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- Using their right hand in preference to, or more skillfully than, their left.
- (not comparable) Intended to be worn on, or used by, the right hand.
- (not comparable) Turning or spiralling from left to right; clockwise.
- (physics) Of a particle for which the direction of its spin is the same as the direction of its motion.
- Of a coordinate system: following the right-hand rule.
right-handed pl (plural only)
- Right-handed people, taken as a whole.
- 1992, Antonio E. Puente & Robert J. McCaffrey, Handbook of Neuropsychological Assessment, →ISBN, page 147:
- Evidence contradicting this principle came from a study of epileptic patients by Penfield and Roberts (1959) who found that dysphasia following surgery on the right hemisphere was not significantly more frequent in the left-handed than in the right-handed.
of one who uses their right hand in preference to, or more skillfully than their left.<|endoftext|>
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# The sides of an isosceles triangle are 5, 5, and 7. How do you find the measure of the vertex angle to the nearest degree?
Dec 7, 2016
89° to the nearest degree.
#### Explanation:
The base of the triangle 7 can be divided in half by a line of symmetry of the isosceles triangle, which will bisect the vertex angle. This creates two right triangles:
Each with a base of 3.5 and a hypotenuse of 5.
The side opposite the half of the vertex angle is 3.5, the hypotenuse is 5.
The sine function can be used to find the angle.
$\sin \theta = \frac{o p p}{h y p}$
$\sin \theta = \frac{3.5}{5} = 0.7$
Use the inverse sin function or a table of trig functions to find the corresponding angle . (Arcsin)
arcsin 0.7 = 44.4°
Remember that this is the value of half of the vertex angle so double the value to find the vertex angle.
2 xx 44.4 = 88.8 °
rounded off to the nearest whole degree = 89°
Dec 9, 2016
theta ~~ 89°
#### Explanation:
As all 3 sides of the triangle are known, the cosine rule can be used to find the vertex angle directly.
$\cos \theta = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$
$\cos \theta = \frac{{5}^{2} + {5}^{2} - {7}^{2}}{2 \times 5 \times 5}$
$\cos \theta = \frac{1}{50} = 0.02$
Using a calculator or tables you can find the angle:
theta = 88.85°
theta ~~ 89°<|endoftext|>
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4.5: Dividing Integers
Difficulty Level: At Grade Created by: CK-12
Introduction
The Scuba Descent
Cameron and his new diving partner Gina are going to be buddies on a 40 foot dive. Gina is a new diver and is still learning to make a descent. Cameron can make a free descent quite easily. This means that he doesn’t hold onto anything as he descends to the appropriate depth. Gina will hold onto the anchor line as she descends. Then they will meet on the bottom.
Cameron has decided to go down with Gina. He will descend freely next to her, while she descends holding onto the rope. He looks at his watch and sets the timer before they descend.
When they reach the bottom, Cameron looks at his watch. He sees that the descent took them 2 minutes. Not bad at all considering that Gina is a beginner. Cameron and Gina meet up with the group and check in with the Dive Master. Then they are off for a beautiful dive!!
How far did Cameron and Gina descend per minute?
To answer this question, you will need to understand how to divide integers. Pay attention and you will be able to answer these questions at the end of the lesson.
What You Will Learn
By the end of this lesson, you will be able to demonstrate the following:
• Analyze patterns of quotients of integers with same and different signs, recognizing division by zero as undefined.
• Divide integers.
• Evaluate variable expressions involving integer division.
• Model and solve real-world problems using simple equations involving integer division.
Teaching Time
I. Analyze Patterns of Quotients of Integers with Same and Different Signs, Recognizing Division by Zero as Undefined
Another important step in learning how to compute with integers is learning how to divide them. You can look for patterns in a sequence of quotients just as you looked for patterns in a sequence of products in an earlier lesson. These patterns will help you to understand the rules for dividing integers.
Let’s look at some integer patterns with division. We are looking at quotients. A quotient is the answer in a division problem.
Example
Use a pattern to find the missing quotients below.
6÷24÷22÷20÷22÷24÷2=3=2=1=0=?=?\begin{align*}6 \div 2 & = 3\\ 4 \div 2 & = 2\\ 2 \div 2 & = 1\\ 0 \div 2 & = 0\\ -2 \div 2 & = ?\\ -4 \div 2 & = ?\\\end{align*}
Look for a pattern among the quotients. Remember that a pattern has a rule that makes it repeat in a certain way. Look at the pattern below.
This pattern uses shapes and not numbers, but there is still a rule that applies, that makes the pattern repeat itself in the way that it does.
Now look at the number pattern.
You will see that you can subtract 1 from the previous quotient to find the next quotient. Remember, subtracting 1 is the same thing as adding its opposite, -1. Try adding -1 to the previous quotients to find the next quotients.
To find the quotient of 2÷2\begin{align*}-2 \div 2\end{align*}, add 0+(1)\begin{align*}0+(-1)\end{align*}
|0|=0\begin{align*}|0|=0\end{align*} and |1|=1\begin{align*}|-1|=1\end{align*}, so subtract the lesser absolute value from the greater absolute value.
10=1\begin{align*}1-0=1\end{align*}
The integer with the greater absolute value is -1, so give the answer a negative sign.
0+(1)=1\begin{align*}0+(-1)=-1\end{align*}, so 2÷2=1\begin{align*}-2 \div 2=-1\end{align*}
To find the quotient of 4÷2\begin{align*}-4 \div 2\end{align*}, add 1+(1)\begin{align*}-1+(-1)\end{align*}
Both integers have the same sign, so add their absolute values.
|1|=1\begin{align*}|-1|=1\end{align*}, so add
1+1=2\begin{align*}1+1=2\end{align*}
Give that answer a negative sign.
1+(1)=2\begin{align*}-1+(-1)=-2\end{align*}, so 4÷2=2\begin{align*}-4 \div 2 =-2\end{align*}.
This shows the completed division facts.
6÷24÷22÷20÷22÷24÷2=3=2=1=0=1=2\begin{align*}6 \div 2 & = 3\\ 4 \div 2 & = 2\\ 2 \div 2 & = 1\\ 0 \div 2 & = 0\\ -2 \div 2 & = -1\\ -4 \div 2 & = -2\\\end{align*}
Each quotient is still 1 less than the previous quotient.
What conclusions can we draw from this pattern?
You may notice the following.
• When a positive integer is divided by a positive integer, 2, the quotient is positive.
• When zero is divided by a positive integer, 2, the quotient is zero.
• When a negative integer is divided by a positive integer, 2, the quotient is negative.
