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A History In Numbers
The wider area within which Stonehenge sits features an unusually large number of earthworks, long barrows and other monuments, dating back as far as 8500 BC. Stonehenge was initially a simpler monument built around 3000 BC which comprised of a circular ditch dug into the earth, with an inner and outer bank. Inside the inner bank were 56 chalk pits.
Around 2,500 BC, the stones arrived. These were set up as a horseshoe of large sarsen trilithons contained within a circle of sarsen stones. Also inside that stone circle were placed a number of smaller bluestones, arranged in a double arc shape. An altar stone was placed inside the horseshoe. A single heel stone was also placed outside the main entrance to the outer ditch.
Then, sometime around 2,200 BC, the bluestones were repositioned to form a complete circle around the horseshoe, so the sarsen horseshoe was now inside an inner bluestone circle and an outer sarsen stone circle. Additionally, some of the bluestones were laid out in an oval within the horseshoe. Still later (date unknown), this oval was reformed into a horseshoe shape as well.
Above: Stonehenge trilithon.
10 Stonehenge Facts
20,000 tons – the estimated amount of chalk dug out to create the Greater Cursus, a causeway enclosure created as much as 500 years before the first Stonehenge monument.
64 – the approximate number of human cremations that have been found at Stonehenge.
31 December 1900 – the date on which one of the uprights of the sarsen stone circle fell over, leading to public pressure to start a major restoration project.
1,000,000+ – the number of visitors the Stonehenge monument receives each year.
c 2500 BC – approximate point that both the stone monument at Stonehenge and the Great Pyramid in Egypt appeared.
10 years – the estimated time it would have taken to drag all of the sarsen stones to the monument site.
How Stonehenge was built
4,000 – the approximate number of people who lived at the Durrington Walls encampment, believed to be the people who built the stone monument.
Who built Stonehenge?
9 m – the approximate length of the tallest stone still standing.
The size of the stones
160 – the approximate number of stones required to build Stonehenge.
The number of stones
56 – the number of chalk pits (the Aubrey Holes) discovered inside the earth bank of Stonehenge.
The Aubrey Holes
Phases of Stonehenge
3000-2920 BC – Stage 1: the earth bank and ditch are built, and the 56 ‘Aubrey Holes’ dug.
2620-2480 BC – Stage 2: the stones arrive – the sarsen stone circle, bluestone circle, and the 5 trilithons.
2480-2280 BC – Stage 3: the Q and R holes appear, the Avenue is built, and the bluestones are rearranged.
2280-2020 BC – Stage 4: the bluestones are rearranged into an oval shape, and then still later into a circle.
1680-1520 BC – Stage 5: the Y and Z holes appear.
Some Useful Definitions
Neolithic – meaning New Stone Age, a period that followed on from the Palaeolithic (Old Stone Age) and the Mesolithic (Middle Stone Age). The people of the Neolithic era used stone tools, made pottery, grew crops and kept domestic animals.
Henge – an enclosure, commonly circular in shape, which has an outer bank and an inner ditch.
Sarsen – derived from “Saracen stone”, Saracen being a name from the Wiltshire dialect which referred to anything considered to be non-Christian (Pagan, Celtic or Muslim).
Bluestone – in general use ‘bluestone’ is the common name for one type of rock, spotted dolerite. However, in the context of Stonehenge, ‘bluestone’ is an informal name used to group all of the smaller stones that are foreign to the area (including spotted dolerites, dolerites, volcanic tuffs, and rhyolites).
Solstice – the longest and shortest days of the year, occurring at midsummer and midwinter.
Equinox – the days of the year on which day and night are exactly the same length, occurring in spring and autumn.
About The Site
There is much written about Stonehenge on the web, much of it excellent, some of it fanciful. StonehengeFacts.net is designed to provide a quick and simple reference to some of the more significant and interesting Stonehenge facts and figures; we hope you find something of value here.
If you reference any material from this site on another website or in other published material please credit StonehengeFacts.net.
All of the facts and figures displayed on StonehengeFacts.net are correct to the best of our knowledge. However, there are disagreements and different interpretations on the history of Stonehenge and so the accuracy of the information presented cannot be guaranteed; in using any of the facts or images from this website you do so entirely at your own discretion.
We do not wish to perpetuate inaccuracies, so if you spot anything that you believe to be inaccurate or incomplete please let us know; you can contact us at ‘hello [at] history in numbers /dot/ com’, quoting ‘Stonehenge’ in the subject line.
A History in Numbers website.<|endoftext|>
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# Algebra-Simplifying Expressions, Solving Linear Equations and Inequalities
I love teaching algebra especially in grades 5 and 6 when students start to learn about variables, expressions, and equations. Although algebra might seem strange to them at first, as they start to understand the rules they begin to enjoy this new chapter in their math adventures.
I cannot stress enough how important it is for students to be fully equipped and prepared before they dive into algebra. Make sure that your students are fluent with
• basic math skills such as adding, subtracting, multiplying, and dividing
• fractions, decimals, percents, their operations, and how they’re all related
• the order of operations
• Integers, the number line, and integer operations
• Probability
• Geometry formulas and vocabulary: perimeter, area, circumference, etc.
• Ratio and proportion
• exponents and their operations
I believe that the misconception that algebra is hard starts when students are not ready to tackle algebra just yet.
### Starting with Algebra
By combining variables with numbers using mathematical operations we form mathematical expressions. A variable is a letter used to stand for a number. The letters most commonly used as variables are x, y, z, a, b, c, m, and n. The letters e and i have special values in algebra and are usually not used as variables. The letter o is usually not used because it can be mistaken for 0 (zero). Below are some examples of expressions.
4-7+5 2x+1 3x2+x-5 5(x-4)
### Terms coefficients, and constant terms
A term is a product of a number and a variable raised to a power. The number in a term is the coefficient of the power of the variable. For example, in the expression picture of 3x2+2x-5 the terms are 3x2, 2x, and 5. The coefficient of x2 is 3 and the coefficient of x is 2. A term that is a number by itself is called a constant or a constant term.
### Simplifying Expressions
As students start with pre-algebra it is important to understand how variables work and the concept of equivalence between two expressions. To find the value of the variable using two equivalent expressions we often need to simplify one or both the expressions. To simplify an expression means to write it as an equivalent expression with as few terms as possible, and write each term as simply as possible.
For example, the expression 2z+7-3z+4 can be simplified to -z+11, and the expression 4(z-6) +3 can be simplified to 4z2 -12z+3.
We can write expressions for word problems to express the relations between the terms.
Example
### Solving Linear Equations
Two expressions are equivalent if they are equal for every value of the variable. An equation is true only for some values of the variable. The solutions to an equation are the values of the variable that make the equation true.
For example, the equation x+4=9 is true only for x=5
We solve an equation we find all the values of the variable that make the equation true. Two of the most important steps we take to solve an equation are:
1. Replace an expression with an equivalent one by simplifying. For example, for the equation 6x-5x+5=16 we can simplify the left-hand side
to x+5, so the equation becomes x+5=16
2. Perform the same mathematical operation on both sides of the equation. For example, starting with the equation x+5, we can subtract 5 from both sides of the equation to get x+5-5=16-5 Simplifying both sides of the equation gives us the solution of the equation which is x=11. Checking with the original equation we see that the equation is true for x=11
### Isolating the variable
When we solve equations with one variable by performing operations on both sides of the equation and then simplifying the expressions until the variable is alone on one side of the equation we say that we isolate the variable.
For example
4z+2z-10=26
4z+2z-10+10=26+10
6z=36 z=6
(to isolate the variable we need to remove -10 from the left-hand side.
By adding 10 we cancel them out. Of course, we need to add 10 to the
right-hand side as well to keep the equivalence.)
Linear equations are equations in which every term is a constant term or a constant times the first power of the variable. 2z+3=5z-4 and a/2+1=7 are linear equations. x2 +1=17 and 2/(a3-1)=56 are not linear equations.
### Multiplying by the reciprocal
When a term or an expression is multiplied by a fraction, we can multiply both sides by the reciprocal of that fraction to eliminate the fraction. Multiplying a fraction by its reciprocal gives answer 1. example below
### Eliminating the denominators
When one of the sides of the equation is divided by a term or an expression we can multiply both the sides with that term or expression to eliminate the denominators and simplify the equation.
For example,
### Cross-multiplying
Rather than performing these multiplications as two separate steps, we can perform both at once. This process is called cross-multiplying.
### Variables on both sides of the equation
When the variable appears on both sides of the equation, we can use addition and subtraction to get all the terms with the variable on the same side. Similarly, we use addition and subtraction to get all the constant terms on the other side of the equation. In the example below we subtracted the variable term of the right side from both sides so that the variable will end up on the left side.
### Using Equations to solve word problems
One of the best things about knowing algebra is that you can use variables and equations to easily solve tricky word problems. We can use equations to solve word problems. Once we decide which quantity the variable should represent we use the known information to find the value of the variable. When we find two different expressions that represent the same quantity we have an equation.
Example
Alice’s dog and Manny’s cat weigh 115 pounds together. Alice’s dog is 27 pounds heavier than Manny’s cat. How much does Alice’s dog weigh?
The unknown quantities are the weights of the dog and the cat so we can name one of these quantities x and use the information we have to calculate it. We need two expressions that represent the same quantity.
If we represent the weight of the dog with x then we have the equation x+(x-27)=115 We simplify the equation to 2x-27=115 2x=115+27 2×142 x=71
The dog’s weight is 71 pounds.
We could have represented the cat’s weight as x and used the equation x+(x+27)=115
Another example
### Inequalities
When we know that an expression is greater than another, we can write an inequality to show the relationship between them. Inequalities can be strict or nonstrict. For example, x+6 > 4 and 6-2z < 8+z are strict inequalities and 2x-5 ≥ 9 and 5x+4 ≤ 8x-5 are nonstrict inequalities. Inequalities have more than one answer. The answer to an inequality is usually a range of numbers/values.
We can graph the values of the variable that satisfy the inequality on a number line. Example below.
We can simplify and solve inequalities following the same steps as with equations. We can also use inequalities to solve word problems.
I have created 3 collections of resources for the concepts above (expressions, linear equations, and inequalities) (slides, print task cards, and worksheets) You can find the individual resources and the bundles below.
The resources include
Google Slides
Print Task Cards
Worksheets
Google slides (52)
Print Task Cards (52)
Worksheets(14)
Google slides
Print Task cards
Worksheets
## Bundles
Google slides (expressions, linear equations, inequalities)
Print Task Cards (expressions, linear equations, inequalities)
Worksheets (expressions, linear equations, inequalities)<|endoftext|>
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Beeline Language Korean is all about learning the Korean language and culture. In our previous lessons, we have learned many different slangs like 품절남, 멘붕 and 솔까말. Also, we have covered a lot of words that would be used by young Korean generations today. In the previous lessons, you have learned how to say different colors in Korean. We covered how to say buildings, fences and streets in Korean too. In the last lesson, we have learned how to say Fire in Korea. Today, we will learn how to say Water in Korean.
How To Say Water In Korean
This is how to say Water in Korean: 물[mul].
Note that the pronunciation must be like mool, not mule.
Sample Sentences Using Water in Korean
이 경기가 끝나고 시원한 물 한잔 마셨으면 좋겠다. [ee-gyeong-gee-gah ggeut-na-goh si-won-han mul-han-jan ma-syeot-eu-myeon joh-kae-dda].
I would be good to drink a cold glass of water after this game.
Note that we described drinking water as 물을 마시다. – > 마셨다 is a past form of the phrase 마시다.
여름에는 물을 많이 마셔서 수분섭취를 해줘야 돼. [yeo-reum-ae-neun mul-eul man-ee ma-shyeo-seo soo-boon-sup-choui-reul hae-joe-ya-doe].
In Summer, you must drink a lot of water to keep yourself hydrated.
Note that the word 수분섭취 simply means hydration.
Different Temperature of Water in Korean
You just have learned how to say water in Korean. How would you say cold water? 시원한 물 [si-won-han mul] as introduced above or you could say 냉수 [neng-soo].
Now, how would you say hot water then? 따듯한 물 [dda-deut-han mul] or 온수 [ohn-soo].
Now you have just learned how to say water in Korean, go and share what you have learned with your friends now.<|endoftext|>
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# let S be the set of all real numbers and let R be a relation in S, defined by $R=\left\{ \left( a,b \right):a\le {{b}^{3}} \right\}$. Show that R satisfies none of reflexivity, symmetry and transitivity.
Last updated date: 12th Jul 2024
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Hint: To show that R is not reflexive on S, find an example such that $\left( a,a \right)\in R$ for $a\in S$. To show that R is not symmetric, find an example such that $\left( a,b \right)\in R$ and $\left( b,a \right)\notin R$ for $a,b\in S$. To show that ‘R’ is not transitive, find an example such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ and $\left( a,c \right)\notin R$ for $a,b,c\in S$.
Let ‘A’ be set then
Reflexivity, symmetry and transitivity of a relation on set ‘A’ is defined as follows.
Reflexive relation: A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself i.e. for every element say (a) in set A, $\left( a,a \right)\in R$.
Thus, R on a set A is not reflexive if there exists an element $a\in A$ such that $\left( a,a \right)\in R$.
Symmetric Relation: A relation R on a set ‘A’ is said to be symmetric relation if $\left( a,b \right)\in R$ then (b, a) must belong to R i.e. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R\ for\ all\ a,b\in A$.
Transitive Relation: A relation R on A is said to be transitive relation if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$ i.e. $\left( a,b \right)\in R\ and\ \left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Let us check one by one.
Let us first check for reflexivity.
To show that R does not satisfy reflexivity, we have to find at least one element in ‘S’ such that $\left( a,a \right)\in R$.
Let us take $a=-2$ (‘-2’ is a real number so, $a\in S$)
${{a}^{3}}={{\left( -2 \right)}^{3}}=-8$
We know, $-8<-2$
For $\left( a,b \right)$ to belong to R, a should be less than or equal to ${{b}^{3}}$i.e. $a\le {{b}^{3}}$
As $-2\ge -8,\left( -2,-8 \right)\notin R$
So, for $a=-2,\left( a,a \right)\notin R$
This, R doesn’t satisfy reflexivity on set ‘S’.
Now, let us check for symmetry.
We have to show that R doesn’t satisfy symmetry.
To show that R doesn’t satisfy symmetry, we have to find at least one example such that $\left( a,b \right)\in R$ and $\left( b,a \right)\notin R$.
We have to show that for any $a,b\in S,\left( a,b \right)\in R$ but $\left( b,a \right)\notin R$
Let a = 1 and b = 2 both ‘1’ and ‘2’ are real numbers.
So $a\in S\ and\ b\in S$
For $\left( 1,2 \right)\to 1\le {{\left( 2 \right)}^{3}}\ so,\ \left( 1,2 \right)\in R$
But for $\left( 2,1 \right)\to 2>{{\left( 1 \right)}^{3}}\ so,\ \left( 2,1 \right)\notin R$
Thus, we have got example in the set ‘S’ which doesn’t symmetry on set S.
So ‘R’ doesn’t satisfy symmetry on set S.
Now let us check for transitivity.
We have to show that R doesn’t satisfy transitivity on S. To show that R doesn’t satisfy on S, we have to find an example such $\left( a,b \right)\in R\ and\ \left( b,c \right)\in R\ but\ \left( a,c \right)\notin R$.
Let us take a = 63, b = 4 and c = 2
As $63\le {{\left( 4 \right)}^{3}}\ i.e.\ a\le {{\left( b \right)}^{3}}\ so,\ \left( a,b \right)\in R$
And $4\le {{\left( 2 \right)}^{3}}\ i.e.\ b\le {{\left( c \right)}^{3}}\ so,\ \left( b,c \right)\in R$
But $63>{{\left( 2 \right)}^{3}}\ i.e.\ a>{{c}^{3}}\ so,\ \left( a,c \right)\notin R$
Thus we have got an example in S which doesn’t satisfy transitivity on set S.
So, R doesn’t satisfy transitivity on set ‘S’.
Hence, we have shown that R satisfies none of reflexivity, symmetry and transitivity on S.
Note: To show that R doesn’t satisfy reflexivity, symmetry and transitivity, we don’t need to prove that $a > {{b}^{3}}$ for all values of set S or for an interval of values of set S. We just need to find at least one example for each which doesn’t satisfy reflexivity, symmetry and transitivity.<|endoftext|>
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# SAT Mathematics : Solving Problems with Exponents
## Example Questions
### Example Question #1 : Solving Problems With Exponents
If , what is ?
2
3
4
6
4
Explanation:
This problem tests your fluency with exponent rules, and gives you a helpful clue to guide you through using them. Here you may see that both 27 and 9 are powers of 3. and . This allows you to express as and as . Then you can simplify those exponents to get . Since when you divide exponents of the same base you subtract the exponents, you now have , and since you really have:
This then tells you that .
Note that had you not immediately seen to express all the numbers in this problem as powers of 3, the fact that the question asks for such a combination of variables, , should be your clue; you're given an exponent problem and asked for a subtraction answer, so that should get you thinking about dividing exponents of the same base to subtract the exponents, and at least give you some fodder for playing with exponent rules until you find a way to make progress.
### Example Question #2 : Solving Problems With Exponents
Which of the following values of is a solution to the equation above?
3
10
9
30
10
Explanation:
This problem involves some creative factoring, and factoring is something you should always look to do whenever you see several exponents amidst some addition or subtraction. You cannot here factor any one thing out of all four terms, but you can group the terms to factor some common elements:
Can become:
And if you factor the common term you have:
The only real number solution available is , so the correct answer is .
### Example Question #3 : Solving Problems With Exponents
If , what is the value of ?
13
15
14
16
15
Explanation:
An important principle of exponents being tested here is that when you multiply/divide exponents of the same base, you add/subtract those exponents. Here you can do the corollary; if you had , you would add together those exponents to get . But in this case you're given the combined exponent and may want to convert it to so that you can factor:
allows you to factor the terms to get:
You can do the arithmetic to simplify , allowing you to then divide both sides by 3 and have:
So .
### Example Question #4 : Solving Problems With Exponents
If , what is the value of ?
6
4
5
3
3
Explanation:
This problem hinges on your ability to recognize 16, 4, and 64 all as powers of 4 (or of 2). If you make that recognition, you can use exponent rules to express the terms as powers of 4:
Since taking one exponent to another means that you multiply the exponents, you can simplify the numerator and have:
And then because when you divide exponents of the same base you can subtract the exponents, you can express this as:
This means that .
### Example Question #5 : Solving Problems With Exponents
Explanation:
With this exponent problem, the key to getting the given expression in actionable form is to find common bases. Since both 9 and 27 are powers of 3, you can rewrite the given expression as:
When you've done that, you're ready to apply core exponent rules. When you take one exponent to another, you multiply the exponents. So the numerator becomes and the denominator becomes . Your new fraction is:
Next deal with the negative exponents, which means that you'll flip each term over the fraction bar and make the exponent positive. This then makes your fraction:
Since when you divide exponents of the same base you subtract their exponents, this simplifies to .
### Example Question #6 : Solving Problems With Exponents
is equivalent to which of the following?
Explanation:
This problem rewards you for being able to factor with exponents. Whenever you're faced with several exponents of the same base separated by addition or subtraction, it is a good idea to factor so that you can get more exponents multiplied together. Here that would mean factoring out the common term to get:
Now you're in a position to do some arithmetic inside the parentheses, since each of those exponents is one you should recognize or be able to quickly calculate by hand. You have:
Which equals:
Here even if you don't recognize as , you should look to the answer choices to see lots of 2s with exponents and that may be your clue. You can simplify this to:
And now you have some options. You might see that with two different bases each taken to the same exponent, you can combine the multiplication to get to or . Or you might go to the answer choices and eliminate the ones that are close but clearly not correct. Either way, you should find the correct answer, .
### Example Question #7 : Solving Problems With Exponents
If , which of the following equations must be true?
Explanation:
You should see on this problem that the numbers used, 2, 4, and 8, are all powers of 2. So to get the exponents in a way to be able to be used together, you can factor each base into a base of 2. That gives you:
Then you can apply the rule that when you take one exponent to another, you multiply the exponents. This then simplifies your equation to:
And now on the left hand side of the equation you can apply another exponent rule, that when you multiply two exponents of the same base, you add the exponents together:
Since the bases here are all the same, you can set the exponents equal. This gives you:
### Example Question #1 : Solving Problems With Exponents
Which of the following represents the average of and ?
Explanation:
While upon first glance it might seem to be a quick problem if you just take the exponent between 61 and 63 and say , a quick test of small numbers should show to you that you can't simply do that. The average of and , for example, is 5, not . So here you'll have to find a way to leverage the rule that Average = (Sum of Terms)/(Number of Terms).
So, algebraically, the average sets up as , but of course those numbers are far too big to calculate and then add. You can, however, use two clues to your advantage: 1) whenever you're adding or subtracting exponents, it's a good idea to factor (remember, exponents are repetitive multiplication, and factoring creates more multiplication). And 2) the answer choices all have powers of 11 in them with no addition, so you should look to factor out a common 11 term so that your math can look more like the answers. If you do so, you'll find that you have:
And from here, you can actually calculate the numbers in parentheses. That gives you:
If you then finish the math, you'll see that you can sum 121 + 1 to get 122, which divides by 2 to give you 61. So your final answer looks like:
### Example Question #9 : Solving Problems With Exponents
is equal to which of the following?
Explanation:
This problem rewards your ability to factor exponents. Here if you factor out common terms in the given equation, you can start to see how the math looks like the correct answer. Factoring negative exponents may feel a bit different from the more traditional factoring that you do more frequently, but the mechanics are the same. Here you can choose to factor out the biggest "number" by sight, , or the number that's technically greatest, . Because all numbers are 2-to-a-power, you'll be factoring out common multiples either way.
If you factor the common , the expression becomes:
Here you can do the arithmetic on the smaller exponents. They convert to:
When you sum the fractions (and 1) within the parentheses, you get:
And since you can express this now as:
, which converts to the correct answer:
Note that you could also have started by factoring out from the given expression. Had you gone that route, the factorization would have led to:
This also gives you the correct answer, as when you sum the terms within parentheses you end up with:
### Example Question #10 : Solving Problems With Exponents
If , then what is the value of ?
6
7
5
8
7
Explanation:
Whenever you are given addition or subtraction of two exponential terms with a common base, a good first instinct is to factor the addition or subtraction problem to create multiplication. Most exponent rules deal with multiplication/division and very few deal with addition/subtraction, so if you're stuck on an exponent problem, factoring can be your best friend.
For the equation can be rewritten as , leveraging the rule that when you multiply exponents of the same base, you add the exponents. This allows you to factor the common term on the left hand side of the equation to yield:
And of course you can simplify the small subtraction problem within parentheses to get:
And you can take even one further step: since everything in the equation is an exponent but that 4, you can express 4 as to get all the terms to look alike:
Now you need to see that can be expressed as or as . So the equation can look like:
You can then divide both sides by and be left with:
This proves that .<|endoftext|>
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Square Root 1 to 20
Created by: Team Maths - Examples.com, Last Updated: June 19, 2024
Square Root 1 to 20
Square roots from 1 to 20 encompass a fundamental aspect of mathematics, bridging algebra, numbers, and geometry. They represent the inverse operation of squaring, finding the number which, when multiplied by itself, equals a given number. In this range, we encounter both rational and irrational numbers, where rational roots yield integer results and irrational roots produce non-integer values. Understanding square and square roots is crucial in various mathematical domains, including statistics, Least square where they play a role in calculating measures of central tendency and dispersion.
Square Root 1 to 20 refers to the list of square roots for numbers ranging from 1 to 20. Square roots can be positive or negative, but in this range, only the positive values are considered, ranging from 1 to approximately 4.47214.
In this sequence, the numbers 1, 4, 9, and 16 are perfect squares, resulting in rational square roots, while the rest are non-perfect squares, yielding irrational square roots. The square root of each number from 1 to 20 can be represented in radical form as √x or in exponential form as (x)½.
Square Root 1 to 20
In radical form: √x
In exponential form: (x)½
Largest Square Root: √20 = 4.4721
Square Root from 1 to 20 Chart
The square root from 1 to 20 chart displays the square roots of numbers ranging from 1 to 20. It highlights both the rational and irrational nature of square roots within this range.
Square Root 1 to 20 for Perfect Squares
In the square root of 1 to 20 for perfect squares, the square roots of numbers 1, 4, 9, and 16 are rational, resulting in whole numbers. This subset highlights the clear and easily identifiable square roots within the range.
Square Root 1 to 20 for Non-Perfect Squares
For non-perfect squares in the range from 1 to 20, their square roots result in irrational numbers, which cannot be expressed as fractions and have non-repeating decimal expansions. These square roots are important in mathematics and various real-world applications, representing values such as lengths, areas, and volumes.
How to Calculate Square Root from 1 to 20?
To calculate the square root of numbers from 1 to 20, you can use various methods such as:
• Calculator: Use a scientific calculator with a square root function to find the square root of each number individually.
• Long Division Method: Apply the long division method to find the square root of each number manually. This involves a step-by-step process of approximation and division.
• Prime Factorization: Decompose each number into its prime factors and then simplify to find the square root.
• Estimation: Make an educated guess of the square root based on the nearest perfect squares and refine the estimate through repeated approximations.
How do you estimate square roots?’
Estimating square roots involves finding the nearest perfect squares above and below the given number, then making an educated guess between them. For example, to estimate √15, we find that 3² = 9 and 4² = 16, so √15 is between 3 and 4.
What are the practical applications of square roots?
Square roots are used in various fields such as mathematics, engineering, finance, and physics. They are essential for calculating distances, areas, volumes, and in algorithms for data analysis.
Why are some square roots irrational?
Square roots of numbers that are not perfect squares result in irrational numbers, meaning they cannot be expressed as fractions. For example, √2 is irrational because it cannot be written as a simple fraction.
Text prompt<|endoftext|>
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Pulses of light are the absolute fastest way to transfer data (because nothing's faster than light), but old school fibre optic cables can only go so many places. Scientists have a new idea: use high-powered lasers to make a column of low-density air that can carry a light signal just as well as a normal cable. Yes, fibre optics made of thin air.
Get ready to dust off your GCSE physics knowledge. Dr. Howard Milchberg and his team at the University of Maryland found that very powerful laser blasts collapse the air around them into a narrow beam called a filament. These filaments of hot air then expand, opening up a tube of low density air. This hole has a lower refractive index than undisturbed air—in other words, it's like a mirrored tube.
In a paper published today, Dr. Milchberg and team showed that four such lasers fired in a square arrangement create a cage of low-density air columns. A fifth laser beam shot down the middle will stay trapped between these four columns (show in red above), since the low density air reflects light. In other words, the four mirrored tubes trap the laser beam in the center, focusing it on its target instead of allowing it to scatter in the air, as diagrammed with black arrows in the above illustration.
All of this happens astonishingly quickly. The filaments themselves disappear after just one-trillionth of a second, but the low-density holes, the reflective tubes that form the boundary of this air-cable, takes a billion times longer to appear. All told, the "air cable" created by the laser pulses survives for a few milliseconds—plenty of time to blast a data beam through. As Dr. Milchberg said in the announcement of the findings, "milliseconds is infinity" when it comes to laser transmission.
In testing, Dr. Milchberg and crew found that a laser fired through this air-cable stayed focused and hardly lost any energy at all over a range of one metre. Perhaps more importantly, the air-cable survives long enough to feasibly allow both an outgoing transmission and a return signal to travel the same tube. "It's like you could just take a physical optical fiber and unreel it at the speed of light, put it next to this thing that you want to measure remotely, and then have the signal come all the way back to where you are," Dr. Milchberg says.
If the technique proves viable over long distances, it could allow cable-like data transmission without those pesky cables. Theoretically, air waveguides could be used where laying physical cable is impossible: running chemical tests on the upper atmosphere, monitoring the inside of nuclear reactors, or perhaps beaming fibre-optic-quality data to orbiting spacecraft. First things first though: Dr. Milchberg's team's next goal is to see if the waveguides will function over a range of 50 metres. Baby steps. [Eurekalert via GigaOM]
Image: Dr. Howard Milchberg<|endoftext|>
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# Two Dice Repetition
### Solution, Question 1
Very much like in the Birthday Problem, but with $36$ distinct equiprobable selections.
A repetition occurs on move $n,~2\le n\le 37$ for the first time if, until then, there were no repetition but then on the $n^{th}$ move there was one. The probability of this event is
\displaystyle \begin{align} p(n)&=\frac{35}{36}\cdot\frac{34}{36}\cdots\frac{36-(n-2)}{36}\cdot\frac{n-1}{36}\\ &=\frac{(n-1)36\cdot 35\cdots (38-n)}{36^n} \end{align}
The expectation for the number of moves then is
$\displaystyle E=\sum_{k=2}^{37}kp(k)\approx 8.203.$
### Solution, Question 2
Now, for $n,~2\le n\le 32,$ the first repetition can occur at move $n$ in one of two ways:
1. there was no $*$ in the first $n-1,$ or
2. there was exactly one $*$ in the first $n-1$ moves, whereas $*$ may or may not appear on move $n.$
Let $q(n)$ is the probability of the first repetitions happening on move $n.$ The, to start with,
\displaystyle\begin{align}q(2)=\frac{30}{36}\cdot\frac{1}{36}+\frac{6}{36}\cdot\frac{6}{36}=\frac{11}{216}. \end{align}
For $3\le n\le 32,$
\displaystyle\begin{align}q(n)&=\frac{30\cdot 29\cdots (32-n)}{36^{n-1}}\cdot\frac{n-1}{36}\\ &\qquad\qquad+(n-1)\cdot\frac{6}{36}\cdot\frac{30\cdot 29\cdots (33-n)}{36^{n-2}}\cdot\frac{6+(n-2)}{36}\\ &=\frac{30\cdot 29\cdots (33-n)}{36^{n}}\cdot (n-1)(5n+56). \end{align}
The expected number of moves at which the first repetition occurs is then
$\displaystyle \sum_{k=2}^{32}kq(k)\approx 6.704.$
### Acknowledgment
This is a paraphrase of problem 409 from Crux Mathematicorum (posed v 5, n 1; solved v 5, n 9, 1979). The problem and solution are by L. F. Meyers of the Ohio State.
The problem is modeled after a popular game of old copyrighted by Pop-o-Mathic.<|endoftext|>
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# How I wish I'd taught... place value
Updated: Feb 24, 2023
Many teachers begin teaching place value in new academic year, and for good reason. Without mastery of the value of digits in numbers, other areas of maths would not be accessible to many.
Zero
Children learn about zero as being a place holder and to show the absence of a value. As we explored here, this allows a familiar patterns when working with multiples of 10 too. The concept of 0 is therefore essential to unlock the curriculum.
Place value
Place value means that digits, singular or when combined, have value and that value differs depending on where the digits are placed in the number. Consider, 2, 20 and 200. Each number contains the digit 2, but the value of the 2 is different.
2 - Two ones (units)
20 - Two tens and zero ones (units)
200 - Two hundreds, zero tens and zero ones (units)
Using a place value chart, you can see how the zero creates places to change the value of each 2.
Decimals
The use of a decimal point differentiates between whole numbers and decimal (fractional) numbers. Decimals are often used in measurements, including money and therefore they are often required to be rounded. Rounding is context driven and dependent to what degree a number needs rounding. Essentially, rounding requires an understanding of which number a value if closest to. A solid understanding of the number system and place value is required to access problems. Using models which reveal the structure of the number system can be powerful.
Using a Gattegno board can allow children to strengthen and master this concept. The links to powers of 10 seems clear and logical.
Negative numbers
Negative numbers found their way into maths relatively recently (in comparison with other areas of the number system) due to a phenomenon familiar to many: debt. Taken at face value, using negative numbers for ideas like debt or temperature can be useful. Things can start to get a little sketchy when we try to apply operations to negative numbers.
Negative numbers - calculation.
When a positive number is added to a negative number, the answer seems logical:
-3 + 1 = -2
-3 - 1 = -4
This concept can take some time for children to master. Is 2 bigger than -5? but isn't 5 bigger than 2? Here, language and models collide to strengthen understanding. 2 is greater than -5 is a much more effective way of describing the relationship. Perhaps having a vertical number line will allow children to strengthen the concept?<|endoftext|>
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# How do you simplify sqrt 2(sqrt2-sqrt 3)?
Mar 25, 2018
$2 - \sqrt{6}$
#### Explanation:
$\sqrt{2} \left(\sqrt{2} - \sqrt{3}\right)$
Expand,
$\sqrt{2} \cdot \sqrt{2} - \sqrt{2} \cdot \sqrt{3}$
Simplify,
$2 - \sqrt{6}$
Mar 25, 2018
$2 - \sqrt{6}$
#### Explanation:
Did you know that if you had, say $\sqrt{a} \times \sqrt{b}$ it is the same as $\sqrt{a \times b}$
Not in this question but also $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: color(blue)(sqrt(2))color(green)((sqrt(2)-sqrt(3))
Multiply everything inside the bracket by the $\textcolor{b l u e}{\sqrt{2}}$ that is outside it.
$\textcolor{g r e e n}{\left[\textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l u e}{\sqrt{2}} \times \sqrt{2} \textcolor{w h i t e}{\frac{2}{2}}\right] - \left[\textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l u e}{\sqrt{2}} \times \sqrt{3} \textcolor{w h i t e}{\frac{2}{2}}\right]}$
color(green)(color(white)("dd")sqrt(2xx2)color(white)("ddddd")-color(white)("d")color(white)("d")sqrt(2xx3)
color(green)(color(white)("dddddddddddd")2-sqrt(6)<|endoftext|>
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- Describe theory of mind, translating and understanding it’s underlying components when considering the everyday behaviour of primary age children with hearing loss.
- Recognise the developmental stages of acquiring theory of mind and how they shape learning, problem solving and socialising.
- Discuss the impact deafness has on the development of theory of mind and recognise specific areas that may be delayed, based on research finding and shared practice.
- Explore how and where theory of mind fits into social thinking, discussing the range of resources available.
- Recall and apply an understanding of theory of mind on a selection of well known children’s books.
- List 4 areas with practical ideas in each for encouraging and developing theory of mind in the primary years.
- Compare current support packages and reflect on how these can be adapted and expanded to include activities that will develop theory of mind.
- Write 3 actions to put into practice, short, medium and long term
Click here for more information and to book a place.<|endoftext|>
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The word essay is derived from the French verb “essayer” meaning “to try” or “to attempt”. In an essay, the author gives their reflections or personal point of view on a certain subject in an attempt to inform, persuade or entertain the reader. As a student, you may be required to write an essay as an assignment or for university admission, or you could decide to enter an essay competition. There are different types of essays, with the most common being descriptive essays, narrative essays, expository essays, personal and persuasive essays. Essay writing helps you develop crucial transferable skills that will come in handy in your professional life, e.g. critical thinking, presentation of ideas, coherent sentence structuring among others. In this guide on how to write an essay we are going to explore how to write a winning essay.
- How to Write an Essay: Understand the Essay Question
- How to Write an Essay: Research on the Essay Topic
- How to Write an Essay: Brainstorming
- While writing my Essay; How do I Write my Thesis Statement?
- When writing my Essay; how do I Create an Essay Outline?
- When writing my Essay; how do I write the Essay Introduction?
- When writing my Essay; how do I write the Essay Paragraphs?
- When writing an Essay; How do I write the Essay Conclusion?
- How to Write an Essay: Referencing and Citation
- How to Write an Essay: Revising Your Essay
- References used to write this article
How to Write an Essay: Understand the Essay Question
The first step to composing a great essay is to have a clear and thorough understanding of what the essay requires. This means analyzing the topic and understanding the key words of the essay question. To comprehend the question:
- Highlight the keywords of the essay question
- Confirm the meaning of these words from a dictionary
For instance if the question is “Describe the effects of standardization on consumer safety”, you might want to highlight and ascertain the meaning of the keywords as follows:
Describe the effects of standardization on consumer safety.
This will give you a clue on few fronts:
- What kind of research do you need? E.g. if the question says “discuss”, that means present arguments for and against an issue, in which case you will need to research for information supporting each sides of the argument.
- Who is the likely audience of the essay?
- How will the audience influence or affect your literary style?
- What type of essay are you writing? Argumentative Essay?
How to Write an Essay: Research on the Essay Topic
Even before putting pen to paper, it is a good idea to get a broad grasp or knowledge base concerning the topic. Start light, and then dig deeper as you get more familiar with the material. When researching, search the internet to acclimatize yourself with the very basic ideas of the topic. (Wikipedia is a good site to acquire broad-based knowledge, but do not cite it or other open web sources. This is because the authenticity of sources of information in such sites is questionable). As you delve deeper, explore scholarly articles on the issue. If you are relying on online research, utilize academic databases such as Google Scholar and JSTOR.
Useful Tips on how to Search for Essay Topic
- Read through your handout notes for relevant background information
- Use the internet to familiarize yourself with the subject material
- Visit your school or local library to read books on the subject matter. Use indices or the list of chapters to find the relevant chapter(s) in books
- Note down quotes from influential thinkers in the field (However do not overuse these as it may be a bit of an overkill. You also want to preserve your own voice)
- Follow a “little from a lot’’ approach, i.e. gather information from many sources as opposed to a lot of information from few sources.
- Refer to many sources to give your essay a balanced view
- Find out the current developments concerning the issue
How to Write an Essay: Brainstorming
At this point, you should have a command of the topic and thus the ability to generate your own ideas. Brainstorming is coming up with original ideas or solutions to a problem. It is a simple technique of gathering your thoughts and working out what you have conceptualized up to this point. Remember your paper must illuminate on the topic from a previously unexplored point of view. Check out these stages on how to brainstorm effectively.
Listing of ideas in essay writing
Write down any idea you can think of. Even ideas that sound stupid may later prove useful. (Use pen and paper or a separate Word document)
Evaluating your essay ideas
- Ask yourself questions about the issue and answer them
- Generate alternative solutions or specific courses of action that you think would be embraced to solve the issue
Scrutinizing of your essay ideas
- Critically examine the ideas. Which ideas do you think are best suited to solve the issue?
- Highlight these ideas
- Using this criteria: “Specific, Balanced, Realistic, Lasting, Fair” , evaluate these ideas
- Spin these ideas into your essay as practicable solutions
While writing my Essay; How do I Write my Thesis Statement?
A thesis statement is an assertion at the introductory stage of your essay that expresses the focus of the essay. It is a strong statement that announces your standpoint on the issue. A thesis is part of your introductory paragraph and is often preceded by the opening statement. It is like the map that guides the reader through the uncharted seas that is your essay.
What Should a good Essay Thesis Statement Have?
Your Essay Thesis Statement Should be Arguable
Your thesis statement must express something that would leave room for agreement or alternate viewpoints. It must be debatable. It is never a fact. For example: Inarguable: “Exercise and a healthy diet are necessary for optimum health.” (Who could refute this? This is not very debatable; it is more like a fact.) Arguable: “Technology has done more harm than good to society and the environment”. (This is subject to debate because many people would not necessarily agree with it. It would make a good thesis statement)
Your Essay Thesis Statement Should be Unambiguous/Precise
Avoid over-generalization. Do not use vague words like good, suitable interesting etc. For example: Coordinating public-transit systems with train travel would be good for American cities. (So what? It does not answer the question “How” or “Why”)
Your Essay Thesis Statement Should be Cohesive
Make your statement coherent by using connecting words such as although, actually, as a result of, etc. Example: “Although Western media depicts Africa as a continent ravaged by war, Africa is actually a fairly peaceful continent.”
When writing my Essay; how do I Create an Essay Outline?
An outline is the design or “blueprint”, which is like a map that is going to direct you as you write your essay. An outline helps you avoid unnecessary material that you will ultimately need to be expunged from the body later. An outline guides you in the writing process and allows you to concentrate on producing quality prose. Do you want to constantly pause in the middle of the process to recollect your thoughts, or will you focus your energies on good vocabulary and literary style? Luckily, well show you to create an outline. There are many ways of constructing an outline but they follow the same formula.
What should an Essay Outline Comprise?
The essay title
An introduction: which consists of an opening sentence(s), a sentence or a couple of sentences building up to the thesis statement, and finally t
he thesis statement.
Paragraphs: which consist of a topic sentence, followed by supporting sentences to back up the topic sentence.
When creating an outline, remember to keep it brief and to the point. When describing a point, keep it short but phrase it in a way that you can understand. Example: If your topic sentence in a paragraph is parents should allow children to keep pets because it helps them learn the values of responsibility and commitment, you could say, “Parents should let children keep pets – responsibility/commitment”.
A Sample Essay Outline
a.) An opening sentence
b.) Sentences leading up to thesis statement
c.) Thesis statement
a.) Paragraph #1
-Main idea i.e. the topic sentence
-Supporting sentences for the main idea
b.) Paragraph #2
-Main idea i.e. the topic sentence
-Supporting sentences for the main idea
c.) Paragraph #3
-Main idea i.e. the topic sentence
-Supporting sentences for the main idea
a.) Opening sentence
b.) A reworking of your thesis
c.) Sentence(s) building up to the end (optional)
d.) Closing sentence
When writing my Essay; how do I write the Essay Introduction?
A good introduction captures the imagination of the reader. Your first goal in the introduction is to grab their attention. Such an introduction should have the effect of delighting, intriguing or even shocking the reader.
How to create attention grabbing Essay Introduction
Use of Surprising statistics or facts in Essay Writing
E.g., 40% of food is wasted in the United States every year.
Remember, this information must not be a wild guess or fiction but rather accurate and verifiable information. It does not necessarily have to be news to the reader, but it serves as a focus point on your argument.
Use of A Paradox in essay writing
E.g., The African continent is endowed with large quantities of natural resources like diamonds, gold, petroleum, cocoa beans, oil and gas deposits, yet it is also the continent with the lowest per capita GDP.
Use of an Anecdote in essay writing
i.e. a short nonfiction story. E.g. when writing a personal or narrative essay.
Use of A rhetorical or thought provoking question in essay writing
e.g., Is strategic patience really a good policy when dealing with aggressive foreign countries?
Use of A Metaphor in essay writing
E.g., The United States is one of the richest countries in the world, but the huge disparity between the wealthy and the poor is the blot of ink on the textbook.
When writing my Essay; how do I write the Essay Paragraphs?
Paragraphs are the building blocks of your essay. They make up the biggest chunk of your essay and hence support your argument. Every new paragraph encompasses a new idea and therefore each paragraph should have a unique focus to it.
An Essay Paragraph Should Comprises
A Topic Sentence in essay paragraph
The first sentence of your paragraph, which not only orientates the reader with the idea of the paragraph but also keeps you, the writer, on track. Example: The discovering of the moon paved way to new inventions. This is a good topic sentence because it allows you to illustrate inventions that were made courtesy of the moon being discovered. In addition, it gives the reader an idea of what to expect next.
An Elaboration in essay paragraph
Support the topic sentence with evidence and examples. Evidence is proof in form of statistics, a study, etc. Transitioning from one paragraph to another should appear natural and seamless. Use transition words such as “another” , “likewise”, ”by the same token” , “ moreover” , “additionally”, “however” , “nevertheless” etc. to switch from one paragraph to another.
For example: “Another factor contributing to crime in America is substance abuse…”
“Additionally, lack of clarity in candidates’ policies leads to low voter turnout…”
“Moreover, lack of collateral prevents many small businesses from accessing loans…”
When writing an Essay; How do I write the Essay Conclusion?
A conclusion is the culmination of your essay. This part takes it home for your reader. The conclusion is no place to introduce new ideas or evidence. In writing a conclusion, reflect on this query: “So what?” Why should the reader care about your argument? How can you convince or persuade them to adopt your point of view? Any conclusion must reference your thesis statement – you are reminding the reader of the main argument of your essay. Render your conclusion with confidence – avoid using phrases such as “I think”, “I am of the opinion that”, “At least that’s what I think” etc., as these poke holes in your credibility. Also avoid such phrases as “to conclude”, “to summarize”, “in conclusion” as these are cliché and redundant.
A good Essay Conclusion Should Have:
- Rearticulates the thesis statement
- Creatively summarizes arguments presented in the essay
- Recommends a solution that could fill the gap in the issue being addressed
- Is authoritative and resolute, as opposed to weak or apologetic
- Leaves the reader with something to think about (a take home message)
- Provides a sense of closure for the essay
How to Write an Essay: Referencing and Citation
Your essay inevitably stands on the shoulders of giants. Your ideas and thoughts are borrowed from other researchers or scholars. This is known as citing.
Guidelines used when Citing Sources in Essay Writing
Quoting Guidelines in essay writing
This is when you borrow from an author word for word. Only quote if the phrase is catchy, or if you feel it is the best rendering of your intended point. Always signal the entry of a quotation, as illustrated below. Example: “It was Nelson Mandela who said, “Education is the most powerful weapon which you can use to change the world.”
Mixed quoting Guidelines in essay writing
This is when you paraphrase (paraphrasing is discussed below) part of an original quote but “quote” the rest. Example: “Nelson Mandela reminded us of the impact of education as being “the most powerful weapon” in changing the world.
Paraphrasing Guidelines in essay writing
Paraphrasing involves using original language and literary style to communicate the same message as somebody else. Do not just use synonyms to paraphrase. Example: “Nelson Mandela knew the indispensability of education in any society. According to him, it was the single most effective way in making a universal difference.”
Referencing allows you to acknowledge the sources of your citation, in addition to granting the reader access to further information on the presented facts. Additionally, referencing provides proof of your research and, importantly, helps you ward off accusations of plagiarism. Plagiarism is using other people’s words or ideas and passing them off as your own. A serious academic and professional offence, it could get your work cancelled, or you could receive worse penalties. Your lecturer or school will specify the referencing style required for your essay.
How to Write an Essay: Revising Your Essay
After you are done drafting your essay, the next step is to review it. Take a day off before revisiting your essay so you can look at it from a renewed perspective.
Follow these steps to polish up your essay
Wording in your Essay
Avoid clichés: Clichés are words or phrases that have been used to the point of losing their original effect. E.g. “when all is said and done”, “avoid it like the plague” “for all intents and purposes” etc. Omit needless words from your draft: Example: Instead of saying, “Before Matt left the country to move abroad, he was working at McDonalds.” Say, “Before Matt left the country for abroad, he worked at McDonalds”
Use gender-neutral language in essay writing
Avoid using words that denote feminine or masculine bias. Example: fireman, waiter/waitress, mankind etc. Instead, say: firefighter, table attendant, humanity etc. Also, avoid the use of third person pronouns i.e. “him/her”
Grammar in essay writing
Reread your essay to check sentence structure, spellings (or misspellings thereof), and punctuation. Make sure transitioning from one idea to another or paragraph to another sound effortless and natural. Refrain from using first person phrases i.e. “In my opinion”, “I really think” etc. as they only serve to weaken your arguments. Make sure you have provided all supporting evidence to your arguments. Provide a list of references for your citations.
Now, you are done with your essay. Congratulations!
If this is still not clear and you still want homework help, please contact me on my Contact Page.
References used to write this article
The following resources were consulted when writing this article:
bd.eduweb.hhs.nl/How to Write an Essay<|endoftext|>
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The Diamond Sutra is an ancient Buddhist sermon that generation of Buddhists have memorized and chanted since at least the fifth century. The sutra, which meditates on the illusory nature of the material world—the central theme of Buddhism, was originally written in Sanskrit in India, from which it was translated to Chinese in 401 AD. It is said that the teachings of The Diamond Sutra “cut like a diamond blade through worldly illusion to illuminate what is real and everlasting.”
A copy of this original translation, printed on an 18-feet-long, yellow-dyed scroll of paper, is housed at the British Library in London. The last few lines of the text, the colophon, identifies who printed it, when and why. It reads: “Reverently made for universal free distribution by Wang Jie on behalf of his two parents, 11 May 868.”
The precise date makes this particular edition of Diamond Sutra the world’s oldest printed and dated book in existence. The explicit public domain dedication is also the first in the history of creative work.
The Diamond Sutra was printed using woodblock printing—a technique where the text to be printed is carefully carved as a relief pattern on a block of wood, and then stamped on paper or fabric after dipping the block in a pool of ink. The scroll comprises of seven panels of paper, each of which was printed from a single block and stuck together to create a single scroll. While individual sheets of woodblock-printed paper have been found dating to the early Tang Dynasty (7th century), the Diamond Sutra is the earliest complete book to be found intact.
Photograph of Aurel Stein with his expedition team in the Tarim Basin, circa 1910.
The scroll was discovered by a monk inside a sealed-up cave at a site called “Caves of the Thousand Buddhas” near Dunhuang, in northwest China. The dry desert air provided the perfect conditions for the preservation of the paper and the silk scrolls inside. The yellow dye used on the scrolls comes from the Amur Cork Tree which has insecticidal properties. This contributed to the preservation of the scrolls.
The Diamond Sutra was among more than 40,000 scrolls and documents hidden in the secret cave about a thousand years ago when the area was threatened by a neighboring kingdom. When the British-Hungarian archaeologist Marc Aurel Stein heard about the scrolls, he bribed the guards in charge of the cave and smuggled away thousands of documents, including The Diamond Sutra. The British hailed him for his effort and even knighted him, but Chinese nationalists called him a thief.
The International Dunhuang Project is now digitizing those documents and thousands of others found on the eastern Silk Road.<|endoftext|>
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Invertebrates with soft bodies covered with shell of one or more pieces with bilateral symmetry are called mollusks. The word mollusks in Latin mean “soft.” Mollusks mostly live in water but some live on land. Almost 110,000 species have been reported.
Mollusks consist of four body parts that include mantle, visceral mass, shell, head, and foot. Each body part has its own specific function.
Mantle (and mantle cavity): Mollusks have a thin layer of visceral tissue called the mantle, which covers the body organs. It contains a gland that helps in the formation of hard shell. Mantle cavity is the space between the soft body and the mantle. It also has a set of gills that helps in breathing, so it does not have to come above water for air. The mantle protects the body if the mollusk does not have a shell. It also serves as the space where the head and foot can retract.
Visceral mass: The visceral mass or the visceral hump, also known as the visceropallium, is located within the shell, and is where the digestive, excretory, reproductive, and respiratory systems are held.
Shell: The shell is primarily made of chitin and conchiolin, which is a hard protein with calcium carbonate deposits. The shell is made up of three layers. The outermost layer is called the periostracum, which is an organic coating made of conchiolin. The middle layer is made of calcite, which is essentially calcium carbonate deposits, and the inner layer is made of laminated calcite, also known as aragonite, and is called nacre.
Head: The head is bilaterally symmetrical, and carries one or two tentacles. The mouth lies in the center of the ventral margin. In some species, the tentacles are able to roll in, while some have tentacles that are contractile. Some have eyes at the tip of the tentacles, while some have the eyes just below the tentacles. Some mollusks have labial palps in their mouth which help in locating the prey. Frequently, the mouth is prolonged to for proboscis.
Foot: The foot is muscular and is located on the underside of the body. It is composed of two statocysts, which also serve to sense balance. In some mollusks, it secretes mucus which helps in movement. The organism moves around by contracting the foot muscles. In certain mollusks, the foot functions as sucker to harbor the animal on a hard surface. In certain others, the foot aids in burrowing and in a few cephalopods, it is used for propulsion.
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In science there are many key concepts and terms that are crucial for students to know and understand. Often it can be hard to determine what the most important science concepts and terms are, and even once you’ve identified them you still need to understand what they mean. To help you learn and understand key science terms and concepts, we’ve identified some of the most important ones and provided detailed definitions for them, written and compiled by Chegg experts.<|endoftext|>
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If you’re struggling with creating a structured lesson plan, this post is for you! I’m going to give you some helpful tips and advice to get you on the right track to make a Killer ESL lesson plan!
