.<|endoftext|>
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Examine the mechanisms of evolution.
Describe how natural selection results in biodiversity.
Explain why biodiversity is important to continued evolution.
Examine mutations and sexual recombination, which are sources of genetic variation.
Evolution refers to gradual changes in the inherited traits of a population of organisms that occur through successive generations. For evolution to occur a force need to be exerted on a population of species. The ultimate goal of evolution can be thought of as 'enhancement of reproduction'. There are two mechanisms by which evolution occurs:
1) Natural selection: This process relies are three simple rules:
- Variations exist within the population of a given species
- Organisms produce more offspring that can actually survive
- Offsprings are not equal in their ability to survive
Having these three factors in mind, one can explain natural selection in words of Darwin who gives this the term " survival of the fittest". In other words the offspring that has 'fittest' genetic make up (these variations happen at random) have the best chance of survival given very specific conditions. For example, consider Pandas. These animals need a 'thumb' on their hand to be able to grab on to bamboo trees (their major source of food) and eat. Now imagine the time that Pandas were evolving from their family of bears (bears don't have a thumb on their hand!). Those Pandas that happened to have a 'thumb' when the weather became really cold and ...
An overview of evolution and the mechanisms by which it occurs.<|endoftext|>
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Home > CCA2 > Chapter Ch1 > Lesson 1.2.1 > Problem1-61
1-61.
Consider the equation $4x−6y=12$. Homework Help ✎
1. Predict what the graph of this equation looks like. Justify your answer.
Are there any exponents? What does this mean?
2. Solve the equation for $y$ and graph the equation.
Get $y$ by itself.
$-6y=-4x+12$
$\frac{-6y}{-6}=\frac{-4x+12}{-6}$
$y=\frac{2}{3}x-2$
3. Explain clearly how to find the $x$- and $y$-intercepts.
The $x$-intercept is the point where $y=0$.
The $y$-intercept is the point where $x=0$.
4. Which form of the equation is best for finding the $x$- and $y$-intercepts quickly? Why?
It’s easier to use the given, standard form of the equation to find the intercepts quickly because there is minimal rewriting involved.
5. Find the $x$- and $y$-intercepts of $2x−3y=−18$. Then use the intercepts to sketch a graph quickly.
Set $x=0$ to find the $y$-intercept, and $y=0$ to find the $x$-intercept.
$2(0)−3y=−18$
$x$-intercept: $(−9,0)$
$y$-intercept: $(0,6)$<|endoftext|>
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# How do you simplify 5sqrt80-12sqrt5?
$8 \sqrt{5}$
$\sqrt{80} = \sqrt{4 \cdot 4 \cdot 5}$ Notice that there are two 4s in there, so, since it is a square root we can write that same thing as $4 \sqrt{5}$ so the equation now becomes:
$5 \cdot \left(4 \sqrt{5}\right) - 12 \sqrt{5}$
$5 \cdot 4 = 20$ so
$20 \sqrt{5} - 12 \sqrt{5}$ From here we can treat $\sqrt{5}$ as $x$ like: $20 x - 12 x = 8 x$ so,
$20 \sqrt{5} - 12 \sqrt{5} = 8 \sqrt{5}$<|endoftext|>
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Humanity faces an immense challenge: providing abundant energy to everyone without wrecking the planet. If we want a high-energy future while protecting the natural world for our children, we must consider the environmental consequences of energy production and use. But money matters too: energy solutions that ignore economic costs are not realistic, particularly in a world where billions of people currently can’t afford access to basic energy services. How can we proceed?
Energy Within Environmental Constraints won’t give you the answer. Instead, we will teach you how to ask the right questions and estimate the consequences of different choices.
This course is rich in details of real devices and light on theory. You won’t find any electrodynamics here, but you will find enough about modern commercial solar panels to estimate if they would be profitable to install in a given location. We emphasizes costs: the cascade of capital and operating costs from energy extraction all the way through end uses. We also emphasize quantitative comparisons and tradeoffs: how much more expensive is electricity from solar panels than from coal plants, and how much pollution does it prevent? Is solar power as cost-effective an environmental investment as nuclear power or energy efficiency? And how do we include considerations other than cost?
This course is intended for a diverse audience. Whether you are a student, an activist, a policymaker, a business owner, or a concerned citizen, this course will help you start to think carefully about our current energy system and how we can improve its environmental performance.
Solar Farm CC-BY Michael Mees on flickr Smokestack CC-BY Patrick on flickr
Week 1: Introduction Meet the instructors and learn what the course is all about. Learn where you’re strongest and weakest, and if you have any commonly-held misconceptions.
Week 2: Energy Overview Forms of energy and common units of measurement. How energy flows through modern and historical economies, including the composition of energy supply, common energy transformations, and which sources are used for which purposes. Prices for energy around the world.
Week 3: Estimating Costs The quantitative techniques at the heart of the course: levelized cost and cost of mitigation. We’ll apply these techniques to energy systems and also to everyday life.
Week 4: Environmental Impacts How severe are air pollution, climate change, and land use impact today, and how severe are they likely to be in the future? How do they affect human health, GDP, and the natural world?
Week 5: Fossil Fuels An abbreviated section focused on the abundance of fossil fuels. Spoiler alert: we won’t run out any time soon.
Week 6: The Electric Grid A brief overview of modern electric grids including major technologies they use, how remarkably reliable and efficient they are, how they’re planned and regulated, and how they’re starting to change.
Week 7: Solar Power What solar power technologies dominate today and which have a chance to in the near future. How to estimate the cost of solar power in different regions, how it compares to other options, and the remarkable decline in its cost in the past 5 years. How we can cope with the intermittent nature of the solar resource. How solar power is regulated and subsidized today.
Week 8: Nuclear Power How nuclear fission works and how it’s harnessed in modern nuclear plants. How much nuclear power costs and how much it’s used, including the stagnation in its use since the 1990’s and the prospects for its revival. Details on the hazards and costs of nuclear waste and power plant accidents. The connection between nuclear power and nuclear weapons.
Week 9: Demand Reduction and Efficiency Reducing energy demand, by changing behavior or making devices more efficient, can reduce environmental harms – sometimes while saving money! But are there limits to this strategy? Can humanity reduce demand and aim towards a lower-energy future?
Week 10: Conclusion Wrap-up and review.Wrap-up and review.<|endoftext|>
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A river is a naturally-winding channel carrying water through the landscape. The sides of a river are called the river banks. They are usually made of fine materials. The top of the riverbank is made of soil.
Small rivers are called brooks and streams. There is no exact definition of when a stream becomes a river. However, usually brooks and streams are small enough to be jumped across, or they can be waded. Rivers are generally too big to jump across, or too deep to be waded.
The river is fed by water seeping out of the banks. Water sinks into the soil as rain, and then drains slowly through the soil (and also sometimes through rock) until it reaches the riverbank. Soil and some rocks are natural sponges. They hold on to water and also only allow the surplus to flow out slowly. This slow flow explains why rivers do not dry up when it is not raining.
However, if rain falls for too long, the soil will fill up and then the rest of the rain will flow over the surface and reach the river much more quickly. Also, if it rains so hard the soil cannot soak it all in fast enough, the rest will flow over the soil surface. When water flows over the surface, it reaches rivers very quickly. Then the river cannot carry it away fast enough and the rest spills out over the banks, causing flooding.
The area that feeds water into a river is called 'the drainage basin' or the 'river basin'.<|endoftext|>
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Sustainable Management of Food Basics
On this page:
- What is Sustainable Management of Food?
- Why is Sustainable Management of Food Important?
- Sources of Statistics
Sustainable Management of Food is a systematic approach that seeks to reduce wasted food and its associated impacts over the entire life cycle, starting with the use of natural resources, manufacturing, sales, and consumption and ending with decisions on recovery or final disposal. EPA works to promote innovation and highlight the value and efficient management of food as a resource. Through the sustainable management of food, we can help businesses and consumers save money, provide a bridge in our communities for those who do not have enough to eat, and conserve resources for future generations. Building on the familiar concept of "Reduce, Reuse, Recycle," this approach shifts the view on environmental protection and more fully recognizes the impacts of the food we waste.
The term “wasted food” describes food that was not used for its intended purpose and is managed in a variety of ways, such as donation to feed people, creation of animal feed, composting, anaerobic digestion, or sending to landfills or combustion facilities. Examples include unsold food from retail stores; plate waste, uneaten prepared food, or kitchen trimmings from restaurants, cafeterias, and households; or by-products from food and beverage processing facilities. EPA uses the overarching term “wasted food” instead of “food waste” for food that was not used for its intended purpose because it conveys that a valuable resource is being wasted, whereas “food waste” implies that the food no longer has value and needs to be managed as waste.
- Excess food refers to food that is recovered and donated to feed people.
- Food waste refers to food such as plate waste (i.e., food that has been served but not eaten), spoiled food, or peels and rinds considered inedible that is sent to feed animals, to be composted or anaerobically digested, or to be landfilled or combusted with energy recovery.
- Food loss refers to unused product from the agricultural sector, such as unharvested crops.
EPA encourages anyone managing wasted food to reference the Food Recovery Hierarchy. When the higher levels of the hierarchy are no longer feasible, then the food waste left over should be put to beneficial use such as composted or sent to be broken down through anaerobic digestion. Additional resources on wasted food can be found at Further with Food: Center for Food Loss and Waste Solutions.Exit
Wasted food is a growing problem in our modern society and an untapped opportunity. In 2015 alone, more than 39 million tons of food waste was generated, with only 5.3 percent diverted from landfills and incinerators for composting. EPA estimates that more food reaches landfills and incinerators than any other single material in our everyday trash, constituting 22 percent of discarded municipal solid waste. Additionally, the U.S. Department of Agriculture (USDA) estimates that in 2010, 31 percent or 133 billion pounds of the 430 billion pounds of food produced was not available for human consumption at the retail and consumer levels (i.e., one-third of the food available was not eaten).1
The Food and Agriculture Organization of the United Nations (FAO) estimated in 2011 that approximately one-third of all food produced for human consumption worldwide is lost or wasted.2 Taking simple steps in your everyday life can make a difference in addressing this issue. Reducing wasted food is a triple win; it's good for the economy, for communities, and for the environment.
When we waste food, we’re not just creating a problem, we’re also missing an opportunity to save businesses and consumers money:
- Pay Less for Trash Pickup – Organizations might pay less for trash pickup by keeping wasted food out of the garbage. Some haulers lower fees if wasted food is separated from the trash and sent to a compost facility instead of the landfill.
- Receive Tax Benefits by Donating – If you donate healthy, safe, and edible food to hungry people, your organization can claim tax benefits. The Bill Emerson Good Samaritan Act protects food donors from legal liability.
Waste Less and Spend Less – If you or your organization can find ways to prevent waste in the first place, you can spend less by buying only the food you will use. Preventing wasted food can also reduce energy and labor costs associated with throwing away good food.
Preventing wasted food and recovering wholesome, nutritious food can help you make a difference in your community:
- Feed People, Not Landfills – Instead of feeding landfills, we should be feeding people in our communities. You can donate a variety of foods to many different types of organizations. Contact Feeding America Exit or your local food rescue organizations for information about where you can donate and what types of food your local organization is able to accept.
- Feed Children – In 2012, the U.S. Department of Agriculture National School Lunch Program provided nutritionally balanced, low-cost or free lunches to more than 31 million children each school day.4 By redirecting food that would otherwise be wasted to homes and schools, we can help feed our country’s children.
- Create Job Opportunities – Recovering and recycling wasted food through donation, salvaging, processing, industrial reuse, and composting strengthens infrastructure and creates jobs. Food recycling in these sectors employs more than 36,000 people, supporting local economies and promoting innovation.5
- Feed the World – According to the Food and Agriculture Organization of the United Nations, from 2012 to 2014 there were about 805 million hungry people on earth. They predict that by eliminating food loss and wasted food we would have enough food to feed all the chronically undernourished. They also expect that we wouldn’t have to increase food production or put additional pressure on our natural resources to do so.3
Reducing wasted food does great things for the environment:
- Reduce Methane from Landfills – When food goes to the landfill, it’s similar to tying food in a plastic bag. The nutrients in the food never return to the soil. The wasted food rots and produces methane gas.
- Save Resources – Wasted food wastes the water, gasoline, energy, labor, pesticides, land, and fertilizers used to make the food. When we throw food in the trash, we’re throwing away much more than food.
- Return Nutrients to the Soil – If you can’t prevent, reduce or donate wasted food, you can compost. By sending food scraps to a composting facility instead of to a landfill or composting at home, you’re helping make healthy soils. Adding compost to gardens, highway construction sites, and poor soils makes great things happen. Properly composted organics (wasted food and yard waste) improve soil health and structure, improve water retention, support more native plants, and reduce the need for fertilizers and pesticides.
- United States Department of Agriculture. The Estimated Amount, Value, and Calories of Postharvest Food Losses at the Retail and Consumer Levels in the United States.
- Food and Agriculture Organization of the United Nations "Global food losses and food waste – Extent, causes, and prevention" (PDF) Exit(38 pp, 1.6 MB, 2011, About PDF).
- Food and Agriculture Organization of the United Nations, International Fund for Agriculture Development and the World Food Programme, The State of Food Insecurity in the World 2014. Strengthening the enabling environment for food security and nutrition (PDF) Exit (57 pp, 3 MB, 2014).
- United States Department of Agriculture, National School Lunch Program Fact Sheet.
- United States Environmental Protection Agency, Advancing Sustainable Materials Management: 2016 Recycling Economic Information (REI) Report Methodology.<|endoftext|>
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# Comparing linear functions: table vs. graph
## Videóátirat
f is a linear function whose table of values is shown below. So they give us different values of x and what the function is for each of those x's. Which graphs show functions which are increasing at the same rate as f? So what is the rate at which f is increasing? When x increases by 4, we have our function increasing by 7. So we could just look for which of these lines are increasing at a rate of 7/4, 7 in the vertical direction every time we move 4 in the horizontal direction. And an easy way to eyeball that would actually be just to plot two points for f, and then see what that rate looks like visually. So if we see here when x is 0, f is negative 1. When x is 0, f is negative 1. So when x is 0, f is negative 1. And when x is 4, f is 6, so 1, 2, 3, 4, 5, 6, so just like that. And two points specify a line. We know that it is a linear function. You can even verify it here. When we increase by 4 again, we increase our function by 7 again. We know that these two points are on f and so we get a sense of the rate of change of f. Now, when you draw it like that, it immediately becomes pretty clear which of these has the same rate of change of f. A is increasing faster than f. C is increasing slower. A is increasing much faster than f. C is increasing slower than f. B is decreasing, so that's not even close. But D seems to have the exact same inclination, the exact same slope, as f. So D is what we would go with. And we could even verify it, even if we didn't draw it in this way. Our change in f for a given change in x is equal to-- when x changed plus 4, our function changed plus 7. It is equal to 7/4. And we can verify that on D, if we increase in the x-direction by 4, so we go from 4 to 8, then in the vertical direction we should increase by 7, so 1, 2, 3, 4, 5, 6, 7. And it, indeed, does increase at the exact same rate.<|endoftext|>
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1. ## Geometry Construction
Construct a triangle, given the measures of two angles and the length of the bisector of the thrid angle.
2. Originally Posted by MATNTRNG
Construct a triangle, given the measures of two angles and the length of the bisector of the thrid angle.
Draw a triangle with the given angles and side connecting them of unit length (I will call this side the base). Bisect the third angle and draw in the bisector.
Produce the bisector and mark the reqired length from the point that it meets the base through the bisected angle. Now construct line paralle to the sides other than the base through this marked point and you are done.
That is you construct a triangle with the given angles but arbitary size, draw in the bisector, mark the required length on the bisector produced and construct a triangle of the required size similar to the one you drew earlier.
CB
3. Originally Posted by MATNTRNG
Construct a triangle, given the measures of two angles and the length of the bisector of the third angle.
Let θ1 and θ2 are the given angles.
The third angle is θ = 180 - ( θ1 + θ2)
Let PO is the angle bisector. Draw a line PQ such that angle OPQ = θ/2
Mark a point Q' on PQ such that PQ'/PO = sin(θ/2)
Taking OQ' as radius draw a circle. Draw PS tangent to the circle which touches the circle at S'.
Draw line OA such that angle Q'OA = π/2- θ1/2 and OB such that angle S'OB = π/2- θ2/2.
If A and B are the points on PQ and PS, then PAB is the required triangle.
4. Hello, MATNTRNG!
Here is Captain Black's excellent construction.
Construct a triangle, given the measures of two angles, $\displaystyle \alpha$ and $\displaystyle \beta$
and the length of the bisector of the thrid angle, $\displaystyle b$.
Draw a horizontal line $\displaystyle PQ.$
At $\displaystyle P$ construct angle $\displaystyle \alpha$, at $\displaystyle Q$ construct angle $\displaystyle \beta,$
. . intersecting at $\displaystyle R.$
Code:
R
o
* *
* *
* *
* *
* *
* *
* *
* *
* α β *
P o * * * * * * * * * o Q
Construct the bisector of angle $\displaystyle R$, intersecting $\displaystyle PQ$ at $\displaystyle S.$
On $\displaystyle SR$, measure off $\displaystyle SC = b.$
Code:
R
o
* *
* * *
* *
* * *
* *
* o C *
* *
* * b *
* α β *
P o * * * * o * * * * o Q
S
Through $\displaystyle C$, construct a line parallel to $\displaystyle PR,$
. . intersecting $\displaystyle PQ$ at $\displaystyle A.$
Through $\displaystyle C$, construct a line parallel to $\displaystyle QR,$
. . intersecting $\displaystyle PQ$ at $\displaystyle B.$
Code:
R
o
* *
* *
* *
* *
* C *
* o *
* * * *
* * * *
* α * * β *
P o * * o * * * o * * o Q
A B
$\displaystyle \Delta ABC$ is the desired triangle.<|endoftext|>
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Pythagorean Triples
Written by
Malcolm McKinsey
Fact-checked by
Paul Mazzola
What is a Pythagorean triple?
Some numbers seem to work perfectly in the Pythagorean Theorem, like 3, 4, and 5, which is ${3}^{2}+{4}^{2}={5}^{2}$. Sets of positive, whole numbers that work in the Pythagorean Theorem are called Pythagorean triples.
For three positive integers to be Pythagorean triples, they must work in the Pythagorean Theorem's formula:
In the Pythagorean Theorem's formula, a and b are legs of a right triangle, and c is the hypotenuse.
Only positive integers can be Pythagorean triples. The smallest Pythagorean triple is our example: (3, 4, and 5).
A quick way to find more Pythagorean triples is to multiply all the original terms by another positive integer:
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Pythagorean triples
Pythagorean triples are relatively prime. Relatively prime means they have no common divisor other than 1, even if the numbers are not prime numbers, like 14 and 15. The number 14 has factors 1, 2, 7, and 14; the number 15 has factors 1, 3, 5, and 15. Their only common factor is 1.
Primitive Pythagorean triples
A set of numbers is considered to be a primitive Pythagorean triple if the three numbers have no common divisor other than 1.
Our first Pythagorean triple is primitive, since (3, 4, and 5) have no common divisors other than 1. Our fifth set from our example above, however, is not primitive (it is imprimitive) because each value for a, b, and c of the right triangle is a multiple of 5.
How to find Pythagorean triples
Here's how to find Pythagorean triples in three easy steps:
1. Pick an even number to be the longer leg's length.
2. Find a prime number one greater than that even number, to be the hypotenuse.
3. Calculate the third value to find the Pythagorean triple.
Pythagorean triples formula
Suppose you pick 12 as the length of a leg, knowing 13 is an adjacent prime number. Use these two as part of the Pythagorean Theorem to complete your primitive Pythagorean triple:
Subtract the value of ${b}^{2}$ from both sides:
So our primitive Pythagorean triple is (5, 12, 13)! You can quickly generate iterations of that (none of which will be primitive) using the same multiplying method we used before: (10, 24, 26); (15, 36, 39); and so on.
Pythagorean triples examples
Generate a primitive Pythagorean triple based on the prime number 41. What even number is adjacent to and smaller than 41?
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Did you get a=9, b=40, and c=41? The primitive Pythagorean triple is (9, 40, 41).
Generating Pythagorean triples
You can come up with your own Pythagorean triples. Label two positive integers m and n, ensuring m>n. Then, for each side of a right triangle:
Formula for generating Pythagorean triples
Generate a set of Pythagorean triples for m=6 and n=5 using our formula:
Let's solve for a:
Now, let's solve for b:
And finally, c:
Our Pythagorean triple is (11, 60, 61). This method catches primitive and imprimitive Pythagorean triples.
Pythagorean triples list
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I know many of you are eager to get your hands on a Pythagorean Triples list already pre-computed for you. Here's a list of some common ones:
• 3, 4, 5
• 5, 12, 13
• 6, 8, 10
• 7, 24, 25
• 8, 15, 17
• 9, 12, 15
• 9, 40, 41
• 10, 24, 26
• 12, 16, 20
• 12, 35, 37
• 14, 48, 50
• 15, 20, 25
• 15, 36, 39
• 16, 30, 34
• 18, 24, 30
• 20, 21, 29
• 21, 28, 35
• 24, 32, 40
• 27, 36, 45
• 30, 40, 50
Lesson summary
By watching the video, reading the lesson, and studying the formulas, you have learned how to identify a Pythagorean triple, apply and use the Pythagorean Theorem (32Â + 42Â = 52), classify a Pythagorean triple as either primitive or imprimitive, and use both the Pythagorean Theorem and another method to find Pythagorean triples.
What you learned:
After reading the lesson and studying the drawings you will be able to:
• Identify a Pythagorean triple
• Apply the Pythagorean Theorem
• Categorize a Pythagorean triple as either primitive or imprimitive (not primitive)
• Use the Pythagorean Theorem and another method to find Pythagorean triples<|endoftext|>
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# CLASS-6EXPONENTS
EXPONENTS
If ‘A’ is any number and m is the natural number and if we wish to multiply unlimited or m times of A each other, then we can write or express like that-
( A x A x A x A x …………… m times taken ) = A, where ‘m’ is the exponents and ‘A’ is the base.
There are some rules of ‘EXPONENTS’ are described below which we have to follow every time in application-
1) If ‘A’ is any number and let x & y are the natural numbers, then-
Aˣ x Aʸ = Aˣʸ
2) If ‘A’ is any number, m & n are the natural number, where m > n then-
A
--------- = A ⁽ ˉⁿ
Aⁿ
3) If ‘A’ is any number, m & n are the natural number, then
( A) ⁿ = A
4) If ‘A’ & ‘B’ are any number and m is a natural number, then
A x B = (AB)
5) If ‘A’ & ‘B’ are any number and m & n are natural numbers, then
A x Bⁿ = Aᵐ Bⁿ
6) If ‘A’ is any number and m is a natural number, then
(-A) = A when ‘m’ is even.
And (-A) = - A when ‘m’ is odd.
7) If ‘A’ is any number, then
A = 1
A¹ = A
Example.1) Please write the value of 7
Ans.) 7⁴ = 7 x 7 x 7 x 7 = 2401
Step.1) First observed in given numbers how much power or natural number is given.
Step.2) Multiply normally the given number 7 as much as time as per given Power or natural numbers (i.e. 4) is provided. Here number 7 is to be multiplied by 4 times and the result is 2401.
Step.3) Value comes with negative (-) sign or value because as per rules-
A = A or (+) A when ‘m’ is even. So, 2401
or (+) 2401 is the actual/final result. (Ans.)
Example.2) Please write the value of (-2)
Ans.) (-2) = (-2) x (-2) x (-2) x (-2) x (-2) x (-2) x (-2) = -128
Step.1) First observed in given numbers how much power or natural number is given.
Step.2) Multiply normally the given number 2 as much as time as per given Power or natural numbers (i.e. 7) is provided. Here number 2 is to be multiplied by 7 times and the result is 128.
Step.3) Value comes with negative (-) sign or value because -
as pr rules (-A) = - A when ‘m’ is odd. 7 is odd number
So, -128 is the actual/final result. (Ans.)
Example.3) Please write the value of (-3)
Ans.) (-3)⁸ = (-3) x (-3) x (-3) x (-3) x (-3) x (-3) x (-3) x (-3) = 6561
Step.1) First observe in given numbers how m power or natural number is given.
Step.2) Multiply normally the given number 3 as much as time as per given Power or natural numbers (i.e. 8) is provided. Here number 3 is to be multiplied by 8 times and the result is 6561.
Step.3) Value comes with negative (-) sign or value because -
as pr rules (-A) = A when ‘m’ is even.
So , 6561 is the actual/final result. (Ans.)
4) Write 625 as a power of 5.
Step.1) First find the prime factorization of the given number.
Step.2) After completing prime factorization of the given number observe how many base has come with their exponents. Clearly shown that only 5 has come 4 times .
Step.3) Clearly shown that 5 has come 4 times, so 5 is the base and 4 its exponents.
So, it can be written as 5.
625 = 5 x 5 x 5 x 5 = 5. (Ans.)
5) Write 38880 in exponential form.
Step.1) First find the prime factorization of the given numbers.
Step.2) After completing prime Factorization of the given number observe how many base and their exponents we can find.
Step.3) Clearly shown that 2 has come 5 times, 5 has come 1 time and 3 has come 5 times. so it can be written as 5.
Step.4) 2 has come 5 times, so 2 is base 5 its exponent. It can be written 2. 5 has come 1 time, so 5 is base and 1 its exponent. It can be written 5¹. 3 has come 5 times, so 3 is base and 5 its exponent. It can be written 3.
Step.5.) Exponential form can be written as multiplication of base with their exponents like
2 x 5¹ x 3
38880 = 2 x 2 x 2 x 2 x 2 x 5 x 3 x 3 x 3 x 3 x 3
= (2 x 2 x 2 x 2 x 2) x 5 x (3 x 3 x 3 x 3 x 3)
= 2 x 5¹ x 3 (Ans.)
6) Write the product of number 3 and 3 as single power.
Ans.) 3 and 3
= 3 x 3 = 3 ⁺⁵ = 3
Step.1) Observe the given two-digit, where the base are the same but exponents are different.
If ‘A’ is any number and let x & y are the natural numbers,
then- Aˣ x Aʸ = A⁽ˣʸ
3 = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 6561 (Ans.)
7) Write the product of number (-8) and (-8) as single power.
Ans.) (-8) and (-8)
= (-8) x (-8) = (-8) ⁵⁺⁷ = (-8)¹²
Step.1) Observe the given two-digit, where the base are the same but exponents are different.
Step.2) We follow the rules- If ‘A’ is any number and let x & y are the natural numbers,
then- Aˣ x Aʸ = A⁽ˣʸ
= (-8) x (-8) x (-8) x (-8) x (-8) x (-8) x (-8) x (-8) x (-8) x (-8) x (-8) x (-8)
= (- 68719476736) (Ans.)
8) Evaluate (3
(3)² = (3)⁴˟² = 3
Step.1) We can observe that, there is one base two exponents
Step.2) We will follow the rules –
If ‘A’ is any number, m & n are the natural number, then-
( A) ⁿ = A
3 = 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 = 6561 (Ans.)
9) Multiply 5 and 2
Ans.) 5 and 2 = 5 x 2 = 10
Step.1) We can observe that, in given numbers base are different but exponents are equal.
Step.2) we follow the rules as discussed below -
If ‘A’ & ‘B’ are any number and m is a natural number, then
A x B = (AB)
= 10 = 10 x 10 x 10 x 10
10000 (Ans.)
10) Explain (-5) = ?
Ans.) (-5)
= (-5) x (-5) x (-5) x (-5) x (-5)
= - 3125 (Ans.)
Step.1) Here we can observe that there are single base and the exponents are shown 5 times, so we have to multiply the given base by 5 times.
Step.2) Value comes with a negative (-) sign or value because -
as pr rules (-A) = - A when ‘m’ is odd. 5 is odd number
(-5) x (-5) x (-5) x (-5) x (-5) - 3125
So , -3125 is the actual/final result. (Ans.)
11) Evaluate 9 / 9
Ans.) 9 / 9
9
= --------- = 9⁷⁻⁴ = 9
9
Step.1) Observe in a given fraction base are the same but exponents are different. so
Step.2) We follow the rules -
If ‘A’ is any number, m & n are the natural number, where m > n then-
A
---------- = A ⁽ᵐˉⁿ
Aⁿ
= 9ᶟ = 9 x 9 x 9 = 729 (Ans.)<|endoftext|>
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There are certain viral infections, you can catch all year round, and among them are influenza and common cold. Although most people think that these are seasonal viruses that spread during cold weathers, but this is not true. Common Cold is a viral infection that attacks the upper respiratory tract, which includes your nose and throat. It is the most common of infections with mild effects over our health. Despite this— common cold results in most number of doctor visits, requires urgent care to control the transmission and in absenteeism from work or school. Researches have shown that in the United States only around 1 billion people are infected by colds every year, resulting in almost 22 million days of school absences per year. The common cold is the most common disease that results in most frequent visits to doctors which is estimated to be approximately 75-100 million doctor visits every year, the economic influence of more than $20 billion annually due to work loss. People who are infected by common cold tend to spend more time in enclosed environment which at time maximises the risk of exposure for others and even for them, because of the enclosed area the infectious virus keeps on growing and spreading.
What causes Common Cold?
Common cold is mostly spread through a number of viruses; almost more than 200 hundred viruses are known to cause common cold. However, the most common among these are “Rhinovirus,” which causes 10% to 40 % of colds and the “Coronaviruses,” from which 20 % of colds are caused. Also, there “Respiratory Syncytial Viruses,” which cause about 10% of common cold. If caused by various kinds rhinovirus, the protective layer of the nose and throat is attacked by the virus, which results in activating an immune system reaction causing sore throat, a congested nose and headaches; thus making it hard to breathe. If the air is dry due to weather conditions or other reasons, then the virus tends to spread more rapidly and gets lower resistance from our body’s immune system. This is because the dryness around the nasal area makes the virus to stick and hold to our nasal passages. The virus is spread in the form of tiny droplets in the air, that gets mixed in the air through a sick person’s sneeze, cough or through their runny nose.