These are the beginnings of our rules for dividing integers.
Let’s look at another pattern to complete these rules.
Example
Look at the number facts below. Analyze the pattern of quotients shown.
9÷(3)=36÷(2)=33÷(1)=30÷0=undefined3÷(1)=36÷(2)=39÷(3)=3\begin{align*}&9 \div (-3) = -3\\ &6 \div (-2) = -3\\ &3 \div (-1) = -3\\ &0 \div 0 = undefined\\ &-3 \div (-1) = 3\\ &-6 \div (-2) = 3\\ &-9 \div (-3) = 3\end{align*}
What do you notice about these facts?
You may notice the following rules.
• When a positive integer is divided by a negative integer, the quotient is negative.
• When zero is divided by zero, the quotient is undefined, not zero. (Note: Any number divided by zero is considered undefined.)
• When a negative integer is divided by a negative integer, the quotient is positive.
Based on the patterns, here are the rules for dividing integers.
Take a few minutes to write these rules into your notebook. Notice that the rules for dividing integers are the same as the rules for multiplying integers.
II. Divide Integers
Now we can use these rules to divide integers. Just like with the rules for multiplying, becoming great at dividing integers will require that you memorize these rules.
Now, let’s apply these rules to dividing integers.
Example
Find the quotient (33)÷(3)\begin{align*}(-33) \div (-3)\end{align*}
To find this quotient, we need to divide two negative integers.
Divide the integers without paying attention to their signs. The quotient will be positive.
(33)÷(3)=33÷3=11\begin{align*}(-33) \div (-3) = 33 \div 3 = 11\end{align*}
The quotient is 11.
Example
Find the quotient (20)÷5\begin{align*}(-20) \div 5\end{align*}.
To find this quotient, we need to divide two integers with different signs.
Divide the integers without paying attention to their signs. Give the quotient a negative sign because the signs are different, indicating a negative answer.
20÷5=4\begin{align*}20 \div 5 =4\end{align*}, so (20)÷5=4\begin{align*}(-20) \div 5 = -4\end{align*}.
The quotient is -4.
These problems used a division sign, but remember we can also show division using a fraction bar where the numerator is divided by the denominator.
Now, it’s time for you to practice applying these rules to figuring out quotients.
4L. Lesson Exercises
1. 12÷3\begin{align*}-12 \div -3\end{align*}
2. 183\begin{align*}\frac{18}{-3}\end{align*}
3. 24÷8\begin{align*}-24 \div 8\end{align*}
Take a few minutes to check your work with a partner.
III. Evaluate Variable Expressions Involving Integer Division
You have used variable expressions with addition, subtraction and multiplication. Now we are going to apply division of integers with variable expressions. Remember that a variable expression is a math sentence that uses numbers, variables and operations.
Example
Find the value of this expression 18x÷(2)\begin{align*}-18x \div (-2)\end{align*}
It may help you to rewrite the problem like this using a fraction bar to divide. Now you can see which values can be divided
18x2\begin{align*}\frac{-18x}{-2}\end{align*}
Then separate out the integers like this.
18x2=18x2=182x\begin{align*}\frac{-18x}{-2} = \frac{-18 \cdot x}{-2} = \frac{-18}{-2} \cdot x\end{align*}
Notice that we can divide the integers. The x\begin{align*}x\end{align*} remains alone because there isn’t another x\begin{align*}x\end{align*}. We separate it out. Then we divide the integer part and add the x\begin{align*}x\end{align*} to the answer.
Since 18÷(2)=18÷2=9\begin{align*}-18 \div (-2) = 18 \div 2 = 9\end{align*} (remembering the rules to determine the sign of the answer), we know that 182x=9x=9x\begin{align*}\frac{-18}{-2} \cdot x= 9 \cdot x = 9x\end{align*}
So, the value of the expression is 9x\begin{align*}9x\end{align*}.
Let’s look at another example where there is a matching variable too.
Example
24y÷2y\begin{align*}-24y \div 2y\end{align*}
Next, we rewrite the expression using a fraction bar.
24y2y\begin{align*}\frac{-24y}{2y}\end{align*}
Now, we can separate the terms.
242yy24÷2=12\begin{align*}&\frac{-24}{2} \cdot \frac{y}{y}\\ &-24 \div 2 = -12\end{align*}
y÷y=1\begin{align*}y \div y =1\end{align*} because the y\begin{align*}y\end{align*}’s cancel each other out
12(1)=12\begin{align*}-12(1) = -12\end{align*}
The value of the expression is -12.
Example
18ab÷9b\begin{align*}-18ab \div 9b\end{align*}
First, rewrite the expression using a fraction bar.
18ab9b\begin{align*}\frac{-18ab}{-9b}\end{align*}
Next, separate out the terms.
189abb18÷9=2\begin{align*}&\frac{-18}{-9} \cdot \frac{ab}{b}\\ &-18 \div -9 = 2\end{align*}
ab÷b=a\begin{align*}ab \div b= a\end{align*} Notice that the b\begin{align*}b\end{align*}’s cancel, but the a doesn’t. It is left as part of the final expression.
Our final answer is 2a\begin{align*}2a\end{align*}.
4M. Lesson Exercises
1. 14a÷7\begin{align*}-14a \div -7\end{align*}
2. 28ab÷7b\begin{align*}28ab \div -7b\end{align*}
3. 6x÷2\begin{align*}-6x \div -2\end{align*}
Take a few minutes to check your answers with a friend.
IV. Model and Solve Real-World Problems Using Simple Equations Involving Integer Division
We can apply these rules for dividing integers to real-world problems.
That’s a great question! To solve a real-world problem, write an expression or an equation that can be used to solve the problem. Then solve.
Example
On 3 consecutive plays, a football team lost a total of 30 yards. The team lost the same number of yards on each play. Represent the number of yards lost on each play as a negative integer.
First, represent the total number of yards lost as an integer.
Since the integer shows a loss of 30 yards, use a negative integer -30.
To represent the loss for each of the 3 plays, divide the integer representing the total number of yards lost by 3.
We write this equation and then fill in the given values
Total yards lost ÷\begin{align*}\div\end{align*} number of plays = yds lost on each play
30÷3=?\begin{align*}-30 \div 3 =?\end{align*}
To find this quotient, we need to divide two integers with different signs.
Divide the integers without paying attention to their signs. Give the quotient a negative sign.
\begin{align*}30 \div 3 = 10\end{align*}, so \begin{align*}(-30) \div 3 = -10\end{align*}.
The integer -10 represents the number of yards lost on each play.
Now let’s apply what we have learned to the problem from the introduction.