Where to Start
Before even thinking about your ESL lesson plan you need a starting point. In other words…the goal of the lesson! This can also be called your lesson objective (or LO for short).
Ask yourself, what do I want to teach my class today? Is it a grammar point, specific vocabulary or phrases around a topic, a skill like listening or reading? The options are endless. Come up with your own or use your textbook to guide you.
The next part is deciding on the ESL lesson plan theme. The theme of the lesson will need to relate to the LO.
- If you want to teach your students vocabulary about movies, you’re the theme is going to be movies!
- If you want to teach the grammar point present simple your theme could be based on daily routines, talking about habits or schedules.
- Want to teach listening? Have your students watch a television show and have students answer questions that help them with their specific and general listening skills.
Make sure the theme related to your students’ lives!! Otherwise, they won’t find it interesting! For example, if you’re teaching University students your theme could be about study habits or drinking habits for present simple.
The lesson Structure
In general, I like to structure my lessons using the Present Practice, Produce method (Otherwise known as the PPP method). I’m not going to go into detail here because I’ve already written about that: English Teaching Methods – PPP Lesson Plan. I use that as a general framework for most of my ESL Lesson plans and change it as required (based on my time and goals).
Always start your lesson with an introductory activity!
The starter activity you choose or create will do 4 very important things for your ESL Lesson plan:
- Introduce the topic and LO
- Connect and relate to the students to increase engagement and interest in the topic
- Allow the students to apply their background knowledge to the LO
- Allow you to see where your students are with the LO (Have they already learned it, are they using it in their speaking, what mistakes are they making etc.)
Example starter activity ideas for a listening lesson:
- Start with a question like: “What do you have the most trouble understanding in English? When do you have to listen to English outside the classroom?
- Have students watch a famous movie or television clip and write down as many words as they can hear and compare it with their classmates (make it into a competition)
This activity should only last about 5 minutes!!
A Great website to help with quick and easy starter ideas is Busy Teacher.
Teaching the Lesson Objective (LO)
Next is the teaching segment of the lesson where you introduce the LO and teach it to the class. This is the most teacher talk time during the lesson, however, make sure to ask concept checking questions, otherwise known as CCQ’s to check that students understand For more information check out this great article by the British Council.
Teaching tips to remember while preparing your ESL lesson plan!
- Make sure to use examples in context, and provide 3-4 examples of the LO in different contexts. Ask questions and check for understanding as you do this! (Get the students involved!)
- Utilize your teachers manual and make sure to cover all the points in your lesson. Be prepared for possible questions or things the students might struggle with.
- Relate it to real life scenarios. When will the students use the word/grammar point/ skill in their lives?
Student’s practice the LO
Students should now have a chance to practice what you’ve taught them!
This is the part of the lesson where students practice the LO in a controlled setting. Use handouts or activities from your textbook for this. (Fill in the blank, matching etc.)
Teaching tips for practice!
- Have students do the exercises with a partner so they can work on it together. Or better yet find a pair work activity.
- Monitor and provide assistance when needed.
- Write down any mistakes or points you want to take up with the students later.
- Make time for correction and feedback – Correct as a class and drill down any mistakes (don’t spend too much time on this!! Pick one or two to correct as a class and check for common mistakes)
The End Phase (Production of the LO)
This is your final activity where students get to practice the LO in context. Make sure that your final activity is something that students can relate to. They should have a chance to produce the LO in context. Some great activities include interviews, surveys, games, short role plays etc. The possibilities are endless!!! Don’t be afraid to get creative with this!
For example: If you’ve taught them vocabulary about the movies, have them do a role-play talking about a movie. Or have them write a movie review and present it to their partner.
- Make sure the activity will actually elicit the LO (If they learn about vocabulary to describe movies make sure the activity has them describing a movie whether it be conversation questions, a survey. The questions themselves have them using the vocabulary in their answer)
- Remind students to USE THE LO in the activity! Sometimes you need to be direct and focussed. This is the time to focus and use what they’ve learned. It should be intentional.
- Be creative with your activity! Use real-life resources like articles, movies or pictures.
Don’t forget the final 5-7 minutes of the class should be for feedback, homework and final questions to make sure they understand the LO and see that they’ve made progress!
Congratulations you’ve successfully created a killer ESL Lesson Plan!
For more online resources to help plan your lesson go to: The Ultimate English Language Resource List
Found this useful? Pin it for later!<|endoftext|>
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# How do you find the derivative of 2cos^2(x)?
Apr 1, 2018
$- 2 \sin 2 x$
#### Explanation:
$\text{differentiate using the "color(blue)"chain rule}$
$\text{Given "y=f(g(x))" then}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$
$y = 2 {\cos}^{2} x = 2 {\left(\cos x\right)}^{2}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 4 \cos x \times \frac{d}{\mathrm{dx}} \left(\cos x\right)$
$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - 4 \sin x \cos x = - 2 \sin 2 x$<|endoftext|>
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In the last session we have seen about arrival of Europeans to India in brief, and this session is a continuation to the earlier one.
Here we are going to see about the permanent settlement and how the remaining powers in India has been defeated by the English East India Company.
Before going further please subscribe to the channel to get latest lessons from us.
OK let’s get started.
The permanent settlement.
Cornwallis’s greatest achievement in India was the reorganization of the land taxation, known as the Permanent Settlement of 1793. Agricultural land in Bengal was cultivated by a large number of small farmers, who paid rent to a group of zamindars (landowners). Under the Mughals, the government had collected taxes from the zamindars. The East India Company, however, had tried to set aside the zamindars, and collect land taxes either directly through company officials, or through revenue-farmers, who collected the rent from peasants and paid a lump sum to the government. The new system led to widespread corruption, and the peasants suffered severely. Cornwallis decided to go back to the old Mughal system.
He granted legal ownership of their land to the zamindars. In return, they had to pay the government 90 per cent of the rent which they collected from the farmers. These arrangements were to last for ever, hence it is known as “permanent settlement.”
The immediate effects of the permanent settlement were not good. In 1769 Bengal was devastated by a terrible scarcity. A large number of rural people died from starvation or fled away from the countryside. As a result, zamindars found it difficult to collect rent from such ruined farmers. Many of them were unable to pay their fixed taxes, and sold their estates. It was not until the beginning of the 1800’s, when the population began to increase once again and land which had gone out of cultivation was brought back under the plough, that the great Bengal zamindars again became prosperous.
The permanent settlement, however, was not extended to the territories later conquered by the East India Company. In Madras Presidency, under the guidance of Thomas Munroe (who later became the governor of Madras), revenue was collected directly from the ryots (peasants) and the system was known as ryotwari. The North Western Provinces (part of present-day Uttar Pradesh) adopted mahalwari settlement, in which the headman of the village collected revenue from individual landholders and remitted the collection to the state. In all the taxation systems, the peasants had to give up a major part of their produce, which led to a series of revolts during the rule of the East India Company.
British territorial expansion. After the conquest of Bengal in 1757, British political influence and territorial control expanded rapidly. The two major Indian powers who challenged the British expansionist plans were Tipu Sultan, the nawab of Mysore, and the Marathas. Tipu Sultan was conscious of European advances in science and technology. He sent a mission to France seeking political alliance and scientific collaboration. He also modernized his state and raised an efficient army. The British regarded him a dangerous enemy, particularly because he was striving to form a confederacy of all major Indian powers against the British. Tipu did not succeed in his efforts. He was killed in the battle of Srirangapatnam (1799).
The Marathas were still a formidable power, with their territorial and political influence extending over western and northern India. Several small Indian principalities accepted their overlordship. Even the Mughal emperor, who by the end of the 1700’s had lost all authority, had sought their protection. Lord Wellesley, governor general from 1798 to 1805, waged two wars against them, in 1798-1800 and 1803-1805, and annexed a major part of their territory. The Marathas were finally defeated in 1818 by Lord Hastings.
Thereafter there was no major military power in India, except the Sikhs in Punjab, who were also brought under British rule in 1848. By the middle of the 1800’s, the whole of India had come under British rule. However, the entire territory was not directly administered by the British. A large area was ruled by Indian princes like the nizam of Hyderabad and the rajas of Travancore, Baroda, and Rajputana.
With this we have come to the end of the session.
We have seen the brief overlook of Europeans arrival, how they expanded their presence and how they played wise politics between Indian rulers to get the whole of India under their rule.<|endoftext|>
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### Compiled by: Sabrina Singh
The prefix quad- means “four” and quadratic expressions are ones that involve powers of x up to the second power (not the fourth power). So why are quadratic equations associated with the number four?
Answer: These equations are intimately connected with problems about squares and quadrangles. (In fact, the word quadratic is derived from the Latin word quadratus for square.) Questions about quadrangles often lead to quadratic equations. For example, consider the problem: A quadrangle has one side four units longer than the other. Its area is 60 square units. What are the dimensions of the quadrangle?If we denote the length of one side of the quadrangle as x units, then the other must be x + 4 units in length. We must solve the equation: x(x+4)+60 , which is equivalent to solving the quadratic equation x^2 +4x-60+0 Solving quadratic equations, even if not derived from a quadrangle problem, still involves the geometry of four-sided shapes. As we shall see, all such equations can be solved by a process of “completing the square.”
Within this unit we moved on from linear relations to quadratics relations. We touched upon the many components of quadratics which ranged from a basic introduction to graphing equations in a variety of forms. Below is the structure of the lessons and activities:
Introduction
• Properties Of A Parabola
• Transformations
Types of Equations
• Factored Form a(x-r)(x-s)
• Vertex Form a(x-h)²+k
• Standard Form ax²+bx+c
Vertex form
• Finding an equation when only given the vertex
• Graphing vertex form
Factored Form
• Factors and zeroes
• Polynomials (factored form to standard form)
• Common factoring
• Factoring simple trinomials
• Factoring complex trinomials
-Special cases
- Difference of squares
• Perfect square
• Factoring by grouping
• Graphing factored form
Standard form
• Completing the square (standard to vertex)
• Graphing from standard form
Word Problems
• Motion
• Revenue
• Numeric
## Introduction
An expression of the form ax²+ bx+ c with x the variable and a, b, and c fixed values (with a ≠ 0 ) is called a quadratic. To solve a quadratic equation means to solve an equation that can be written in the form ax²+ bx+ c=0 .
Explain the meaning of the term function, and distinguish a function from a relation that is not a function, through investigation of linear and quadratic relations using a variety of representations (i.e., tables of values, mapping diagrams, graphs, function machines, equations) and strategies.
• Substitute into and evaluate linear and quadratic functions represented using function notation, including functions arising from real-world applications.
• Explain the meanings of the terms domain and range, through investigations using numeric, graphical, and algebraic representations of linear and quadratic functions, and describe the domain and range of a function appropriately.
• Explain any restrictions on the domain and the range of a quadratic function in contexts arising from real-world applications.
• Determine, through investigation using technology, the roles of a, h, and k in quadratic functions of the form f(x)=a(x – h) 2 + k, and describe these roles in terms of transformations on the through investigation with and without technology,from primary sources, using a variety of tools, or from secondary sources, and graph the data.
• Determine, through investigation using a variety of strategies, the equation of the quadratic function that best models a suitable data set graphed on a scatter plot, and compare this equation to the equation of a curve of best fit generated with technology.
• Solve problems arising from real-world applications, given the algebraic representation of a quadratic function.
While the first derivative can tell us if the function is increasing or decreasing, the second derivative tells us if the first derivative is increasing or decreasing. If the second derivative is positive, then the first derivative is increasing, so that the slope of the tangent line to the function is increasing as x increases. We see this phenomenon graphically as the curve of the graph being concave up, that is, shaped like a parabola open upward. Likewise, if the second derivative is negative, then the first derivative is decreasing, so that the slope of the tangent line to the function is decreasing as x increases. Graphically, we see this as the curve of the graph being concave down, that is, shaped like a parabola open downward. At the points where the second derivative is zero, we do not learn anything about the shape of the graph: it may be concave up or concave down, or it may be changing from concave up to concave down or changing from concave down to concave up
This table indicates that the equation y=x² is a quadratic relation because the first differences are not equal while the second differences are.
## Parts Of Parabola
The Vertex of a Parabola
• The vertex of a parabola is the point where the parabola crosses its axis of symmetry. If the coefficient of the x^2 term is positive, the vertex will be the lowest point on the graph, the point at the bottom of the “U”-shape. If the coefficient of the x^2 term is negative, the vertex will be the highest point on the graph, the point at the top of the “U”-shape.
The Axis of Symmetry
• The axis of symmetry of a parabola is the vertical line through the vertex. For a parabola in standard form, y =ax^2 + bx + c, the axis of symmetry has the equation.Note that –b/2a is also the x-coordinate of the vertex of the parabola.
Intercepts
• You can find the y-intercept of a parabola simply by entering 0 for x. If the equation is in standard form, then you can just take c as the y-intercept. For instance, in the above example: y =2(0)2 + (0) – 1=–1 So the y-intercept is –1.
• The x-intercepts are a bit trickier. You can use factoring, or completing the square, or the quadratic formula to find these (if they exist!).
## Graphical Transformations of Functions
In this section you will learn how to draw the graph of the quadratic function defined by the equation f(x) = a(x − h) 2 + k. You will quickly learn that the graph of the quadratic function is shaped like a "U" and is called a parabola. The form of the quadratic function in equation is called vertex form, so named because the form easily reveals the vertex or “turning point” of the parabola. Each of the constants in the vertex form of the quadratic function plays a role. As you will soon see, the constant a controls the scaling (stretching or compressing of the parabola), the constant h controls a horizontal shift and placement of the axis of symmetry, and the constant k controls the vertical shift. Let’s begin by looking at the scaling of the quadratic.
The graph of the basic quadratic function f(x) = x 2 is called a parabola. We say that the parabola “opens upward.” The point at (0, 0), the “turning point” of the parabola, is called the vertex of the parabola. We've tabulated a few points for reference in the table and then superimposed these points on the graph of f(x) = x 2.
Now that we know the basic shape of the parabola determined by f(x) = x 2 , let’s see what happens when we scale the graph of f(x) = x 2 in the vertical direction. For example, let’s investigate the graph of g(x) = 2x 2 . The factor of 2 has a doubling effect. Note that each of the function values of g is twice the corresponding function value of f in the table
When the points in the table are added to the coordinate system , the resulting graph of g is stretched by a factor of two in the vertical direction. It’s as if we had put the original graph of f on a sheet of rubber graph paper, grabbed the top and bottom edges of the sheet, and then pulled each edge in the vertical direction to stretch the graph of f by a factor of two. Consequently, the graph of g(x) = 2x 2 appears somewhat narrower in appearance, as seen in comparison to the graph of f(x) = x 2 . Note, however, that the vertex at the origin is unaffected by this scaling. In like manner, to draw the graph of h(x) = 3x 2 , take the graph of f(x) = x 2 and stretch the graph by a factor of three, tripling the y-v
## Vertical Translations
The graph of f(x) = x ^2 + 1 is shifted one unit upward from the graph of g(x) = x 2 . This is easy to see as both equations use the same x-values in the table , but the function values of g(x) = x^ 2 + 1 are one unit larger than the corresponding function values of f(x) = x 2 .
Note that the vertex of the graph of g(x) = x 2 + 1 has also shifted upward 1 unit and is now located at the point (0, 1).
## Horizontal Translations
The graph of g(x) = (x + 1) shows a basic parabola that is shifted one unit to the left. Examine the table and note that the equation g(x) = (x + 1)2 produces the same y-values as does the equation f(x) = x 2 , the only difference being that these y-values are calculated at x-values that are one unit less than those used for f(x) = x 2 . Consequently, the graph of g(x) = (x + 1)2 must shift one unit to the left of the graph of f(x) = x 2 , as is evidenced. Note that this result is counter intuitive. One would think that replacing x with x + 1 would shift the graph one unit to the right, but the shift actually occurs in the opposite direction. Finally, note that this time the vertex of the parabola has shifted 1 unit to the left and is now located at the point (−1, 0). We are led to the following conclusion.
A similar thing happens when you replace x with x − 1, only this time the graph is shifted one unit to the right.
## Vertical Reflections
Let’s consider the graph of g(x) = ax^2 , when a < 0. For example, consider the graphs of g(x) = −x^ 2 and h(x) = (−1/2)x^ 2
When the table is compared with the other table , it is easy to see that the numbers in the last two columns are the same, but they've been negated. The result is easy to see. The graphs have been reflected across the x-axis. Each of the parabolas now “opens downward.” However, it is encouraging to see that the scaling role of the constant a in g(x) = ax^2 has not changed. In the case of h(x) = (−1/2)x^2 , the y-values are still “compressed” by a factor of two, but the minus sign negates these values, causing the graph to reflect across the x-axis. Thus, for example, one would think that the graph of y = −2x ^2 would be stretched by a factor of two, then reflected across the x-axis. Indeed, this is correct, and this discussion leads to the following property
• Vertex form: a(x-h)²+k
• Factored form : a(x-r)(x-s)
• Standard form: ax²+bx+c
## Vertex Form of Quadratic Functions
The vertex form of a quadratic function is given by f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola.
When written in "vertex form":
(h, k) is the vertex of the parabola, and x = h is the axis of symmetry.
• the h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0).
• the k represents a vertical shift (how far up, or down, the graph has shifted from y = 0).
• notice that the h value is subtracted in this form, and that the k value is added.
If the equation is y = 2(x - 1)2 + 5, the value of h is 1, and k is 5.
If the equation is y = 3(x + 4)2 - 6, the value of h is -4, and k is -6.
## Finding an equation when only given the vertex
Many real world situations that model quadratic functions are data driven. What happens when you are not given the equation of a quadratic function, but instead you need to find one? In order to obtain the equation of a quadratic function, some information must be given. Significant data points, when plotted, may suggest a quadratic relationship, but must be manipulated algebraically to obtain an equation.
3.4 Finding the Equation given Vertex
• When do I use each form?
• When you are given the vertex and at least one point of the parabola, you generally use the vertex form.
• When you are given points that lie along the parabola, you generally use the general form.
Let's use a vertex that you are familiar with: (0,0).
Use the following steps to write the equation of the quadratic function that contains the vertex (0,0) and the point (2,4).
1. Plug in the vertex.
2. Simplify, if necessary.
3. Plug in x & y coordinates of the point given.
4. Solve for "a."
5. Now substitute "a" and the vertex into the vertex form.
Our final equation looks like this:
f(x)= x^2
## Graphing Quadratic Function In Vertex Form
Vertex form of a quadratic function: f(x)= a(x-h)^2+k
• The graph f(x)= a(x-h)^2+k is the parabola y=x^2 translated an h united horizontally and k units vertically.
• y = a(x - h)2 + k ( y = 3(x - 2)2 - 4)
• Pull out the values for h and k.
If necessary, rewrite the function so you can clearly see the h and k values.
(h, k) is the vertex of the parabola.
Plot the vertex. y = 3(x - 2)2 + (-4)
h = 2; k = -4
Vertex: (2, -4)
3.2 Graphing from Vertex Form
Find two or three points on one side of the axis of symmetry, by substituting your chosen x-values into the equation.
For this problem, we chose (to the left of the axis of symmetry):
x = 1; y = 3(1 - 2)2 - 4 = -1
x = 0; y = 3(0 - 2)2 - 4 = 8
Plot (1, -1) and (0,8)
Plot the mirror images of these points across the axis of symmetry, or plot new points on the right side. Draw the parabola. Remember, when drawing the parabola to avoid "connecting the dots" with straight line segments. A parabola is curved, not straight, as its slope is not constant.
## Factored Form
An essential skill in many applications is the ability to factorise quadratic expressions. In this unit you will see that this can be thought of as reversing the process used to ‘remove’ or ‘multiply-out’ brackets from an expression. You will see a number of worked examples followed by a discussion of special cases which occur frequently with which you must become familiar. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that all this becomes second nature, and you eventually carry out the process of factorising simply by inspection. To help you to achieve this, the unit includes a substantial number of such exercises.
## Finding polynomial functions ( factors and zeros)
As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned addition and then you learned subtraction or first you learned multiplication and then you learned division. Each operation being the opposite of the other. Yet another example is one in which you learned the properties of exponents and then you learned about radicals or rational exponents.
Well, finding polynomials is the reverse of finding factors. In the previous lesson, you were given a polynomial and asked to find its factors and zeros. In this lesson, you will be given factors or zeros and asked to find the polynomial of lowest degree with real coefficients.
Example 1: Given the factors ( x - 3) 2 (2x + 5), find the polynomial of lowest degree with real coefficients.Let's analyze what we already know. ( x - 3) 2 is a repeated factor, thus the zero 3 has multiplicity of two. Did you know that the linear factorization theorem states that a polynomial of degree n has precisely n linear factors. And since we h been given three linear factors, the lowest degree of this polynomials must be three.
STEPS
• Rewrite the factors to show the repeating factor. ( x - 3)( x - 3)(2x + 5)
• Write factors in polynomial notation. The a is the leading coefficient of the polynomial P(x) = a( x - 3)( x - 3)(2x + 5)
• Multiply ( x - 3)( x - 3)
• Then multiply ( x 2 - 6x + 9)(2x + 5)
• For right now let an = 1
• Thus, the polynomial of lowest degree with real coefficients, and factors ( x - 3)2 (2x + 5) is: P(x) = 2x 3 - 7x 2 + 8x + 45.
The main concept is simply to multiply the factors in order to determine the polynomial of lowest degree with real coefficients.
## Multiplying-out brackets, and quadratic expressions
In this unit you will learn how many quadratic expressions can be factorised. Essentially, this is the reverse process of removing brackets from expressions such as (x + 2)(x + 3).
You will be familiar already with the well-known process of multiplying-out brackets. For example, you will have seen expressions like (x + 2)(x + 3) and then expanded these terms to arrive at a quadratic expression. Let us see how this works. The arrows in the figure below show how each term in the first pair of brackets multiplies each term in the second pair.
Then the like-terms, 3x and 2x, are collected together to give 5x. The result of multiplying-out the brackets is the quadratic expression x^2 + 5x + 6.
Like many processes in mathematics, it is useful to be able to go the other way. That is, starting with the quadratic expression x^2+5x+6, can we carry out a process which will result in the form (x + 2)(x + 3) ? The answer is: yes we can! This process is called factorising the quadratic expression. This would help, for example, if we wanted to solve a quadratic equation. Such an equation is formed when we set a quadratic expression equal to zero, as in x^2 + 5x +6=0 This is because the equation can then be written (x + 2)(x + 3) = 0 and if we have two expressions multiplied together resulting in zero, then one or both of these must be zero. So, either x + 2 = 0, or x + 3 = 0, from which we can conclude that x = −2, or x = −3. We have found the solutions of the quadratic equation x^2 + 5x + 6 = 0.
So, the ability to factorise a quadratic is a useful basic skill, which you will learn about in this unit. However, be warned, not all quadratics will factorise, but a lot do and so this is a process you have got to get to know!
To learn how to factorize let us study again the removal of brackets from (x + 3)(x + 2). (x + 3)(x + 2) = x^2 + 2x + 3x +6= x2 + 5x + 6
Clearly the number 6 in the final answer comes from multiplying the numbers 3 and 2 in the brackets. This is an important observation. The term 5x comes from adding the terms 2x and 3x. So, if we were to begin with x^2 + 5x + 6 and we were going to reverse the process we need to look for two numbers which multiply to give 6 and add to give 5.
? × ?=6
?+? = 5
What are these numbers ? Well, we know that they are 3 and 2, and you will learn with practice to find these simply by inspection. Using the two numbers which add to give 5 we split the 5x term into 3x and 2x. We can set the calculation out as follows.
x^2 + 5x +6
= x2 + 3x + 2x + 6
= x(x + 3) + 2x + 6 by factorizing the first two terms
= x(x + 3) + 2(x + 3) by factorizing the last two terms
= (x + 3)(x + 2) by noting the common factor of x + 3
The quadratic has been factorized. Note that you should never get these wrong, because the answer can always be checked by multiplying-out the brackets again!
Example 1:
Suppose we want to factorize the quadratic expression x^2 − 7x + 12. Starting as before we look for two numbers which multiply together to give 12 and add together to give −7. Think about this for a minute and you will realize that the two numbers we seek are −3 and −4 because
−4 × −3 = 12, and − 4 + −3 = −7
So, using the two numbers which add to give −7 we split the −7x term into −4x and −3x. We set the calculation out like this:
x^2 − 7x + 12 = x2 − 4x − 3x + 12
= x(x − 4) − 3x + 12 by factorizing the first two terms
= x(x − 4) − 3(x − 4) by factorizing the last two terms extracting a factor of −3 in order to leave x − 4
= (x − 4)(x − 3) by noting the common factor of x − 4 Once again, note that the answer can be checked by multiplying out the brackets again.
Example 2:
Suppose we wish to factories the quadratic expression x2 − 5x − 14. Starting as before we look for two numbers which multiply together to give −14 and add together to give −5. Think about this for a minute and you will realize that the two numbers we seek are −7 and 2 because
−7 × 2 = −14, and − 7+2= −5
So, using the two numbers which add to give −5 we split the −5x term into −7x and +2x. We set the calculation out like this:
x2 − 5x − 14 = x2 − 7x + 2x − 14
= x(x − 7) + 2x − 14 by factorizing the first two terms
= x(x − 7) + 2(x − 7) by factorizing the last two terms
= (x − 7)(x + 2) by noting the common factor of x − 7
So the factorization of x2 − 5x − 14 is (x − 7)(x + 2).
## Factorising by inspection
Clearly, when you have some experience of this method, it is possible to avoid writing out all these stages. This is rather long-winded. What you need to do now is carry out the process by inspection. That is, simply by looking at the given expression, decide in your head which numbers need to be placed in the brackets to do the job. Look at the following two examples and see if you can do this.
Example :
Suppose we want to factorize the quadratic expression x^2 − 9x + 20 by inspection. We write x^2 − 9x + 20 = ( )( )
and try to place the correct quantities in brackets. Clearly we will need an x in both terms:
x^2 − 9x + 20 = (x )(x )
We want two numbers which multiply to give 20 and add to give −9. With practice you will be able to do this in your head. The two numbers are −4 and −5. x^2 − 9x + 20 = (x − 4)(x − 5) The answer should always be checked by multiplying-out the brackets again.
Example:
Suppose we want to factorise the quadratic expression x^2 − 9x − 22 by inspection. We write
x^2 − 9x − 22 = ( )( )
and try to place the correct quantities in brackets. Clearly we will need an x in both terms:
x^2 − 9x − 22 = (x )(x )
We want two numbers which multiply to give −22 and add to give −9. The two numbers are −11 and +2
x^2 − 9x − 22 = (x − 11)(x + 2)
The answer should always be checked by multiplying-out the brackets again. If you can’t manage to do these by inspection yet do not worry. Do it the way we did it before. Your ability to do this will improve with practice and experience. All of these examples have involved quadratic expressions where the coefficient of x^2 was 1. When the coefficient is a number other than 1 the problem is more difficult. We will look at examples like this in the next section.
## Expressions where the coefficient of x^2 is not 1.
Suppose we wish to factorize the expression 3x^2 + 5x − 2. As we did before we look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. However, instead of looking for two numbers which multiply to give −2 we must look for two numbers which multiply to give −6, (that is, the coefficient of x^2 multiplied by the constant term, 3 × −2). This is entirely consistent with the method we applied before because in those examples the coefficient of x^2 was always 1.
So
? × ? = −6
? + ?=5
By inspection, or trial and error, two such numbers are 6 and −1.
6 × −1 = −6
6+ −1=5
We use these two numbers to split the 5x term into 6x and −1x.
3x^2 + 5x − 2
=3x2 + 6x − x − 2
= 3x(x + 2) − x − 2 by factorizing the first two terms
= 3x(x + 2) − (x + 2) factorizing the last two terms by extracting −1
= (x + 2)(3x − 1) by noting the common factor of x + 2
Example 1:
Suppose we wish to factorize 2x^2 + 5x − 7. As we did before we look for two numbers which add to give the coefficient of x; so we seek two numbers which add to give 5. We look for two numbers which multiply to give −14, (that is, the coefficient of x^2 multiplied by the constant term, 2 × −7).
? × ? = −14
? + ? = 5
By inspection, or trial and error, two such numbers are 7 and −2.
7 × −2 = −14
7 + −2=5
We use these two numbers to split the 5x term into 7x and −2x. 2x^2 + 5x − 7
=2x^2 + 7x − 2x − 7
= x(2x + 7) − 2x − 7 by factorising the first two terms
= x(2x + 7) − (2x + 7) factorising the last two terms by extracting −1
= (2x + 7)(x − 1) by noting the common factor of 2x + 7
## Common Factoring
Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved
Finding the GCF of a List of Integers or Terms
1)Prime factor the numbers.
2)Identify common prime factors.
3)Take the product of all common prime factors.
If there are no common prime factors, GCF is 1
## Examples
1)6, 8 and 46
6 = 2 · 3
8 = 2 · 2 · 2
46 = 2 · 23
So the GCF is 2.
2)144, 256 and 300
144 = 2 · 2 · 2 · 3 · 3
256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
300 = 2 · 2 · 3 · 5 · 5
So the GCF is 2 · 2 = 4.
a3b2, a2b5 and a4b7
a3b2 = a · a · a · b · b
a2b5 = a · a · b · b · b · b · b
a4b7 = a · a · a · a · b · b · b · b · b · b · b
So the GCF is a · a · b · b = a2b2
Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable.
The first step in factoring a polynomial is to find the GCF of all its terms.
Then we write the polynomial as a product by factoring out the GCF from all the terms.
The remaining factors in each term will form a polynomial.
Factor out the GCF in each of the following polynomials.
6x3 – 9x2 + 12x =
3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 =
3x(2x2 – 3x + 4)
2) 14x3y + 7x2y – 7xy =
7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 =
7xy(2x2 + x – 1)
Remember that factoring out the GCF from the terms of a polynomial should always be the first step in factoring a polynomial.
This will usually be followed by additional steps in the process.
Factor 90 + 15y2 – 18x – 3xy2.
90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6xxy2) =
3(5 · 6 + 5 · y2 – 6 · x – x · y2) =
3(5(6 + y2) – x (6 + y2)) =
3(6 + y2)(5x)
Factoring Simple Trinomials
• A simple trinomial is a polynomial with three terms of the form x2+bx+c.
• It is called a simple trinomial because the squared term has a coefficient of 1.
• These trinomials are formed from multiplying two binomials together. eg. Multiply (x+3)(x+4)
• Factoring is the reverse of this process, we will go from x^2+7x+12 to (x+3)(x+4).
• Note that 3 and 4 multiply to 12 and add to 7.
## Complex Trinomials
To work backward and factor a complex trinomial, we must determine the two values, p and q, with a sum of b and a product of ac. To make things work, we decompose the middle term into separate two terms with coefficients p and q. This process is often referred to as decomposition. The resulting expression will look like this: ax2 + bx+ c Since ac, b and c all share the same factors, it will be possible to use grouping to find common factors. Some examples should make this clearer.
Factor 2x ^2 + 13x + 15.
We must determine two numbers, p and q, that have a product of 2 × 15 = 30 and a sum of 13. Since the product and sum are both positive, both p and q will be positive. Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. The two numbers that meet the requirements are 3 and 10.
Decompose the middle term into two terms with p and q as coefficients. 2x^ 2 + 13x + 15 = 2x^ 2 + 3x + 10x + 15 Factor by grouping. 2x^ 2 + 3x + 10x + 15
= x(2x + 3) + 5(2x + 3)
= (2x + 3)(x + 5) Therefore, 2x 2 + 13x + 15
= (2x + 3)(x + 5).
Factor 6x^ 2 + 19x − 7. We must determine two numbers, p and q, that have a product of 6(−7) = −42 and a sum of 19. Since the product is negative and the sum is positive, p will be positive and |p| > |q|. Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42. The two numbers that meet the requirements are 21 and −2.
Decompose the middle term into two terms with p and q as coefficients. 6x^ 2 + 19x − 7 = 6x 2 + 21x − 2x − 7 Factor by grouping. 6x 2 + 21x − 2x − 7
= 3x(2x + 7) − 1(2x + 7)
= (2x + 7)(3x − 1) Therefore, 6x 2 + 19x − 7
= (2x + 7)(3x − 1).
Factor 25^ 2 − 20x + 4. We must determine two numbers, p and q, that have a product of 25 × 4 = 100 and a sum of −20. Since the product is positive and the sum is negative, both p and q will be negative. Factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50 and 100. In this case, the two numbers that meet the requirements have the same value: −10 and −10.
Decompose the middle term into two terms with p and q as coefficients. 25x^ 2 − 20x + 4 = 25x 2 − 10x − 10x + 4 Factor by grouping. 25x 2 − 10x − 10x + 4
= 5x(5x − 2) − 2(5x − 2)
= (5x − 2)(5x − 2) Therefore, 25x 2 − 20x + 4
= (5x − 2)(5x − 2), or (5x − 2)2 . While decomposition works for perfect squares like the one above, we will look at a more efficient method in the next lesson
## Difference Of Two Squares
A special case concerns what is known as the difference of two squares. What does this look like ? Typically, x^2 − 9. Note that there is no x term and that the number 9 is itself a square number. A square number is one which has resulted from squaring another number. In this case 9 is the result of squaring 3, (3^2 = 9), and so 9 is a square number. Hence x^2 − 9 is the difference of two squares, x^2 − 3^2. When we try to factorize x^2 − 9 we are looking for two numbers which add to zero (because there is no term in x), and which multiply to give −9. Two such numbers are −3 and 3 because
−3 × 3 = −9, and − 3+3=0
We use these two numbers to split the 0x term into −3x and 3x
x^2 − 9
= x^2 − 3x + 3x − 9
= x(x − 3) + 3x − 9 by factorizing the first two terms
= x(x − 3) + 3(x − 3) factorizing the last two terms by extracting +3
= (x − 3)(x + 3) by noting the common factor of x − 3
x^2 − 9=(x − 3)(x + 3) Recall that 9 is a square number, (9 = 3^2). So this example is a difference of two square. The result of this example is true in more general cases of the difference of two squares. It is always the case that x^2 − a^2 factorizes to (x − a)(x + a).
Example:
Suppose we want to factorizes x^2 − 25.
Note that x^2 − 25 is the difference of two squares because 25 is a square number (25 = 52). So we need to factorize x^2 − 52. We can do this by inspection using the formula above. x^2 − 52 = (x − 5)(x + 5)
## complete squares
A second important special case is when we need to deal with complete squares. This happens when the answer can be written in the form ( )^2, that is as a single term, squared.
Suppose we wish to factorize x^2 + 10x + 25. As before we want two numbers which multiply to give 25 and add to give 10.
Two such numbers are 5 and 5 because 5 × 5 = 25, and 5 + 5 = 10 We use these two numbers to split the 10x term into 5x and 5x.
x^2 + 10x + 25
= x^2 + 5x + 5x + 25
= x(x + 5) + 5x + 25 by factorizing the first two terms
= x(x + 5) + 5(x + 5) factorizing the last two terms by extracting +5
= (x + 5)(x + 5) by noting the common factor of x + 5 The result can be written as (x + 5)2, a complete square.
Example:
Suppose we want to factories 25x^2 − 20x + 4. Note that 25x^2 can be written as (5x)^2, a squared term.
Note also that 4 = 22. In this case, by inspection,
25x^2 − 20x + 4 = (5x − 2)(5x − 2)
The result can be written as (5x − 2)^2, a complete square
Do not worry if you have difficulty with this last example. The skill will come with practice. Even if you did not recognize the complete square, you could still use the previous method: To factorize 25x^2 − 20x + 4
As before we want two numbers which multiply to give 100 (the coefficient of x^2 multiplied by the constant term) and add to give −20.
Two such numbers are −10 and −10 because −10 × −10
= 100, and − 10 + −10
= −20 We use these two numbers to split the −20x term into −10x and −10x.
25x2 − 20x + 4
= 25x2 − 10x − 10x + 4
= 5x(5x − 2) − 10x + 4 by factorising the first two terms
= 5x(5x − 2) − 2(5x − 2) factorising the last two terms by extracting −2
= (5x − 2)(5x − 2) by noting the common factor of 5x − 2
So 25x^2 − 20x + 4
= (5x − 2)^2 as before.
## The constant term is missing
There is one more special case that you ought to be familiar with. This arises in examples where the constant term is absent, as in 3x^2 − 8x. How do we proceed ? The way forward is to look for common factors. In the case of 3x^2 −8x note that each of the two terms has a common factor of x. (Note that this is not the case when dealing with a quadratic expression containing a constant term). The common factor, x, is written outside a bracket, and the contents of the bracket are chosen by inspection so that they multiply-out to give the correct terms. 3x^2 − 8x = x(3x − 8)
## Factoring My Grouping
Note that the polynomial is factored by grouping the first and second terms and the third and fourth terms. You could just as easily have grouped the first and third terms and the second and fourth terms, as follows.
## Graphing Factored Form
A quadratic equation is a polynomial equation of degree of 2. The standard form of a quadratic equation is
0 = ax^2 + bx + c
where a, b, and c are all real numbers and a ≠ 0.
If we replace 0 with y , then we get a quadratic Function
y = ax^2 + bx + c
whose graph will be a parabola.
The points where the graph intersects the x-axis will be the solutions to the equation, ax^2 + bx + c = 0. That is, if the polynomial ax^2 + bx + c can be factored to ( xp )( xq ), w e know by the zero product principle that if ( xp )( x q ) = 0, either ( x p ) = 0 or ( x q ) = 0. Then p and q are the solutions to the equation ax^2 + bx + c = 0 and therefore the x-intercepts of the quadratic equation.
Since the x-coordinate of the vertex is exactly the midpoint of the x-intercept, the x-coordinate of the vertex will be .
You can use the x-coordinate of the vertex to find the y-coordinate.
Now you have the vertex and 2 other points on the parabola (namely, the x-intercepts). You can use these three points to sketch the graph
## Example 1:
Graph the function using factoring.
Compare the equation with the standard form,y = ax^2 + bx + c . Since the value of a is positive, the parabola opens up.
Factor the trinomial, . Identify 2 numbers whose sum is –8 and the product is 12. The numbers are –2 and –6. That is,
.
So, by the zero product property, either x – 2 = 0 or x – 6 = 0. Then the roots of the equation are 2 and 6.
Therefore, the x-intercepts of the function are 6 and 2.
The x-coordinate of the vertex is the midpoint of the x -intercepts. So, here the x-coordinate of the vertex will be .
Substitute x = 4 in the equation to find the y-coordinate of the vertex.
That is, the coordinates of the vertex are (4, –4).
Now we have 3 points (4, –4), (2,0) and (6,0) which are on the parabola. Plot the points. Join them by a smooth curve and extend the parabola.
## Standard Form Of A Quadratic function
The quadratic equation is a formula which is used to find the zeroes(x-intercepts). It is equivalent to factoring and I recommend it if the equation is not easily factor-able
## Completing The Square
In this unit we consider how quadratic expressions can be written in an equivalent form using the technique known as completing the square. This technique has applications in a number of areas, but we will see an example of its use in solving a quadratic equation. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that all this becomes second nature. To help you to achieve this, the unit includes a substantial number of such exercises.
Completing The Square ~ Sabrina Singh
## Introduction To Standard
In this unit we look at a process called completing the square. It can be used to write a quadratic expression in an alternative form. Later in the unit we will see how it can be used to solve a quadratic equation.
Some simple equations
Example Consider the quadratic equation x^2 = 9. We can solve this by taking the square root of both sides:
x = 3 or − 3 remembering that when we take the square root there will be two possible answers, one positive and one negative. This is often written in the briefer form x = ±3. This process for solving x^2 = 9 is very straightforward, particularly because: • 9 is a ‘square number’, or ‘complete square’. This means that it is the result of squaring another number, or term, in this case the result of squaring 3 or −3. • x^2 is a complete square - it is the result of squaring x. So simply square-rooting both sides solves the problem.
Example Consider the equation x^2 = 5.
Again, we can solve this by taking the square root of both sides: x = √ 5 or − √ 5 In this example, the right-hand side of x^2 = 5, is not a square number. But we can still solve the equation in the same way. It is usually better to leave your answer in this exact form, rather than use a calculator to give a decimal approximation.
Suppose we wish to solve the equation (x − 7)^2 = 3
Again, we can solve this by taking the square root of both sides. The left-hand side is a complete square because it results from squaring x − 7. x − 7
= √ 3 or − √ 3 By adding 7 to each side we can obtain the values for x: x
= 7 + √ 3 or 7 − √ 3
We could write this in the briefer form x = 7 ± √ 3.
Example
Suppose we wish to solve (x + 3)^2 = 5
Again the left-hand side is a complete square. Taking the square root of both sides: x + 3 = √ 5 or − √ 5 By subtracting 3 from each side we can obtain the values for x: x = −3 + √ 5 or − 3 − √ 5
The basic technique
Now suppose we wanted to try to apply the method used in the three previous examples to x^2 + 6x = 4 In each of the previous examples, the left-hand side was a complete square. This means that in each case it took the form (x + a)^2 or (x − a)^2 . This is not the case now and so we cannot just take the square-root. What we try to do instead is rewrite the expression so that it becomes a complete square - hence the name completing the square. Observe that complete squares such as (x + a)^2 or (x − a)^2 can be expanded as follows:
We will use these expansions to help us to complete the square in the following examples.
Example
Consider the quadratic expression x^2 + 6x − 4
We compare this with the complete square x^2 + 2ax + a^2
Clearly the coefficients of x^2 in both expressions are the same. We would like to match up the term 2ax with the term 6x. To do this note that 2a must be 6, so that a = 3.
Recall that
(x + a)^2
= x^2 + 2ax + a^2
Then with a = 3 (x + 3)^2
= x^2 + 6x + 9 This means that when trying to complete the square for x^2 +6x −4
we can replace the first two terms, x^2 + 6x, by (x + 3)62 − 9.
So x^ 2 + 6x − 4
= (x + 3)^2 − 9 − 4
= (x + 3)^2 − 13
We have now written the expression x^ 2 + 6x − 4 as a complete square plus or minus a constant. We have completed the square. It is important to note that the constant term, 3, in brackets is half the coefficient of x in the original expression.
Example
Suppose we wish to complete the square for the quadratic expression x^2 − 8x + 7.
We want to try to rewrite this so that it takes the form of a complete square plus or minus a constant.
We compare x^ 2 − 8x + 7 with the standard form x ^2 − 2ax + a 2
The coefficients of x 2 are the same. To make the coefficients of x the same we must choose a to be 4.
Recall that (x − a) ^2
= x ^2 − 2ax + a 2 Then with a
= 4 (x − 4)^2 = x 2 − 8x + 16
This means that when trying to complete the square for x^ 2 −8x +7 we can replace the first two terms, x 2 − 8x, by (x − 4)^2 − 16.
So x^ 2 − 8x + 7
= (x − 4)^2 − 16 + 7
= (x − 4)^2 − 9 We have now written the expression x^ 2 − 8x + 7 as a complete square plus or minus a constant. We have completed the square. Again note that the constant term, −4, in brackets is half the coefficient of x in the original expression.
Example
Suppose we wish to complete the square for the quadratic expression x^ 2 + 5x + 3. This means we want to try to rewrite it so that it has the form of a complete square plus or minus a constant. In the examples we have just worked through we have seen how this can be done by comparing with the standard forms (x + a)^ 2 and (x − a) ^2 . We would like to be able to ‘complete the square’ without writing down all the working we did in the previous examples. The key point to remember is that the number in the bracket of the complete square is half the coefficient of x in the quadratic expression.
We have now written the expression x^ 2 + 5x + 3 as a complete square plus or minus a constant. We have completed the square. Again note that the constant term, 5/2 , in brackets is half the coefficient of x in the original expression. The explanation given above is really just an outline of our thought process; when we complete the square in practice we would not write it all down. We would probably go straight to equation (1). The ability to do this will come with practice.
Cases in which the coefficient of x 2 is not 1
We now know how to complete the square for quadratic expressions for which the coefficient of x ^2 is 1. When faced with a quadratic expression where the coefficient of x^ 2 is not 1 we can still use this technique but we put in an extra step first - we factor out this coefficient
Suppose we wish to complete the square for the expression 3x ^2 − 9x + 50. We begin by factoring out the coefficient of x ^2 , in this case 3. It does not matter that 3 is not a factor of 50; we can still do this by writing the expression as
and we have completed the square. This is the ‘completing the square’ form for a quadratic expression for which the coefficient of x ^2 is not 1.
Summary of the process
Solving a quadratic equation by completing the square
Let us return now to a problem posed earlier. We want to solve the equation x^ 2 + 6x = 4. We write this as x ^2 + 6x − 4 = 0. Note that the coefficient of x^ 2 is 1 so there is no need to take out any common factor. Completing the square for quadratic expression on the left-hand side:
We have solved the quadratic equation by completing the square. To produce equation (1) we have noted that the coefficient of x in the quadratic expression is 6 so the number in the ‘complete square’ bracket must be 3; then we have balanced the constant by subtracting the square of this number, 3 ^2 , and putting in the constant from the quadratic, −4. To get equation (2) we just do the arithmetic which in this example is quite straightforward.
We are now ready to derive and use the quadratic formula, which will allow us to solve all quadratic equations. We derive the formula by using the method of completing the square. To use the quadratic formula, the quadratic equation you want to solve must be in standard form. That form is ax^2+ bx+ c= 0 in which a is not 0
This unit is about the solution of quadratic equations. These take the form ax^2 +bx+c = 0. We will look at four methods: solution by factorization, solution by completing the square, solution using a formula, and solution using graphs In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
• solve quadratic equations by factorisation
• solve quadratic equations by completing the square
• solve quadratic equations using a formula
• solve quadratic equations by drawing graphs
This unit is about how to solve quadratic equations. A quadratic equation is one which must contain a term involving x^ 2 , e.g. 3x ^2 , −5x^ 2 or just x ^2 on its own. It may also contain terms involving x, e.g. 5x or −7x, or 0.5x. It can also have constant terms - these are just numbers: 6, −7, 1/ 2 . It cannot have terms involving higher powers of x, like x^ 3 . It cannot have terms like 1 x in it. In general a quadratic equation will take the form ax^2 + bx + c = 0 a can be any number excluding zero. b and c can be any numbers including zero. If b or c is zero then these terms will not appear.
In this section we will assume that you already know how to factorize a quadratic expression. If this is not the case you can study other material in this series where factorization is explained.
Example
Suppose we wish to solve 3^x 2 = 27. We begin by writing this in the standard form of a quadratic equation by subtracting 27 from each side to give 3x^ 2 − 27 = 0.
We now look for common factors. By observation there is a common factor of 3 in both terms. This factor is extracted and written outside a pair of brackets. The contents of the brackets are adjusted accordingly: 3x^ 2 − 27
= 3(x^ 2 − 9) = 0 Notice here the difference of two squares which can be factorized as 3(x ^2 − 9) = 3(x − 3)(x + 3) = 0
If two quantities are multiplied together and the result is zero then either or both of the quantities must be zero. So either x − 3 = 0 or x + 3 = 0
so that x = 3 or x = −3
These are the two solutions of the equation.
Solving quadratic equations by completing the square
Again we have two answers. These are exact answers. Approximate values can be obtained using a calculator.
Solving quadratic equations using a formula
Consider the general quadratic equation ax^2 + bx + c = 0.
There is a formula for solving this: x = −b ± √ b^ 2 − 4ac /2a . It is so important that you should learn it.
We will illustrate the use of this formula in the following example.
Example
Suppose we wish to solve x^ 2 − 3x − 2 = 0. Comparing this with the general form ax^2 + bx + c = 0 we see that a = 1, b = −3 and c = −2. These values are substituted into the formula
These solutions are exact
Suppose we wish to solve 3x ^2 = 5x − 1. First we write this in the standard form as 3x^ 2 − 5x + 1 = 0 in order to identify the values of a, b and c. We see that a = 3, b = −5 and c = 1. These values are substituted into the formula.
Again there are two exact solutions. Approximate values could be obtained using a calculator
Solving quadratic equations by using graphs
In this section we will see how graphs can be used to solve quadratic equations. If the coefficient of x 2 in the quadratic expression ax^2 + bx + c is positive then a graph of y = ax^2 + bx + c will take the form shown. If the coefficient of x ^2 is negative the graph will take the form .
Graphs of y = ax^2 + bx + c have these general shapes We will now add x and y axes. Figure 2 shows what can happen when we plot a graph of y = ax^2 + bx + c for the case in which a is positive.
Graphs of y = ax^2 + bx + c when a is positive
The horizontal line, the x axis, corresponds to points on the graph where y = 0. So points where the graph touches or crosses this axis correspond to solutions of ax^2 + bx + c = 0. In Figure 2, the graph in (a) never cuts or touches the horizontal axis and so this corresponds to a quadratic equation ax^2 + bx + c = 0 having no real roots. The graph in (b) just touches the horizontal axis corresponding to the case in which the quadratic equation has two equal roots, also called ‘repeated roots’. The graph in (c) cuts the horizontal axis twice, corresponding to the case in which the quadratic equation has two different roots
## Graphing Standard Form
STANDARD FORM: y = ax^2 + bx + c
Vertex: find the x-value of the vertex by using the handy dandy formula –b/2a Once you find the x-value, plug it into the equation to find the y-value.
Axis of symmetry: This is the line that is smack in the middle of the parabola – it is the mirror for a parabola. In standard form it is a vertical line in the form: x = –b/2a
y-intercept: This is an easy one – like always we find the y-intercept by plugging in x = 0. In this case that makes the y-intercept (0, c)
Other points: Plug in an x-value on one side or another of the vertex into the equation. If you know where the axis of symmetry is then you get some points for free by finding reflections over the axis of symmetry. Plot a point, count horizontally to the axis of symmetry, then keep on going the same number and place a point.
Ex: y = x^2 – 4x - 5
a = 1, b = -4, c = 5
Vertex x = 4 /2(1) = 2 plug in x = 2 Y = (2)2 -4(2) - 5 = 4 - 8 – 5 = -9 (2, -9)
axis of symmetry is x = 2 (through the vertex)
y – intercept (0, -5)
Plug in another point like x = -1 Which gives us (-1, 0)
Now we can count from the axis of symmetry From our two points to get two additional points: From (0,-5) to the axis you must travel 2 spaces to the right, so keep going two more spaces to get the point (4, -5). From (-1, 0) you must travel 3 spaces to the right, continue on 3 more spaces and come to (5, 0). Now sketch the graph through those points.
## Motion Problem
The height, , in feet of an object above the ground is given by h=-16t^2+64+190 where t is the time in seconds. Find the time it takes the object to strike the ground and find the maximum height of the object.
Let’s first find the time it takes for the object to hit the ground. Since represents the height above the ground, we would like to know at what time h=0. So in the equation h=-16t^2+64+190 we will set h=0 and solve for t.
h=-16t^2+64+190
0=-16t^2+64+190
So we simply want to solve this quadratic equation. It is easiest to use the quadratic formula in this situation. So we get
Therefore you get 5.98, -1.98
However, since t represents time, we must throw out –1.98. Therefore, it takes 5.98 seconds for the object to strike the ground.
The other part of the question is we want to know that maximum height that the object reaches. Since we can see that the function is clearly a quadratic function which opens down, we know that this maximum must occur at the vertex. So let’s find the vertex
t= -b/2a
so we get - 64/2(16)
=-16/-32
=2
So at t=2 seconds the object reaches its maximum height. However, we wanted to know what that maximum height is. Therefore, we must find the value of the vertex, in this case it will be the value h of t=2 when . So we plug this in to get
h=-16(2)^2+64(2)+190
= 254
Therefore the maximum is 254 feet.