Common cold is considered to be highly contagious as it transmits from person to person. There are chances that a sick person at home infects others around him. As the virus spreads in the air through sneezing and coughing, other people making contact with areas or objects contaminated by the cold germs are more likely to catch a common cold and get sick. It is not necessary that a person gets a cold because of getting wet in the cold weather or being exposed to cold, people with nasal or throat allergies can catch a common cold at any time. At times emotional stress or fatigue can also be a cause of common cold.
There may have been several times when your children came home from school, badly sneezing and coughing or you woke up in the middle of night due to an irritating scratchy throat or nose congestion. The initial symptoms of common cold: a scratchy throat, congested or runny nose, watery eyes, sneezing, coughing, mucus drainage and headaches; are known almost by everyone. However severe symptoms along with these such as high fever or cramps are also a sign of influenza or flu. Other symptoms may include decreased appetite and aggression due to headache.
The common colds are usually mild in nature and will eventually go away after the virus becomes less active and the immune system restore functionality. However, there are a few things that can be done to get well and take good care of yourself.
- When infected by common cold, you will mostly feel tired and exhausted due to the headache or breathing problems. In this case, the best cure for your common cold would be a lot of rest. The more you’ll rest, the sooner you will recover.
- Due to the drainage of mucus, watery eyes and loss of appetite; it is recommended that you increase your fluid intake. You can have soups, juices and all forms of warm fluids that will help in unblocking the congestion in the respiratory tract and will also restore health.
- Consult a doctor or physician to suggest you the right medicines which will help you.
- Over the counter medicines can also be used to treat common cold. However, it is suggested that to only use medicines you know and which can be used to treat cold. In any case, it is do not use over the counter medicines to treat common cold in children less than 4 years.
- In cases of extreme congestion or stuffines, saline nasal drops and sprays can help release the pain and unblocks your congested nose.
- If you are considering using antibiotics it is recommended that you consult a doctor. As in many cases, antibiotics don’t generally work and result in more dryness leading to severe congestion.
- Natural remedies like honey or salt water gargles can help with the scratchy irritation in your throat.
- Using a humidifier can help overcome the dryness and will provide moisture to a dry congested upper respiratory tract. When using a humidifier it is recommended that you clean it properly and change the water of your humidifier daily.
- The best thing to do when infected by a common cold is to prevent it from spreading. If you have a sick person at home or at work avoid touching the objects he/she has been using.
- If you are nursing a sick person it is best to keep yourself protected by washing your hands or sanitizing them more often. Do not touch your eyes and nose if you are in close contact with an infected person to avoid the risk of exposure.
The common cold although being the common of viral infection throughout the world is not something to be considered dangerous. If taken care of properly and preventing the germs from spreading can minimize the spread of this virus.
Christine Rudolph is a content developer at Centra Care, a Florida Urgent Care Hospital. Follow @CentraCare for more updates.<|endoftext|>
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Basic Calculus Teaching Guide for Senior High School
The Pre-Calculus course bridges Basic Mathematics and Calculus. This course completes the foundational knowledge on Algebra, Geometry, and Trigonometry of students who are planning to take courses in the STEM track. It provides them with conceptual understanding and computational skills that are crucial for Basic Calculus and future STEM courses.
Based on the Curriculum Guide for Pre-Calculus of the Department of Education, the primary aim of this Teaching Guide is to give Math teachers adequate stand-alone material that can be used for each session of the Grade 11 Pre-Calculus course.
The Guide is divided into three units: Analytic Geometry, Summation Notation and Mathematical Induction, and Trigonometry. Each unit is composed of lessons that bring together related learning competencies in the unit. Each lesson is further divided into sublessons that focus on one or two competencies for effective teaching and learning. Each sublesson is designed for a one-hour session, but the teachers have the option to extend the time allotment to one-and-a-half hours for some sub-lessons.
Each sub-lesson ends with a Seatwork/ Homework, which consists of exercises related to the topic being discussed in the sub-lesson. As the title suggests, these exercises can be done in school (if time permits) or at home. Moreover, at the end of each lesson is a set of exercises (simply tagged as Exercises) that can be used for short quizzes and long exams. Answers, solutions, or hints to most items in Seatwork/ Homework and Exercises are provided to guide the teachers as they solve them.
Some items in this Guide are marked with a star. A starred sub-lesson is optional and it is suggested that these be taken only if time permits. A starred example or exercise requires the use of a calculator.
To further guide the teachers, Teaching Notes are provided on the margins. These notes include simple recall of basic definitions and theorems, suggested teaching methods, alternative answers to some exercises, quick approaches and techniques in solving particular problems, and common errors committed by students.
We hope that Pre-Calculus teachers will find this Teaching Guide helpful and convenient to use. We encourage the teachers to study this Guide carefully and solve the exercises themselves. Although great effort has been put to this Guide for technical correctness and precision, any mistake found and reported to the Team is a gain for other teachers. Thank you for your cooperation.
Basic Calculus – Specialized Subject / Academic-STEM
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. No changes have been made.
- Media and Information Literacy: Senior High School Teaching Guide
- Physical Science: Senior High School SHS Teaching Guide<|endoftext|>
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Two rhombuses have sides with lengths of 9 . If one rhombus has a corner with an angle of pi/12 and the other has a corner with an angle of (7pi)/12 , what is the difference between the areas of the rhombuses?
Aug 8, 2016
$57.28$
Explanation:
Area of the rhombus with angle $\theta = \frac{\pi}{12}$ and Side $a = 9$ is
$= {a}^{2} \sin \theta$
$= {9}^{2} \sin \left(\frac{\pi}{12}\right)$
$= 81 \left(0.2588\right)$
$= 20.96$
Area of the rhombus with angle $\theta = 7 \frac{\pi}{12}$ and Side $a = 9$ is
$= {a}^{2} \sin \theta$
$= {9}^{2} \sin \left(\frac{7 \pi}{12}\right)$
$= 81 \left(0.966\right)$
$= 78.24$
So difference in Area$= 78.24 - 20.96 = 57.28$<|endoftext|>
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# Are all rhombuses also rectangles?
## Are all rhombuses also rectangles?
A square is a special case of rectangle with all equal sides. Since, we know, a rhombus has all equal sides. The sets of rectangles and rhombuses only intersect in the case of squares. Therefore, the rectangle is not a rhombus.
Is rhombus a square True or false?
A rhombus is a square. This is sometimes true. It is true when a rhombus has 4 right angles. It is not true when a rhombus does not have any right angles.
Is every rhombus a square True or false?
A rhombus is a quadrilateral with all sides equal in length. A square is a quadrilateral with all sides equal in length and all interior angles right angles. Thus a rhombus is not a square unless the angles are all right angles. A rhombus that is not a square.
### Why is a rhombus a rectangle?
The dual polygon of a rhombus is a rectangle: A rhombus has all sides equal, while a rectangle has all angles equal. A rhombus has opposite angles equal, while a rectangle has opposite sides equal.
Are all rhombuses square?
Therefore, the square is always a rhombus but a rhombus may not necessarily be a square. A square is always a rhombus since all the sides of a square are equal in length. In addition to this, the diagonals of both the closed figures, square and rhombus are perpendicular to each other and bisect the opposite angles.
Are all rhombuses parallelograms?
Not every parallelogram is a rhombus, though any parallelogram with perpendicular diagonals (the second property) is a rhombus. In general, any quadrilateral with perpendicular diagonals, one of which is a line of symmetry, is a kite.
#### Is every square is a rectangle?
All squares are rectangles, but not all rectangles are squares.
Is a rhombus a rectangle always sometimes or never?
The diagonals of a rhombus are equal. There is one right angle in a parallelogram and it is not a rectangle. An equiangular rhombus is a square. The opposite angles of a parallelogram are supplementary.
Is a rhombus always a square?
## Is a rhombus never a square?
A rhombus is a quadrilateral with all sides equal in length. A square is a quadrilateral with all sides equal in length and all interior angles right angles. Thus a rhombus is not a square unless the angles are all right angles.
Are all squares rhombuses true or false?
Hence, every square is a rhombus but the opposite is not true. Why all squares are rhombus? Square is a rhombus because as rhombus all the sides of a square are equal in length. Even, the diagonals of both square and rhombus are perpendicular to each other and bisect the opposite angles. Therefore, we can say the square is a rhombus.
Are all squares a rhombus?
It is often said that a square is a rhombus, but a rhombus isn’t always a square because a square fulfills all the properties of a rhombus. When a rhombus satisfies all the properties of a square, it becomes a square. Let us learn more about the difference between square and rhombus.
### What is the difference between a rhombus and a rectangle?
Properties of a Rhombus and a Rectangle. All four sides are equal in length.
• Rhombus and Rectangle are Parallelograms. A parallelogram is a two-dimensional closed shape with opposites sides being equal and parallel to each other.
• Difference Between a Rhombus and a Rectangle.
• Rhombus and Rectangle Formulas.
• What is an example of a rhombus in real life?
Ice Cubes. As soon as summers arrive,we begin to stock our freezers with ice cube trays.…
• Dice. Dice are used all over the world for various games.…
• Sugar Cubes. Two cubes of sugar,please!…
• Rubik Cube.…
• Old Iron Lockers.…<|endoftext|>
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# Square Root of 11
Finding the square of numbers is easy. You have to multiply the number against itself to get its square. Mathematicians have invented some methods of finding the square root of a given number. Now finding the square root of 11 and other numbers is easy with square root formula and long division method.
## Square root of 11, = 3.31662479036
To find the square root of any number, you must know what squares are and how to find the square of any number.
## What is the square of a number?
If we multiply an integer by itself, the product we get is the square of the number.
For example 3 x 3 = 9, so the square of 3 is 9.
Below are some more examples to see how the squares are formed.
## table of squares of numbers
The square of a number can be represented in many ways.
For example, if you want to square 4;
You can say, 16 is the square of 4 or the square of 4 is 16 or 4 is the square of 16 or 16 is a square number or 16 is a perfect square of 4.
#### square root
Finding the square root of any number, such as the square root of 2, is the exact opposite of computing squares.
## What is the square root of 11?
A square root is represented by a symbol: ‘ ‘.
• The square root of a number is a number such that in mathematics b² = a, or a number b whose square is a. Example, 3 and -3 are square roots of 9 because 3² = (-3)² = 9.
### square root of 11 by long division method
To find the square root of 11, use the long division method to get the approximate value.
Scroll to Top<|endoftext|>
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21st Century Fact Finds Using Online Research Tools to Reinforce Common Core Skills Grade 5
Fascinating stories help students develop skills not only in comprehension and vocabulary, but also in critical thinking since they have to do some research to determine which "facts" are true and which aren't facts at all!
Students find this information by using online research tools including:
- Metric Converters
- Image Galleries
Follow-up questions require students to synthesize what they researched and apply their knowledge.
A related writing exercise challenges them to think critically about what they have read, form opinions using information they inferred or determined to be true or false, an combine logical reasoning with writing mechanics.
Correlated to the Common Core State Standards.
Product Code TCR 3494<|endoftext|>
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Strategic Cybersecurity, Module 5: Vulnerabilities and Vectors.
This lecture describes the classification of the Internet into five distinct layers (geographic, physical network, logic, cyber persona, and persona) and the vulnerabilities of each of those layers. The module also looks at common types of cyber attacks.
Once you have completed the readings, lecture, activity, and assessment, you will be able to:
- Describe the five layers of the Internet from which all vulnerabilities arise
- Describe the following: DDoS attack, spear phishing attack, zero-day exploit.
As your readings for this module have noted, the Internet was originally designed to be an open system for sharing information, but securing its infrastructure was largely an afterthought.
The 5 Layers of The Internet
Rosenzweig provides a relatively easy-to-understand classification of the Internet into five distinct layers:
- a geographic layer,
- a physical network layer,
- a logic layer,
- a cyber persona layer,
- and a persona layer.
Each of these layers has its own specific vulnerabilities, which we will discuss in this module.
The geographic layer of the Internet comprises the physical location of its components, such as computers, servers, routers, and cables that are linked throughout the world.
Due to the diverse locations of these components, they are subject to various legal and political jurisdiction, making open, unadulterated access difficult to ensure.
One vulnerability of the geographic layer, then, might be an authoritarian government that filters information it deems subversive.
Physical Network Layer
The physical network layer of the Internet comprises its actual equipment and components, such as the computers, servers, routers, and cables just mentioned.
One example of a vulnerability to the physical layer is a major geological event, such as the 2006 earthquake off the coast of Taiwan. That earthquake damaged several communications cables, disrupting the Internet service in Japan, Taiwan, South Korea, China, and other Asian countries.
Another example of a physical layer vulnerability is the accidental cut of a fiber-optic cable during construction work. This error can disrupt Internet service for an entire building or neighborhood.
The logic layer represents the binary system of the Internet's integrated circuits, storing and transmitting the ones and zeros as needed. Vulnerability in this layer includes disruption of the computer code for illicit purposes, a primary focus of malware.
Another example of vulnerability to the logic layer is a zero-day exploit, in which a user identifies a weakness in the computer program's code and sells or uses this information to gain unlawful access to the program or even to crash it.
Once the vulnerability is realized by the owners of the computer code, it is generally immediately patched.
Cyber Persona Layer
The cyber persona layer represents merely how the users of the Internet are identified. The cyber persona layer includes such elements as your email addresses and IP addresses associated with your communication devices.
Let's say you receive a message from an email address that, at first glance, appears to be from your bank, but it is, in fact, slightly different. It asks you for confidential information, such as your username and password. You have now just experienced a vulnerability of the cyber persona layer of the Internet.
The fifth layer of the Internet, the persona layer, comprises the actual people using the Internet, not just the representations with email or IP addresses. In many ways, the persona layer is most vulnerable to malicious actions such as socially engineered phishing emails.
When we are tired or overworked, we may unwittingly open a spoofed email, not realizing what we thought was our bank's email address is actually one character off and from a criminal instead.
In addition to the description of the five layers of the Internet, this module's readings highlighted several common types of cyber attacks. One of the more common attacks you will see is a distributed denial of service, or DDoS, attack. A DDoS attack occurs when hundreds or thousands of computers, infected with a specific form of malware, barrages a specific server with request for information. This barrage of requests can overburden a server to a point that it crashes.
Quiz Question 1: Which of the following is not part of Rosenzweig's five layers of the Internet?
A: geographic layer.
B: BIOS layer.
C: physical network layer.
D: logic layer.
E: persona layer.
The answer is B: BIOS layer.
Quiz Question 2: 101 True or false: The geographic layer of the Internet comprises its actual equipment and components, such as computers, servers, routers, and cables.
The answer is False. The physical network layer comprises its actual equipment and components, such as computers, servers, routers, and cables.
Quiz Question 3: Which type of cyber attack is defined as when hundreds or thousands of computers infected with a specific form of malware barrage a specific server with request for information, resulting in a server crash?
A: zero-day exploit.
B: phishing email.
C: DDoS attack.
The answer is C: DDoS attack.
The activity for this module asks that using a paper and pen, or any type of drawing app, illustrate the five distinct layers of the Internet as well as one vulnerability that might occur in that layer. You may use images, text, or anything else to communicate your thoughts. When finished, consider if you are more vulnerable in some layers versus others.<|endoftext|>
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10.5 Conic sections in polar coordinates (Page 6/8)
Page 6 / 8
$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1$
$\frac{{\left(y-1\right)}^{2}}{49}-\frac{{\left(x+1\right)}^{2}}{4}=1$
${x}^{2}-4{y}^{2}+6x+32y-91=0$
$2{y}^{2}-{x}^{2}-12y-6=0$
For the following exercises, find the equation of the hyperbola.
Center at $\text{\hspace{0.17em}}\left(0,0\right),$ vertex at $\text{\hspace{0.17em}}\left(0,4\right),$ focus at $\text{\hspace{0.17em}}\left(0,-6\right)$
Foci at $\text{\hspace{0.17em}}\left(3,7\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(7,7\right),$ vertex at $\text{\hspace{0.17em}}\left(6,7\right)$
$\frac{{\left(x-5\right)}^{2}}{1}-\frac{{\left(y-7\right)}^{2}}{3}=1$
The Parabola
For the following exercises, write the equation of the parabola in standard form. Then give the vertex, focus, and directrix.
${y}^{2}=12x$
${\left(x+2\right)}^{2}=\frac{1}{2}\left(y-1\right)$
${\left(x+2\right)}^{2}=\frac{1}{2}\left(y-1\right);\text{\hspace{0.17em}}$ vertex: $\text{\hspace{0.17em}}\left(-2,1\right);\text{\hspace{0.17em}}$ focus: $\text{\hspace{0.17em}}\left(-2,\frac{9}{8}\right);\text{\hspace{0.17em}}$ directrix: $\text{\hspace{0.17em}}y=\frac{7}{8}$
${y}^{2}-6y-6x-3=0$
${x}^{2}+10x-y+23=0$
${\left(x+5\right)}^{2}=\left(y+2\right);\text{\hspace{0.17em}}$ vertex: $\text{\hspace{0.17em}}\left(-5,-2\right);\text{\hspace{0.17em}}$ focus: $\text{\hspace{0.17em}}\left(-5,-\frac{7}{4}\right);\text{\hspace{0.17em}}$ directrix: $\text{\hspace{0.17em}}y=-\frac{9}{4}$
For the following exercises, graph the parabola, labeling vertex, focus, and directrix.
${x}^{2}+4y=0$
${\left(y-1\right)}^{2}=\frac{1}{2}\left(x+3\right)$
${x}^{2}-8x-10y+46=0$
$2{y}^{2}+12y+6x+15=0$
For the following exercises, write the equation of the parabola using the given information.
Focus at $\text{\hspace{0.17em}}\left(-4,0\right);\text{\hspace{0.17em}}$ directrix is $\text{\hspace{0.17em}}x=4$
Focus at $\text{\hspace{0.17em}}\left(2,\frac{9}{8}\right);\text{\hspace{0.17em}}$ directrix is $\text{\hspace{0.17em}}y=\frac{7}{8}$
${\left(x-2\right)}^{2}=\left(\frac{1}{2}\right)\left(y-1\right)$
A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 5 feet across at its opening and 1.5 feet deep.
Rotation of Axes
For the following exercises, determine which of the conic sections is represented.
$16{x}^{2}+24xy+9{y}^{2}+24x-60y-60=0$
${B}^{2}-4AC=0,$ parabola
$4{x}^{2}+14xy+5{y}^{2}+18x-6y+30=0$
$4{x}^{2}+xy+2{y}^{2}+8x-26y+9=0$
${B}^{2}-4AC=-31<0,$ ellipse
For the following exercises, determine the angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that will eliminate the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term, and write the corresponding equation without the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term.
${x}^{2}+4xy-2{y}^{2}-6=0$
${x}^{2}-xy+{y}^{2}-6=0$
$\theta ={45}^{\circ },{{x}^{\prime }}^{2}+3{{y}^{\prime }}^{2}-12=0$
For the following exercises, graph the equation relative to the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ system in which the equation has no $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term.
$9{x}^{2}-24xy+16{y}^{2}-80x-60y+100=0$
${x}^{2}-xy+{y}^{2}-2=0$
$\theta ={45}^{\circ }$
$6{x}^{2}+24xy-{y}^{2}-12x+26y+11=0$
Conic Sections in Polar Coordinates
For the following exercises, given the polar equation of the conic with focus at the origin, identify the eccentricity and directrix.
Hyperbola with $\text{\hspace{0.17em}}e=5\text{\hspace{0.17em}}$ and directrix $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ units to the left of the pole.
Ellipse with $\text{\hspace{0.17em}}e=\frac{3}{4}\text{\hspace{0.17em}}$ and directrix $\text{\hspace{0.17em}}\frac{1}{3}\text{\hspace{0.17em}}$ unit above the pole.
For the following exercises, graph the conic given in polar form. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse or a hyperbola, label the vertices and foci.
For the following exercises, given information about the graph of a conic with focus at the origin, find the equation in polar form.
Directrix is $\text{\hspace{0.17em}}x=3\text{\hspace{0.17em}}$ and eccentricity $\text{\hspace{0.17em}}e=1$
Directrix is $\text{\hspace{0.17em}}y=-2\text{\hspace{0.17em}}$ and eccentricity $\text{\hspace{0.17em}}e=4$
Practice test
For the following exercises, write the equation in standard form and state the center, vertices, and foci.
$\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1$
$\frac{{x}^{2}}{{3}^{2}}+\frac{{y}^{2}}{{2}^{2}}=1;\text{\hspace{0.17em}}$ center: $\text{\hspace{0.17em}}\left(0,0\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(3,0\right),\left(–3,0\right),\left(0,2\right),\left(0,-2\right);\text{\hspace{0.17em}}$ foci: $\left(\sqrt{5},0\right),\left(-\sqrt{5},0\right)$
$9{y}^{2}+16{x}^{2}-36y+32x-92=0$
For the following exercises, sketch the graph, identifying the center, vertices, and foci.
$\frac{{\left(x-3\right)}^{2}}{64}+\frac{{\left(y-2\right)}^{2}}{36}=1$
center: $\text{\hspace{0.17em}}\left(3,2\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(11,2\right),\left(-5,2\right),\left(3,8\right),\left(3,-4\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(3+2\sqrt{7},2\right),\left(3-2\sqrt{7},2\right)$
$2{x}^{2}+{y}^{2}+8x-6y-7=0$
Write the standard form equation of an ellipse with a center at $\text{\hspace{0.17em}}\left(1,2\right),$ vertex at $\text{\hspace{0.17em}}\left(7,2\right),$ and focus at $\text{\hspace{0.17em}}\left(4,2\right).$
$\frac{{\left(x-1\right)}^{2}}{36}+\frac{{\left(y-2\right)}^{2}}{27}=1$
A whispering gallery is to be constructed with a length of 150 feet. If the foci are to be located 20 feet away from the wall, how high should the ceiling be?
For the following exercises, write the equation of the hyperbola in standard form, and give the center, vertices, foci, and asymptotes.
$\frac{{x}^{2}}{49}-\frac{{y}^{2}}{81}=1$
$\frac{{x}^{2}}{{7}^{2}}-\frac{{y}^{2}}{{9}^{2}}=1;\text{\hspace{0.17em}}$ center: $\text{\hspace{0.17em}}\left(0,0\right);\text{\hspace{0.17em}}$ vertices $\text{\hspace{0.17em}}\left(7,0\right),\left(-7,0\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(\sqrt{130},0\right),\left(-\sqrt{130},0\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=±\frac{9}{7}x$
$16{y}^{2}-9{x}^{2}+128y+112=0$
For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes.
$\frac{{\left(x-3\right)}^{2}}{25}-\frac{{\left(y+3\right)}^{2}}{1}=1$
center: $\text{\hspace{0.17em}}\left(3,-3\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(8,-3\right),\left(-2,-3\right);$ foci: $\text{\hspace{0.17em}}\left(3+\sqrt{26},-3\right),\left(3-\sqrt{26},-3\right);\text{\hspace{0.17em}}$ asymptotes: $\text{\hspace{0.17em}}y=±\frac{1}{5}\left(x-3\right)-3$
${y}^{2}-{x}^{2}+4y-4x-18=0$
Write the standard form equation of a hyperbola with foci at $\text{\hspace{0.17em}}\left(1,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,6\right),$ and a vertex at $\text{\hspace{0.17em}}\left(1,2\right).$
$\frac{{\left(y-3\right)}^{2}}{1}-\frac{{\left(x-1\right)}^{2}}{8}=1$
For the following exercises, write the equation of the parabola in standard form, and give the vertex, focus, and equation of the directrix.
${y}^{2}+10x=0$
$3{x}^{2}-12x-y+11=0$
${\left(x-2\right)}^{2}=\frac{1}{3}\left(y+1\right);\text{\hspace{0.17em}}$ vertex: $\text{\hspace{0.17em}}\left(2,-1\right);\text{\hspace{0.17em}}$ focus: $\text{\hspace{0.17em}}\left(2,-\frac{11}{12}\right);\text{\hspace{0.17em}}$ directrix: $\text{\hspace{0.17em}}y=-\frac{13}{12}$
For the following exercises, graph the parabola, labeling the vertex, focus, and directrix.
${\left(x-1\right)}^{2}=-4\left(y+3\right)$
${y}^{2}+8x-8y+40=0$
Write the equation of a parabola with a focus at $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ and directrix $\text{\hspace{0.17em}}y=-1.$
A searchlight is shaped like a paraboloid of revolution. If the light source is located 1.5 feet from the base along the axis of symmetry, and the depth of the searchlight is 3 feet, what should the width of the opening be?
Approximately $\text{\hspace{0.17em}}8.49\text{\hspace{0.17em}}$ feet
For the following exercises, determine which conic section is represented by the given equation, and then determine the angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ that will eliminate the $\text{\hspace{0.17em}}xy\text{\hspace{0.17em}}$ term.
$3{x}^{2}-2xy+3{y}^{2}=4$
${x}^{2}+4xy+4{y}^{2}+6x-8y=0$
parabola; $\text{\hspace{0.17em}}\theta \approx {63.4}^{\circ }$
For the following exercises, rewrite in the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ system without the $\text{\hspace{0.17em}}{x}^{\prime }{y}^{\prime }\text{\hspace{0.17em}}$ term, and graph the rotated graph.
$11{x}^{2}+10\sqrt{3}xy+{y}^{2}=4$
$16{x}^{2}+24xy+9{y}^{2}-125x=0$
${{x}^{\prime }}^{2}-4{x}^{\prime }+3{y}^{\prime }=0$
For the following exercises, identify the conic with focus at the origin, and then give the directrix and eccentricity.
Hyperbola with $\text{\hspace{0.17em}}e=\frac{3}{2},\text{\hspace{0.17em}}$ and directrix $\text{\hspace{0.17em}}\frac{5}{6}\text{\hspace{0.17em}}$ units to the right of the pole.
For the following exercises, graph the given conic section. If it is a parabola, label vertex, focus, and directrix. If it is an ellipse or a hyperbola, label vertices and foci.
Find a polar equation of the conic with focus at the origin, eccentricity of $\text{\hspace{0.17em}}e=2,$ and directrix: $\text{\hspace{0.17em}}x=3.$
Questions & Answers
how can are find the domain and range of a relations
austin Reply
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply
i want to sure my answer of the exercise
meena Reply
what is the diameter of(x-2)²+(y-3)²=25
Den Reply
how to solve the Identity ?
Barcenas Reply
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
Shakeena Reply
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
Rhudy Reply
what is a complex number used for?
Drew Reply
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim
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## What you need to know
Two straight lines are parallel if they are always the same distance away from each other, no matter how long the lines are extended. In other words, they’re going the exact same direction and will never meet. Additionally, two lines are perpendicular if, when they meet, they form a right-angle. Both of these words appear all over in maths so are worth getting used to.
Before going into this topic, you should be familiar with the equation of a straight line (https://mathsmadeeasy.co.uk/gcse-maths-revision/ymxc-gcse-maths-revision-worksheets/) as well as drawing straight-line graphs (https://mathsmadeeasy.co.uk/gcse-maths-revision/drawing-straight-line-graphs-gcse-maths-revision-worksheets/).
If two lines are parallel, then they have the same gradient. This means two things, 1) you can tell if two straight lines are parallel by looking at their equations, and 2) if you are told that a line is parallel to a different line whose gradient you know, then you also know the gradient of the first line.
Example: Is the line $y=3x-4$ parallel to the line $3y-9x=21$?
So, we need to know the gradient of both lines. The first line equation is given to us in the desired form of $y=mx+c$, so we know its gradient (the coefficient of $x$) is 3.
The second line equation requires some rearranging before we can know its gradient. Firstly, add $9x$ to both sides of the equation $3y-9x=21$ to get
$3y=9x+21$.
Then, if we divide both sides by 3 we get
$y=3x+7$.
This is now in the right form, and we can see that its gradient is 3 so must be parallel to the 1st line.
Example: Find the equation of the straight line that passes through $(1, 3)$ and is parallel to the line $y=5x-1$. Plot this straight line.
The gradient of the given line is 5, which means the gradient of the perpendicular line must also be 5. We’re given that it passes through $(1, 3)$, and we now know the gradient to be 5, so we can substitute these values into $y=mx+c$ in order to find $c$. Doing so, we get
$3 = 1\times 5 +c = 5+c, \text{ therefore }c =3-5=-2$.
So, the equation of this line is
$y=5x-2$.
We now have plenty of information to plot the graph. The result looks like the figure on the right.
The remainder of this topic is only relevant for the higher course.
If two lines are perpendicular, then the product of their two gradients (i.e., the result of multiplying them together) is -1. Another way of putting this is: if you have a straight line with gradient $m$, then a line which is perpendicular to it will have gradient $-\frac{1}{m}$. This is often referred to as the negative reciprocal of $m$.
Note: to find the reciprocal of a fraction, simply flip the fraction over.
Example: Is the line $x+4y=8$ perpendicular to the line $y=4x-13$?
The second line equation is in the desired form, but the first is not. So, subtracting $x$ from both sides of the first equation, we get
$4y = -x+8$.
Then, dividing both sides by 4, we get
$y=-\dfrac{x}{4}+2$
So, the gradient of this line is $-\frac{1}{4}$, and the gradient of the other line is 4. Multiplying these two values together, we get
$4 \times \left(-\dfrac{1}{4}\right) = -1$
Their product is -1, so the lines are perpendicular.
Example: Find the equation of the straight line that passes through $(-9, -2)$ and is perpendicular to $y=-3x+10$. Plot the straight line.
The gradient of the given line is -3, so the gradient of the perpendicular line must
$-\left(\dfrac{1}{-3}\right) = \dfrac{1}{3}$
The two minus signs cancel, so the result is a positive number. The gradient of one line and another line perpendicular to it will always have opposite signs – if one is negative the other will always be positive and vice versa.
We’re given that the line passes through $(-9, -2)$, and we now know the gradient is $\frac{1}{3}$, so we can substitute these values into $y=mx+c$ in order to find $c$. Doing so, we get
$-2=(-9)\times\dfrac{1}{3}+c=-3+c,\text{ therefore }c=-2+3=1$.
So, the equation of this line is
$y=\dfrac{1}{3}x+1$.
We now have plenty of information to plot the graph. The result looks like the figure below.