Real-Life Example Completed
The Diving Descent
Here is the original problem once again. Reread it and underline any important information.
Cameron and his new diving partner Gina are going to be buddies on a 40 foot dive. Gina is a new diver and is still learning to make a descent. Cameron can make a free descent quite easily. This means that he doesn’t hold onto anything as he descends to the appropriate depth. Gina will hold onto the anchor line as she descends. Then they will meet on the bottom.
Cameron has decided to go down with Gina. He will descend freely next to Gina, while she descends holding onto the rope. He looks at his watch and sets the timer before they descend.
When they reach the bottom, Cameron looks at his watch. He sees that the descent took them 2 minutes. Not bad at all considering that Gina is a beginner. Cameron and Gina meet up with the group and check in with the Dive Master. Then they are off for a beautiful dive!!
How far did Cameron and Gina descend per minute?
First, we need to write integers to represent the depth and the time.
-40 feet is the depth
2 minutes is the time
We divide the depth by the time to find out the number of feet per minute.
\begin{align*}-40 \div 2 = -20\end{align*}
Cameron and Gina descended -20 feet per minute.
Vocabulary
Here are the vocabulary words that are found in this lesson.
Quotient
the answer in a division problem.
Undefined
when an integer is divided by 0, the answer is undefined.
Variable Expression
a math sentence with numbers, operations and variables
Time to Practice
Directions: Analyze the patterns and find the missing quotients.
1. \begin{align*}24 \div 8 = 3\!\\ 16 \div 8 = 2\!\\ 8 \div 8 = 1\!\\ 0 \div 8 = 0\!\\ -8 \div 8 = ?\!\\ -16 \div 8 = ?\end{align*}
2. \begin{align*}21 \div (-3) = -7\!\\ 14 \div (-2) = -7\!\\ 7 \div (-1) = ?\!\\ 0 \div 0 = ?\!\\ -7 \div (-1) = ?\!\\ -14 \div (-2) = 7\!\\ -21 \div (-3) = 7\end{align*}
Directions: Find each quotient.
3. \begin{align*}48 \div 8\end{align*}
4. \begin{align*}64 \div (-8)\end{align*}
5. \begin{align*}-28 \div (-4)\end{align*}
6. \begin{align*}-35 \div 7\end{align*}
7. \begin{align*}-80 \div (-4)\end{align*}
8. \begin{align*}-50 \div 10\end{align*}
9. \begin{align*}-18 \div -2\end{align*}
10. \begin{align*}42 \div -6\end{align*}
11. \begin{align*}-72 \div 9\end{align*}
12. \begin{align*}-48 \div -12\end{align*}
13. \begin{align*}-16 \div 4\end{align*}
14. \begin{align*}-22 \div -11\end{align*}
15. \begin{align*}72 \div -12\end{align*}
Directions: Find each quotient with variable expressions.
16. \begin{align*}36t \div (-9)\end{align*}
17. \begin{align*}-56n \div (-7)\end{align*}
18. \begin{align*}-22n \div -11n\end{align*}
19. \begin{align*}-28n \div 7\end{align*}
20. \begin{align*}18xy \div 2x\end{align*}
Directions: Solve the following real-world example.
21. Company X lost a total of \$1,200 during its first 4 months in business. Suppose the company lost the same number of dollars each month. Represent the number of dollars lost each month as a negative integer.
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CK.MAT.ENG.SE.1.Math-Grade-7.4.5<|endoftext|>
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Gneiss is a medium to course grained metamorphic rock.
Shale is the typical parent rock. It is made up of clay minerals. Shale can metamorphose into slate, phyllite, schist or gneiss depending on the degree of heat and pressure. Gneiss has the greatest degree of metamorphism in this group. It is classed as a high-grade metamorphic rock and is associated with regional metamorphism do to major mountain building.
Igneous rocks especially granite, can also make up the parent rock for gneiss.
This is a foliated dense rock that has light and dark colored banding. It is typically composed of feldspars, quartz, micas, and amphiboles. Because of the high heat and pressure, the minerals contained in gneiss have been mostly recrystalized.
Gneiss is a very common metamorphic rock.Back to Examples of Metamorphic Rocks<|endoftext|>
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# Introduction to Estimation
Estimation is a very important skill. It isn't just important in mathematics. It can come in handy in our day to day lives
Have you ever entered a jelly bean guessing competition? If you haven't, the idea is to guess the number of jelly beans in a jar. Each person writes down their guess, and the person with the guess that's closest to the number of jelly beans in the jar wins the jar of jelly beans. If you could get very good at estimating things, you would have a real advantage in a jelly bean guessing competition!
• if a bill is correct.
• how far something is away, how big something is, how heavy something is or the angle between things.
• how much your grocery bill is likely to be.
If you learn to estimate, you might be able to get a good idea of how many flowers will fit in a flower bed, how many people will fit in a room, or how many people are at a cricket match.
When we do mathematics, we often worry about getting the answer exactly right. However, when we want to check whether an answer is possible, or we want to work out how far something is away, it doesn't really matter if the estimate we make isn't quite right. It really doesn't make much difference if something is 13.121 kilometres away or 13 kilometres away. An approximate answer is good enough.
Estimation gives you a quick answer that is near enough to the correct answer. It does not give the exact answer that you can sometimes get by doing calculations.
## Estimation and Bills
Sam, Al and I went to a restaurant the other day. We ordered three dishes that were $\21.99$ each and three drinks that were $2.99$ each. When we received our bill, it came to $\93.47$.
This seemed a bit high to me, so I used estimation to check. The dishes were about $\22$ each, so three dishes should have been about $\66$, and the drinks were about $\3$ each, so they should have come to about $\9$. Our total bill should have been about $\75$ dollars. It was quite a bit more than that, so I asked to have it checked. When it was rechecked, the bill came to $\74.94$, almost $\ 20$ less than the original bill. This was pretty close to my estimate. Being able to estimate saved me quite a bit of money!
## Saving Time Using Estimation
Gus the snail is working on his garden again. He has decided to plant a new cabbage garden that measures 63 centimetres by 49 centimetres. He is at the garden centre, hoping to buy some new punnets of cabbage plants. Each punnet contains $10$ plants, and cabbages have to be planted $12$ centimetres away from other cabbages. Gus needs to work out how many punnets of cabbages to buy.
We can use estimates to help Gus out. Each row is about $60$ centimetres long, so Gus will be able to plant about $5$ cabbages in each row ($12 \times 5 = 60$). The garden is about $48 = 4 \times 12$ centimetres wide, so Gus can plant $4$ rows of cabbages. All together, Gus can plant about $5 \times 4 = 20$ cabbages, so he needs to buy $2$ punnets of cabbages.