## Revenue Problem
Revenue = (number of items sold) x (price of per item)
Castlebrooke's Athletic Council is selling tickets for their first ever FNL Football game.The current price of an amateur theater tickets is \$20, and the venue typically sells 500 tickets. A survey found that for each \$1 increase in ticket price, 10 fewer tickets are sold.
(a) What is the number of \$1 increases in price that will maximize the revenue?
(b) What price per ticket will maximize the revenue?
(c) What is the maximum revenue?
These parts asks for the number of \$1 increases, so, let x be the number of \$1 increases. The current revenue before the survey is R = (500) (20) because there are 500 items sold at \$20 each. Now add the survey results into the equation using x as the number of \$1 increases
• The price increases for each \$1 increments, so the price changes from (20) to (20 + 1x)
• For each \$1 increase, 10 fewer tickets are sold. Tickets sold changes from (500) to (500 - 10x).
The revenue equation based on the survey is R = (500 - 10x) (20 + 1x). Because the question says “maximize” we need to find the vertex of R = (500 - 10x) (20 + 1x) . Since the equation is already factored (although not fully factored) the fastest way to do this is by using the midpoint of the zeros. Solve each bracket for zero just like any other quadratic equation. The first zero is 500÷10=50, and the second is -20. The midpoint of these zeros is (50-20)÷2=15. This means that h = 15, and the x-value of the vertex is 15. In short, x = 15.
(a): The theatre should have 15, \$1 increases to maximize their revenue.
(b): The new price will be 20 + 1×15, or \$35 per ticket.
For part (c) we need to find the k value of the vertex. To do this, substitute x = 15 in the R equation, and calculate R. R=(500-10×15) (20+1×15)=12250
(c): maximum revenue will be \$12250.
## Numeric Problems
The sum of the squares of two consecutive integers is 365. What are the integers?
Let one number be x. Since the two numbers are consecutive, the other number is x + 1.
The sum of the squares is (x)^2 + (x + 1)^ 2 and it is equal to 365. The equation is (x) 2 + (x + 1)^2 = 365.
If you look at the question, it does not say “maximum”, “minimum” so we do not need the vertex. What we need to do, is to solve the equation (x)^2 + (x + 1)^2 = 365, by factoring or quadratic formula.
Expand and simplify the equation into standard form: 2x^2 + 2x – 364 = 0 Factor or use the quadratic formula to find the two solutions: x = -14, x = 13.
The two solutions to the equation will give us two possible sets of integers which are answers to the question: The first set of integers is -14 and -13, and the second set of integers is 13 and 14.
## Reflection
I don't really have any of my test of quizzes from quadratics but that doesn't mean I can't tell you what I've learned throughout the Quadratics from where I started to where I ended off. When coming into quadratics I had never even heard of the word. Now I know about the different degrees of functions. Which is when a polynomial is expressed as a sum or difference of terms, the term with the highest degree is the degree of the polynomial. The degree of a term is the sum of the powers of each variable in the term. In functions of a single variable, the degree of a term is simply the exponent. From learning about degree of difference of functions to the difference between successive terms of a sequence form another sequence of numbers. If these numbers are constant, the sequence is a 1st level difference. If these numbers are not constant and their difference form a sequence of constant numbers, then the sequence is a 2nd level difference. After we passed this and we went onto graphing I was able to understand how to do so because each for had a specific format which you had to follow and if you did that you will graph it right all the time. This is something I liked because finding the x-intercepts, axis of symmetry, vertex, and optimal value was very easy in my opinion.
## Math Learning Goals ( mini reflection of each part of quadratics)
Using Algebra tiles
• Connect the algebraic representations to the graphical representations of quadratics of the forms y = x 2 + bx + c and y = (x – r) (x – s).
• Expand and simplify second-degree polynomial expressions involving one variable that consist of the product of two binomials or the square of a binomial, using algebra tiles.
Multiply a Binomial
• Expand and simplify second-degree polynomial expressions involving one variable that consist of the product of two binomials or the square of a binomial, using the chart method, and the distributive property.
Find the y-Intercept of Quadratic Equations
• Determine the connections between the factored form of a quadratic relation and the x-intercepts of its graph.
• Factor simple trinomials of the form x 2 + bx + c, using a variety of tools, e.g., algebra tiles, paper and pencil, and patterning strategies.<|endoftext|>
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Sales Toll Free No: 1-800-481-2338
How to Find Open Intervals of A Function?
TopEvery function has a Domain which contains the values at which the function is defined. The domain can be called as a collection of various closed and open intervals. If the interval is closed then it is represented as: [x, y] and if it is open then we write it as: (x, y). An interval can be closed – opened like [x, y) and opened – closed(x, y]. We generally define a function not defined at critical points by finding their Derivatives. So, let us learn how to find open intervals of a function.
Example: Suppose we have a function x2 + 4 x + 3 < 0. Find its domain?
Solution: In the function we have a less than inequality. We have to keep in consideration this inequality as:
x2 + 4 x + 3 < 0
or x2 + 3 x + x + 3 < 0,
or x (x + 3) + (x + 3) < 0,
or (x + 1) (x + 3) < 0,
Now there can be two possibilities as follows:
Either x < -1 and x < - 3 or x > -1 and x > - 3
If we take the first case x < -1 and x < - 3 we get the domain as:
'x' belongs to: Intersection ((- infinity, -1), (- infinity, -3)) = (- infinity, -3) in open intervals and
If we take the second case x > -1 and x > - 3 we get the domain as:
'x' belongs to: Intersection ((-1, infinity), (-3, infinity)) = (-1, infinity) in open intervals.
Thus the open intervals of the function x2 + 4 x + 3 < 0 are defined as: (- infinity, -3) or (-1, infinity).
These open intervals means that the value of the function is not defined at x = -infinity, -3, 1 and infinity.<|endoftext|>
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The national curriculum for computing aims to ensure that all pupils:
- can understand and apply the fundamental principles and concepts of computer science
- can analyse problems in computational terms, and have repeated practical experience of writing computer programs to solve such problems
- can evaluate and apply information technology analytically to solve problems
- are responsible, competent, confident and creative users of information and communication technology.
At Golcar Junior, Infant and Nursery School, we seek to prepare our pupils for a world where the use of digital technology is regarded as an integral part of everyday practices.
Our aims are:
- To enable all our pupils to be confident, competent, and independent users of Information and Communication Technology (ICT). We aim to use ICT where appropriate to motivate and inspire pupils and raise standards across the curriculum.
- To develop every pupil’s ICT skills, knowledge, understanding and capability through taught ICT lessons and to provide opportunities for pupils to apply and consolidate their ICT capability across all curriculum contexts.
- To provide an environment where access to and use of ICT resources is natural, commonplace and enjoyable.
- To keep pace with educational developments and emerging technologies in ICT.
- To use ICT to extend and develop communication skills.
- To make children aware of inappropriate use of ICT and how to use ICT responsibly.
At Golcar Junior, Infant and Nursery School, Computing is taught through our cross-curricular topics, as well as discretely, to ensure thorough curriculum coverage. We use the Discovery Coding package to systematically develop pupils’ depth of knowledge and understanding of how computer programs are created and used. We aim to develop pupils coding abilities further by using the Scratch programme to stretch their capabilities. We teach the essential skills of using word processing, presentation and analytical programs within Maths, Science and other subjects when opportunities present themselves.
In our school, we take the online safety of our pupils as seriously as all other aspects of their safety and wellbeing. We inculcate the responsible use of technology from the earliest ages by using specially targeted, age-appropriate resources.<|endoftext|>
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It’s important for today’s young people to recall the history of the civil rights movement in America. Yet too many Millennials and members of their younger cohort, Generation Z, consider civil rights history as ancient history at the dawn of a new millennium.
However, as Americans and others pause on Monday to honor the life and legacy of Dr. Martin Luther King, Jr. (MLK), there are profound and poignant lessons which today’s young people need to learn. The most important lesson is how to make major changes in society through the type of peaceful means championed by Dr. King and his fellow civil rights leaders of the time.
A term of significance for young people to comprehend is: “civil disobedience.”
Dr. King staked his life and legacy on preaching non-violence, similar to that of Mahatma Gandhi during the independence movement of India, a country then controlled by British rule. In fact, Dr. King is said to have greatly admired and closely studied Gandhi’s successful strategy of non-violent opposition, which MLK emulated via the civil rights movement across the South.
MLK promoted civil disobedience in the face of vicious police brutality and mass jailings of peaceful demonstrators — including himself — which were commonplace back then.
Similarly and tragically, both Dr. King and Gandhi met their untimely deaths at the hand of an assassin’s bullet. This is the ultimate price to pay for fostering peace and freedom on a grand scale. Moreover, unlike some black leaders of the 1960s who heeded calls for violence from militant groups, like the Black Panthers, Dr. King persevered with a solid strategy of civil disobedience.
Dr. King’s steadfastness and perseverance paid off through the enactment of groundbreaking civil rights laws that altered the course of American history.
Therefore, more young people should be taught to leverage peaceful means of protest championed by Dr. King, via the constitutional guarantees of free speech, freedom of expression, and freedom of assembly. These lawful tactics of non-violent resistance are what ultimately resulted in historic gains via the Civil Rights Act of 1964 and the Voting Rights Act of 1965.
These landmark civil rights laws changed America for the better and ushered in a new era of increased equality and opportunity for minority groups.
While the sweeping civil rights laws of the 1960s obviously did not cure all societal ills, they have certainly had a long-term positive impact on the fabric of America. Thus, today’s teens and 20-somethings who might be prone to violence and knee-jerk reactions during police confrontations need to recall, abide by and honor the legacy of non-violence taught by Dr. King.
Dr. King referred to non-violence as “a sword that heals.” He said, for example: “Nonviolence is a powerful and just weapon which cuts without wounding and ennobles the man who wields it.”
Dr. King, current Congressman John Lewis, and other civil rights leaders of the 1960s persistently tested the nation’s conscience about racism, bias and bigotry. Their unwavering discipline and fortitude through strict adherence to non-violence is why minds were changed and historic progress was made.
Graphic TV video and news photos of peaceful protesters being beaten bloody by police, hosed down by water cannons, and attacked by police dogs caused most whites to take a hard look in the mirror when pondering such outrageous over reactions by law enforcement — actions which ultimately backfired.
Congressman Lewis, then a young civil rights leader, was nearly beaten to death by police during a pivotal civil rights march in Alabama that became known as “Bloody Sunday” — a common story exemplifying the unjust times.
That’s why more Millennials and Gen Z need to realize that non-violence was the core foundation of Dr. King’s effective leadership and ability to alter the course of American history for the better.
Despite recent racial progress made — such as the election and reelection of America’s first black president — the civil rights struggle is far from over. There’s still too much discrimination based on race, color and a host of other factors, from the workplace to every place in America.
In hindsight, many citizens of every race, color and creed had sincerely hoped and believed that the 2008 election of President Barack Obama would result in a post-racial society. But this promise has failed to materialize, despite a new generation of young people who tend to look beyond the lens of race.
In fact, some leading black scholars and influencers, such as Tavis Smiley for example, have pointed out that blacks are allegedly no better off economically today than they were in 2008 when President Obama took office.
Mr. Smiley, a best-selling author who hosts his own talk show on PBS, has said: “Sadly, and it pains me to say this, over the last decade black folk in the era of Obama have lost ground in every major economic category.”
The ugly truth is that the scourge of racial bias is still a persistent problem in too many aspects of modern day society.
Perhaps what has changed most is that racism, bias and bigotry are more subtle and less overt today compared to prior times. Many point to so-called “unconscious” or “unintentional” discrimination at the heart of some in white America.
But there’s nothing unconscious or unintentional about burning down black churches or police killing unarmed black youth, among other things grotesquely witnessed nationwide in recent years.
Let’s also recall that there’s nothing lawful or morally right about minorities discriminating against whites based on race, especially considering that race is supposed to be “color blind” under the law.
Moreover, racial discrimination is equally abhorrent whether it’s directed at light skinned blacks by darker skinned blacks, whites against blacks, Hispanics against blacks, and/or blacks and other traditionally known minority groups against whites.
I would note that as racial and ethnic diversity greatly increases among the U.S. population, the demographic of white Americans remains mostly stagnant. And the population of white men is actually shrinking. While this might be reason enough for some to cheer, there’s never a justifiable reason for discrimination against anyone.
In fact, most major U.S. cities currently have a so-called “minority-majority” population. This means the combined number of traditional minority groups now outnumbers that of whites. Yet this should never be a purported justification for so-called “reverse discrimination.”
True equality means not discriminating against any individual based on race — period!
Thus, the big question arises:
This is a critically important question to consider as we observe the federal holiday honoring the life and legacy of MLK.
In essence, we must all ask ourselves: where do we go from here, and how?
What strategies should a new generation of young leaders leverage to create the kind of society in which all people are judged on the content of their character and not the color of their skin, as Dr. King spoke of half a century ago?
Are the answers simply too elusive in today’s increasingly diverse multiracial, multiethnic and multicultural world?
Note: This blog post first appeared on beBee.com Affinity Social Network — “Successful Personal Branding”
ABOUT THE AUTHOR: I’m an independent writer and strategic communications advisor for large employers. My background of 20+ years of experience in the public and private sectors includes work in the White House, Congress and national news media. I’m also a Brand Ambassador for beBee Affinity Social Network. You can also find me buzzing on beBee Twitter LinkedIn.
NOTE: All views and opinions are those of the author only and not official statements or endorsements of any public sector employer, private sector employer, organization or political entity.
Originally published at medium.com<|endoftext|>
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Child Development: Activities for Everyday Brain Development and Building
What fosters healthy brain development in children? Learn fun brain building activities for babies, toddlers, and preschoolers.
The first five years of life are when a child’s brain development is the fastest and when more than 700 neural connections are being formed every single second. What fosters healthy development during these critical early years?
Positive interaction is essential between children and the adults who care for them. Every time we connect with children, it’s not just their eyes that light up—it’s their brains, too. Positive early experiences with adults strengthen the connections that a child builds up and help children to be eager, engaged, and ready for a lifetime of learning. This is called brain building, which doesn’t require extra money or extra resources. So, how can you help your children have productive interactions? That’s where Vroom comes in.
Child Brain Development and Building Basics
Bright Horizons has collaborated with Vroom to bring you fun activities to do with your children. Vroom offers more than 1,000 activities and tips for families with children ages 0-5 that fit into everyday routines. Each activity is based on what Vroom calls the five Brain Building Basics. look, follow, chat, take turns, and stretch.
Think about the experience of parenting as slowly releasing responsibility over time, starting when your child is in preschool. Your child should learn from an early age that you are her best advocate and cheerleader. At the same time, it's your job to keep her safe and healthy by setting reasonable limits. Mutual respect, understanding, and cooperation guide every interaction.
Look: Even before babies can talk, they’re showing you what they’re interested in. Look into their eyes, or what catches their eye, and begin to build with your child.
Follow: Powerful moments are created when you let children lead the way and you follow by responding to their words, sounds, actions, and ideas.
Chat: It may not seem like it, but the sounds and gestures young children make are their way of communicating with you. So, talk out loud together—even if they can’t talk yet—and keep chatting as your children grow to engage them in learning about the world around them.
Take Turns: Back and forth interactions between you and your child are one of the most important ways to help development. So be sure to take turns while you’re talking, playing, or exploring with your children.
Stretch: Make the moment last longer by building on what your child says, or asking follow-up questions that expand your child’s thinking and learning. When you stretch the conversation with questions like, “What do you think about that?” or “How does that make you feel?” you’re stretching the building moments as well.
Everyday Brain Development & Building Activities for Children
Here are a few Vroom activities you can try with your children.
Infant and Toddler Brain Development & Building Activities
Stair Count: When your child is learning to walk up and down stairs, hold hands, and count each step you take together. This will help your child become familiar with numbers and think it's a fun game at the same time! By making connections that numbers aren’t just words to memorize, your child is beginning to learn math.
Food Rhymes: During a meal or snack, create a rhyme or a rap about what your child is eating: “No slice, no dice, we eat rice!” or “(Your child's name) is no rookie, eating her cookie.” Your child will enjoy the sound of the words, and if she responds, make rhymes from her words, too. Listening to rhymes—and making up their own—makes mealtime fun while helpings your child develop communication skills.
Water Works: Hand-washing or bath time? Say to your child, “This is the HOT water (point to the faucet); this is the COLD water. Together they make warm water! This is the soap. Soap and water make BUBBLES that clean our hands. Now, let’s rinse off the bubbles. Can you help me dry my hands with this towel?” Through sharing the science of how things work, you’re helping your child learn a routine to stay healthy!
Preschooler & Pre-K Brain Development & Building Activities
Fruit Facts: When picking out fruit for a snack, play a guessing game. "Is this orange sweet or sour? Is this apple crisp or mushy?" Use words your child might not understand yet and ask him to use his own words to describe tastes. When you eat the fruit later, talk about the guesses. By using new words and asking questions, you help build your child’s vocabulary and their reasoning skills.
Beat & Repeat: Create a beat with two claps, and repeat. Can your child copy it? Have your child create beats with two claps. Anytime someone misses, try again. Then make it three beats. How many beats can you get to? Four? Five? Six? This fun activity can be done anytime, anywhere! As your child copies your patterns and creates their own, they’re developing self-control and learning from their mistakes!
Guess Who: Ask your child to think of a family member or friend without telling you who. Have him give you hints until you guess who the mystery person is. Then it’s your turn to play the same game with him. This guessing game helps your child develop critical thinking skills by figuring out the important clues.
Webinar: How Does My Child Learn?
Do you know how growth, learning, and development happen in your child? Watching your little one recognize letters or count fingers and toes will make you proud, but how much do you know about what’s behind that learning?
Access this parenting webinar to get the scoop on the science of learning, and come away with simple activities and ideas you can do to encourage your child’s developing mind.
Want more tips and activities for promoting child development everyday through play? Visit joinvroom.org or download the Daily Vroom app—available on the Apple Store, the Google Play Store, and the Amazon App Store—to receive customized Vroom tips for your child.
Vroom, an initiative of the Bezos Family Foundation, translates the latest brain science research into everyday language and makes it actionable for parents. It takes the science out of the lab and puts it directly into the hands of those most poised to act: parents. Vroom shows parents that the moments they already share with their children—like bath time and meal time—can be brain building moments.
More on Brain Development & Building Activities for Children:
- Access our parenting webinar, The Essentials of Learning Through Play, to find out and learn all about the value and importance of play.
- Find tips and opportunities for encouraging your toddler’s growth and development.
- Discover ways to foster imagination, creativity, and ingenuity in your child.
- Browse our at-home learning activities and find fun ways to extend your child's learning opportunities beyond the classroom.<|endoftext|>
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INTRODUCTIONEvery day men and women are confronted with identical problems like how to get to the store. The way they go about solving these problems can be totally different. For example, when looking for a store, a woman may be more prone to call a friend and ask for directions, while a man may stubbornly drive around for hours until he finally arrives at the store. These differences in the ways in which males and females solve problems have lead to much confusion. Simply stated, neither side thinks like the other, nor understands how or why they think that way. Many studies have been conducted in attempts to clarify this gap between the sexes, and most have found that the differences stem from one main idea: Different languages exist between men and women (Sachs, 2001). This dilemma was escalated by the stereotypes produced by our societies and our communication styles (Sachs, 2001), both of which worked together to increase the distances and differences in help seeking behavior between genders. Men and women were born into the same world, but from the beginning society showed us different ways to live (Sachs, 2001). We were taught ideas about ourselves and others regarding the ways we should act and think, which were often referred to as stereotypes (Tannen, 1993). Women like to gossip and men like to talk about sports were examples of stereotypes that appeared in our everyday view so often that we just began to accept them without question, which could have both positive and negative effects (Wisch, Mahalik, Hayes, & Nutt, 1995). In a recent study, researchers attempted to discover the influence these positive and negative sides of stereotypes had on males regarding their help seeking behavior (Wisch et al., 1995). They hypothesized that men who accepted the traditional male role with its restrictions on the display of emotion and affectionate behavior among men would be less likely to seek psychological help (Wisch et al., 1995). The researchers first surveyed the men to see how strongly they believed in the traditional male gender role, and then observed them to discover how much time would need to elapse and what circumstances would have to be present for the men to seek psychological help. The results not only confirmed their original ideas, but also found that men who strongly believed in their gender roles felt uncomfortable even witnessing another male weakly express his feelings to a psychologist (Wisch et al., 1995).If society helped produce different stereotypes, then stereotypes helped produce different communication styles. Researchers say that we spend 70 percent of our awake time communicating, and 30 percent of our communication is talking (Sachs, 2001, p.1). For men, this talking time was usually defined as report, with the purpose of information. Conversations usually included minimal, direct facts that had a goal in mind. Women, on the other hand, usually talked using rapport, with the purpose of connection. A conversational piece was more often a person than an object with the attempt of effecting relationships (Sachs, 2001). To document how these communication variations played into help seeking behavior, a researcher gave preschoolers puzzles and recorded their reactions (Thompson, 1999). He found that boys used more object-orientated language, and girls used more self-orientated language (Thompson, 1999). As expected, the study also showed that the girls asked for help significantly more than the boys. Two other very interesting discoveries were made. The boys requested assistance less than the girls, but the rate at which they did increased greatly as the difficulty of the tasks increased. Also, even with the differences in help seeking behavior, the average boy and girl had extremely close finish times for the puzzles (Thompson, 1999).These studies, along with our knowledge of society, stereotypes, and communication styles, helped explain and support the differences between males and females when it came to help seeking behavior. The most obvious form of help seeking behavior was asking questions. It has been documented on many occasions that women were more willing to seek help on anything from directions to interior designing (Thompson, 1999). In fact, women asked questions nearly twice as much as men did during an average conversation (Tannen, 1994). For women, though, this was socially acceptable and normal since they used questioning as a connection rather than a tool. Men, on the other hand, were supposed to be the independent ones, so they asked less questions because their public faces were important to them (Tannen, 1999).We have substantial amounts of information on stereotypes and communication ranging across different times and different cultures. Regarding help seeking behavior, many studies have been done like the previously discussed ones on preschoolers and males. But, to better complete psychological knowledge of the way we work, more modern studies need to be conducted comparing adult men and adult women and their differences in help seeking behavior. This more modern research is especially important since we are in an era where society and stereotypes that define communication styles are rapidly changing.In this study, we attempted to look at gender differences in help seeking behavior. We did an archival study using episodes of the primetime show “Who Wants to Be a Millionaire.” This research expanded on past studies because it not only compared adult men and women, but did so using an entirely distinct and modern tool (“Who Wants to Be a Millionaire”) that reflects where our society is today in its technology and goals. Our study went a step further than putting together a puzzle. Because we did archival research, we were able to carefully review and document this help seeking behavior on charts that noted sex, hometown, which lifelines they used on which questions, how far they went, and how much money they won. Based on all the research on society, stereotypes, and communication styles in reference to help seeking behavior, we expected that women would use their lifelines earlier and be more likely to phone a friend first, while men would use their lifelines later and be more likely to use the 50/50 and poll the audience first.
The participants were 40 contestants, 20 males and 20 females, that we watched on the television show “Who Wants to Be a Millionaire.” These contestants were all above the age of eighteen, usually ranging from mid 20s to mid 40s. Our research was slightly biased in the fact that the contestants who played had to be intelligent in order to even appear on the show. We had no control over this, though, since the screening was conducted by the television network before the contestants participated in the show. Besides the intelligence factor, the contestants were all average people. In other words, no celebrities were included in our study. We omitted them because they were playing for charities, and therefore allowed to cheat. This cheating would inevitably affect the use of their lifelines, and, in the end, contaminate our results.
The materials used were 15 episodes of the television show “Who Wants to Be a Millionaire” and a chart we designed. “Who Wants to Be a Millionaire” was a game show where the contestants answered a series of questions in an attempt to win 1 million dollars. The questions escalated in difficulty and the amounts of money they were worth. Along with each question came 4 multiple-choice answers. In case the contestant did not know the answer, they were provided with 3 lifelines (50/50, ask the audience, and phone a friend) to assist them. Our other main material was the chart we constructed. At the top of the page on the left side were blanks for the sex of the contestant, their name, and their hometown. At the top on the right side were blanks for the tape number (for organizational purposes) and the amount of money the contestant won. The bottom half of the page contained a chart with numbers 1-15 down the left side, which represented the levels of the game that could be reached. Across the top of the chart were columns that designated spaces for the type of lifeline used (50/50, ask the audience, or phone a friend), a copy of the exact question they were used on, the category of the question (entertainment, sports, science and nature, history, literature, or current events), and whether they answered the question correctly. The chart also helped promote inter-rater reliability since both observers viewed the episodes and recorded the contestants’ actions on the charts. For an example of the chart, view the appendix.
DESIGN AND PROCEDURE
Our design was quasi-experimental since the main variables we used were sex differences, which cannot be actively manipulated by the experimenters. Our research was also considered archival since we used previously documented data, episodes of “Who Wants to Be a Millionaire,” to conduct our study. Recording of this information took place during a two-month period at the researchers’ convenience. The variables noted from the show included the sex of the contestant, name, hometown, award received, lifelines used (50/50, ask the audience, and phone a friend), the questions they were used on, the categories of the questions (entertainment, sports, science and nature, history, literature, and current events), and whether they answered the questions correctly. Our main focus was the sex of the contestants. Based on the gender differences in help seeking behavior noted in previous research, we concluded that males and females would ask for help differently on the questions they had difficulty answering on “Who Wants to Be a Millionaire.” Specifically, we said that since women were more likely to ask for help, they would use their lifelines earlier than men who were more likely not to ask for help, and therefore would use their lifelines later. When the participants utilized their lifelines, this represented them asking for help, since this forced them to admit they could not do it alone and needed to lean on someone else’s knowledge to answer the question correctly. We also used the types of lifelines to reflect the types of help the contestants asked for. The 50/50 option represented an anonymous type of help seeking because the source was a computer who could not hold the contestant accountable. The ask the audience option represented an impersonal type of help because the sources were humans, but they were strangers that the contestant did not interact with. The phone a friend option, on the other hand, represented a personal request for help, since the source was a close friend whom the contestant probably interacted with frequently. We also noted some other variables to see if they provided any other patterns and insights into our results. We included hometown in the chance that maybe the area one lived in had an effect on the time it took for a lifeline to be used. For example, maybe people from the south are more comfortable with asking for help, and would therefore use their lifelines before someone from the north. We also included the question the lifelines were used on and the category they fell into. Many connections could be made with this information, like maybe males know more than females about current events so questions in this area encouraged less lifeline usage for the males. Or the categories could be linked to hometown, where southerners might use more lifelines on questions involving the stock exchange while northerners might use more on questions involving farming. The amount of money the contestants won was also documented because a connection might be found between the bigger winners and the timing and types of lifelines used. Along with the many variables, there was a control that concerned the contestants we used in our study. We decided in the beginning to omit celebrities from our research. Since the celebrities were usually playing for money to donate to charities, the show allowed them to cheat. We felt that this cheating would cause the contestants to change their lifeline usage by utilizing them much later or maybe not utilizing them at all, and inevitably confound our results. The researchers viewed the episodes of “Who Wants to Be a Millionaire” together and documented the information on the charts. When a contestant was introduced, their sex, name, and hometowns were written in the appropriate blanks at the top of the page. As the contestant played, checks were placed in the boxes in the charts next to which questions they answered right at which level and which lifelines they used at which level. After a lifeline was used, the exact question it was used on was written down along with the corresponding category. After the contestant’s turn was over, the amount of money they won was recorded at the top of the page. Due to the pre-selected contestants who were not in any risk or any way affected by our study, there was no need for informed consent or debriefing.
RESULTS We conducted a study to see if gender differences existed in the help seeking behavior of participants on the television show “Who Wants to Be a Millionaire.” The relationship between sex and the earliest signs of help seeking behavior was not significant. The males (M=9.74, SD=2.57) and the females (M=9.02, SD=2.58) did not differ significantly on the level they first used their lifelines t(13)=5.11, p=.618 (two-tailed). The relationship between sex and the type of help asked for first was also not significant. The males (M=11.14, SD=1.35) and the females (M=9.575, SD=2.14) did not differ significantly on whether the phone a friend option was used before the poll the audience and the 50/50 options t(11)=1.63, p=.131 (two-tailed). Overall, there was no significant difference by gender on help seeking behavior involving the contestants on “Who Wants to Be a Millionaire.” We were able to note, though, that the number of men contestants outnumbered the amount of women greatly (2:1), and the difference in the dollar winnings earned by males (M=92,900, SD=88,150) and females (M=52,200, SD=45,300) was not statistically significant t(13)=.958, p=.355 (two-tailed). Also, the difference in the level of difficulty reached for males (M=11.70, SD=1.25) and females (M=10.80, SD=1.92) was not statistically significant t(13)=1.10, p=.290 (two-tailed).
DISCUSSION Major differences in communication styles and the sexes still exist today. The purpose of this study was to explore gender differences in the area of help seeking behavior, a type of communication, as seen on the television show “Who Wants to Be a Millionaire.” It was hypothesized that women would use their lifelines earlier and be more prone to utilize their phone a friend option first, while men would use their lifelines later and be more prone to utilize their 50/50 and ask the audience options first. The results found, though, did not support this proposed hypothesis. A difference in sex did not have a significant effect on who used their lifelines first nor which type of lifeline was used first. These findings were inconsistent with the literature produced by Thompson’s research (1999) regarding the preschoolers, the puzzles and help seeking behavior. This difference could be a result of validity. Thompson conducted his experiment using a more normal situation that the preschoolers encountered daily, putting together a puzzle. Also, his tests were done in a more controlled environment with more similar participants that allowed the help seeking behavior variable to be the focus of his research. In our study, the setting was the show “Who Wants to Be a Millionaire,” which was hardly normal or a situation that people encountered daily, so many outside variables could have contributed to the difference between our findings and Thompson’s. Also, our participants varied greatly in age, ethnicity, occupation, and background, which could have greatly affected our insignificant results regarding our hypothesis. In the future, experimenters might consider conducting the study in slightly more controlled environments for the sake of internal validity. When our research was also compared with Tannen’s literature (1994), it was inconsistent. Tannen said that men and women have different communication styles and preferences towards the type of help they use. In our study we used the types of lifelines used by each contestant on “Who Wants to Be a Millionaire” to reflect these ideas. But we did not find a significant difference between men and women. This could be because the lifelines were not sufficient representations of the concepts that Tannen wrote about. Maybe the contestants did not view the 50/50 as help from the computer that could not hold them accountable, nor view the phone a friend as direct help from a peer who could hold them accountable. This difference in our views and the contestants could have kept the participants from utilizing the lifelines in the manners that we predicted because they did not see them in the light that we based our study on. The next experimenters should try to use a tool that is more defined and understood by both sides of the study so the predictions and findings can be more accurate.This research did find “Who Wants to Be a Millionaire” to be a fair game show in the sense that there was no statistically significant difference between the amount of money men and women won and the levels of difficulty they reached. Although the means for the money the men and women won did differ a good amount, the standard deviations from these means also differed by a lot. In other words, although the means were far apart, they were not good representations of the groups as wholes because the scores were not normally distributed and covered such a wide range of scores. Our study did have some strength regarding the reliability of our measuring instrument. The chart we constructed was very easy to record the information on and read the information from. Because both experimenters watched the episodes together, recorded what we saw on the charts, and discussed our observations to make sure there were no discrepancies, our inter-rater reliability was high and not biased. Also, our simple chart contributed to our simple procedure, which would make this study very easy for other experimenters to replicate in the future. If other experimenters did decide to replicate our study, a few dilemmas were found that could be improved upon. The first problem arose with the initial acquiring of the episodes of “Who Wants to Be a Millionaire.” We attempted to tape them off of the television whenever they came on. We soon realized that this was a bad idea when the station stopped airing the episodes on a regular schedule and began to just play them at random. Praying that the show would be aired on a night when we could tape it was not a sufficient way to obtain an asset so essential to our experiment. Along with this first dilemma came our second major problem, a lack of participants. We originally estimated that we needed at least 40 participants (20 males and 20 females) to conduct our study. Due to the trouble we had recording “Who Wants to Be a Millionaire,” we only obtained 5 episodes, which in turn meant only 15 participants (10 males and 5 females). Not only were our numbers few, but they were unbalanced. With more episodes and more participants we could have had more of a variety and a larger subject pool to mix with our variables in the hope of finding something statistically significant. In the future, if experimenters did attempt to replicate our study, they should find a more reliable way of acquiring the episodes like contacting the television station.Communication differences do still exist between men and women today. The more studies that are conducted regarding help seeking behavior and subjects like it, the more likely we will be able to uncover these differences and better understand each other. Bridging these gaps will promote healthier relationships on all levels. Although no statistically significant information was found through this study, we hope that other experimenters will be able to read it and learn from our mistakes so that more advances can be made in the ever-prominent field of communication.
REFERENCES Sachs, M. A. (2001). Ohio state university fact sheet: Male/female communication styles. Retrieved September 11, 2001, from http://www.ag.ohio-state.edu/~ohioline/hyg-fact/5000/htmlTannen, D. (1993). Gender and conversational interaction. New York: Oxford University Press.Tannen, D. (1994). Gender and discourse. New York: Oxford University Press.Tannen, D. (1999). Gender styles. Retrieved September 24, 2001, from http://www.usm.maine.edu/com/genderlect/index/htmThompson, R. B. (1999). Gender differences in preschoolers’ help-eliciting communication. Journal of Genetic Psychology, 160, 357-368. Wisch, A. F., Mahalik, J. R., Hayes, J. A., & E. A. Nutt. (1995). The impact of gender role conflict and counseling technique on psychological help seeking in men. Sex Roles: A Journal of Research, 33, 77-85.
APPENDIX Data Sheet
Sex: Tape #: Name: Award: Hometown: Life Line Used Question Category Correct Incorrect1 2 3 4 5 6 7 8 9 10 11 12 13 14 15<|endoftext|>
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## Number Series Questions With Solution For SBI Clerk
Number Series Questions With Solution For SBI Clerk.If you are preparing for Bank and other competitive exams, you will come across Number series section on Quantitative Aptitude. If you are preparing for Bank and other competitive exams, you will come across Number series section on Quantitative Aptitude.
Today, we are providing you with Quant Quiz on Number Series based on the latest pattern for your daily practice. This Quant subject quiz based number series is important for other banking exams such as SBI Clerk, SBI PO, IBPS PO, IBPS Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, and other competitive exams.
## Number Series Questions With Solution Set – 23
1.The following numbers form a series. Find the odd one out.
1, 6, 21, 52, 104, 186
A 6
B 52
C 104
D 186
E 21
C. 104
13 + 1 – 12 = 1 + 1 – 1 = 1
23 + 2 – 22 = 8 + 2 – 4 = 6
33 + 3 – 32 = 27 + 3 – 9 = 21
43 + 4 – 42 = 64 + 4 – 16 = 52
53 + 5 – 52 = 125 + 5 – 25 = 105
63 + 6 – 62 = 216 + 6 – 36 = 186
Therefore, 105 should be in place of 104.
2.The following numbers form a series. Find the odd one out.
14, 30, 48, 68, 90, 118
A 118
B 30
C 68
D 90
E 48
A. 118
13 × 1 + 12 = 13 + 1 = 14
13 × 2 + 22 = 26 + 4 = 30
13 × 3 + 32 = 39 + 9 = 48
13 × 4 + 42 = 52 + 16 = 68
13 × 5 + 52 = 65 + 25 = 90
13 × 6 + 62 = 78 + 36 = 114
Therefore, 114 should be in place of 118.
3.The following numbers form a series. Find the odd one out.
1152, 448, 288, 112, 96, 30, 18
A 448
B 112
C 18
D 30
E 288
D. 30
210 + 27 = 1024 + 128 = 1152
29 – 26 = 512 – 64 = 448
28 + 25 = 256 + 32 = 288
27 – 24 = 128 – 16 = 112
26 + 23 = 64 + 32 = 96
25 – 22 = 32 – 4 = 28
24 + 21 = 16 + 2 = 18
Therefore, 28 should be in place of 30.
4.The following numbers form a series. Find the odd one out.
-7, 37, 15, 131, 105, 337
A 37
B 15
C 131
D 105
E 337
B. 15
(13 + 22) – (3 × 4) = 5 – 12 = -7
(23 + 32) + (4 × 5) = 17 + 20 = 37
(33 + 42) – (5 × 6) = 43 – 30 = 13
(43 + 52) + (6 × 7) = 89 + 42 = 131
(53 + 62) – (7 × 8) = 161 – 56 = 105
(63 + 72) + (8 × 9) = 265 + 72 = 337
Therefore, 13 should be in place of 15.
5.The following numbers form a series. Find the odd one out.
0, 1, 5, 23, 119, 720
A 23
B 1
C 119
D 720
E 5
D. 720
0 + 1 × 0 = 0
0 + (0 + 1) × 1 = 0 + 1 = 1
1 + (1 + 1) × 2 = 1 + 4 = 5
5 + (5 + 1) × 3 = 5 + 18 = 23
23 + (23 + 1) × 4 = 23 + 96 = 119
119 + (119 + 1) × 5 = 119 + 600 = 719
Therefore, 719 should be in place of 720
6.What should come in place of the question mark (?) in the following number series?
14, 16, 22, 34, 54, ?
A 34
B 54
C 22
D 84
E None of these
D. 84
The series follows the pattern:
14 + (12 + 1) = 16,
16 + (22 + 2) = 22,
22 + (32 + 3) = 34,
34 + (42 + 4) = 54,
54 + (52 + 5) = 84
7.What should come in place of the question mark (?) in the following number series?
125, 126, 135, 160, 209, ?, 411
A 290
B 266
C 302
D 360
E None of these
A. 290
The series follows the pattern:
125 + 12 = 126,
126 + 32 = 135,
135 + 52 = 160,
160 + 72 = 209,
209 + 92 = 290,
290 + 112 = 411
8.What should come in place of the question mark (?) in the following number series?
140, 145, 153, 165, 183, ?
A 248
B 194
C 207
D 212
E None of these
C. 207
The series follows the pattern:
140 + (2 + 3) = 145,
145 + (3 + 5) = 153,
153 + (5 + 7) = 165,
165 + (7 + 11) = 183,
183 + (11 + 13) = 207
9.What should come in place of the question mark (?) in the following number series?
12, 23, 49, 93, 193, ?, 765
A 558
B 340
C 608
D 377
E None of these
D. 377
The series follows the pattern:
12 × 2 – 1 = 23,
23 × 2 + 3 = 49,
49 × 2 – 5 = 93,
93 × 2 + 7 = 193,
193 × 2 – 9 = 377,
377 × 2 + 11 = 765
Hence, option d.
10.What should come in place of the question mark (?) in the following number series?
-1, 5, 19, 41, ?, 109
A 98
B 50
C 82
D 81
E None of these
D. 81
The series follows the pattern:
12 – 2 = -1,
32 – 4 = 5,
52 – 6 = 19,
72 – 8 = 41,
92 – 10 = 71,
112 – 12 = 109<|endoftext|>
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Most of our posts include causal loop diagrams because some things are better expressed with a visual model than in words alone. Systems thinking takes on complex, dynamic systems and how they behave over time, which calls for a different sort of language. This quick tutorial will teach you the basics about reading causal loop diagrams through a Population model.
Causal loop diagrams consist of variables (things, actions or feelings) connected by causal links (arrows) with polarities (+ and – signs) and delays (||). Together, these create positive and negative feedback loops that describe the circles of cause and effect that take on a life of their own. Let’s hop into our example to make this more concrete.
The two things that cause the Population to change are Births and Deaths, so we use arrows to represent these causal links. We know that more Births lead to a greater Population, and fewer Births will lead to a lower Population, all else equal. We would say this relationship has a positive polarity, meaning that the two variables move in the same direction: more leads to more, or less leads to less. We indicate that a causal relationship has a positive polarity by placing a + sign next to the arrow head.
We also know that more deaths lead to a lower Population, and fewer deaths lead to a greater population. The variables move in the opposite direction, more leads to less, or less leads to more, so we would say that this relationship has a negative polarity. We represent this by labeling the arrow head with a – sign.
These causal links are true independently, and they are also both true at the same time. On their own, they don’t tell us what’s actually happening to the population. The direction of change in population is determined by whichever of these two relationships is dominant. As long as births exceed deaths population will grow, and whenever deaths exceed births, the population will shrink.
Now let’s introduce some Feedback into the model. While more Births lead to a greater Population, a greater Population also leads to more Births since more people make more babies, given a birth rate stays constant (this is why we say “all else equal”, because we only consider the two variables that we are linking when we think about polarity). Therefore, we draw a positive causal link from Population back to Births.
This link forms our first feedback loop, shown on the left side of the image above. A feedback loop is what we call a set of relationships where one variable leads to a change in another variable that eventually leads to a change in the original variable. To read a feedback loop, you pick a variable to start with and arbitrarily pick a direction – either More or Less.
So let’s read this one starting with Population and More. More population leads to more births which leads to more population. This is called a Reinforcing Feedback Loop (marked with an R) because more births today lead to more births in the future – births reinforce births. Similarly, less Births would lead to a lower Population which would lead to less births in the future; the reinforcing process works in the opposite direction too. If this were the only Feedback Loop in the Population system and people did not die, then we would see exponential growth in the number of people.
We see a different type of Feedback Loop when we examine Deaths. More Deaths today leads to fewer Deaths in the future. This is because more deaths today will cause the population to fall, which means less people will be around to die later. These types of loops are called Balancing Feedback Loops (marked with a B) since more leads to less or less leads to more – the original change is balanced by a change in the opposite direction.
Feedback loops take on a life of their own. We see a set of relationships that are always happening over and over again generating behavior that unfold over time. These two feedback loops can cause a few different behaviors based on the the birth rate and life expectancy – we will observe Population growing and growing ever faster as long as the reinforcing Births loop dominates, and leveling off if the Deaths balancing loop is dominant.
Notice in the image above that there are two hash marks, || , on the causal links between Population and Births and between Population and Deaths. Hash marks represent a Delay, a situations where it takes time before the effect plays out. It takes time for an individual to be of-age to have a child which is why there is a delay between Population and Births. This delay is longer in some countries like New Zealand where the average woman has children at age 29, wherease it’s less than 20 years of age in certain developing countries.
The delay in deaths is one where we see huge differences across various countries. In Japan, the life expectancy is over 80 years of age, while it’s only 49 years in Afghanistan. Delays have important implications, so whenever you run into one, think to your self, “how long is this delay?” If delays are relatively long, that could lead to a lag in responsiveness or inability to adapt (i.e. you simply can’t change the population instantly), whereas if delays are very short or non-existent, the system might be more sporadic.
The delay it takes for people to change their opinions is very short in a young child, and very long in adults (some adults never change their worldviews after a certain age).
A Resource-Constrained Poor Country
We’ll begin with the new balancing loop on the bottom right. As Population increases, the number of Resources per Person falls, and when this happens, the average Life Expectancy will also fall since fewer resources means less food, a weaker economy, fewer doctors, and fewer jobs. As Life Expectancy falls, the rate of Deaths increases, which causes the Population to fall. This balancing loop makes sense, but it will only come into play given that resource constraints are a serious issue.
Another interesting thing plays out as related to Life Expectancy in the new reinforcing loop on the bottom left. When Life Expectancy falls and infant mortality rates increases, people may desire to have larger families. This ultimately leads to more children in each household, which increases the Population size, exacerbates resource constraints and decreases Life Expectancy further. This reinforcing loop represents a vicious cycle where people essentially get what they want in the present at the expense of the future. Does this mechanism make sense? It certainly wouldn’t apply in every context, but in some situations, you could imagine how a mother, expecting several of her children to die before they reach a ripe old age, would want to have more children in anticipation of early deaths. The model is true in the context of a prevailing set of factors (resource constraints matter) and beliefs (having many children is the best way to ensure that you have family in the future).
Context Is Key
Keep in mind that this is just one simplified population model of a hypothetical population. It may represent some countries more than others. For example, some would argue that the link between Resources and Life Expectancy is a weak one as long as technological progress and innovations allow us to support our consumption habits without extracting resources at too high of a rate. But others argue that technology can only do so much, and that even the U.S. will eventually reach its limits. Some believe that we are using oil as if we’re fetching water from a well – we have no idea how much is left, so we behave as if it’s bottomless.
The particular problem and context of a model should always be clear. Models are used to frame problems and answer questions. They are explicit theories of why something behaves the way it does. They should help to clarify what is being considered and what is being excluded and present opportunities to suggest corrections and additions and improvements.
So next time you see a causal loop diagram:
- Ask what problem is this describing
- Walk through the major feedback loops, identify what type they are, and boil them down to the process that they are capturing
- Estimate the delays to get an idea of the timescale of each feedback loop
- Identify which variables and loops are dominant
- Think about what is missing
Now practice what you’ve learned in this short Causal Loop Diagram Quiz!
Practice Reading More Causal Loop Diagrams (or see full list of essays with CLDs):
Learn the Foundations:<|endoftext|>
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Eric Chapdelaine
Student at Northeastern University Studying Computer Science.
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# Introduction and Fundamentals
## Definitions, Theorems, and Proofs
Mathematical Reasoning: Rigorous math based on logic
Statement: Declarative sentence that has a definite truth value
3 Types of Statements
1. Definition: Gives precise meaning. Always true.
2. Theorem: Have to prove to be true (8 is an even number).
3. Proof: Logical arguments to show a theorem is true
Proposition (more modest than theorem) = result. Another word is fact.
Undefined terms/Axioms: Don’t need to explicit state in proof.
1. Numbers: Integer, Real Number (We will define even, odd, rational, etc.)
2. Operations: Addition, Subtraction, Multiplication, and Division. (We will prove even + even = even)
3. Basic Properties of Arithmetic
4. Ordering of Real Numbers
When in doubt, prove it (or ask).
## Logical Connectives
NOTE:
The rows of a truth table can be determined by $2^{\text{number of compnent statements}}$
### Not
Takes one component statement
Notation: $\sim p$
Truth Table:
$p$ $\sim p$
T F
F T
### And
Conjunction
Takes 2 component statements
Notation: $p \wedge q$
Truth Table:
$p$ $q$ $p \wedge q$
T T T
T F F
F T F
F F F
### Or
Disjunction
Takes 2 component statements
Notation: $p \vee q$
Truth Table:
$p$ $q$ $p \vee q$
T T T
T F T
F T T
F F F
### If then
Conditional statement
Another way to say it is $p$ implies $q$.
• In this case, $p$ is the condition/hypothesis/premise/assumption and $q$ is the conclusion.
Truth Table:
$p$ $q$ $p \rightarrow q$
T T T
T F F
F T T
F F T
Warning:
It doesn’t say anything about the condition when $\sim p$ is true.
$p \rightarrow q$
• Inverse: $\sim p \rightarrow \sim q$
• Converse: $q \rightarrow p$
Note, the inverse and the converse have the same truth table and therefore they are the same.
### if and only if
Also called iff or biconditional
Notation: $p \leftrightarrow q$
• = $(p \rightarrow q) \wedge (q \rightarrow p)$
• = $(p \rightarrow q) \wedge (\sim p \rightarrow \sim q)$
Truth Table:
$p$ $q$ $p \leftrightarrow q$
T T T
T F F
F T F
F F T
NOTE:
All definitions are bidirectional. That is, if a definition has “if … then”, it means “if and only if”.<|endoftext|>
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The cerebral peduncle is made of a mass of nerve fibers, and there is one peduncle on each side of the brain. The term 'cerebral' means it is related to the brain. A 'peduncle' is a stem-like connector.
The cerebral peduncles are connected to the pons, which is a part of the frontal brain stem that looks like a swelling. Many other nerve bundles also connect to the pons. Cerebral peduncles help transport nerve impulses from the higher part of the brain (cortex) and the brain stem, or lower part of the brain, to other areas of the central nervous system.
The cerebral peduncles help refine our movements. If body movement impulses came straight from the cortex, the movements would seem erratic and clumsy. The peduncles adjust the commands by taking into account where the body parts currently are located before directing the movement, and they sometimes slow down the movement. When there is an injury to cerebral peduncles, the symptoms of the injury show up in the part of the body related to the injured peduncle.<|endoftext|>
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Course: JavaScript
Progress (0%)
# Exercise: Euclidean Algorithm
Exercise 36 Easy
Prerequisites for the exercise
1. JavaScript Function Recursions
2. All previous chapters
## Objective
Create a recursive function to implement the Euclidean algorithm.
## Description
Thousands of years back, an extremely ingenious method was devised to find the greatest common divisor of two integers by the great Greek mathematician, Euclid. Today, it's commonly known as the Euclidean algorithm.
The method goes as follows:
Consider two positive integers $a$ and $b$. Let $r_1 = a$ and $r_2 = b$. The greatest common divisor of $a$ and $b$, denoted as $\text{gcd}(a, b)$, can likewise also be expressed as $\text{gcd}(r_1, r_2)$.
To find $\text{gcd}(r_1, r_2)$, we apply the rule,
$\text{gcd}(r_1, r_2) = \text{gcd}(r_2, r_3)$
where $r_3 = r_1 \bmod r_2$ (i.e. the remainder when $r_1$ is divided by $r_2$).
Similarly,
$\text{gcd}(r_2, r_3) = \text{gcd}(r_3, r_4)$
where $r_4 = r_2 \bmod r_3$ (i.e. the remainder when $r_2$ is divided by $r_3$).
This goes on until $r_{i + 1}$ in $\text{gcd}(r_i, r_{i + 1})$ becomes equal to $0$. At that point, the greatest common divisor becomes equal to $r_i$. That is:
$\text{gcd}(r_i, r_{i + 1}) = \text{gcd}(r_i, 0) = r_i$
Thus, the answer to the original question, of finding $\text{gcd}(a, b)$, becomes $r_i$.
Let's consider an example.
Suppose we want to find $\text{gcd}(104, 16)$. Since $104 \bmod 16 = 8$, $\text{gcd}(104, 16) = \text{gcd}(16, 8)$. Next, since $16 \bmod 8 = 0$, $\text{gcd}(16, 8) = \text{gcd}(8, 0)$. And finally, $\text{gcd}(8, 0) = 8$. Ultimately, $\text{gcd}(104, 16) = 8$.
The best thing about the Euclidean algorithm is that each subsequent computation is smaller than the previous one. Thus, it's guaranteed that we'll eventually get to a termination point.
That exactly why Euclidean algorithm works is out of the scope of this exercise. The good news is that the proof is extremely elementary, relying on the very basic ideas from number theory and naive set theory.
Anyways, in this exercise, you have to create a function gcd() to compute the greatest common divisor of two positive integers using the aforementioned Euclidean algorithm.
You shall assume that the provided arguments are valid integers. There is no need to implement an argument-validation mechanism (though, if you want to you could).
Note that your solution MUST use recursion, NOT iteration.
## New file
Inside the directory you created for this course on JavaScript, create a new folder called Exercise-36-Euclidean-Algorithm and put the .html solution files for this exercise within it.
## Solution
As stated in the description above, if the second argument to gcd(a, b) is 0, the greatest common divisor is simply a. Otherwise, the gcd is the value of gcd(a, a % b).
This is accomplished below:
function gcd(a, b) {
if (b === 0) {
return a;
}
else {
return gcd(b, a % b);
}
}
Note that another way to accomplish the same idea above is to stop right when a % b is 0 instead of passing that 0 value to gcd().
The code below illustrates this approach:
function gcd(a, b) {
var r = a % b;
if (r === 0) {
return b;
}
else {
return gcd(b, r);
}
}
The variable r is created in order to prevent repeating the expression a % b twice, first in the if conditional and then in the second return statement.