## Example Questions
#### 1) State which, if any, of the following 3 lines are parallel.a) $4y - 1 = 2x$b) $2y + 4x = 5$c) $y - \dfrac{1}{2}x = 45$
We need to write all 3 equations in the form $y=mx+c$ and see which ones have the same gradient.
1. a) Add 1 to both sides to get
$4y = 2x + 1$
Then, divide both sides by 4 to get
$y=\dfrac{1}{2}x + \dfrac{1}{4}$
1. b) Subtract $4x$ from both sides to get
$2y = -4x+5$
Then, divide both sides by 2 to get
$y = -2x + \dfrac{5}{2}$
1. c) Add $\frac{1}{2}x$ to both sides to get
$y=\dfrac{1}{2}x+45$
With all 3 equations written in the desired form, we can see that whilst b) has gradient -2, both a) and c) have gradient $\frac{1}{2}$, therefore a) and c) are parallel.
#### 2) Plot the graph, from $x=0$ to $x=4$, of the straight line that is parallel to the line $5y=-10x-3$ and passes through the point $(1, 6)$.
We need to find the gradient of the line given in the question by writing it in the form $y=mx+c$. Dividing both sides by 5, we get
$y = 2x - \dfrac{3}{5}$
Now, the line we need to draw is parallel to this one so must have the same gradient: -2. We’re not asked to work out the equation (although you’re welcome to do that if it helps) and knowing that the line has gradient -2 and passes through $(1, 6)$ is enough to draw it. The correct plotting is shown below.
#### 3) (HIGHER ONLY) Find the equation of the line that is perpendicular to $y=\dfrac{3}{7}x+9$ and passes through the point $(5, -4)$.
The line given in the question has gradient $\frac{3}{7}$. The negative reciprocal of this (and therefore the gradient of the perpendicular line) is
$-\dfrac{7}{3}$.
Now we have the gradient, and we know the lines passes through $(5, -4)$, we can substitute these values into $y=mx+c$ in order to find $c$. So, we get
$-4 = \left(-\dfrac{7}{3}\right)\times 5+c = -\dfrac{35}{3}+c$
The most important thing here is to be careful with your fraction operations. Adding $\frac{35}{3}$ to both sides, we get
\begin{aligned}c=-4+\dfrac{35}{3}&=-\dfrac{4}{1}+\dfrac{35}{3}\\&=-\dfrac{12}{3}+\dfrac{35}{3}=\dfrac{23}{3}\end{aligned}
Therefore, the equation of the line is
$y=-\dfrac{7}{3}x+\dfrac{23}{3}$
If you are looking for perpendicular lines revision notes and practice questions then you have arrived on the right page. The perpendicular and parallel line resources on this page can be used by Maths teachers and tutors everywhere to support your pupils learning. If you are interested in the other GCSE Maths resources we also offer then visit our GCSE Maths revision page.<|endoftext|>
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On June 1, 1676 the Battle of Öland was raging, as the Swedish navy grappled with a Danish-Dutch fleet for control of the southern rim of the Baltic Sea. Amid bad weather, Kronan—Sweden’s naval flagship in the region and one of the largest warships of its kind at the time—made a sudden left turn. Its sails began to take too much wind. The ship tipped over as water gushed into its gun ports. Kronan soon lay horizontal on the water. It was then that an explosion rung out, tearing off a large chunk of the vessel’s front side. Kronan’s gunpowder storage room had lit ablaze. The ship—along with around 800 men, loads of military equipment, and piles of valuable coins—sunk to the bottom of the sea, 85 feet down. Sweden lost the battle.
From 1679 to 1686, Swedish divers using diving bells recovered over 60 valuable cannons from the wreck. After that Kronan’s precise location was forgotten, the ship left alone in its watery resting place for almost three centuries. In August 1980, however, a team located the old warship once again. Since then over 30,000 artifacts have been retrieved from the wreck in one of the most elaborate archaeological projects in Sweden’s history. But materials recovered from Kronan also have important implications on Sweden’s future, as well.
One of the artifacts from the ship—a bronze cannon—has caught the attention of the Swedish Nuclear Fuel and Waste Management Company (SKB) and Finland’s nuclear-waste-management company Posiva Oy. These companies are working to build what will likely become, in the early 2020s, the world’s first two working, deep-underground repositories for spent nuclear fuel. As part of planning the repositories, SKB experts studied how the cannon fared in a harsh seawater environment over the centuries.
The cannon was selected because it contains a lot of copper, was long partially surrounded by seafloor clay, and sat for ages in in abrasive seawater. Such factors, Posiva experts suggest, makes it a good analog for the nuclear-waste canisters to be used in the repositories, which will be made of copper and surrounded by clay and groundwater for thousands of years. Studying the cannon may help predict how and whether the copper nuclear-waste containers might corrode over the long term. The experts also studied other analogs to help predict how and whether repository components will persevere, break down, or transform in futures near and distant.
The study of these analogs is part of SKB’s and Posiva’s broader endeavors to forecast Nordic regions’ radically long-term futures. In Finland, for instance, experts now work to augur changes in the geological, ecological, and climatic conditions that the repositories’ surrounding region of Olkiluoto might see over the coming millennia. Some of these experts study the possible effects that environmental changes—including those caused by humans and the natural waxing and waning of future ice ages—might have on the long-term durability of the underground facilities. Others assess potential health risks to future humans and other living things that could arise if radionuclides escape from the repository and then, in worst-case scenarios, disperse near or on Earth’s surface.
These farsighted scientific and engineering ambitions were what led me to spend 32 months living in Northern Europe, doing an anthropological study of these experts’ lives and mentalities. The specific band of experts I came to know was developing Posiva’s safety case: an enormous portfolio of technical models, data, and forecasts that aim to gauge the safety of Finland’s planned repository’s over the coming millennia.
As an anthropologist, my interest is in exploring how these experts envision the far future and how that can help us develop a more farsighted view of our world—a skill we must acquire in our current moment of global environmental crisis.
Tomorrow on Facts So Romantic, Vincent Ialenti looks more closely at how researchers used analogs to try to predict the future of the nuclear repositories.
Vincent Ialenti is a U.S. National Science Foundation Graduate Research Fellow and a PhD Candidate in Cornell University’s department of anthropology. He holds an MSc in law, anthropology & society from the London School of Economics.<|endoftext|>
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This post is also available in: Português
To the many moral and practical reasons for protecting the creatures with whom we share this blue marble, add one more. They might help Earth safely store more carbon.
Writing in the journal Nature Ecology and Evolution, ecologists led by Mar Sobral and José Fragoso of Stanford University describe a three-year-long study of mammals and carbon cycles in southeast Guyana. Previous research suggested links between biodiversity and carbon storage; the more species of plants and animals there are in a given ecosystem, the more CO2 it seems to absorb. Exactly how this works, though, is still being revealed.
Sobral and Fragoso’s team wanted to learn more about the roles played by mammals. And so, across three years and 18,500 square miles, they conducted more than 10,000 surveys, counting some 218,000 individual mammals belonging to 28 species. They noted more than 43,000 feeding events and identified more than one million ground-deposited food remains. While in the field, they also measured soil carbon concentrations and tree size at hundreds of sites. Out of the number-crunching emerged a pattern: more mammal species meant more carbon ending up in soil or the bodies of trees.
This pattern reflects a vast, unceasing, little-appreciated animal labor — “moving plant matter across the landscape and processing it in ways that make it available to a larger diversity of invertebrates, fungi and microbes,” write the researchers. Eating, digesting, defecating. Spreading nutrients and, crucially, the seeds of trees that eventually breathe atmospheric CO2 and lock it in cellulose. Eventually dying and decomposing, nourishing yet more growth. Each action of minute significance but, like raindrops, adding up.
The researchers didn’t calculate exactly how much extra carbon a mammal-rich forest might store, but Sobral offered this rough estimation: a forest with 30 mammal species should sequester an extra 10,000 kilograms per hectare in above-ground tree biomass alone. Extrapolated to the entire Amazon, that translates to some 5.5 billion tons of carbon, roughly equivalent to U.S. emissions in 2015 — and that doesn’t even account for soil-bound carbon, which would drive the total much higher. It’s a back-of-the-envelope calculation, but it hints at the potential.
Far more remains to be learned, write the researchers. The relationships are not simple. They involve many scales and processes, and “cannot be easily deconstructed into individual function–service relationships.” Yet some implications are already clear: human-induced defaunation is potentially a climate problem, and conservation a climate solution.
Near roads and communities, where habitat was fragmented and disturbance higher and hunting more intense, trees were smaller and soils stored less carbon. “Mammal diversity should indeed be considered in management policies seeking to maximize carbon storage,” says Sobral. And while the study focused specifically on mammals and Guyana, she believes the findings hold for other creatures — reptiles, amphibians, birds, invertebrates — in other types of forest, though this this needs to be tested. “I believe the richness of any animal group would affect the biogeochemical cycles,” she says.
Source: Sobral et al., “Mammal diversity influences the carbon cycle through trophic interactions in the Amazon.” Nature Ecology & Evolution, 2017.
Image: Marissa Strniste / Flickr<|endoftext|>
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## #math trivia #107 solution
Assuming that a is non-zero, the square root must be a three-digit number. Let xyz be the digits of a number N, so that
N = 100*x + 10*y + z
The square of N is then
N^2 = 10000*x^2 + 1000*(2*x*y) + 100*(2*x*z+y^2) + 10*(2*y*z) + z^2.
When y = 0, the equation simplifies to:
N^2 = 10000*x^2 + 100*(2*x*z) + z^2.
To get the pattern aabbc, we need:
• x = 1, 2, or 3, so that x^2 is a single digit
• 2*x*z has the same 10’s digit as x^2
• 2*x*z has the same 1’s digit as z^2
Let’s consider the three values of x in turn.
If x = 1, then we need 2*x*z to be between 10 and 19, which means that z must be between 5 and 9. The possible values of 2*x*z are thus 10, 12, 14, 16, and 18. In the last three cases, 2*x*z has the same 1’s digits as z^2, corresponding to the values
107^2 = 11449
108^2 = 11664
109^2 = 11881
If x = 2, then we need 2*x*z to be between 20 and 29, which means that z must be between 5 and 7. The possible values of 2*x*z are thus 20, 24, and 28. None of these cases matches z^2.
If x = 3, z must be either 5 or 6. None of these cases matches z^2 either.
So, the other numbers (so far) are 108 and 109.
Are there any with y > 0? A quick spreadsheet calculation reveals two more: 235^2 = 55225 and 315^2 = 99225.
If we allow a be zero, then we can also consider 1^1 = 1, 2^2 = 4, 3^2 = 9, 15^2 = 225, and 21^2 = 441.<|endoftext|>
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Earlier this year, the largest ever iceberg, named A-68, broke off from Antarctica. Now, Nasa has captured incredible aerial shots of the massive trillion tonne iceberg. Nasa’s Operation IceBridge flew a special aircraft over the Larsen C ice shelf, capturing several photos of the massive iceberg, which despite having broken off into smaller sections, remains one of the largest bergs on Earth.
Nasa’s photos mark the first time that the A-68 iceberg was seen by human eyes, having only been previously spotted by satellites. As the berg floats father away from Antarctica, it continues splintering into smaller chunks. Although the iceberg does not pose much of an immediate danger, the calving event is expected to eventually exacerbate the issue of rising sea levels.
Mashable reported that with the rise in air and sea temperatures, the region’s Antarctic ice sheet is now considered to be more vulnerable to melting. In fact, just weeks after A-68’s so-called birth, scientists spotted new cracks on the Antarctic ice shelf.
“I was aware that I would be seeing an iceberg the size of Delaware, but I wasn’t prepared for how that would look from the air,” science writer Kathryn Hansen, who was on the flight that flew over A-68, wrote in a piece penned for Nasa’s Earth Observatory. “Most icebergs I have seen appear relatively small and blocky, and the entire part of the berg that rises above the ocean surface is visible at once. Not this berg. A-68 is so expansive it appears if it were still part of the ice shelf.”
In addition to taking photos of the massive berg, Nasa scientists also measured the depth of the water below A-68 using radar and a gravimeter. Operation IceBridge’s mission provided scientists with the clearest images of A-68, which in turn will help them track and study its progress.
Last month, scientists discovered that the A-68 iceberg has an entire hidden ecosystem that has remained frozen for nearly 6 120,000 years.<|endoftext|>
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Main content
## MAP Recommended Practice
### Course: MAP Recommended Practice>Unit 34
Lesson 27: Multiplying fractions word problems
# Multiplying fractions word problem: bike
This video is all about understanding how to multiply fractions and mixed numbers. Watch as the steps are explained in a simple and fun way. Created by Sal Khan.
## Want to join the conversation?
• isn't 3 1/3's improper fraction supposed to be 10/3
(22 votes)
• Hey, the answer is 10/3 but he just did 9/3 + 1/3 to simplify how to find ten thirds here
(2 votes)
• I didn't get really why we should multiply! I do on paper though. But Sal understand that right way! Can someone explain that why should we multiply?
(9 votes)
• As you learn word problems, you will find that there are a variety of formulas that occur and that you need to learn. This problem uses one of those formulas. Specifically, it used the formula: Distance = Rate (Time).
Sal knows this formula, so he knows he needs to multiply the speed/ rate (the 1/5 miles per minutes) times the time (the 3 1/3 minutes).
Hope this helps.
(8 votes)
• umm, why did he not simplify the 10 and the 5 before multiplying them? that would have make things much easier
(7 votes)
• He could but maybe it would be easier for some people if he does the other way.
(5 votes)
• Is there a quicker way and easier way ?
(8 votes)
• You can ride your bike 1/5 of a mile per minute. If it takes you 3 and 1/3 minutes to get to your friend's house, how many miles away does your friend live? And this here is pictures of these guys on bicycles. It's pretty clear they're not riding to work, or some of these guys aren't even riding a bicycle. But let's focus on the question. So you can ride your bike 1/5 of a mile per minute. And you're going to do this for 3 and 1/3 minutes-- times 3 and 1/3. So we really have to figure out, how do we multiply 1/5 times 3 and 1/3? So there's a couple of ways to think about it. You could literally view a 3 and 1/3 as this is the same thing as 1/5 times 3 plus 1/3. That's exactly what 3 and 1/3 is. And then we can just apply the distributive property. This would be 1/5 times 3-- I'm going to keep the colors the same-- plus 1/5 times 1/3. And this is going to be equal to-- well, we could rewrite 1/5 times 3 as 1/5 times 3/1. That's what 3 really is if we wrote it as a fraction. And then, of course, we're going to have plus 1/5 times 1/3. And let's just think about what each of these evaluate to. Here you multiplied the numerators, and you multiplied the denominators. So this is going to be equal to 1 times 3 over 5 times 1. And this business right over here is going to be-- and remember, order of operations. We want to do our multiplication first. So this is going to be 1 times 1 over 5 times 3. And so that's going to be equal to 3/5 plus 1/15. And now we have different denominators here. But lucky for us, 3/5, if we multiplied the numerator and the denominator by 3, we're going to get a denominator of 15. And so that's equal to 9/15 plus 1/15, which equals 10/15. And if you divide the numerator and the denominator both by 5, you're going to get 2/3. So your friend lives 2/3 miles away from your house. Well, that's kind of interesting. And this was kind of a long way to do it. Let's think about if there's a simpler way to do it. So this is the same thing as 1/5 times-- and I'm just going to write 3 and 1/3 as a mixed number. So it's 1/5 times 3 and 1/3 can be rewritten as 9/3-- sorry, I'm going to rewrite 3 and 1/3 as an improper fraction. So this is the same thing as 9/3-- that's 3-- plus 1/3, which is the same thing as 1/5-- well, I switched colors arbitrarily-- which is the same thing-- I'm still on the same color-- as 1/5 times 9/3 plus 1/3 is 10/3. And now we can just multiply the numerator and multiply the denominator-- or multiply the numerators. So this is 1 times 10-- I'm trying to stay good with the color coding-- over 5 times 3, which is exactly equal to what we just got. 1 times 10 is equal to 10. 5 times 3 is 15. 10/15, we already established, is the same thing as 2/3. So your friend lives 2/3 of a mile away from you.
(6 votes)
• If I'm being honest the way he explains things is very confusing. I couldn't understand the lesson.
(5 votes)
• There is a much easier way and less complicated way. You don’t have to do all the complex steps that sal has shown. He is just showing different ways to do it so hopefully you might get it. Start with 1/5 x 3 1/3. 3 1/3 can be changed to 10/3 * 1/5 =10/15. Then you can simplify to 2/3 which gives you the answer. Hope this helps!
(4 votes)
• at to about he multiplies 1/5X 3 + 3X like 1/3. How is this possible?
(4 votes)
• Isn't 3 times 3 plus 1 10/3? Why'd he right 9/3 + 1/3?
(3 votes)
• not gonna lie I din't understand most of that
(2 votes)
## Video transcript
You can ride your bike 1/5 of a mile per minute. If it takes you 3 and 1/3 minutes to get to your friend's house, how many miles away does your friend live? And this here is pictures of these guys on bicycles. It's pretty clear they're not riding to work, or some of these guys aren't even riding a bicycle. But let's focus on the question. So you can ride your bike 1/5 of a mile per minute. And you're going to do this for 3 and 1/3 minutes-- times 3 and 1/3. So we really have to figure out, how do we multiply 1/5 times 3 and 1/3? So there's a couple of ways to think about it. You could literally view a 3 and 1/3 as this is the same thing as 1/5 times 3 plus 1/3. That's exactly what 3 and 1/3 is. And then we can just apply the distributive property. This would be 1/5 times 3-- I'm going to keep the colors the same-- plus 1/5 times 1/3. And this is going to be equal to-- well, we could rewrite 1/5 times 3 as 1/5 times 3/1. That's what 3 really is if we wrote it as a fraction. And then, of course, we're going to have plus 1/5 times 1/3. And let's just think about what each of these evaluate to. Here you multiplied the numerators, and you multiplied the denominators. So this is going to be equal to 1 times 3 over 5 times 1. And this business right over here is going to be-- and remember, order of operations. We want to do our multiplication first. So this is going to be 1 times 1 over 5 times 3. And so that's going to be equal to 3/5 plus 1/15. And now we have different denominators here. But lucky for us, 3/5, if we multiplied the numerator and the denominator by 3, we're going to get a denominator of 15. And so that's equal to 9/15 plus 1/15, which equals 10/15. And if you divide the numerator and the denominator both by 5, you're going to get 2/3. So your friend lives 2/3 miles away from your house. Well, that's kind of interesting. And this was kind of a long way to do it. Let's think about if there's a simpler way to do it. So this is the same thing as 1/5 times-- and I'm just going to write 3 and 1/3 as a mixed number. So it's 1/5 times 3 and 1/3 can be rewritten as 9/3-- sorry, I'm going to rewrite 3 and 1/3 as an improper fraction. So this is the same thing as 9/3-- that's 3-- plus 1/3, which is the same thing as 1/5-- well, I switched colors arbitrarily-- which is the same thing-- I'm still on the same color-- as 1/5 times 9/3 plus 1/3 is 10/3. And now we can just multiply the numerator and multiply the denominator-- or multiply the numerators. So this is 1 times 10-- I'm trying to stay good with the color coding-- over 5 times 3, which is exactly equal to what we just got. 1 times 10 is equal to 10. 5 times 3 is 15. 10/15, we already established, is the same thing as 2/3. So your friend lives 2/3 of a mile away from you.<|endoftext|>
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# Solving an equation with parentheses Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Solving an equation with parentheses. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Solve the following equation:
7(p + 6) = 28
### Explanation
Step 1:
Given $7(p + 6) = 28$
Dividing both sides by 7
$\frac{7(p + 6)}{7} = \frac{28}{7}; \: (p + 6) = 4$
Step 2:
Subtracting 6 from both sides
$p + 6 − 6 = 4 − 6 = −2$
So, $p = −2$
Q 2 - Solve the following equation:
5(m + 7) = 35
### Explanation
Step 1:
Given $5(m + 7) = 35$
Dividing both sides by 5
$\frac{5(m + 7)}{5} = \frac{35}{5}; \: (m + 7) = 7$
Step 2:
Subtracting 7 from both sides
$m + 7 − 7 = 7 − 7 = 0$
So, $m = 0$
Q 3 - Solve the following equation:
13(g − 2) = 52
### Explanation
Step 1:
Given $13(g − 2) = 52$
Dividing both sides by 13
$\frac{13(g − 2)}{13} = \frac{52}{13}; \: (g − 2) = 4$
Step 2:
$g + 2 − 2 = 4 + 2 = 6$
So, $g = 6$
Q 4 - Solve the following equation:
27 = 9(w − 5)
### Explanation
Step 1:
Given $27 = 9(w − 5)$
Dividing both sides by 9
$\frac{9(w − 5)}{9} = \frac{27}{9}; \: (w − 5) = 3$
Step 2:
$w + 5 − 5 = 3 + 5 = 8$
So, $w = 8$
Q 5 - Solve the following equation:
7(k − 13) = 0
### Explanation
Step 1:
Given $7(k − 13) = 0$
Dividing both sides by 7
$\frac{7(k − 13)}{7} = \frac{0}{7}; \: (k − 13) = 0$
Step 2:
$k + 13 − 13 = 0 + 13 = 13$
So, $k = 13$
Q 6 - Solve the following equation:
2(n − 5) = −8
### Explanation
Step 1:
Given $2(n − 5) = −8$
Dividing both sides by 2
$\frac{2(n − 5)}{2} = \frac{−8}{2}; \: (n − 5) = −4$
Step 2:
$n + 5 − 5 = −4 + 5 = 1$
So, $n = 1$
Q 7 - Solve the following equation:
−12 = 3(c + 4)
### Explanation
Step 1:
Given $−12 = 3(c + 4)$
Dividing both sides by 3
$\frac{−12}{3} = \frac{3(c + 4)}{3}; \: (c + 4) = −4$
Step 2:
Subtracting 4 from both sides
$(c + 4 − 4) = −4 −4 = −8$
So, $c = −8$
Q 8 - Solve the following equation:
35 = 7(w − 2)
### Explanation
Step 1:
Given $35 = 7(w − 2)$
Dividing both sides by 7
$\frac{35}{7} = \frac{7(w − 2)}{7}; \: (w − 2) = 5$
Step 2:
$w + 2 − 2 = 5 + 2 = 7$
So, $w = 7$
Q 9 - Solve the following equation:
2(y − 4) = −10
### Explanation
Step 1:
Given $2(y − 4) = −10$
Dividing both sides by 2
$\frac{2(y − 4)}{2} = \frac{−10}{2}; \: (y − 4) = −5$
Step 2:
$y + 4 − 4 = −5 + 4 = −1$
So, $y = −1$
Q 10 - Solve the following equation:
4(p + 5) = 28
### Explanation
Step 1:
Given $4(p + 5) = 28$
Dividing both sides by 4
$\frac{4(p + 5)}{4} = \frac{28}{4}; \: (p + 5) = 7$
Step 2:
Subtracting 5 from both sides
$(p + 5 − 5) = 7 − 5 = 2$
So, $p = 2$
solving_an_equation_with_parentheses.htm<|endoftext|>
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# Unit 3 hw 2 - solving 1 step equations
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• ### Unit 3 hw 2 - solving 1 step equations
1. 1. Unit 3 - Homework 2 Solving 1-Step Equations
2. 2. Solving EquationsSolving means to get the variable alone.Think OPPOSITE SIDE = OPPOSITEOPERATION
3. 3. Solve for x.x+3= 5
4. 4. Solve for x. Circle the term with thex+3= 5 variable you need to isolate.
5. 5. Solve for x. Circle the term with thex+3= 5 variable you need to isolate.
6. 6. Solve for x. Circle the term with thex+3= 5 variable you need to isolate. Anything outside the circle needs to move to the other side by doing the opposite operation.
7. 7. Solve for x. Circle the term with thex+3= 5 variable you need to isolate. Anything outside the circle needs to move to the other side by doing the opposite operation. Opposite of addition is Subtraction.
8. 8. Solve for x. Circle the term with thex+3= 5 variable you need to isolate. −3 −3 Anything outside the circle needs to move to the other side by doing the opposite operation. Opposite of addition is Subtraction.
9. 9. Solve for x. Circle the term with thex+3= 5 variable you need to isolate. −3 −3 Anything outside the circle needs to move to the other side by doing the opposite x=2 operation. Opposite of addition is Subtraction.
10. 10. Solve for x. Circle the term with thex+3= 5 variable you need to isolate. −3 −3 Anything outside the circle needs to move to the other side by doing the opposite x=2 operation. Opposite of addition is Subtraction. x is alone so you are done.
11. 11. Solve for x.x − 8 = −2
12. 12. Solve for x. Circle the term with thex − 8 = −2 variable you need to isolate.
13. 13. Solve for x. Circle the term with thex − 8 = −2 variable you need to isolate.
14. 14. Solve for x. Circle the term with thex − 8 = −2 variable you need to isolate. Move 8 by opposite operation.
15. 15. Solve for x. Circle the term with thex − 8 = −2 variable you need to isolate. Move 8 by opposite operation. Opposite of subtraction is addition.
16. 16. Solve for x. Circle the term with thex − 8 = −2 variable you need to +8 +8 isolate. Move 8 by opposite operation. Opposite of subtraction is addition.
17. 17. Solve for x. Circle the term with thex − 8 = −2 variable you need to +8 +8 isolate. Move 8 by opposite x=6 operation. Opposite of subtraction is addition.
18. 18. Solve for x. Circle the term with thex − 8 = −2 variable you need to +8 +8 isolate. Move 8 by opposite x=6 operation. Opposite of subtraction is addition. x is alone so you are done.
19. 19. Solve for x.−3x = −8
20. 20. Solve for x. Circle the term with the−3x = −8 variable you need to isolate.
21. 21. Solve for x. Circle the term with the−3x = −8 variable you need to isolate.
22. 22. Solve for x. Circle the term with the−3x = −8 variable you need to isolate. There is nothing outside the circle so look inside now.
23. 23. Solve for x. Circle the term with the−3x = −8 variable you need to isolate. There is nothing outside the circle so look inside now. Opposite of multiplication is division.
24. 24. Solve for x. Circle the term with the−3x = −8 variable you need to isolate. − 3 −3 There is nothing outside the circle so look inside now. Opposite of multiplication is division.
25. 25. Solve for x. Circle the term with the−3x = −8 variable you need to isolate. − 3 −3 There is nothing outside the circle so look inside 8 now. x= Opposite of multiplication is 3 division.
26. 26. Solve for x. Circle the term with the−3x = −8 variable you need to isolate. − 3 −3 There is nothing outside the circle so look inside 8 now. x= Opposite of multiplication is 3 division. x is alone so you are done.
27. 27. Solve for x. x =7 5
28. 28. Solve for x. x Circle the term with the =7 variable you need to 5 isolate.
29. 29. Solve for x. x Circle the term with the =7 variable you need to 5 isolate.
30. 30. Solve for x. x Circle the term with the =7 variable you need to 5 isolate. There is nothing outside the circle so look inside.
31. 31. Solve for x. x Circle the term with the =7 variable you need to 5 isolate. There is nothing outside the circle so look inside. Opposite of division is multiplication.
32. 32. Solve for x. x Circle the term with the5 ⋅ = 7⋅ 5 variable you need to 5 isolate. There is nothing outside the circle so look inside. Opposite of division is multiplication.
33. 33. Solve for x. x Circle the term with the5 ⋅ = 7⋅ 5 variable you need to 5 isolate. There is nothing outside x = 35 the circle so look inside. Opposite of division is multiplication.
34. 34. Solve for x. x Circle the term with the5 ⋅ = 7⋅ 5 variable you need to 5 isolate. There is nothing outside x = 35 the circle so look inside. Opposite of division is multiplication. x is alone so you are done.
35. 35. Solve for x.7 − x = −1
36. 36. Solve for x. Circle the term with the7 − x = −1 variable you need to isolate.
37. 37. Solve for x. Circle the term with the7 − x = −1 variable you need to isolate.
38. 38. Solve for x. Circle the term with the7 − x = −1 variable you need to isolate. Anything outside the circle needs to move to the other side by doing the opposite operation.
39. 39. Solve for x. Circle the term with the7 − x = −1 variable you need to isolate. Anything outside the circle needs to move to the other side by doing the opposite operation. The 7 is positive so Subtract.
40. 40. Solve for x. Circle the term with the 7 − x = −1 variable you need to isolate.−7 −7 Anything outside the circle needs to move to the other side by doing the opposite operation. The 7 is positive so Subtract.
41. 41. Solve for x. Circle the term with the 7 − x = −1 variable you need to isolate.−7 −7 Anything outside the circle needs to move to the other −x = −8 side by doing the opposite operation. The 7 is positive so Subtract.
42. 42. Solve for x. Circle the term with the 7 − x = −1 variable you need to isolate.−7 −7 Anything outside the circle needs to move to the other −x = −8 side by doing the opposite operation. The 7 is positive so Subtract. x is NOT alone. Divide by negative one.
43. 43. Solve for x. Circle the term with the 7 − x = −1 variable you need to isolate.−7 −7 Anything outside the circle needs to move to the other −x = −8 side by doing the opposite operation. −1 −1 The 7 is positive so Subtract. x is NOT alone. Divide by negative one.
44. 44. Solve for x. Circle the term with the 7 − x = −1 variable you need to isolate.−7 −7 Anything outside the circle needs to move to the other −x = −8 side by doing the opposite operation. −1 −1 The 7 is positive so Subtract. x is NOT alone. Divide by x=8 negative one.
45. 45. Solve for x. Circle the term with the 7 − x = −1 variable you need to isolate.−7 −7 Anything outside the circle needs to move to the other −x = −8 side by doing the opposite operation. −1 −1 The 7 is positive so Subtract. x is NOT alone. Divide by x=8 negative one. x is alone so you are done.<|endoftext|>
| 4.65625 |
245 |
Associated Press 2008 election map at NBC News headquarters
Red States and Blue States: History of Presidential Electoral Maps
The use of color on maps is a key component in cartographic design. Perhaps the best known colors on a map are the “red states” and “blue states” on a presidential electoral map.
According to GeoLounge, the practice of identifying a political party on a map by color dates back to 1883, when the first map of red and blue political affiliations was produced. And since the 2000 election, the practice has been standardized with red representing Republicans and blue representing Democrats.