You might be thinking that you have a calculator or computer, so you don't need to worry about learning to estimate. But the fact is, you still do! Sometimes you might press the wrong buttons on your calculator, or the calculator or computer might simply make a mistake because of a programming error. Learning to estimate will help you to catch those errors quickly. You'll know what sort of answer to expect, and will be able to catch it quickly when something is wrong.
For example, today I used the calculator on my phone to add up the numbers $1,2,3,4,5,6$ and $7$. It gave me the answer of $88$. I knew, immediately, (because I was able to estimate) that this answer was wrong. The correct answer is actually $28$, which is quite a bit less than $88$. It's always a good idea to use your brain to double check everything.
## Estimating Things Other Than the Results of Calculations
We can use estimation for things other than the results of calculations. For example, the picture shows someone holding some grapes in their hand to estimate their weight. You might use your thumb, your hand or paces to estimate the length of something. Scouts learn to use sticks to estimate how far something is away from them. There are lots of different estimation techniques that can help you out in different situations.
## How do I get good at estimating things?
I'm glad you asked! Like any skill, the only way to get good at estimation is to do lots of practice:
• Every time you do a calculation, check whether your answer is possible by doing estimations.
• When you go to the supermarket, try to estimate the bill by adding up estimates for each of the prices. Compare your answer to the total bill to see how good your estimate was.
• Try and estimate the time it might take you to do things, like read a chapter of a book, get home from school or do your maths homework.
### Description
This mini book covers the core of Math for Foundation, Grade 1 and Grade 2 mathematics including
1. Numbers
3. Subtraction
4. Division
5. Algebra
6. Geometry
7. Data
8. Estimation
9. Probability/Chance
10. Measurement
11. Time
12. Money
13. and much more
This material is provided free of cost for Parent looking for some tricks for their Prekinder, Kinder, Prep, Year 1 and Year 2 children
### Learning Objectives
These lessons are for kids aged 4-8 with the core objective to expose their brains to concepts of addition, subtraction, division, algebra and much more.
Author: Subject Coach
You must be logged in as Student to ask a Question.<|endoftext|>
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The Pacific War Online Encyclopedia
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Naval History and Heritage Command #NH 44324
Military decorations are awards made to soldiers, sailors, and airmen for heroism or for outstanding performance of their duties. They are an ancient custom that was perpetuated by medieval orders of chivalry and became institutionalized in modern armies. By the time of the Pacific War, a decoration typically took the form of a medal (a pendant draped from a ribbon) worn with the dress uniform, but could it take other forms. When in service or combat dress, a small ribbon was usually worn in place of the medal, usually over the left breast pocket.
Decorations had a formalized order of precedence,
reflecting the criteria for each decoration. At the top of the
hierarchy were awards reserved for great acts of heroism beyond the
call of duty, such as the American Medal of Honor or the Commonwealth Victoria Cross.
Lesser decorations were awarded for exceptional military skills such as
marksmanship, for length of service, or for participation in a
particular military campaign. In some cases, a decoration came in
several classes for various levels of merit, such as the seven classes
of the Japanese Order of the Golden Kite. Decorations were considered important for building morale,
and each military service had to strike a balance between awarding
enough decorations to create a sense of pride while not awarding so
many decorations that their value was diluted.
Lesser decorations could be awarded by commanding officers
according to straightforward criteria, but the higher awards typically
required nomination by a high-ranking officer and the concurrence of a
review board. Sledge (1981) voiced a common complaint that enlisted men
often did not receive a well-deserved decoration simply because no
officer of sufficient rank witnessed the act of heroism.
National Archives #80-G-K-12122
The highest American military decoration was the Medal of Honor, awarded for "gallantry and intrepidity at the risk of his life above and beyond the call of duty." Over half of the awards were posthumous. Lesser acts of heroism could result in an award of the Navy Cross (Navy and Marines) or the Distinguished Service Cross (Army). The Silver Star was awarded for "gallantry and intrepidity in action, not sufficient to justify the award of the Medal of Honor or Navy [or Distinguished Service] Cross".
The Distinguished Service Medal was awarded for outstanding performance, not necessarily in combat. It was often awarded to flag and general officers who directed important operations.
The Bronze Star could be awarded for any meritorious service in
combat. It could be spot awarded by senior officers to enlisted men,
and Perret (1991) claims it was awarded so frequently that its value
was significantly diluted.
The Purple Heart was awarded to soldiers, sailors, or airmen who were killed or wounded as a result of enemy action. The wound had to be serious enough to require medical treatment, but in practice this was not a tight restriction. During the torpedoing of Saratoga, Frank Fletcher "banged his head against something" (Lundstrom 2006) and suffered a bloody cut that forced him to call a pharmacist's mate to bandage his forehead. He was duly entered into the casualty list and, to his considerable embarrassment, he was later awarded the Purple Heart.
There were numerous lesser awards for meritorious or gallant service, as well as a bewildering variety of campaign, service, and skills decorations. An exhaustive list of these lesser awards is beyond the scope of this Encyclopedia.
The highest Commonwealth decoration was the Victoria Cross, which, like the Medal of Honor, was awarded only for outstanding heroism and very often posthumously. Lesser awards included the Distinguished Service Order and the Military Medal.
A soldier might also be Mentioned in Despatches, meaning that he was
commended for gallantry in a formal report by a senior commander published in the London Gazette, the Crown's official journal of record.
As with the Americans, the Commonwealth nations awarded a bewildering variety of lesser awards for meritorious or gallant service and a variety of campaign, service, and skills decorations.
Though it was not a military decoration in the strict sense, a
senior officer might be knighted for outstanding service. Few of the
senior officers of the Commonwealth nations escaped this honor.
The Bukoshu has sometimes been described as Japan's Medal of Honor. It came in two classes and was awarded only for heroism in combat. It could apparently be spot awarded by a division commander (normally a lieutenant general.)
The chief Japanese military decoration was the Order of the Golden Kite, which had seven classes. Enlisted men could be awarded only the sixth and seventh classes. Army commanders could spot award the fifth, sixth, or seventh classes, although this had to be subsequently ratified by the War Ministry. A total of about 630,000 awards of this decoration were made during the Pacific War.
The Japanese also awarded a Military Wound Badge that corresponded to the American Purple Heart, and the usual variety of campaign, service, and skills medals.