"I created Codeguage to save you from falling into the same learning conundrums that I fell into."
— Bilal Adnan, Founder of Codeguage<|endoftext|>
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## Sum of an Arithmetic Sequence Formula Proof
The sum of the first n terms of an arithmetic sequence is given by:
$S_n=\frac{n}{2} (2a+(n-1)d)$
where the first term is a and the common difference is d. Alternatively, we can write this as:
$S_n=\frac{n}{2} (a+L)$
where L is the last term.
## Proof
Suppose we have a general arithmetic sequence with first term a and common difference d. Then the sum of the first n terms, $$S_n$$, is given by:
$S_n=a+(a+d)+(a+2d)+ \cdots + (a+(n-3)d) + (a+(n-2)d) + (a+(n-1)d)$
We can write $$2S_n$$ as:
2S_n=(( a,+(a+d),+(a+2d),+, \cdots, + (a+(n-3)d), + (a+(n-2)d), + (a+(n-1)d) ),(+ a,+(a+d),+(a+2d),+, \cdots, + (a+(n-3)d), + (a+(n-2)d) ,+ (a+(n-1)d) ))
We can write the terms in reverse order the second time we sum them, giving:
2S_n=(( a,+(a+d),+(a+2d),+ ,\cdots, + (a+(n-3)d), + (a+(n-2)d), + (a+(n-1)d) ),(+ (a+(n-1)d),+ (a+(n-2)d),+ (a+(n-3)d),+ ,\cdots, +(a+2d), +(a+d), +a ))
Now we can group the term in the top row of the sum with the term below it, giving:
$2S_n=[a+ (a+(n-1)d)]+[(a+d)+ (a+(n-2)d)]+[(a+2d)+ (a+(n-3)d)]+ \cdots + [(a+(n-3)d)+(a+2d)] + [(a+(n-2)d)+(a+d)] + [(a+(n-1)d)+a]$
$2S_n=[2a+(n-1)d)]+[2a+(n-1)d)]+[2a+(n-1)d)]+ \cdots + [2a+(n-1)d)] + [2a+(n-1)d)] + [2a+(n-1)d)]$
$2S_n=n[2a+(n-1)d)]$
$\therefore S_n=\frac{n}{2}[2a+(n-1)d)]$
$Q.E.D.$
This trick of summing the rth and (n-r+1)th term to get n was famously said to be used by Gauss at the age of 8, when his teacher asked the class to add the numbers from 1 to 100 in the hope that this would take up the whole lesson and he wouldn't have to teach the class. However, Gauss quickly noticed that 1+100=101, 2+99=101 and so on for 50 pairs of numbers, so the sum is equal to 5050.<|endoftext|>
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Who was she?
Born in Trinidad and Tobago on September 7, 1925, Walcott was a trade unionist and political activist.
Best known as…
An ardent champion for the working class.
Starting as a member of the Union of Commercial and Industrial Workers at the age of 40, she continued her fight for the plight of the working class in her country by supporting political campaigns and candidates as well as joining the local Black power movement.
How she was a Caribbean Catalyst for Change:
When newspapers of the time refused to publish her opinions, she took it upon herself to learn how to type and publish her own materials, selling them at political rallies. Some of her earliest writings dealt with the exploitation of women in the workplace, such as: The exploitation of Working-Class Women – v Cannings Ltd. Guilty?, A Woman’s Fight – An exploitation of the Working-Class Woman, Women’s Aim Now is to End Exploitation and Working-Class Woman Speaks Out. These pieces were then published by the Institute of Social Studies in Netherlands in a 1980’s booklet titled “Fight Back Says a Woman.”
Her work and campaigns also resulted in the passing of the Minimum Wages and Terms and Conditions for Household Assistants Order under the Minimum Wages Act as well as the Unremunerated Work Act, 1995, making Trinidad and Tobagoone of the first countries in the world to pass such legislation and the Trinidad and Tobago language being used as the model for the Beijing Declaration on Women.<|endoftext|>
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Geo-engineering can stop the Earth warming, at least in theory, scientists say, but doubts persist over the possible risks.
LONDON, 10 November, 2017 – Climate scientists now know that geo-engineering – in principle at least – would halt global warming and keep the world at the temperatures it will reach by 2020.
It is simple: inject millions of tons of sulphate aerosols into the stratosphere at carefully chosen locations, and keep on doing so for as long as humans continue to burn fossil fuels and release greenhouse gases into the atmosphere.
The desired effect: global temperatures will be contained because the pollutants in the upper atmosphere will dim the sun’s light and counteract the greenhouse effect of all the carbon dioxide pumped from power stations, vehicle exhausts, factory chimneys and burning forests.
It won’t be the perfect answer. The oceans will go on becoming more acidic, and the skies will become subtly darker. Rainfall patterns could be affected. Repairs to the ozone layer – an invisible shield against dangerous ultraviolet radiation – would be slowed.
“For decision makers to accurately weigh the pros and cons of geo-engineering against those of human-caused climate change, they need more information. Our goal is to better understand what geo-engineering can do – and what it cannot”
The volumes of sulphate aerosols that would need to be flown to stratospheric heights and released each year would continue to grow as humans went on burning ever more fossil fuels.
The technical and energy demands of such an operation would be colossal. There could be serious geopolitical problems about the impacts and responsibility for such decisions. But, at least in principle, researchers now believe geo-engineering could be made to work.
“For decision makers to accurately weigh the pros and cons of geo-engineering against those of human-caused climate change, they need more information,” said Ben Kravitz, of the Pacific Northwest National Laboratory, and one of a consortium which has published a succession of five studies in the Journal of Geophysical Research – Atmospheres. “Our goal is to better understand what geo-engineering can do – and what it cannot.”
Climate scientists have repeatedly investigated the so-called techno-fix. By burning maybe 50 million years of fossil fuel deposits in just two centuries, humans have raised global temperatures and inadvertently engineered climate change.
So perhaps science and technology could come to the rescue, and deliberately engineer the climate to a new kind of stability. The consensus is that the ideal solution would be to stop burning fossil fuels and to start restoring the planet’s forests, the great absorbers of atmospheric carbon. But despite promises by the world’s nations in Paris in 2015, global temperatures continue to rise.
Geo-engineering is an idea that won’t go away. Research teams have repeatedly examined ideas for countering global warming, instead of reducing the cause, and found them wanting: such action world ultimately fail, or it would make the world’s problems worse, or at best it would take the heat out of the hurricane season.
But scientists from the US National Centre for Atmospheric Research, along with other US institutions and international colleagues, chose a different approach: what could geo-engineering achieve?
Famously, an eruption of Mt Pinatubo in the Philippines in 1991 cooled the planet by dumping 20 million tons of sulphur dioxide in the stratosphere. The researchers used computer simulations to test the effect of what might be called artificial eruptions: how would stratospheric winds spread these sulphate aerosols, and how would this diffuse global dust cloud cool the planet, and for how long, and to what extent?
They played with the idea of injecting sulphates at 14 different sites at seven latitudes and two altitudes, to find that the idea worked best if injections happened at 30 degrees latitude, north and south. They experimented with varying levels of sulphur dioxide: up to 12 million metric tons at a time.
They found out how to contain overall global temperature rise to the predicted 2020 average: some regions however became – in their computer models – hotter or cooler than the citizens might appreciate.
But the challenge of keeping the world cool became more and more demanding. By the end of the century, if humans went on burning fossil fuels in the notorious business-as-usual scenario, their model demanded the equivalent of almost five Mt Pinatubo eruptions a year. The research goes on.
The Carnegie Climate Geoengineering Governance Initiative aims ”to encourage the development of governance for research on climate geo-engineering that is balanced between enabling and regulatory aspects.”
One of its priorities is to put solar geo-engineering deployment on hold until the risks and potential benefits are better known and governance frameworks are agreed.
”We are still a long way from understanding all the interactions in the climate system that could be triggered by geo-engineering, which means we don’t yet understand the full range of possible side effects,” said scientist Simone Tilmes, of the National Centre for Atmospheric Research.
“But climate change also poses risks. Continuing research into geo-engineering is critical to assess benefits and side effects and to inform decision makers and society.” – Climate News Network<|endoftext|>
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# Out of 60W and 40W lamps, which one has a higher electrical resistance when in use.
Last updated date: 09th Sep 2024
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Views today: 7.18k
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418.5k+ views
Hint:The power is defined as the rate of energy transferred and it is expressed in terms of watts (W). Resistance is defined as the hurdles or obstacles in the motion current flow. There is a direct relation between the power and resistance of a device.
Formula used:The formula of the power of the device is given by,
$\Rightarrow P = \dfrac{{{V^2}}}{R}$
Where power is P. the potential difference is V and the resistance is R.
Complete step by step solution:
It is asked in the problem that out of 60W and 40W lamps, which one has a higher electrical resistance. Let us consider that both the lamps are used in the circuit having the same power source but one by one.
The formula of the power of the device is given by,
$\Rightarrow P = \dfrac{{{V^2}}}{R}$
Where power is P, the potential difference is V and the resistance is R.
Here we can observe that the power is inversely proportional to the resistance of the circuit.
$\Rightarrow P \propto \dfrac{1}{R}$
Where power is P and resistance is R.
Here we can see that if the power is more than the resistance will be less.
Here we can see that if the power is more than the resistance will be less and if the power is less than the resistance is more. So out of both the lamps the lamp having less power is 40 W and therefore it will have more resistance when used.
The lamp which will have higher resistance in use is the lamp with power 40 W.
Note:The resistance of a device should be small otherwise the losses in the circuit will be very large and then the power drainage will be also high. The Ohm’s law gives us the relation between the potential difference, the current and the resistance of the circuit.<|endoftext|>
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# A and B undertake a piece of work for Rs. 300. A can do it in 20 days and B can do it in 60 days. With the help of C, they finish it in 10 days. How much should C be paid for her contribution?
1. Rs. 100
2. Rs. 200
3. Rs. 300
4. Rs. 250
5. Rs. 150
Option 1 : Rs. 100
## Detailed Solution
Given:
A and B undertake a piece of work for Rs. 300
A can do it in 20 days
B can do it in 60 days
Calculation:
A’s efficiency = 60/20 = 3
B’s efficiency = 60/60 = 1
Combined Efficiency of A, B, and C = 60/10= 6
C’s efficiency = 6 – 3 -1 = 2
C contributed 1/3 of the total work done.
∴ C should be paid 1/3 of the total amount = 300/3 = Rs. 100<|endoftext|>
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The Primary Spanish Programme is designed to address key issues encountered by schools when delivering the KS2 MFL curriculum to Learn Spanish.
1. Most KS2 teachers are not MFL specialists. The Primary Spanish programme shifts focus away from topic vocabulary lists to the structures required for ‘practical communication’. This means more progression with less new language for non-specialists to learn.
2. “The focus of study in modern languages will be on practical communication”(NC 2014). A focus on vocabulary lists to learn Spanish such as colours and numbers does not meet the requirement for progression. Primary Spanish focusses on the essential structures required for practical communication.
3. Most Year 7 teachersbeginfrom scratchbecause pupils arrive with such disparatelevels ofMFL competence, which is demotivating for learners.AsPrimary Spanish focusses on the key structures and not topic vocabulary, pupils who have been taught with the programme will avoid repeating already learnt language. They will have the advantage of being able to apply the new language to the structures they have already learnt, leading to accelerated progress.
4. There are many time pressures inKS2.As the focus is on quality not quantity, substantial progress can be made in 25-35 minutes direct teaching per week.
Step by Step Instructions
It is recommended that you watch the 10-minute Quick Guide video. The information included can be found below.
1. Refer to the script and the 3-minute teachers’ video, which provides the new language, pronunciation and how to ensure progression. After the first two lessons the non-specialist will be able to use the scripts with confidence.
2. As you can successfully teach the content in around 25-35 minutes, you will have the flexibility to incorporate practical activities and games to extend the lesson as required.
3. Initially, each lesson can be taught using the corresponding full lesson video, with the teacher following the instructions. Alternatively, in the early stages, the non-specialist teacher may wish to view the video as preparation and then deliver the lesson using the script.
4. While there is no requirement to teach additional topic vocabulary, any extra vocabulary is always beneficial as it can be applied to the communication structures already taught by the programme. For example, if animals are taught, pupils can incorporate this vocabulary into the structures creating statements and questions such: es un león – it’s a lion, es mi perro – it’s my dog, tengo un gato, pero no quiero un perrro – I have a cat, but I want a dog etc…
Presenting New Language
There is a tendency to assume that all new language should be taught with visuals, but flashcards will have more impact when used with discretion and when most appropriate. Below are some other highly effective methods to introduce new language.
“Something very important to learn in any language is to say I want. For Christmas, birthday, dinner… you need to be able to say I want an Xbox, I want a book, I want pizza etc.. And to say I want in Spanish we say…” Lean in and lower your voice as if a secret, “QUIERO … I want is QUIERO.”
The opposite of above. Prompt pupils at the beginning of the class that you will be casually dropping a new word at some point in the lesson and checking to see who catches it. Use this with something easy to pronounce.
Then, at the appropriate moment: “By the way, the word for something in Spanish is algo“. After 15-20 seconds check to see if anyone can tell you the word for something and make a big deal if anyone did. Do this regularly, always making a fuss of those who catch the new word and you will notice an increase in attentiveness.
Some language, say opposites, stick better when taught as a pair. “We’re now going to learn two words together, good and bad, good bad. The Spanish for good bad is bueno malo, bueno malo etc…
“To eat and to drink, comer beber …”
Using a hook
Some language items are easy to remember when presented with an aide memoir which acts as a hook on the learner’s memory. “Bueno means good. Has anyone had the chocolate bar BUENO? It’s called that because it’s goooood …”
Flashcards are useful when presenting related items, e.g. animals, actions etc… However, it is important to vary the usage.
In addition learn Spanish top 3 vocabulary tricks.
Practising New Language
There is no right way to do this, the only test is whether all pupils can correctly recall the new language at the end of the lesson. Below are some tried and tested suggestions.
Repetition is very important to establish correct and confident pronunciation, but too much can be boring and kills the pace of the lesson. So, keep it short and varied by limiting how much language is practised at any one time. It helps, if you want to teach, say, four colours, include one which is easy to remember, e.g. blanco – white.
After a round of normal repetition, begin to introduce variety: whisper, loud, fast, slow, just the boys, just the girls, just the pupils with an A in their name etc…
Lightning Flash Cards
After some repetition using flashcards, “I’m going to show you a picture, but lightning fast, so get ready”. Flash one of the cards impossibly fast in front of them and ask them to say what is was in Spanish. The guesses will be good practice. Then flash the picture again, but slower so they can be successful if attentive.
Only repeat if correct
This is a very effective drill as pupils need to be switched on. “Only repeat if I say what’s on the picture”. After a few rounds, increase the fun by trying to catch them out.
Simon says, pretend you’re un gato. Simon says, saltar…”
Heads Down Thumbs Up
All time favourite and a fantastic motivator as a class reward at the end of the lesson. When guessing who has pinched their thumb, pupils identify by the picture the ‘pinchers’ are holding (e.g. un lápiz), not their name.
For older, Year 5 and Year 6, pupils. This is not in the official repertoire of the MFL teacher, but from twenty years of teaching, it is by far the most effective method. “The sheet in front of you has 10 new words.” Briefly familiarise pupils with the words – it is not necessary to dwell on the pronunciation as this is done subsequently. “You have 5 minutes to learn as many of these as you can. You can choose how you do it, but it must be in silence and by yourself. You are not expected to learn all the words.” Make sure the easier words are at the beginning of the list to ensure adequate differentiation This was used very successfully in a lesson which Ofsted rated Outstanding.
One original way for teachers to learn Spanish is this video.<|endoftext|>
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Image: Eric H. Cline, George Washington University
In November of 2013, a team from George Washington University was conducting an archaeological excavation in Israel when they unearthed a 3,700-year-old wine cellar. The excavation site, in the ruins of a Canaanite palace in northern Israel, is close to modern wineries.
The team found 40 3-foot-tall jars in what was possibly a storage area. Each jar held wine once enjoyed by the Canaaanites. The scientists were able to collect the substance left in the bottom of the jars and analyze it, which helped them understand what kind of wine the ancient peoples of the area consumed.
The contents of the wine residue included tree resins, honey, mint, juniper berries, cinnamon and cedar. Age analysis of the find led researchers to believe that the craft of winemaking was developed in that region prior to spreading to the Mediterranean and Egypt.
Read about more recent and fascinating archaeological discoveries that lent insight into the past in this article at Listverse.<|endoftext|>
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College Physics for AP® Courses 2e
# Section Summary
## 8.1Linear Momentum and Force
• Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.
• In symbols, linear momentum $pp$ is defined to be
$p=mv,p=mv,$
where $mm$ is the mass of the system and $vv$ is its velocity.
• The SI unit for momentum is $kg·m/skg·m/s$.
• Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes.
• In symbols, Newton’s second law of motion is defined to be
$F net = Δp Δt , F net = Δp Δt ,$
$FnetFnet$ is the net external force, $ΔpΔp$ is the change in momentum, and $ΔtΔt$ is the change time.
## 8.2Impulse
• Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts:
$Δp=FnetΔt.Δp=FnetΔt.$
• Forces are usually not constant over a period of time.
## 8.3Conservation of Momentum
• The conservation of momentum principle is written
$p tot = constant p tot = constant$
or
$p tot = p′ tot ( isolated system ) , p tot = p′ tot ( isolated system ) ,$
$ptotptot$ is the initial total momentum and $p′totp′tot$ is the total momentum some time later.
• An isolated system is defined to be one for which the net external force is zero $Fnet=0.Fnet=0.$
• During projectile motion and where air resistance is negligible, momentum is conserved in the horizontal direction because horizontal forces are zero.
• Conservation of momentum applies only when the net external force is zero.
• The conservation of momentum principle is valid when considering systems of particles.
## 8.4Elastic Collisions in One Dimension
• An elastic collision is one that conserves internal kinetic energy.
• Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions.
## 8.5Inelastic Collisions in One Dimension
• An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).
• A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.
• Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.
## 8.6Collisions of Point Masses in Two Dimensions
• The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the $xx$-axis parallel to the velocity of the incoming particle.
• Two-dimensional collisions of point masses where mass 2 is initially at rest conserve momentum along the initial direction of mass 1 (the $xx$-axis), stated by $m1v1=m1v′1 cosθ1+m2v′2 cosθ2m1v1=m1v′1 cosθ1+m2v′2 cosθ2$ and along the direction perpendicular to the initial direction (the $yy$-axis) stated by $0=m1v′1y+m2v′2y0=m1v′1y+m2v′2y$.
• The internal kinetic before and after the collision of two objects that have equal masses is
$1 2 mv 1 2 = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 + mv ′ 1 v ′ 2 cos θ 1 − θ 2 . 1 2 mv 1 2 = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 + mv ′ 1 v ′ 2 cos θ 1 − θ 2 .$
• Point masses are structureless particles that cannot spin.
## 8.7Introduction to Rocket Propulsion
• Newton’s third law of motion states that to every action, there is an equal and opposite reaction.
• Acceleration of a rocket is $a= v e mΔmΔt−ga= v e mΔmΔt−g$.
• A rocket’s acceleration depends on three main factors. They are
1. The greater the exhaust velocity of the gases, the greater the acceleration.
2. The faster the rocket burns its fuel, the greater its acceleration.
3. The smaller the rocket's mass, the greater the acceleration.<|endoftext|>
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### Combinatorics
Wikipedia defines combinations as:
In combinatorial mathematics, a combination is an un-ordered collection of unique elements. (An ordered collection is called a permutation.) Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important (two lists with the same elements in different orders are considered to be the same combination). Also, the elements cannot be repeated in a combination (every element appears uniquely once); this is often referred to as “without replacement/repetition”. This is because combinations are defined by the elements contained in them, s the set {1, 1, 1} is the same as {1}. For example, from a 52-card deck any 5 cards can form a valid combination (a hand). The order of the cards doesn’t matter and there can be no repetition of cards.
Mathworld provides a more terse definition:
The number of ways of picking k unordered outcomes from n possibilities.
The combinations of n elements chosen as k is the number of unique ways of selecting k elements from a set of n.
From now on, by set of n I always mean one of the form {1, 2, 3, …, n}.
So, what are the ways of choosing 2 elements from a set of 4, {1, 2, 3, 4}?
```{1, 2} {1, 3} {1, 4} {2, 3} {2, 4} {3, 4}```
That’s 6 ways, but what is the general formula?
This is easily proved: for a set of n, there are n ways of choosing the first element, n * (n – 1) ways of choosing the first two elements, …, n * (n – 1) * … * (n – k + 1) ways of choosing the first k elements. Unfortunately, this will generate duplicate subsets: for every subset of k elements, this will generate all the k! permutations of the subset. So, we have to divide the total number of subsets (n * (n – 1) * … * (n – k + 1)) by the number of repetitions (k!). This yields exactly the formula noted above.
Combinations are an astoundingly wide-spread concept, and are used in every branch of mathematics and especially in the analysis of algorithms. This said, there’s only one thing you really need to know: how to apply the formula.
Look at the formula above, notice that there are exactly k factors in the nominator and k factors in the denominator. So, to remember the formula and easily apply it:
```P1. Draw the fraction line. P2. Above the line, write k terms of the form: n, n - 1, n - 2, ... P3. Below the line, write k terms of the form: 1, 2, 3, ...```
Here are a few examples:
And now for the fun part. How do you generate combinations? Look closely at the example above. First thing to note is that every combination is an array of k elements. Next, the first digit in every set is, basically, every digit between 1 and n. What about the other digits? They’re always between 1 and n and they’re always in ascending order. Now it should be obvious what the algorithm is:
```P1. Start of with (1, 2, ..., k); this is the first combination. P2. Print it. P3. Given the combination (c0, c1, ..., cn), start from the back and for ci, if it is larger than n - k + 1 + i then increment it and go on to the next indice i. After this, if c0 > n - k, then this is not a valid combination so we stop. Otherwise give ci+1, ci+2, ... the values of ci + 1, ci+1 + 1, .... Jump to P2.```
Here’s the sourcecode in C (comb1.c):
NOTE: Source is mangled by WordPress. Download the source file, or copy-paste it from here or remember to replace the amp-s with ampersands and the lt-s with “less then” signs.
```#include <stdio.h>
/* Prints out a combination like {1, 2} */
void printc(int comb[], int k) {
printf("{");
int i;
for (i = 0; i < k; ++i)
printf("%d, ", comb[i] + 1);
printf("\\b\\b}\\n");
}
/*
next_comb(int comb[], int k, int n)
Generates the next combination of n elements as k after comb
comb => the previous combination ( use (0, 1, 2, ..., k) for first)
k => the size of the subsets to generate
n => the size of the original set
Returns: 1 if a valid combination was found
0, otherwise
*/
int next_comb(int comb[], int k, int n) {
int i = k - 1;
++comb[i];
while ((i >= 0) && (comb[i] >= n - k + 1 + i)) {
--i;
++comb[i];
}
if (comb[0] > n - k) /* Combination (n-k, n-k+1, ..., n) reached */
return 0; /* No more combinations can be generated */
/* comb now looks like (..., x, n, n, n, ..., n).
Turn it into (..., x, x + 1, x + 2, ...) */
for (i = i + 1; i < k; ++i)
comb[i] = comb[i - 1] + 1;
return 1;
}
int main(int argc, char *argv[]) {
int n = 5; /* The size of the set; for {1, 2, 3, 4} it's 4 */
int k = 3; /* The size of the subsets; for {1, 2}, {1, 3}, ... it's 2 */
int comb[16]; /* comb[i] is the index of the i-th element in the
combination */
/* Setup comb for the initial combination */
int i;
for (i = 0; i < k; ++i)
comb[i] = i;
/* Print the first combination */
printc(comb, k);
/* Generate and print all the other combinations */
while (next_comb(comb, k, n))
printc(comb, k);
return 0;
}
```
Always open to comments. Have fun.
Wikipedia defines the partition of a set as:
In mathematics, a partition of a set X is a division of X into non-overlapping “parts” or “blocks” or “cells” that cover all of X. More formally, these “cells” are both collectively exhaustive and mutually exclusive with respect to the set being partitioned.
A more succinct definition is given by Mathworld:
A set partition of a set S is a collection of disjoint subsets of S whose union is S.
Simply put, the partitions of a set S are all the ways in which you can choose disjoint, non-empty subsets of S that unioned result in S.
From now on, when I say a set of n elements, I mean {1, 2, …, n}. So, what are the subsets of {1, 2, 3}?
```{1, 2, 3} {2, 3} {1} {1, 3} {2} {3} {1, 2} {3} {2} {1}```
It’s obvious that these verify the definition: {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3} and {1, 2, 3} are all subsets of {1, 2, 3}. They’re all non-empty and, in any partition, the same element never appears twice. Finally, in a partitioning, the union of the partitions is the original set.
In how many ways can you partition a set of n elements? There are many ways to calculate this, but as far as I can tell, the easiest is using Catalan numbers:
If you check the formula for 3 you’ll see that it does give the correct answer: 5.
A reader pointed out that what we may need here are not Catalan numbers, but Bell numbers. Wikipedia’s definition seems to agree with him.
Ok. We know what a partitioning is, we know how many there are, but how do you generate them? This is the first algorithm I could think of. It may not be clear from the explanation why it works but try it on a piece of paper for n=3 and it will become obvious. Here’s how I came up with it:
First of all, how do you represent a partitioning of a set of n elements? The straight-forward way would be using a vector of n integers, each integer representing the number of the subset in which the corresponding element is in. If the corresponding element of 3 is 2, that means that 3 is in the 2nd subset. So, given the set {1, 2, 3}:
```Partitioning -> Encoding {1, 2, 3} -> (1, 1, 1) {1} {2, 3} -> (2, 1, 1) {2} {1, 3} -> (1, 2, 1) {1, 2} {3} -> (2, 2, 1) {1} {2} {3} -> (3, 2, 1) ```
Notice that the encodings, written backwards are: 111, 112, 121, 122 and 123. From this you can guess how the generator works: more or less, generate all the numbers between 111 and 123 using only the digits 1, 2 and 3:
``` 111 112 113 121 122 123 ```
That’s almost right. The encodings (1, 1, 2) and (1, 1, 3) translate into the same partitioning: {1} {2, 3}. If you do the same thing for a larger n you’ll notice this happening again and again. Fortunately, there’s an easy solution: never use a digit that’s more than 1 larger than any other digit in the encoding. i.e. You can’t use (1, 1, 3) because 3 is larger by 2 than the other digits in the encoding (1 and 1).
To do this, I use another vector m with the following significance: m[i] is the largest of the first i elements in the encoding. This makes it very easy not to generate any duplicate partitionings.
Here’s the code in C (part.c):
```#include <stdio.h>
/*
printp
- print out the partitioning scheme s of n elements
as: {1, 2, 4} {3}
*/
void printp(int *s, int n) {
/* Get the total number of partitions. In the exemple above, 2.*/
int part_num = 1;
int i;
for (i = 0; i < n; ++i)
if (s[i] > part_num)
part_num = s[i];
/* Print the p partitions. */
int p;
for (p = part_num; p >= 1; --p) {
printf("{");
/* If s[i] == p, then i + 1 is part of the pth partition. */
for (i = 0; i < n; ++i)
if (s[i] == p)
printf("%d, ", i + 1);
printf("\\b\\b} ");
}
printf("\\n");
}
/*
next
- given the partitioning scheme represented by s and m, generate
the next
Returns: 1, if a valid partitioning was found
0, otherwise
*/
int next(int *s, int *m, int n) {
/* Update s: 1 1 1 1 -> 2 1 1 1 -> 1 2 1 1 -> 2 2 1 1 -> 3 2 1 1 ->
1 1 2 1 ... */
/*int j;
printf(" -> (");
for (j = 0; j < n; ++j)
printf("%d, ", s[j]);
printf("\\b\\b)\\n");*/
int i = 0;
++s[i];
while ((i < n - 1) && (s[i] > m[i] + 1)) {
s[i] = 1;
++i;
++s[i];
}
/* If i is has reached n-1 th element, then the last unique partitiong
has been found*/
if (i == n - 1)
return 0;
/* Because all the first i elements are now 1, s[i] (i + 1 th element)
is the largest. So we update max by copying it to all the first i
positions in m.*/
int max = s[i];
for (i = i - 1; i >= 0; --i)
m[i] = max;
/* for (i = 0; i < n; ++i)
printf("%d ", m[i]);
getchar();*/
return 1;
}
int main(int argc, char *argv[]) {
int s[16]; /* s[i] is the number of the set in which the ith element
should go */
int m[16]; /* m[i] is the largest of the first i elements in s*/
int n = 3;
int i;
/* The first way to partition a set is to put all the elements in the same
subset. */
for (i = 0; i < n; ++i) {
s[i] = 1;
m[i] = 1;
}
/* Print the first partitioning. */
printp(s, n);
/* Print the other partitioning schemes. */
while (next(s, m, n))
printp(s, n);
return 0;
}
```
The code is heavily commented, but I’ll happily respond to any questions. This is also what I used to generate all the above listings. Try decommenting some of the code to see how the programme works. Good luck!
P.S. Every encoding after (3, 2, 1) yields a duplicate partitioning. For fun, try proving this mathematically.
There quite a few definitions of what a set is, but it all boils down to this:
A set defined as a collection of distinct elements, in which order is not important.
So {1, 2, 3}, {3, 4}, {} and {5, 99, -1} are all sets. Because the order of the elements is ignored, {1, 2, 3} and {3, 2, 1} is the same set. In case you’re wandering, there are exactly n! diffrent ways to write a set of n elements.
For the rest of the discussion, I’ll use sets of the form {1, 2, …, n}, so when I say a set of 3 elements, I mean {1, 2, 3}. Just remember that is not a property of sets. They can contain anything as elements, not necessarily consecutive numbers.
The set S1 is said to be the subset of the set S2, if all the elements of S1 also belong to S2.
Knowing this, it’s easy to figure out the subsets of {1, 2, 3}:
```{ } { 1 } { 2 } { 1, 2 } { 3 } { 1, 3 } { 2, 3 } { 1, 2, 3 }```
How many subsets are there? For a set of one element, there are 2 subsets: {} and {1}. For a set of 2 elements, there are 4 subsets: {}, {1}, {2}, {1, 2}. For a set of 3 elements, there are 8 subsets. Notice the pattern?
```n = 1: 21 n = 2: 22 n = 3: 23```
For a set of n there are 2n subsets. This is easily proved: Any subset of the set can either contain or not contain an element; so, for a subset, there are 2 states for the first element, 2 for the second element, …, 2 for the nth element; so, there are 2 states for the first element, 2 * 2 = 22 states for the first two, 2 * 2 * 2= 23 states for the first three, …, 2 * 2 * 2 * … * 2 = 2n states for all the n elements.
The problem here is how to generate all the subsets of a given set. There are a few algorithms for doing this, but in the end, only two are worth considering.
The first is this: given all the subsets of S and the element y, you can generate all the subsets of S U {y} by taking each subset of S, once adding to it y and once leaving it as it is. i.e. Knowing that {1, 3} is a subset of S, you obtain the following two subsets of S U {y}: {1, 3, y} and {1, 3}.
This does what it’s supposed to – it generates all the subsets of S, and it wastes no time. It can also be used as another way to prove that there are 2n subsets for any set of n elements. The only problem is that you need the subsets from the previous step to generate those of this step. This means that just before the end, you must have 2n – 1 subsets in memory. Considering how much memory computers have this days, it’s not particularly wasteful, but still, there’s a better way.
The better way involves using a mask. If you have the a set of n elements, a valid mask would be an array of n boolean (true/false; 1/0) elements. When you apply a mask to a set, you check each element (e) in the set and the corresponding one in the mask (m): if m is true(1), you add e to the result, otherwise, you ignore it. After applying the mask (0, 1, 0, 0, 1) to {1, 2, 3, 4, 5}, you get {2, 5}.
So, to generate all the subsets of a set of n elements, you first have to generate all the possible 2n masks of the set and then apply them.
Generating the masks is a simple problem. Basically, you just have to implement a binary counter, i.e. something that generates:
```000 001 010 011 100 101 110 111```
Here’s the code in C (sub.c):
```#include <stdio.h>
/* Applies the mask to a set like {1, 2, ..., n} and prints it */
void printv(int mask[], int n) {
int i;
printf("{ ");
for (i = 0; i < n; ++i)
if (mask[i])
printf("%d ", i + 1); /*i+1 is part of the subset*/
printf("\\b }\\n");
}
/* Generates the next mask*/
int next(int mask[], int n) {
int i;
for (i = 0; (i < n) && mask[i]; ++i)
mask[i] = 0;
if (i < n) {
mask[i] = 1;
return 1;
}
return 0;
}
int main(int argc, char *argv[]) {
int n = 3;
int mask[16]; /* Guess what this is */
int i;
for (i = 0; i < n; ++i)
mask[i] = 0;
/* Print the first set */
printv(mask, n);
/* Print all the others */
while (next(mask, n))
printv(mask, n);
return 0;
}
```
Note: The `next()` function generates the bits in reverse order.
Always open to comments.
Last time, we defined what permutation is and gave a few basic properties.
In a few minutes we’ll see another algorithm for generating them, but first a little theory.
Lexicographical order is defined by Wikipedia as:
In mathematics, the lexicographic or lexicographical order, (also known as dictionary order, alphabetic order or lexicographic(al) product), is a natural order structure of the Cartesian product of two ordered sets.
Given two partially ordered sets A and B, the lexicographical order on the Cartesian product A × B is defined as
(a,b) ≤ (a′,b′) if and only if a < a′ or (a = a′ and b ≤ b′).
The result is a partial order. If A and B are totally ordered, then the result is a total order also.
More generally, one can define the lexicographic order on the Cartesian product of n ordered sets, on the Cartesian product of a countably infinite family of ordered sets, and on the union of such sets.
Mathworld adds the following regarding permutations and sets:
When applied to permutations, lexicographic order is increasing numerical order (or equivalently, alphabetic order for lists of symbols; Skiena 1990, p. 4). For example, the permutations of {1,2,3} in lexicographic order are 123, 132, 213, 231, 312, and 321.
When applied to subsets, two subsets are ordered by their smallest elements (Skiena 1990, p. 44). For example, the subsets of {1,2,3} in lexicographic order are {}, {1}, {1,2}, {1,2,3}, {1,3}, {2}, {2,3}, {3}.
An easy way to determine if a set is lexicographically after another is to interpret them as numbers in base n, where n is the largest element the set contains. So, (2, 1, 3) is after (1, 2, 3) because 213 < 123. Note: You may also choose n as any number greater than the largest element of the set. This is particularly convenient as most would rather use numbers in base 10 and not base 3.
Ok, but what does this have to do with permutations? Well, generating permutations in any order isn’t enough; you must generate them in lexicographic order.
Now, if you run last times’ algorithm, you find that, for n = 3, it prints:
```1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1```
Now, 123 < 132 < 213 < 231 < 312 < 321. So, the permutations are in lexicographic order!
The worst algorithm for any problem is usually called naive, but a more adequate adjective for the last algorithm would be retarded. It’s the slowest one I can think of, but it’s extraordinarily easy to explain.
This algorithm is slightly faster (about twice as fast) than the last one. It’s quite complex and harder to understand. It does the same thing as the last one, but where the naive algorithm just generated all possible sets, this one generates only valid permutations.
Here it is:
``` P1. Given n, we start with the first imaginable permutation p = (1, 2, ..., n) from the lexicographic point of view.```
``` P2. Print the the permutation p or use it for something else. ```
```P3. Let's say we have already build the permutation p = (p1, p2, ..., pn). In order to obtain the next permutation, we must first find the largest index i so that Pi<Pi + 1. Then, the element, Pi will be swapped with the smallest of the elements after Pi, but not larger than Pi. Finally, the last n - i elements will be reversed so that they appear in ascending order. Then, jump to P2. ```
That’s it for the algorithm, here’s the code in C (lexicoPerm.c):
```#include <stdio.h>
void printv(int v[], int n) {
int i;
for (i = 0; i < n; i++)
printf("%d ", v[i]);
printf("\\n");
}
/*!
This just swaps the values of a and b
i.e if a = 1 and b = 2, after
SWAP(a, b);
a = 2 and b = 1
*/
#define SWAP(a, b) a = a + b - (b = a)
/*!
Generates the next permutation of the vector v of length n.
@return 1, if there are no more permutations to be generated
@return 0, otherwise
*/
int next(int v[], int n) {
/* P2 */
/* Find the largest i */
int i = n - 2;
while ((i >= 0) && (v[i] > v[i + 1]))
--i;
/* If i is smaller than 0, then there are no more permutations. */
if (i < 0)
return 1;
/* Find the largest element after vi but not larger than vi */
int k = n - 1;
while (v[i] > v[k])
--k;
SWAP(v[i], v[k]);
/* Swap the last n - i elements. */
int j;
k = 0;
for (j = i + 1; j < (n + i) / 2 + 1; ++j, ++k)
SWAP(v[j], v[n - k - 1]);
return 0;
}
int main(int argc, char *argv[]) {
int v[128];
int n = 3;
/* The initial permutation is 1 2 3 ...*/
/* P1 */
int i;
for (i = 0; i < n; ++i)
v[i] = i + 1;
printv(v, n);
int done = 1;
do {
if (!(done = next(v, n)))
printv(v, n); /* P3 */
} while (!done);
return 0;
}
```
The code is commented and it does nothing but implement the algorithm. Have fun!
A permutation of n objects is an arrangement of n distinct objects.
Wikipedia gives a slightly more detailed definition:
Permutation is the rearrangement of objects or symbols into distinguishable sequences. Each unique ordering is called a permutation. (For cases wherein the ordering of elements is irrelevant, compare combination and set.) For example, with the numerals one to six, each possible ordering consists of a complete list of the numerals, without repetitions. There are 720 total permutations of these numerals, one of which is: “4, 5, 6, 1, 2, 3”.
And Mathworld gives the standard mathematical definition:
A permutation, also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself.
Permutations are crucial to studying the behaviour of many algorithms and we’ll find a lot of intresting things about them.
For starters, what are the permutations of {1, 2, 3}? The definition says a permutation is a rearrangement of the list’s elements. So, the permutations (plural) are all the possible rearrangements of the list’s elements. This gives us six permutations:
``` 123, 132, 213, 231, 312, 321 ```
For convenience, we’ll only work with sets like {1, 2, 3, …, n}. In computer science, the permutations of this set is called the permutations of n. In mathematics the permutations of n means the number of permutations of the given set.
How many permutations of n are there? This is easily solved, to create a permutation one element at a time: there are n ways in which to choose the first element; then, there are n – 1 ways in which to choose the second element, so that no element repeats itself; then, there are n – 2 ways to choose the third element; …; finally, there is only one way to choose the nth element. How many posibilities does this give us? So, n ways to choose the 1st element, n(n – 1) ways for the first 2 elements, n(n – 1)(n – 2) for the first 3 elements, …, n(n – 1)(n – 2)…(n – k + 1) for the first k elements, …, n(n – 1)(n – 2)…(1) ways for all the n elements.
Now we can calculate that there are 1 * 2 * 3 = 6 permutations of 3. And … that’s right! By the way, the value 1 * 2 * 3 * … * n is usually written as n! and is called n factorial.
Great. We know what a permutation is. We know how many there are for a given set. But how do we generate them?
There are quite a few (more like dozens) methods, and I’ll describe a few here. The simplest one I can think of is this:
Let’s say you want to generate all the permutations of 3. So, you want to generate the permutations of the set {1, 2, 3}. We’ll generate the list:
``` 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233 311, 312, 313, 321, 322, 323, 331, 332, 333 ```
That’s all the numbers you can make of length 3 using only the digits 1, 2 and 3. I start from 123 and not from 111 because there’s no permutation between 111 and 123.
Then we’ll filter the results using the rule: “A valid permutation cannot contain the same digit twice“.
Then we’ll print out what’s left.
Here’s the code in C (naiveperm.c):
```#include <stdio.h>
/*!
Generates the next try.
If v is 1 2 1 2, after calls to
next(v, 4);
v will be 1 2 1 3
1 2 1 4
1 2 2 1
1 2 2 2
@return 0, if there are no more valid tries
@return 1, otherwise
*/
int next(int v[], int n) {
int i = n - 1;
v[i] = v[i] + 1;
while ((i >= 0) && (v[i] > n)) {
v[i] = 1;
i--;
if(i >= 0)
v[i]++;
}
if (i < 0)
return 0;
return 1;
}
void printv(int v[], int n) {
int i;
for (i = 0; i < n; i++)
printf("%d ", v[i]);
printf("\\n");
}
/*!
@return 1, if v is a valid permutation (no digits repeat)
@return 0, otherwise
*/
int is_perm(int v[], int n) {
int i, j;
for (i = 0; i < n; i++)
for (j = i + 1; j < n; j++)
if (v[i] == v[j])
return 0;
return 1;
}
int main(int argc, char *argv[]) {
int v[128];
int n = 8;
/* The initial permutation is 1 2 3 ...*/
int i;
for(i = 0; i <= n; i++)
v[i] = i + 1;
while (next(v,n))
if (is_perm(v,n))
printv(v,n);
return 0;
}
```
The code’s commented and it’s fairly simple, so there shouldn’t be any problems understanding it. Of course, I’m open to suggestions.
The next article in this series is Generating permutations: 2.<|endoftext|>
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A healthy diet is one that helps maintain or improve general health. A healthy diet provides the body with essential nutrition: fluid, adequate essential amino acids from protein, essential fatty acids, vitamins, minerals, and adequate calories. The requirements for a healthy diet can be met from a variety of plant-based and animal-based foods. A healthy diet supports energy needs and provides for human nutrition without exposure to toxicity or excessive weight gain from consuming excessive amounts. Where lack of calories is not an issue, a properly balanced diet (in addition to exercise) is also thought to be important for lowering health risks, such as obesity, heart disease, type 2 diabetes, hypertension and cancer.
Various nutrition guides are published by medical and governmental institutions to educate the public on what they should be eating to promote health. Nutrition facts labels are also mandatory in some countries to allow consumers to choose between foods based on the components relevant to health.
- 1 World Health Organization
- 2 American Heart Association / World Cancer Research Fund / American Institute for Cancer Research
- 3 Harvard School of Public Health
- 4 For specific conditions
- 5 Diet and possible reduced disease risk
- 6 Unhealthy diets
- 7 Public health
- 8 Cultural and psychological factors
- 9 See also
- 10 References
- 11 External links
World Health Organization
- Eat roughly the same amount of calories that your body is using. A healthy weight is a balance between energy consumed and energy that is 'burnt off'.
- Increase consumption of plant foods, particularly fruits, vegetables, legumes, whole grains and nuts.
- Limit intake of fats, and prefer less unhealthy unsaturated fats to saturated fats and trans fats.
- Limit the intake of sugar. A 2003 report recommends less than 10% simple sugars.
- Limit salt / sodium consumption from all sources and ensure that salt is iodized.
Other recommendations include:
- Essential micronutrients such as vitamins and certain minerals.
- Avoiding directly poisonous (e.g. heavy metals) and carcinogenic (e.g. benzene) substances.
- Avoiding foods contaminated by human pathogens (e.g. E. coli, tapeworm eggs).
American Heart Association / World Cancer Research Fund / American Institute for Cancer Research
The American Heart Association, World Cancer Research Fund, and American Institute for Cancer Research recommends a diet that consists mostly of unprocessed plant foods, with emphasis a wide range of whole grains, legumes, and non-starchy vegetables and fruits. This healthy diet is replete with a wide range of various non-starchy vegetables and fruits, that provide different colors including red, green, yellow, white, purple, and orange. They note that tomato cooked with oil, allium vegetables like garlic, and cruciferous vegetables like cauliflower, provide some protection against cancer. This healthy diet is low in energy density, which may protect against weight gain and associated diseases. Finally, limiting consumption of sugary drinks, limiting energy rich foods, including “fast foods” and red meat, and avoiding processed meats improves health and longevity. Overall, researchers and medical policy conclude that this healthy diet can reduce the risk of chronic disease and cancer.
Harvard School of Public Health
- Choose good carbohydrates: whole grains (the less processed the better), vegetables, fruits and beans. Avoid white bread, white rice, and the like as well as pastries, sugared sodas, and other highly-processed food.
- Pay attention to the protein package: good choices include fish, poultry, nuts, and beans. Try to avoid red meat.
- Choose foods containing healthy fats. Plant oils, nuts, and fish are the best choices. Limit consumption of saturated fats, and avoid foods with trans fat.
- Choose a fiber-filled diet which includes whole grains, vegetables, and fruits.
- Eat more vegetables and fruits—the more colorful and varied, the better.
- Calcium is important, but milk is not its best source. Good sources of calcium are collards, bok choy, fortified soy milk, baked beans, and supplements which contain calcium and vitamin D.
- Water is the best source of liquid. Avoid sugary drinks, and limit intake of juices and milk. Coffee, tea, artificially-sweetened drinks, 100-percent fruit juices, low-fat milk and alcohol can fit into a healthy diet but are best consumed in moderation. Sports drinks are recommended only for people who exercise more than an hour at a stretch to replace substances lost in sweat.
- Limit salt intake. Choose more fresh foods, instead of processed ones.
- Moderate alcohol drinking has health benefits, but is not recommended for everyone.
- Daily multivitamin and extra vitamin D intake has potential health benefits.
For specific conditions
In addition to dietary recommendations for the general population, there are many specific diets that have primarily been developed to promote better health in specific population groups, such as people with high blood pressure (as in low sodium diets or the more specific DASH diet), or people who are overweight or obese (in weight control diets). However, some of them may have more or less evidence for beneficial effects in normal people as well.
A low sodium diet is beneficial for people with high blood pressure. A Cochrane review published in 2008 concluded that a long term (more than 4 weeks) low sodium diet in Caucasians has a useful effect to reduce blood pressure, both in people with hypertension and in people with normal blood pressure.
The DASH diet (Dietary Approaches to Stop Hypertension) is a diet promoted by the National Heart, Lung, and Blood Institute (part of the NIH, a United States government organization) to control hypertension. A major feature of the plan is limiting intake of sodium, and it also generally encourages the consumption of nuts, whole grains, fish, poultry, fruits and vegetables while lowering the consumption of red meats, sweets, and sugar. It is also "rich in potassium, magnesium, and calcium, as well as protein". Evidence shows that the Mediterranean diet improves cardiovascular outcomes.
WHO recommends few standards such as an intake of less than 5 grams per person per day so as to prevent one from cardiovascular disease. Unsaturated fatty acids with polyunsaturated vegetable oils, on the other hand plays an essential role in reducing coronary heart disease risk as well as diabetes.
Diets to promote weight loss are divided into four categories: low-fat, low-carbohydrate, low-calorie, and very low calorie. A meta-analysis of six randomized controlled trials found no difference between the main diet types (low calorie, low carbohydrate, and low fat), with a 2–4 kilogram weight loss in all studies. At two years, all calorie-reduced diet types cause equal weight loss irrespective of the macronutrients emphasized.
Diet and possible reduced disease risk
Template:Further There may be a relationship between lifestyle including food consumption and potentially lowering the risk of cancer or other chronic diseases. A healthy diet may consist mostly of whole plant foods, with limited consumption of energy-dense foods, red meat, alcoholic drinks and salt while reducing consumption of sugary drinks, and processed meat. A healthy diet may contain non-starchy vegetables and fruits, including those with red, green, yellow, white, purple or orange pigments. Tomato cooked with oil, allium vegetables like garlic, and cruciferous vegetables like cauliflower "probably" contain compounds which are under research for their possible anti-cancer activity.
A healthy diet is low in energy density, lowering caloric content, thereby possibly inhibiting weight gain and lowering risk against chronic diseases. Chronic Western diseases are associated with pathologically increased IGF-1 levels. Findings in molecular biology and epidemiologic data suggest that milk consumption is a promoter of chronic diseases of Western nations, including atherosclerosis, carcinogenesis and neurodegenerative diseases.
The WHO estimates that 2.7 million deaths are attributable to a diet low in fruit and vegetable every year. Globally it is estimated to cause about 19% of gastrointestinal cancer, 31% of ischaemic heart disease, and 11% of strokes, thus making it one of the leading preventable causes of death worldwide.
Fad diet usually refers to idiosyncratic diets and eating patterns. They are diets that claim to promote weight loss or treat obesity by various mechanisms, provide little to no scientific reasoning behind their purported health benefits, and have little or no proof to support them.
Fears of high cholesterol were frequently voiced up until the mid-1990s. However, more recent research has shown that the distinction between high- and low-density lipoprotein ('good' and 'bad' cholesterol, respectively) must be addressed when speaking of the potential ill effects of cholesterol. Different types of dietary fat have different effects on blood levels of cholesterol. For example, polyunsaturated fats tend to decrease both types of cholesterol; monounsaturated fats tend to lower LDL and raise HDL; saturated fats tend to either raise HDL, or raise both HDL and LDL; and trans fat tend to raise LDL and lower HDL. Dietary cholesterol itself is only found in animal products such as meat, eggs, and dairy, but studies have shown that even large amounts of dietary cholesterol only have negligible effects on blood cholesterol.
Vending machines in particular have come under fire as being avenues of entry into schools for junk food promoters. However, there is little in the way of regulation and it is difficult for most people to properly analyze the real merits of a company referring to itself as "healthy." Recently, the United Kingdom removed the rights for McDonald's to advertise its products, as the majority of the foods that were seen have low nutrient values and high fat counts were aimed at children under the guise of the "Happy Meal". The British Heart Foundation released its own government-funded advertisements, labeled "Food4Thought", which were targeted at children and adults displaying the gory nature of how fast food is generally constituted.
Cultural and psychological factors
From a psychological and cultural perspective, a healthier diet may be difficult to achieve for people with poor eating habits. This may be due to tastes acquired in childhood and preferences for sugary, salty and/or fatty foods.
In addition to the above factors, physical activity in itself is identified as the best way to keep onself healthy. WHO develops a global recommendation for health basically for three age groups namely 5–17 years old; 18–64 years old; and 65 years old and above. For 5–17 years old, at least 60 minutes of physical activity such as games, recreation daily is highly recommended. For the ones between 18–64 years and above 65 years old age group, at least 150 minutes of moderate-intensity aerobic exercise or at least 75 minutes of vigorous-intensity aerobic physical activity is recommended.
Affects one in three adults
Affecting about 35 percent of all adults in the United States according to the CDC, metabolic syndrome contributes to weight gain, by causing a state of internal starvation called metabolic starvation. This in turn leads to increases hunger, sugar cravings and increased portions leading to overeating and weight gain.
Cause and effect misunderstood
Since we traditionally thought that the portion control (which in turn was attributed wrongly to poor will power)is the cause of weight gain, rather than the effect of this metabolic starvation, all our traditional ideas about cause and effect of obesity were not only wrong but lead to the “blame the victim” attitude when it comes to obesity.
Secret of weight gain revealed
Secret of weight gain, and metabolic syndrome revealed - it has been recently proven that metabolic syndrome, and the weight gain itself are caused by a process called insulin resistance. Check your metabolic syndrome risk using the free Metabolic syndrome meter. Watch this amazing Ted Med video that reveals the secret of weight loss - Stop blaming the victim for obesity
- Diet, Nutrition and the prevention of chronic diseases, by a Joint WHO/FAO Expert consultation (2003)
<ref> tags exist, but no
<references/> tag was found<|endoftext|>
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# How do you compute -19-8?
Mar 3, 2018
$- 27$
#### Explanation:
To avoid confusion between signs, we can factor out a $- 1$. We get:
$- 1 \left(19 + 8\right)$
We can simplify the terms in the parenthesis to get:
$- 1 \cdot 27 = - 27$
Mar 3, 2018
The answer is -27, see explanation for mini-lesson.