The complete historical timeline of all 57 presidential elections can be tracked on 270 to Win. Metrocosm, a unique Web page for current maps and statistical analysis, has created maps showing all of the colors of the early presidential elections.
An exhibit depicting the results of the presidential elections since 1952 is currently displayed in the front windows of the GIS Research and Map Collection (GRMC) on the second floor of Bracken Library. Sixty Years of Red States and Blue States will be on display through Election Day. And maps of the electoral results are available for circulation from the GRMC.<|endoftext|>
| 3.671875 |
490 |
As the new school year approaches, most students want to start off on a good note. They’ll reflect on what they might do to improve and increase their academic success… and “improved study skills” might just be at the top of that list.
To help students start the semester on a path to success, we’ve listed several key study skills, along with links to previous Engaging Minds posts that offer suggestions that help them build those skills and achieve their academic goals!
Which study skills would you add to this list? Share them in the comments.
⇒ Take thorough and thoughtful notes. Effective note-taking practices help you focus while you listen to your instructors’ lectures. And as a bonus: those notes make for useful study and review tools come test time!
⇒ Read your texts with a careful and critical eye. The way you read your course materials should differ from the way you read your favorite novel. To get the most out of your reading time, review our suggestions on how to read a textbook.
To make the time spent reading even more effective, learn how to take critical notes on your reading assignments. The process of taking thoughtful notes can help you remember what you’ve read, and it helps you think more critically and analytically about the material.
⇒ Brush up on your test-taking strategies. Everyone gets a bit nervous at test time, but by adopting some key study and test-taking strategies, you can build your confidence.
If you’re taking an online test, consider the practices that will help you “make the grade” in that arena.
After the test, take the time to review your study strategies and evaluate whether or not they contributed to your level of success.
⇒ Manage your time and your resources. Create a schedule to plan your time and stay focused on achieving your short- and long-term goals. Prioritize your responsibilities, and endeavor to avoid procrastination.
⇒ Maintain a positive attitude. Though not strictly a “study skill,” a positive mindset will help you stay motivated on the path towards academic success. Develop the habit of keeping a positive, yet realistic perspective on life’s challenges, and make an effort to manage your stress levels, especially when you have a number of responsibilities on your plate.<|endoftext|>
| 3.953125 |
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# Vector Algebra MCQ Quiz - Objective Question with Answer for Vector Algebra - Download Free PDF
Last updated on Aug 10, 2023
## Latest Vector Algebra MCQ Objective Questions
#### Vector Algebra Question 1:
Let a̅, b̅, and c̅ be three unit vectors such that $$\overline a \times \left( {\overline b \times \overline c } \right) = \frac{1}{{\sqrt 2 }}\left( {\overline b + \overline c } \right)$$ and b̅ is not parallel to c̅ . If α and β are the angles between a̅, b̅, and a̅, c̅ respectively then α - β =
1. $$\frac{3\pi}{4}$$
2. $$\frac{\pi}{4}$$
3. $$\frac{\pi}{2}$$
4. 0
5. None of the above/More than one of the above.
#### Answer (Detailed Solution Below)
Option 3 : $$\frac{\pi}{2}$$
#### Vector Algebra Question 1 Detailed Solution
Concept:
Vector Triple Product:
If $$\vec{a},\vec{b},\vec{c}$$ are three vectors, then the vector triple product is given by $$\vec{a} \times ( \vec{b} \times \vec{c} ) = (\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$$
Calculation:
Given α and β are the angles between $$\vec{a},\vec{b}$$ and $$\vec{a},\vec{c}$$ respectively.
Also, $$\vec{a} \times ( \vec{b} \times \vec{c} ) = \frac{1}{{\sqrt 2 }}( \vec{ b} + \vec{ c} )$$
⇒ $$(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c} = \frac{1}{{\sqrt 2 }}( \vec{ b} + \vec{ c} )$$
⇒ $$(|\vec{a}||\vec{c}|\cos\beta)\vec{b}-(|\vec{a}||\vec{b}|\cos\alpha)\vec{c} = \frac{1}{{\sqrt 2 }}( \vec{ b} + \vec{ c} )$$
⇒ $$\vec{b}\cos\beta-\vec{c}\cos\alpha = \frac{1}{{\sqrt 2 }}\vec{ b} + \frac{1}{{\sqrt 2 }}\vec{ c}$$
⇒ cos β = $$\frac{1}{\sqrt{2}}$$ and cos α = $$-\frac{1}{\sqrt{2}}$$
⇒ α = $$\frac{3\pi}{4}$$ and β = $$\frac{\pi}{4}$$
∴ α - β = $$\frac{3\pi}{4}$$ - $$\frac{\pi}{4}$$ = $$\frac{\pi}{2}$$
The correct answer is option 3.
#### Vector Algebra Question 2:
Let $$\overrightarrow a = \widehat i + 2\widehat j - 3\widehat k$$ and $$\overrightarrow b = 2\widehat i - 3\widehat j + 5\widehat k$$. If $$\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow r$$$$\overrightarrow r .(α \widehat i + 2\widehat j + \widehat k) = 3$$ and $$\overrightarrow r .(2\widehat i + 5\widehat j - α \widehat k) = - 1$$, α ∈ R, then the value of α + $${\left| {\overrightarrow r } \right|^2}$$ is equal to:
1. 9
2. 11
3. 13
4. 15
5. None of the above/More than one of the above.
Option 4 : 15
#### Vector Algebra Question 2 Detailed Solution
Concept:
If two vectors $$\overrightarrow a$$ and $$\overrightarrow b$$are parallel, then $$\overrightarrow a \times \overrightarrow b = 0$$
Explanation:
Given, $$\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow r$$
⇒ $$\overrightarrow r \times \overrightarrow a = -\overrightarrow r \times \overrightarrow b$$
⇒ $$\overrightarrow r \times \overrightarrow a +\overrightarrow r \times \overrightarrow b =0$$
⇒ $$\overrightarrow r \times (\overrightarrow a+\overrightarrow b)=0$$
$$\overrightarrow r$$ and $$(\overrightarrow a+\overrightarrow b)$$ are parallel because $$\overrightarrow r \times (\overrightarrow a+\overrightarrow b)=0$$
Therefore, $$\overrightarrow r= λ (\overrightarrow a+\overrightarrow b)$$ ---(1)
From the question, $$\overrightarrow a = \widehat i + 2\widehat j - 3\widehat k$$ and $$\overrightarrow b = 2\widehat i - 3\widehat j + 5\widehat k$$
$$\overrightarrow a+\overrightarrow b$$= ($$\widehat i + 2\widehat j - 3\widehat k$$ )+( $$2\widehat i - 3\widehat j + 5\widehat k$$)
⇒ $$\overrightarrow a+\overrightarrow b$$ = $$3 \widehat i -\widehat j +2\widehat k$$
Substitute the value of $$\overrightarrow a+\overrightarrow b$$ in equation (1), we have
⇒ $$\overrightarrow r= λ$$ ($$3 \widehat i -\widehat j +2\widehat k$$) ---(2)
∵ $$\overrightarrow r .(α \widehat i + 2\widehat j + \widehat k)$$ = 3
⇒ λ ( $$3 \widehat i -\widehat j +2\widehat k$$)$$⋅(α \widehat i + 2\widehat j + \widehat k)$$ =3
⇒ λ (3α -2+2) =3
⇒ λ (3α )=3
⇒ λα = 1 ---(3)
∵ $$\overrightarrow r .(2\widehat i + 5\widehat j - α \widehat k)$$ =-1
⇒ λ($$3 \widehat i -\widehat j +2\widehat k$$) ⋅ $$(2\widehat i + 5\widehat j - α \widehat k)$$ = -1
⇒ λ(6 - 5 - 2α )= -1
⇒ λ (1 - 2α)= -1
⇒ λ - 2λα =-1
⇒ λ -2(1) =-1 [∵ λα = 1]
⇒ λ - 2 = -1
⇒ λ = 1
Substitute the value of λ in equation (3), α = 1
Now, from equation (2), $$\overrightarrow r=$$ ( $$3 \widehat i -\widehat j +2\widehat k$$)
$${\left| {\overrightarrow r } \right|}$$ = √ 32+(-1)2+22
$${\left| {\overrightarrow r } \right|}$$ =√ 9+1+4
$${\left| {\overrightarrow r } \right|}$$ =√ 14
⇒ $${\left| {\overrightarrow r } \right|^2}$$ = 14
Now, α + $${\left| {\overrightarrow r } \right|^2}$$ = 1 + 14 =15
The value of α + $${\left| {\overrightarrow r } \right|^2}$$ is 15.
Hence, the correct answer is option (4).
#### Vector Algebra Question 3:
If $$\rm(\vec {a} + \vec{b})$$ is perpendicular to $$\rm\vec {a}$$ and magnitude of $$\rm\vec {b}$$ is twice that of $$\rm\vec {a}$$, then what is the value of $$\rm(4\vec {a} + \vec{b})\cdot \vec{b}$$ equal to?
1. 0
2. 1
3. $$8|\vec{a}|^2$$
4. $$8|\vec{b}|^2$$
5. None of the above/More than one of the above.
Option 1 : 0
#### Vector Algebra Question 3 Detailed Solution
Concept:
If $$\rm(\vec {a} + \vec{b})$$ is perpendicular to $$\rm\vec {a}$$, then
$$\rm(\vec {a} + \vec{b})$$.$$\vec {a}$$ = 0
Calculation:
$$|\vec{a}|^{2}$$ + $$\rm \vec {a}.\rm \vec {b}$$ = 0
Given: $$\rm |\vec{b}| = 2\times |\vec{a}|$$
So, we can write
$$\rm \frac{|\vec{b}|^{2}}{4}$$ + $$\rm \vec {a}.\rm \vec {b}$$ = 0
$$\rm \vec {a}.\rm \vec {b}$$ = - $$\rm \frac{|\vec{b}|^{2}}{4}$$ -----(1)
To find: $$\rm (4\vec {a} + \vec{b})\cdot \vec{b}$$
$$\rm (4\vec {a} + \vec{b})\cdot \vec{b}$$ = 4$$\rm \vec {a}.\rm \vec {b}$$ + $$\rm |\vec{b}|^{2}$$ ----(2)
From equations (1) & (2) we can write,
$$\rm (4\vec {a} + \vec{b})\cdot \vec{b}$$ = -$$\rm |\vec{b}|^{2}$$ + $$\rm |\vec{b}|^{2}$$ = 0
∴ The value of $$\rm(4\vec {a} + \vec{b})\cdot \vec{b}$$ equal to 0.
#### Vector Algebra Question 4:
If $$\rm \vec{a}= 2\hat i + \hat j +3\hat k, \rm \vec{b}= \hat i + \hat j- 2\hat k \ and \ \rm \vec{c}= 2\hat i + 3\hat k$$ are three vectors such that $$\rm \vec{a}+λ\vec{b}$$ is perpendicular to $$\rm \vec{c}$$, then find the value of λ ?
1. 13/8
2. 5/4
3. 13/4
4. 5/8
5. None of the above/More than one of the above.
Option 3 : 13/4
#### Vector Algebra Question 4 Detailed Solution
Concept:
If $$\rm \vec{a} \ and \ \vec b$$ are two vectors then $$\rm \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos\theta$$
Note: If vectors $$\rm \vec{a} \ and \ \vec b$$ are perpendicular to each other then $$\rm \vec{a}.\vec{b}=0$$
Calculation:
Given: $$\rm \vec{a}= 2\hat i + \hat j +3\hat k, \rm \vec{b}= \hat i + \hat j- 2\hat k \ and \ \rm \vec{c}= 2\hat i + 3\hat k$$
$$\Rightarrow \rm \vec{a}+λ\vec{b}=( 2\hat i+\hat j +3\hat k ) +λ(\hat i+\hat j-2\hat k)$$
$$\rm \vec{a}+λ\vec{b}=( 2+λ)\vec i+(1+λ)\vec j +(3-2λ)\vec k$$
Now, $$\rm \vec{a}+λ\vec{b}$$ and $$\rm \vec{c}$$ are perpendicular
$$\Rightarrow (\rm \vec{a}+λ\vec{b}) \cdot \vec c = 0$$
⇒ $$\rm [( 2+λ)\hat i+(1+λ)\hat j +(3-2λ)\hat k].(2\hat i+3\hat k)=0$$
⇒ 2(2 + λ) + 3(3 - 2λ) = 0
⇒ 4 + 2λ + 9 - 6λ = 0
⇒ 13 - 4λ = 0
⇒ λ = 13/4
Hence, option 3 is correct.
#### Vector Algebra Question 5:
The value of x if $$\rm x(\widehat i + \widehat j + \widehat k)$$ is a unit vector is
1. $$\pm \frac{1}{3}$$
2. ± 3
3. $$\pm \sqrt 3$$
4. $$\pm \frac{1}{{\sqrt 3 }}$$
5. None of the above/More than one of the above.
#### Answer (Detailed Solution Below)
Option 4 : $$\pm \frac{1}{{\sqrt 3 }}$$
#### Vector Algebra Question 5 Detailed Solution
Concept:
The modulus of a unit vector $$\rm \hat A$$ = $$\rm |\hat A|$$ = 1
Calculation:
Let $$\rm \vec A$$ = $$\rm x(\widehat i + \widehat j + \widehat k)$$
$$\rm \vec A$$ = $$\rm x\widehat i + x\widehat j + x\widehat k$$
Given: $$\rm x(\widehat i + \widehat j + \widehat k)$$ is a unit vector
So, $$\rm |\vec A| =1$$
$$\rm \sqrt{x^2 + x^2 + x^2} = 1$$
$$\rm \sqrt{3x^2 } = 1$$
|x|$$\rm \sqrt{3} = 1$$
|x| = $$1\over\sqrt3$$
x = $$\boldsymbol{\pm{1\over\sqrt3}}$$
## Top Vector Algebra MCQ Objective Questions
#### Vector Algebra Question 6
What is the value of p for which the vector p(2î - ĵ + 2k̂) is of 3 units length?
1. 1
2. 2
3. 3
4. 6
Option 1 : 1
#### Vector Algebra Question 6 Detailed Solution
Concept:
Let $$\rm {\rm{\vec a}} = {\rm{x\;\vec i}} + {\rm{y\;\vec j}} + {\rm{z\;\vec k}}$$ then magnitude of the vector of a = $$\left| {{\rm{\vec a}}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {\rm{\;}}{{\rm{y}}^2} + {{\rm{z}}^2}}$$
Calculation:
Let $$\rm \vec{a}$$ = p(2î - ĵ + 2k̂)
Given, $$\left| {{\rm{\vec a}}} \right| = 3$$
⇒ $$\rm \sqrt{4p^2 + p^2+4p^2} = 3$$
⇒ $$\rm \sqrt{9p^2} = 3$$
⇒ 3p = 3
∴ p = 1
#### Vector Algebra Question 7
If the vectors $$\widehat i + 2\widehat j + 3\widehat k$$$$λ \widehat i + 4\widehat j + 7\widehat k$$$$- 3\widehat i - 2\widehat j - 5\widehat k$$ are collinear if λ equals
1. 3
2. 4
3. 5
4. 6
Option 1 : 3
#### Vector Algebra Question 7 Detailed Solution
Concept:
Conditions of collinear vector:
• Three points with position vectors $$\vec a,\;\vec b\;and\;\vec c$$ are collinear if and only if the vectors $$\left( {\vec a - \vec b} \right)$$ and $$\left( {\vec a\; - \vec c} \right)$$ are parallel. ⇔ $$\left( {\vec a - \vec b} \right) = λ \left( {\vec a\; - \vec c} \right)$$
• If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then $$\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0$$
Solution:
We know that, If the points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) be collinear then $$\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{{z_1}}\\ {{x_2}}&{{y_2}}&{{z_2}}\\ {{x_3}}&{{y_3}}&{{z_3}} \end{array}} \right| = 0$$
Given $$\widehat i + 2\widehat j + 3\widehat k$$$$λ \widehat i + 4\widehat j + 7\widehat k$$$$- 3\widehat i - 2\widehat j - 5\widehat k$$ are collinear
∴ $$\left| {\begin{array}{*{20}{c}} { 1}&{ 2}&3\\ λ&4&7\\ -3&-2&-5 \end{array}} \right| = 0$$
⇒ 1 (-20 + 14) – (2) (-5λ + 21) + 3 (-2λ + 12) = 0
⇒ -6 + 10λ – 42 - 6λ + 36 = 0
⇒ 4λ = 12
∴ λ = 3
#### Vector Algebra Question 8
Find the value of $$\rm \vec{a} \times \vec{a}$$
1. 1
2. 0
3. $$\rm |\vec{a}|$$
4. $$\rm |\vec{a}|^2$$
Option 2 : 0
#### Vector Algebra Question 8 Detailed Solution
Concept:
Dot product of two vectors is defined as:
$${\rm{\vec A}}{\rm{.\vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times cos}}\;{\rm{\theta }}$$
Cross/Vector product of two vectors is defined as:
$${\rm{\vec A \times \vec B = }}\left| {\rm{A}} \right|{\rm{ \times }}\left| {\rm{B}} \right|{\rm{ \times sin}}\;{\rm{\theta }} \times \rm \hat{n}$$
where θ is the angle between $${\rm{\vec A}}\;{\rm{and}}\;{\rm{\vec B}}$$
Calculation:
To Find: Value of $$\rm \vec{a} \times \vec{a}$$
Here angle between them is 0°
$${\rm{\vec a \times \vec a = }}\left| {\rm{a}} \right|{\rm{ \times }}\left| {\rm{a}} \right|{\rm{ \times sin}}\;{\rm{0 }} \times \rm \hat{n}=0$$
#### Vector Algebra Question 9
If A = $$\rm 5 \hat i-2\hat j +4\hat k$$ and B = $$\rm \hat i+3\hat j -7\hat k$$ , then what is the value of $$\rm |\vec{AB}|$$?
1. 6√2
2. 7√2
3. 8√2
4. 9√2
Option 4 : 9√2
#### Vector Algebra Question 9 Detailed Solution
Concept:
If $$\rm \vec A = x \hat i-y\hat j +z\hat k$$, then $$\rm |\vec A| = \sqrt {x^2 +y^2+z^2}$$
Calculation:
Given A = $$\rm 5 \hat i-2\hat j +4\hat k$$ and B = $$\rm \hat i+3\hat j -7\hat k$$
$$\rm \vec{AB} = \vec B - \vec A$$
$$\rm \vec{AB}$$ = $$\rm \hat i+3\hat j -7\hat k - (5 \hat i-2\hat j +4\hat k)$$
$$\rm \vec{AB}$$ = $$\rm -4\hat i+5\hat j -11\hat k$$
Now $$\rm |\vec {AB}| = \sqrt{(-4)^2 +5^2+(-11)^2}$$
$$\rm |\vec {AB}| = \sqrt{16 +25+121}$$
$$\rm |\vec {AB}| = \sqrt{162}$$ = 9√2
#### Vector Algebra Question 10
The point with position vectors 5î - 2ĵ, 8î - 3ĵ, aî - 12ĵ are collinear if the value of a is
1. 31
2. 51
3. 42
4. 35
Option 4 : 35
#### Vector Algebra Question 10 Detailed Solution
Concept:
Three or more points are collinearif slope of any two pairs of points is same.
The slope of a line passing through the distinct points (x1, y1) and (x2, y2) is $$\rm \frac{y_2 -y_1 }{x_2-x_1}$$
Calculation:
Here, $$\rm 5\hat i-2\hat j, 8\hat i-3\hat j, a\hat i-12\hat j$$
Let, A = (5, -2), B = (8, -3), C = (a, -12)
Now, slope of AB = Slope of BC = Slope of AC ....(∵ points are collinear)
$$\rm \frac{-3-(-2)}{8-5}=\frac{-12-(-3)}{a-8}\\ ⇒ \frac{-1}{3}=\frac{-9}{a-8}$$
⇒ a - 8= 27
⇒ a = 27 + 8 = 35
Hence, option (4) is correct.
#### Vector Algebra Question 11
The sine of the angle between vectors $$\vec a = 2\hat i - 6\hat j - 3\hat k$$ and $$\vec b = 4\hat i + 3\hat j - \hat k$$ is
1. $$\frac{1}{{\sqrt {26} }}$$
2. $$\frac{5}{{\sqrt {26} }}$$
3. $$\frac{5}{{26}}$$
4. $$\frac{1}{{26}}$$
#### Answer (Detailed Solution Below)
Option 2 : $$\frac{5}{{\sqrt {26} }}$$
#### Vector Algebra Question 11 Detailed Solution
Concept:
If $$\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\;and\;\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$$ then $$\vec a \cdot \;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos \theta$$
Calculation:
Given: $$\vec a = 2\hat i - 6\hat j - 3\hat k$$ and $$\vec b = 4\hat i + 3\hat j - \hat k$$
$$\left| {\vec a} \right| = 7,\;\left| {\vec b} \right| = \sqrt {26} \;and\;\vec a \cdot \;\vec b = - 7$$
$$\Rightarrow \;\cos \theta = \frac{{\vec a \cdot \;\vec b}}{{\left| {\vec a} \right| \times \left| {\vec b} \right|}} = \frac{{ - \;7}}{{7 \times \sqrt {26} }} = - \frac{1}{{\sqrt {26} }}$$
$$\Rightarrow \;{\sin ^2}\theta = 1 - {\cos ^2}\theta = 1 - \frac{1}{{26}} = \frac{{25}}{{26}}$$
$$\Rightarrow \;\sin \theta = \frac{5}{{\sqrt {26} }}$$
#### Vector Algebra Question 12
If $$4\hat i + \hat j - 3\hat k$$ and $$p\hat i + q\hat j - 2\hat k$$ are collinear vectors, then what are the possible values of p and q respectively?
1. 4, 1
2. 1, 4
3. $$\frac{8}{3}, \frac{2}{3}$$
4. $$\frac{2}{3}, \frac{8}{3}$$
#### Answer (Detailed Solution Below)
Option 3 : $$\frac{8}{3}, \frac{2}{3}$$
#### Vector Algebra Question 12 Detailed Solution
Concept:
For two vectors $$\vec m \ and \ \vec n$$ to be collinear, $$\vec m\; = \;λ \vec n$$ where λ is a scalar.
Calculation:
Given that, the vectors $$4\hat i + \hat j - 3\hat k$$ & $$p\hat i + q\hat j - 2\hat k$$ are collinear.
Since two vectors $$\vec m \ and \ \vec n$$ are collinear then $$\vec m\; = \;λ \vec n$$ where λ is a scalar.
⇒ $$4\hat i + \hat j - 3\hat k\;\ = λ × (\;p\hat i + q\hat j - 2\hat k)$$
⇒ $$4\hat i + 1\hat j - 3\hat k\;\ = λ p \hat i + λq \hat j - 2λ \hat k$$
⇒ λp = 4, λq = 1 and -2λ = -3
⇒ λ = 3/2
So, by substituting λ = 3/2 in λp = 4 and λq = 1, we get
⇒ (3/2)p = 4 and (3/2)q = 1
⇒ p = 8/3 and q = 2/3
∴ $$\frac{8}{3}, \frac{2}{3}$$is the correct answer.
#### Vector Algebra Question 13
If $$\vec a + \vec b + \vec c = \vec 0,\;|\vec a| = 3,\;|\vec b| = 5$$ and $$|\vec c| = 7$$, find the angle between $$\vec a$$ and $$\vec b$$.
1. π / 2
2. π / 3
3. π / 6
4. π / 4
Option 2 : π / 3
#### Vector Algebra Question 13 Detailed Solution
Concept:
Let the angle between $$\vec a$$ and $$\vec b$$is $$\rm \theta$$
$$\rm \vec a.\vec b = 2ab cos\;\theta$$
Calculations:
consider, the angle between $$\vec a$$ and $$\vec b$$is $$\rm \theta$$
Given, $$\vec a + \vec b + \vec c = \vec 0$$
$$\vec a + \vec b = - \vec c$$
$$\rm |\vec a + \vec b| = |- \vec c |$$
Squaring on both side, we get
$$\rm |\vec a + \vec b|^2 = |- \vec c |^2$$
$$\rm |\vec a|^2 +2\;\vec a.\vec b+ |\vec b|^2 = |- \vec c |^2$$
$$\rm |\vec a|^2 +|\vec b|^2+2\;ab\cos\;\theta = |- \vec c |^2$$
$$\rm (3)|^2 +(5)^2+2\;(3)(5)\cos\;\theta = (7)^2$$
$$\rm 30\cos\;\theta = 15$$
$$\rm \cos\;\theta = \dfrac 12$$
⇒ $$\rm \theta$$ = π / 3
Hence, If $$\vec a + \vec b + \vec c = \vec 0,\;|\vec a| = 3,\;|\vec b| = 5$$ and $$|\vec c| = 7$$, then the angle between $$\vec a$$ and $$\vec b$$is π / 3
#### Vector Algebra Question 14
Let $$\rm \vec a =\hat i +\hat j +\hat k,\; \vec b =\hat i -\hat j + \hat k$$ and c = î - ĵ - k̂ be three vectors. A vector $$\rm \vec v$$ in the plane of $$\rm \vec a$$ and $$\rm \vec b$$ whose projection on $$\rm \frac {\vec c} {|\vec c|}$$ is $$\frac 1 {\sqrt 3},$$ is
1. 3î - ĵ + 3k̂
2. î - 3ĵ + 3k̂
3. 5î - 2ĵ + 5k̂
4. 2î - ĵ + 3k̂
#### Answer (Detailed Solution Below)
Option 1 : 3î - ĵ + 3k̂
#### Vector Algebra Question 14 Detailed Solution
Calculation:
$$\rm \vec a =\hat i +\hat j +\hat k,\; \vec b =\hat i -\hat j + \hat k$$ and c = î - ĵ - k̂
Given: vector $$\rm \vec v$$ in the plane of $$\rm \vec a$$ and $$\rm \vec b$$
Therefore, $$\rm \vec v = \vec a + λ \vec b$$
⇒ $$\rm \vec v =(\hat i +\hat j +\hat k ) \; + λ (\hat i -\hat j + \hat k)$$
= (1 + λ)î + (1 - λ)ĵ + (1 + λ)k̂ .... (1)
Projection of $$\rm \vec v$$ on $$\rm \frac {\vec c} {|\vec c|}=\frac 1 {\sqrt 3}$$
⇒ $$\rm \vec v=\rm \frac {\vec c} {|\vec c|}=\frac 1 {\sqrt 3}$$
⇒ $$\frac {(1 + λ) - (1 - λ) - (1 + λ)}{\sqrt3} = \frac {1}{\sqrt 3}$$
⇒ -(1 - λ) = 1
∴ λ = 2
Now, put the value of λ in equation (1), we get
$$\rm \vec v$$ = 3î - ĵ + 3k̂
#### Vector Algebra Question 15
If $$\rm \vec{i} - a\vec{j} + 5\vec{k}$$and $$\rm 3\vec{i} - 6\vec{j} + b\vec{k}$$ are parallel vectors then b is equal to?
1. 5
2. 10
3. 15
4. 20
Option 3 : 15
#### Vector Algebra Question 15 Detailed Solution
Concept:
If $${\rm{\vec a\;and\;\vec b}}$$ are two vectors parallel to each other then $${\rm\vec{a} = λ \vec{b}}$$ or $$\rm \vec{a} × \vec{b} =0$$
Calculation:
Given:
$$\rm \vec{i} - a\vec{j} + 5\vec{k}$$ and $$\rm 3\vec{i} - 6\vec{j} + b\vec{k}$$ are parallel vectors,
Therefore, $$\rm \vec{i} - a\vec{j} + 5\vec{k}= λ (\rm 3\vec{i} - 6\vec{j} + b\vec{k})$$
Equating the coefficient of $$\rm \vec{i},\vec{j} \;and\; \vec{k}$$
⇒ 1 = 3λ, ∴ λ = 1/3
⇒ -a = -6λ
⇒ 5 = bλ .... (1)
Put the value of λ in equation (1), we get
5 = b × (1/3)
So, b = 15<|endoftext|>
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## Saturday, May 23, 2020
### limit comparison test
The Limit Comparison Test
If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges. If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
this section we will discuss using the Comparison Test and Limit Comparison Tests to determine if an infinite series converges or diverges.
In mathematics, the limit comparison test (LCT) is a method of testing for the convergence of an infinite series. Contents. 1 Statement; 2 Proof; 3 Example .
## Statement
Suppose that we have two series ${\displaystyle \Sigma _{n}a_{n}}$ and ${\displaystyle \Sigma _{n}b_{n}}$ with ${\displaystyle a_{n}\geq 0,b_{n}>0}$ for all ${\displaystyle n}$.
Then if ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ with ${\displaystyle 0, then either both series converge or both series diverge.[1]
## Proof
Because ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ we know that for all ${\displaystyle \varepsilon >0}$ there is a positive integer ${\displaystyle n_{0}}$ such that for all ${\displaystyle n\geq n_{0}}$ we have that ${\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon }$, or equivalently
${\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }$
${\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}
${\displaystyle (c-\varepsilon )b_{n}
As ${\displaystyle c>0}$ we can choose ${\displaystyle \varepsilon }$ to be sufficiently small such that ${\displaystyle c-\varepsilon }$ is positive. So ${\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}}$ and by the direct comparison test, if ${\displaystyle \sum _{n}a_{n}}$ converges then so does ${\displaystyle \sum _{n}b_{n}}$.
Similarly ${\displaystyle a_{n}<(c+\varepsilon )b_{n}}$, so if ${\displaystyle \sum _{n}a_{n}}$ diverges, again by the direct comparison test, so does ${\displaystyle \sum _{n}b_{n}}$.
That is, both series converge or both series diverge.
## Example
We want to determine if the series ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}}$ converges. For this we compare with the convergent series ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}$.
As ${\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0}$ we have that the original series also converges.
## One-sided version
One can state a one-sided comparison test by using limit superior. Let ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n}$. Then if ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}$ with ${\displaystyle 0\leq c<\infty }$ and ${\displaystyle \Sigma _{n}b_{n}}$ converges, necessarily ${\displaystyle \Sigma _{n}a_{n}}$ converges.