A peculiarity of the Japanese regulations for decorations was that
decorations were to be returned to the War Ministry upon the death of
References"Handbook of Japanese Military Forces" (1944-9-15)
The Pacific War Online Encyclopedia ©2011, 2014 by Kent G. Budge. Index
Comment on this article<|endoftext|>
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It’s a big, bad, lava world – and according to what we know about astronomy, it simply shouldn’t exist.
Kepler-78b circles its star every 8 and 1/2 hours, featuring one of the tightest known orbtits. According to currently accepted theories on planetary formation, it couldn’t have formed so close to its star, nor could it have migrated there.
“This planet is a complete mystery,” says astronomer David Latham of the Harvard-Smithsonian Center for Astrophysics (CfA). “We don’t know how it formed or how it got to where it is today. What we do know is that it’s not going to last forever. “Kepler-78b is going to end up in the star very soon, astronomically speaking,” agrees CfA astronomer Dimitar Sasselov.
To make this planet even more interesting, it is the first known Earth-sized planet with an Earth-like density. Kepler-78b has a radius approximately 20 percent larger than Earth, and weighs almost twice as much – which means that the density is pretty much the same as Earth’s. But there is no apparent connection between this and the planet’s formation.
“It couldn’t have formed in place because you can’t form a planet inside a star. It couldn’t have formed further out and migrated inward, because it would have migrated all the way into the star. This planet is an enigma,” explains Sasselov.
According to Latham, Kepler-78b is part of a new class of stars, which researchers use to characterize worlds who orbit their stars in less than 12 hours. Kepler-78b is the first of these planets to have its mass measured.
“Kepler-78b is the poster child for this new class of planets,” notes Latham.
But, for all its fame and resemblance to Earth, the planet is doomed. Gravitational tides will draw it even closer to its star and eventually, it will move so close to the star that the gravity will tear it apart. But astronomers estimate that Kepler-78b still has some 3 billion years before this happens.<|endoftext|>
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# ACT Math : How to find the length of the side of a quadrilateral
## Example Questions
### Example Question #1 : How To Find The Length Of The Side Of A Quadrilateral
Find the length of a rectangle if the area is , and the width is .
Explanation:
Write the formula to solve the area of a rectangle.
Substitute the dimension and area to solve for the length.
### Example Question #1 : How To Find The Length Of The Side Of A Quadrilateral
If the perimeter of a square is and the area is , what is the side length of the square?
Explanation:
The square has 4 equal sides. Let's assume a side is .
Each side of the square has a length of .
The area of the square is:
Substitute into the area to solve for .
Since one side of the square is , substitute the value of to determine the length of the square side.
### Example Question #1 : How To Find The Length Of The Side Of A Quadrilateral
Find the side of a square if the area is .
Explanation:
The area of a square is:
Substitute the area and solve for the side.
The quantity inside the square root may not be factorized in attempt to eliminate the square root!
The side length of the square is:
### Example Question #4 : How To Find The Length Of The Side Of A Quadrilateral
Find the perimeter of the rhombus above.
Explanation:
By definition, a rhombus is a quadrilateral with four equal sides whose angles do not all equal 90 degrees. To find the perimeter, we must find the values of x and y. In order to do so, we must set up a system of equations where we set two sides equal to each other. Any two sides can be used to create these systems.
Here is one example:
Eq. 1
Eq. 2
Now we plug into the first equation to find the value of :
Plugging these values into any of the three equations will give us the length of one side equaling 11.
Since there are four sides, .<|endoftext|>
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# Engineering Graphs
Dive into the world of Engineering Graphs. This comprehensive guide will provide an in-depth understanding of various graph types including Cartesian, polar, and logarithmic graphs prevalent in engineering mathematics. Learn the nuances of graph interpretation and discover the practical applications of these graphs in day-to-day life as well as theoretical research. The guide finally unveils the key tools and techniques that aid in drawing these critical graphs. Harness the power of graph theory concepts for engineering purposes and learn to avoid common pitfalls.
#### Create learning materials about Engineering Graphs with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
## Understanding Engineering Graphs
Engineering Graphs are fundamental to interpreting and analysing data in the field of engineering. These graphs provide a visual representation of data and can assist in identifying underlying patterns, trends, and relationships. Understanding this form of data representation can significantly aid in designing effective solutions to engineering problems.
### The Basics of Engineering Graph Types
Engineering mainly employs three types of graphs. To comprehend these better, you'll find each prominently discussed below.
A Cartesian plot is a type of graph that depicts relationships between two variables where values for each are plotted along X and Y axes.
#### Cartesian Plots in Engineering Mathematics
Cartesian plots, sometimes referred to as Cartesian coordinates or a Cartesian grid, are incredibly valuable graph types in Engineering Mathematics. On these graphs, you plot data points according to their x- and y- coordinates. The coordinates represent the intersection of the lines drawn from the point vertically (Y-axis) and horizontally (X-axis) to the axes. One example of a Cartesian plot is a straight line graph represented by the formula $y = mx + c$, where $$m$$ is the gradient of the line and $$c$$ is the y-intercept. You can characterise these plots into:
• 1D (One Dimensional)
• 2D (Two Dimensional)
• 3D (Three Dimensional)
#### Polar Graphs in Engineering Mathematics
A Polar Graph provides an alternative means to designate a data point's location using the property of direction and distance from a fixed point.
In polar graphs, a point's location is determined by its distance from the origin (the pole or zero) and its angle relative to a reference direction (the polar axis) in a two-dimensional space. It is represented by the formula $r = \sqrt{x^2 + y^2}$. These graphs are firmly rooted within areas of engineering such as signal processing, where they serve a central role in phase and magnitude plotting.
#### Logarithmic Graphs in Engineering Mathematics
A logarithmic graph, or log plot, is a specialised graph that can assist in visualising and interpreting data that spans several orders of magnitude. Logarithmic plots have a logarithmic scale (either log10, natural log, etc.) on either one or both axes. It enables you to capture both big and small movements in the data more clearly. The most common application you'll find is for Bode plots in control systems, earthquake magnitude interpretation, as well as RF (Radio Frequency) applications.
### Interpreting Engineering Graphs
Engineering graphs provide you with a powerful tool for data interpretation. However, it's important not only to read them correctly but to also avoid common errors that could potentially distort the data analysis.
#### Reading and Analysing Engineering Graphs
A systematised approach to reading any kind of graph includes recognising the type of graph used, understanding the scales used on the axes, identifying key data points, and interpreting the overall trend. For instance, when analysing a logarithmic plot, if the x-axis (horizontal) operates on a logarithmic scale, each increment can represent a tenfold increase. On the other hand, if the y-axis (vertical) is logarithmic, it requires a relative comparison rather than an absolute one.