#### Explanation:
In order to find the answer to these types of problems, you must know how to keep the negative and positive signs straight.
Else the problem is completely messed up.
First off, you have a $- 19$.
Make sure to take notice of the negative here.
Then, you have another negative, $- 8$.
This problem is the same as doing $- 19 + - 8$.
Basically, you are adding two negative together.
So, for the sake of keeping the signs in order, I will temporarily move both the negatives out to show how easy the problem is.
$- \left(19 + 8\right)$ is another way of writing this.
By adding $19$ and $8$, we get 27.
So, take the negative of this, and get the answer.
Simply, really.<|endoftext|>
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This year, Fontbonne is a proud partner of FabLab. As such, we have built a state-of-the-art facility that provides students with a modern means of invention. Look through all of the different parts below!
3D printing technology positions students as creators. Instead of consuming the creations of someone else, they become the inventors. With 3D printing, students become designers and creators using cutting-edge technology to visualize and create their own imaginations. Not to mention the growth students will have in their spatial reasoning skills and 2D to 3D conversion understanding. Seriously, everything can be hands on. Learning about ancient societies? Design and print a model of a mummy or a pyramid. Learning about land forms? Design and print a peninsula or a mountain range. Learning about natural disasters? Design and print tools to protect you during a disaster. The list goes on and on…
CNC machines are programmed with a design which can then be manufactured hundreds or even thousands of times. Each manufactured product will be exactly the same. CNC milling centers are ideal solutions to everything, ranging from prototyping and short-run production of complex parts, to the fabrication of unique precision components. A single machine can do all kinds of different tasks, such as drilling, panning, routing, and more, and all without any hands on human help.
Using drones in the classroom opens up a new set of opportunities to make classes more relevant and engaging for students. Drones are an important step towards introducing technology in classrooms. Together with similar gadgets, they have given teachers the opportunity to use these rather simple and affordable tools to teach children verbal skills, languages and even math. However, drones can be used to help enhance orientation skills, motor skills, and even give students a better understanding of how the world around us works.
By going through the process of building a robot, students explore many different learning pathways. When students interact with robots in the classroom and make them perform various motions and tasks, their’ different strengths will start to shine. The STEM skills that robotics teach are great for inspiring tomorrow’s engineers. However, not all students are going to work for NASA or even work in a science and math-related field. Yet some of the teamwork skills they learn through robotics are ones they will use for the rest of their lives.
Laser and vinyl systems are generally some of the easier pieces of equipment a student can learn. Designs are produced in a vector drawing package, sent to the machine via a printer driver, and the system does the rest. Perhaps the biggest advantage of a cutter is its versatility. Because a cutter’s power allows it to cut through a wide range of materials, it’s one of the most versatile pieces of fabrication equipment a school can have. The applications, from K-12 to community college to university, are endless.
Virtual reality technology can bring a benefit to the education of new generations. Virtual reality is great because it lets us explore different realities and alternate our experiences. By wearing a VR headset, students encounter high-quality visualizations that can change them in a positive way. No matter what age they are, students will usually prefer to sit and watch something instead of reading it. The VR technology is quite interesting, as it can create amazing experiences that could never be “lived” in real life. Students will definitely feel more motivated to learn with the use of this technology.<|endoftext|>
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Question Video: Finding an Interval given the Function’s Average Value over It | Nagwa Question Video: Finding an Interval given the Function’s Average Value over It | Nagwa
# Question Video: Finding an Interval given the Functionβs Average Value over It Mathematics • Higher Education
The average value of π(π₯) = β6π₯Β² + 6π₯ β 1 on the interval [0, π] is 0. Find all possible values of π.
01:47
### Video Transcript
The average value of π of π₯ equals negative six π₯ squared plus six π₯ minus one on the closed interval zero to π is zero. Find all possible values of π.
Remember, the formula for the average value of a function π over a closed interval π to π is one over π minus π times the integral evaluated between π and π of π of π₯ with respect to π₯. We can see that π of π₯ is equal to negative six π₯ squared plus six π₯ minus one. And we see that π is equal to zero, and we donβt know the value of π. So we begin by substituting what we know about the average value of our function into the formula.
We get one over π minus zero, which is of course one over π. And we multiply that by the definite integral of negative six π₯ squared plus six π₯ minus one evaluated between zero and π. We do, however, know that the average value of our function is equal to zero. So we can set this equal to zero.
Letβs look to evaluate our integral. The integral of negative six π₯ squared is negative six π₯ cubed over three. The integral of six π₯ is six π₯ squared over two. And the integral of negative one is negative π₯. Letβs substitute π and zero into this expression. We see that zero is equal to one over π times negative two π cubed plus three π squared minus π. And then we divide through by π. And so we obtain that negative two π squared plus three π minus one is equal to zero. And we see we have a quadratic equation.
So letβs multiply through by negative one. And then weβll factor the expression two π squared minus three π plus one. When we do, we see that two π minus one times π minus one must be equal to zero. And for this statement to be true, either two π minus one must be equal to zero or π minus one must be equal to zero. And we solve for π. And we see that π must be equal to one-half or one.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions<|endoftext|>
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Because a frog lives on land and in water, he has a number of sensory organs that make him well-suited to both environments. Most of these sensory organs are on the frog’s head, because a frog needs to be vigilant -- he is both predator and prey. He can keep his head just above the surface of the water to take in sights, sounds, smells, tastes and other sensations, while his body remains hidden underwater.
A tadpole’s eyes are on the side of his head. As he develops into an adult frog, the eyes move to the top of his head. Most frogs have large eyes, placed to let them see to the sides and behind without moving. A frog also has excellent night vision and depth perception, and he can detect the slightest movement.
A frog does not have outer ears, but he does have an eardrum on each side of his head, called a tympanum. The tympanum transfers sound vibrations to the inner ear. The tympanum also allows the frog to maintain a sense of balance. A frog listens for a variety of calls from other frogs. Males listen for the calls of rival males, and females listen for the calls of potential mates. Frogs also listen for distress calls from other frogs that warn when a threat is present.
Like us, a frog uses his two nostrils to sample odors in the air. He also has a second type of olfactory organ between the nostrils, called the Jacobson’s organ. It is used to detect chemicals in the water. Because all of his olfactory organs are on the top of his head, the frog can sample air and water odors simultaneously by putting his nostrils just above the water’s surface.
A frog will eat just about any living thing he can fit in his mouth, but he also has sensitive taste buds. He will occasionally spit out hastily grabbed prey if the taste is unpleasant. The taste buds are on the surface of his tongue and the inside of his mouth. Completely aquatic frogs in the Pipidae family do not have tongues, but they have taste discs in their mouth tissue to receive taste sensations.
Through his skin, a frog can learn a lot about himself and his environment. He can detect temperature, pressure, touch and pain. An aquatic frog has one additional feature, making him uniquely adapted for life underwater -- the lateral line. The lateral line receptors are present not only on the head and around the eyes, but on the body and neck as well. They detect vibrations through the water, giving the frog an idea of the shape and direction of a prey item in the water.
- Herpetology: An Introductory Biology of Amphibians and Reptiles, George R. Zug et al.
- Frogs: A Chorus of Colors; John L. Behler et al.
- Biology Teaching and Learning Resources: Frogs - an Introduction
- Britannica Online Encyclopedia – Jacobson’s organ (zoology)
- Physics News Update - Good Vibrations Help a Frog Locate Tasty Prey
- Sensory Evolution on the Threshold: Adaptations in Secondarily Aquatic Vertebrates; J. G. M. Thewissen, Sirpa Nummela
- Digital Vision/Digital Vision/Getty Images<|endoftext|>
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We solve quadratic equations by either factoring or using the quadratic formula.
Definition of the Discriminant We define the discriminant of the quadratic ax2 + bx + c as D = b2 - 4ac
The discriminant is the number under the square root in the quadratic formula. We immediately get
D # of Roots > 0 2 < 0 0 0 1
Example
How many roots does
1045456564x2 + 3x + 2134534265256
have?
Solution
It is clear that 4ac is larger than b2 = 9.
Hence
D = 9 - 4ac < 0
So that the quadratic has no real roots.
Example:
Solve
x2 - x - 6 > 0
Solution:
First we solve the equality by factoring:
(x - 3)(x + 2) = 0
Hence
x
= -2 or x = 3
Next we cut the number line into three regions:
x < -2, -2 < x < 3, and x > 3
On the first region (test x = -3), the quadratic is positive, on the second region (test x = 0) the quadratic is negative, and on the third region (test x = 5) the quadratic is positive.
Region Test Value y-Value Sign x < 2 x = -3 y = 6 + -2 < x < 3 x = 0 y = -6 - x > 3 x = 5 y = 14 +
We are after the positive values since the equation is "> 0".Hence our solution is region 1 and region 2.
x < -2 or x > 3
We will see how to verify this on a graphing calculator by noticing that
y = x2 - x - 6
stays above the x-axis when x < -2 and when x > 3.
3. Applications
A 4 ft walkway surrounds a circular flower garden, as shown in the sketch. The area of the walk is 44% of the area of the garden. Find the radius of the garden.
Solution:
Area of the walk = p(4 + r)2 - p( r)2 = .44(p)( r)2
Dividing by p we have,
(4 + r)2 - r2 = .44r2
multiplying out, we get,
16 + 8r + r2 -r2 = .44r2
or
.44r2 -8r -16
a = .44, b = -8, c = -16
so
r = 1.1 or r = -.1
since -.1 does not make sense, we can say that the radius of the garden is 1.1feet.
Example:
The profit function for burgers at Heavenly is given by
P = 35x - x2/25,000,000 - 40,000.
Where x represents the number of skiers that come on a given day. How many skiers paying for Heavenly will produce the maximal profit?<|endoftext|>
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Social Reform and Issues of Race and Class
In this activity students explore how Progressive Era reforms did not apply universally, but rather varied depending on issues like race and class. Students watch the 30-minute film Heaven Will Protect the Working Girl and read an article that explains tensions among immigrants and African Americans in the Progressive Era.
Students will examine the experiences of African Americans during the Uprising of the 20,000.
Students will analyze the ways that race and class affected the goals and impacts of social reform movements.
This activity supports the following Common Core Literacy Standards in History/Social Studies:
WHSS.6-8.2. Write informative/explanatory texts.
Step 1: Have students watch the 30-minute film Heaven Will Protect the Working Girl. Alternatively, students can read relevant passages from the viewing guide for the film.
Step 2: Give each student a copy of the excerpt from Meredith Tax's The Rising Women and the triple-entry journal form. Each student should read the excerpt and take notes with the form:
In column A, note key facts, words or phrases
In column B, note your reactions to the reading
At the bottom of the page, identify areas or issues you wanted to know more about as a result of this reading.
Step 3: Now have students choose a partner (or divide students into pairs). With their partners, students should exchange their notes on the reading and write their responses to the notes in Column C. After noting their responses, students should pass the journal form back to their partners for them to read. Partners should discuss the issues raised with each other.
Step 4: Lead the entire group in a discussion of the reading and the questions it raised. Discussion questions include:
What has the reading informed you about issues of race, class and gender?
Do similar issues resonate today?
What do you still want to know more about?
Step 5: Ask students to imagine that they are journalists in 1909. Using the information provided in the film, the viewer's guide and Meredith Tax's excerpt, they should write an editorial about the Uprising of the 20,000, focusing on the situation of black women.
As millions of immigrants flooded into turn-of-the-century New York City, they encountered a smaller but growing African-American community, and America's heritage of racial inequality. In many areas of life, immigrants and African-Americans found themselves in competition; in some areas they struggled to find common ground.
Creator | American Social History Project/Center for Media and Learning
Rights | Copyright American Social History Project/Center for Media and Learning This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
Item Type | Teaching Activity
Cite This document | American Social History Project/Center for Media and Learning, “Social Reform and Issues of Race and Class,” HERB: Resources for Teachers, accessed May 26, 2019, https://herb.ashp.cuny.edu/items/show/1496.<|endoftext|>
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[ Home ]
How to add, subtract, multiply and divide complex numbers.
This topic is part of the TCS FREE high school mathematics 'How-to Library', and shows you how to add, subtract, multiply and divide complex numbers. (See the index page for a list of all available topics in the library.) To make best use of this topic, you need to download the Maths Helper Plus software. Click here for instructions.
Theory:
This topic shows you how to add, subtract, multiply and divide complex numbers.
If 'A' and 'B' are two complex numbers, so that:
A = a + bi and B = c + di
Then we can sum A and B like this:
A + B = a + bi + c + di
= (a + c) + (b + d)i
For example, if A = 1 + 2i, and B = 3 - 4i, then:
A + B = (1 + 3) + (2 - 4)i
= 4 - 2i
2 Subtracting complex numbers
If 'A' and 'B' are two complex numbers, so that:
A = a + bi and B = c + di
Then we can subtract B from A like this:
A - B = a + bi - ( c + di)
= (a - c) + (b - d)i
For example, if A = 1 + 2i, and B = 3 - 4i, then:
A - B = (1 - 3) + (2 - -4)i
= -2 + 6i
3 Multiplying complex numbers
If 'A' and 'B' are two complex numbers, so that:
A = a + bi and B = c + di
Then we can multiply A and B like this:
A × B = ( a + bi ) × ( c + di )
= a × c + a ×di + bi × c + bi × di
= ac + adi + bci + bdi²
But i²-1
So A × B = ( ac - bd ) + ( ad + bc )i
For example, if A = 1 + 2i, and B = 3 - 4i, then:
A × B = (1×3 - 2×(-4)) + (1×(-4) + 2×3)i
= 11 + 2i
4 Dividing complex numbers
If 'A' and 'B' are two complex numbers, so that:
A = a + bi and B = c + di
Then we can divide A by B like this:
For example, if A = 1 + 2i, and B = 3 - 4i, then:
It should be noted that the denominator terms: ( c2 + d2 ), are the square of the magnitude of complex number 'B'. ( c2 + d2 ) = |B|², which leads to the more compact version of the division formula:
where |B|² = ( c2 + d2 )
The 'Method' section below shows you how Maths Helper Plus can display all of the working steps for adding, subtracting, multiplying or dividing complex numbers.
Method:
In the following steps, we will use the examples from the 'Theory' section above. To solve your own complex number problems, insert your own numbers.
If you have just launched the software then you already have an empty document, otherwise hold down ‘Ctrl’ while you briefly press the ‘N’ key.
Step 2 Create a complex calculator
1. Press the F3 key to activate the 'input box' for typing (see below):
2. Type: com into the input box:
3. Press Enter to complete the entry
The 'options' box for the complex calculator will display immediately.
Click the 'Complex Calculations' tab (see below) at the top of the options box to display the calculation options:
Click to select the operations that you want to see working steps for.
Step 3 Enter the complex numbers A and B
Click the 'Complex Editor' tab at the top of the options box. This displays the 'Complex Editor' tab where you enter the complex numbers A and B to create the working steps. See below:
Click on the 'A' edit box and type complex number A like this:
real part , imaginary part
( So, to enter A = 1 + 2i, type: 1,2 )
The edit box displays in yellow to show that the values have been changed.
Similarly, click on the 'B' edit box and type complex number B.
( So, to enter B = 3 - 4i, type: 3,-4 )
Now, click the 'Apply' button at the bottom of the edit box. The text view will display the complex number calculations.
IMPORTANT: Each time you change either 'A' or 'B', remember to click the 'Apply' button to update the calculations.
The text view displays the results. To see the results, you will probably need to close the complex calculator options box by clicking the OK button.
Here are the results displayed by Maths Helper Plus for the 'A' and 'B' values entered above:
Complex calculations
Complex Sum:
Let A = a + bi, and B = c + di, then
A + B = (a + c) + (b + d)i
= (1 + 3)
+ (2 + -4)i
= 4 + -2i
|A + B| = 4.47214, arg(A+B) = -26.5651°
Complex Difference:
Let A = a + bi, and B = c + di, then
A - B = (a - c) + (b - d)i
= (1 - 3)
+ (2 - -4)i
= -2 + 6i
|A - B| = 6.32456, arg(A-B) = 108.435°
Complex Product:
Let A = a + bi, and B = c + di, then
A × B = (ac - bd) + (ad + bc)i
= (1 × 3 - 2 × -4)
+ (1 × -4 + 2 × 3)i
= 11 + 2i
|A × B| = 11.1803, arg(A × B) = 10.3048°
Complex Division:
Let A = a + bi, and B = c + di,
so |B|² = c² + d²
= 3² + -4²
= 25
A ÷ B = [(ac + bd)/|B|²] + [(bc - ad)/|B|²]i
= [ (1 × 3
+ 2 × -4)/25 ]
+ [ (2 × 3
- 1 × -4)/25 ]i
= -0.2 + 0.4i
|A ÷ B| = 0.447214, arg(A ÷ B) = 116.565°
How to make changes to your complex settings
To display the complex calculator options box at any time, double click on the text view to the left of the results display, OR if there are no results displayed, to the left of the words 'Complex Calculator'.
This picture (right) shows you where to double click... If there are results displayed, double click in the area we have shaded blue. Otherwise, double click in the area we have shaded pink, then choose the required tab at the top of the options box when it is displayed.<|endoftext|>
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Post provided by MATT MALISHEV (@DARWINANDDAVIS)
Changes in temperature and available food determine where and when animals move, reproduce, and survive. Our understanding of how environmental change impacts biodiversity and species survival is well-established at the landscape, country and global scales. But, we know less about what could happen at finer space and time scales, such as within habitats, where behavioural responses by animals are crucial for daily survival.
Simulating Movement and Daily Survival with Individual-Based Movement Models
Key questions at these scales are how the states of individuals (things like body temperature and nutritional condition) influence movement decisions in response to habitat change, and how these decisions relate to patchiness in microclimates and food. So we need tools to make reliable forecasts of how fine-scale habitat use will change under future environments. Individual-based movement simulation models are powerful tools for these kinds of studies. They let you construct habitats that vary in temperature and food conditions in both space and time and ask ‘what if’ questions. By populating these models with activity, behaviour, and movement data of animals, we can simulate different habitat conditions and predict how animals will respond to future change. Continue reading<|endoftext|>
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When thinking about Ebola, Australia is not the first country that comes to mind. Africa and some of the other developing nations are more in line with community feelings and assumptions, and while it’s true that (particularly in the latest outbreak) Liberia, Sierra Leone and Guinea are the victims, Australia and our neighbours must be mindful of the threat – and be prepared.
What is Ebola? Ebola is a disease transmitted by one of the five Ebola virus strains. It is passed from person to person via contact; that is, you must contact an infected person’s bodily fluids (such as blood, secretions or organs). It is not possible to contract Ebola by breathing the same air as an infected person, drinking water or consuming food that has been properly prepared and cooked, from mosquitoes or from uninfected people. Ebola transmission is from touching a person or surface that has the virus on it.
Outbreaks are thought to begin with something like an infected wild animal being consumed, although the fruit bat population in the affected countries are carriers of the virus. Outbreaks occur after a person is infected by the virus and it spreads throughout a community and beyond, and unfortunately in Africa, conditions are ripe for swift transmission (poor healthcare infrastructure, poverty and ignorance, testing regimes not being adhered to and false beliefs surrounding ‘outsiders’ helping all contribute to fast and seemingly unstoppable outbreaks). There is no cure for Ebola at the present time, and it has a mortality rate of up to 80%.
The first symptoms of Ebola are very much like those of the flu: the sufferer displays a high fever, malaise and a sore throat. The viruses cause clots to form, which causes infarction. The virus also attacks collagen in connective tissue and internal organs, causing both internal and external bleeding. Sufferers pass away within days of contracting the disease, from kidney failure or blood loss, amongst other things.
But there is some good news – for every person infected with Ebola, it is estimated that they will affect only two others, much like Swine Flu. However measles, for example, can affect 18 people for every infected person. And Ebola prevention can be as simple as thoroughly cooking any bushmeat you want to consume, washing (or sanitising) your hands thoroughly and frequently, and not touching anyone or anything that may have come into contact with the virus.
If you have recently travelled to West Africa and are experiencing symptoms similar to those of Ebola, it’s best to have yourself examined as soon as possible. At 1stAvailable, you can book an appointment online at any time of the day or night. It’s a free service; you don’t have to wait until morning to make the call for an appointment. Put your mind at rest and have yourself checked out, but bear in mind most cases of flu-like symptoms in people who’ve not travelled to infected areas are just that – the flu.<|endoftext|>
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# How do you find lim (2-sqrt(x+2))/(4-x^2) as x->2 using l'Hospital's Rule?
Jan 8, 2017
${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}} = \frac{1}{16}$
#### Explanation:
First we observe that:
${\lim}_{x \to 2} 2 - \sqrt{x + 2} = 0$
${\lim}_{x \to 2} 4 - {x}^{2} = 0$
So the limit:
${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}}$
is in the indeterminate form $\frac{0}{0}$ and we can use l'Hospital's rule:
${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}} = {\lim}_{x \to 2} \frac{\frac{d}{\mathrm{dx}} \left(2 - \sqrt{x + 2}\right)}{\frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right)}$
Calculate the derivatives:
$\frac{d}{\mathrm{dx}} \left(2 - \sqrt{x + 2}\right) = - \frac{1}{2 \sqrt{x + 2}}$
$\frac{d}{\mathrm{dx}} \left(4 - {x}^{2}\right) = - 2 x$
So:
${\lim}_{x \to 2} \frac{2 - \sqrt{x + 2}}{4 - {x}^{2}} = {\lim}_{x \to 2} \frac{1}{4 x \left(\sqrt{x + 2}\right)} = \frac{1}{16}$<|endoftext|>
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Education Projectors Buyer's Guide
Aspect ratio defines the relationship between the width and the height of an image. For example, if your old television screen was 40 inches wide and 30 inches high it would have an aspect ratio of 40/30, which is equivalent to 4/3 and commonly shown as 4:3. So the aspect ratio defines the relationship of the width and height of the display, but not its size. It gives a sense of how square or rectangular the image is.
Have you ever borrowed a projector for your class and seen black bars on either the top and bottom or sides of the projected image? There are two ways this can happen: 1) if the projector has a different aspect ratio than the screen, or 2) if the material you're projecting has a different aspect ratio than the projector.
Nearly all projectors today allow you to electronically stretch or shrink an image to fill all or most of the screen; however, there is always some loss of detail when doing this and it is more noticeable on text than video. The finer the text the more evident the problem.
If you intend to use HDTV, its aspect ratio is 16:9 meaning it is 16 units wide for every nine units high. This is similar to what you might find at a movie theater or even your home television. So 16:9 is proportionally wider for a more panoramic experience while the 4:3 of regular TV looks almost square by comparison. There are many aspect ratios in both the computer and video worlds. The emerging standard for many laptop computers is 16:10, which is close enough to 16:9 to hardly notice the difference.
In choosing your projector, you need to determine which aspect ratio is most prevalent in the multimedia curriculum you’ve developed for your students and, perhaps more importantly, which way are you headed in the future. Consider the DVDs, tapes, computers, and televisions you already have to help you decide which aspect ratio to choose. If standard definition is the bulk of your content and you are not transitioning to high definition, then a 4:3 projector may be best. If you have high definition content or a mix of standard definition and high definition content, then 16:9 may be best.<|endoftext|>
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## Announcements:
Today we finish 4.3 and start 4.4. Continue reading Section 4.4 for Wednesday. Work through recommended homework questions.
Tutorials: Quiz 8 covers 4.2, 4.3, and the parts of Appendix D that we covered in class.
Help Centers: Monday-Friday 2:30-6:30 in MC 106.
The final exam will take place on Tuesday, April 22, 2-5pm. All students write in AH201 (Alumni Hall). The final exam will cover all the material from the course, but will emphasize the material after the midterm. See the course home page for final exam conflict policy. You should immediately notify the registrar or your Dean (and your instructor) of any conflicts!
### Partial review of Section 4.3
The eigenvalues of a square matrix $A$ can be computed as the roots (also called zeros) of the characteristic polynomial $$\det ( A - \lambda I)$$
Theorem D.2 (The Factor Theorem): Let $f$ be a polynomial and let $a$ be a constant. Then $a$ is a root of $f(x)$ (i.e. $f(a) = 0$) if and only if $x - a$ is a factor of $f(x)$ (i.e. $f(x) = (x - a) g(x)$ for some polynomial $g$).
The largest $k$ such that $(x-a)^k$ is a factor of $f$ is called the multiplicity of the root $a$ in $f$.
Example: Let $f(x) = x^2 - 2x + 1$. Since $f(1) = 1 - 2 + 1 = 0$, $1$ is a root of $f$. And since $f(x) = (x-1)^2$, $1$ has multiplicity $2$.
In the case of an eigenvalue, we call its multiplicity in the characteristic polynomial the algebraic multiplicity of this eigenvalue.
We also define the geometric multiplicity of an eigenvalue $\lambda$ to be the dimension of the corresponding eigenspace $E_\lambda$.
Theorem 4.15: The eigenvalues of a triangular matrix are the entries on its main diagonal (repeated according to their algebraic multiplicity).
Theorem D.4 (The Fundamental Theorem of Algebra): A polynomial of degree $n$ has at most $n$ distinct roots. In fact, the sum of the multiplicities is at most $n$.
Therefore:
Theorem: An $n \times n$ matrix $A$ has at most $n$ distinct eigenvalues. In fact, the sum of the algebraic multiplicities is at most $n$.
### Partial review of Appendix C
A complex number is a number of the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is a symbol such that $i^2 = -1$.
Addition: $(a+bi)+(c+di) = (a+c) + (b+d)i$, like vector addition.
Multiplication: $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$. (Explain.)
The conjugate of $z = a+bi$ is $\bar{z} = a-bi$. Reflection in real axis. We learned the properties of conjugation.
The absolute value or modulus $|z|$ of $z = a+bi$ is $$\kern-4ex |z| = |a+bi| = \sqrt{a^2+b^2}, \qtext{the distance from the origin.}$$ Note that $$\kern-7ex z \bar{z} = (a+bi)(a-bi) = a^2 -abi+abi-b^2 i^2 = a^2 + b^2 = |z|^2$$ This means that for $z \neq 0$ $$\kern-4ex \frac{z \bar{z}}{|z|^2} = 1 \qtext{so} z^{-1} = \frac{\bar{z}}{|z|^2}$$ This can be used to compute quotients of complex numbers: $$\kern-4ex \frac{w}{z} = \frac{w}{z} \frac{\bar{z}}{\bar{z}} = \frac{w \bar{z}}{|z|^2}.$$ Example: $$\kern-8ex \frac{-1+2i}{3+4i} = \frac{-1+2i}{3+4i} \frac{3-4i}{3-4i} = \frac{5+10i}{3^2+4^2} = \frac{5+10i}{25} = \frac{1}{5} + \frac{2}{5}i$$
We learned the properties of absolute value. One of them was $|w z| = |w| |z|$.
A complex number $z = a + bi$ can also be expressed in polar coordinates $(r, \theta)$, where $r = |z| \geq 0$ and $\theta$ is such that $$\kern-6ex a = r \cos \theta \qqtext{and} b = r \sin \theta$$ Then $$\kern-6ex z = r \cos \theta + (r \sin \theta) i = r(\cos \theta + i \sin \theta)$$
Let $$\kern-7ex z_1 = r_1(\cos \theta_1 + i \sin\theta_1) \qtext{and} z_2 = r_2(\cos \theta_2 + i \sin\theta_2) .$$ Then \kern-9ex \begin{aligned} z_1 z_2 &= r_1 r_2 (\cos \theta_1 + i \sin\theta_1) (\cos \theta_2 + i \sin\theta_2) \\ &= r_1 r_2 [(\cos \theta_1 \cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1 \cos\theta_2 + \cos\theta_1\sin\theta_2)] \\ &= r_1 r_2 [\cos(\theta_1 + \theta_2) +i \sin(\theta_1+\theta_2)] \end{aligned} So $$\kern-8ex |z_1 z_2| = |z_1| |z_2| \qtext{and} \Arg(z_1 z_2) = \Arg z_1 + \Arg z_2$$ (up to multiples of $2\pi$).
In particular, if $z = r (\cos \theta + i \sin \theta)$, then $z^2 = r^2 (\cos (2 \theta) + i \sin (2 \theta))$. It follows that the two square roots of $z$ are $$\pm \sqrt{r} (\cos (\theta/2) + i (\sin \theta/2))$$
### New material: complex eigenvalues and eigenvectors
This material isn't covered in detail in the text.
Example 4.7: Find the eigenvalues of $A = \bmat{rr} 0 & -1 \\ 1 & 0 \emat$ (a) over $\R$ and (b) over $\C$.
Solution: We must solve $$0 = \det(A-\lambda I) = \det \bmat{cc} -\lambda & -1 \\ 1 & -\lambda \emat = \lambda^2 + 1 .$$ (a) Over $\R$, there are no solutions, so $A$ has no real eigenvalues. This is why the Theorem above says "at most $n$". (This matrix represents rotation by 90 degrees, and we also saw geometrically that it has no real eigenvectors.)
(b) Over $\C$, the solutions are $\lambda = i$ and $\lambda = -i$. For example, the eigenvectors for $\lambda = i$ are the nonzero complex multiples of $\coll i 1$, since $$\bmat{rr} 0 & -1 \\ 1 & 0 \emat \coll i 1 = \coll {-1} i = i \coll i 1 .$$ In fact, $\lambda^2 + 1 = (\lambda - i)(\lambda + i)$, so each of these eigenvalues has algebraic multiplicity 1. So in this case the sum of the algebraic multiplicities is exactly 2.
The Fundamental Theorem of Algebra can be extended to say:
Theorem D.4 (The Fundamental Theorem of Algebra): A polynomial of degree $n$ has at most $n$ distinct complex roots. In fact, the sum of their multiplicities is exactly $n$.
Another way to put it is that over the complex numbers, every polynomial factors into linear factors.
### Real matrices
Notice that $i$ and $-i$ are complex conjugates of each other.
If the matrix $A$ has only real entries, then the characteristic polynomial has real coefficients. Say it is $$\kern-6ex \det(A - \lambda I) = a_n \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_1 \lambda + a_0 ,$$ with all of the $a_i$'s real numbers. If $z$ is an eigenvalue, then so is its complex conjugate $\bar{z}$, because \kern-8ex \begin{aligned} &a_n \bar{z}^n + a_{n-1} \bar{z}^{n-1} + \cdots + a_1 \bar{z} + a_0 \\[5pt] &\quad= \overline{a_n {z}^n + a_{n-1} {z}^{n-1} + \cdots + a_1 {z} + a_0} = \bar{0} = 0. \end{aligned}
Theorem: The complex eigenvalues of a real matrix come in conjugate pairs.
### Complex matrices
A complex matrix might have real or complex eigenvalues, and the complex eigenvalues do not have to come in conjugate pairs.
Examples: $\bmat{rr} 1 & 2 \\ 0 & i \emat$, $\bmat{rr} 1 & i \\ 0 & 2 \emat$.
### General case
In general, don't forget that the quadratic formula $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$ gives the roots of $a x^2 + b x + c$, and these can be real (if $b^2 - 4ac \geqslant 0$) or complex (if $b^2 - 4ac < 0$). This formula also works if $a$, $b$ and $c$ are complex.
Also don't forget to try small integers first.
Example: Find the real and complex eigenvalues of $A = \bmat{rrr} 2 & 3 & 0 \\ 1 & 2 & 2 \\ 0 & -2 & 1 \emat$.
Solution: \kern-8ex \begin{aligned} \bdmat{ccc} 2-\lambda & 3 & 0 \\ 1 & 2-\lambda & 2 \\ 0 & -2 & 1-\lambda \edmat &= (2 - \lambda) \bdmat{cc} 2 - \lambda & 2 \\ -2 & 1-\lambda \edmat - 3 \bdmat{cc} 1 & 2 \\ 0 & 1-\lambda \edmat \\ &= (2 - \lambda) ( \lambda^2 - 3 \lambda + 6 ) - 3 (1-\lambda) \\ &= - \lambda^3 + 5 \lambda^2 - 9 \lambda + 9 . \end{aligned} By trial and error, $\lambda = 3$ is a root. So we factor: $$- \lambda^3 + 5 \lambda^2 - 9 \lambda + 9 = (\lambda - 3)(\query{-} \lambda^2 + \query{2} \lambda \toggle{+ \text{?}}{-3}\endtoggle)$$ We don't find any obvious roots for the quadratic factor, so we use the quadratic formula: \kern-6ex \begin{aligned} \lambda &= \frac{-2 \pm \sqrt{2^2 - 4(-1)(-3)}}{-2} = \frac{-2 \pm \sqrt{-8}}{-2} \\ &= \frac{-2 \pm 2 \sqrt{2} \, i}{-2} = 1 \pm \sqrt{2} \, i . \end{aligned} So the eigenvalues are $3$, $1 + \sqrt{2} \, i$ and $1 - \sqrt{2} \, i$.
Note: Our questions always involve real eigenvalues and real eigenvectors unless we say otherwise. But there will be problems where we ask for complex eigenvalues.
### More review: Eigenvalues of powers and inverses
Theorem 4.18: If $\vx$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $\vx$ is an eigenvector of $A^k$ with eigenvalue $\lambda^k$. This holds for each integer $k \geq 0$, and also for $k < 0$ if $A$ is invertible.
We saw that this was useful computationally. We also saw:
Theorem 4.20: If $\vv_1, \vv_2, \ldots, \vv_m$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_m$, then $\vv_1, \vv_2, \ldots, \vv_m$ are linearly independent.
We saw that sometimes the eigenvectors span $\R^n$, and sometimes they don't.
### Section 4.4: Similarity and Diagonalization
We're going to introduce a new concept that will turn out to be closely related to eigenvalues and eigenvectors.
Definition: Let $A$ and $B$ be $n \times n$ matrices. We say that $A$ is similar to $B$ if there is an invertible matrix $P$ such that $P^{-1} A P = B$. When this is the case, we write $A \sim B$.
It is equivalent to say that $AP = PB$ or $A = PBP^{-1}$.
Example 4.22: Let $A = \bmat{rr} 1 & 2 \\ 0 & -1 \emat$ and $B = \bmat{rr} 1 & 0 \\ -2 & -1 \emat$. Then $A \sim B$, since $$\bmat{rr} 1 & 2 \\ 0 & -1 \emat \bmat{rr} 1 & -1 \\ 1 & 1 \emat = \bmat{rr} 1 & -1 \\ 1 & 1 \emat \bmat{rr} 1 & 0 \\ -2 & -1 \emat.$$ We also need to check that the matrix $P = \bmat{rr} 1 & -1 \\ 1 & 1 \emat$ is invertible, which is the case since its determinant is $2$.
It is tricky in general to find such a $P$ when it exists. We'll learn a method that works in a certain situation in this section.
Theorem 4.21: Let $A$, $B$ and $C$ be $n \times n$ matrices. Then:
a. $A \sim A$.
b. If $A \sim B$ then $B \sim A$.
c. If $A \sim B$ and $B \sim C$, then $A \sim C$.
Proof: (a) $I^{-1} A I = A$
(b) Suppose $A \sim B$. Then $P^{-1}AP = B$ for some invertible matrix $P$. Then $PBP^{-1} = A$. Let $Q = P^{-1}$. Then $Q^{-1}BQ = A$, so $B \sim A$.
(c) Exercise.$\quad\Box$
Similar matrices have a lot of properties in common.
Theorem 4.22: Let $A$ and $B$ be similar matrices. Then:
a. $\det A = \det B$
b. $A$ is invertible iff $B$ is invertible.
c. $A$ and $B$ have the same rank.
d. $A$ and $B$ have the same characteristic polynomial.
e. $A$ and $B$ have the same eigenvalues.
Proof: Assume that $P^{-1}AP = B$ for some invertible matrix $P$.
We discussed (a) last time: \begin{aligned} \det(B) &= \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P)\\ &= \frac{1}{\det (P)} \det(A) \det(P) = \det A . \end{aligned} (b) follows immediately.
(c) takes a bit of work and will not be covered.
(d) follows from (a): since $B - \lambda I = P^{-1} A P - \lambda I = P^{-1} (A - \lambda I) P$ it follows that $B - \lambda I$ and $A - \lambda I$ have the same determinant.
(e) follows from (d).$\quad\Box$
Question: Are $\bmat{rr} 1 & 2 \\ 3 & 4 \emat$ and $\bmat{rr} 1 & 1 \\ 2 & -1 \emat$ similar?
Question: Are $\bmat{rr} 1 & 1 \\ 0 & 1 \emat$ and $\bmat{rr} 1 & 0 \\ 0 & 1 \emat$ similar?
### Diagonalization
Definition: $A$ is diagonalizable if it is similar to some diagonal matrix.
Example 4.24: $A = \bmat{rr} 1 & 3 \\ 2 & 2 \emat$ is diagonalizable. Take $P = \bmat{rr} 1 & 3 \\ 1 & -2 \emat$. Then $$P^{-1} A P = \cdots = \bmat{rr} 4 & 0 \\ 0 & -1 \emat$$ If $A$ is similar to a diagonal matrix $D$, then $D$ must have the eigenvalues of $A$ on the diagonal. But how to find $P$?
On board: notice that the columns of $P$ are eigenvectors for $A$!
Theorem 4.23: Let $A$ be an $n \times n$ matrix. Then $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.
More precisely, there exist an invertible matrix $P$ and a diagonal matrix $D$ with $P^{-1}AP = D$ if and only if the columns of $P$ are $n$ linearly independent eigenvectors of $A$ and the diagonal entries of $D$ are the corresponding eigenvalues in the same order.
This theorem is one of the main reasons we want to be able to find eigenvectors of a matrix. Moreover, the more eigenvectors the better, so this motivates allowing complex eigenvectors. We're going to say a lot more about diagonalization.<|endoftext|>
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Aggregate Expenditure, Economic Output, and Inflation
Potential output is what the economy can produce when its resources are used at normal rates, which is also considered the full-employment Gross Domestic Product (GDP). This is considered the maximum sustainable output by an economy. Aggregate output is the amount actually produced. In spite of its name, potential output is not the greatest output by an economy, but it is the greatest output possible without straining economic resources, such as requiring overtime labor, and, therefore, it is the output that the economy gravitates to after an output gap is created, such as by changes in inflation or in the cost of resources. Over time, the amount of labor and capital increases, and technology improves, with the concomitant increase in economic efficiency, resulting in increases in economic output.
However, over the short term, inflation, interest rates, and other factors can cause aggregate output to fall below potential output. Hence, a main objective of modern monetary policy is to adjust inflation, usually accomplished by adjusting the target interest rate, to return economic output to its potential output. However, monetary policy can only be successful if the relationships between economic output, inflation, and interest rates are understood.
Inflation and Aggregate Expenditure
Aggregate expenditure is the total amount spent for the economy's output by all households, firms, foreigners, and the government. Prices are determined by the equilibrium between aggregate demand and aggregate supply, but aggregate expenditure is the amount actually spent, revealing actual demand at current prices and aggregate supply.
When aggregate expenditure is less than aggregate output, inventories will increase and prices will fall. Firms will lower production and lay off workers to save on costs and lower prices to sell their inventory. If the resulting aggregate output is less than potential output, then this will cause a recessionary output gap (aka recessionary gap), leading to a fall in employment and inflation.
By contrast, when aggregate expenditure exceeds aggregate output, inventories fall, causing firms to hire more workers and to invest in other factors of production, such as land or machinery, to produce more. If aggregate expenditure exceeds the potential output of the economy, then firms will have to pay higher prices for its factors of production, including overtime to its workers, and pay higher variable costs when using existing facilities for longer time periods to increase production. This increased demand leads to an expansionary output gap (aka inflationary gap), which, in turn, increases prices sharply — demand-pull inflation.
Higher inflation will eventually cause aggregate expenditures to decrease, because higher prices reduces the wealth of consumers, thus leading to lower spending. This is particularly true for poorer consumers, since they tend to spend all the money that they have, to pay for essentials. Generally, the marginal propensity to consume, which is the proportion of income spent for products and services, is inversely proportional to the wealth of the consumer, so inflation will curtail the spending of poorer consumers compared to wealthier consumers, which will lower aggregate expenditures even more than if the effects of inflation were spread evenly across the population.
Greater risk and cheaper imports also decrease aggregate expenditures. Higher inflation causes increased risk, motivating people to save more, rather than spend. Moreover, imports become cheaper relative to domestically produced goods and services, so a greater proportion of the money spent is for imports.
Components of Aggregate Expenditure
The components of aggregate expenditure consist of household consumption, business investment, government purchases, and net exports, equal to exports minus imports. Business investment is the purchase of real capital used in the production of its goods or services. Although most people consider investments to be the purchase of securities, such as stocks or bonds, this money is ultimately used by firms to buy products and services to produce final goods and services. Otherwise, financial instruments would not earn a return on investment unless that money can be put to work to create income from real goods or services.
Since aggregate expenditure is the total spending on the economy's total output, imports are subtracted from exports, since the amount spent on imports is the amount not spent on the economy's output.
Aggregate Expenditure = Consumption + Investment + Government Purchases + Net Exports.
Net Exports = Exports – Imports
The aggregate expenditure equation is usually written as:
Y = C + I + G + (X – M) = GDP
Note that aggregate expenditure equals the GDP.
Long-Run Real Interest Rates and Aggregate Expenditure
Interest rates are determined by the equilibrium between the supply and demand of loanable funds. Lenders want to be compensated for the risk of credit defaults, the opportunity cost of deferring purchases or investments for the term of the loan period (what economists call time preference or the marginal rate of time preference, which is expressed as an interest rate), and the expected rate of inflation. The nominal interest rate is the market rate of interest that includes compensation for the expected rate of inflation; the real interest rate equals the nominal rate minus the inflation premium. Central banks can manage the real interest rate by changing the supply of loanable funds, by changing the nominal interest rate. Real interest rates are considered rather than nominal interest rates because, over the long-term, the economy compensates for the inflation by simply changing nominal prices.
How does the real interest rate change the level of economic output?
Higher real interest rates reduces aggregate expenditure by increasing the cost of loans while increasing the earnings from savings. Both factors reduce expenditures by reducing consumption and investments, and therefore, aggregate expenditure. Higher real interest rates also increase the foreign exchange rate with other currencies, increasing the cost of exports while decreasing the cost for imports. Hence, more of the national income is spent for imports, while net exports decline. Lower real interest rates have the opposite effects.
Only government purchases are not sensitive to the interest rate.
If the supply of money grows only as fast as the economy, then there will be a long-run real interest rate that equates aggregate expenditure with the potential output quantity. An interest rate higher than the long-run real interest rate will cause the economy to contract, to produce less than its potential output, which will increase unemployment and lessen inflation, while an interest rate lower than the long-run rate will increase inflation.
Real interest rates are not the only thing that can change aggregate expenditure. Factors less sensitive to the interest rate, such as changes in consumer or business confidence or changes in government purchases or taxation, will also affect aggregate expenditure. Nonetheless, real interest rates will be a major factor in altering the economic output due to these other factors — either enhancing or blunting their effect.
Hence, interest rates are the major monetary policy tool used by central banks to keep economic output close to potential output, using inflation as a measure of the deviation from potential output. By keeping inflation low and steady, economic output will equal potential output.
Monetary Policy Reaction
The long-run real interest rate is the ideal rate because it is the rate that would prevail if economic output was a sustainable maximum, which maximizes the wealth of society. When the money supply grows with the economy, then the inflation rate would remain constant when the economy is producing its potential output. Therefore, changes in the inflation rate will reveal when the economy is deviating from its potential output.
When the real interest rate is plotted with respect to inflation, it forms the monetary policy reaction curve (MPRC). On this curve, there is an inflation rate that corresponds to the long-run real interest rate. So one monetary policy tool to achieve potential output is to adjust interest rates to achieve a target inflation rate that corresponds to the long-run real interest rate. These adjustments are often based on a rule, such as Taylor's rule, which provides some guidance as to how much the nominal interest rate should be changed to return the inflation rate to the desired target. Rules are used because it takes time to get feedback from changes in monetary policy, so using a rule that has produced the desired changes in the past is better than just changing monetary policy and waiting to see what effect it will have; the sooner that an effective monetary policy is implemented, the sooner the economy recovers.
When the inflation rate rises or falls, then the central bank will either increase or decrease nominal interest rates by slightly more than the change in inflation rate so that the economy can be returned to the target inflation rate.
Because the nominal interest rate equals the real interest rate plus expected inflation and because expected inflation takes time to change, changing the target nominal interest rate has the effect of changing the real interest rate, which changes the inflation rate.
Nominal Interest Rate = Real Interest Rate + Expected Inflation
Real Interest Rate = Nominal Interest Rate – Expected Inflation
Hence, when the central bank changes nominal interest rates, the real interest rate changes by the same amount. Since expected inflation does not change in the short term, changing the nominal interest rate directly changes the real interest rate.
When aggregate expenditure changes because of other factors that are not sensitive to the interest rate, such as changes in consumer or business confidence or changes in government purchases or taxation, then this has the effect of shifting the monetary policy reaction curve, causing the long-term real interest rate to correspond to a different inflation rate, or vice versa. Nonetheless, monitoring and adjusting the inflation rate is an effective means of monitoring and adjusting the economy.<|endoftext|>
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## During the summer, you work 30 hours per week at a gas station and earn \$8.75 per hour. You also work as a landscaper for \$11 per
Question
During the summer, you work 30 hours per week at a gas station and earn \$8.75 per hour. You also work as a landscaper for \$11 per hour and can work as many hours as you want. You want to earn a total of \$400 per week. How many hours, \$t\$ , must you work as a landscaper?
in progress 0
8 mins 2023-01-13T07:50:53+00:00 1 Answer 0 views 0
1. The number of hours to work as a landscaper is 12.5 hours.
Given,
you work 30 hours per week at a gas station and earn \$8.75 per hour.
You also work as a landscaper for \$11 per hour and can work as many hours as you want.
You want to earn a total of \$400 per week.
We need to find how many hours to work in landscraper to earn \$400 a week.
We have,
At the gas station:
\$8.75 = 1 hour
Find the dollars in 30 hours:
\$8.75 = 1 hour
Multiply 30 on both sides
30 x \$8.75 = 30 x 1 hour
\$262.5 = 30 hours
At landscaping:
1 hour = \$11
We need to have \$400 a week.
We already have \$262.5 earned from the gas station so we need to earn the remaining dollars from the landscaper.
Find the remaining dollars:
= \$400 – \$262.5
= \$137.5
Find the number of hours worked as a landscaper:
We have,
1 hour = \$11
Multiplying 137.5/11 on both sides
137.5/11 x 1 hour = 137.5/11 x \$11
12.5 hours = \$137.5
We see that to earn \$137.5 he needs to work in the landscraper for 12.5 hours.
We can also crosscheck:
30 x \$8.75 + 12.5 x \$11
\$262.5 + \$137.5
\$400
Thus the number of hours to work as a landscaper is 12.5 hours.<|endoftext|>
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Calculus 10.3 day 2 - University of Houston
10.3 day 2 Calculus of Polar Curves Lady Bird Johnson Grove, Redwood National Park, California Photo by Vickie Kelly, 2007 Greg Kelly, Hanford High School, Richland, Washington Try graphing this on the TI-89. r 2sin 2.15 0 16 To find the slope of a polar curve: dy dy d dx dx
d d r sin d d r cos d r sin r cos r cos r sin We use the product rule here. To find the slope of a polar curve: dy dy d dx dx d
d r sin d d r cos d r sin r cos r cos r sin dy r sin r cos dx r cos r sin Example: r 1 cos r sin
sin sin 1 cos cos Slope sin cos 1 cos sin 2 2 sin cos cos sin cos sin sin cos 2 2 sin cos cos 2sin cos sin cos 2 cos sin 2 sin Area Inside a Polar Graph:
The length of an arc (in a circle) is given by r. when is given in radians. For a very small , the curve could be approximated by a straight line and the area could be found using the triangle formula: A 1 bh 2 r r d 1 1 2 dA rd r r d 2 2 1 2 dA r d 2 We can use this to find the area inside a polar graph.
1 dA r 2 d 2 1 2 A r d 2 Example: Find the area enclosed by: r 2 1 cos 1 2 0 2 r d 2 1 2 4 1 cos d 0 2 2
2 2 1 2 cos cos d 2 0 2 0 1 cos 2 2 4 cos 2 d 2 2 0
1 cos 2 2 4 cos 2 d 2 2 3 4 cos cos 2 d 0 1 3 4sin sin 2 2 2 0 6 0 6 Notes:
To find the area between curves, subtract: 1 2 2 A R r d 2 Just like finding the areas between Cartesian curves, establish limits of integration where the curves cross. When finding area, negative values of r cancel out: r 2sin 2 Area of one leaf times 4: 2 1 2 A 4 2sin 2 d 2 0 A 2 Area of four leaves:
2 1 2 A 2sin 2 d 2 0 A 2 To find the length of a curve: Remember: ds dx 2 dy 2 For polar graphs: x r cos y r sin If we find derivatives and plug them into the formula, we (eventually) get: 2
dr ds r d d 2 So: Length 2 dr r d d 2
2 dr r d d 2 Length There is also a surface area equation similar to the others we are already familiar with: When rotated about the x-axis: 2 dr S 2 y r d
d 2 2 dr S 2 r sin r d d 2
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Question of Exercise 1
Question The average of 15 numbers is 18. The average of first 8 is 19 and that last 8 is 17, then the 8th number is
Option 1 15
Option 2 16
Option 3 18
Option 4 20
Represent Root 9 point 3 on the number line
Solution:
Explanation:
Step 1: Draw a line segment AB of length 9.3 units.
Step 2: Now, Extend the line by 1 unit more such that BC=1 unit .
Step 3: Find the midpoint of AC.
Step 4: Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.
Step 5: Draw an arc DE such that BE=BD.
Hence, Number line of √ 9.3 is attached below.
Which one of the following statement is true
A: Only one line can pass through a single point.
B: There are an infinite number of lines which pass through two distinct points.
C: Two distinct lines cannot have more than one point in common.
D: If two circles are equal, then their radii are not equal.
Solution:
Explanation:
From one point there is an uncountable number of lines that can pass through.
Hence, the statement “ Only one line can pass through a single point” is false.
We can draw only one unique line passing through two distinct points.
Hence, the statement “There are an infinite number of lines which pass through two distinct points” is false.
Given two distinct points, there is a unique line that passes through them.
Hence, the statement “Two distinct lines cannot have more than one point in common” is true.
If circles are equal, which means the circles are congruent. This means that circumferences are equal and so the radii of two circles are also equal.
Hence, the statement “If two circles are equal, then their radii are not equal” is false.
The correct option is (C) Two distinct lines cannot have more than one point in common.
The class mark of the class 90-120 is
A: 90
B: 105
C: 115
D: 120
Solution:
Explanation:
To find the class mark of a class interval, we find the sum of the upper limit and lower limit of a class and divide it by 2 Thus,
Class -mark=Upper limit + Lower limit/2
Here, the lower limit of 90-120=90
And the upper limit of 90-120=120
So,
Class -mark=120+90/2
=210/2
=105
Hence, the class mark of the class 90-120 is 105
That is, option (B) is correct.
Option (B) 105 is correct.
ABCD is a parallelogram and AP
and CQ are perpendiculars from vertices A and C on diagonal BD Show that i)ΔAPBΔCQD ii) AP = CQ
Solution:
Find the roots of the following equation
A: - 1, - 2
B: - 1, - 3
C: 1,3
D: 1,2
Solution:
Explanation:-<|endoftext|>
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This false-color mosaic of Saturn shows deep-level clouds silhouetted against Saturn's glowing interior. The image was made with data from Cassini's visual and infrared mapping spectrometer, which can image the planet at 352 different wavelengths.