## Example
Let ${\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}}$ and ${\displaystyle b_{n}={\frac {1}{n^{2}}}}$ for all natural numbers ${\displaystyle n}$. Now ${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }(1-(-1)^{n})}$ does not exist, so we cannot apply the standard comparison test. However, ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )}$ and since ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$ converges, the one-sided comparison test implies that ${\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}}$ converges.
## Converse of the one-sided comparison test
Let ${\displaystyle a_{n},b_{n}\geq 0}$ for all ${\displaystyle n}$. If ${\displaystyle \Sigma _{n}a_{n}}$ diverges and ${\displaystyle \Sigma _{n}b_{n}}$ converges, then necessarily ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty }$, that is, ${\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0}$. The essential content here is that in some sense the numbers ${\displaystyle a_{n}}$ are larger than the numbers ${\displaystyle b_{n}}$.
## Example
Let ${\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}}$ be analytic in the unit disc ${\displaystyle D=\{z\in \mathbb {C} :|z|<1\}}$ and have image of finite area. By Parseval's formula the area of the image of ${\displaystyle f}$ is ${\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}}$. Moreover, ${\displaystyle \sum _{n=1}^{\infty }1/n}$ diverges. Therefore, by the converse of the comparison test, we have ${\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0}$, that is, ${\displaystyle \liminf _{n\to \infty }n|a_{n}|=0}$.
The limit comparison test shows that the original series is divergent. The limit comparison test does not apply because the limit in question does not exist. The ever useful Limit Comparison Test will save the day! using the comparison tests to determine convergence of an infinite series. Comparison Test Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Limit Comparison Test Example Example Example Example Example.
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limit comparison test with square roots
Labels:<|endoftext|>
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# Angles: What are Angles? Categories of Angles
### What are Angles?
Angles, triangles, squares, circles, cubes, cones and cylinders are all part of geometry in mathematics.
Geometry is the branch of mathematics that deals with the study of shapes, sizes and dimensions of a variety of things.
As a mathematics student, understanding angles is very important because it is vital in buildings, constructions and designs.
Sailors use angles to navigate their path along the sea while a carpenter can apply them to make beautiful designs of his job.
Amazingly, sport men and women can utilize angles in making an excellent performance. For instance, the use of angles can be applied when taking free throws in basket ball. Similarly, a footballer may utilize it to curl a free-kick into the football net.
By definition, an angle is a figure formed by two sides sharing a common endpoint.
As an instance to the above definition, an angle is formed where two walls in the corner of our homes meet or at the point where the two arms of a scissors meet.
The two sides that forms the angle are called arms or rays while the endpoint where both sides meet is called the vertex.
The figure below illustrates the arms and vertex of an angle:
The angle formed will depend on the amount of turn between each arm.
Below are some notable points from the above diagram:
• Most times, an angle is represented by the figure .
• The angle in the above figure can be represented as ABC.
• The unit of an angle is in degree. This implies that angles are measured in degrees and a protractor can be used to achieve this.
### Representation of Angles
Angles can be represented mainly in two ways. These are:
1. By using the three letters on the shape of the specific angle. Understand that the angle is formed on the middle letter, (vertex).
In tha above diagram, the angle will be represented as ABC
2. By giving the angle a name. The names are often denoted with a Greek letter like α (alpha) or θ (theta). Also, we can also name the angle with lowercase letters like 'p or q'.
Still using the above diagram as a basis, the angle will be labeled α
### Categories of Angles
An angle can be categorized as either interior or exterior based on where they are formed with regards to a shape. They are shown in the image below:
• Interior angles are formed inside a shape. In the above diagram, the angle is formed inside a triangle. (A triangle has 3 sides and 3 angles).
• Exterior angles are formed outside a shape via a line extending from the adjoining sides.
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Suggested languages for you:
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Q9E
Expert-verified
Found in: Page 556
### Essential Calculus: Early Transcendentals
Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280
# To find a dot product between $${\rm{a}}$$ and $${\rm{b}}$$.
The dot product $$a \cdot b$$ is $$- 15$$
See the step by step solution
## Step 1: Concept of Dot Product
Formula used:
The dot product of two vectors $${\rm{a}}$$ and $${\rm{b}}$$ in terms of $$\theta$$ is $${\rm{a}} \cdot {\rm{b}} = |{\rm{a}}||{\rm{b}}|\cos \theta$$.
## Step 2: Calculation of the dot product
The given magnitude of two vectors are $$|{\rm{a}}| = 6$$ and $$|{\rm{b}}| = 5$$ with angle $$\frac{{2\pi }}{3}$$ between them.
Substitute $$|{\rm{a}}| = 6,|\;{\rm{b}}| = 5$$, and $$\theta = \frac{{2\pi }}{3}$$ in above formula and obtain the dot product as follows.
\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (6)(5)\cos \left( {\frac{{2\pi }}{3}} \right)\\{\rm{a}} \cdot {\rm{b}} &= 30( - 0.5)\\{\rm{a}} \cdot {\rm{b}} &= - 15\end{aligned}
Thus, the dot product between two vector $${\rm{a}}$$ and $${\rm{b}}$$ is $$a \cdot b = - 15$$.<|endoftext|>
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The Supreme Court is the highest court of The Indian Republic. Judiciary, the third organ of the government, has an important role to play in the governance. It settles the disputes, interprets laws, protects fundamental rights and acts as guardian of the Constitution. India has a single unified and integrated judicial system and that the Supreme Court is the highest court in India.
The promulgation of Regulating Act of 1773 by the King of England paved the way for establishment of the Supreme Court of Judicature at Calcutta. The Letters of Patent was issued on 26 March 1774 to establish the Supreme Court of Judicature at Calcutta, as a Court of Record, with full power & authority to hear and determine all complaints for any crimes and also to entertain, hear and determine any suits or actions against any of His Majesty’s subjects in Bengal, Bihar and Orissa. The Supreme Courts at Madras and Bombay was established by King George – III on 26 December 1800 and on 8 December 1823 respectively.
Federal Court of India was established under the Government of India Act 1935. The Federal Court had jurisdiction to solve disputes between provinces and federal states and hear appeal against Judgements from High Courts.
After India attained independence in 1947, the Constitution of India came into being on 26 January 1950. The Supreme Court of India also came into existence and its first sitting was held on 28 January 1950.
The Chief Justice and other judges of the Supreme Court are appointed by the President of India. While appointing the Chief Justice, the President is constitutionally required to consult such other judges of the Supreme Court as he deems proper, but outgoing Chief Justice is always consulted. Normally, the senior most judge of the Supreme Court is appointed as the Chief Justice of India, although there is no constitutional requirement to do so. While appointing other judges, the President is bound to consult the Chief Justice and other senior judges, if he deems proper.
The original Constitution of 1950 envisaged a Supreme Court with a Chief Justice and 7 puisne Judges – leaving it to Parliament to increase this number.
According to the Constitution of India, the role of the Supreme Court is that of a federal court, guardian of the Constitution and the highest court of appeal. Articles 124 to 147 of the Constitution of India lay down the composition and jurisdiction of the Supreme Court of India. Primarily, it is an appellate court which takes up appeals against judgments of the High Courts of the states and territories.
The Supreme Court is a Court of Record. It has two implications. All its decisions and judgments are cited as precedents in all courts of the country. They have the force of law and are binding on all lower Courts, and indeed the High Courts. As a Court of Record, the Supreme Court can even send a person to jail who may have committed contempt of the court.
As a Federal Court: Supreme Court is the Federal Court of India, India being a federation; powers are divided between the Union and State governments. The Supreme Court of India is the final authority to see to it that the division of powers as specified in the constitution is obeyed by both the Union and the State governments. So, Article 131 of the Indian Constitution vests the Supreme Court with original and exclusive jurisdiction to determine the justiciable disputes between the Union and the States or between the States.
Interpreter of the Constitution and Law: The responsibility of interpreting the constitution rests on the Supreme Court. The interpretation of the constitution which the Supreme Court shall make must be accepted by all. It interprets the constitution and preserves it. Where a case involves a substantial question of law as to the interpretation of the constitution either certified by the High Court or being satisfied by the Supreme Court itself, an appeal shall lie to the Supreme Court for interpretation of the question of law raised.
As a Court of Appeal: The Supreme Court is the highest court of appeal from all courts in the territory of India. Appeal lies to the Supreme Court of the cases involving interpretation of the constitution. Appeals in respect of civil and criminal cases also lie to the Supreme Court irrespective of any constitutional question.
Advisory Role: The Supreme Court has an advisory jurisdiction in offering its opinion an any question of law or fact of public importance as may be referred to it for consideration by the President.
Guardian of the Constitution: The Supreme Court of India is the guardian of the constitution. There are two points of significance of the Supreme Court’s rule as the protector and guardian of the constitution.
- First, as the highest Federal Court, it is within the power and authority of the Supreme Court to settle any dispute regarding division of powers between the Union and the States.
- Secondly, it is in the Supreme Court’s authority to safeguard the fundamental rights of the citizens.
In order to discharge these two functions it is sometimes necessary for the Supreme Court to examine or review the legality of the laws enacted by both the Union and the State Governments. This is known as the power of Judicial Review. Indian Supreme Court enjoys limited power of Judicial Review.
Writ Jurisdictions: Under Article 32 of the constitution of Supreme Court can issue Writs for the enforcement of fundamental rights. These writs are in the nature of Habeas Corpus, Mandamas, Prohibition, and Quo-warranto Certiorari.
Power of Judicial Review and Supreme Court: The power of the Judiciary to examine the validity of such law is called Judicial Review. The Supreme Court of India enjoys limited power of Judicial Review. Judicial Review empowers the courts to invalidate laws passed by the legislature. Supreme Court of India also enjoys the power of Judicial Review. If it occurs to the Supreme Court that any law enacted by Parliament or by a State Legislature curbs or threatens to curb the citizen’s fundamental rights, the Supreme Court may declare that law as unlawful or unconstitutional.JPSC Notes brings Prelims and Mains programs for JPSC Prelims and JPSC Mains Exam preparation. Various Programs initiated by JPSC Notes are as follows:-
- JPSC Mains Tests and Notes Program
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# Learning Machines
### Week 2
#### Linear Algebra Primer
##### Notation
Scalar: x (lowercase, regular)
Vector: $$\mathbf{u}$$ (lowercase, bold)
Vector: $$\overrightarrow{u}$$ (lowercase, w/ arrow)
Matrix: $$\mathbf{A}$$ (uppercase, bold)
Summation: $$\sum$$
Product: $$\prod$$
##### Vector Definition
Formal Definition
n-tuple of values (usually real numbers) where n is the dimension of the vector and can be any positive integer $$\ge$$ 1.
A vector can be thought of as...
• a point in space
• a directed line segment with a magnitude and direction
Formatting
• Vectors can be written in column form or row form
• Column form is conventional
• Vector elements are referenced by subscript
Column Vector
$$\mathbf{x} = \begin{bmatrix} x_1 \\ \vdots \\ x_n \\ \end{bmatrix}$$
Row Vector
$$\mathbf{x} = \begin{bmatrix} x_1 \cdots x_n \end{bmatrix}$$
Transpose Row Vector to Column Vector
$$\begin{bmatrix} x_1 \cdots x_n \end{bmatrix}^\text{T} = \begin{bmatrix} x_1 \\ \vdots \\ x_n \\ \end{bmatrix}$$
Transpose Column Vector to Row Vector
$$\begin{bmatrix} x_1 \\ \vdots \\ x_n \\ \end{bmatrix}^\text{T} = \begin{bmatrix} x_1 \cdots x_n \end{bmatrix}$$
##### Vector Properties
• Vectors are commutative: ( $$\overrightarrow{u}$$ + $$\overrightarrow{v}$$ ) is equal to ( $$\overrightarrow{v}$$ + $$\overrightarrow{u}$$ )
• Vectors are associative: ( $$\overrightarrow{u}$$ + ( $$\overrightarrow{v}$$ + $$\overrightarrow{w}$$ ) ) is equal to ( ( $$\overrightarrow{u}$$ + $$\overrightarrow{v}$$ ) + $$\overrightarrow{w}$$ )
##### Working with Vectors and Matrices in Python and Numpy
Importing Numpy library
import numpy as np
Array Creation
>>> np.array( [ 0, 2, 4, 6, 8 ] )
array([0, 2, 4, 6, 8])
>>> np.zeros( 5 )
array([ 0., 0., 0., 0., 0.])
>>> np.ones( 5 )
array([ 1., 1., 1., 1., 1.])
>>> np.zeros( ( 5, 1 ) )
array([[ 0.],
[ 0.],
[ 0.],
[ 0.],
[ 0.]])
>>> np.zeros( ( 1, 5 ) )
array([[ 0., 0., 0., 0., 0.]])
>>> np.arange( 5 )
array([0, 1, 2, 3, 4])
>>> np.arange( 0, 1, 0.1 )
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
>>> np.linspace( 0, 1, 5 )
array([ 0. , 0.25, 0.5 , 0.75, 1. ])
>>> np.random.random( 5 )
array([ 0.22035712, 0.89856076, 0.46510509, 0.36395359, 0.3459122 ])
Add corresponding elements. Result is a vector.
$\overrightarrow{z} = \overrightarrow{x} + \overrightarrow{y} = \begin{bmatrix} x_1 + y_1 \cdots x_n + y_n \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> y = np.array( [ 10.0, 20.0, 30.0, 40.0, 50.0 ] )
>>> z = x + y
>>> z
array([ 11., 22., 33., 44., 55.])
Vector Subtraction
Subtract corresponding elements. Result is a vector.
$\overrightarrow{z} = \overrightarrow{x} - \overrightarrow{y} = \begin{bmatrix} x_1 - y_1 \cdots x_n - y_n \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> y = np.array( [ 10.0, 20.0, 30.0, 40.0, 50.0 ] )
>>> z = x - y
>>> z
array([ -9., -18., -27., -36., -45.])
Multiply corresponding elements. Result is a vector.
$\overrightarrow{z} = \overrightarrow{x} \circ \overrightarrow{y} = \begin{bmatrix} x_1 y_1 \cdots x_n y_n \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> y = np.array( [ 10.0, 20.0, 30.0, 40.0, 50.0 ] )
>>> z = x * y
>>> z
array([ 10., 40., 90., 160., 250.])
Vector Dot Product
Multiply corresponding elements, then add products. Result is a scalar.
$a = \overrightarrow{x} \cdot \overrightarrow{y} = \sum_{i=1}^n x_i y_i$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> y = np.array( [ 10.0, 20.0, 30.0, 40.0, 50.0 ] )
>>> a = np.dot( x, y )
>>> a
550.0
Add scalar to each element. Result is a vector.
$\overrightarrow{y} = \overrightarrow{x} + a = \begin{bmatrix} x_1 + a \cdots x_n + a \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> a = 3.14
>>> y = x + a
>>> y
array([ 4.14, 5.14, 6.14, 7.14, 8.14])
Vector-Scalar Subtraction
Subtract scalar from each element. Result is a vector.
$\overrightarrow{y} = \overrightarrow{x} - a = \begin{bmatrix} x_1 - a \cdots x_n - a \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> a = 3.14
>>> y = x - a
>>> y
array([-2.14, -1.14, -0.14, 0.86, 1.86])
Vector-Scalar Multiplication
Multiply each element by scalar. Result is a vector.
$\overrightarrow{y} = \overrightarrow{x} \ a = \begin{bmatrix} x_1 a \cdots x_n a \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> a = 3.14
>>> y = x * a
>>> y
array([ 3.14, 6.28, 9.42, 12.56, 15.7 ])
Vector-Scalar Division
Divide each element by scalar. Result is a vector.
$\overrightarrow{y} = \frac{\overrightarrow{x}}{a} = \begin{bmatrix} \frac{x_1}{a} \cdots \frac{x_n}{a} \end{bmatrix}^\text{T}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> a = 3.14
>>> y = x / a
>>> y
array([ 0.31847134, 0.63694268, 0.95541401, 1.27388535, 1.59235669])
Vector Magnitude
Compute vector length. Result is a scalar.
$a = || \overrightarrow{x} || = \sqrt{ x_1^2 + \cdots + x_n^2 } = \sqrt{ \overrightarrow{x} \cdot \overrightarrow{x} }$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> a = np.linalg.norm( x )
>>> a
7.416198487095663
Vector Normalization
Compute unit vector. Result is a vector.
$\hat{x} = \frac{\overrightarrow{x}}{|| \overrightarrow{x} ||}$
>>> x = np.array( [ 1.0, 2.0, 3.0, 4.0, 5.0 ] )
>>> a = np.linalg.norm( x )
>>> x = x / a
>>> x
array([ 0.13483997, 0.26967994, 0.40451992, 0.53935989, 0.67419986])
Matrix Transposition
Swaps the row and column index for each element. For an m x n matrix, result is an n x m matrix.
$\mathbf{Y} = \mathbf{X}^\text{T}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> Y = X.T
>>> Y
array([[ 1., 4.],
[ 2., 5.],
[ 3., 6.]])
Add corresponding elements. Result is a matrix.
$\mathbf{Z} = \mathbf{X} + \mathbf{Y} = \begin{bmatrix} x_{11} + y_{11} & x_{12} + y_{12} & \cdots & x_{1n} + y_{1n} \\ x_{21} + y_{21} & x_{22} + y_{22} & \cdots & x_{2n} + y_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} + y_{m1} & x_{m2} + y_{m2} & \cdots & x_{mn} + y_{mn} \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> Y = np.array( [ [ 10.0, 20.0, 30.0 ], [ 40.0, 50.0, 60.0 ] ] )
>>> Z = X + Y
>>> Z
array([[ 11., 22., 33.],
[ 44., 55., 66.]])
Matrix Subtraction
Subtract corresponding elements. Result is a matrix.
$\mathbf{Z} = \mathbf{X} - \mathbf{Y} = \begin{bmatrix} x_{11} - y_{11} & x_{12} - y_{12} & \cdots & x_{1n} - y_{1n} \\ x_{21} - y_{21} & x_{22} - y_{22} & \cdots & x_{2n} - y_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} - y_{m1} & x_{m2} - y_{m2} & \cdots & x_{mn} - y_{mn} \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> Y = np.array( [ [ 10.0, 20.0, 30.0 ], [ 40.0, 50.0, 60.0 ] ] )
>>> Z = X - Y
>>> Z
array([[ -9., -18., -27.],
[-36., -45., -54.]])
Multiply corresponding elements. Result is a matrix.
$\mathbf{Z} = \mathbf{X} \circ \mathbf{Y} = \begin{bmatrix} x_{11} y_{11} & x_{12} y_{12} & \cdots & x_{1n} y_{1n} \\ x_{21} y_{21} & x_{22} y_{22} & \cdots & x_{2n} y_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} y_{m1} & x_{m2} y_{m2} & \cdots & x_{mn} y_{mn} \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> Y = np.array( [ [ 10.0, 20.0, 30.0 ], [ 40.0, 50.0, 60.0 ] ] )
>>> Z = X * Y
>>> Z
array([[ 10., 40., 90.],
[ 160., 250., 360.]])
Matrix Multiplication
See Understanding Matrix Multiplication section.
\begin{align} \mathbf{Z} & = \mathbf{X} \cdot \mathbf{Y} \\ \\ & = \begin{bmatrix} x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} & x_{m2} & \cdots & x_{mn} \\ \end{bmatrix} \begin{bmatrix} y_{11} & y_{12} & \cdots & y_{1p} \\ y_{21} & y_{22} & \cdots & y_{2p} \\ \vdots & \vdots & \ddots & \vdots \\ y_{n1} & y_{n2} & \cdots & y_{np} \\ \end{bmatrix} \\ \\ & = \begin{bmatrix} x_{11} y_{11} + x_{12} y_{21} + \cdots + x_{1n} y_{n1} & x_{11} y_{12} + x_{12} y_{22} + \cdots + x_{1n} y_{n2} & \cdots & x_{11} y_{1p} + x_{12} y_{2p} + \cdots + x_{1n} y_{np} \\ x_{21} y_{11} + x_{22} y_{21} + \cdots + x_{2n} y_{n1} & x_{21} y_{12} + x_{22} y_{22} + \cdots + x_{2n} y_{n2} & \cdots & x_{21} y_{1p} + x_{22} y_{2p} + \cdots + x_{2n} y_{np} \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} y_{11} + x_{m2} y_{21} + \cdots + x_{mn} y_{n1} & x_{m1} y_{12} + x_{m2} y_{22} + \cdots + x_{mn} y_{n2} & \cdots & x_{m1} y_{1p} + x_{m2} y_{2p} + \cdots + x_{mn} y_{np} \\ \end{bmatrix} \\ \\ \end{align}
>>> X = np.array( [ [ 2, -4, 6 ], [ 5, 7, -3 ] ] )
>>> Y = np.array( [ [ 8, -5 ], [ 9, 3 ], [ -1, 4 ] ] )
>>> Z = np.dot( X, Y )
>>> Z
array([[-26, 2],
[106, -16]])
Add scalar to each element. Result is a matrix.
$\mathbf{Y} = \mathbf{X} + a = \begin{bmatrix} x_{11} + a & x_{12} + a & \cdots & x_{1n} + a \\ x_{21} + a & x_{22} + a & \cdots & x_{2n} + a \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} + a & x_{m2} + a & \cdots & x_{mn} + a \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> a = 3.14
>>> Y = X + a
>>> Y
array([[ 4.14, 5.14, 6.14],
[ 7.14, 8.14, 9.14]])
Matrix-Scalar Subtraction
Subtract scalar from each element. Result is a matrix.
$\mathbf{Y} = \mathbf{X} - a = \begin{bmatrix} x_{11} - a & x_{12} - a & \cdots & x_{1n} - a \\ x_{21} - a & x_{22} - a & \cdots & x_{2n} - a \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} - a & x_{m2} - a & \cdots & x_{mn} - a \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> a = 3.14
>>> Y = X - a
>>> Y
array([[-2.14, -1.14, -0.14],
[ 0.86, 1.86, 2.86]])
Matrix-Scalar Multiplication
Multiply each element by scalar. Result is a matrix.
$\mathbf{Y} = \mathbf{X} a = \begin{bmatrix} x_{11} a & x_{12} a & \cdots & x_{1n} a \\ x_{21} a & x_{22} a & \cdots & x_{2n} a \\ \vdots & \vdots & \ddots & \vdots \\ x_{m1} a & x_{m2} a & \cdots & x_{mn} a \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> a = 3.14
>>> Y = X * a
>>> Y
array([[ 3.14, 6.28, 9.42],
[ 12.56, 15.7 , 18.84]])
Matrix-Scalar Division
Divide each element by scalar. Result is a matrix.
$\mathbf{Y} = \frac{\mathbf{X}}{a} = \begin{bmatrix} \frac{x_{11}}{a} & \frac{x_{12}}{a} & \cdots & \frac{x_{1n}}{a} \\ \frac{x_{21}}{a} & \frac{x_{22}}{a} & \cdots & \frac{x_{2n}}{a} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{x_{m1}}{a} & \frac{x_{m2}}{a} & \cdots & \frac{x_{mn}}{a} \\ \end{bmatrix}$
>>> X = np.array( [ [ 1.0, 2.0, 3.0 ], [ 4.0, 5.0, 6.0 ] ] )
>>> a = 3.14
>>> Y = X / a
>>> Y
array([[ 0.31847134, 0.63694268, 0.95541401],
[ 1.27388535, 1.59235669, 1.91082803]])
##### Understanding Matrix Multiplication
The term matrix multiplication can be a bit confusing because its meaning is not consistent with matrix addition and matrix subtraction.
In matrix addition, we take two $$m$$ x $$n$$ matrices and add their corresponding elements, which results in another $$m$$ x $$n$$ matrix. Matrix subtraction works similarly.
So, it would seem to follow that matrix multiplication would have a similar meaning - that you multiply the corresponding elements. This operation, however, is not called matrix multiplication. It is instead called the Hadamard Product.
So what is matrix multiplication?
Matrix multiplication is a row-by-column operation in which the elements in the $$i^{th}$$ row of the first matrix are multiplied by the corresponding elements of the $$j^{th}$$ column of the second matrix and the results are added together.
For example, if we want to multiply these two matrices:
$$\begin{bmatrix} \color{blue}{2} & \color{blue}{-4} & \color{blue}{6} \\ 5 & 7 & -3 \\ \end{bmatrix} \begin{bmatrix} \color{green}{8} & -5 \\ \color{green}{9} & 3 \\ \color{green}{-1} & 4 \\ \end{bmatrix} \\ \\$$
To compute the first element of the resulting matrix, we perform:
$$( 2 * 8 ) + ( -4 * 9 ) + ( 6 * -1 ) = -26$$
And insert this value into the resulting matrix:
$$\begin{bmatrix} -26 & ? \\ ? & ? \\ \end{bmatrix} \\ \\$$
We follow this same form for each subsequent row-column pairing:
\begin{align} \begin{bmatrix} 2 & -4 & 6 \\ 5 & 7 & -3 \\ \end{bmatrix} \begin{bmatrix} 8 & -5 \\ 9 & 3 \\ -1 & 4 \\ \end{bmatrix} & = \begin{bmatrix} \mathbf{X}_{row1} \cdot \mathbf{Y}_{col1} & \mathbf{X}_{row1} \cdot \mathbf{Y}_{col2} \\ \mathbf{X}_{row2} \cdot \mathbf{Y}_{col1} & \mathbf{X}_{row2} \cdot \mathbf{Y}_{col2} \\ \end{bmatrix} \\ \\ & = \begin{bmatrix} (2)(8)+(-4)(9)+(6)(-1) & (2)(-5)+(-4)(3)+(6)(4) \\ (5)(8)+(7)(9)+(-3)(-1) & (5)(-5)+(7)(3)+(-3)(4) \\ \end{bmatrix} \\ \\ & = \begin{bmatrix} (16)+(-36)+(-6) & (-10)+(-12)+(24) \\ (40)+(63)+(3) & (-25)+(21)+(-12) \\ \end{bmatrix} \\ \\ & = \begin{bmatrix} -26 & 2 \\ 106 & -16 \\ \end{bmatrix} \\ \\ \end{align}
General Rules of Matrix Multiplication
$$\mathbf{X}_{mn} \cdot \mathbf{Y}_{np} = \mathbf{Z}_{mp}$$
• The number of columns in $$\mathbf{X}$$ must equal the number of rows in $$\mathbf{Y}$$.
(Their inner dimensions must be the same)
• The order of $$\mathbf{Z}$$ is the number of rows in $$\mathbf{X}$$ by the number of columns in $$\mathbf{Y}$$.
(The dimensions of $$\mathbf{Z}$$ are the outer dimensions)
• Each element in row $$i$$ of $$\mathbf{X}$$ is paired with the corresponding element in column $$j$$ of $$\mathbf{Y}$$.
• The element in row $$i$$, column $$j$$ of $$\mathbf{Z}$$ is formed by multiplying these paired elements and summing the results.
• For each element in $$\mathbf{Z}$$, there will be $$n$$ products that are summed.
Why is Matrix Multiplication Important?
At first glance, matrix multiplication seems to have a very specific definition, the value of which may not be obvious. Yet, matrix multiplication is one of the most commonly used operations in machine learning. Why? What does this seemingly obscure operation represent?
Matrix multiplication provides a natural mechanism for representing linear transformations.
For example, let's say we have a coordinate in two-dimensional space $$( x, y )$$ that we wish to transform with the following formula:
$$Transform( x, y ) = (2x + 3y, 4x - 5y)$$
If $$( x, y ) = ( 7, 9 )$$,
then $$Transform( 7, 9 ) = (2*7 + 3*9, 4*7 - 5*9) = (41, -17)$$
To represent $$Transform$$ in matrix form, we create a matrix containing the coefficients of $$x$$ and $$y$$ like so:
$$Transform = \begin{bmatrix} 2 & 3 \\ 4 & -5 \\ \end{bmatrix} \\ \\$$
We want to use this to produce our transformation: $$Transform( x, y ) = (2x + 3y, 4x - 5y)$$
Using matrix multiplication, we can write it like this:
$$\begin{bmatrix} 2 & 3 \\ 4 & -5 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 2x + 3y \\ 4x - 5y \\ \end{bmatrix} \\ \\$$
In this form, we could replace $$\begin{bmatrix} x \\ y \\ \end{bmatrix}$$ with any specific $$(x, y)$$ pair in order to apply the transformation to that point. For example:
$$\begin{bmatrix} 2 & 3 \\ 4 & -5 \\ \end{bmatrix} \begin{bmatrix} 7 \\ 9 \\ \end{bmatrix} = \begin{bmatrix} 2*7 + 3*9 \\ 4*7 - 5*9 \\ \end{bmatrix} = \begin{bmatrix} 41 \\ -17 \\ \end{bmatrix} \\ \\$$
#### Getting Started with Plotting in Python and Matplotlib
Importing Numpy library
import numpy as np
Importing Pyplot library
import matplotlib.pyplot as plt
Plot y-axis data
# Note: x-axis is automatically generated as [ 0, 1, 2, 3 ]
plt.plot( [ 1, 4, 9, 16 ] )
plt.show()
Plot x-axis and y-axis data
plt.plot( [ 1, 2, 3, 4 ], [ 1, 4, 9, 16 ] )
plt.show()
Plot x-axis and y-axis data with per-axis extents
# Note: axis() formatted as [ xmin, xmax, ymin, ymax ]
plt.axis( [ 0, 6, 0, 50 ] )
plt.plot( [ 1, 2, 3, 4 ], [ 1, 4, 9, 16 ] )
plt.show()
Customize axis labels
plt.xlabel('X-AXIS DATA')
plt.ylabel('Y-AXIS DATA')
plt.plot( [ 1, 2, 3, 4 ], [ 1, 4, 9, 16 ] )
plt.show()
Customize plot stylization
plt.plot( [ 1, 2, 3, 4 ], [ 1, 4, 9, 16 ], 'ro--')
plt.show()
Additional documentation of stylization options can be found here: Pyplot Lines and Markers and Pyplot Line Properties
Plot functions
def sigmoid(x):
return 1.0 / ( 1.0 + np.exp( -x ) )
def dsigmoid(x):
y = sigmoid( x )
return y * ( 1.0 - y )
def tanh(x):
return np.sinh( x ) / np.cosh( x )
def dtanh(x):
return 1.0 - np.square( tanh( x ) )
xData = np.arange( -10.0, 10.0, 0.1 )
ySigm = sigmoid( xData )
ySigd = dsigmoid( xData )
yTanh = tanh( xData )
yTand = dtanh( xData )
plt.axis( [ -10.0, 10.0, -1.1, 1.1 ] )
plt.plot( xData, ySigm, 'r', xData, ySigd, 'r--' )
plt.plot( xData, yTanh, 'g', xData, yTand, 'g--' )
plt.show()
Working with multiple figures and axes
def f(t):
return np.exp(-t) * np.cos(2*np.pi*t)
t1 = np.arange(0.0, 5.0, 0.1)
t2 = np.arange(0.0, 5.0, 0.02)
plt.figure(1)
plt.subplot(211)
plt.plot(t1, f(t1), 'bo', t2, f(t2), 'k')
plt.subplot(212)
plt.plot(t2, np.cos(2*np.pi*t2), 'r--')
plt.show()
Working with text
mu, sigma = 100, 15
x = mu + sigma * np.random.randn(10000)
# the histogram of the data
n, bins, patches = plt.hist(x, 50, normed=1, facecolor='g', alpha=0.75)
plt.xlabel('Smarts')
plt.ylabel('Probability')
plt.title('Histogram of IQ')
plt.text(60, .025, r'$\mu=100,\ \sigma=15$')
plt.axis([40, 160, 0, 0.03])
plt.grid(True)
plt.show()
ax = plt.subplot(111)
t = np.arange(0.0, 5.0, 0.01)
s = np.cos(2*np.pi*t)
line, = plt.plot(t, s, lw=2)
plt.annotate('local max', xy=(2, 1), xytext=(3, 1.5),
arrowprops=dict(facecolor='black', shrink=0.05),
)
plt.ylim(-2,2)
plt.show()
Plotting in 3D
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
fig = plt.figure()
ax = fig.gca(projection='3d')
X = np.arange(-5, 5, 0.25)
Y = np.arange(-5, 5, 0.25)
X, Y = np.meshgrid(X, Y)
R = np.sqrt(X**2 + Y**2)
Z = np.sin(R)
surf = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm,
linewidth=0, antialiased=False)
ax.set_zlim(-1.01, 1.01)
ax.zaxis.set_major_locator(LinearLocator(10))
ax.zaxis.set_major_formatter(FormatStrFormatter('%.02f'))
fig.colorbar(surf, shrink=0.5, aspect=5)
plt.show()
#### Classification as Spatial Partitioning
Classification can be a somewhat arbitrary process. It forces us to draw a line in the sand even though conceptual categories often have fuzzy boundaries. Presumably no one would say that one grain of sand makes a pile, but everyone would say one million grains of sand do. Somewhere between these, we must draw a line. In other words, categories are perceptual - their existence is contingent upon our looking for them. Let's see what this means for clustering algorithms.