#### Misconceptions and Common Errors in Interpreting Engineering Graphs
A deep understanding of engineering graphs is not complete without the knowledge and awareness of potential pitfalls and errors in interpretation. Some common mistakes include:
• Misreading the axes, for example, confusing linear with log scales.
• Failing to note the units of measurement used in the graph.
• Not considering the context of the data and the manner in which it was gathered.
• Misinterpreting trend lines or best fit lines.
Avoid these common missteps to ensure an accurate and comprehensive understanding of the information the graph presents. Always approach charts and graphical data with a critical eye and check your interpretations meticulously.
## Practical Applications of Engineering Graphs
Engineering graphs further their use considerably beyond the traditional education setting. These crucial visual tools are vital to the everyday operations within both common and complex spaces, making activities easier to understand and execute.
### Engineering Graph Applications in Everyday Life
Engineering graphs facilitate vital data interpretation and analysis in numerous everyday life areas. They bring clarity to complex data sets, making them understandable and more interpretable.
#### The Role of Engineering Graphs in Complex Calculations
Engineering Graphs profoundly influence complex computations. For instance, in physics, to calculate acceleration – a change of speed over time – a velocity-time graph proves handy. These readings may be impossible to decipher numerically, but with graphs, the computational visualization becomes relatively straightforward. In the field of civil engineering as well, forces on structures such as bridges or buildings can be visually represented through force diagrams. Stress-strain graphs are utilised in material science to determine a material's elastic limit and ultimate tensile strength. Through such engineering graphs, the calculations are represented in ways that practical impacts are far more perceivable.
#### Engineering Graphs in Project Planning and Design
In project planning and design phases, engineering graphs like Gantt charts assist engineers in tracking the project progress. These charts represent the project timeline, individual task durations, and their relations. The use of network diagrams further provides valuable insights into resource management and the possible paths to project completion. Frequency histograms and bar graphs, meanwhile, help monitor defects or failures, aid in quality control, and allow performance forecasts.
Engineering graphs even pervade the realm of urban planning. Population distribution, land use, and transportation patterns are analysed via different engineering graphs to create efficient urban designs. From roads to housing complexes, these graphs lay the foundation of systematic planning.
### Influence of Engineering Graphs in Theoretical Research
Engineering graphs are not confined to the practical aspects, they have a significant role in theoretical research too. They empower scholars to conceptualize complex phenomena, theories, and mathematical calculations effectively.
#### Implementation of Graph Theory in Engineering Solutions
One of the predominant uses of engineering graphs in theoretical research is the application of Graph Theory. Graph theory, a branch of discrete mathematics, is extensively employed in areas like network analysis, computer programming, and topological studies. Nodes and edges represented in graphs can model various engineering problems, including telecommunication networks, transportation networks, electrical circuits, and much more. Typically, nodes represent specific entities (like telecommunication towers or cities) with edges symbolising the relationships or connections between these entities. It's essential to appreciate the impact of graph theory in theoretical research as it enables the application of mathematics in creating practical engineering solutions.
#### How Engineering Graphs Simplify Complex Engineering Mathematics
For a better understanding of complex engineering mathematics, engineering graphs play a critical role. They provide an easy-to-understand medium, especially for concepts that can be hard to comprehend numerically. Take, for example, Fourier Transformations, widely used to analyse the main frequency components of a signal in Electrical Engineering. Traditionally, these are incredibly complex mathematical functions. But with the help of an engineering graph, you can visually extract and interpret this information. Laplace Transformations, complex calculations used in system modelling and differential equations, are made more navigable using engineering graphs. Output response and stability investigations of a system control theory are determined visually through Nyquist and Bode plots. The blend of visual representation, clarity, and data analysis in these graphical tools significantly simplifies the understanding of complex engineering mathematics.
## Tools and Techniques for Engineering Graphs
The effective creation and interpretation of Engineering Graphs often demand the use of various tools and technologies. From conventional methods like graph paper and compasses to sophisticated software applications, these tools are crucial in inscribing and deciphering intricate graphical data.
### Essential Tools for Drawing Engineering Graphs
Engineering graphs can be drawn and analysed using an array of tools. Each of them serves distinct functions and are critical to producing different types of graphs. A basic set of essential tools used in drawing Engineering Graphs include:
• Graph Paper: This patterned paper is gridded into small squares to ensure precision and uniformity in plotting data points.
• Compass: A classic instrument used to draw circles or arcs in polar graphs or similar circles in other application areas.
• Protractors: These semi-circular tools assist in measuring and drawing angles especially helpful in constructing polar graphs.
• Ruler: Essential for drawing straight lines, establishing scale, and measuring distances.
• Pencil and Eraser: Necessary for initial drawings and adjustments.
While these physical tools are suitable for manual graphing, digital applications like Microsoft Excel, Google Sheets, and Tableau offer advanced features to generate sophisticated graphs.
#### Examples of Popular Engineering Graph Tools
Introducing digital tools into your Engineering Graphs' development can aid in creating more precise, complex, and visually appealing graphical data. Here are a few examples of popular digital tools:
• Microsoft Excel: An excellent tool for constructing and analysing various types of graphs. Its wide array of pre-constructed graph types and customisable options make it a go-to choice for many engineers.
• Matplotlib: A popular Python Library extensively used in producing high quality 2D graphs. With its customisable feature, it becomes an excellent tool for engineering mathematics.
• Tableau: An advanced data visualisation tool that excels in creating interactive charts, making data analysis and interpretation intuitive. It's especially beneficial for larger datasets.
• AutoCAD: A software application extensively used for creating 2D and 3D Engineering Graphs in fields like architectural and mechanical engineering.
• Graphing Calculator: An electronic calculator capable of plotting graphs, solving simultaneous equations, and performing numerous other tasks.
### Plotting in Engineering Mathematics using Tools
When constructing graphs for engineering mathematics, physical tools have limited use due to the complexity and precision demanded by these graphs. Modern digital tools mentioned above offer the required advanced features to conveniently plot these graphs. These tools notably have a wide range of built-in functionalities that include, but are not limited to, customising data series, recalculating automatically when data is modified, applying a trend line to your data, among others.
#### Key Tips and Tricks for Effective Plotting in Engineering Mathematics
While these digital tools ease plotting graphs, certain tips can make the process more efficacious:
• Proper Labelling: Always label the axes accurately and provide units.
• Scale Selection: Select your scale carefully to ensure the data is neither squashed nor stretched.
• Data Segregation: When working with multiple data sets, colour code for easy distinction.
• Accuracy: Ensure precise plotting of data points and drawing of lines between them.