This mosaic shows the entire planet, including features like Saturn's ring shadows and the terminator, the boundary between day and night.
The data were obtained in February 2006 at a distance of 1.6 million kilometers (1 million miles) from directly over the plane of Saturn's rings, which appear here as a thin, blue line over the equator. The image was constructed from images taken at wavelengths of 1.07 microns shown in blue, 2.71 microns shown in green, and 5.02 microns shown in red.
The blue-green color (lower right) is sunlight scattered off clouds high in Saturn's atmosphere and the red color (upper left) is the glow of thermal radiation from Saturn's warm interior, easily seen on Saturn's night side (top left), within the shadow of the rings, and with somewhat less contrast on Saturn's day side (bottom right). The darker areas within Saturn show the strongest thermal radiation. The bright red color indicates areas where Saturn's atmosphere is relatively clear. The great variety of cloud shapes and sizes reveals a surprisingly active planet below the overlying sun-scattering haze.
The brighter glow of the northern hemisphere versus the southern indicates that the clouds and hazes there are noticeably thinner than those in the south. Scientists speculate that this is a seasonal effect, and if so, it will change as the northern hemisphere enters springtime during the next few years.
The Cassini-Huygens mission is a cooperative project of NASA, the European Space Agency and the Italian Space Agency. The Jet Propulsion Laboratory, a division of the California Institute of Technology in Pasadena, manages the mission for NASA's Science Mission Directorate, Washington, D.C. The Cassini orbiter was designed, developed and assembled at JPL. The Visual and Infrared Mapping Spectrometer team is based at the University of Arizona where this image was produced.
Image credit: NASA/JPL/University of Arizona<|endoftext|>
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# Tangent Curiosity in Equilateral Triangle
### Problem
In an equilateral triangle $ABC,$ two circles are tangent to the base $BC,$ the circumcircle $(ABC)$ and to each other. $AP$ is tangent to one, $AQ$ to the other. $M$ and $N$ are their respective points of tangency with the base.
Prove that $AP+AQ=2MN.$
### Solution 1
As we shall see, the requirement that the two circles are tangent to each other is a red herring: the required identity holds even if they are not. The purpose of the applet above was to suggest an even stronger result.
The proof is analytic. Let $B=(-3,0),$ $C=(3,0),$ $A(0,3\sqrt{3}).$ This makes the side length of $\Delta ABC$ $6,$ the circumradius $AO=R=2\sqrt{3}$ and the inradius $OH=r=\sqrt{3}.$ $H$ the midpoint of the base is at the origin. We focus on the left circle as shown.
There is a good deal of right triangles that, through a repeated use of the Pythagorean theorem, lead to the relation
$AP=2MH$
which clearly solves the problem because a similar identity holds for the right triangle. (Note that in fact two configurations are possible: one shown above, the other for the case where $S$ is above $O.$ We stick with the one shown; the other submits to an only slightly different treatment.)
So, $S$ is the center of the left circle whose radius is denoted $s,$ and $MH=a.$
In $\Delta EOS,$ $EO^{2}+ES^{2}=OS^{2}.$ In other words, $a^{2}+(r-s)^{2}=(R-s)^{2}$ which leads to the relation $\displaystyle s=\frac{9-a^{2}}{2\sqrt{3}}.$
In $\Delta APS,$ $AP^{2}+PS^{2}=AS^{2}.$ Since $\displaystyle AS^{2}=(a-0)^{2}+(s-3\sqrt{3})^{2}=a^{2}+\frac{(9+a^{2})^{2}}{12}$ and $\displaystyle s^{2}=\frac{(9-a^{2})^{2}}{12},$
$\displaystyle AP^{2}= a^2 + \frac{(9+a)^{2}}{12} - \frac{(9-a)^{2}}{12} = a^{2}+\frac{18\cdot 2a^{2}}{12}=4a^{2}.$
It follows that $AP=2a=2MH,$ as promised. For the second (right) circle, we similarly get $AQ=2HN,$ so that for the two of them
$AP+AQ=2MH+2HN=2(MH+HN)=2MN.$
### Solution 2
This solution too depends on a
Lemma
Two circles are tangent internally at point $T.$ From points $D$ and $G$ on the big circle, tangents $DE$ and $GF$ are drawn to the small one.
Then $\displaystyle\frac{DT}{GT}=\frac{DE}{GF}.$
Proof of Lemma
Let $D'$ and $G'$ be the intersections of $TD$ and $TG,$ respectively, with the small circles. Then, a homothety from $T,$ $\displaystyle\frac{D'T}{DT}=\frac{G'T}{GT}=t,$ where $t$ is the raio od the radii of the two circles.
By the Power of a Point Theorem,
$GF^{2}=GT\cdot GG'=GT\cdot (GT-G'T)=GT^{2}(1-t).$
Similarly, $DE^{2}=DT\cdot DD'=DT\cdot (DT-D'T)=DT^{2}(1-t).\space$ This proves the lemma.
Proof of Statement
Introduce the notations as shown below.
By van Schooten's theorem, $BS=AS+CS,$ or $BS=u+v.$ By Lemma, $\displaystyle\frac{a}{n}=\frac{v}{u},$ such that also $\displaystyle\frac{a+n}{n}=\frac{u+v}{u}=\frac{m+x}{n}.$ The latter stems from a property of a circle inscribed into a circular segment and a property of angle bisectors.
It follows that $AQ = a = m+x-n,$ where $x=MN.$ Similarly, $AP=n+x-m,$ so that $AP+AQ=2x=2MN.$
### Acknowledgment
This problem has been posted by Miguel Ochoa Sanchez (Peru) at the CutTheKnotMath facebook page. A solution by Leo Giugiuc (Romania) followed almost immediately. Leo also suggested that the two circles did not need to be tangent to each other. The solution above follows in Leo's footsteps.<|endoftext|>
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The Oxford Dictionary's first record of the word workhouse dates back to 1652 in Exeter — 'The said house to bee converted for a workhouse for the poore of this cittye and also a house of correction for the vagrant and disorderly people within this cittye.' However, workhouses were around even before that — in 1631 the Mayor of Abingdon reported that "wee haue erected wthn our borough a workehouse to sett poore people to worke"
State-provided poor relief is often dated from the end of Queen Elizabeth's reign in 1601 when the passing of an Act for the Relief of the Poor made parishes legally responsible for looking after their own poor. This was funded by the collection of a poor-rate tax from local property owners (a tax that survives in the present-day "council tax"). The 1601 Act made no mention of workhouses although it provided that materials should be bought to provide work for the unemployed able-bodied — with the threat of prison for those who refused. It also proposed the erection of housing for the "impotent poor" — the elderly, chronic sick, etc.
Parish poor relief was dispensed mostly through "out-relief" — grants of money, clothing, food, or fuel, to those living in their own homes. However, the workhouse gradually began to evolve in the seventeenth century as an alternative form of "indoor relief", both to save the parish money, and also as a deterrent to the able-bodied who were required to work, usually without pay, in return for their board and lodging. The passing of the Workhouse Test Act in 1723, gave parishes the option of denying out-relief and offering claimants only the workhouse.
Parish workhouse buildings were often just ordinary local houses, rented for the purpose. Sometimes a workhouse was purpose-built, like this one erected in 1729 for the parishes of Box and Ditteridge in Wiltshire.
In some cases, the poor were "farmed" — a private contractor undertook to look after a parish's poor for a fixed annual sum; the paupers' work could be a useful way of boosting the contractor's income. The workhouse was not, however, necessarily regarded as place of punishment, or even privation. Indeed, conditions could be pleasant enough to earn some institutions the nickname of "Pauper Palaces".
Gilbert's Act of 1782 simplified and standardized the procedures for parishes to set up and run workhouses, either on their own, or by forming a group of parishes called a Gilbert Union. Under Gilbert's scheme, able-bodied adult paupers would not be admitted to the workhouse, but were to be maintained by their parish until work could be found for them. Although relatively few workhouses were set up under Gilbert's scheme, the practice of supplementing labourers' wages out of the poor rate did become widely established. The best known example of this was the "Speenhamland System" which supplemented wages on a sliding scale linked to the price of bread and family size. By the start of the nineteenth century, the nationwide cost of out-relief was beginning to spiral. It was also believed by some that parish relief had become seen as an easy option by those who did not want to work. There was also growing civil unrest during this period, culminating in the Captain Swing riots whose targets included workhouses. In 1832, the Government set up a Royal Commission to investigate the problems and propose changes.
In 1834, the Commission's report resulted in the Poor Law Amendment Act which was intended to end to all out-relief for the able bodied. The 15,000 or so parishes in England and Wales were formed into Poor Law Unions, each with its own union workhouse. A similar scheme was introduced in Ireland in 1838, while in 1845 Scotland set up a separate and somewhat different system.
Each Poor Law Union was managed by a locally elected Board of Guardians and the whole system was administered by a central Poor Law Commission. In the late 1830s, hundreds of new union workhouse buildings were erected across the country. The Commission's original proposal to have separate establishments for different types of pauper (the old, the able-bodied, children etc.) was soon abandoned and a single "general mixed workhouse" became the norm. The new buildings were specially designed to segregate the different categories of inmate. The first purpose-built workhouse to be erected under the new scheme was at Abingdon in 1835.
Under the new Act, the threat of the Union workhouse was intended to act as a deterrent to the able-bodied pauper. This was a principle enshrined in the revival of the "workhouse test" — poor relief would only be granted to those desperate enough to face entering the repugnant conditions of the workhouse. If an able-bodied man entered the workhouse, his whole family had to enter with him.
Life inside the workhouse was was intended to be as off-putting as possible. Men, women, children, the infirm, and the able-bodied were housed separately and given very basic and monotonous food such as watery porridge called gruel, or bread and cheese. All inmates had to wear the rough workhouse uniform and sleep in communal dormitories. Supervised baths were given once a week. The able-bodied were given hard work such as stone-breaking or picking apart old ropes called oakum. The elderly and infirm sat around in the day-rooms or sick-wards with little opportunity for visitors. Parents were only allowed limited contact with their children — perhaps for an hour or so a week on Sunday afternoon.
By the 1850s, the majority of those forced into the workhouse were not the work-shy, but the old, the infirm, the orphaned, unmarried mothers, and the physically or mentally ill. For the next century, the Union Workhouse was in many localities one of the largest and most significant buildings in the area, the largest ones accommodating more than a thousand inmates. Entering its harsh regime and spartan conditions was considered the ultimate degradation.
The workhouse was not, however, a prison. People could, in principle, leave whenever they wished, for example when work became available locally. Some people, known as the "ins and outs", entered and left quite frequently, treating the workhouse almost like a guest-house, albeit one with the most basic of facilities. For some, however, their stay in the workhouse would be for the rest of their lives.
In the 1850s and 60s, complaints were growing about the conditions in many London workhouses. Figures such as Florence Nightingale, Louisa Twining, and the medical journal The Lancet, were particularly critical of the treatment of the sick in workhouses which was frequently in insanitary conditions and with most of the nursing care provided by untrained and often illiterate female inmates. Eventually, parliament passed the Metropolitan Poor Act which required workhouse hospitals to be on sites separate from the workhouse. The Metropolitan Asylums Board (MAB) was also set up to look after London's poor suffering from infectious diseases or mental disability. The smallpox and fever hospitals set up by the MAB were eventually opened up to all London's inhabitants and became the country's first state hospitals, laying the foundations for the National Health Service which began in 1948.
Towards the end of the nineteenth century, conditions gradually improved in the workhouse, particularly for the elderly and infirm, and for children. Food became a little more varied and small luxuries such as books, newspapers, and even occasional outings were allowed. Children were increasingly housed away from the workhouses in special schools or in cottage homes which were often placed out in the countryside.
On 1st April 1930, when the 643 Boards of Guardians in England and Wales were abolished and their responsibilities passed to local authorities. Some workhouse buildings were sold off, demolished, or fell into disuse. Most, however, became Public Assistance Institutions and continued to provide accommodation for the elderly, chronic sick, unmarried mothers and vagrants. For inmates of these institutions, life often changed relatively little during the 1930s and 40s. Apart from the abolition of uniforms, and more freedom to come and go, things improved only slowly. With the introduction of the National Health Service in 1948, many former workhouse buildings continued to house the elderly and chronic sick. With the reorganisation of the NHS in the 1980s and 90s, the old buildings were often turned over for use as office space or demolished to make way for new hospital blocks or car parks. More recently, the survivors have increasingly been sold off for redevelopment, ironically, in some cases, as luxury residential accommodation.
Increasingly little remains of these once great and gloomy edifices. What does survive often passes unnoticed. But even now, more than seventy years after its official abolition, the mere mention of the workhouse can still send a shiver through those old enough to remember its existence. In Kendal, the location of a long-gone workhouse is modestly marked in a now renamed side-road. However, some local residents clearly feel this is an institution they would rather not commemorate...
For more information about every facet of the workhouse, please explore the rest of this web-site. If you can't find what you need, try typing a word or a phrase (in quotes) into the Search box near the top of any page.
Unless otherwise indicated, this page () is copyright Peter Higginbotham. Contents may not be reproduced without permission.<|endoftext|>
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Knarr is the Old Norse term for a type of ship built for long sea voyages and used during the Viking expansion. The knarr was a cargo ship; the hull was wider, deeper and shorter than a longship, and could take more cargo and be operated by smaller crews. They were built with a length of about 16 m (54 ft), a beam of 5 m (15 ft), and a hull capable of carrying up to 24 tons.[ambiguous] It was primarily used to transport trading goods like walrus ivory, wool, timber, wheat, furs and pelts, armour, slaves, honey, and weapons. It was also used to supply food, drink, weapons and armour to warriors and traders along their journeys across the Baltic, the Mediterranean and other seas. Knerrir routinely crossed the North Atlantic carrying livestock such as sheep and horses, and stores to Norse settlements in Iceland, Greenland and Vinland as well as trading goods to trading posts in the British Isles, Continental Europe and possibly the Middle East. They may have been used in colonising, although a similar small cargo vessel (the byrthing) is another possibility.
Only one well-preserved knarr has been found, discovered in a shallow channel in Roskilde Fjord in Denmark in 1962. Known as Skuldelev 1, it was placed among two warships, a Baltic trader, and a ferryboat. Archaeologists believe that the ships were placed there to block the channel against enemy raiders. Today all five ships, known as the Skuldelev ships, are exhibited at the Viking Ship Museum in Roskilde.
- Greenhill, Basil (1976) Archaeology of the Boat (London: Adam and Charles Black Publishers Ltd) ISBN 978-0-7136-1645-3
- Harrison, Mark (2006) The Vikings, Voyagers of Discovery and Plunder (Osprey Publishing) ISBN 978-1-84603-340-7
- Crumlin-Pedersen, Ole (1997) Viking-Age Ships and Shipbuilding in Hedeby (Viking Ship Museum, Roskilde) ISBN 978-87-85180-30-8
|Wikimedia Commons has media related to Knarr.|<|endoftext|>
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1996 AHSME Problems/Problem 13
Problem
Sunny runs at a steady rate, and Moonbeam runs $m$ times as fast, where $m$ is a number greater than 1. If Moonbeam gives Sunny a head start of $h$ meters, how many meters must Moonbeam run to overtake Sunny?
$\text{(A)}\ hm\qquad\text{(B)}\ \frac{h}{h+m}\qquad\text{(C)}\ \frac{h}{m-1}\qquad\text{(D)}\ \frac{hm}{m-1}\qquad\text{(E)}\ \frac{h+m}{m-1}$
Solution
If Sunny runs at a rate of $s$ for $h$. Then the distance covered is $sh$. Now we know that Moonbeam runs $m$ times as fast than Sunny, so Moonbeam runs at the rate of $ms$. Now Moonbeam gave Sunny a headstart of $h$ meters, so he will catch on Sunny at the rate of $s(m-1)$ . At time $\frac{h}{m-1}$ Moon beam will catch on Sunny. Now we are asked how much in meters he have to run to catch on Sunny. That is $\frac{hm}{m-1}$.
Solution 2
Note that $h$ is a length, while $m$ is a dimensionless constant. Thus, $h$ and $m$ cannot be added, and $B$ and $E$ are not proper answers, since they both contain $h+m$.
Thus, we only concern ourselves with answers $A, C, D$.
If $m$ is a very, very large number, then Moonbeam will have to run just over $h$ meters to reach Sunny. Or, in the language of limits:
$\lim_{m\rightarrow \infty} d(m) = h$, where $d(m)$ is the distance Moonbeam needs to catch Sunny at the given rate ratio of $m$.
In option $A$, when $m$ gets large, the distance gets large. Thus, $A$ is not a valid answer.
In option $C$, when $m$ gets large, the distance approaches $0$, not $h$ as desired. This is not a valid answer. (In fact, this is the distance Sunny runs, which does approach $0$ as Moonbeam gets faster and faster.)
In option $D$, when $m$ gets large, the ratio $\frac{m}{m-1}$ gets very close to, but remains just a tiny bit over, the number $1$. Thus, when you multiply it by $h$, the ratio in option $D$ gets very close to, but remains just a tiny bit over, $h$. Thus, the best option out of all the choices is $\boxed{D}$.
Solution 3
Assume that Sunny originally runs at a unit speed, and thus Moonbeam runs at a rate of $m$.
Choose a new reference frame where Sunny is still, and Moonbeam runs at a rate of $m-1$. In this new reference frame, the distance to be run is still $h$.
Moonbeam runs this distance $h$ in a time of $\frac{h}{m-1}$
Returning to the original reference frame, if Moonbeam runs for $\frac{h}{m-1}$ seconds, Moonbeam will cover a distance of $\frac{hm}{m-1}$, which is option $\boxed{D}$.<|endoftext|>
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ABC Alphabet Worksheet | Letter B Tracing PDF
Shall we continue learning the alphabet with the letter "B"? Here is one more ABC writing worksheet.
Put the pencil on the red dot, trace and write the capital letter "B" then trace and write the lowercase letter "b". After that have a look at the pictures. Catch the bee and cross the bridge, and don't forget to write the letters to complete the words. You can find other printable ABC worksheets for kindergarten.Grades: More Worksheets<|endoftext|>
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This collection of resources is all about the diverse lessons you can create to teach contour drawing for an understanding of line, outline, form, space, and many other visual devices and tools for art-making.
What is contour drawing?
A contour is the line which defines a form or an edge. A contour may describe the outermost edges of a form, as well as dramatic changes of plane within the form. ‘Cross contours’ are drawn to describe imaginary lines that cut across the form and they can give great detail.
‘Blind contour drawing’ is when contour drawing is done without looking at the paper.
‘Continuous line drawing’ is when the artist does not let the drawing implement leave the page – the image is created by following contours in one continuous line.
Another iPad Art Room iMovie introduction to this activity – contour drawing.
A good prezi on contour drawing that includes stepping into cross-contour work and integrating colour to take imagery further.
Here’s some ideas for mixing media, including the iPad, to produce interesting contour drawings, or works that use line drawing as a starting point.
Pierre Emmanuel Godet is a French self-taught artist who creates incredible drawings with a single continuous line. Click the link to see more of his portraits.
Ian Sklarsky and his blind contour drawings…
The Contour Drawings of David Habben start “with a simple free form shape. The llustrator lets his pencil squiggle whichever way it wants to. Once a shape is achieved, he fills it in with a unique character one step at a time. Each spirit evolves to have its own narrative as he slowly fills in the space. Most of his drawings are created using India ink on Arches Aquarelle pigment paper without any sketching.” See the rest of this article and lots more of his work, Shapes of Consciousness, here.
‘A line is a dot that went for a walk’ Paul Klee
Check out these great lesson seeds…
Still want more ideas? Check out Pinterest on contour drawing – lots of great lesson ideas here.<|endoftext|>
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Weave – it is one of the oldest human inventions. In ancient times, people without a loom, weaving fabric by hand. With the development of weaving people, for beautiful patterns, are joining in a single several types of fabric weaves. Quick Reference Dictionary of Weaving – is the order of overlapping warp threads weft. Weave – it is one of the key indicators of its structure.
Affordable weave in the weaving join the warp and weft and get woven fabric – fabrics and textiles. Here two interwoven strands of arranged vertically. Foundation called threads, which go along fabric weft – threads are located across, horizontally. Way to connect these threads determines the type of weave. A plain weave is the most ancient weave the fabric fibers is – plain (it is call – taffeta) interlacing. This is the simplest and most common way to weave. Plain weave produce cotton fabrics, chintz, calico, linen and platevye tissue, almost all linen tissue, and many fabrics from natural silk and chemical fiber.
Less commonly, plain weave is used to create wool fabrics. Plain weave – this is an exact alternation of major and weft yarns in a ratio of 1:1. In this case, if the first warp thread left on the surface, the second weft thread is closed, etc. Of all existing plain weave, characterized the smallest repeats: two threads in warp and two weft yarns. Each main thread interwoven with each weft thread through one. This netting provides excellent fusion of tissue structure and, consequently, most of its strength. Plain weave fabrics are characterized by flat matt surface and identical facial appearance and the wrong side. Fabric goes way, because the warp and weft are distributed equally on both sides. In plain weave the increase density on the basis leads to compression of yarns in the vertical direction, the weft – the horizontal. Because of this, the cells formed by weaving, are no longer symmetrical and elongated in one direction or another. When significant difference in linear density of basic and weft yarns in the fabric of plain weave are longitudinal or transverse ridges, appears rep effect. Thin filaments are bent around a thick and close them. Therefore, the thick filaments are located within the tissue, and thin – on its surface. This structure allows even to create a plain weave fabric surface from one system threads. If you are using yarn increased twisting, the fabric is formed crapy effect, the surface becomes 'grainy'.<|endoftext|>
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Discovery of the Island
When Christopher Columbus arrived in Cuba on October 27, 1492 and their ships traveled forty days through the north eastern coast of the island, it was observed, the highlights of the exuberant nature, the presence of peaceful and innocent people who offered cotton, yarn and small pieces of gold for trinkets. Two years later, to explore the southern coast of Cuba during his second voyage, the admiral would realize the diversity of the indigenous people, as the natives of the eastern region who accompanied him, could not get along with the people of the west. Indeed, the population of the island had begun four millennia before the arrival of various migration flows: the first probably from northern continent through Florida and subsequent arrivals in successive waves from the mouth of the Orinoco along the arc of the Antilles.
Colonial Era and brief occupation of the English
The conquest of the island for Spain nearly two decades after the first voyage of Columbus, as part of the occupation that radiated to various Caribbean lands begins. In a few years the Spanish founded the first towns such as Havana, Santiago de Cuba, Camaguey and other small human concentrations. Economic activity was based on the work of the Indians, given to settlers by the Crown through the system of “encomiendas”, a kind of personal concession, whereby the settler undertook to clothe, feed and Christianize the native in exchange to have the right to make them work for them. The dominant economic sector in the early years of the colony was mining, specifically gold mining activity in which Indians and black slaves were used. Before this inevitable development, the Taino and Ciboney were unable to adapt to the new situation which caused there death, as in most of the territories conquered by the Spaniards. Thanks to its strategic location, Havana quickly consolidated as major port freight traffic, attracting pirates and private business men. The rapid exhaustion of the gold mines and the drastic reduction of the population converted to livestock in the main source of wealth in Cuba. A lack of gold, salted meat and skins would be almost exclusive to the few settlers of the island and would be incorporated into the channels of commerce of goods for the Spanish empire. In the sixteenth century the most important activity was livestock on large farms called herds. Sugarcane was introduced, resulting in the first “geniuses”. It was the cultivation of sugar cane which caused the introduction of African slaves from the year 1595. In the seventeenth century a number of circumstances made Cuba the largest producer of sugar in the world. In 1762 the island was occupied by the English for a short time of 10 months, prompting a large importation of slaves and laying the groundwork for a new policy that would continue liberalizing the Spanish. At that time Cuba was the largest black population colony.
With the independence movements in other Latin American countries, Cubans fought to negotiate with the crown, settling for some concessions on free trade with other countries. However, this situation reinforced the national consciousness, and in 1868, Carlos Manuel de Céspedes a proclaimed republic, granting freedom to slaves. This war lasted 10 years and ended with the Spanish victory, signing peace in Zanjón in 1878. Two years later, slavery was abolished. However, the troubled situation and problems continued and in 1892 José Martí founded the Cuban Revolutionary Party, reconciling the interests of social classes and social forces. In 1895 Martí ordered the start of the war that lasted three years. General Weyler was sent to suppress the movement but failed. The situation worsened by what the United States, with strong interests in the island, intervened militarily, annihilating the Spanish fleet, and occupied territories by signing the Treaty of Paris (1898), by which the United States gave Cuba, Puerto Rico and the Philippines.
The dictatorship of Fulgencio Batista
Over the next forty years of the twentieth century, various military and civilian governments, characterized by corruption, causing deterioration of the political system followed. In 1925 General Machado takes the power and keeps the regime in a repressive state. The problems are widespread on the island, and the price of sugarcane in international markets dropped significantly; the cities make a general strike. Later the communist party that is fighting against Machado, are trying to adopt urgent social measures. This period lasts a short time as Colonel Fulgencio Batista who controls the politicians, decides to take power of Cuba in 1940, where he remains for four years until he was defeated in an election. However, in 1952, Batista again rose to power after a military coup under the government mismanagement, launching a harsh dictatorship that would last until 1959, with the arrival of Fidel Castro Rus, Ernesto “Che “Guevara, Raúl Castro, Camilo Cienfuegos and other guerrillas who began their conquest from Sierra Maestra.
The Cuban revolution and socialism
The first attempt by Fidel Castro to overthrow Batista was in 1953, but without success. Exiled in Mexico and in the company of Che Guevara and Raul Castro Rus. Castro prepares to assault the island, landing in 1956 in the province of Oriente. From the Sierra Maestra he began guerrilla attacks that culminated after three years, with the flight of Batista and Fidel’s triumphant entry into Havana. An interim head of the state was appointed “Manuel Urrutia” and in 1960 Cuba confiscated American property without compensation. This started the crisis between Cuba and U.S.A. and the United States imposed an embargo, which continues to this day. The economy of the island has suffered severe crisis, especially after the demise of the Soviet Union. The politics of Fidel Castro, seen by some as undemocratic and by others as heroic resistance. During the last few years Cuba has opened to new ways of thinking with the economic crisis: Private people are now allowed to do certain business and it is possible for a Cuban to sell their properties and cars.<|endoftext|>
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The definition of an alliance system is a formal agreement or treaty between two or more nations to cooperate for specific purposes. An alliance system can also be defined as an agreement between individuals, families or corporations.
The alliance system was one of the main causes of World War One.The alliance system was made up of two groups, the Central powers (Germany, Austria- Hungary, Italy(1914), and Turkey).The second group was the Allied powers (Russia, France, Great Britain, and United states).The alliance system is when countries join forces or worked together to achieve a certain goal.
ALLIANCE SYSTEM bismarck's alliance system rival alliance systems the alliance system and the outbreak of war bibliography. The European alliance system that was in place prior to World War I is often seen as one of the long-term causes for the outbreak of war in 1914. On the eve of war, Europe was divided into two opposing camps, with Germany, Austria-Hungary, and Italy on one side and France...
Overview. The Concert of Europe was founded by the powers of Austria, Prussia, Russia and the United Kingdom, which were the members of the Quadruple Alliance that defeated Napoleon and his First French Empire.In time, France was established as a fifth member of the Concert, following the restoration of the Bourbon monarchy. At first, the leading personalities of the system were British ...
The European Alliance (EA) is a political group in the European Committee of the Regions composed of a mix of regionalist parties and independent members. It was established in 1996, with strong influence of the European Free Alliance, and has since then existed in several incarnations.
Alliance, confederation, league, union all mean the joining of states for mutual benefit or to permit the joint exercise of functions. An alliance may apply to any connection entered into for mutual benefit. League usually suggests closer combination or a more definite object or purpose.
He wanted to make sure he est. alliances to keep the peace in Europe, so he started making alliances w/ other countries, which led to war. He felt France was the biggest threat, so aimed to isolate France. Formed "dual alliance" w/ Austria-Hungary. 13 yrs. later Italy joins the alliance to form the "triple alliance".
The alliance system was made up of two groups, the central powers (germany, austria 20 may 2014 european systems are often seen as a major cause world war one. Alliance system dictionary ...
1. The alliance system was a network of treaties, agreements and ententes that were negotiated and signed prior to 1914. 2. National tensions and rivalries have made alliances a common feature of European politics, however, the alliance system became particularly extensive in the late 1800s. 3.
the war aims outlined by President Wilson in 1918, which he believed would promote lasting peace; called for self-determination, freedom of the seas, free trade, end to secret agreements, reduction of arms and a league of nations.<|endoftext|>
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# More properties of the definite integral
Here we will look at some properties of the Riemann integral that are not directly related to its evaluation, but are more of a theoretical interest.
Theorem.
Let f be a Riemann integrable function on [a,b]. Then the equalities
and
are true.
Note that the integrals in limits make sense. Consider the first equality: We assume that f is Riemann integrable on [a,b]. If we pick some B between a and b (note that in the limit we approach b from the left, now we can see why), then by the Theorem here part (i), f is also Riemann integrable on [a,B] and so we can integrate inside the limit. The situation is shown in the following picture.
If we cut away a part of the region along the right edge and make this cut-away part smaller and smaller, the resulting areas should converge to the whole area.
Similarly, f is Riemann integrable on [A,b], and in the second equality we cut away along the left edge.
The second property we will cover here is a modification of the standard Mean Value Theorem (see Derivatives - Theory - MVT) for the function F defined in the Fundamental Theorem of Calculus:
The Fundamental theorem then says
When we substitute for F, we get the following thorem.
Theorem (The Mean Value Theorem for integrals, Lagrange theorem for integrals).
Let f be a continuous function on [a,b]. Then there is a number c in (a,b) such that
If we recall the definition of an average of f (see Applications - Average), we can restate the Mean Value Theorem as follows:
Theorem.
Let f be a continuous function on [a,b]. Then there is a number c in (a,b) such that f (c) is equal to the average of f over the interval [a,b].
Note that the continuity in the assumption of the Mean Value Theorem is crucial. Indeed, recall the example of a jump function we saw before. Its average over [0,2] is 3/2, but the function is never equal to 3/2.
The Mean Value Theorem for integrals has many versions. The one we stated above is probably the most popular, but also the weakest, merely a reformulation of the good old MVT. There are much stronger statements, we will show one of the more popular here.
Theorem (The Mean Value Theorem for integrals).
Let f be a continuous function on [a,b] and g an integrable function on [a,b] that is positive there. Then there is a number c in (a,b) such that
Note that if you use this theorem with the constant function g(x) = 1, you get the first, weaker statement.<|endoftext|>
| 4.5 |
449 |
Expand negative two 𝑥 times five 𝑥 cubed minus five 𝑥 squared.
To expand here, we’ll need to use the distributive property. The distributive property tells us that if we’re multiplying 𝑎 times 𝑏 plus 𝑐, it will be equal to multiplying 𝑎 times 𝑏 and then adding that to 𝑎 times 𝑐. Let’s see how we would use that here.
In our case, the 𝑎 that we’re by multiplying by is negative two 𝑥. The 𝑏 would be five times 𝑥 cubed and the 𝑐 would be five times 𝑥 squared. To distribute our 𝑎 here to multiply the 𝑎 times 𝑏, we’ll need to multiply negative two 𝑥 times five 𝑥 cubed. After that, we multiply the 𝑎 term by the 𝑐 term, which in our case would be multiplying negative two 𝑥 by five 𝑥 squared. Just to know before we do that, remember that you need to keep your signs the same. We’re subtracting five 𝑥 squared from five 𝑥 cubed, so we wanna subtract here.
We’ve now added negative two 𝑥 times five 𝑥 squared. In our next step, we need to multiply negative two 𝑥 times five 𝑥 cubed equals negative 10𝑥 to the fourth. Bring down our subtraction then multiply negative two 𝑥 times five 𝑥 squared. When we multiply those together, we get negative 10𝑥 cubed. But because we’re subtracting a negative, we can write addition here and then bring down our 10𝑥 cubed. We expanded the original problem using the distributive property to come to the final answer of negative 10𝑥 to the fourth plus 10𝑥 cubed.<|endoftext|>
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By all rights, NASA’s Kepler spacecraft should have stopped working long ago. The observatory launched in 2009 with the aim of finding out how common exoplanets are throughout the universe. It turns out: very common. Mechanical failures forced NASA to change Kepler’s mission profile, but it still kept working until a few months ago. NASA was unsure if Kepler could go on, but the spacecraft just woke up to begin a new observational campaign.
Kepler uses the transit method to detect exoplanets. That means it has to watch vast swaths of the sky for weeks in order to track small light dips that could indicate an exoplanet passing in front of its host star. That became much more difficult in 2013 when a second of four reaction wheels failed in the spacecraft. At that point, Kepler could no longer maintain its orientation to point at the same area over time.
Since 2014, Kepler has been engaged in its secondary “K2” mission. This allows NASA to balance Kepler for several weeks a few times per year using the remaining reaction wheels and pressure from the solar wind. The original Kepler mission identified 2,244 candidate exoplanets with 2,327 of them confirmed. Kepler’s K2 phase resulted in 479 candidates exoplanets and 323 confirmed. Together, that’s most of the roughly 4,000 known exoplanets.
This part of the mission doesn’t require fuel, but Kepler needs to use its thrusters to orient its antenna toward Earth for beaming data back. NASA placed the telescope into hibernation mode over the summer to ensure it would have enough fuel left to send back data from its 18th observation. NASA got all the data back late last month and started a new observational campaign on August 29th.
NASA originally only expected 10 campaigns in K2, but there’s no fuel gauge on the spacecraft. Engineers are merely estimating what’s left in the tank based on past usage. It’s possible Kepler won’t have enough fuel to send back the data it’s currently collecting in the 19th campaign, but NASA isn’t giving up.
In the most recent mission update, NASA reports that one of Kepler’s thrusters isn’t working correctly, so its ability to reorient could be limited. NASA says it will continue monitoring the spacecraft throughout the current campaign. Even if Kepler can’t manage another full campaign, the agency’s Transiting Exoplanet Survey Satellite (TESS) is ready to pick up where it left off.
New Horizons Probe Is Awake and Ready to Explore the Kuiper Belt
According to NASA's Alan Stern, New Horizons has successfully woken up from hibernation.
NASA Issues Arbitrary Deadline to Reawaken Mars Rover
NASA has set a 45-day limit for attempting to reactivate the Opportunity rover, and science team members that work on the project are speaking out against that decision.<|endoftext|>
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Autism is commonly understood as a condition that causes a person to withdraw from and avoid social interaction.
What if that isn’t true?
There is a growing amount of scientific evidence that shows that traditional definitions of autism need to be revised.
Even people with severe autism want to connect, if we can connect with them in a way that feels safe and predictable.
When brains can predict, and even control, the rhythm of an interaction with another person, then that interaction feels safe. The pattern is recognisable and familiar. This explanation is the basis for interventions like Intensive Interaction, which have been used to connect with people who cope with very severe autism.
The effectiveness of this approach leads us to ask new questions about what exactly autism is and how might it be affect a person’s ability to display the inescapable capacity for connection with which they were born?
These are important questions to ask, because autism rates are quickly rising throughout the world. We need to better understand it if we are to provide the best care and support for people with autism.
For more insights into dementia and connection, see our connected baby guides.
Go back to The science page.<|endoftext|>
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# Asymptotic Efficiency of Recurrences by ert554898
VIEWS: 17 PAGES: 35
• pg 1
``` Asymptotic Efficiency of Recurrences
• Find the asymptotic bounds of recursive
equations.
– Substitution method
• domain transformation
• Changing variable
– Recursive tree method
– Master method (master theorem)
• Provides bounds for: T(n) = aT(n/b)+f(n) where
– a 1 (the number of subproblems).
– b>1, (n/b is the size of each subproblem).
– f(n) is a given function.
1
Recurrences
• MERGE-SORT
– Contains details:
• T(n) = (1) if n=1
T(n/2)+ T(n/2)+ (n) if n>1
• Ignore details, T(n) = 2T(n/2)+ (n).
– T(n) = (1) if n=1
2T(n/2)+ (n) if n>1
2
The Substitution Method
• Two steps:
1. Guess the form of the solution.
• By experience, and creativity.
• By some heuristics.
– If a recurrence is similar to one you have seen before.
» T(n)=2T(n/2+17)+n, similar to T(n)=2T(n/2)+n, , guess O(nlg n).
– Prove loose upper and lower bounds on the recurrence and then reduce the
range of uncertainty.
» For T(n)=2T(n/2)+n, prove lower bound T(n)= (n), and prove upper
bound T(n)= O(n2), then guess the tight bound is T(n)=O(nlg n).
• By recursion tree.
2. Use mathematical induction to find the constants and show
that the solution works.
3
Solve T(n)=2T(n/2)+n
• Guess the solution: T(n)=O(nlg n),
– i.e., T(n) cnlg n for some c.
• Prove the solution by induction:
– Suppose this bound holds for n/2, i.e.,
• T(n/2) cn/2 lg (n/2).
– T(n) 2(cn/2 lg (n/2))+n
• cn lg (n/2))+n
• = cn lg n - cn lg 2 +n
• = cn lg n - cn +n
• cn lg n (as long as c1)
Question: Is the above proof complete? Why?
4
Boundary (base) Condition
• In fact, T(n) =1 if n=1, i.e., T(1)=1.
• However, cnlg n =c1lg 1 = 0, which is odd with
T(1)=1.
• Take advantage of asymptotic notation: it is
required T(n) cnlg n hold for n n0 where n0 is a
constant of our choosing.
• Select n0 =2, thus, n=2 and n=3 as our induction
bases. It turns out any c 2 suffices for base cases
of n=2 and n=3 to hold.
5
Subtleties
• Guess is correct, but induction proof not work.
• Problem is that inductive assumption not strong
enough.
• Solution: revise the guess by subtracting a lower-order
term.
• Example: T(n)=T(n/2)+T(n/2)+1.
– Guess T(n)=O(n), i.e., T(n) cn for some c.
– However, T(n) c n/2+c n/2+1 =cn+1, which does not
imply T(n) cn for any c.
– Attempting T(n)=O(n2) will work, but overkill.
– New guess T(n) cn – b will work as long as b 1.
– (Prove it in an exact way).
6
Avoiding Pitfall
• It is easy to guess T(n)=O(n) (i.e., T(n)
cn) for T(n)=2T(n/2)+n.
• And wrongly prove:
– T(n) 2(c n/2)+n
• cn+n
• =O(n). wrongly !!!!
• Problem is that it does not prove the exact
form of T(n) cn.
7
Find bound, ceiling, floor, lower term–
domain transformation
• Find the bound: T(n)=2T(n/2)+n (O(nlogn))
– T(n)2T(n/2+1)+n
– Domain transformation
• Set S(n)=T(n+a) and assume S(n) 2S(n/2)+O(n) (so S(n)=O(nlogn))
• S(n) 2S(n/2)+O(n) T(n+a) 2T(n/2+a)+O(n)
• T(n)2T(n/2+1)+n T(n+a) 2T((n+a)/2+1)+n+a
• Thus, set n/2+a=(n+a)/2+1, get a=2.
• so T(n)=S(n-2)=O((n-2)log(n-2)) = O(nlogn).
– Set S(n)=T(n+a) and get a=38.
• As a result, ceiling, floor, and lower terms will not affect.
– Moreover, the master theorem also provides proof for this.
8
Changing Variables
• Suppose T(n)=2T(n)+lg n.
• Rename m=lg n. so T(2m)=2T(2m/2)+m.
• Domain transformation:
– S(m)=T(2m), so S(m)=2S(m/2)+m.
– Which is similar to T(n)=2T(n/2)+n.
– So the solution is S(m)=O(m lg m).
– Changing back to T(n) from S(m), the solution is T(n)
=T(2m)=S(m)=O(m lg m)=O(lg n lg lg n).
9
The Recursion-tree Method
• Idea:
– Each node represents the cost of a single subproblem.
– Sum up the costs with each level to get level cost.
– Sum up all the level costs to get total cost.
• Particularly suitable for divide-and-conquer
recurrence.
• Best used to generate a good guess, tolerating
“sloppiness”.
• If trying carefully to draw the recursion-tree and
compute cost, then used as direct proof.
10
Recursion Tree for T(n)=3T(n/4)+(n2)
T(n) cn2 cn2
T(n/4) T(n/4) T(n/4) c(n/4)2 c(n/4)2 c(n/4)2
T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16) T(n/16)
(a) (b) (c)
cn2 cn2
c(n/4)2 c(n/4)2 (3/16)cn2
c(n/4)2
log 4n (3/16)2cn2
c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2
T(1)T(1)T(1) T(1)T(1)T(1) (nlog 43)
3log4n= nlog 43 Total O(n2)
(d) 11
Solution to T(n)=3T(n/4)+(n2)
• The height is log 4n,
• #leaf nodes = 3log 4n= nlog 43. Leaf node cost: T(1).
• Total cost T(n)=cn2+(3/16) cn2+(3/16)2 cn2+
+(3/16)log 4n-1 cn2+ (nlog 43)
=(1+3/16+(3/16)2+ +(3/16)log 4n-1) cn2 + (nlog 43)
<(1+3/16+(3/16)2+ +(3/16)m+ ) cn2 + (nlog 43)
=(1/(1-3/16)) cn2 + (nlog 43)
=16/13cn2 + (nlog 43)
=O(n2).
12
Prove the above Guess
• T(n)=3T(n/4)+(n2) =O(n2).
• Show T(n) dn2 for some d.
• T(n) 3(d (n/4)2) +cn2
3(d (n/4)2) +cn2
=3/16(dn2) +cn2
dn2, as long as d(16/13)c.
13
One more example
• T(n)=T(n/3)+ T(2n/3)+O(n).
• Construct its recursive tree (Figure 4.2,
page 71).
• T(n)=O(cnlg3/2n) = O(nlg n).
• Prove T(n) dnlg n.
14
Recursion Tree of T(n)=T(n/3)+ T(2n/3)+O(n)
15
Master Method/Theorem
• Theorem 4.1 (page 73)
– for T(n) = aT(n/b)+f(n), n/b may be n/b or n/b.
– where a 1, b>1 are positive integers, f(n) be a non-
negative function.
1. If f(n)=O(nlogba-) for some >0, then T(n)= (nlogba).
2. If f(n)= (nlogba), then T(n)= (nlogba lg n).
3. If f(n)=(nlogba+) for some >0, and if af(n/b) cf(n)
for some c<1 and all sufficiently large n, then T(n)=
(f(n)).
16
Implications of Master Theorem
• Comparison between f(n) and nlogba (<,=,>)
• Must be asymptotically smaller (or larger) by a
polynomial, i.e., n for some >0.
• In case 3, the “regularity” must be satisfied, i.e.,
af(n/b) cf(n) for some c<1 .
• There are gaps
– between 1 and 2: f(n) is smaller than nlogba, but not
polynomially smaller.
– between 2 and 3: f(n) is larger than nlogba, but not
polynomially larger.
– in case 3, if the “regularity” fails to hold.
17
Application of Master Theorem
• T(n) = 9T(n/3)+n;
– a=9,b=3, f(n) =n
– nlogba = nlog39 = (n2)
– f(n)=O(nlog39-) for =1
– By case 1, T(n) = (n2).
• T(n) = T(2n/3)+1
– a=1,b=3/2, f(n) =1
– nlogba = nlog3/21 = (n0) = (1)
– By case 2, T(n)= (lg n).
18
Application of Master Theorem
• T(n) = 3T(n/4)+nlg n;
– a=3,b=4, f(n) =nlg n
– nlogba = nlog43 = (n0.793)
– f(n)= (nlog43+) for 0.2
– Moreover, for large n, the “regularity” holds for
c=3/4.
• af(n/b) =3(n/4)lg (n/4) (3/4)nlg n = cf(n)
– By case 3, T(n) = (f(n))= (nlg n).
19
Exception to Master Theorem
• T(n) = 2T(n/2)+nlg n;
– a=2,b=2, f(n) =nlg n
– nlogba = nlog22 = (n)
– f(n) is asymptotically larger than nlogba , but not
polynomially larger because
– f(n)/nlogba = lg n, which is asymptotically less
than n for any >0.
– Therefore,this is a gap between 2 and 3.
20
Where Are the Gaps
f(n), case 3, at least polynomially larger
n Gap between case 3 and 2
c1
nlogba f(n), case 2: within constant distances
c2
n Gap between case 1 and 2
f(n), case 1, at least polynomially smaller
Note: 1. for case 3, the regularity also must hold.
2. if f(n) is lg n smaller, then fall in gap in 1 and 2
3. if f(n) is lg n larger, then fall in gap in 3 and 2
4. if f(n)=(nlogbalgkn), then T(n)=(nlogbalgk+1n). (as exercise)
21
Proof of Master Theorem
• The proof for the exact powers, n=bk for k1.
• Lemma 4.2
– for T(n) = (1) if n=1
– aT(n/b)+f(n) if n=bk for k1
– where a 1, b>1, f(n) be a nonnegative function,
– Then logbn-1
– T(n) = (nlogba)+ ajf(n/bj)
j=0
• Proof:
– By iterating the recurrence
– By recursion tree (See figure 4.3)
22
Recursion tree for T(n)=aT(n/b)+f(n)
23
Proof of Master Theorem (cont.)
• Lemma 4.3:
– Let a 1, b>1, f(n) be a nonnegative function defined on
exact power of b, then
logbn-1
– g(n)= ajf(n/bj) can be bounded for exact power of b as:
j=0
1. If f(n)=O(nlogba-) for some >0, then g(n)= O(nlogba).
2. If f(n)= (nlogba), then g(n)= (nlogba lg n).
3. If f(n)= (nlogba+) for some >0 and if af(n/b) cf(n) for
some c<1 and all sufficiently large n b, then g(n)= (f(n)).
24
Proof of Lemma 4.3
• For case 1: f(n)=O(nlogba-) implies f(n/bj)=O((n /bj)logba-), so
logbn-1 logbn-1
• g(n)= ajf(n/bj) =O( aj(n /bj)logba- )
j=0 j=0
logbn-1 logbn-1
• = O(nlogba- aj/(blogba-)j ) = O(nlogba- aj/(aj(b-)j))
j=0 j=0
logbn-1
• = O(nlogba- (b)j ) = O(nlogba- (((b ) logbn-1)/(b-1) )
j=0
• = O(nlogba- (((blogbn) -1)/(b-1)))=O(nlogba n- (n -1)/(b-1))
• = O(nlogba )
25
Proof of Lemma 4.3(cont.)
• For case 2: f(n)= (nlogba) implies f(n/bj)= ((n /bj)logba), so
logbn-1 logbn-1
• g(n)= ajf(n/bj) = ( aj(n /bj)logba)
j=0 j=0
logbn-1 logbn-1
• = (nlogba aj/(blogba)j ) = (nlogba 1)
j=0 j=0
• = (nlogba logbn) = (nlogbalg n)
26
Proof of Lemma 4.3(cont.)
• For case 3:
– Since g(n) contains f(n), g(n) = (f(n))
– Since af(n/b) cf(n), ajf(n/bj) cjf(n) , why???
logbn-1 j logbn-1 j
– g(n)= a f(n/bj) c f(n) f(n) cj
j=0 j=0 j=0
– =f(n)(1/(1-c)) =O(f(n))
– Thus, g(n)=(f(n))
27
Proof of Master Theorem (cont.)
• Lemma 4.4:
– for T(n) = (1) if n=1
– aT(n/b)+f(n) if n=bk for k1
– where a 1, b>1, f(n) be a nonnegative function,
1. If f(n)=O(nlogba-) for some >0, then T(n)= (nlogba).
2. If f(n)= (nlogba), then T(n)= (nlogba lg n).
3. If f(n)=(nlogba+) for some >0, and if af(n/b) cf(n)
for some c<1 and all sufficiently large n, then T(n)=
(f(n)).
28
Proof of Lemma 4.4 (cont.)
• Combine Lemma 4.2 and 4.3,
– For case 1:
• T(n)= (nlogba)+O(nlogba)=(nlogba).
– For case 2:
• T(n)= (nlogba)+(nlogba lg n)=(nlogba lg n).
– For case 3:
• T(n)= (nlogba)+(f(n))=(f(n)) because f(n)=
(nlogba+).
29
Floors and Ceilings
• T(n)=aT(n/b)+f(n) and T(n)=aT(n/b)+f(n)
• Want to prove both equal to T(n)=aT(n/b)+f(n)
• Two results:
– Master theorem applied to all integers n.
– Floors and ceilings do not change the result.
• (Note: we proved this by domain transformation too).
• Since n/bn/b, and n/b n/b, upper bound for
floors and lower bound for ceiling is held.
• So prove upper bound for ceilings (similar for lower
bound for floors).
30
Upper bound of proof for T(n)=aT(n/b)+f(n)
• consider sequence n, n/b, n/b/b, n/b /b/b,
…
• Let us define nj as follows:
• nj = n if j=0
• = nj-1/b if j>0
• The sequence will be n0, n1, …, nlogbn
• Draw recursion tree:
31
Recursion tree of T(n)=aT(n/b)+f(n)
32
The proof of upper bound for ceiling
logbn -1
– T(n) = (nlogba)+ ajf(nj)
j=0
– Thus similar to Lemma 4.3 and 4.4, the upper
bound is proven.
33
The simple format of master theorem
• T(n)=aT(n/b)+cnk, with a, b, c, k are
positive constants, and a1 and b2,
O(nlogba), if a>bk.
• T(n) = O(nklogn), if a=bk.
O(nk), if a<bk.
34
Summary
Recurrences and their bounds
– Substitution
– Recursion tree
– Master theorem.
– Proof of subtleties
– Recurrences that Master theorem does not
apply to.
35
```
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Voices from the Working Class
The Industrial Revolution in England is defined as the era where major changes and transformations took place in almost all areas of manufacturing, a prime source of business, as it completely eliminated the traditional way—that of pure manual labor. The advancements during this period were represented by the use of machinery, which abruptly changed all working structures, but still requires human hands (Montagna, 1981). Jobs all over the country sprouted, giving more opportunities to more people in more towns.
However, the accompanying progress also had its own evils. The works of famed poet Elizabeth Barrett Browning and novelist Elizabeth Gaskell both clearly resonate the plight of the poor working in mines and factories in England in the mid-1800s—the rise of the Industrial age. In particular, Browning’s The Cry of the Children (1843), captures the sorrow and yearning of young children forced to labor from dawn till late evening; while Mary Barton (1848) by Gaskell is focused on the excruciating way of live experienced by Manchester’s poor.During the 1842 to 1843, at the height of the Industrial Revoluition, parliamentary hearings were held that exposed the hard truth of the state of the working class, many through testimonies of children. The commission in charge of these hearings documented the findings in a report written by R.
H. Horne, known to be a close friend and ally of Browning. The poet’s access to said information allowed her to create a poem that illustrated the conditions of child workers, and sympathized strongly with their loss of youth.The length of the piece, the graphic descriptions of the treatment of children, and the decidedly emotional tone that emulates, quite successfully, their ‘cry’. This makes the poem effective and correct in fulfilling its objective of portraying the exact state of mind, body and soul. What is primarily being defined here is the ultimate loss of the joys of childhood, as exemplified by the last line in the poem: But the child’s sob in the silence curses deeper/Than the strong man in his wrath.