#### A Brief Look at k-means Clustering
Concept
• Assign data points to a set number of clusters:
• Find the nearest "cluster center" for each data point
• Adjust each cluster center position to be the centroid of its associated data points
• Repeat until no data point changes its assigned cluster from previous iteration
Algorithm
• Let $$\mathbf{X} = \{ \mathbf{x_1}, \mathbf{x_2} \cdots \mathbf{x_n} \}$$ be the set of data points
• Let $$\mathbf{V} = \{ \mathbf{v_1}, \mathbf{v_2} \cdots \mathbf{v_c} \}$$ be the set of cluster centers
1. Randomly select 'c' cluster centers
2. Compute the distance between each data point and each cluster center
3. Assign each data point to its nearest cluster
4. Recompute the new cluster centers using the formula:
$$\mathbf{v_i} = \frac{1}{c_i}\sum_{j=1}^{c_i} \mathbf{x_j}$$
where $$c_i$$ is the number of data points in the $$i^{th}$$ cluster
5. Recompute the distance between each data point and the new cluster centers
6. If no data point was reassigned then stop, otherwise repeat from Step 3
Limitations
• Requires user to specify the number of clusters.
Therefore, we cannot really probe how many distinct categories exist.
• Does not perform well on highly overlapping data.
• Euclidean distance can skew weighting of underlying factors.
• Algorithm is not invariant with respect to non-linear transformations.
With different representations of the data - polar coordinates vs Cartesian coordinates, e.g. - we get different results.
#### Homework
Assignment
• Install Python, Pip, NumPy and Matplotlib.
• Implement K-Means Clustering in Python.
• Spend some time experimenting with vectors and matrices in Python / NumPy:
• Try out the operations we've discussed.
• Plug-in different values and dimensions.
• Consider the results, tweak the values and repeat.
• The more you play with these operations, the more intuitive they will become.<|endoftext|>
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Pristine quantum light source created at the edge of silicon chip
The smallest amount of light you can have is one photon, so dim that it's pretty much invisible to humans. While imperceptible, these tiny blips of energy are useful for carrying quantum information around. Ideally, every quantum courier would be the same, but there isn't a straightforward way to produce a stream of identical photons. This is particularly challenging when individual photons come from fabricated chips.
Now, researchers at the Joint Quantum Institute (JQI) have demonstrated a new approach that enables different devices to repeatedly emit nearly identical single photons. The team, led by JQI Fellow Mohammad Hafezi, made a silicon chip that guides light around the device's edge, where it is inherently protected against disruptions. Previously, Hafezi and colleagues showed that this design can reduce the likelihood of optical signal degradation. In a paper published online on Sept. 10 in Nature, the team explains that the same physics which protects the light along the chip's edge also ensures reliable photon production.
Single photons, which are an example of quantum light, are more than just really dim light. This distinction has a lot to do with where the light comes from. "Pretty much all of the light we encounter in our everyday lives is packed with photons," says Elizabeth Goldschmidt, a researcher at the US Army Research Laboratory and co-author on the study. "But unlike a light bulb, there are some sources that actually emit light, one photon at time, and this can only be described by quantum physics," adds Goldschmidt.
Many researchers are working on building reliable quantum light emitters so that they can isolate and control the quantum properties of single photons. Goldschmidt explains that such light sources will likely be important for future quantum information devices as well as further understanding the mysteries of quantum physics. "Modern communications relies heavily on non-quantum light," says Goldschmidt. "Similarly, many of us believe that single photons are going to be required for any kind of quantum communication application out there."
Scientists can generate quantum light using a natural color-changing process that occurs when a beam of light passes through certain materials. In this experiment the team used silicon, a common industrial choice for guiding light, to convert infrared laser light into pairs of different-colored single photons.
They injected light into a chip containing an array of miniscule silicon loops. Under the microscope, the loops look like linked-up glassy racetracks. The light circulates around each loop thousands of times before moving on to a neighboring loop. Stretched out, the light's path would be several centimeters long, but the loops make it possible to fit the journey in a space that is about 500 times smaller. The relatively long journey is necessary to get many pairs single photons out of the silicon chip.
Such loop arrays are routinely used as single photon sources, but small differences between chips will cause the photon colors to vary from one device to the next. Even within a single device, random defects in the material may reduce the average photon quality. This is a problem for quantum information applications where researchers need the photons to be as close to identical as possible.
The team circumvented this issue by arranging the loops in a way that always allows the light to travel undisturbed around the edge of the chip, even if fabrication defects are present. This design not only shields the light from disruptions—it also restricts how single photons form within those edge channels. The loop layout essentially forces each photon pair to be nearly identical to the next, regardless of microscopic differences among the rings. The central part of the chip does not contain protected routes, and so any photons created in those areas are affected by material defects.
The researchers compared their chips to ones without any protected routes. They collected pairs of photons from the different chips, counting the number emitted and noting their color. They observed that their quantum light source reliably produced high quality, single-color photons time and again, whereas the conventional chip's output was more unpredictable.
"We initially thought that we would need to be more careful with the design, and that the photons would be more sensitive to our chip's fabrication process," says Sunil Mittal, a JQI postdoctoral researcher and lead author on the new study. "But, astonishingly, photons generated in these shielded edge channels are always nearly identical, regardless of how bad the chips are."
Mittal adds that this device has one additional advantage over other single photon sources. "Our chip works at room temperature. I don't have to cool it down to cryogenic temperatures like other quantum light sources, making it a comparatively very simple setup."
The team says that this finding could open up a new avenue of research, which unites quantum light with photonic devices having built-in protective features. "Physicists have only recently realized that shielded pathways fundamentally alter the way that photons interact with matter," says Mittal. "This could have implications for a variety of fields where light-matter interactions play a role, including quantum information science and optoelectronic technology."<|endoftext|>
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|Fig. 1 Typical dew-point meter for panel-mounting, with a gas moisture probe, the tip of which is placed in the atmosphere flow pipe. (Photo courtesy of Michell Instruments)
On a warm, moist day, our earth’s atmosphere will contain a significant amount of moisture in it. During the night, when the sun has gone down, this atmosphere will become cooled, and will not be able to hold onto the amount of moisture (water) that it could when it was warm, and so, some of that moisture will condense out onto the grass in the form of “dew”. Then, during the following day, when the sun heats the air up once again, the dew will evaporate from the ground.
It is well known that the warmer the gas, the greater will be the amount of moisture that gas can hold.
At any given point in time, all gases will have what is called a “dewpoint”. The “dewpoint” of any gas is the temperature to which that gas must be cooled to get the first droplet of moisture to condense out of that gas (assumed to be at one standard atmosphere of pressure). The less the amount of moisture in that gas, the cooler must be the temperature to which that gas must be cooled in order to get the first condensation to occur. Based on that fact then, it will be understood that the lower the dewpoint of a gas, the drier (lacking moisture) is that gas.
Thus, the dewpoint of a gas tells you how much moisture (water) is present in a gas.
|Fig. 2 Some dewpoint meters are portable, and can moved from location to location on a cart. (photo courtesy of Shaw Moisture Meters)|
Please note, however, that water (H2O) also represents the presence of oxygen, and oxygen is an element that can react with metals to form oxides on the metal surface being heated up in that furnace. Thus, to prevent oxidation of metals during a furnace brazing process, it is important to keep oxygen away from the part being brazed, which means controlling the amount of moisture present in any gaseous atmosphere being used in a vacuum-furnace during that brazing process. Gaseous atmospheres are indeed introduced into vacuum furnaces for rapid-cooling after brazing, or may be introduced into the vacuum chamber in order to build up a partial-pressure to prevent outgassing of certain elements from the metals being brazed, or from the brazing-filler-metal (BFM) itself. Gases may also be introduced into “multi-bar” vacuum furnaces for purposes of perhaps heat-treating parts during or following a brazing run.
The dewpoint of a gas can be measured fairly easily by the use of a dewpoint-meter (also called a hygrometer), which, as the name suggests, is an instrument that will measure the amount of moisture in a gas and provide that information via readouts, either as part of the furnace control instrumentation (as shown in Fig. 1), or perhaps as a separate, portable dewpoint-meter that can be connected to a special port in the atmosphere-piping system to the vacuum furnace (as shown in Fig. 2).
Three very important things to know about dewpoint control:
1. Dewpoint must be measured at the furnace, not at the gas source.
Receiving a certification from your gas-supplier stating that the atmosphere (gaseous or liquid) they have delivered to you has “such and such” dewpoint, is actually quite meaningless to your brazing operations. What is important, instead, is the dewpoint of that atmosphere when it is inside the furnace chamber itself where the brazing is taking place – after it’s been piped from the outside tanks all the way to the furnace. The piping and pipe-fittings can have a major effect (either negative or positive) on the dewpoint of an atmosphere gas being piped from a liquid tank or other external source of gas. This will be the topic of another article soon.
2. If you are, in fact, putting a gaseous atmosphere into your vacuum chamber, you MUST know its dewpoint.
If a gaseous atmosphere is being used in your vacuum furnace brazing process, perhaps for outgassing-suppression, or for rapid cooling, etc., or for multi-bar HT/brazing, then the dewpoint of that gas MUST be measured – at the furnace — and provided to the furnace operator, or the brazing process is NOT being properly controlled! I am often amazed by brazing shop personnel who do NOT know the dewpoint of the gaseous-atmosphere being used inside their furnace. They merely “assume” its okay. That is a very poor brazing practice.
Remember, dewpoint represents moisture. Moisture represents oxygen. Oxygen at brazing temp causes oxidation of metals. Unfortunately, brazing filler metals (BFMs) don’t like to bond to, or flow over, oxides on metals! Thus, if you don’t accurately monitor, and control, the dewpoint of the gaseous atmosphere you are using in your brazing operations, you can easily ruin your brazing.
Therefore, the dewpoint of any brazing-atmosphere used must be known in order for the brazing process to be “in control”. There are a number of good manufacturers of dewpoint equipment available in the marketplace. Remember, if a gaseous atmosphere touches a brazed component at any time during the entire vacuum-brazing cycle, from loading to unloading, know (and control) the dewpoint of that gas.
3. If the dewpoint is not acceptable, it can be “dried” further.
Okay, so if you’ve now installed, and are using, a good dewpoint-meter, and it’s showing you that the atmosphere is not sufficiently “dry” for use in your vacuum-brazing operations, what can you do about it? The answer may very well be to add a desiccant-drier, placed right next to the furnace, to further dry the atmosphere prior to its entering the furnace. An example of such a gas-drier system is shown in Figure 3.
A regenerative desiccant dryer can automatically dry the atmosphere-gas just prior to its entering the vacuum chamber. The dual drying chambers are alternately cycled through drying and regeneration cycles, so that one chamber is drying the atmosphere-gas at all times.
The compressed atmosphere-gas flows through one drying tower where any residual water vapor is adsorbed into the desiccant (the desiccant is an inert, hygroscopic material that has an enormous surface-to-mass ratio). The off-steam drying tower is regenerated by taking a small amount of the dried process gas (purge) and passing it though the off-stream tower’s desiccant, thus grabbing any moisture molecules trapped in that desiccant. These moisture molecules are released and carried off by the purge flow and exhausted to atmosphere.
Is the atmosphere-gas (hydrogen, helium, nitrogen, and argon) that you are using in your vacuum furnace for partial-pressure, or for rapid cooling, or for multi-bar operations, truly a dry gas of at least -80°F (-65°C) dewpoint or drier when it is introduced into your vacuum furnace? If not, then your brazing process is not fully in control, and you will need to implement some of the recommendations described in this article regarding dewpoint control via dewpoint-meters, perhaps also with some desiccant-driers at the furnace, if needed. Dewpoint control is absolutely essential whenever industrial gases are used as part of anyone’s brazing operations.
And, as mentioned earlier, you cannot depend on certifications from your gas-suppliers that the gas they delivered to you is very dry, since the “dryness” of that gas can change dramatically (for the worse) when you pipe that gas to your furnace, especially if that piping system is leaky. We’ll look at that topic next month.
Dan Kay – Tel: 860-651-5595 860-651-5595: – Dan Kay operates his own brazing consulting/training company, and has been involved full-time in brazing for 40-years. Dan regularly consults in areas of vacuum and atmosphere brazing, as well as in torch (flame) and induction brazing. His brazing seminars, held a number of times each year help people learn how to apply the fundamentals of brazing to improve their productivity and lower their costs. Contact information for Dan Kay (e-mail, phone, fax, etc.), can be found by visiting his company’s website at: http://www.kaybrazing.com/
To View a listing of all Dan’s articles please click here
Read Dan Kay’s Biography
© Copyright Dan Kay 2014<|endoftext|>
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Lesson Plan #:AELP-SPA0022
Author: Jane Whaling, St. Michael, AK Date: 1994
Grade Level(s): 3, 4, 5
- Science/Space Sciences
Overview: This lesson is designed for review of the solar system. It is fun for students and involves cooperative learning. The children should have a good working knowledge of the information being reviewed.
The students will:
Activities and Procedures: Divide the class into two groups. Have each group designate a spokesperson for their group. Give the groups ten to 15 minutes to make up five to ten questions about the solar system, using their notes. They must know the correct answers to the questions. When the groups are ready with their questions you may begin the game.
Determine which group goes first by flipping a coin. The teams will take turns asking each other their questions. A team gets 30 seconds to find the correct answer to the question in their notes and hand-outs. The answer must be given by the spokesperson. If the group gets the question correct they receive one point. They also get a chance to earn an extra point by making a basket with the nerf ball. Have the group pick a person to shoot the nerf ball into the trash can. Continue taking turns until all the questions have been asked. The winning group has the most points.
Tying it all Together:
Evaluate the students on their performance in the group. They should work cooperatively.<|endoftext|>
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The following worksheets are designed to help your younger beginners associate notes belonging to the key of C major on the stave with notes on the fingerboard in first position. They combine music theory (learning to write neatly and accurately on the stave) with basic cello theory (learning the notes and fingering of first position). Each document is arranged in the order of an ascending C Major scale (2 octaves) – one page per note. By mixing up the pages you can make the worksheet more challenging. Kept in their current order they will be much easier to do, but a good way to introduce the C major scale.
These worksheets are free to download and print out, but please observe the copyright: no selling, no incorporating into other works or documents. Feedback welcome – especially from those who try them out!
© D C Cello Studio<|endoftext|>
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History.com is one of the best online sources for history material. For example:
The American Revolution (1775-83) is also known as the American Revolutionary War and the U.S. War of Independence. The conflict arose from growing tensions between residents of Great Britain’s 13 North American colonies and the colonial government, which represented the British crown. Skirmishes between British troops and colonial militiamen in Lexington and Concord in April 1775 kicked off the armed conflict, and by the following summer, the rebels were waging a full-scale war for their independence. France entered the American Revolution on the side of the colonists in 1778, turning what had essentially been a civil war into an international conflict. After French assistance helped the Continental Army force the British surrender at Yorktown, Virginia, in 1781, the Americans had effectively won their independence, though fighting would not formally end until 1783.
For more than a decade before the outbreak of the American Revolution in 1775, tensions had been building between colonists and the British authorities. Attempts by the British government to raise revenue by taxing the colonies (notably the Stamp Act of 1765, the Townshend Tariffs of 1767 and the Tea Act of 1773) met with heated protest among many colonists, who resented their lack of representation in Parliament and demanded the same rights as other British subjects. Colonial resistance led to violence in 1770, when British soldiers opened fire on a mob of colonists, killing five men in what was known as the Boston Massacre. After December 1773, when a band of Bostonians dressed as Mohawk Indians boarded British ships and dumped 342 chests of tea into Boston Harbor, an outraged Parliament passed a series of measures (known as the Intolerable, or Coercive Acts) designed to reassert imperial authority in Massachusetts.
Did You Know?: Now most famous as a traitor to the American cause, General Benedict Arnold began the Revolutionary War as one of its earliest heroes, helping lead rebel forces in the capture of Fort Ticonderoga in May 1775.
In response, a group of colonial delegates (including George Washington of Virginia, John and Samuel Adams of Massachusetts, Patrick Henry of Virginia and John Jay of New York) met in Philadelphia in September 1774 to give voice to their grievances against the British crown. This First Continental Congress did not go so far as to demand independence from Britain, but it denounced taxation without representation, as well as the maintenance of the British army in the colonies without their consent, and issued a declaration of the rights due every citizen, including life, liberty, property, assembly and trial by jury. The Continental Congress voted to meet again in May 1775 to consider further action, but by that time violence had already broken out. On April 19, local militiamen clashed with British soldiers in Lexington and Concord, Massachusetts, marking the first shots fired in the Revolutionary War.
To check out both text and a video in orcder to learn more about the American Revolutionary War, please click here.<|endoftext|>
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What is the meaning of angle?
Angle in planar geometry is nothing but a figure that is formed with the help of 2 rays & these are well – known as the sides of an angle. In addition to this, they tend to share a common endpoint which is popularly known as the vertex of the angle. Another aspect which we should be aware of is that the angles which are formed by 2 rays are said to be on the same plane however, this plane does not have to be the Euclidean plane.
How are angles formed?
Angles are essentially formed when 2 planes intersect in not only Euclidean but also in the other spaces. Hence, these are known as the dihedral angles. An individual will also have to remember the fact that these angles which are formed by the intersection of 2 curves in a plane can be defined as that of the angle which is determined by tangent rays at point of intersection. Another example can be the spherical angle which is stated to be formed by 2 great circles on sphere is treated to be the dihedral angle between those planes which have been determined by great circles. Moreover, angle is used to determine measure of angle or rotation.
Degrees to radians conversion table / Radians to degrees conversion table
|Radians (rad)||Radians (rad)||Degrees (°)|
|rad||0. 5235987756 rad||30°|
Formula used in an angle converter
The formula that is put in to use in an angle converter is stated to be one which has been given below –
\(S \times C=E\)
In the above equation, S is said to be the starting value while the conversion factor is represented by C & the converted value that is the result is got by E.
How to use an angle conversion tool?
When you require to make use of the online angle converter tool, an individual would be required to make use of the simple steps listed below. First of all, you would be required to enter the value which is to be converted. Remember that the value should be a real number or it can also be a scientific notation.
Later you will have to select the unit in which the value is before selecting the unit in which it is to be converted in to. And when the details have been entered, you can click on the convert tab to know the answer. However, if you wish to change the values then you can also click on reset to re – enter the values before finally clicking on the convert tab. Hence, with this angle converter tool, you would be able to convert the angles easily & without any trouble.<|endoftext|>
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# Word Problems on H.C.F.
Let us consider some of the word problems on H.C.F. (highest common factor).
1. Two wires are 12 m and 16 m long. The wires are to be cut into pieces of equal length. Find the maximum length of each piece.
To find the maximum length, we find the H.C.F. of 12 and 16
Factors of 12: 1, 2, 3, 4, 6, 12Factors of 16: 1, 2, 4, 8, 16 1 × 12, 2 × 6, 3 × 41 × 16, 2 × 8, 4 × 4
H.C.F. = 4
The maximum length of each piece is 4 m.
2. Find the greatest number which is less by 2 to divide 24, 28 and 64 exactly.
First we find the H.C.F. of 24, 28 and 64.
H.C.F. = 4
The required number is 4 - 2 = 2.
(We see that 2 is less than 4 by 2.)
3. Find the greatest number that divides 90 and 126 exactly.
Solution:
The greatest number that divides both 90 and 126 exactly is their HCF.
We can find HCF by long division method.
HCF of 90 and 126 is 18.
4. Rachel has 24 red candies and Maya has 18 green candies. They want to arrange the candies in such a way that each row contains equal number of candies and also each row should have only red candies or green candies. What is the greatest number of candies that can be arranged in each row?
Solution:
In order to find the greatest number of candies that can be arranged in equal rows, we find the HCF of two numbers.
Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Factors of 18 are 1, 2, 3, 6, 9 and 18.
The highest common factor is 6.
So, the greatest number of candies that can be arranged in each row is 6.
## You might like these
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Divisible by 10 is discussed below. A number is divisible by 10 if it has zero (0) in its units place. Consider the following numbers which are divisible by 10, using the test of divisibility by 10:
• ### Divisible by 5 | Test of divisibility by 5| Rules of Divisibility by 5
Divisible by 5 is discussed below: A number is divisible by 5 if its units place is 0 or 5. Consider the following numbers which are divisible by 5, using the test of divisibility by
• ### Divisible by 9 | Test of Divisibility by 9 |Rules of Divisibility by 9
A number is divisible by 9, if the sum is a multiple of 9 or if the sum of its digits is divisible by 9. Consider the following numbers which are divisible by 9, using the test of divisibility by 9:
• ### Divisible by 6 | Test of Divisibility by 6| Rules of Divisibility by 6
Divisible by 6 is discussed below: A number is divisible by 6 if it is divisible by 2 and 3 both. Consider the following numbers which are divisible by 6, using the test of divisibility by 6: 42
• ### Divisible by 4 | Test of Divisibility by 4 |Rules of Divisibility by 4
A number is divisible by 4 if the number is formed by its digits in ten’s place and unit’s place (i.e. the last two digits on its extreme right side) is divisible by 4. Consider the following numbers which are divisible by 4 or which are divisible by 4, using the test of
• ### Divisible by 3 | Test of Divisibility by 3 |Rules of Divisibility by 3
A number is divisible by 3, if the sum of its all digits is a multiple of 3 or divisibility by 3. Consider the following numbers to find whether the numbers are divisible or not divisible by 3: (i) 54 Sum of all the digits of 54 = 5 + 4 = 9, which is divisible by 3.
• ### Relationship between H.C.F. and L.C.M. |Highest Common Factor|Examples
The product of highest common factor (H.C.F.) and lowest common multiple (L.C.M.) of two numbers is equal to the product of two numbers i.e., H.C.F. × L.C.M. = First number × Second number or, LCM × HCF = Product of two given numbers
• ### Divisibility Rules | Divisibility Test|Divisibility Rules From 2 to 18
To find out factors of larger numbers quickly, we perform divisibility test. There are certain rules to check divisibility of numbers. Divisibility tests of a given number by any of the number 2, 3, 4, 5, 6, 7, 8, 9, 10 can be perform simply by examining the digits of the
• ### Method of H.C.F. |Highest Common Factor|Factorization &Division Method
We will discuss here about the method of h.c.f. (highest common factor). The highest common factor or HCF of two or more numbers is the greatest number which divides exactly the given numbers. Let us consider two numbers 16 and 24.
• ### Prime and Composite Numbers | Prime Numbers | Composite Numbers
What are the prime and composite numbers? Prime numbers are those numbers which have only two factors 1 and the number itself. Composite numbers are those numbers which have more than two factors.
• ### Multiples | Multiples of a Number |Common Multiple|First Ten Multiples
What are multiples? ‘The product obtained on multiplying two or more whole numbers is called a multiple of that number or the numbers being multiplied.’ We know that when two numbers are multiplied the result is called the product or the multiple of given numbers.
• ### 4th Grade Factors and Multiples Worksheet | Factors & Multiples
In 4th grade factors and multiples worksheet we will find the factors of a number by using multiplication method, find the even and odd numbers, find the prime numbers and composite numbers, find the prime factors, find the common factors, find the HCF(highest common factors
• ### Examples on Multiples | Different Types of Questions on Multiples
Examples on multiples on different types of questions on multiples are discussed here step-by-step. Every number is a multiple of itself. Every number is a multiple of 1. Every multiple of a number is either greater than or equal to the number. Product of two or more numbers
• ### Worksheet on Word Problems on H.C.F. and L.C.M. |Highest Common Factor
In worksheet on word problems on H.C.F. and L.C.M. we will find the greatest common factor of two or more numbers and the least common multiple of two or more numbers and their word problems. I. Find the highest common factor and least common multiple of the following pairs
• ### Word Problems on L.C.M. | L.C.M. Word Problems | Questions on LCM
Let us consider some of the word problems on l.c.m. (least common multiple). 1. Find the lowest number which is exactly divisible by 18 and 24. We find the L.C.M. of 18 and 24 to get the required number.<|endoftext|>
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Written by Cathryn Sill
Illustrated by John Sill
What are forests and why are they important? What kind of wildlife lives in a forest? These questions and more are answered in the latest offering from the About Habitats series.
Cathryn and John Sill have created a wonderful beginner’s guide, presented in a clean, organized fashion. On the left-hand side of the page spread, a simple sentence with easy language appears over a white background. To the right, readers are treated to a painting, vivid with color and detail. With clear labeling for easy cross-reference, different levels of information are available for different reader levels. A kindergartener may pore over the picture of the cute chipmunk, while a second or third grader can expand their knowledge and literacy skills from the plant and animal identifiers (e.g., an Eastern Chipmunk in a deciduous forest, surrounded by morel mushrooms and Christmas ferns).
Young readers will learn about the different types of forests like deciduous, tropical, or boreal and how each has their own distinct characteristics. The illustrations show animals and plants thriving within each habitat: a snowshoe hare blends into the snowy landscape of the boreal forest; an orangutan finds shelter in the tropical forest canopy; a black bear, broad-winged hawk, and box turtle live side by side in the deciduous forest.
Extras include a global map of the major forest areas of the world, a glossary, bibliography, and additional websites for further information. For more titles in the About Habitats series, visit the publishers website: www.peachtree-online.com.<|endoftext|>
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#### If the perimeter of a circle is equal to that of a square, then the ratio of their areas is(A) 22 : 7 (B) 14 : 11 (C) 7 : 22 (D) 11: 14
Solution
According to question
$2 \pi r=4a$ (Because perimeter of circle = 2πr Perimeter of square =4 $\times$ side)
$\pi r=2a$ (here side of square =a)
$a=\frac{\pi r}{2}\\\\ \therefore \frac{Area \; of \; circle}{Area \; of\; square}=\frac{\pi r^{2}}{\left (\frac{\pi r}{2} \right )^{2}}=\frac{\pi r^{2}}{\pi r^{2}}\times \frac{4}{\pi}=\frac{4 \times 7}{22}=\frac{14}{11}$ (Using area of square = a2)
Hence ratio of their areas is 14: 11<|endoftext|>
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# How to Calculate the Area, Perimeter, and Radius of Quarter Circles
Today, we're focusing on quarter circles, a particular type of sector of a circle. We'll learn how to calculate their area and perimeter, and even reverse-engineer the radius if we know these properties!
## 1. Understanding Quarter Circles
A quarter circle is a sector of a circle that represents one-fourth of the circle. Because it’s a quarter of a circle, the central angle that generates a quarter circle is $$90$$ degrees.
## 2. Calculating the Area, Perimeter, and Radius
The area and perimeter (also called the circumference in full circles) are fractions of the respective measures of a full circle. The radius of a quarter circle is the same as that of the full circle it is a part of.
## Step-By-Step Guide to Calculating the Area, Perimeter, and Radius of Quarter Circles
Let’s break down the process:
### Step 1: Calculate the Area
The area of a quarter circle with radius r is given by: Area $$= 0.25\times π\times r^2$$.
### Step 2: Calculate the Perimeter
The perimeter of a quarter circle is a bit tricky. It includes the arc length (one-fourth of the circumference of the whole circle) and twice the radius (the two straight edges). So, the formula is: Perimeter $$= 0.5\times π\times r + 2r$$.