• Legend Usage: A valuable tool for helping readers understand the plotted data better.
Matplotlib library from Python is quite a potent tool in plotting engineering mathematics graphs. For instance, to plot a simple 2D Cartesian graph using Matplotlib, this is a basic example code:
import matplotlib.pyplot as plt
x = [1, 2, 3, 4, 5]
y = [1, 4, 9, 16, 25]
plt.plot(x, y)
plt.xlabel('X - Axis')
plt.ylabel('Y - Axis')
plt.title('Cartesian Plot Example')
plt.show()
Here, 'x' and 'y' are the respective coordinates. This code will generate a Cartesian graph with 'x' and 'y' as the axes.
#### Avoiding Common Pitfalls in Plotting Engineering Graphs
While plotting graphs, it's also important to be aware of some of the common errors that can lead to misinterpretations.
• Neglecting Negative Values: Don't forget to include negative numbers if your data range includes them.
• Non-uniform Scale: Always have a uniform scale for accurate representation of the data.
• Missing Outliers: Ensure to include all data points, particularly the outliers, for a complete picture.
• Starting From Zero: Starting the y-axis from zero can often help give an accurate representation of the data.
These potential pitfalls can distort the picture the data presents and hence, they should be avoided for accurate and meaningful interpretation.
### Using Graph Theory Tools in Engineering
Graph theory is a branch of discrete mathematics that studies graphs or networks, providing a mathematical structure that simplifies many engineering problems.
Graph Theory is a mathematical study of graphs, which are mathematical structures used to model pairwise relations between objects.
The core components of a graph are vertices (or nodes) and edges (or arcs). Graph theory tools are extensively used in various fields of engineering, including computer science, electrical engineering, and telecommunications.
#### Decoding Graph Theory Concepts for Engineering Purposes
The basic terminology of Graph Theory includes:
• Vertice or Node: These represent the entities.
• Edge or Arc: These denoting the relationship or connections between entities.
• Degree: It's the number of edges incident on a vertex.
• Path: It's a sequence of vertices such that each adjacent pair is connected by an edge.
Having a basic understanding of these fundamental concepts is crucial in utilising them effectively in engineering solutions. Consider graph theory's application in computer networks. The nodes represent the computers while the edges represent the network cables. Here, the degree represents the number of computers each computer is directly connected to.
#### Advantages of Implementing Graph Theory Tools in Engineering Practice
Integrating Graph Theory techniques to engineering problems offers several advantages:
• Modelling complex systems becomes easier.
• Optimal solutions can be obtained for different problems like Network Routing, Traveling Salesman Problem.
• System resources can be effectively utilised.
• Controls system complexity in domains like control systems, circuits etc.
Recognising and harnessing these benefits can prove valuable in making your engineering practice more efficient and effective. It allows you to grasp the complexity of engineering problems better and seek innovative and optimal solutions.
## Engineering Graphs - Key takeaways
• A Cartesian plot is graph represented by the formula y = mx + c, where m is the gradient and c is the y-intercept, this can be characterised as 1D, 2D or 3D type.
• Polar Graphs provide an alternative way to represent a data point's location using direction and distance from a fixed point. These graphs are widely used in engineering, especially in signal processing.
• Logarithmic graph or log plot assists in visualising and interpreting data that spans several orders of magnitude. These graphs are most commonly used for Bode plots in control systems and earthquake magnitude interpretation.
• Engineering Graphs are important for data interpretation, their correct reading and analysis is important which includes recognising the type of the graph used, understanding the scales used on the axes, identifying key data points, and interpreting the overall trend.
• Graph Theory, a branch of discrete mathematics, is extensively used in network analysis, computer programming, and topological studies. Nodes and edges in graphs model various engineering problems.
• Digital tools like Microsoft Excel, Matplotlib, Tableau, AutoCAD and Graphing Calculator are extensively used for construction and analysis of various types of graphs.
• Engineering Graph tools for graphing include graph paper, compasses, protractors, rulers, pencils, and erasers.
• Plotting in Engineering Mathematics using these tools includes customising data series, recalculating automatically when data is changed, application of trend line to data.
• In Graph Theory, the core components of a graph are vertices (or nodes) and edges (or arcs). Graph theory tools are extensively used in engineering for problem solving.
#### Flashcards in Engineering Graphs 27
###### Learn with 27 Engineering Graphs flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What is engineering graph paper?
Engineering graph paper is a form of graph paper which is printed with a grid of fine lines representing fixed intervals, often used by engineers for drawing diagrams and plotting mathematical functions. It's usually divided into multiple scales for precision.
When should one use a scatter plot versus a line chart in engineering?
Use a scatter plot in engineering when you want to display and assess relationships between two numerical variables, especially if the data is not continuous. A line graph, on the other hand, is ideal for showing trends over time or progression of a variable in a simple, clear manner.
Can you explain the role of Graph Theory in Engineering Mathematics?
Graph theory in engineering mathematics is used to model and analyse networks. These networks can represent anything from communication and transportation systems to hydraulic circuits or structures. Graph theory facilitates the understanding of network connectivity, pathway options, and system optimisation.
What types of graphs are commonly used in Engineering Mathematics?
In Engineering Mathematics, graphs commonly used include bar graphs, pie charts, line graphs, scatter plots, histograms, function graphs, and control charts. Computational tools are often utilised for complex graph plotting.
Which tools are commonly used to create engineering graphs?
Common tools used to create engineering graphs include mathematical software like MATLAB, computer-aided design (CAD) software such as AutoCAD, spreadsheet programs like Microsoft Excel, and specialised graphing utilities like Grapher and Minitab.
## Test your knowledge with multiple choice flashcards
What role do engineering graphs play in complex computations in areas such as physics and civil engineering?
What are some key tips and tricks for effective plotting in Engineering Mathematics using digital tools?
What is a Cartesian plot in Engineering Mathematics?
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There are two species of elephant seal. Southern elephants seals are the larger of the two and live in Antarctica; northern elephant seals breed off the California coast. These large seals are named for the inflatable trunk-like nose found on males of the species. In addition to the elephantine facial appendage, male elephant seals also weigh eight to 10 times more than females -- the greatest size difference between males and females of any mammalian species.
The largest southern elephant seals weigh nearly 9 tons and are 20 feet long. Northern elephant seals are still massive, weighing up to 5 tons and reaching 16 feet in length. Aside from their distinctive facial features, elephant seals have hind limbs that are used as tail fins when swimming. The two fins on either side of their bodies are not used for swimming, but propel their bulky bodies forward on land.