Elizabeth Gaskell, wife to a Manchester clergyman, wrote the novel Mary Barton as a way to veer from the depression she went through after the death of her son. In the novel, she also tackled the concerns of the poor working class, but centered on the anguish of the uneducated workers. Because of the increasing success brought about by faster and more efficient methods, the gap between the masters (those who owned the mines and factories) and the poor (the employed workers) grew even more substantially—giving the workers more reason to begrudge their employers.They complained of neglect, poor conditions, and their unchanging economic state as individuals. Gaskell’s original goal to merely create a more romantic piece showcasing the lives and character of the hardworking miners and factory men grew to produce a novel that contained concepts and ideas more vivid, more apt to the suffering and injustice. In this manner, Gaskell was victorious in depicting the working man’s life and tribulations, for soon after Mary Barton was published, several rebellions against the ruling class took place all over Europe.
The testimony of Elizabeth Bentley, a 23-year-old worker summoned at a parliamentary hearing, validated all the theses, symbols and contexts used by Gaskell and Browning. What was most disturbing was the manner by which Bentley would answer the questions thrown at her—it had many indications of poor education, simplistic comprehension, and a sureness of the abuse committed, even if she had to recall events from ten years before. All these elements—suffering, sorrow, injustic and loss of youth—came through in both literary works, making them suitable documents of the conditions of the working class during the Industrial Revolution.Uncannily enough, the three main subjects in this project are all named Elizabeth—a curious yet symbolic detail that may connote the only level of equality existing at the time.Works Cited:The Norton Anthology of English Literature: Norton Topics Online. W.
W. Norton and Company, 2003-2008. http://www. wwnorton. com/college/english/nael/victorian Montagna, Joseph A.
The Industrial Revolution. Yale-New Haven Teachers Institute, 1981. http://www. yale. edu/ynhti/curriculum/units/1981/2/81. 02.
06. x. html
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The only medieval whorl with an inscription found in Poland comes from Czermno
The only early medieval whorl (a small, biconical object, a moving part of a spindle) with an inscription found in Poland has been discovered by archaeologists in Czermno (Lublin province).
Spindle whorl adds weight to a spindle, prevents the threads from sliding off and maintains or increases the spin. Whorls come in different shapes - they can be spherical and biconical. They have a hole in the middle. They are usually made of clay, less often of stone.
The object in question was found during excavations over 60 years ago. However, Iwona Florkiewicz from the Institute of Archeology of the University of Rzeszów only now noticed the inscription during the analysis of the artefacts discovered at that time.
"Interestingly, archaeologists who conducted research in Czermno in 1952 did not notice that the seemingly chaotic lines formed an inscription. This is an unprecedented situation - it is the only spindle whorl from this period covered with writing known in Poland" - says Florekiewicz. The archaeologist notes that there could be other objects of this type in the collections of Polish museums that have not yet been identified. "Archaeologists probably did not expect spindle whorls to have inscriptions, so these objects were not analysed in this respect" - she concludes.
The object from Czermno was made of the so-called Owrucz slate, which is not a local raw material - its outcrops are located in present-day Ukraine, near Owrucz (hence the name).
"It is hard to say whether the whorl was been made in Czermno or imported from the east, but it is certainly of Russian origin. Similar whorls covered with letters are known from the same period from the former Kievan Rus' territory, that is present Ukraine, Russia and Belarus" - the archaeologist says.
Palaeographic research (on the form of writing) carried out by Dr. Adrian Jusupović, historian from the Institute of History PAS in Warsaw, shows that the inscription was made in the second half of the 12th century or in the 13th century.
The spindle whorl bears an inscription composed of 6 letters in Cyrillic, which can be transcribed as Hoten\'. It is a masculine name that means a lover, volunteer or master. It is also possible that the inscription was a toponym. In that case it should be associated with the name of a town or village. But researchers find this second possibility is less likely.
According to archaeologists, today's Czermno in the Lublin region was the historical Cherven - the administrative centre of the Cherven Cities, over which the Piasts had been fighting with the rulers of Kievan Rus' since the 10th century. Bolesław the Brave managed to recapture them in 1018, but after his death Yaroslav the Wise retook the territory in 1031. It was not until the 14th century that Casimir III the Great reincorporated the former Cherven Cities into Poland.
"Thus, the spindle whorl probably comes from the time when this area was a part of Kievan Rus'. Remember that Czermno was a borderland town, where cultural influences from the east and the west mixed" - adds Florkiewicz.
"We are not sure whether the spindle whorl was used for its original purpose - spinning - or had a secondary function as an amulet. The latter possibility should not be ruled out" - Florkiewicz believes. She adds that according to the researchers, similar inscriptions could be signs of ownership. There are known spindle whorls that bear the names of women and men. "It is also possible that in this case it is the name of the object's maker" - she concludes.
The spindle whorl was discovered inside the settlement but outside the buildings, which may suggest that it was lost. The object is currently in the collection of the Zamość Museum.
Scientists stumbled upon the spindle whorl while doing an inventory of archival finds from Czermno (discovered over several decades) as part of the scientific project "Golden Apple of Polish Archeology...", funded by the National Program for the Development of Humanities and supervised by Prof. Marcin Wołoszyn.
PAP - Science in Poland, Szymon Zdziebłowski
szz/ agt/ kap/<|endoftext|>
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Common Core: 7th Grade Math : Finding Volume of a Rectangular Prism
Example Questions
Example Question #1 : Finding Volume Of A Rectangular Prism
An aquarium is shaped like a perfect cube; the perimeter of each glass face is meters. If it is filled to the recommended capacity, then, to the nearest hundred cubic liters, how much water will it contain?
Insufficient information is given to answer the question.
Note:
Explanation:
A perfect cube has square faces; if a face has perimeter meters, then each side of each face measures one fourth of this, or meters. The volume of the tank is the cube of this, or
cubic meters.
Its capacity in liters is liters.
of this is
liters.
This rounds to liters, the correct response.
Example Question #17 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6
Calculate the volume of the provided figure.
Explanation:
In order to solve this problem, we need to recall the volume formula for a rectangular prism:
Now that we have the correct formula, we can substitute in our known values and solve:
Example Question #1 : Finding Volume Of A Rectangular Prism
Calculate the volume of the provided figure.
Explanation:
In order to solve this problem, we need to recall the volume formula for a rectangular prism:
Now that we have the correct formula, we can substitute in our known values and solve:
Example Question #2 : Finding Volume Of A Rectangular Prism
Calculate the volume of the provided figure.
Explanation:
In order to solve this problem, we need to recall the volume formula for a rectangular prism:
Now that we have the correct formula, we can substitute in our known values and solve:
Example Question #3 : Finding Volume Of A Rectangular Prism
A rectangular prism has the following dimensions:
Length:
Width:
Height:
Find the volume.
Explanation:
Given that the dimensions are: , , and and that the volume of a rectangular prism can be given by the equation:
, where is length, is width, and is height, the volume can be simply solved for by substituting in the values.
This final value can be approximated to .
Example Question #981 : Act Math
Solve for the volume of a prism that is 4m by 3m by 8m.
Explanation:
The volume of the rectangle
so we plug in our values and obtain
.
Example Question #21 : Prisms
The dimensions of Treasure Chest A are 39” x 18”. The dimensions of Treasure Chest B are 16” x 45”. Both are 11” high. Which of the following statements is correct?
Treasure Chest A and B can hold the same amount of treasure.
Treasure Chest A has the same surface area as Treasure Chest B.
Treasure Chest B can hold more treasure.
Treasure Chest A can hold more treasure.
There is insufficient data to make a comparison between Treasure Chest A and Treasure Chest B.
Treasure Chest B can hold more treasure.
Explanation:
The volume of B is 7920 in3. The volume of A is 7722 in3. Treasure Chest B can hold more treasure.
Example Question #1 : Finding Volume Of A Rectangular Prism
A rectangular prism has a width of 3 inches, a length of 6 inches, and a height triple its length. Find the volume of the prism.
Explanation:
A rectangular prism has a width of 3 inches, a length of 6 inches, and a height triple its length. Find the volume of the prism.
Find the volume of a rectangular prism via the following:
Where l, w, and h are the length width and height, respectively.
We know our length and width, and we are told that our height is triple the length, so...
Now that we have all our measurements, plug them in and solve:
Example Question #6 : Finding Volume Of A Rectangular Prism
The above diagram shows a rectangular solid. The shaded side is a square. In terms of , give the volume of the box.<|endoftext|>
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BULLYING STATISTICS IN 2017
Bullying can be defined as a type of aggressive behaviour which is intentional, repeated, and usually involves imbalance of power between the bully and the victim (Olweus, 1993). It’s usually seen among school children, but also among adults in relationships or in workplaces.
This type of aggressive behaviour must be repeated and involve an imbalance of power to be considered bullying.
It’s usually occurring during or after school hours, in school building or playground, but can also happen travelling to or from school, in the neighbourhood or on the Internet.
TYPES OF BULLYING
Bullying includes actions like spreading rumours, making threats, physically or verbally attacking someone, or excluding someone from a group. There are many types of bullying present today; some are obvious, but some are more subtle and can remain unnoticed for a long time.
There are 4 common types of bullying: physical, verbal, social, and a new, modern type - cyber bullying.
Physical bullying, or bullying with aggressive physical intimidation, is manifested as hitting, kicking, pinching, tripping, pushing, blocking, touching in unwanted or inappropriate ways, or damaging property. The most common example for this type of bullying is a scenario where a child is being pulled down on the playground during lunchtime in school.
Verbal bullying, or bullying with cruel spoken words, includes name-calling, threatening, insulting, and making disrespectful comments about someone’s attributes like appearance, religion, disability, sexual orientation, ethnicity…). Common example is bullying of heavier kids in school, calling them “fat” or other harmful names.
Social or relational bullying, or sometimes called bullying with exclusionary tactics, involves lying or spreading rumours, playing nasty jokes to embarrass and humiliate, negative facial of physical gestures, encouraging others to socially exclude someone and damaging someone’s reputation and social acceptance. This type of bullying is often harder to recognize and can often be carried out behind the victim’s back.
Cyber bullying, or bullying in cyberspace, is bullying using digital technologies (computers, smartphones) over social media, texts, websites and other various online platforms. Internet trolling and cyber stalking are also common form of cyber bullying. Cyber bullying is often similar to traditional bullying, but with some distinctions. Victims sometimes don’t know the identity of their bully, or even why he is targeting them.
WHAT TO DO IF YOUR CHILD IS BEING BULLIED
As a parent, there is a good chance you will have to deal with bullying at some point, whether your child is a perpetrator, victim or witness. If your child is being bullied, listen to him, putting your feelings aside. Reassure him that it’s not their fault and encourage them to stay strong and to appear confident. Never tell your child to fight back. The children are always dreading the things can get worse, so don’t just rush off angrily to the bully’s parents. All school are legally required to have an anti-bullying policy. List all the facts about the bullying situations your child has been in and seek school’s help. Encourage your child to develop new skills, join a club or train elf-defence. That can help them restore confidence, make new friends and keep the bullying in perspective.
You can also help by raising awareness of this ever growing problem, by creating a custom wristband here.
Recent U.S. studies have found that 28% of students in grades 6-12 and 20% of students in grades 9-12 have experienced bullying. That’s between 1 in 4 and 1 in every 3 students in the U.S.
But, the UK Annual Bullying Survey of 2017 has showed more alarming results. The survey was conducted in secondary schools and colleges all across the United Kingdom. 54% of all respondents said that they have been bullied at some point in their lives – that’s every other child! 1 in 5 said that they’ve been bullied within the past year, and 1 in 10 has been bullied at least one in the past week.
Number one motive for bullying was attitude towards victim’s appearance – 50% of all bullying motives. 40% were attitudes towards interest and hobbies, followed by attitudes towards high grades, household income, low grades, family issues, disabilities, race, cultural identity, religion, sexuality and gender identity.
The most common type of bullying is reported to be verbal bullying, followed by physical, cyber and social.
PHYCHOLOGICAL EFFECTS OF BULLYING
As the years pass, and bullying gains more and more awareness from the general public, more studies show that bullying can have serious psychological effects, particularly for the victim.
According to the 2017 survey, 37% of all victims developed social anxiety and 36% developed depression. The alarming number is that almost ¼ of victims had suicidal thoughts!
67% of victims reported being bullying, mostly to the teachers and parents, but only 57% of those who reported to teachers were satisfied with support. On the other hand, 89% of those who reported to parents were satisfied with support. The victims who didn’t report to anyone mostly said that it didn’t affect them enough or that they can deal with it themselves, as a reason for not reporting the bullying. The rest said that they’d be called a snitch, that they won’t be taken seriously, or feared that it would get worse. 17% of victims even claimed that they reported it in the past and that nothing happened.
Cyber bullying had devastating impact on the lives of victims over the recent years. According to the study, 17% of young people have experienced cyber bullying in some way.
Respondents mostly reported that they’d been sent a nasty private message, had rumours about them posted online or had a nasty comment posted on their profiles of pictures.
69% of all respondents have done something abusive towards another person online: sent a screenshot of someone’s status or photo to laugh at them in a group chat, trolled somebody in an online game, liked or shared something online that openly mocks another person, sent a nasty message, either privately or publicly to somebody or created a fake profile and used it to annoy or upset somebody.
Awareness is the number one way to prevent and fight bullying. Educating children, students, parents and teachers about bullying is the first step. Everyone should know how to recognize bullying and how they can help stop it.
For the victims, the best strategy when being bullied is to stand up, and calmly say to the bully to stop, walk away and tell an adult what happened.
You can support bullying prevention in many ways, starting with awareness. There are many schools that promote anti-bullying prevention with silicone wristbands. Silicone wristbands are a popular way to raise awareness about a number of causes. Create your custom wristband here and help raise awareness of this growing problem. Don’t be afraid to step up.<|endoftext|>
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# How Do You Divide 2 Fractions
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## How Do You Divide 2 Fractions
Dividing fractions by a whole number is not as difficult as it may seem. To divide a fraction by a whole number, all you need to do is convert the whole number into a fraction, find the reciprocal of that fraction, and multiply the result by the first fraction. To learn how to do this, follow these steps:
### Dividing Mixed Number By Fractions Worksheets
To divide a fraction by a whole number, first write the whole number down to 1 to make it look like a fraction. Then, by changing the numerator and denominator, find the reciprocal of the whole number. Your reciprocity must be equal to 1 in the whole number. When you “divide” fractions by whole numbers, you are actually multiplying the fraction by the reciprocal of the whole number. To do this, multiply the numerator and denominator of the two fractions. Finally, simplify the result as much as possible. If you want to learn more, like how to simplify a subtraction after you’re done, read on! Welcome to this free step-by-step guide to fraction division. This guide will teach you how to use a simple three-step method called Keep-Change-Flip to easily divide fractions by fractions (as well as fractions by whole numbers).
Below are several examples of dividing fractions using the Keep-Change-Flip method, along with an explanation of why this method works for any math problem involving division of fractions. Plus, this free guide includes an animated video lesson and a free practice sheet with answers!
#### Dividing Fractions: 3/5 ÷ 1/2 (video)
Before you learn how to divide fractions using the Keep-Change-Flip method, you need to know how to multiply fractions (which is even easier than division!).
Because fraction multiplication is usually taught before fraction division, you may already know how to multiply two fractions together. If so, you can move on to the next section.
Rule for multiplying fractions: Whenever you multiply fractions together, multiply the numerators and then multiply the denominators as follows.
## Using Subtraction To Divide Fractions
Now that you know how to multiply fractions, you’re ready to learn how to divide fractions using the simple 3-step Keep-Change-Flip method.
To solve this example (and any problem where you need to divide fractions, we’ll use the Keep-Change-Flip method)
If we think of 1/2 ÷ 1/4 in the form of a question, how many 1/4 are there in 1/2?
### Unit 3 Fractions Preview
And then if we visualize 1/4 and 1/2, we clearly see that there’s 2 1/4 in 1/2, and therefore the final answer is 2.
As in example 01, you can solve this problem with the keep change flip method as follows:
What if you want to divide a fraction by a whole number? The process appears to be exactly the same as the previous examples!
## Ex 2.7, 5
Note that in this example you are dividing a fraction by a whole number. But actually converting a whole number to a fraction is very simple. All you need to do is rewrite the number as a fraction, where the numerator and denominator are 1.
Now that you’ve rewritten the whole number as a fraction, you can use the Keep-Change-Flip method to solve the problem.
Watch the video lesson below to learn how to divide fractions by fractions and fractions by whole numbers:
## Dividing Fractions Maze 1
Looking for extra fraction division practice? Click on the links below to download free worksheets and answer keys: This just means that we will flip the fraction so that the numerator is converted to the denominator and the denominator is converted to the numerator.
We care because it helps us create a sense of identity. In other words, when a number is multiplied by its reciprocal, it always equals one!
Throughout this lesson we will learn that the reciprocal of a whole number is always a unit fraction. The multiplicative inverse of a complex number will always be a true fraction.
### Yr6 Progression In Mastery Pack 031
The converse is key to fraction division because the only operations we are allowed to do on fractions are:
So we need to have a way to convert division into a reverse multiplication operation, and the way we do that is reverse our subtraction!
Before we get into the steps, let’s see a visual representation of how diving fractions work by looking at an area model.
And as we saw with multiplication, although an area model represents the process of division, its use is not always practical.
Now we evaluate our numerical expression by multiplying the first fraction by three-fifths by the reciprocal of the second fraction.
We’ll also see that our rules will be useful for multiplying fractions too, because after converting our division problems as mentioned on the prodigy blog, we want to reduce our fractions before multiplying. The math you did in grade school sounds scary. for adults because there are many special rules and words. And dividing fractions is no different: you need to flip fractions and know words like divisor and divisor and reciprocal. It may seem difficult to remember, but with a little practice you won’t succeed.
#### Multiplying And Dividing Fractions Exercise
Because math is all about remembering rules and terms, and if you can do that, dividing fractions is pretty easy. Division is the inverse of multiplication, so one thing to remember when dividing fractions is that the answer is always greater than any part of the problem. You are actually trying to figure out how many divisors (the second number in the problem) can be found in the dividend (the first number).
The first step in dividing fractions is to look at both fractions, take a deep breath, and tell yourself that if a sixth grader can do it, you can probably do it too.
The first step is that simple. Suppose you are trying to find the answer 2/3 ÷ 1/6. Do nothing! Keep these numbers as they are.
#### How To Visualise Dividing With Fractions
The second step is to multiply two fractions. So you just need to change the ÷ sign to the x sign: 2/3 ÷ 1/6 becomes 2/3 x 1/6.
The third step is to do the reciprocal of the divisor – but don’t be afraid! It just means you’re returning the numerator (top number) and denominator (bottom number) of the fraction to the right of the division sign, called the divisor.
For example, if you divide 2/3 by 1/6, you start working on the problem by flipping the divisor: 2/3 x 6/1 = 12/3.
#### How To Divide Fractions In 3 Easy Steps With Examples, Worksheets & More
The fraction may no longer be a true fraction, in which case it will be less than the denominator. This is an improper fraction, meaning that the number represented by the fraction is greater than 1.
No, it’s close, but not quite your definitive answer. All you have to do is simplify the fraction 12/3. You do this by finding the largest number that is divisible by the numerator and denominator, which in this case is 3, meaning the fraction simplifies to 1/4, or just 4.
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A 15th-century suit of armour, inlaid with gold The first knights wore chainmail coats, made from thousands of metal rings joined together. Plate armour offered much more protection than mail, and so by the 1400s complete suits of armour were worn. They were designed to make the sharp edges or points of weapons glance off their smooth surfaces. Armour was no heavier than the mail suits and quite flexible to wear. The suits were jointed together with metal rivets or leather straps so the knight could still bend his legs and swing his sword or mace. On his hands he wore gauntlets, metal gloves that were also jointed so he could move his fingers. A shield was no longer needed to fend off the blows of a hand-held weapon, although a crossbow bolt or metal-tipped arrow from a longbow could still penetrate the armour.
Knights in the 1100s and 1200s wore a hauberk, a tunic made of chainmail—small iron rings linked together—over a padded jacket called an aketon. The combination of the two gave some protection from sword blows, but a direct hit from an arrow or a lance could still be fatal. For a helmet, knights of this period wore a pointed metal cap with a noseguard.
Armour of the 1200s (left) and the 1300s (right)
A full suit of armour of the 1400s weighed about 20–25 kg (45–55 lb). A fit man would have no trouble mounting his horse unaided while wearing armour.
Find the answer<|endoftext|>
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Calculating With Fractions
Adding, Subtracting, Multiplying and Dividing Fractions
Example 1 In this example the common dominator is 12 and both original fractions had to be converted. We have to apply the conversion operation to both the numerator and the denominator in order for the new fraction to be equivalent. We could simplify this fraction to make it 5/6.
Example 2 In this example, we already have a common denominator. We only have to add the two numerators together to get the solution.
Example 3 In this example, we were able to get the common denominator (8) by converting just 1 fraction. We could simplify this fraction to make it 3/4.
Example 4 This example is a bit more complicated because we are adding 3 fractions together. However, the only difference is that we have to convert 3 fractions and add 3 numerators together to get the answer.
Example 1 In this example, we convert the first fraction so that it has the common denominator (9) before subtracting the second fraction from it.
Example 2 In this example, we have the 3 fractions in the expression. Fractions 2 and 3 are subtracted from fraction one after they have been converted to have the common denominator (8).
Example 1 In this example, the numerators and denominators were times together to get the answer which was then simplified. (Remember to always try to cancel instead of simplifying as it is easier.)
Example 2 In this example, the numerators and denominators were again multiplied. However, this time we are unable to cancel or simplify as the fraction is already i its simplest form.
Example 1 In this division example, we invert the second fraction and then multiply the numerators and denominators. Finally we simplify it. (Always try to cancel when doing multiplication questions.)
Example 2 In this example, we have again inverted the second fraction and multiplied both fractions together. As a result we have gotten a mixed number (or improper fraction). The solution was already in its simplest form so we are unable to cancel or simplify.
ULTIMATE MATHS
WHERE MATHS IS AT YOUR FINGERTIPS!
ULTIMATE MATHS Becoming an Accomplished Mathematician Ultimate Maths is a professional maths website that gives students the opportunity to learn, revise and apply different maths skills. We provide a wide range of lessons and resources...
Contact Get in touch by writing us an e-mail to: [email protected] or by using the Contact Us button.<|endoftext|>
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From 11:00PM PDT on Friday, July 1 until 5:00AM PDT on Saturday, July 2, the Shmoop engineering elves will be making tweaks and improvements to the site. That means Shmoop will be unavailable for use during that time. Thanks for your patience!
# At a Glance - Adding and Subtracting Constants
### Sample Problem
Consider the equation x = 5. This equation claims that x is equal to 5. That means the "weight" of x is the same as the "weight" of 5, so let's get visual and draw a picture.
We can create equivalent equations to x = 5 by adding or subtracting the same amount of weight to or from each side of the scale. If we add a hacky sack with weight 1 to each pan, the scale must still balance. (Yes, those are hacky sacks. Just go with it.) It now represents the equation x + 1 = 6...although it's much more difficult to play hacky sack this way.
If we add another hacky sack with weight 1 to each pan, the scale must still balance, now representing the equation x + 2 = 7. Where are we getting all these things?
If we remove two hacky sacks from each side, we're back where we started, representing the equation x = 5. Plus, we get our hacky sacks back. We were starting to worry.
Now we know how to solve equations such as x + 2 = 7. We want to isolate x on one side of the equation, so we take 2 from each side and see the scale balance at = 5. The solution to the equation is 5.
Although the hacky sack analogy breaks down a bit with negative numbers, unless you have one of those newfangled "negative space hacky sacks," the idea is the same. We can add or subtract any number we like, as long as we add or subtract it from both sides of the equation. Otherwise, we'd be unbalancing the scale. The goal is to keep the scale balanced while getting the variable all by itself on one side of the equation.
### Sample Problem
What's the solution to the equation x – 3 = -1?
All we've gotta do is add 3 to both sides of the equation, which keeps everything balanced:
x – 3 = -1
x – 3 + 3 = -1 + 3
x = 2
Boom, done. We got x by itself on one side, so we know x is 2.
### Sample Problem
What's the solution to x + 5 = -2?
Subtract 5 from each side of the equation to get that x all alone:
x + 5 = -2
x + 5 – 5 = -2 – 5
x = -7
One way to keep track of what we're doing is to write the operation we're performing under each side of the equation. Doctors will often do this sort of thing when they're performing operations:
#### Example 1
Solve the equation z + 6 = 3.
#### Example 2
Solve the equation x + 5 – 2 = 3(4).
#### Example 3
Solve the equation z – 6 = 5.
#### Example 4
Solve the equation 9 – 8 + 2 = x + 4.
#### Exercise 1
Solve the equation x + 1 = 1.
#### Exercise 2
Solve the equation 4 + y = 100.
#### Exercise 3
Solve the equation .
#### Exercise 4
Solve the equation 24 ÷ 8 = y + 3.
#### Exercise 5
What's the solution to 5(6) – 4 = z + 11(2)?
#### Exercise 6
The following equation is solved incorrectly. Identify the mistake and then correctly solve the equation:
x + 5 = 14
Incorrect solution:
#### Exercise 7
The following equation is solved incorrectly. Identify the mistake and then correctly solve the equation:
x + 4 = 16
Incorrect solution:
#### Exercise 8
The following equation is solved incorrectly. Identify the mistake and then correctly solve the equation:
y – 1 = -3
Incorrect solution:<|endoftext|>
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Formation of questions and negatives (beginner level)
Sentences in the simple present and simple past tenses do not contain an auxiliary verb. Therefore, we use the auxiliary verbs do, does and did to change them into negatives and questions.
Do and does are used in the simple present tense. Did is used in the simple past tense. Note that do is used with plural nouns and pronouns. The first person pronoun I also take the verb do. Does is used with singular nouns and pronouns.
Study the sentence given below.
Little children learn something new every day. (Affirmative)
Little children do not learn something new every day. (Negative)
Do little children learn something new every day? (Question)
Here the noun children is a plural noun. Therefore we use the verb do.
Change the following sentences into negatives and questions.
1. Beggars sit outside the station all day long.
2. She works at a bank.
3. My son writes with his left hand.
4. She earns a six figure salary.
5. She wants to talk to the manager.
6. The baker called later this afternoon.
1. Beggars do not sit outside the station all day long. / Do beggars sit outside the station all day long?
2. She does not work at a bank. / Does she work at a bank?
3. My son does not write with his left hand. / Does my son write with his left hand?
4. She does not earn a six figure salary. / Does she earn a six figure salary?
5. She does not want to talk to the manager. / Does she want to talk to the manager?
6. The baker did not call later this afternoon. / Did the baker call later this afternoon?<|endoftext|>
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The Nervous System
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The Nervous System
The nervous system controls all body functions.
It receives information from inside and outside the body and then sends out messages to appropriate organs in the body in response to this information.
The organs of the body that receive such information are known as receptor organs and include the eyes, ears, skin and nose.
Organs in the muscles and joints provide further information.
The nervous system has three main parts:
- the brain
- the spinal cord
- the merves
The brain is the control centre for every activity of the body, conscious as well as involuntary actions. It is suspended in clear (cerebrospinal) fluid and surrounded by the skull (cranium) for its protection.
The spinal cord is about 1 cm wide when it leaves the base of the brain through an opening at the base of the skull.The vertebrae of the vertebral column protect it.
It is made up of nerve cells and fibres and carries messages to and from all parts of the body through a series of branching nerves situated in gaps between the vertebrae.
Nerves are thread-like substances made up of a large number of neurones, which are enclosed in an outer coat.
Neurones consist of three parts:
- nucleus: the main cell body
- dendrites: pick up and receive messages or impulses
- axons: transmit messages or impulses
The nerve impulses or messages are carried along neurones that are not directly connected to each other.
The gap between them is known as the synapse or synaptic joint.
Neurones can be divided into two types:
- Sensory neurones carry information to the brain and central nervous system.
- Motor neurones carry information from the brain or the central nervous system to other parts of the body.
There are three types of receptor or sense organs depending on where they are found.
Exteroceptors receive information from outside the body. Examples of these are the eyes, ears and skin.
Interoceptors receive information from organs inside the body including the lungs and digestive system.
Proprioceptors are found mainly in the muscles, tendons and joints. They respond to the degree of stretching in their particular body part and so give information about relative positions of different parts of the body. Proprioceptors enable us to move our limbs with great accuracy and speed without the need to actually watch them.
Log in here<|endoftext|>
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# Lim X → 7 4 − √ 9 + X 1 − √ 8 − X - Mathematics
$\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}$
#### Solution
$\lim_{x \to 7} \left[ \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}} \right]$ It is of the form $\frac{0}{0}$
Rationalising the numerator and the denominator:
$\lim_{x \to 7} \left[ \frac{\left( 4 - \sqrt{9 + x} \right)}{1} \times \frac{\left( 4 + \sqrt{9 + x} \right)}{\left( 4 + \sqrt{9 + x} \right)} \times \frac{1}{\left( 1 - \sqrt{8 - x} \right)} \times \frac{\left( 1 + \sqrt{8 - x} \right)}{\left( 1 + \sqrt{8 - x} \right)} \right]$
= $\lim_{x \to 7} \left[ \frac{16 - \left( 9 + x \right)}{\left( 4 + \sqrt{9 + x} \right)} \times \frac{\left( 1 + \sqrt{8 - x} \right)}{1 - \left( 8 - x \right)} \right]$
= $\lim_{x \to 7} \left[ \frac{- 1\left( - 7 + x \right)\left( 1 + \sqrt{8 - x} \right)}{\left( 4 + \sqrt{9 + x} \right)\left( - 7 + x \right)} \right]$
= $\lim_{x \to 7} \left[ \frac{- \left( 1 + \sqrt{8 - x} \right)}{4 + \sqrt{9 + x}} \right]$
= $- \left( \frac{1 + \sqrt{8 - 7}}{4 + \sqrt{9 + 7}} \right)$
= $\frac{- 2}{4 + 4}$
= $\frac{- 1}{4}$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.4 | Q 15 | Page 28<|endoftext|>
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You are doing a science fair project. Good for you. Your project could be chosen to enter Into your school fair and move on to regional, state, or national competition. You willhave a chance to show off your work at the science fair. You could Possibly Even Receive a award. Science fair projects Can be a lot of fun, or They Can Be a nightmare. It all Depends on how you go about doing it. The most Important Benefit of Entering a science fair is that you Will Learn something cool about science. You Can Learn a lot by just watching and sharing ideas with the other Participants at the science fair.
Do you like Mysteries? Science fair projects are a lot like a mystery where you are the detective looking for the answers to your question. You Will get the chance to test your detective skills When you take on a science project. You Will become the creator of your own scientific experiment That Will Help You Solve your mystery. You Will search for clues Which Will lead you to the answer to your question or mystery. Here is a list of the names of the subjects you Will Need to read to get the guidance and the ideas you need. They Will also tell you all the things you need to know for your science fair project to be a success.
Scientific Method – Research, Problem, Hypothesis, Experiment, CONCLUSION
This section Will help you with tips on how to research your science project. It Will Help You Decide on the problem, and you Will Learn how to suggest a hypothesis, It Will Teach you about experiments and the CONCLUSION.
Topic Research – Project types, Three Steps to a Topic, a Research Topic, Topic Ideas
The topic research section is the section That Will Help You Decide on a topic for your science project
Research Project – Primary Research, Secondary Research
You Will Learn the difference of primary and secondary research. You Will Learn how to write a bibliography and you Will get tips on how to get an interview with people Who Know a lot about the topic you have chosen.
A Sample Project – Starting Your Project, Procedures, Results, explaining Your Results, Hypothesis Content, Abstract, Introduction, Experiment and Data, CONCLUSION, Sources, Acknowledgements. This section Will walk you step by step through a sample project so you Learn everything you need to know Before starting your own project. You Will Learn how to write an Abstract, the Introduction, the Experiment and Data and the CONCLUSION. These are all Important pieces of information you Will Need to Know When doing the written report for your project.
The Display – The display is important in getting your project noticed and remembered. You Will Learn all the tips and tricks to making a great looking display for your science fair project.
Presentation and Evaluation – Have you ever given a presentation in front of all of your classmates beforehand? This section Will give you all the dos and don’ts for give a good presentation to your classmates. After your class mates Will you be ready for the science fair judges at the science fair.
After reading all of this information You will be ready to start on your project. Work hard and follow the steps written in this article and your project Will be a success. Lots of kids just like you have done great science fair projects. I’ve put together a whole series of articles That go Into detail on Each step of your science fair project. You Can look for links to more articles here, or you can find themself all together at the web site below.<|endoftext|>
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## Geometry: Common Core (15th Edition)
Yes, the triangles are similar. $\triangle AZE$ ~ $\triangle RBE$ using the AA Similarity Postulate.
We are given the measures of several angles in the two triangles. Let's see if the Angle-Angle Similarity Postulate can be applied here: $\angle REB ≅ \angle AEZ$, so if $m \angle REB = 35^{\circ}$, then $m \angle AEZ = 35^{\circ}$. Now we have the measures of $\angle REB$ and $\angle AEZ$. We need to find the measure of another pair of corresponding angles to be able to use the AA Similarity Postulate to compare the two triangles. We will now use the triangle sum theorem to find $m \angle Z$ in $\triangle AZE$, which should be congruent to its corresponding angle, $\angle B$ in $\triangle RBE$: $\angle Z = 180 - (100 + 35)$ Evaluate parentheses first: $\angle Z = 180 - (135)$ Subtract to solve: $\angle Z = 45^{\circ}$ $m \angle Z ≅ m \angle B$, so we have two congruent angles. Thus, $\triangle AZE$ ~ $\triangle RBE$ using the AA Similarity Postulate.<|endoftext|>
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# Expressing a number as a difference of two squares
Any whole number can be expressed as a product of two whole numbers. If the number is prime, it can be expressed as a product of itself and 1. The algebraic formula a2 – b2 = (a + b)(a – b) shows that the product of the sum (a + b) and their difference (a – b) is a difference of two squares.
Case A is easy to illustrate. 17, a prime number can be expressed only as a product of 17 and 1. So let (a – b) = 1; that is, let a and b be consecutive numbers adding up to 17. a and b are thus 9 and 8 respectively.
Therefore, ( 9 + 8) ( 9 – 1) = (92 – 82) = 81 – 64 = 17
In case B, 24 can be expressed as 24 x 1, 12 x 2 , 8 x 3 and 6 x 4. We can immediately see that we can not apply the method we used in case B. We need a more general solution. In his book Vedic Mathematics, Sri Bharati Tithaji showed how we can do that.
If we expand (a + b)2 and (a – b)2 and take their difference, we would get
(a2 + 2ab + b2) – (a2 – 2ab + b2) = 4ab or
(a + b)2 – (a – b)2 = 4 ab and
[(a + b)/2]2 – [(a – b)/2]2 = ab
Now if we take 24 as 12 x 2, we would have (12 + 2)/2 = 7 and (12 – 2)/2 = 5 and
72 – 52 = 49 – 25 = 24.
For 24 = 6 x 4, we have (6 + 4)/2 = 5 and (6 – 4)/2 = 1 ; 52 – 12 = 25 – 1 = 24.
We can also apply it for 3 x 8. (8 + 3)/2 = 11/2 and (8 – 3)/2 = 5/2;
(11/2)2 – (5/2)2 = 121/4 – 25/4 = 96/4 = 24
For 1 x 24, we have (24 + 1)/2 = 25/2 and (24 – 1)/2 = 23/2;
(25/2)2 – (23/2)2 = 625/4 – 529/4 = 96/4 = 24
Let us now apply this formula in case B. (17 + 1)/2 = 9 and (17 – 1)/2 = 8. We will again have 92 – 82 = 81 – 64 = 17.
Later we will see how this formula can be used in determining the sides of a right triangle.
Suggested readings: Sri Bharati Tirthaji, Vedic Mathematics, pp 281-284.<|endoftext|>
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This is a true statement because we are dependent on the use of antibiotics to treat infections. More resistance means it is harder to treat infections and infections that we no longer think of as harmful will once again become deadly. Surgical procedures will become much more dangerous because post-operative infections will become more common and harder to treat. Many non-surgical therapies also rely on the ability of antibiotics to defeat infections. Modern clinical medicine is essentially built on the effectiveness of antibiotics.
Recall that mutations can occur randomly in bacteria and are also influenced by environmental and chemical conditions. Mutations are a mechanism that allow for survival in the face of antibiotics. When a resistance gene is present in a population subjected to antibiotics, the bacteria that have it will survive and rapidly reproduce so that all remaining organisms have the resistant gene. These genes are often on plasmids that can easily be shared with other bacteria thereby promoting resistance in multiple species through horizontal gene transfer.
The mechanisms of antibiotic resistance are inactivation of the antibiotic, efflux pumping of the antibiotic, modification of the antibiotic target, and alteration of the targeted metabolic pathway. Inactivation of the antibiotic usually involves bacterial production of an enzyme, such as beta lactamase, that breaks down the antibiotic molecules. This requires expression of, and may involve acquisition of, the gene that codes for the modifying enzyme. Efflux pumping does not require changes in bacterial cell structure or function as efflux pumps are part of the normal cellular apparatus that removes molecules from the cell. Target modification and destruction of the antibiotic require special cellular activity.<|endoftext|>
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# How do you differentiate f(x)=ln(x^x)?
Nov 14, 2015
Use logarithm rules to rewrite, then use the Product Rule.
#### Explanation:
Remember that $\log \left({a}^{b}\right) = b \log \left(a\right)$. You can use this identity to rewrite $\ln \left({x}^{x}\right)$ as $x \ln \left(x\right)$.
Now, in order to find $\frac{d}{\mathrm{dx}} \left[x \ln \left(x\right)\right]$, we must use the Product Rule.
The rule dictates that f'(x)=color(red)(d/(dx)[x])*ln(x)+x*color(blue)(d/(dx)[ln(x)]
Let's take a moment to determine the derivatives.
color(red)(d/(dx)[x]=1),color(blue)(d/(dx)[ln(x)]=1/x
Plug back in.
$f ' \left(x\right) = \textcolor{red}{1} \cdot \ln \left(x\right) + x \cdot \textcolor{b l u e}{\frac{1}{x}}$
color(green)(f'(x)=ln(x)+1<|endoftext|>
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The universe owes its existence, not to reality, (of which we know precious little), but to human imagination and mathematical equations. Only through the medium of conscious observation does the universe become a bona fide representation of itself, and even then, only through the eyes of the beholder. Imagine then, what the universe must look like to other creatures, a frog for instance, or birds that see the ultra violet spectrum, or an insect with compound eyes? The rendering of the universe by human imagination can only be one account of thousands. But of course, we have methods of proof at our disposal, which makes our image of the universe at least plausible. Is it not always the case, that we first imagine something, and then set about trying to create it?
Of late it is noticeable that our sciences are struggling. Imagination has run too fast for science and technology to keep pace. In fact it is precisely because of rampant imagination and advanced mathematics that science does not possess the practical means to confirm the existence of Dark Matter and Dark Energy. These phenomena have been imagined and mathematically proven, but have thus far eluded all methods of detection. Consequently Dark Matter and Dark Energy are presently hypothetical phenomena, and definitive proof of their existence is for the moment ‘work in progress’. In the meantime, science entertains us with pie charts, artist’s impressions, and enthusiastic announcements.
Thus, according to the Planck mission team and based on the standard model of cosmology, the universe consists of 27% Dark matter, 68% Dark energy, and 5% recognisable matter. (For convenience, we will lump Dark Matter and Dark Energy together, under the label; ‘Dark Stuff’, which hypothetically constitutes 95% of the imagined mathematical universe).
So what is Dark Stuff? Astrophysicists and scientists believe that something must exist for the imagined mathematical universe to behave as it does. This inexplicable something they have called ‘Dark Stuff’, because it does not absorb or emit light and is therefore invisible. However; according to the mathematical scientific model of the imagined universe, Dark Stuff must exist.
The basic argument is that the universe exists in a state of delicate balance.
The 5% known must be balanced by the 95% unknown. Therefore the universe as proposed by modern science is ‘asymmetrical’; that is to say, the 5% observable stuff is equal to the 95% invisible stuff, allowing the imagined universe to exist in a state of mathematical / gravitational equilibrium.
It is this concept of an asymmetric universe that offers us a tantalising glimpse of reality. ‘Dark Stuff’ does not absorb or emit light, which means that to all intents and purposes ‘Dark Stuff’ is insensible, impenetrable, and invisible. We might therefore suggest that because ‘Dark Stuff’ is incapable of absorbing or emitting anything at all, it is fundamentally ‘cataleptic’.
Thus the universe according to the mathematical and scientific model is effectively 95% inert and ignorant, leaving only 5% of the universe responsive and illuminated.
If we now apply to these observations the principle of quantum entanglement, it becomes obvious that this asymmetry must be fundamental to all aspects of the interconnected universe, and therefore applies to all – life – matter – and formless phenomena.
This becomes particularly conspicuous within the conscious psyche of human beings, where it correlates exactly with the 95% / 5% asymmetric split proposed by the standard model of science. In fact; if we look objectively at the human condition, we recognise that; while wisdom (Illumination) is limited (5%), stupidity and ignorance (Darkness) is almost unlimited (95%). Thus; by recognising that the human condition itself is subject to the laws of universal asymmetry and quantum entanglement, the existence of ‘Dark Stuff’ becomes verified beyond all doubt.
Einstein came tantalisingly close to this discovery.
Contained within a discarded version of his theory of gravity, he proposed what he called a “Cosmological constant”. This unknown constant (or missing ingredient) has now come to light in the form of ‘Dark Stuff’ or cosmic mindlessness. Einstein was also on the right path when he famously said –
“Only two things are infinite – The universe, and human stupidity. And I’m not so sure about the universe”!
Thus; further research into ‘Dark Stuff’ need not be focused into the depths of outer space, or require enormous amounts of equipment and funding. It has been confirmed here that ‘Dark Stuff’ undeniably exists in the form of utter stupidity and ignorance within the very fabric of the imagined mathematical universe, and is unmistakably present (And obvious) within the consciousness of humanity at large.
We therefore conclude that: ‘Dark Stuff’ should be given the same scientific status as Gamma radiation and Gravity. In that, we know for certain these invisible and formless phenomena exist, but they are detectable only by their effects.
We also propose that ‘Dark Stuff’ should be recognised in proper scientific terminology as ‘Cretinite’ consisting of hitherto unknown sub – atomic particles, identified here as ‘Morons’.
The ‘Moron’ takes a similar form to the atom, containing a central core; ‘Numbtron’, orbited by one or more ‘Ignorons’.
Thus; as the number of ‘Ignorons’ orbiting the central ‘Numbtron’ increase, the dim-wittedness and insensibility (Density) of ‘Dark Stuff’, or ‘Cretinite’, increases proportionately; eventually attaining a level of stupidity so dense, that it becomes incapable of absorbing or emitting anything at all, and thereby becomes an absolute nonentity. (For this reason it cannot be assigned an atomic weight)
We may further speculate that ‘Morons’ consist of lesser particles – ‘Quirks’ and ‘Dolts’ which at present constitute the smallest theoretical unobservable particles within the insentient imagined mathematical universe. Once again; although the ‘Quirk’ and its inconceivably obtuse counterpart, the ‘Dolt’, are undetectable in the physical sense, their presence is clearly recognizable in the human condition.
Thus; the ‘Quirk’ and ‘Dolt’ provide further support to this coherent and well-substantiated hypothesis, which we hereby submit for peer review and mathematical corroboration as: The theory of Cosmic Stupidity.
by John Melhuish<|endoftext|>
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University of Missouri scientists have played a key role in developing new technology that takes the guesswork out of deciding how much nitrogen to apply to crops. The technology has the potential to keep money in farmers’ pockets and help protect the environment.
With “on-the-go sensing,” optical sensors mounted in front of a tractor or fertilizer applicator measure the color and size of plants. An on-board computer uses this data to control the rate that fertilizer is released as the farmer drives through a field.
Efficient nitrogen use has both economic and environmental benefits, said David Dunn, manager of the soil testing labs at MU’s Delta Research Center, a 1,000-acre research farm in the Missouri Bootheel.
“If excess fertilizer is applied to the field, that fertilizer can end up in watersheds, lakes and streams and cause problems there,” he said.
Moreover, producing commercial nitrogen fertilizer consumes fossil fuels, so using less nitrogen shrinks a farm’s carbon footprint.
Traditionally, producers rely on testing soil or plant tissue to measure a field’s nutrient needs. But collecting samples from fields can take hours and getting results back from a lab can take days-time farmers can ill-afford to lose if plants aren’t getting enough nitrogen.
In addition, nutrient needs can vary considerably within a field, Dunn said.
“Traditional methods would give you a broad shotgun approach where you would manage 20 acres at a time with one test,” Dunn said. “Obviously, within those 20 acres, some areas are receiving too much nitrogen and some areas are not receiving enough.”
MU Extension soil scientist Peter Scharf collaborated with researchers in seven Corn Belt states on studies showing that plant color can be a better predictor of nitrogen needs than soil testing.
Today, some producers use aerial surveys, satellite imagery or hand-held sensors to map their fields’ nutrient needs.
“The downside is that you have to take those images back to a computer lab, do a lot of analysis on them, take them to a fertilizer applicator and load them into his machine,” said Gene Stevens, MU Extension plant scientist at the Delta Center. “If a sensor is showing deficient nitrogen, the crop needs fertilizer right now, not a week from now.”
On-the-go sensors reduce that delay to a matter of seconds. “It allows us to take the lab to the field,” Dunn said.
Since 2004, MU researchers have tested on-the-go nitrogen sensing on almost 100 plots of corn at farms and agriculture research stations throughout Missouri. During 2004-2007, the use of on-the-go sensing saved an average of 24 pounds of nitrogen per acre.
Last year, persistent rainfall washed away a lot of the nitrogen that farmers applied before planting. In this case, cornfields treated by on-the-go sensing got more nitrogen than fields managed conventionally, but bigger yields more than made up for that, Scharf said.
Researchers at the Delta Center are adapting this technology for cotton. “In cotton production it’s very important to have the right amount,” Dunn said. Both too little and too much nitrogen will hurt yield.
In 2008, researchers worked with farmer Jim Stuever in Malden, Mo. Delta Center staff treated half of an 80-acre field with on-the-go sensing.
“On our half, we applied 45 pounds less nitrogen per acre and ended up with equivalent yields,” Dunn said. “The area we treated was much more uniform, making it easier to manage-easier to deal with insects, easier to irrigate, easier to harvest.”
The equipment for on-the-go nitrogen sensing isn’t yet widely available and won’t be cheap when it is, but the savings can be significant.
“That kind of efficiency keeps money in my pocket,” Stuever said. Treating Stuever’s field with on-the-go sensing saved about $15 per acre in reduced fertilizer costs.
“With a thousand acres, that’s $15,000 it’s possible to save,” Dunn said.
Corn producers are impressed when they see the system in action, Scharf said. “It is a huge selling point when producers see the application rates go down while driving through dark green corn, then back up when they get into light green, stressed corn.”
The USDA’s Environmental Quality Incentive Program offers cost-sharing agreements to qualifying corn producers who want to adopt certain nutrient-management practices, including on-the-go nitrogen sensing, Scharf said. Similar incentives may become available to cotton farmers as well.<|endoftext|>
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Intermediate Algebra
# Intermediate Algebra
## Intermediate Algebra
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Intermediate Algebra Summer School Credit Recovery
2. Welcome! • Expectations • Earning Credit • Passes • Supplies (student packet, folders, paper)
3. Day 1: Solving Equations Goal: To solve equations in one variable that contain more than one operation Standard: Prior Standard Guiding Question: How do I solve an equation for a variable? Materials: Pencil, Folder, Student Packet
4. Math Review Day 1 Adding and Subtracting Decimals 13.34 + 12 “When you add or subtract decimals, make sure you line the decimals up.” Dividing Fractions “To give dividing fractions a try, flip the second and multiply.” Mental Math Adding and Subtracting Fractions “When you add or subtract fractions, you need a common denominator” Reflection Starters: “I know……” or “I need to work on……”
5. Access: Apply the correct order of operations: • 7 x 4 + 3 = B) (1 + 3)2 – 9 ÷ 3 + 6 = • 12 – 6 x 2 + 7 = D) 24 – 12 ÷ 2 x 3 + 7 =
6. One-Step Equations: • 3 + x = 7 B) -10 = x – 4 Try: C) X – 9 = 11 D) -5 – x = 10 E) -13 = x – 4 F) 17 = 6 - x
7. One-Step Equations: G) 5x = -30 H) 6x - 42 Try: I) 16 = -2x J) 24 =5x K) L)
8. Two-Step Equations: A) 2x – 9 = 18 B) 3x + 6 = -8 Try: C) 4 – 3x = 10 D) E) 13 + 2x = 9 F) 2(5x + 3) = 20
9. Word Problems: Josie bought 4 cases of sports drinks for an upcoming meet. After talking to her coach she bought three more cases spending an additional \$6.95 on additional items. Her receipts totaled \$74.15. Write and solve an equation to find out how much each case of sports drink costs.
10. Work Time: Work through pages 3 and 4 in your packet Multiplication test at: ______ Exit Slip at: _________
11. Multiplication Timed Test: -Page 5 of your packet – tear in half and remove one from the packet You have five minutes to fill in as much as you can Go = Start, Stop = hands up! Highlight the ones you did not know
12. Exit Slip: (on a half-sheet of scratch paper) • 5x – 9 = 17 Make sure it has your name and turn it in!
13. Day 2: Solving Equations Goal: To solve equations that have variables on both sides Standard: Prior Standard Guiding Question: How do I solve an equation for a variable? Materials: Pencil, Folder, Student Packet
14. Math Review Day 2 Adding and Subtracting Decimals 45 – 9.867 “When you add or subtract decimals, make sure you line the decimals up.” Dividing Fractions “To give dividing fractions a try, flip the second and multiply.” Mental Math Adding and Subtracting Fractions “When you add or subtract fractions, you need a common denominator” Reflection Starters: “I know……” or “I need to work on……”
15. Access: Solve the equation: • 3x – 9 = 11 B) C)
16. Solve the equation: • 3d + 8 = 2d – 17 B) – t + 5 = t – 19 C) 5 – (t – 3) = -1 + (2 – 3) D) x + 4 – 6x = 6 - 5x E)-8x + 6 + 9x = -17 + x
17. Try: F) 2y + 3 = 3(y + 7) G) 10 - y + 5 + 6y = 1 + 5y + 3
18. Try: H) 4(x – 3) = 2x + 3x – 9 I) 3(2x – 5) = 2(3x – 2)
19. Work Time: Work through pages 7 and 8 in your packet Multiplication test at: ______ Exit Slip at: _________
20. Multiplication Timed Test: -Page 5 of your packet – tear out of packet You have five minutes to fill in as much as you can Go = Start, Stop = hands up! Highlight the ones you did not know
21. Exit Slip: (on a half-sheet of scratch paper) Previous Material (PM): • 5y – 9 = 16 New Material (NM): B) 3x – 8 = 6 – 2x C) 6x = 5x – 10 Make sure your name is on it, and turn it in!