### Step 3: Find the Radius
If you know the area $$(A)$$ or perimeter $$(P)$$ of the quarter circle, you can rearrange the formulas to solve for the radius:
• From the area: $$r = \sqrt(\frac{(4A)}{(π)})$$
• From the perimeter: $$r = P / (2 + 0.5π)$$
For example, let’s say you have a quarter circle with radius 4 units:
1. Find the area: Area $$= 0.25\times π\times 4^2 = 4π$$ square units.
2. Find the perimeter: Perimeter $$= 0.5\times π\times 4 + 2\times 4 = 2π + 8$$ units.
As always, keep practicing, keep exploring, and enjoy your mathematical journey!
In this blog post, we’ve explained how to calculate the area, perimeter, and radius of quarter circles. This practical geometric skill can help you with a variety of math problems and real-world applications. Keep practicing, and you’ll master this in no time. Happy calculating!
### What people say about "How to Calculate the Area, Perimeter, and Radius of Quarter Circles - Effortless Math: We Help Students Learn to LOVE Mathematics"?
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Note: Because it’s National Pollinator’s Week, this post has been updated and republished.
We maintain quite a few honey bee hives, and a common question is why? The obvious answer would be the honey, but it’s actually more complex – honey bees are very important, but their numbers have been declining rapidly in the last several years.
Why They’re Important
Honey bees are important because their pollination is estimated to be responsible for producing one-third of the food we eat worldwide.
According to a recent United Nations study, 70 out of the 100 most important food crops in the world must be pollinated by bees. But the bees are in trouble; since 2006 they have been dying off, in part due to exposure to systemic and other pesticides.
Without enough pollinators, crop yields decline and the varieties of foods that can be produced diminish.
In order to produce food, plants require a few things. For some plants, pollinators are required because they’re incapable of producing fruit without them. Popular plants that require bee pollination include apple, blueberry, cucumber, pumpkin, raspberry, squash, and watermelon (we were amazed at how much our raspberry yield benefitted from having honey bees).
Other plants self-pollinate, but honey bee pollination significantly increases the abundance and size of the plant yield. Plants that benefit from pollination include beans, eggplant, peppers, potatoes, strawberries, and tomatoes.
How They Work
Honey bees live in hives organized around a queen, and cannot survive without one. She’s the only female capable of laying viable eggs and lays up to 2000 eggs per day. The other honey bees we generally observe are also females; when they first mature they work as nurse bees, caring for the eggs and feeding the larvae as they hatch. As they grow older, they transition to guard bees, protecting the hive from animals or other insects. Then, they serve as worker bees, flying up to five miles to find food. Male bees are few, and only leave the hive to mate with the queen during her yearly mating flight.
The food (nectar and pollen) collected by the worker bees is fed to hive bees and turned into honey to sustain the hive during the winter when it’s too cold to forage and there’s little food available. While foraging, the worker bees pollinate plants by carrying pollen from one plant to another. This enables the plants to fruit and produce viable seed.
What’s Happening to Them
The rapid decline in bee numbers has mainly been attributed to two phenomena – colony collapse disorder and neonicotinoid pesticides. Colony collapse disorder was first identified in the winter of 2006 when unprecedented honey bee hive losses (between 30% and 90%) were documented across the United States.
A single cause for colony collapse disorder hasn’t been identified; rather it’s believed to be a combination of factors such as pesticide exposure, pathogens, parasite, and nutritional deficiencies. Whatever the cause, winter hive losses have averaged 30 – 40% in the US since 2006.
Neonicotinoids are systemic pesticides (introduced in the 1990’s) that are neurotoxic to many insects and are especially toxic to honey bees. They are one of the fastest growing classes of pesticides used today and are commonly used to coat seeds and flea collars as well as treat lawns and garden compost. Because systemic pesticides are applied to the plant root (as a seed coating or a pour-on), they are taken in through the plant’s internal system and are given off in the plant’s nectar and pollen.
The use of these pesticides has been connected to numerous instances of major honey bee die-offs, and recent research has shown that even if they don’t immediately kill the bee, they undermine their immune systems making them more susceptible to pathogens and parasites. Four European countries have begun restricting the use of systemic pesticides because of the impact on honey bees.
What Can We Do?
Things that can be done to help the honey bees include:
♦ Become knowledgeable about the pesticides that are being used, and make sure you’re not using any of the systemic pesticides that are harming the honey bees
♦ Eat pesticide free foods and support farmers who don’t use pesticides
♦ Plant a garden with plants that bees like – even in cities, a plant container on a balcony can make a difference to urban bees and their keeper
♦ Plant flowers with a mix of colors, shapes and scents to create seasonal buffets
♦ Cut flowers to encourage re-blooming and prolong the foraging window
♦ Grow native plants like echinacea, coreopsis, sunflowers and butterfly weed
♦ Become a beekeeper
SUPPORT BEES IN YOUR GARDEN!<|endoftext|>
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# If tan A = n tan B and sin A = m sin B, then cos2 A is equal to :
1. $$\frac{m^2 + 1}{n^2 - 1}$$
2. $$\frac{m^2 - 1}{n^2 + 1}$$
3. $$\frac{m^2 - 1}{n^2 - 1}$$
4. $$\frac{n^2 - 1}{m^2 - 1}$$
Option 3 : $$\frac{m^2 - 1}{n^2 - 1}$$
## Detailed Solution
Given:
tan A = n tan B and sin A = m sin B
Formula used:
tan A = 1/cotA
sin A = 1/cosecA
cosec2A - cot2A = 1
Calculation:
As tan A = n tan B
⇒ cot B = n cot A
Squaring both sides, we get
⇒ cot2 B = n2 cot2A ----(1)
Also, sin A = m sin B
⇒ cosec B = m cosec A
Squaring both sides, we get
⇒ cosec2 B = m2 cosec2 A ----(2)
Subtracting (1) from (2), we get
⇒ cosec2 B - cot2B = m2 cosec2 A - n2 cot2A
⇒ $$1 = {m^2 \over sin^2 A} - {m^2 cos^2A \over sin^2 A}$$
⇒ $$1 = {m^2 - n^2 cos^2A \over sin^2 A}$$
⇒ sin2A = m2 - n2cos2A
⇒ 1 - cos2A = m2 - n2cos2A
⇒ n2cos2A - cos2A = m2 - 1
⇒ cos2A(n2 - 1) = m2 - 1
⇒ cos2A = (m2 - 1)/(n2 - 1)
∴ The value of cos2A is (m2 - 1)/(n2 - 1).<|endoftext|>
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Biology explains biological concepts and explores related fields as well as the history of biological science. Discusses critical issues such as embryogenesis and commercial applications of research in biology and addresses ethical issues. Provides information on careers in biology and contains color illustrations, sidebars, bibliographies, timelines, charts, and a glossary.
Chemical Elements: From Carbon to Krypton. Provides in-depth information on 112 known chemical elements. For younger and middle school students, but also appropriate for high school students.
Environmental Science Experiments
Environmental Science Experiments: Offers students and teachers the tools to explore various environmental issues; includes hands-on activities to learn more about environmental problems and what can be done to solve them.
Forensic Science Experiments
Forensic Science Experiments: Features experiments that will help students get a firm grasp on concepts in the field of forensic science. Each experiment is a self-contained unit with its own further resources, tips for the teacher, and links to the National Science Education Standards and allows students to think independently and build on their own base of science knowledge.
Grzimek's Animal Life Encyclopedia
Grzimek's Animal Life Encyclopedia: A completely revised and updated version of the original work published in Germany in 1960, this edition incorporates recent developments in the animal world as noted by prominent advisors and contributors from the scientific community.
Real-Life Math: Provides an understanding of commonly studied math concepts by illustrating their use in everyday life in everyday tasks, such as buying insurance, constructing a budget, reading graphs, adjusting cooking recipes or planning for retirement. Topics are designed to support the modern mathematics curriculum and contain examples related to the global economy. 2 volumes.
World of Earth Science
World of Earth Science: This series is a comprehensive guide to the concepts, theories, discoveries, pioneers and issues relating to topics in earth science. World of Earth Science's encyclopedic approach offers approximately 650 entries in a convenient A-Z format, cross-reference headings and is written in easy to understand language.
Beacham's Guide to the Endangered Species of North America
Beacham's Guide to the Endangered Species of North America: This guide presents extensive data on the habitats and ecosystems of more than 1,200 species identified as endangered or threatened by the U.S. Fish and Wildlife service. Entries include introductory information with an image of the species; a summary section detailing information on each species; a concluding summary of recovery efforts; and a bibliography of contacts for further information.<|endoftext|>
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This is “The Empirical Rule and Chebyshev’s Theorem”, section 2.5 from the book Beginning Statistics (v. 1.0). For details on it (including licensing), click here.
Has this book helped you? Consider passing it on:
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.
## 2.5 The Empirical Rule and Chebyshev’s Theorem
### Learning Objectives
1. To learn what the value of the standard deviation of a data set implies about how the data scatter away from the mean as described by the Empirical Rule and Chebyshev’s Theorem.
2. To use the Empirical Rule and Chebyshev’s Theorem to draw conclusions about a data set.
You probably have a good intuitive grasp of what the average of a data set says about that data set. In this section we begin to learn what the standard deviation has to tell us about the nature of the data set.
## The Empirical Rule
We start by examining a specific set of data. Table 2.2 "Heights of Men" shows the heights in inches of 100 randomly selected adult men. A relative frequency histogram for the data is shown in Figure 2.15 "Heights of Adult Men". The mean and standard deviation of the data are, rounded to two decimal places, $x-=69.92$ and s = 1.70. If we go through the data and count the number of observations that are within one standard deviation of the mean, that is, that are between $69.92−1.70=68.22$ and $69.92+1.70=71.62$ inches, there are 69 of them. If we count the number of observations that are within two standard deviations of the mean, that is, that are between $69.92−2(1.70)=66.52$ and $69.92+2(1.70)=73.32$ inches, there are 95 of them. All of the measurements are within three standard deviations of the mean, that is, between $69.92−3(1.70)=64.822$ and $69.92+3(1.70)=75.02$ inches. These tallies are not coincidences, but are in agreement with the following result that has been found to be widely applicable.
Table 2.2 Heights of Men
68.7 72.3 71.3 72.5 70.6 68.2 70.1 68.4 68.6 70.6 73.7 70.5 71 70.9 69.3 69.4 69.7 69.1 71.5 68.6 70.9 70 70.4 68.9 69.4 69.4 69.2 70.7 70.5 69.9 69.8 69.8 68.6 69.5 71.6 66.2 72.4 70.7 67.7 69.1 68.8 69.3 68.9 74.8 68 71.2 68.3 70.2 71.9 70.4 71.9 72.2 70 68.7 67.9 71.1 69 70.8 67.3 71.8 70.3 68.8 67.2 73 70.4 67.8 70 69.5 70.1 72 72.2 67.6 67 70.3 71.2 65.6 68.1 70.8 71.4 70.2 70.1 67.5 71.3 71.5 71 69.1 69.5 71.1 66.8 71.8 69.6 72.7 72.8 69.6 65.9 68 69.7 68.7 69.8 69.7
Figure 2.15 Heights of Adult Men
### The Empirical Rule
If a data set has an approximately bell-shaped relative frequency histogram, then (see Figure 2.16 "The Empirical Rule")
1. approximately 68% of the data lie within one standard deviation of the mean, that is, in the interval with endpoints $x-±s$ for samples and with endpoints $μ±σ$ for populations;
2. approximately 95% of the data lie within two standard deviations of the mean, that is, in the interval with endpoints $x-±2s$ for samples and with endpoints $μ±2σ$ for populations; and
3. approximately 99.7% of the data lies within three standard deviations of the mean, that is, in the interval with endpoints $x-±3s$ for samples and with endpoints $μ±3σ$ for populations.
Figure 2.16 The Empirical Rule
Two key points in regard to the Empirical Rule are that the data distribution must be approximately bell-shaped and that the percentages are only approximately true. The Empirical Rule does not apply to data sets with severely asymmetric distributions, and the actual percentage of observations in any of the intervals specified by the rule could be either greater or less than those given in the rule. We see this with the example of the heights of the men: the Empirical Rule suggested 68 observations between 68.22 and 71.62 inches but we counted 69.
### Example 19
Heights of 18-year-old males have a bell-shaped distribution with mean 69.6 inches and standard deviation 1.4 inches.
1. About what proportion of all such men are between 68.2 and 71 inches tall?
2. What interval centered on the mean should contain about 95% of all such men?
Solution:
A sketch of the distribution of heights is given in Figure 2.17 "Distribution of Heights".
1. Since the interval from 68.2 to 71.0 has endpoints $x-−s$ and $x-+s$, by the Empirical Rule about 68% of all 18-year-old males should have heights in this range.
2. By the Empirical Rule the shortest such interval has endpoints $x-−2s$ and $x-+2s.$ Since
$x-−2s=69.6−2(1.4)=66.8 and x-+2s=69.6+2(1.4)=72.4$
the interval in question is the interval from 66.8 inches to 72.4 inches.
Figure 2.17 Distribution of Heights
### Example 20
Scores on IQ tests have a bell-shaped distribution with mean μ = 100 and standard deviation σ = 10. Discuss what the Empirical Rule implies concerning individuals with IQ scores of 110, 120, and 130.
Solution:
A sketch of the IQ distribution is given in Figure 2.18 "Distribution of IQ Scores". The Empirical Rule states that
1. approximately 68% of the IQ scores in the population lie between 90 and 110,
2. approximately 95% of the IQ scores in the population lie between 80 and 120, and
3. approximately 99.7% of the IQ scores in the population lie between 70 and 130.
Figure 2.18 Distribution of IQ Scores
Since 68% of the IQ scores lie within the interval from 90 to 110, it must be the case that 32% lie outside that interval. By symmetry approximately half of that 32%, or 16% of all IQ scores, will lie above 110. If 16% lie above 110, then 84% lie below. We conclude that the IQ score 110 is the 84th percentile.
The same analysis applies to the score 120. Since approximately 95% of all IQ scores lie within the interval form 80 to 120, only 5% lie outside it, and half of them, or 2.5% of all scores, are above 120. The IQ score 120 is thus higher than 97.5% of all IQ scores, and is quite a high score.
By a similar argument, only 15/100 of 1% of all adults, or about one or two in every thousand, would have an IQ score above 130. This fact makes the score 130 extremely high.
## Chebyshev’s Theorem
The Empirical Rule does not apply to all data sets, only to those that are bell-shaped, and even then is stated in terms of approximations. A result that applies to every data set is known as Chebyshev’s Theorem.
### Chebyshev’s Theorem
For any numerical data set,
1. at least 3/4 of the data lie within two standard deviations of the mean, that is, in the interval with endpoints $x-±2s$ for samples and with endpoints $μ±2σ$ for populations;
2. at least 8/9 of the data lie within three standard deviations of the mean, that is, in the interval with endpoints $x-±3s$ for samples and with endpoints $μ±3σ$ for populations;
3. at least $1−1∕k2$ of the data lie within k standard deviations of the mean, that is, in the interval with endpoints $x-±ks$ for samples and with endpoints $μ±kσ$ for populations, where k is any positive whole number that is greater than 1.
Figure 2.19 "Chebyshev’s Theorem" gives a visual illustration of Chebyshev’s Theorem.
Figure 2.19 Chebyshev’s Theorem
It is important to pay careful attention to the words “at least” at the beginning of each of the three parts. The theorem gives the minimum proportion of the data which must lie within a given number of standard deviations of the mean; the true proportions found within the indicated regions could be greater than what the theorem guarantees.
### Example 21
A sample of size n = 50 has mean $x-=28$ and standard deviation s = 3. Without knowing anything else about the sample, what can be said about the number of observations that lie in the interval (22,34)? What can be said about the number of observations that lie outside that interval?
Solution:
The interval (22,34) is the one that is formed by adding and subtracting two standard deviations from the mean. By Chebyshev’s Theorem, at least 3/4 of the data are within this interval. Since 3/4 of 50 is 37.5, this means that at least 37.5 observations are in the interval. But one cannot take a fractional observation, so we conclude that at least 38 observations must lie inside the interval (22,34).
If at least 3/4 of the observations are in the interval, then at most 1/4 of them are outside it. Since 1/4 of 50 is 12.5, at most 12.5 observations are outside the interval. Since again a fraction of an observation is impossible, x (22,34).
### Example 22
The number of vehicles passing through a busy intersection between 8:00 a.m. and 10:00 a.m. was observed and recorded on every weekday morning of the last year. The data set contains n = 251 numbers. The sample mean is $x-=725$ and the sample standard deviation is s = 25. Identify which of the following statements must be true.
1. On approximately 95% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m. to 10:00 a.m. was between 675 and 775.
2. On at least 75% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m. to 10:00 a.m. was between 675 and 775.
3. On at least 189 weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m. to 10:00 a.m. was between 675 and 775.
4. On at most 25% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m. to 10:00 a.m. was either less than 675 or greater than 775.
5. On at most 12.5% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m. to 10:00 a.m. was less than 675.
6. On at most 25% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m. to 10:00 a.m. was less than 675.
Solution:
1. Since it is not stated that the relative frequency histogram of the data is bell-shaped, the Empirical Rule does not apply. Statement (1) is based on the Empirical Rule and therefore it might not be correct.
2. Statement (2) is a direct application of part (1) of Chebyshev’s Theorem because $(x-−2s,x-+2s)=(675,775).$ It must be correct.
3. Statement (3) says the same thing as statement (2) because 75% of 251 is 188.25, so the minimum whole number of observations in this interval is 189. Thus statement (3) is definitely correct.
4. Statement (4) says the same thing as statement (2) but in different words, and therefore is definitely correct.
5. Statement (4), which is definitely correct, states that at most 25% of the time either fewer than 675 or more than 775 vehicles passed through the intersection. Statement (5) says that half of that 25% corresponds to days of light traffic. This would be correct if the relative frequency histogram of the data were known to be symmetric. But this is not stated; perhaps all of the observations outside the interval (675,775) are less than 75. Thus statement (5) might not be correct.
6. Statement (4) is definitely correct and statement (4) implies statement (6): even if every measurement that is outside the interval (675,775) is less than 675 (which is conceivable, since symmetry is not known to hold), even so at most 25% of all observations are less than 675. Thus statement (6) must definitely be correct.
### Key Takeaways
• The Empirical Rule is an approximation that applies only to data sets with a bell-shaped relative frequency histogram. It estimates the proportion of the measurements that lie within one, two, and three standard deviations of the mean.
• Chebyshev’s Theorem is a fact that applies to all possible data sets. It describes the minimum proportion of the measurements that lie must within one, two, or more standard deviations of the mean.
### Basic
1. State the Empirical Rule.
2. Describe the conditions under which the Empirical Rule may be applied.
3. State Chebyshev’s Theorem.
4. Describe the conditions under which Chebyshev’s Theorem may be applied.
5. A sample data set with a bell-shaped distribution has mean $x-=6$ and standard deviation s = 2. Find the approximate proportion of observations in the data set that lie:
1. between 4 and 8;
2. between 2 and 10;
3. between 0 and 12.
6. A population data set with a bell-shaped distribution has mean μ = 6 and standard deviation σ = 2. Find the approximate proportion of observations in the data set that lie:
1. between 4 and 8;
2. between 2 and 10;
3. between 0 and 12.
7. A population data set with a bell-shaped distribution has mean μ = 2 and standard deviation σ = 1.1. Find the approximate proportion of observations in the data set that lie:
1. above 2;
2. above 3.1;
3. between 2 and 3.1.
8. A sample data set with a bell-shaped distribution has mean $x-=2$ and standard deviation s = 1.1. Find the approximate proportion of observations in the data set that lie:
1. below −0.2;
2. below 3.1;
3. between −1.3 and 0.9.
9. A population data set with a bell-shaped distribution and size N = 500 has mean μ = 2 and standard deviation σ = 1.1. Find the approximate number of observations in the data set that lie:
1. above 2;
2. above 3.1;
3. between 2 and 3.1.
10. A sample data set with a bell-shaped distribution and size n = 128 has mean $x-=2$ and standard deviation s = 1.1. Find the approximate number of observations in the data set that lie:
1. below −0.2;
2. below 3.1;
3. between −1.3 and 0.9.
11. A sample data set has mean $x-=6$ and standard deviation s = 2. Find the minimum proportion of observations in the data set that must lie:
1. between 2 and 10;
2. between 0 and 12;
3. between 4 and 8.
12. A population data set has mean μ = 2 and standard deviation σ = 1.1. Find the minimum proportion of observations in the data set that must lie:
1. between −0.2 and 4.2;
2. between −1.3 and 5.3.
13. A population data set of size N = 500 has mean μ = 5.2 and standard deviation σ = 1.1. Find the minimum number of observations in the data set that must lie:
1. between 3 and 7.4;
2. between 1.9 and 8.5.
14. A sample data set of size n = 128 has mean $x-=2$ and standard deviation s = 2. Find the minimum number of observations in the data set that must lie:
1. between −2 and 6 (including −2 and 6);
2. between −4 and 8 (including −4 and 8).
15. A sample data set of size n = 30 has mean $x-=6$ and standard deviation s = 2.
1. What is the maximum proportion of observations in the data set that can lie outside the interval (2,10)?
2. What can be said about the proportion of observations in the data set that are below 2?
3. What can be said about the proportion of observations in the data set that are above 10?
4. What can be said about the number of observations in the data set that are above 10?
16. A population data set has mean μ = 2 and standard deviation σ = 1.1.
1. What is the maximum proportion of observations in the data set that can lie outside the interval $(−1.3,5.3)$?
2. What can be said about the proportion of observations in the data set that are below −1.3?
3. What can be said about the proportion of observations in the data set that are above 5.3?
### Applications
1. Scores on a final exam taken by 1,200 students have a bell-shaped distribution with mean 72 and standard deviation 9.
1. What is the median score on the exam?
2. About how many students scored between 63 and 81?
3. About how many students scored between 72 and 90?
4. About how many students scored below 54?
2. Lengths of fish caught by a commercial fishing boat have a bell-shaped distribution with mean 23 inches and standard deviation 1.5 inches.
1. About what proportion of all fish caught are between 20 inches and 26 inches long?
2. About what proportion of all fish caught are between 20 inches and 23 inches long?
3. About how long is the longest fish caught (only a small fraction of a percent are longer)?
3. Hockey pucks used in professional hockey games must weigh between 5.5 and 6 ounces. If the weight of pucks manufactured by a particular process is bell-shaped, has mean 5.75 ounces and standard deviation 0.125 ounce, what proportion of the pucks will be usable in professional games?
4. Hockey pucks used in professional hockey games must weigh between 5.5 and 6 ounces. If the weight of pucks manufactured by a particular process is bell-shaped and has mean 5.75 ounces, how large can the standard deviation be if 99.7% of the pucks are to be usable in professional games?
5. Speeds of vehicles on a section of highway have a bell-shaped distribution with mean 60 mph and standard deviation 2.5 mph.
1. If the speed limit is 55 mph, about what proportion of vehicles are speeding?
2. What is the median speed for vehicles on this highway?
3. What is the percentile rank of the speed 65 mph?
4. What speed corresponds to the 16th percentile?
6. Suppose that, as in the previous exercise, speeds of vehicles on a section of highway have mean 60 mph and standard deviation 2.5 mph, but now the distribution of speeds is unknown.
1. If the speed limit is 55 mph, at least what proportion of vehicles must speeding?
2. What can be said about the proportion of vehicles going 65 mph or faster?
7. An instructor announces to the class that the scores on a recent exam had a bell-shaped distribution with mean 75 and standard deviation 5.
1. What is the median score?
2. Approximately what proportion of students in the class scored between 70 and 80?
3. Approximately what proportion of students in the class scored above 85?
4. What is the percentile rank of the score 85?
8. The GPAs of all currently registered students at a large university have a bell-shaped distribution with mean 2.7 and standard deviation 0.6. Students with a GPA below 1.5 are placed on academic probation. Approximately what percentage of currently registered students at the university are on academic probation?
9. Thirty-six students took an exam on which the average was 80 and the standard deviation was 6. A rumor says that five students had scores 61 or below. Can the rumor be true? Why or why not?
1. For the sample data
$x26272829303132f341612621$
$Σx=1,256$ and $Σx2=35,926.$
1. Compute the mean and the standard deviation.
2. About how many of the measurements does the Empirical Rule predict will be in the interval $(x-−s,x-+s)$, the interval $(x-−2s,x-+2s)$, and the interval $(x-−3s,x-+3s)$?
3. Compute the number of measurements that are actually in each of the intervals listed in part (a), and compare to the predicted numbers.
2. A sample of size n = 80 has mean 139 and standard deviation 13, but nothing else is known about it.
1. What can be said about the number of observations that lie in the interval (126,152)?
2. What can be said about the number of observations that lie in the interval (113,165)?
3. What can be said about the number of observations that exceed 165?
4. What can be said about the number of observations that either exceed 165 or are less than 113?
3. For the sample data
$x12345f8429331$
$Σx=168$ and $Σx2=300.$
1. Compute the sample mean and the sample standard deviation.
2. Considering the shape of the data set, do you expect the Empirical Rule to apply? Count the number of measurements within one standard deviation of the mean and compare it to the number predicted by the Empirical Rule.
3. What does Chebyshev’s Rule say about the number of measurements within one standard deviation of the mean?
4. Count the number of measurements within two standard deviations of the mean and compare it to the minimum number guaranteed by Chebyshev’s Theorem to lie in that interval.
4. For the sample data set
$x4748495051f131821$
$Σx=1224$ and $Σx2=59,940.$
1. Compute the sample mean and the sample standard deviation.
2. Considering the shape of the data set, do you expect the Empirical Rule to apply? Count the number of measurements within one standard deviation of the mean and compare it to the number predicted by the Empirical Rule.
3. What does Chebyshev’s Rule say about the number of measurements within one standard deviation of the mean?
4. Count the number of measurements within two standard deviations of the mean and compare it to the minimum number guaranteed by Chebyshev’s Theorem to lie in that interval.
1. See the displayed statement in the text.
2. See the displayed statement in the text.
1. 0.68.
2. 0.95.
3. 0.997.
1. 0.5.
2. 0.16.
3. 0.34.
1. 250.
2. 80.
3. 170.
1. 3/4.
2. 8/9.
3. 0.
1. 375.
2. 445.
1. At most 0.25.
2. At most 0.25.
3. At most 0.25.
4. At most 7.
1. 72.
2. 816.
3. 570.
4. 30.
1. 0.95.
1. 0.975.
2. 60.
3. 97.5.
4. 57.5.
1. 75.
2. 0.68.
3. 0.025.
4. 0.975.
2. By Chebyshev’s Theorem at most 1∕9 of the scores can be below 62, so the rumor is impossible.
1. Nothing.
2. It is at least 60.
3. It is at most 20.
4. It is at most 20.
1. $x-=48.96$, s = 0.7348.
2. Roughly bell-shaped, the Empirical Rule should apply. True count: 18, predicted: 17.
3. Nothing.
4. True count: 23, guaranteed: at least 18.75, hence at least 19.<|endoftext|>
| 4.75 |
6,556 |
Below are answer explanations to the full-length Math test of the previously released ACT from the 2021-2022 “Preparing for the ACT Test” (form 2176CPRE) free study guide available here for free PDF download.
The ACT Math test explained below begins on page 22 of the guide (24 of the PDF). Other answer explanations in this series of articles:
When you’re finished reviewing this official practice ACT test, start practicing with our own 10 full-length practice ACT tests.
## ACT Practice Test Math Answer Explanations 2021-2022
Question 1 “paper in a jar” the answer is E
This question tests your understanding of probability.
1. There are 15 pieces of paper in the jar. Eight of those pieces are less than 9.
2. The probability of drawing desired pieces is
3. The probability is therefore .
Question 2 “” the answer is J
This question tests your understanding of manipulating algebraic expressions.
1. First we add the like terms () and ()
2. We then end up with . This is the correct answer.
Question 3 “10 + 3(12 ÷ (3x)) the answer is 16
This question tests your understanding of simplifying an expression using PEMDAS.
1. The first step is to plug in x = 2.
2. This would simplify the equation to 10 + 3(12 ÷(3⋅2))
3. Simplifying from inside the parenthesis we would get 10 + 3 (12 ÷(6))
4. Which would be equal to 10+3(2) = 10+6 = 16.
Question 4 “⎪6 − 4⎪ − ⎪3 − 8⎪” the answer is G
This question tests your understanding of simplifying absolute value expressions.
1. The straight brackets | are used to indicate the absolute value of the expression in between two brackets.
2. The first step would be to simplify the two expressions inside the straight brackets. This would simplify the equation to |2| – |-5|.
3. The next step is to calculate the absolute value of each term. This would simplify the expression to 2 – 5.
4. Finally performing the final subtraction would yield an answer of -3.
Question 5 “(4c-3d)(3c+d)” the answer is C
This question is testing that we understand the distributive property.
1. To calculate we begin by multiplying all the terms in the second parenthesis with the term 4c. This would yield 4c(3c) + 4c(d). This would simplify to + 4cd.
2. The next step is to multiply all the terms in the second parenthesis with the term (-3d). This would yield -3d(3c) – 3d(d). This would simplify to .
3. Combine the terms from the first and second steps, we get . Then by combining the like terms we end up with the expression . This is the final answer.
This question tests your understanding of applying proportions in the setting of word problems.
1. We know that there are 180 students, so if students earned an A, that would mean that 45 students earned an A.
2. We also, know that students earned a B, so that would mean of the 180 students, 60 students earned a B.
3. Hence, we can calculate the students who earned a C by subtracting those that earned an A or a B from the total. This would be calculated as follows: 180-45-60 = 75.
4. So the number of students that earned a C for the course is 75.
Question 7 “Skipper’s Pond” the answer is B
This question tests your understanding of applying proportions in the setting of word problems.
1. We’re told that the equation to represent the number of fish in the pond is .
2. Hence in the year 2006, the value of x is 2006 – 2000 = 6.
3. Plugging this value of x into the equation would leave the equation . This would mean that there are 192 fish in the Pond at the beginning of the year 2006.
Question 8 “Speed” the answer is H
This question tests your understanding of speed and distance.
1. Since we know that Manish was 510 km away from Baton Rouge at 8:00 a.m., and 105 km away from Baton Rouge at 1:00 p.m., we can conclude that Manish covered 510 – 105 km in this time. This is 405 km.