22. Day 3: Solving Inequalities Goal: To solve multi-step inequalities AND to solve inequalities that contain variables on both sides. Standard: Prior Standard Guiding Question: How do I solve an inequality for a variable? Materials: Pencil, Folder, Student Packet
23. Math Review Day 3 Adding and Subtracting Decimals 15.87+ 1.9 “When you add or subtract decimals, make sure you line the decimals up.” Dividing Fractions “To give dividing fractions a try, flip the second and multiply.” Mental Math Adding and Subtracting Fractions “When you add or subtract fractions, you need a common denominator” Reflection Starters: “I know……” or “I need to work on……”
24. Access: Solve the equation: • 3x + 9 = x – 8 B) 7 – 4x = 6x + 2 C) 7x = 10x - 1
25. Solve: - 3x > 9 Check a Number. What is the rule when solving inequalities?
26. Solve the inequality and graph the solution: • 2m + 1 > 13 B) 2d + 21 ≤ 11 C) D) 4 – X > 3(4 – 2)
27. Solve the inequality and graph the solution: E) 4r – 9 > 7 F) 3 ≤ 5 – 2x G)-4x – 8 > 16 H) I) 12 (x – 3) + 2x ≥ 6
28. Solve the inequality and graph the solution: J) 2x > 4x – 6 K) 5(4 – x) ≤ 3(2 + x)
29. Solve and graph the solution: L) 27x + 33 > 58x – 29 M) 5c – 4> 8c + 2 N) 2(6 – x) < 4x O) 4(y+1)< 4y +2 P) -3(n + 4) ≤ 6( 1 – n)
30. Work Time: Work through pages 9 and 10 in your packet Multiplication test at: ______ Exit Slip at: _________
31. Multiplication Timed Test: -Page 11 of your packet – tear out of packet You have five minutes to fill in as much as you can Go = Start, Stop = hands up! Highlight the ones you did not know
32. Exit Slip: (on a half-sheet of scratch paper) Previous Material (PM): • 2r + 20 = 200 B) 3(2x – 5) = 2(3x – 2) New Material (NM): C) 2 + (-6) > -8p D) 3(1-x) ≥ 3(x + 2) Make sure your name is on it, and turn it in!
33. Day 4: Graphing Linear Functions Goal: To solve for a variable AND To graph linear functions using tables or equations Standard: 9.2.1.8 – Make Qualitative statements about the rate of change of a function based on its graph or table of values 9.2.2.3 – Sketch graphs of linear, quadratic and exponential functions and translate between graphs, tables and symbolic representations. Know how to use graphing technology to graph these functions. Guiding Question: What and how are the many ways I can graph a line? Materials: Pencil, Folder, Student Packet
34. Math Review Day 4 Adding and Subtracting Decimals 1.309+ 134.8 “When you add or subtract decimals, make sure you line the decimals up.” Dividing Fractions “To give dividing fractions a try, flip the second and multiply.” Mental Math Adding and Subtracting Fractions “When you add or subtract fractions, you need a common denominator” Reflection Starters: “I know……” or “I need to work on……”
35. Access: Graph the points on a coordinate plane: A (5, 6) B (-1, -3) C (4, -9) D (-1.5, 0)
36. Solve for a variable: • 2x - 3y = 12 B) 2x + y = 8 Try: C) 5y = 5x - 10 D) 2y - 6y = -8
37. What is a function? What makes a function linear? How can I graph a line? Table, Slope and Intercept, x-and y- intercepts, and slope-intercept form
38. Graph: • Slope = y-intercept = 4 B) Slope = 4, y-intercept =
39. Try: C) Slope = y-intercept = 4 D) slope = 3, y-intercept = 2
40. Graph A) B) C) y = x + 6
41. Try: D) E) y = 3x - 1 F) y = -2x + 4
42. Graph: • 6x + 3y = 12 B) 2x + y = 8
43. Try: C) 2x - 6y = 6 D) 2x + 3y = -12 E) 5x - 2y = 10
44. Work Time: Work through pages 13 and 14 in your packet Multiplication test at: ______ Exit Slip at: _________
45. Multiplication Timed Test: -Page 11 of your packet – tear out of packet You have five minutes to fill in as much as you can Go = Start, Stop = hands up! Highlight the ones you did not know
46. Exit Slip: (on a half-sheet of scratch paper) New Material (NM): Solve for y: • 7y + 14x = 28 B) -5y = 2x + 7 Graph: A) B) y = -3x C) y = 2 D) 3x - 2y = 6 Make sure your name is on it, and turn it in!
47. Day 5: Graphing Linear Inequalities Goal: To graph linear inequalities using tables or equations. AND To write equations to describe lines. Standard: 9.2.4.4 – Represent relationships in various contexts using systems of linear inequalities; solve them graphically. Indicate which parts of the boundary are included in and excluded from the solution set using solid and dotted lines. Guiding Question: How do I graph a linear inequality? AND How can I write equations of lines? Materials: Pencil, Folder, Student Packet
48. Math Review Day 5 QUIZ Adding and Subtracting Decimals • 67.8 + 5.23 • 71 – 8.09 Dividing Fractions Adding and Subtracting Fractions
49. Access: Graph: • y = 3x - 2 B) C) y = -2x + 5
50. Write an equation with the following information: • Slope = , y-intercept = 4 B) Slope = 4, y-intercept = Try: C) Slope = , y-intercept = 4 D) slope = 3, y-intercept = 2<|endoftext|>
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#### Passage
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$
Mathematics
# To evaluate $\displaystyle\int{\frac{dx}{(x-1)\sqrt{-x^2+3x-2}}}$ one of the most suitable substitution could be
$\sqrt{-x^2+3x-2}=u(1-x)$
$\sqrt{-x^2+3x-2}=u(x-2)$
##### SOLUTION
$-{ x }^{ 2 }+3x-2=-{ x }^{ 2 }+2x+x-2=-x\left( x-2 \right) +1\left( x-2 \right) =\left( 1-x \right) \left( x-2 \right)$
has real roots $1$ and $2$
So from case III: $\sqrt { -{ x }^{ 2 }+3x-2 } =u(1-x)$ or $u(x-2)$
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Multiple Correct Hard Published on 17th 09, 2020
Mathematics
# $\displaystyle\int{\frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx}$ is equal to
$\displaystyle\frac{{(x+\sqrt{1+x^2})}^{15}}{15}+C$
##### SOLUTION
$\displaystyle\int{\frac{{(x+\sqrt{1+x^2})}^{15}}{\sqrt{1+x^2}}dx}$
$\displaystyle=\int { \frac { { { (x+\sqrt { 1+x^{ 2 } } ) }(x+\sqrt { 1+x^{ 2 } } ) }^{ 14 } }{ \sqrt { 1+x^{ 2 } } } dx }$
Substitute $(x+\sqrt { 1+x^{ 2 } } )=t$
$\displaystyle (1+\frac { 2x }{ 2\sqrt { 1+x^{ 2 } } } )dx=dt$
$\displaystyle \frac { { { (x+\sqrt { 1+x^{ 2 } } ) } } }{ \sqrt { 1+x^{ 2 } } } dx=dt$
So,
$I=\displaystyle \int{{ t }^{ 14 }dt }$
$\displaystyle =\frac { { { { t }^{ 15 } } } }{ 15 } +C$
$I=\displaystyle \frac { { { { (x+\sqrt { 1+x^{ 2 } } ) }^{ 15 } } } }{ 15 } +C$
Its FREE, you're just one step away
Single Correct Hard Published on 17th 09, 2020
Mathematics
# If $\displaystyle I=\int{\frac{dx}{x-\sqrt{9x^2+4x+6}}}$ to evaluate $I$, one of the most proper substitution could be
$\sqrt{9x^2+4x+6}=u\pm3x$
##### SOLUTION
As $9=m>0$ then from case 1
$\sqrt{9x^2+4x+6}=u\pm \sqrt{m}x=u\pm 3$
Its FREE, you're just one step away
Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111
#### Realted Questions
Q1 Single Correct Medium
Let $f:R \to R,g:R \to R$ be continuous functions. Then the value of indtegral
$\displaystyle \int\limits_{In\lambda }^{InI/\lambda } {\frac{{f\left( {\dfrac{{{x^2}}}{4}} \right)\left[ {f\left( x \right) - f\left( { - x} \right)} \right]}}{{g\left( {\dfrac{{{x^2}}}{4}} \right)\left[ {g\left( x \right) + g\left( { - x} \right)} \right]}}} dx$
• A. a non-zero constant
• B. Zero
• C. None of these
• D. depend on $\lambda$
1 Verified Answer | Published on 17th 09, 2020
Q2 Subjective Medium
Evaluate: $\int (1-x)\sqrt {x}\ dx$
1 Verified Answer | Published on 17th 09, 2020
Q3 Passage Hard
Integral of form $\int f\left ( x, \sqrt{ax^{2}+bx+c} \right )dx$ can be evaluated with the help of Euler's substitution in the following manners:
(a) If $a> 0$, we put $\sqrt{ax^{2}+bx+c}=t\pm x\sqrt{a}$ or $ax^{2}+bx+c=t^{2}\pm tx\sqrt{a}+ax^{2}$ i.e. $bx+c=t^{2}\pm 2tx\sqrt{a}$
(b) If $c> 0$, we put $\sqrt{ax^{2}+bx+c}=tx\pm \sqrt c$ or $ax+b=t^{2}x\pm 2tx\sqrt{c}$
(c) If the trinomial $ax^{2}+bx+c$ has $\alpha$, $\beta$ as its real zero's
i.e. $ax^{2}+bx+c=a\left ( x-\alpha \right )\left ( x-\beta \right )$ then we put $\sqrt{ax^{2}+bx+c}=t\left ( x-\alpha \right )$ or $t\left ( x-\beta \right )$
On the basis of above information answer the following question:
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Medium
If $y=\sqrt [ 4 ]{ { x }^{ 4 }-1 }$ then $\int { \frac { dx }{ \sqrt [ 4 ]{ { x }^{ 4 }-1 } } }$ is equal to
• A. $-\frac { 3 }{ 4 } [log|\frac { y+1 }{ y-1 } |-{ Tan }^{ -1 }(y)]+c$
• B. $-\frac { 1 }{ 2 } [log|\frac { y+x-1 }{ y-x+1 } |+{ Tan }^{ -1 }(y)]+c$
• C. $-\frac { 3 }{ 8 } [log|\frac { y-x }{ y+x } |-{ Tan }^{ -1 }(\frac { y }{ x } )]+c$
• D. $\frac { 1 }{ 4 } log|\frac { y-x }{ y+x } |-\frac { 1 }{ 2 } { Tan }^{ -1 }(\frac { y }{ x } )+c$
1 Verified Answer | Published on 17th 09, 2020
Q5 Single Correct Hard
The value of the integral $\displaystyle \int_{\frac {1}{3}}^1\frac {(x-x^3)^{\frac {1}{3}}}{x^4}dx$
• A. $0$
• B. $3$
• C. $4$
• D. $6$<|endoftext|>
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# Proof: By Euclid
• Let $A$ and $B$ be square numbers, and let $C$ be the side of $A$, and $D$ (the side) of $B$.
• I say that there exists one number in mean proportion to $A$ and $B$, and that $A$ has to $B$ a squared ratio with respect to (that) $C$ (has) to $D$.
• For let $C$ make $E$ (by) multiplying $D$.
• And since $A$ is square, and $C$ is its side, $C$ has thus made $A$ (by) multiplying itself.
• And so, for the same (reasons), $D$ has made $B$ (by) multiplying itself.
• Therefore, since $C$ has made $A$, $E$ (by) multiplying $C$, $D$, respectively, thus as $C$ is to $D$, so $A$ (is) to $E$ [Prop. 7.17].
• And so, for the same (reasons), as $C$ (is) to $D$, so $E$ (is) to $B$ [Prop. 7.18].
• And thus as $A$ (is) to $E$, so $E$ (is) to $B$.
• Thus, one number (namely, $E$) is in mean proportion to $A$ and $B$.
• So I say that $A$ also has to $B$ a squared ratio with respect to (that) $C$ (has) to $D$.
• For since $A$, $E$, $B$ are three in (continued) proportion numbers, $A$ thus has to $B$ a squared ratio with respect to (that) $A$ (has) to $E$ [Def. 5.9] .
• And as $A$ (is) to $E$, so $C$ (is) to $D$.
• Thus, $A$ has to $B$ a squared ratio with respect to (that) side $C$ (has) to (side) $D$.
• (Which is) the very thing it was required to show.
Thank you to the contributors under CC BY-SA 4.0!
Github:
non-Github:
@Fitzpatrick<|endoftext|>
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How to Complete the Square: A Comprehensive Guide
Baca Cepat
Introduction:
Welcome to our guide on how to complete the square! Completing the square is an essential skill in algebra that opens doors to solving more complex equations. Whether you’re a student or just looking to brush up on your math skills, this guide will provide you with all the tools you need to master this important concept.
But before we dive into the details, let’s first define what it means to complete the square in algebra. Essentially, to complete the square means to rewrite a quadratic equation in a certain form that makes it easier to solve. This form is called the vertex form, and it is written as:
y = a(x – h)^2 + k
Where a, h, and k are constants that depend on the coefficients of the quadratic equation. By completing the square, we can find the values of h and k, which are the x and y coordinates of the vertex of the parabola defined by the quadratic equation.
In the rest of this guide, we’ll go through the steps required to complete the square for any quadratic equation, as well as some common mistakes to avoid. Let’s get started!
Step 1: Convert the Equation to Standard Form
The first step to completing the square is to convert the quadratic equation into standard form, which is written as:
y = ax^2 + bx + c
To do this, we need to collect all the terms on one side of the equation and simplify as much as possible. For example, let’s say we have the equation:
x^2 + 6x – 5 = 0
To put this equation in standard form, we need to move the constant term to the other side:
x^2 + 6x = 5
Now, we can complete the square by adding a constant term to both sides of the equation.
Step 2: Add and Subtract the Correct Constant
To complete the square, we need to add and subtract a certain constant to the equation to create a perfect square trinomial. The constant we add is half the coefficient of the x term squared:
c = (b/2)^2
For example, in the equation:
x^2 + 6x = 5
The coefficient of the x term is 6, so we need to add and subtract 9 to make it a perfect square trinomial:
x^2 + 6x + 9 – 9 = 5
Notice that we added 9 to the left side of the equation, but we also subtracted 9 to keep the equation balanced.
Step 3: Rewrite as a Perfect Square Trinomial
Now that we have added and subtracted the correct constant, we can rewrite the left side of the equation as a perfect square trinomial:
(x + 3)^2 – 9 = 5
Notice that we rewrote x^2 + 6x + 9 as (x + 3)^2, which is a perfect square trinomial. We also simplified the expression on the right side of the equation.
Step 4: Solve for x
Now that we have rewritten the equation as a perfect square trinomial, we can easily solve for x. Continuing with our example:
(x + 3)^2 – 9 = 5
First, we add 9 to both sides of the equation:
(x + 3)^2 = 14
Next, we take the square root of both sides:
x + 3 = ±√14
Finally, we subtract 3 from both sides:
x = -3 ±√14
And there you have it! We have successfully completed the square of the quadratic equation x^2 + 6x – 5 = 0 and found the values of x.
Table of Common Quadratic Equations and Their Solutions
Equation Solution
x^2 + 4x + 4 = 0 x = -2
x^2 – 6x + 9 = 0 x = 3
x^2 + 5x + 6 = 0 x = -2, -3
FAQs
1. What is completing the square?
Completing the square is a technique used to rewrite a quadratic equation in a certain form that makes it easier to solve.
2. Why is completing the square useful?
Completing the square is useful because it allows us to find the values of x that make the quadratic equation equal to zero.
3. How do I know when to complete the square?
You should complete the square when you have a quadratic equation that you need to solve.
4. What is the vertex form of a quadratic equation?
The vertex form of a quadratic equation is written as y = a(x – h)^2 + k, where a, h, and k are constants.
5. Can you complete the square for a cubic equation?
No, completing the square is only applicable to quadratic equations. For cubic equations, you would need to use a different technique.
6. What is the discriminant of a quadratic equation?
The discriminant of a quadratic equation is the part of the equation under the square root symbol, which is written as b^2 – 4ac.
7. How do I know if a quadratic equation has real solutions?
A quadratic equation has real solutions if its discriminant is greater than or equal to zero.
8. What is a perfect square trinomial?
A perfect square trinomial is a trinomial that can be factored into two factors that are identical.
9. What is the quadratic formula?
The quadratic formula is a formula that can be used to solve any quadratic equation. It is written as x = (-b ± √(b^2 – 4ac)) / 2a.
10. What is the difference between factoring and completing the square?
Factoring and completing the square are both techniques used to solve quadratic equations, but they involve different methods.
11. How can I check my work when completing the square?
You can check your work by verifying that the equation is in vertex form and that the values of h and k are correct.
12. What happens if I make a mistake when completing the square?
If you make a mistake when completing the square, your final answer will be incorrect. Make sure to double-check your work and avoid common mistakes.
13. Can completing the square be used to find the maximum or minimum value of a quadratic equation?
Yes, completing the square can be used to find the maximum or minimum value of a quadratic equation, which is also the value of the vertex.
Conclusion
Congratulations, you have now learned how to complete the square of any quadratic equation! By following the steps outlined in this guide and avoiding common mistakes, you can now confidently solve quadratic equations using this powerful technique.
Remember to practice completing the square on a variety of equations to solidify your understanding. And if you ever get stuck, come back to this guide for a refresher.
We hope this guide has been helpful, and we wish you the best of luck in your math studies!
Disclaimer
The information contained in this article is for educational purposes only and should not be construed as professional advice. While we have taken great care to ensure the accuracy of the information presented, we make no warranties or representations as to the completeness or accuracy of the content. The reader assumes full responsibility for any actions taken based on the information presented in this article.<|endoftext|>
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Click on any wind turbine technology to read more about it.
To study detailed technical constraints in design of these wind mills based on Aerodynamic, Electrical, Transmission, Control and Safety designs, click here.
Horizontal Axis Wind Turbine (HAWT)
Horizontal axis wind turbines have the main rotor shaft and electrical generator at the top of a tower, and they must be pointed into the wind. Small turbines are pointed by a simple wind vane placed square with the rotor (blades), while large turbines generally use a wind sensor coupled with a servo motor. Most large wind turbines have a gearbox, which turns the slow rotation of the rotor into a faster rotation that is more suitable to drive an electrical generator. Since a tower produces turbulence behind it, the turbine is usually pointed upwind of the tower. Wind turbine blades are made stiff to prevent the blades from being pushed into the tower by high winds. Additionally, the blades are placed a considerable distance in front of the tower and are sometimes tilted up a small amount. Downwind machines have been built, despite the problem of turbulence, because they don't need an additional mechanism for keeping them in line with the wind, and because in high winds, the blades can be allowed to bend which reduces their swept area and thus their wind resistance. Since turbulence leads to fatigue failures, and reliability is so important, most HAWTs are upwind machines. Types of HAWT wind farm installations are:
- On-Shore: Mountains and hilly areas have been the original choice to setup these farms. Individual wind turbines at these farms contribute towards power generation of 100 MW or more. The land occupied by the wind parks are often used for agriculture or animal grazing. Denmark, Spain and Portugal are some of the leading countries in the onshore wind farm electricity production.
- Offshore wind farms are the results of revolutionary technology that has encouraged man to set up wind energy harvesting farms on the water surface. Apart from oceans, lakes also act as sites for the installation of wind parks. An advantage of offshore wind farm is that it makes use of powerful winds blowing over the water surface. Moreover, it is easy to transport huge parts of a wind turbine to the offshore sites using big ships and vessels. Some of the other advantages of these farms include mitigation of noise due to distance from land and higher capacity factors. The United Kingdom is the nation that leads in electricity generation using offshore wind parks. Denmark, Sweden and Netherlands are other countries that follow.
Off-shore Wind in India
- A near-shore wind farm is the third type of farms used for harvesting wind power. As suggested by its name, a near-shore wind farm is installed near the shore, thus making use of land and see breezes to turn the turbines. In future, you might come across air-borne wind farms, with wind turbines requiring no towers for installation.
- The tall tower base allows access to stronger wind in sites with wind shear. In some wind shear sites, every ten meters up the wind speed can increase by 20% and the power output by 34%.
- High efficiency, since the blades always move perpendicularly to the wind, receiving power through the whole rotation. In contrast, all vertical axis wind turbines, and most proposed airborne wind turbine designs, involve various types of reciprocating actions, requiring airfoil surfaces to backtrack against the wind for part of the cycle. Backtracking against the wind leads to inherently lower efficiency.
- Massive tower construction is required to support the heavy blades, gearbox, generator and an additional yaw control mechanism to turn the blades toward the wind.
- Downwind variants suffer from fatigue and structural failure caused by turbulence when a blade passes through the tower's wind shadow (for this reason, the majority of HAWTs use an upwind design, with the rotor facing the wind in front of the tower).
- HAWTs generally require a braking or yawing device in high winds to stop the turbine from spinning and destroying or damaging itself.
Vertical Axis Wind Turbine (VAWT)
VAWTs, have the main rotor shaft arranged vertically. The main advantage of this arrangement is that the wind turbine does not need to be pointed into the wind. This is an advantage on sites where the wind direction is highly variable or has turbulent winds. With a vertical axis, the generator and other primary components can be placed near the ground, so the tower does not need to support it, also makes maintenance easier. The main drawback of a VAWT is that it generally creates drag when rotating into the wind. It is difficult to mount vertical-axis turbines on towers, meaning they are often installed nearer to the base on which they rest, such as the ground or a building rooftop. Hence these models are not frequently used for off-shore installations which if used might require water proof casings that adds to extra costs. Also, offshore installations require very high height towers. The wind speed is slower at a lower altitude, so less wind energy is available for a given size turbine. Air flow near the ground and other objects can create turbulent flow, which can introduce issues of vibration, including noise and bearing wear which may increase the maintenance or shorten its service life. However, when a turbine is mounted on a rooftop, the building generally redirects wind over the roof and these can double the wind speed at the turbine. If the height of the rooftop mounted turbine tower is approximately 50% of the building height, this is near the optimum for maximum wind energy and minimum wind turbulence. Various types of VAWT are:
- Darrieus wind turbine: "Eggbeater" turbines, or Darrieus turbines, were named after the French inventor, Georges Darrieus. They have good efficiency, but produce large torque ripple and cyclical stress on the tower, which contributes to poor reliability. They also generally require some external power source, or an additional Savonius rotor to start turning, because the starting torque is very low. The torque ripple is reduced by using three or more blades which results in greater solidity of the rotor. Solidity is measured by blade area divided by the rotor area. Newer Darrieus type turbines are not held up by guy-wires but have an external superstructure connected to the top bearing.
- Giromill: It is a subtype of Darrieus turbine with straight, as opposed to curved, blades. The cycloturbine variety has variable pitch to reduce the torque pulsation and is self-starting. The advantages of variable pitch are: high starting torque; a wide, relatively flat torque curve; a lower blade speed ratio; a higher coefficient of performance; more efficient operation in turbulent winds; and a lower blade speed ratio which lowers blade bending stresses. Straight, V, or curved blades may be used.
- Savonius wind turbine: These are drag-type devices with two (or more) scoops that are used in anemometers, Flettner vents (commonly seen on bus and van roofs), and in some high-reliability low-efficiency power turbines. They are always self-starting if there are at least three scoops.
- Twisted Savonius: Twisted Savonius is a modified savonius, with long helical scoops to give a smooth torque, this is mostly used as roof windturbine or on some boats (like the Hornblower Hybrid)
- No yaw mechanism is needed because they have lower wind startup speeds than the typical the HAWTs.
- A VAWT can be located nearer the ground, making it easier to maintain the moving parts.
- VAWTs situated close to the ground can take advantage of locations where rooftops, mesas, hilltops, ridgelines, and passes funnel the wind and increase wind velocity.
- Most VAWTs have an average decreased efficiency from a common HAWT, mainly because of the additional drag that they have as their blades rotate into the wind.
- Having rotors located close to the ground where wind speeds are lower due and do not take advantage of higher wind speeds above.
Wind Turbine Technology Based on Applications
Wind power can be divided into three size ranges, which are used for different applications. The size is chosen differently depending on the turbine’s purpose. The height/span of wind turbines can be anticipated by comparing with some other large objects and installations. Typical sizes in the three ranges available are:
Residential: below 30 kW
- Choose a size based on electrical load
- Diameter: 1 - 13 m (4 - 43 ft)
- Height: 18 - 37 m (60 - 120 ft)
- Example: 20,000 kWh/year
Medium: 30 - 500 kW
- May be sized to a load. Typically used when there is a large electrical load.
- Diameter: 13 - 30 m (43 - 100 ft)
- Height: 35 - 50 m (115 - 164 ft)
- Example: 600,000 kWh/year
Commercial scale: 500 kW - 2 MW
- Usually fed into the grid, not sized to a single load
- Diameter: 47 - 90 m (155 - 300 ft)
- Height: 50 - 80 m (164 - 262 ft)
- Example: 4,000,000 kWh/year<|endoftext|>
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# 71000 in words
71000 in words is written as Seventy One Thousand. In 71000, 7 has a place value of ten thousand and 1 has a place value of thousand. The article on Place Value gives more information. The number 71000 is used in expressions that relate to money, distance, length, social media view, and many more. For example, “Seventy One Thousand is the number of subscribers to a newly started Youtube channel.” “There were Seventy One Thousand orders for newly launched smartphones on day 1.”
71000 in words Seventy One Thousand Seventy One Thousand in Numbers 71000
## How to Write 71000 in Words?
We can convert 71000 to words using a place value chart. The number 71000 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 7 1 0 0 0
Thus, we can write the expanded form as:
7 × Ten thousand + 1 × Thousand + 0 × Hundred + 0 × Ten + 0 × One
= 7 × 10000 + 1 × 1000 + 0 × 100 + 0 × 10 + 0 × 1
= 71000
= Seventy One Thousand.
71000 is the natural number that is succeeded by 70999 and preceded by 71001.
71000 in words – Seventy One Thousand.
Is 71000 an odd number? – No.
Is 71000 an even number? – Yes.
Is 71000 a perfect square number? – No.
Is 71000 a perfect cube number? – No.
Is 71000 a prime number? – No.
Is 71000 a composite number? – Yes.
## Solved Example
1. Write the number 71000 in expanded form
Solution: 7 x 10000 + 1 x 1000 + 0 x 100 + 0 x 10 + 0 x 1
Or Just 7 x 10000 + 1 x 1000
We can write 71000 = 70000 + 1000 + 0 + 0 + 0
= 7 x 10000 + 1 x 1000 + 0 x 100 + 0 x1 0 + 0 x 1
## Frequently Asked Questions on 71000 in words
### How to write 71000 in words?
71000 in words is written as Seventy One Thousand.
### State whether True or False. 71000 is divisible by 3?
False. 71000 is not divisible by 3.
### Is 71000 divisible by 10?
Yes. 71000 is divisible by 10.Â<|endoftext|>
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Genealogists researching Jewish ancestry in Eastern Europe face several difficulties. Differing official policies across the regions of Jewish settlement produced great variability in vital records. Laws and customs varied by nationality, and borders were constantly in flux. A single town may have gone through many border changes and governing entities during its history. The Jewish history timeline below will help you sort through the region’s complicated past.
The century and a half from the first partition of Poland (1772) until the end of World War I (1918) was a time of profound changes for East European Jewry. Poland ceased to exist as an independent entity as three continental empires—the Austrian Hapsburgs, tsarist Russia, and Teutonic Prussia—carved up the territory in a series of three partitions, leaving a meandering border that arbitrarily divided Jewish communities.
Following the defeat of Napoleon in 1815, the Congress of Vienna locked in a map of Central Europe that prevailed with minor adjustments until the end of World War I (1918). Under the Treaty of Versailles, Poland regained its independence (see the map above).
Each empire administered its acquired Jewish population differently:
- In the Polish territories that Prussia annexed, Napoleonic-era reforms led to the emancipation of some Prussian Jews as early as 1812.
- Austria acquired a tier of southern Poland on the north slope of the Carpathian Mountains called Galicia, which it administered with some autonomy for its local populations. Jews gradually obtained legal rights until achieving full emancipation in 1869.
- Russia was the most restrictive, designating a region called the Pale of Settlement in which Jews could live. The Pale comprised the acquired Polish lands as well as southern, formerly Ottoman provinces. Russia administered the central region of Poland, called Congress Poland after the Congress of Vienna, separately from the Pale. After a period of liberalization in the mid-19th century, a crackdown in 1882 following the assassination of the Tsar set off a wave of anti-Jewish violence. The pogroms and widespread poverty led to an earlier wave of migration from Russia than occurred elsewhere. Jews in the Pale were not granted citizenship rights until the Pale’s demise during the Russian Revolution.
Ashkenazi family researchers who have family branches from more than one of the imperial territories will have to contend with the mixture of laws, cultural values and beliefs that combined to determine everything from an ancestor’s chosen occupation to the family kugel recipe.
Regardless of which region your ancestors lived in, a few key dates affected Jews all over Eastern European. Here are some of the most important dates in the Jewish history timeline:
The first partition of Poland creates Austrian Galicia and cedes Polish lands to Russia and Prussia.
The Russian empress Catherine the Great establishes the Pale of Settlement, which dictated where the government allowed Jews to live.
The second partition of Poland further expands Russian and Prussian holdings.
The third partition of Poland brings the demise of the Polish-Lithuanian Commonwealth.
The Congress of Vienna encodes European borders, which stand for 100 years.
Prussian Jews are emancipated.
Austrian Jews are emancipated.
Prussia and various German states unite to form the German Empire.
The Russian government enacts the “May Laws,” which restrict Jewish settlement and undo previous progressive policies. Meant to be temporary, the laws remain in place for more than 30 years.
Jewish nationalism rises and the First Zionist Congress meets.
Russian Jews are emancipated following World War I and the Russian Revolution.
The Treaty of Versailles restores Poland as an independent entity.
For more on researching your Jewish roots in Eastern Europe, check out the May/June 2019 issue of Family Tree Magazine. The issue includes this timeline, plus a guide to the best resources for Jewish ancestry in modern Russia, Poland, Belarus and more.<|endoftext|>
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# Thread: Two Exponential Equations, and Two Triq Questions!
1. ## Two Exponential Equations, and Two Triq Questions!
1. $9^x+4*3^x - 3 = 0$
2. $5^x = 4^x+5$ <-----its 4^x+5, 4 is raised to the x+5!!! Can't figure out how to make it look that way.. =/
3. A formula for cos(3X) in terms of cosX alone.
4. The period of 3cos(5pix+7)
2. _______________________________________
#1:
\begin{aligned}9^x + 4 \cdot 3^x - 3 & = 0 \\ \left(3^2\right)^x + 4 \cdot 3^x - 3 & = 0 \\ \left(3^x\right)^2 + 4 \cdot 3^x - 3 & = 0 \end{aligned}
If it helps, imagine $y = 3^x$. Then we get an easier problem: $y^2 + 4y - 3 = 0$
_______________________________________
#2: Start by taking the natural logarithm of both sides ..
\begin{aligned} {\color{red}\ln \ }5^x & = {\color{red}\ln \ }4^{x+5} \\ x \ln 5 & = (x+5) \ln 4 \qquad \text{ Since: } \ln a^b = b\ln a\\ & \ \ \vdots \end{aligned}
_______________________________________
#3: $\cos (3x) = \cos (x + 2x)$
Now these will come in handy:
• $\cos (a + b) = \cos a \cos b - \sin a \sin b$
• $\cos (2a) = 2\cos^2 a - 1$
• $\sin (2a) = 2\sin a \cos a$
• $\sin^2 (a) = 1 - \cos^2 a$
_______________________________________
#4: $3\cos \left(5\pi x + 7\right) \ = \ 3\cos \left( 5\pi \left(x + \frac{7}{5\pi}\right)\right)$
If you remember, if you have a sinusoidal wave in the form: $y = A\cos (B (x + C)) + D$
The period is given by: $\frac{2\pi}{B}$
_______________________________________<|endoftext|>
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The kidneys are two reddish-brown, bean-shaped organs located just above the waist. One kidney is just to the left and the other just to the right of the backbone. Both are protected by the lower ribcage. The kidneys do not re-grow like the liver does. When kidney tissue is lost, which can happen in surgery, kidney function decreases. In addition, adults lose kidney function as they age.
The kidneys' main job is to filter the blood and rid the body of excess water, salt and waste products. The filtered waste products are concentrated into urine. Urine passes from the kidneys into the bladder through two tubes called ureters. Urine leaves the bladder through another tube called the urethra.
Several types of cancer can develop in the kidney. Renal cell cancer is the most common type of kidney cancer. It occurs more often in men than in women.
Like all cancers, renal cell cancer begins small and grows larger over time. Although renal cell cancer usually grows as a single mass, a kidney may contain more than one tumor. Tumors may also be found in both kidneys at the same time. Some renal cell cancers are noticed only after they have become quite large, but most are found before they metastasize (spread) to other organs through the bloodstream or lymph vessels.
Transitional cell cancer, which affects the lining of the renal pelvis, is a less common form of kidney cancer. It is similar to cancer that occurs in the lining of the bladder and is often treated like bladder cancer.
Treatment for Kidney Cancer at MD Anderson Cooper
Different types of treatment are available for kidney cancer. Some treatments are standard – the currently used treatment – and some are being tested in clinical trials. Before starting treatment, patients may want to think about taking part in a clinical trial.
A treatment clinical trial is a research study meant to help improve current treatments or obtain information on new treatments for patients with cancer. When clinical trials show that a new treatment is better than the standard treatment, the new treatment may become the standard treatment.
Choosing the most appropriate cancer treatment is a decision that ideally involves the patient, family and health care team.
Five Types of Standard Treatment
There are five types of standard treatment for kidney cancer.
Surgery to treat kidney cancer is called nephrectomy. Depending on the tumor size, location and stage, the surgeon may choose to remove the entire kidney (radical nephrectomy) or just the portion affected by cancer (partial nephrectomy). If surgery is your best treatment option, your doctor will discuss the procedure with you.
Radiation treatment uses high-energy X-rays or other types of radiation to kill cancer cells or keep them from growing. Radiation has a limited role in treating kidney cancer. Kidney tumors often do not shrink or stop growing when treated with radiation, so this as a frontline treatment is usually not advised. In some cases, radiation may be used to ease pain and other symptoms of advanced kidney cancer that has spread to bone or other areas of the body.
Chemotherapy is usually not an effective treatment for kidney tumors, but is sometimes used to treat tumors that have spread to the lung, bones, brain or lymph nodes. In these cases, chemotherapy would be combined with surgery or another type of treatment.
Immunotherapy (sometimes called biologic therapy) is a treatment that uses the patient’s immune system to fight cancer. Substances made by the body or made in a laboratory are used to boost, direct or restore the body’s natural defenses against cancer.
Targeted therapy uses drugs or other substances that can find and attack specific cancer cells without harming normal cells. Antiangiogenic agents are a type of targeted therapy now frequently used to treat advanced kidney cancer. They keep blood vessels from forming in a tumor, causing it to starve and stop growing or to shrink.
New types of treatment are tested in clinical trials. Ask your doctor if a clinical trial is available for your type of cancer.<|endoftext|>
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1. ## Probability Problem
In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
2. Originally Posted by mathhomework
In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
Form a list of 7 empty spaces:
_ _ _ _ _ _ _
If it rains on Sunday, Monday, Tuesday, Wednesday, and Friday you write:
X X X X X _ _
There are $\displaystyle {7\choose 5} = {7\choose 2} = \frac{7\cdot 6}{2} = 21$ different ways to put five X's.
The probability that it does not rain on the last two days (I assume that is what you mean by 'weekend') is getting the first five X's and the last two empty. There is only one way to do that. Therefore, the probability that it would not rain is $\displaystyle \tfrac{1}{21}$. Thus, the probability that it would rain is $\displaystyle \tfrac{20}{21}$.
3. Originally Posted by mathhomework
In a mountain village it rains on average 5 days a week. Find the probability that over a weekend it rains on exactly one of the two days.
My take on this is the following:
Let X be the random variable number of days on the weekend that it rains.
X ~ Binomial(n = 2, p = 5/7).
Calculate Pr(X = 1).
4. Originally Posted by mr fantastic
My take on this is the following:
Let X be the random variable number of days on the weekend that it rains.
X ~ Binomial(n = 2, p = 5/7).
Calculate Pr(X = 1).
5. Originally Posted by ThePerfectHacker
Form a list of 7 empty spaces:
_ _ _ _ _ _ _
If it rains on Sunday, Monday, Tuesday, Wednesday, and Friday you write:
X X X X X _ _
There are $\displaystyle {7\choose 5} = {7\choose 2} = \frac{7\cdot 6}{2} = 21$ different ways to put five X's.
The probability that it does not rain on the last two days (I assume that is what you mean by 'weekend') is getting the first five X's and the last two empty. There is only one way to do that. Therefore, the probability that it would not rain is $\displaystyle \tfrac{1}{21}$. Thus, the probability that it would rain is $\displaystyle \tfrac{20}{21}$.
You are correct. However, that doesn't answer the question. The question is what is the probability that it will rain exactly ONE day on the weekend. You have determined the probability that it will rain on the weekend period. There are of course, three ways for that to happen: it can rain on Saturday, on Sunday, or on both. The number of total possible outcomes is simply 7!/5!/2! = 21 Fixing a rain day on Saturday and dry day on Sunday, we have 5!/4! = 5 outcomes. Doing the same for a rain day on Sunday and dry day on Sat. we have 5!/4!=5 outcomes. The probability is 10/21
6. Originally Posted by Bilbo Baggins
You are correct. However, that doesn't answer the question. The question is what is the probability that it will rain exactly ONE day on the weekend. You have determined the probability that it will rain on the weekend period. There are of course, three ways for that to happen: it can rain on Saturday, on Sunday, or on both. The number of total possible outcomes is simply 7!/5!/2! = 21 Fixing a rain day on Saturday and dry day on Sunday, we have 5!/4! = 5 outcomes. Doing the same for a rain day on Sunday and dry day on Sat. we have 5!/4!=5 outcomes. The probability is 10/21
I would dispute this answer. The probability of rain on any given day is 5/7. It's not a matter of listing outcomes .....
7. I think Bilbo is right. If it rains on saturday you have 4 pegs, let's say, to put on 5 different days, mon, tues, weds, thurs, friday.
there are 5 ways to do it as you can only miss one day out each time, and there are 5 days.
Then you double this as you could also put the first peg on sunday.
so you have the total number of ways you can get the outcome, divide it by the total number of outcomes and you have 10/21
8. Originally Posted by Bruce
I think Bilbo is right. If it rains on saturday you have 4 pegs, let's say, to put on 5 different days, mon, tues, weds, thurs, friday.
there are 5 ways to do it as you can only miss one day out each time, and there are 5 days.
Then you double this as you could also put the first peg on sunday.
so you have the total number of ways you can get the outcome, divide it by the total number of outcomes and you have 10/21
5 days out of 7 is only an average, it does not mean that it does rain 5 days a week. It only means that the probability of rain on a given day is 5/7. There's a big difference.
9. Originally Posted by mr fantastic
5 days out of 7 is only an average, it does not mean that it does rain 5 days a week. It only means that the probability of rain on a given day is 5/7. There's a big difference.
True. However, I think that you are looking into the problem too deeply. A problem of this type usually doesn't require that type of scrutiny. I think that saying "rains an average of 5 times a week" is just code for "rains 5 times a week." Perhaps, because this is a statistics problem they didn't want to sound too unequivocal and added "on average"
10. ## Thank You
Thanks everybody for helping me!
11. Originally Posted by Bilbo Baggins
True. However, I think that you are looking into the problem too deeply. A problem of this type usually doesn't require that type of scrutiny. I think that saying "rains an average of 5 times a week" is just code for "rains 5 times a week." Perhaps, because this is a statistics problem they didn't want to sound too unequivocal and added "on average"
Then I suppose we will agree to disagree until the answer is given by the OP.
12. My friend at school gave his take on the question:
there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49
the same for rain on sunday and none on saturday, so you end up with 20/49
13. Originally Posted by Bruce
My friend at school gave his take on the question:
there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49 the same for rain on sunday and none on saturday, so you end up with 20/49. This seems a plausible answer.
I absolutely agree with that interpretation of this problem.
And I think that $\displaystyle \frac{10}{49}$ is correct.
14. Originally Posted by Bruce
My friend at school gave his take on the question:
there is a 5/7 chance for it to rain on saturday. For the outcome to be correct, then there needs to be no rain on sunday. So you need 5/7 x 2/7 = 10/49
the same for rain on sunday and none on saturday, so you end up with 20/49
Which exactly follows from what I posted way back in post #3 since $\displaystyle \Pr(X = 1) = ^2C_1 \left(\frac{5}{7}\right)^1 \left(\frac{2}{7}\right)^1 = \frac{20}{49}$.<|endoftext|>
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# Fraction math solver
Here, we will be discussing about Fraction math solver. We can solve math word problems.
## The Best Fraction math solver
Looking for Fraction math solver? Look no further! How to solve factorials? There are a couple different ways to do this. The most common way is to use the factorial symbol. This symbol looks like an exclamation point. To use it, you write the number that you want to find the factorial of and then put the symbol after it. For example, if you wanted to find the factorial of five, you would write 5!. The other way to solve for factorials is to use multiplication. To do this, you would take the number that you want to find the factorial of and multiply it by every number below it until you reach one. Using the same example from before, if you wanted to find the factorial of five using multiplication, you would take 5 and multiply it by 4, 3, 2, 1. This would give you the answer of 120. So, these are two different ways that you can solve for factorials!
Algebra can be a helpful tool for solving real-world problems. In many cases, algebraic equations can be used to model real-world situations. Once these equations are set up, they can be solved to find a solution that meets the given constraints. This process can be particularly useful when solving word problems. By taking the time to carefully read the problem and identify the relevant information, it is often possible to set up an equation that can be solved to find the desired answer. In some cases, multiple equations may need to be written and solved simultaneously. However, with a little practice, solving word problems using algebra can be a straightforward process.
Doing math homework can be a fun and rewarding experience. It can also be frustrating and time-consuming. However, there are ways to make the process more enjoyable and efficient. First, it is important to have a good attitude. Doing math homework should be seen as an opportunity to learn, not a chore. Second, it is important to be organized. Having all of the necessary materials close at hand will make the process go more smoothly. Third, it is important to take breaks. Doing too much at once can lead to mistakes and frustration. Finally, it is important to ask for help when needed. Don't be afraid to reach out to a teacher or tutor if you are having trouble with a particular concept. By following these tips, doing math homework can be an enjoyable and productive experience.
Once the equation has been factored, you can solve each factor by setting it equal to zero and using the quadratic formula. Another method for solving the square is to complete the square. This involves adding a constant to both sides of the equation so that one side is a perfect square. Once this is done, you can take the square root of both sides and solve for the variable. Finally, you can use graphing to solve the square. To do this, you will need to plot the points associated with the equation and then find the intersection of the two lines. Whichever method you choose, solving the square can be a simple process as long as you have a strong understanding of algebra.
Composite functions can be used to model real-world situations. For example, if f(t) represents the temperature in degrees Celsius at time t, and g(t) represents the number of hours since midnight, then the composite function (fog)(t), which represents the temperature at a certain hour of the day, can be used to predict how the temperature will change over the course of 24 hours. To solve a composite function, it is important to understand the individual functions that make up the composite function and how they interact with each other. Once this is understood, solving a composite function is simply a matter of plugging in the appropriate values and performing the necessary calculations.
## More than just an app
It helps so much. It has a special calculator in app that you can use to figure out fractions and percentages plus it tells you how to do the work so you can actually learn it instead of just looking up the answers and still not knowing what to do<|endoftext|>
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Generation of fault lines
How to get mountain chains for our map? Collision and torque of tectonic plates generate characteristic linear patterns. Perlin or fractal noise alone cannot generate such structures, so let's think in a procedure on top of it for obtaining some physical-like chains. My favorite is a procedure where the full tectonic dynamics are simulated (online soon) but let's work here on a simpler and cheaper alternative. I hope your algebra skills are tuned!
Let's start with defining the box where we will work and the central points of the plates. These points will divide the box in pieces, each of them representing a plate. We can use voronoi cells indeed. We also assign a velocity to each plate, that is, a vector with certain length and direction (it can be randomly generated). The idea is to somewhat mimic this.
According to relative velocities between neighboring plates we will have:
- Convergent boundary: when they collide. Mountains, some earthquakes and eventual volcanoes are expected.
- Divergent boundary: when they slide apart. Valleys or new seafloor are expected, and some volcanoes as well but earthquakes are not.
- Transform boundary: when they move tangentially. No mountains but earthquakes are expected.
What we need then is to obtain:
(A) For every of the boundaries in the box
(B) selecting one by one together with the couple of plates involved,
(C) obtain relative velocities of the plates
(D) and get the parallel and perpendicular components to the fault line.
When perpendicular components point each other (as in the example) we will deal with a convergent boundary, being the modulus of the component the intensity of the collision (to be used for defining the height of the mountains or the scale of the earthquakes). In the contrary case we will deal with a divergent boundary instead, and modulus will be used for the deep of the valley/sea. However, if the dominant component is the parallel one, then we deal with a transform boundary. You can see that in fact every boundary will have a component of transform and another of divergent/convergent.
For coherence in directions and signs, it is important to keep a global criteria on the perpendicular and parallel components. For instance, this is how I will define the unitary basis for every boundary:
Note that every parallel unit vector (blue) points following a hypothetical flow from left to right, and every perpendicular unit vector (red) is oriented in such a way that the vector for the third dimension would point to you following the right-hand rule. Easy, right? (Rewording it: if parallel unit vector points 3 o'clock, perpendicular unit vector points 12 o'clock.) This criteria will be quite relevant for next steps and actually simplifies the algorithm: Indeed, once I have defined these basis vectors, I just need to follow:
1) For every boundary, I got the relative velocity as a single vector defined as the difference between the velocity vector of the plate at the 'top' (the one that the red unit vector is pointing) and the velocity vector of the plate at the 'bottom' (the one that the red unit vector is not pointing).
2) I project the relative velocity into the basis of the boundary (remember, red a blue unit vectors) just by the dot product between the velocity and each of the unit vectors. Transform forces are always given by the dot product with the parallel unit vector (blue) and sign or direction (left or right) doesn't really matters. However:
- If the dot product with the perpendicular unit vector (red) is positive, we deal with a divergent boundary.
- If the dot product with the perpendicular unit vector (red) is negative, we deal with a convergent boundary.
I'm showing below the convergent boundaries in red and the divergent in blue following this method. Cool. With this information I know where to put mountains, where to put valleys, and the scale of it according to the value obtained from the dot products.
However, it would look a bit orthopedic... lets' introduce some noise to the lines: first, I keep fixed the vertices of the boundaries but introduce some distortion in the lines as show below. I do this in a discrete fashion: I fix the density number of points per boundary so they are composed by small segments.
For every of these segments, I define the unitary basis following the criteria defined before. I show them below for this example but with less density of points (for making it easy to see). Note that all red and blue arrows keep pointing coherently: blue vectors follow the boundary as a streamline and red vectors always point to the same plate along the same boundary.
Applying the same rule as before for every individual segment, we got something that looks somewhat less rigid. In the plot below, the width represents the strength of the tectonic forces and the color the divergent or convergent nature of the fault.
Combining convergent and transform faults we can create a map for earthquakes to mimic this:
Ok, with the data generated like this we can now create our mountains! Here you can use your favorite procedure. For instance, this is how it looks in a flat surface with some lorentzian-shaped mountains/valleys along the lines:
But it would look cooler if we include some background Perlin noise and erosion/deposition:
...or focusing in a region, with a bit of more detail defining the sea level, including some rivers and biomes according to the elevation:
That it. It's a simple method that can be improved in many way but can also save your day. Anyway, I'm moving land for a more sophisticated procedure, updates soon! (or late).<|endoftext|>
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Become familiar with roots for the ASVAB. A root is the opposite of a power or an exponent. There are infinite kinds of roots. You have the square root, which means “undoing” a base to the second power; the cube root, which means “undoing” a base raised to the third power; a fourth root, for numbers raised to the fourth power; and so on.
## Square roots
A math operation requiring you to find a square root is designated by the radical symbol
The number underneath the radical line is called the radicand. For example, in the operation
the number 36 is the radicand.
A square root is a number that, when multiplied by itself, produces the radicand. Take the square root of 36
If you multiply 6 by itself (6 × 6), you come up with 36, so 6 is the square root of 36.
When you multiply two negative numbers together, you get a positive number. For example, –6 × –6 also equals 36, so –6 is also the square root of 36.
When you take a square root, the results include two square roots — one positive and one negative.
Computing the square roots of negative numbers, such as
is also possible, but it involves concepts such as imaginary numbers that you won’t be asked about.
Square roots come in two flavors:
• Perfect squares: Only a few numbers, called perfect squares, have exact square roots.
• Irrational numbers: All the rest of the numbers have square roots that include decimals that go on forever and have no pattern that repeat, so they’re called irrational numbers.
### Perfect squares
Because you can’t use a calculator during the test, you’re going to have to use your mind and some guessing methods. Make an educated guess and then verify your results.
The radical symbol indicates that you’re to find the principal square root of the number under the radical. The principal square root is a positive number. But if you’re solving an equation such as x2 = 36, then you give both the positive and negative roots: 6 and –6.
To use the educated-guess method, you have to know the square roots of a few perfect squares. One good way to do so is to study the squares of the numbers 1 through 12:
• 1 and –1 are both square roots of 1.
• 2 and –2 are both square roots of 4.
• 3 and –3 are both square roots of 9.
• 4 and –4 are both square roots of 16.
• 5 and –5 are both square roots of 25.
• 6 and –6 are both square roots of 36.
• 7 and –7 are both square roots of 49.
• 8 and –8 are both square roots of 64.
• 9 and –9 are both square roots of 81.
• 10 and –10 are both square roots of 100.
• 11 and –11 are both square roots of 121.
• 12 and –12 are both square roots of 144.
### Irrational numbers
If you have to find the square root of a number that isn’t a perfect square, the ASVAB usually asks you to find the square root to the nearest tenth.
Suppose you run across this problem:
• The square root of 49 is 7, and 54 is slightly greater than 49. You also know that the square root of 64 is 8, and 54 is slightly less than 64.
• So if the number 54 is somewhere between 49 and 64, the square root of 54 is somewhere between 7 and 8.
• Because 54 is closer to 49 than to 64, the square root will be closer to 7 than to 8, so you can try 7.3 as the square root of 54:
1. Multiply 7.3 by itself.
7.3 × 7.3 = 53.29, which is very close to 54.
2. Try multiplying 7.4 by itself to see whether it’s any closer to 54.
7.4 × 7.4 = 54.76, which isn’t as close to 54 as 53.29.
3. So 7.3 is the square root of 54 to the nearest tenth.
## Cube roots
A cube root is a number that when multiplied by itself three times equals the number under the radical. For example, the cube root of 27 is 3 because 3 × 3 × 3 = 27. A cube root is expressed by the radical sign with a 3 written on the left of the radical.
You may see one or two cube-root problems on the math subtests of the ASVAB, but probably not more than that.
Unlike square roots, numbers only have one possible cube root. If the radicand is positive, the cube root will be a positive number.
Also, unlike square roots, finding the cube root of a negative number without involving advanced mathematics is possible. If the radicand is negative, the cube root will also be negative. For example,
Just like square roots, you should memorize a few common cube roots:
• 1 is the cube root of 1, and –1 is the cube root of –1.
• 2 is the cube root of 8, and –2 is the cube root of –8.
• 3 is the cube root of 27, and –3 is the cube root of –27.
• 4 is the cube root of 64, and –4 is the cube root of –64.
• 5 is the cube root of 125, and –5 is the cube root of –125.
• 6 is the cube root of 216, and –6 is the cube root of –216.
• 7 is the cube root of 343, and –7 is the cube root of –343.
• 8 is the cube root of 512, and –8 is the cube root of –512.
• 9 is the cube root of 729, and –9 is the cube root of –729.
• 10 is the cube root of 1,000, and –10 is the cube root of –1,000.<|endoftext|>
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