2. The amount of time it took to cover 405 km is 5 hours, which is the difference between 1:00 p.m., and 8:00 a.m.
3. Based on this we could calculate the speed by dividing 405 km by the 5 hours, which would yield a speed of 81 km per hour.
Question 9 “triangle with parallel lines” the answer is D
This question tests your understanding of similar triangles and parallel lines.
1. Since we know that is constant, and that is parallel to we can say that the ADE is similar to the ABC.
2. Hence, we can conclude the ratios of the sides are similar.
3. Specifically, the ratio of is equal to the ratio of .
4. The value of is which simplifies .
5. We know that = = . Solving for we get that = 42.
Question 10 “Katerina” the answer is G
This question tests your understanding of speed and how it is calculated.
1. Katerina average minutes for running one mile can be calculated by dividing the total number of minutes by the total miles ran.
2. The total number of minutes requires us to convert hours into minutes by multiplying by 60. Performing this step we get that the number of minutes ran is 150 minutes.
3. The total number of miles is given at 15.
4. The average number of minutes per mile can be calculated by dividing 150 by 15. This would yield 10 minutes per mile.
Question 11 “marbles” the answer is B
This question tests your understanding of probability and algebraic manipulations.
1. There are currently 8 red marbles out of 24 total marbles. If we add x new red marbles, the new number of red marbles is 8 + x and the total number of marbles is 24 + x.
2. The probability of drawing a red marble is equal to . We know this is equal to .
3. Setting the two equations equal to each other, we can simplify to the equation 3 (24+x) = 5 (8 + x).
4. Solving for x we end up with 16, which is the number of additional red marbles that must be added to the bag.
Question 12 “midpoint” the answer is G
This question tests your understanding of probability and algebraic manipulations.
1. We know that the midpoint of is (2,1).
2. From point C (6,8), we travel 4 units to the left, and 7 units down to get to the midpoint.
3. Hence, to get to point D, we have to travel the same distance from the midpoint.
4. So point D would be represented by the point (2-4, 1-7). This is equal to (-2,-6).
Question 13 “overtime” the answer is D
This question tests your understanding of word problems and algebraic manipulations.
1. We know that for the first 40 hours, Thomas gets paid \$15.
2. After the first 40 hours, Thomas gets paid \$15 x 1.5 = \$22.5.
3. Since Thomas worked 46 hours, his total pay prior to deductions is 40 x \$ 15 + 6 x \$22.5 = \$600 + \$135 = \$735.
4. Since \$117 is removed for deductions, the amount of money that Thomas is left with is \$735-\$117 = \$618.
Question 14 “Sweet Stuff Fresh Produce” the answer is J
This question tests your understanding of money based expressions.
1. In 2 trips, Janelle purchased 3 bags at \$3 a bag and 4 bags at \$2.80 a bag.
2. The amount she spent was \$9 on Monday and \$11.20 on Wednesday, for a total of \$20.20.
3. If she had bought 7 bags at once, she would have paid \$2.60 per bag, for a total of \$18.20.
4. She would have saved \$2.00.
Question 15 “3% of 4.14 x the answer is A
This question tests your understanding of simplifying expressions with percentages.
1. 3% of 4.14 x = 0.03 x 4.14 x = 3 x 4.14 x 100 = 12.42 x 100 = 1242.
Question 16 “value of x” the answer is K
This question tests your understanding of solving algebraic equations.
1. To find the value of x that satisfies this equation, -3 ( 4x – 5) = 2 (1 – 5x), we begin by distributing the values on both parentheses.
2. We end up with the equation -12x + 15 = 2 – 10x. Combining like terms we end up with 13 = 2x.
3. x = .
Question 17 “right triangle” the answer is D
This question tests your understanding of sine of an angle in a triangle.
1. Sin A = .
2. Sin A =
Question 18 “ ” the answer is J
This question tests your understanding of simplifying exponents.
1. We begin by taking the cube root of the fraction to eliminate the 3 in the exponent. Doing this we end up with the term: .
2. To eliminate the negative sign in the exponent, we take the reciprocal of the fraction. We end up with the term .
3. Squaring the numerator and denominator we end up with the value: .
Question 19 “Loto’s walk” the answer is A
This question tests your understanding of the Pythagorean Theorem.
1. Adding up the number of yards Loto walked east and north separately, we end up with 20 yards east and 11 yards north.
2. To find the distance Loto would walk if he could walk directly, we apply the Pythagorean theorem to get the equation:
3. Solving for x we end up with x = 22.83.
4. The total difference in distance Loto would have saved is 20 + 11 – 22.83 = 8.17 yards which is approximately 8 yards.
Question 20 “standard score” the answer is F
This question tests your understanding of solving algebraic equations.
1. We know that z = 2 = .
2. Solving for x we get x = 90.
Question 21 “circle” the answer is E
This question tests your understanding of triangles in the setting of a circle.
1. In approaching this we can begin by looking at the answer choices and seeing if they exist in the circle centered at O.
2. The ABO is an acute triangle. Since is the radius and is also the radius we know the A = B. Since we know the AOB, is 60 degrees we can deduce that A and B are both 60 degrees.
3. Hence, the AOB is also an equilateral triangle.
4. Since we know that and are both radii of the circle, we can conclude that the DOC is an isosceles triangle. The DOC is also a right triangle since the DOC is 90 degrees.
5. Hence the only triangle that does not appear in the figure is the scalene triangle.
Question 22 “slope” the answer is G.
This question tests your understanding of parallel lines and slopes.
1. If the line is parallel to x + 5y = 9, we can conclude they have the same slope.
2. Solving the equation for y we get, y = . So the slope is .
Question 23 “y = ” the answer is E
This question tests your understanding of solving equations.
1. Since we know that x > 1, we know that y cannot be negative or 0. Hence, the only possible answer choices that are possible are 0.9 and 1.9
2. Plugging 0.9 for y; we get x = – 9. Since, x has to be greater than 1, we can conclude that only 1.9 works for y.
3. Plugging in 1.9 for y we get x = 2.111. Since x is greater than 1 we can conclude that this is the correct answer.
Question 24 “all positive integers” the answer is H.
This question tests your understanding of prime factors.
1. For a number to be divisible by 15 and 35 it has to be a multiple of both of the numbers’ factors.
2. 15 has the prime factors 5 and 3. 35 has the prime factors 5 and 7.
3. The least common multiple is 5 x 3 x 7. This is 105.
Question 25 “ triangle ABC” the answer is D
This question tests your understanding of angles.
1. If = , then B = C. Since A is 58 degrees, we can calculate the sum of B and C by subtracting 58 from 180.
2. 180 – 58 = 122 degrees. Since B is equal to C, both angles have to 61 degrees.
Question 26 “Earth’s surface” the answer is G
This question tests your understanding of probability.
1. To find the probability of landing on water, we have to find the area covered by water and divide that value by the total area on Earth’s surface.
2. The desired equation would be P = . Factoring out and dividing both the numerator and denominator by yields the equation
3. P =
Question 27 “statistical tests” the answer is E
This question tests your understanding of mean, median and mode of a series of numbers.
1. Prior to his 8th test, Jamal’s mean, median, and mode were 79, 80, and 80 respectively.
2. Following his score of 90 on his 8th test, Jamal’s mean was 80.13. His median was still 80, and his mode was still 80. Hence, the only value that changed was his mean, which was greater.
Question 28 “solid rectangular prism” the answer is H
This question tests your understanding of three-dimensional figures.
1. The solid rectangular prism has sides of length 5, 6, and 7 units. Hence, there are a total of 210 cubes.
2. Since the black cubes and white cubes are alternating equally, we can conclude that half of the cubes are black and half are white.
3. Half of 210 is 105.
Question 29 “square ABCD” the answer is C
This question tests your ability to calculate the area of squares and rectangles.
1. If one side of the square ABCD has length 12 meters, then the area of the square is 144 .
2. Since, we know the width of the rectangle is 8 meters, dividing 144 by 8 meters, yields the length which is 18 meters.
Question 30 “average” the answer is H
This question tests your understanding of averages for a series of numbers.
1. We know that the average of the numbers w, x, y, and z is 92.0. Hence the sum of w, x, y, and z can be represented by the equation. w + x + y + z = 4 x 92 = 368.
2. We know that z is 40, so the sum of w, x and y is 368 – 40 = 328.
3. We also know that the 4th number of the new list is 48, so the sum of the numbers of the new list is equal to the expression w + x + y + 48. Plugging in 328 for the sum of w, x and y gives the expression 328 + 48 = 376.
4. Dividing 376 by 4, yields the average 94.0.
Question 31 “vector” the answer is B
This question tests your understanding of velocity and vectors.
1. Since the vector j represents 1 mile per hour north, -j would represent 1 mile per hour south.
2. Maria’s velocity is 12 miles per hour south which would be represented by -12j.
Question 32glasses” the answer is K.
1. To begin we add the volume of water in each of the glasses.
2. To distribute this water equally among the 4 glasses, we divide by 4. This would result in each glass being full.
Question 33 “carpet tiles” the answer is D
This question tests your understanding of area of a surface.
1. To begin we convert his living room floor length and width to inches. feet = 100 inches. Likewise, 10 feet = 120 inches.
2. Hence the area in square inches of the living room floor is 100 inches x 120 inches = 12,000
3. Each carpet tile has an area of 20 inches x 20 inches = 400 . Dividing the area of the living room floor by the 400 of each carpet tile, would yield 30 carpet tiles needed to cover this area.
Question 34 “coordinate plane” the answer is F
This question tests your understanding of the equation of a circle.
1. A circle with a center of () and a radius r would have be mapped out by the equation .
2. Plugging the relevant () = (8,5) and r = 9, into the equation we end up with the answer .
Question 35 “total cost” the answer is E
This question tests your understanding of applying algebra to word problems.
1. Since it costs \$2,500 per each Yq test and there were 1000 tests administered, the cost of the Yq test was \$2,500,000.
2. Likewise, it costs \$50 pear each Sam77 test and there were 1000 tests administered the total cost of the Sam77 test was \$50,000.
3. Adding these two numbers yields a total cost of \$2,550,000.
Question 36 “percent of the volunteers” the answer is J
This question tests your understanding of percentages.
1. The number of people who carry the Yq77 gene can be calculated by adding up all the people who had a positive Yq test. This is equal to 590 + 25 = 615.
2. Since there were 1000 people tested total, or 61.5% carried the Yq77 gene.
Question 37 “how many volunteers” the answer is C
This question tests your understanding of understanding tables.
1. To calculate how many volunteers had an incorrect result from the Sam77 test, we have to add up the people who had a Negative Sam77 test but had a Positive Yq test and the people who had a positive Sam77 test but a negative Yq test.
2. These two numbers are 25 and 10 respectively, so the total number of volunteers who received an incorrect result from the Sam77 test are 25 + 10 = 35.
Question 38 “does NOT possess Yq77” the answer is F
This question tests your understanding of probability.
1. The number of volunteers who had a positive Sam77 test was 590 + 10 = 600.
2. Of these volunteers, only 10 did not have the Yq77 gene.
3. Hence the probability is .
Question 39 “matrices” the answer is B
This question tests your understanding of matrix manipulations.
1. To calculate the product of X and Y, we have to multiply the matrices.
2. This can be done as follows .
Question 40 “symmetry” the answer is F
This question tests your understanding of lines of symmetry.
1. A scalene triangle is one where all three sides have different lengths. This would not have a vertical line of symmetry.
2. A line would have a verticle line of symmetry. A square would have a verticle line of symmetry. A pentagon would have a verticle line of symmetry. A parallelogram would have a verticle line of symmetry.
Question 41 “” the answer is A.
1. To solve this equation, we can begin by subtracting 15 from both sides. We get the equation .
2. We can factor this to yield (6x+5)(4x-3) = 0. Setting 6x+5 = 0 and 4x -3 = 0, we get the two solutions for this equation as and .
3. The greater of these two solutions is .
Question 42 “” the answer is J
This question tests your understanding of sine and cosine of common angles.
1. To solve this we have to know that . Hence, we can use this to replace the with .
2. Likewise, we can replace the from the second term with .
3. Hence the original equation simplifies to = 1.
4. Of the answer choices, the only one whose value = 1 is .
Question 43 “area of circle” the answer is D
This question tests your understanding of circumference and area of a circle.
1. We know the circumference is = . Solving for r we get r = 6.
2. The area of circle is . Plugging in r = 6, we get the area as .
Question 44 “solvent mixture” the answer is G.
This question tests your understanding of percentages and mixtures.
1. We know that of the 25 liters, 40% is solvent and 60% is water. Hence, we can conclude that there is 0.40 x 25 liters of solvent = 10 liters.
2. So if x liters of solvent is added, the amount of solvent would be 10 + x. Furthermore, the new mixture would have a volume of 25 + x.
3. So the percent of solvent in the new mixture would be equal to .
4. Since we know this is 50% or 0.50 we can set these two values equal to each other and solve for x.
5. Simplifying, we get 10 + x = (25 + x )(0.5). Further simplification shows us that 0.5x = 2.5, so x = 5.
Question 45 “” the answer is E
This question tests your understanding of simplifying expressions with fractions.
1. To simplify this equation we find the common multiple of the two denominators which would be .
2. Hence, we would have to multiple the numerator of the first term by (x-y) and the numerator of the second term by (x+y).
3. The new expression after combining the two terms would be .
4. Factoring out the numerator and denominator we get .
This question tests your understanding of solving word problems with algebra.
2. In total, they sold c + 2c + 6c = 9c advertisements.
Question 47 “flower shop” the answer is D
This question tests your understanding of permutations.
1. For the first spot, Emily has 6 plants to choose from.
2. After planting the first plant, Emily only has 5 plants to choose from for the second spot.
3. Finally, after planting the first two plants, Emily only has 4 plants to choose from for the third spot.
4. Multiplying these options out we get 6 x 5 x 4 = 120 different arrangements for the plants.
This question tests your understanding of the area of a triangle.
1. To calculate the area of we begin by using the formula for the area of a triangle = .
2. Based on this, the base of the triangle is the distance between point M and point P which is 6a-2a = 4a.
3. The height of the triangle is the distance from point Q to the line segment MP, which is 5b.
4. Hence the area is .
Question 49 “Point M” the answer is D
This question tests your understanding of slope of a line on a coordinate plane.
1. The slope of can be represented by the equation .
2. We know that the numerator is negative because point Q has a smaller y value than point M.
3. As point Q moves to the right the difference gets larger.
4. Hence, the slope will be negative but will increase as the value of Q gets larger.
Question 50 “f(5a)” the answer is K
This question tests your understanding of a function plotted on a coordinate plane.
1. To find the value of f(5a) we look at the quadratic function and find where it crosses the line x = 5a.
2. Looking at the curve, we see that this happens at 8b which is the correct answer.
Question 51 “Jurors” the answer is D
This question tests your understanding of fractions and decimals.
1. We know that if we invite x people only 0.4x of the people actually appear.
2. Of this, one-third are excused so two-thirds remain. Specifically, of the initial x people invited, remain.
3. We know, that this expression should be equal to 60 people for the jury pool. Hence we get the equation .
4. Solving for x we get x = 225 people to summon.
Question 52 “275th digit” the answer is K
This question tests your understanding of repeating digits for decimals.
1. Since we know the decimal 0.6295 repeats every 4 digits, we should divide 275 by 4.
2. Dividing 275 by 4 we end up with 68 with a remainder of 3.
3. This would mean that the the 272nd digit would be a 5. Then the next three digits would be a 6, 2 and 9 in that order.
4. Based on this, the 275th digit would be a 9.
Question 53 “ ” the answer is A
1. Since we know that we know the solution to f(x) can be found by factoring out f(x). f(x) = (x+2) (x-2).
2. Hence, the two solutions for f(x) are x = 2, and x = -2.
3. Since we know that these two solutions of x make f(x) equal to 0, we can set g(x) equal to these values.
4. Hence we end up with two equations for g(x). x +3 = 2 and x + 3 = -2.
5. Solving both equations for x we end up with x = -1, and -5.
Question 54 “p and n” the answer is G
This question tests your understanding of integers and absolute values.
1. Since we know that p is a positive number, n is a negative number and |p| > |n|, we can assign values for both p and n that satisfy these conditions.
2. Specifically we can assign p = 5, and n = -3. Then we can solve all of the equations;
3. ;
4. ;
5. Of these answer choices, the greatest expression is
Question 55 “” the answer is B
This question tests your understanding of imaginary numbers and calculations using i.
1. Since , then , , and .
2. Plugging in those values into the equation , we get
3. . Simplifying this equation we have which is equal to -1.
Question 56 “coordinate plane” the answer is K
This question tests your understanding of slopes and lines in a coordinate plane.
1. The first equation is x + 2y 6. Rearranging this equation to solve for y, we end up with y -0.5x + 3. This would mean that the slope of the line is negative, and the shaded area has to be below this line.
2. Of all the answer choices only F and K have a line with a negative slope with the shaded area below the line.
3. The second equation is . Rearranging this equation and dividing by 3, we end up with the equation .
4. To choose between F and K we can find the two y-intercepts of the circle and if they lie under the line y -0.5x + 3, we know the answer is K.
5. The two y intercepts of the circle can be calculated by setting x = 0 in the equation of the circle.
6. We end up with the equation , so y = 2 or y = -2 .
7. Plugging in the positive y-intercept (x,y) = (0,2) into the equation for the line we have 2 -0.5(0) + 3. Simplifying we have 2 3, which is true so the correct answer is K.
Question 57 “real numbers” the answer is D
This question tests your understanding of your understanding of multiplying and dividing by 0.
1. Since , we can conclude that d = 0.
2. Since, abc = d, we know that abc = 0.
3. Since ac = 1, we know that b(1) = 0. So, b = 0.
Question 58cosine function” the answer is K
This question tests your understanding of the cosine function and functions.
1. We know that the y-intercept is 3, so that means if x = 0, y has to be 3. Since cos(0) = 1, we know that the constant multiplying the cosine has to be 3.
2. We also know that , based on the graph. Plugging in for all the values of x we end up with y = 3cos(2x) as the solution that yields a y of -3.
Question 59 “flying kite” the answer is B
This question tests your understanding of triangles, their sides and angles.
1. In the triangle shown, the missing angle would be 60 degrees.
2. We then could set an equation such that . Solving this for the length of the string we end up with = length of string. Plugging in values of and , we can simplify to .
Question 60 “publisher” the answer is F
This question tests your understanding of using algebra to solve word problems.
1. We know that the publisher charges \$15 first the first book and \$12 for every additional copy.
2. If there are y books ordered, the publisher charges \$15 for the first book and \$12 for y-1 books.
3. The equation for the cost of the books becomes \$15 + \$12(y-1). Simplifying this we get cost = 15-12 + 12y = 12y + 3.<|endoftext|>
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Class 10 Maths MCQ – Real Numbers
This set of Class 10 Maths Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on “Real Numbers”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. Which of the following is a composite number?
a) 2
b) 3
c) 9
d) 7
Explanation: A prime number has two factors the number itself and 1. In case of 9 there are three factors i.e. 3 × 3 × 1. Hence, it is not a prime number.
2. Which of the following is a prime number?
a) 31
b) 52
c) 21
d) 32
Explanation: A prime number has two factors the number itself and 1. In case of 31 there are two factors i.e. 31 and 1. Hence, it is a prime number.
3. The fundamental theorem of arithmetic states that, every composite number can be factorized as product of primes and this factorization is unique.
a) False
b) True
Explanation: Let us consider a composite number, say 25
25 can be factorized as 5 × 5 × 1. This factorization is unique for 25 and no other number can be represented in the same manner.
4. The least common multiple of 135 and 24 is _________
a) 90
b) 360
c) 240
d) 1080
Explanation: 135 can be written as 3 × 3 × 3 × 5 × 1 and 24 can be written as 2 × 2 × 2 × 3 × 1.
LCM is the product of greatest power of each prime factor involved in the numbers.
Therefore, LCM = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 1080
5. The highest common factor of 21 and 90 is _________
a) 3
b) 2
c) 1
d) 4
Explanation: 21 can be written as 3 × 7 × 1 and 90 can be written as 3 × 3 × 2 × 5.
HCF is the product of smallest power of each prime factor involved in the numbers.
Therefore, HCF = 3 × 1 = 3
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6. The LCM of two numbers is 7991 and the two numbers are 61 and 131. What will be their HCF?
a) 2
b) 1
c) 3
d) 4
Explanation: For two numbers a and b, we know that
(a × b) = HCF of (a, b) × LCM of (a, b)
Here a = 61 and b = 131, and LCM is 7991
61 × 131 = HCF × 7991
HCF = $$\frac {7991}{7991}$$ = 1
7. What will be the largest number that divides 100 and 25, and leaves 3 as remainder in each case?
a) 7
b) 5
c) 1
d) 4
Explanation: The required number divides (100-3) i.e. 97 and (25-3) i.e. 22 exactly.
Now, 97 = 97 × 1 and 22 = 2 × 11
HCF of 97 and 22 is 1.
Hence, the required number is 1.
8. A bakery sells cookies in three boxes. The three boxes contain 60, 84 and 108 number of cookies. The baker wants to sells all the cookies in any of the three boxes. The least number of cookies that he can bake, in a day, so that he is able to sell all his cookies in any of the three boxes is _______
a) 5467
b) 2243
c) 1123
d) 3780
Explanation: The three boxes contain 60, 84 and 108 number of cookies.
60 can be factorized as 2 × 2 × 3 × 5, 84 as 2 × 2 × 3 × 7 and 108 as 2 × 2 × 3 × 3 × 3
To find the least number of cookies that can be filled in the container, we have to find the LCM of the three numbers
LCM of 60, 84 and 108 = 2 × 2 × 3 × 3 × 3 × 5 × 7 = 3780
Hence, the least number of cookies that can be filled in the container is 3780.
9. Two buckets contain 546 and 764 liters of water respectively. What will be maximum capacity of container which can measure the water of either buckets exact number of times?
a) 108
b) 54
c) 34
d) 456
Explanation: The two buckets contain 546 and 764 liters of water.
546 can be factorized as 2 × 2 × 3 × 3 × 3 × 5 and 764 as 2 × 2 × 3 × 3 × 3 × 7.
To find the maximum capacity of the container which can measure the water of either buckets exact number of times, we have to find the HCF of the two numbers.
HCF of 546 and 764 = 22 × 33=108
Hence, the maximum capacity of the container is 108 liters.<|endoftext|>
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The Rapid Prototyping is abbreviated RP. It enables a quick fabrication of models in physical form, using a three dimensional CAD (Computer Aided Design) data. It is used in many industries, because RP allows companies to turn their innovative ideas into end product that is efficiently and rapidly successful.
A very useful technology in 3D printing is SLA. SLA is an abbreviation of Stereolithography. It is a main technology that also include photopolymerization being used for producing a solid part from liquid material. It requires using supporting structure, as it is attaching the part to the platform that elevate the layers.
There is practically no limit to the objects that can be printed with 3D printer. But, there are different ideas that can be introduced. A 3D firearms is an idea that has been printed from a 3D printer. This is an idea that can digitally be designed and prepared for printing. Another idea is acoustic guitar that can be printed, using materials like plastic, in combination with metal and other relevant materials. This will result in a working musical instrument.
How 3D printing works is a process that ultimately result in printing directly from a digital file. This process starts with using the computer to virtually design the object that you would like to create.<|endoftext|>
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# One-step inequalities
One-step inequalities are very easy to solve. They are very similar to one-step equations when it comes to the solving procedure. The biggest difference is that, while the solution of an equation is one number, the solution of an inequality is often a large group of numbers. For instance, let us say you have to solve an inequality that looks like this:
x + 4 > 8
The first thing you have to do is to move the number 4 to the right side of the inequality. Do not forget to change the sign of the number you are moving to the other side. After that, just perform the subtraction. Like this:
x > 8 – 4
x > 4
As you can see, the inequation states that x is greater than 4. That means that the solution of the inequation are all numbers greater than 4.
Also, a very important thing to remember is to change the sign of inequality when multiplying or dividing the whole expression with a negative number. For example, let us say we have an expression that looks like this:
-4x >16
To get to the value of x, we have to divide the whole expression with (-4). Like this:
-4x > 16 |: (-4)
Now we have to change that sign, or else we will get the wrong result. The final result will be:
x < -4
You can check that result by inserting any value smaller than (-4) as the value of x. If you wish to practice solving one-step inequalities, feel free to use the free math worksheets below.<|endoftext|>
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Chapter 8 covers what to do when connecting the parts of your essay together. They explain that creating connections between sentences and ideas increases sentence variety and helps construct a more convincing argument. Consider transitions both within a paragraph and between paragraphs. Look at pages 108-109 for transition words.
They give four principles:
- Using transition terms (like “therefore” and “as a result”)
- Adding pointing words (like “this” and (such”)
- Developing a set of key terms and phrases for each text you write
- Repeating yourself, but with a difference
Let’s come up with a list of transition words to use in our essays. What words do you use to transition?
“Never Marry a Mexican”
“Never Marry a Mexican” short story by Sandra Cisneros. Let’s break it down. Who are the characters? What themes do you notice? What point of view did the author use? What are some key scenes that stick out to you?
Pick a scene that you think is particularly memorable or important. You can pick one from whichever story you will be writing about.
Draw a comic of the scene. Make sure to add text, or dialogue to the scene like a comic book.
Now, describe the scene you drew in words. A picture is worth a thousand words. Explain the significance of the scene to your thesis or the story.<|endoftext|>
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Part of the challenge of learning the past simple is the pronunciation. First of all, it’s not easy to remember the past simple forms of all the irregular verbs, but even the regular verbs can cause problems when it comes to the pronunciation of their endings.
In order to help our learners get to grips with this issue, we need to understand the rules that lie behind the different pronunciations.
Consider the following words:
When you say those words out loud you realise that though they all end in –ed, they exhibit three different pronunciations. Looked ends with a /t/, enjoyed ends with a /d/ and wanted ends with a /Id/.
Now consider these words:
Can you see the rule?
For words which end in an unvoiced sounds – /p/, /k/, /ʃ/, /ʈʃ/, /f/ or /s/ – the word ends in /t/.
Words which end in any voiced sound end with /d/.
Words which end with /t/ or /d/ are pronounced with an /Id/ sound.
Though learners may naturally figure out the difference in pronunciation between the /d/ and /t/ endings, it is usually the /Id/ sound which causes trouble. One activity which can help your students to think about this issue is to incorporate a categorisation activity in which they must categorise words according to the pronunciation of their endings.
For younger learners there are numerous songs or stories which can be utilised to familiarise learners with the rules and let them practise the pronunciation. For older learners there are many different online games which allow the learner to hear the word being pronounced correctly and then asks the learner to categorise the sound.
Playing these games will help learners feel more comfortable using the past simple in their own language production. With time and with practise, learners will pick up the rules and integrate them into their pronunciation.<|endoftext|>
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# A note on basic set theory
This is a short note listing some basic facts on set theory and set theory notations, mostly about cardinality of sets. The discussion in this note is useful for proving theorems in topology and in many other areas. For more information on basic set theory, see [2].
Let $A$ and $B$ be sets. The cardinality of the set $A$ is denoted by $\lvert A \lvert$. A function $f:A \rightarrow B$ is said to be one-to-one (an injection) if for $x,y \in A$ with $x \ne y$, $f(x) \ne f(y)$. A function $f:A \rightarrow B$ is said to map $A$ onto $B$ (a surjection) if $B=\left\{f(x): x \in A\right\}$, i.e. the range of the function $f$ is $B$. If the function $f:A \rightarrow B$ is both an injection and a surjection, then $f$ is called a bijection, in which case, we say both sets have the same cardinality and we use the notation $\lvert A \lvert = \lvert B \lvert$. When the function $f:A \rightarrow B$, we denote this condition by $\lvert A \lvert \le \lvert B \lvert$. The Cantor–Bernstein–Schroeder theorem states that if $\lvert A \lvert \le \lvert B \lvert$ and $\lvert B \lvert \le \lvert A \lvert$ then $\lvert A \lvert = \lvert B \lvert$ (see 1.12 in [2]).
For the functions $f:X \rightarrow Y$ and $g:Y \rightarrow Z$, we define the function $g \circ f$ by $(g \circ f)(x)=g(f(x))$ for each $x \in X$. The function $g \circ f$ is denoted by $g \circ f:X \rightarrow Z$ and is called the composition of $g$ and $f$.
We use the notation $B^A$ to denote the set of all functions $f:A \rightarrow B$. It follows that if $\lvert A \lvert \le \lvert B \lvert$ then $\lvert A^C \lvert \le \lvert B^C \lvert$. To see this, suppose we have a one-to-one function $f:A \rightarrow B$. We define a one-to-one function $H: A^C \rightarrow B^C$ by $H(h)=f \circ h$ for each $h \in A^C$. Since $f:A \rightarrow B$, it follows that $H$ is one-to-one.
By $\omega$, we mean the first infinite ordinal, which can be viewed as the set of all nonnegative intergers. By $\omega_1$ we mean the first uncountable ordinal. The notation $2^\omega$ has dual use. With $2=\left\{0,1\right\}$, the set $2^\omega$ denotes all functions $f:\omega \rightarrow 2$. It can be shown that $2^\omega$ has the same cardinality as the real line $\mathbb{R}$ and the unit interval $[0,1]$ and the middle third Cantor set (see The Cantor set, I). Thus we also use $2^\omega$ to denote continuum, the cardinality of the real line.
If $\lvert A \lvert=2^\omega$, then the set $\lvert A^{\omega} \lvert=2^\omega$ where $A^{\omega}$ is the set of all functions from $\omega$ into $A$. Since $\omega_1$ is the first uncountable ordinal, we have $\omega_1 \le 2^\omega$. The Continuum Hypothesis states that $\omega_1 = 2^\omega$, i.e. the cardinality of the real line is the first uncountable cardinal number.
The union of $2^\omega$ many sets, each of which has cardinality $2^\omega$, has cardinality $2^\omega$. Furthermore, the union of $\le 2^\omega$ many sets, each of which has cardinality $\le 2^\omega$, has cardinality $\le 2^\omega$.
Reference
1. Kunen, K. Set Theory, An Introduction to Independence Proofs, 1980, Elsevier Science Publishing, New York.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.<|endoftext|>
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