Elephant seals are solitary in the ocean, but on land they form large colonies around their breeding grounds. These heavy animals are best adapted to the sea and clumsy on land, but can move at speeds up to 5 miles per hour if threatened. In practice, elephant seals only charge if vocalizations and posturing do not scare off their rival. Because males routinely fast for months while defending their breeding grounds, they prefer not to expend energy charging after challengers. While great white sharks and killer whales prey on elephant seals in open water, they have no predators on land apart from humans.
Elephant seals only come on land during breeding and molting seasons, spending 85 to 90 percent of their lives deep underwater. Southern elephant seals routinely travel thousands of miles away from their breeding grounds. These marine mammals feed exclusively at sea, hunting squid, crustaceans, fish, smaller sharks and other oceanic life. Their longest dives last two hours, with only a couple minutes between dives spent breathing at the surface.
Once heavily hunted for its blubber, the northern elephant seal was nearly extinct in the 1800s. The southern elephant seal was hunted as well, but never as extensively as the northern given the isolation of its Antarctic breeding grounds. With national and international laws in place to protect the northern elephant seal and prohibit commercial hunting, the species population has rebounded to an estimated 175,000 adults in 2013, according to the Wildlife Conservation Society.
- Jupiterimages/Photos.com/Getty Images<|endoftext|>
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The incredible colonization of Pacific islands by the Polynesians presents a fascinating conundrum for scholars. How, exactly, did anyone manage to cross thousands of miles of open ocean to land on tiny islands? Researchers have tried to answer the question for decades, by analyzing the lore passed down through generations and, occasionally, attempting the journey themselves.
Now, two new studies published in the Proceedings of the National Academy of Sciences provide additional insight into how this ancient people managed to navigate such long distances.
The first study focused on the 2012 discovery of a nearly 20-foot-long section of a sailing canoe, carved from a single timber. To find such a large section of a canoe preserved is rare in and of itself. But what made the find more extraordinary was that it shared features with Polynesian artifacts not normally found in New Zealand, including the carved sea turtle (pictured above). The study authors dated the canoe to around the year 1400. The Los Angeles Times reports:
A few features, including four transverse ribs carved into the hull, haven’t been known historically in New Zealand, but have been featured in canoes in the Southern Cook Islands, described in 1913. The New Zealand canoe also shares some design elements with a canoe found about 30 years ago on Huahine in the Society Islands. It's thought to be from around the same time period as the New Zealand canoe, even though it was discovered roughly 2,500 miles away. The canoes “could have come from the same design tradition,” the authors wrote. Clearly, the Polynesians knew how to get around.
Finding similar cultural artifacts indicates that there was a connection between the early Polynesians and New Zealand. But how would they have made it there? The South Pacific's current wind patterns would have made sailing between Polynesia and New Zealand difficult with the canoe technology in use at the time New Zealand was colonized. In the second paper, a different group of researchers found that the Polynesian colonists actually had the weather on their side. Science:
Because of shifting climate conditions, there were several decades-long windows of opportunity in which Polynesian seafarers could have sailed with the wind at their backs to travel east and other times when winds favored travel between the Central Pacific islands and New Zealand. "Our reconstructed sailing conditions during the period of East Polynesian colonization would have enabled all of the known colonizing routes, and others,” to have been successfully navigated by canoes that couldn’t sail into the wind
So, the Polynesians came to New Zealand in canoes during periods of good climate conditions, and everything gets tied up with a neat little bow, right? Unfortunately, it’s not quite that simple. While the canoe found in the first paper was dated to 1400, the friendly weather anomaly shut down nearly 100 years earlier, around 1300. The researchers interviewed in Science suggested one possible explanation: people who settled in New Zealand just kept on building the same kinds of canoes for a while. Another possibility: the dates found by researchers in the first paper might be off by a bit.<|endoftext|>
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Just the facts. On the surface, this is all that news seems to be. What happened? Where did it happen? Why? Who was involved? Unfortunately, simple questions don’t always beget simple answers. The more important the event, the less likely that a simple answer will do. In some cases, simple answers can be downright dangerous.
This problem can be compounded by language. Words are rarely objective. The way we define a word is a reflection of our experiences. This explains why words like Liberal or Conservative, Democrat or Republican, can elicit wildly different responses based on the nature of the audience.
Sometimes, the impact of language on our perception of a news story is more obvious:
- are protesters citizens exercising their democratic rights or are they law-breakers challenging authority?
- is a devastating hit in hockey an example of hard-nosed play by the home team or a dirty hit that changed the momentum of the game?
- are stories circulating about a politician important issues that he refuses to address, or unsubstantiated rumors?
Let’s take a look at this site which compares the use of language in news sources.
Consider the following graph that compares the use of language on FOX News and CNN during the conflicts in Gaza. Though you have not watched any of the broadcasts, what can you infer from the language use?
Now, it is your turn to examine two versions of the same story.
- Find two stories about something in the news (sports? music? politics?) from two different sources. For example, choose the coverage of an NFL game in newspapers from the cities represented in the game, or the coverage of a tense international situation in newspapers of two countries involved.
- Compare the language used in the two news stories. Where is the language different? How does this language use impact meaning and understanding? If you wish, you can arrange your work in a way that mimics the chart about language in coverage of conflict in Gaza. Don’t worry about percentages; instead, use the extreme left and right of the chart to track the strongest language used by either source.
- In a few sentences, explain the value messages (intended or unintended) conveyed by the language used in the sources.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Geometry Concepts Go to the latest version.
# 2.3: Converse, Inverse, and Contrapositive
Difficulty Level: At Grade Created by: CK-12
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Practice Converse, Inverse, and Contrapositive
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What if your sister told you "if you do the dishes, then I will help you with your homework"? What's a statement that is logically equivalent to what your sister said? After completing this Concept, you will know how to answer this question as you discover converses, inverses, and contrapositives, and how changing a conditional statement affects truth value.
### Guidance
Consider the statement: If the weather is nice, then I will wash the car. This can be rewritten using letters to represent the hypothesis and conclusion:
If p,then qwherep=the weather is nice andq=I will wash the car.Or,pq.
In addition to these positives, we can also write the negations, or “not”s of p\begin{align*}p\end{align*} and q\begin{align*}q\end{align*}. The symbolic version of not p\begin{align*}p\end{align*}, is p\begin{align*}\sim p\end{align*}.
p=the weather is not niceq=I will not wash the car
Using these negations and switching the order of p\begin{align*}p\end{align*} and q\begin{align*}q\end{align*}, we can create three more conditional statements.
ConverseInverseContrapositiveqppqqpIf I wash the carq, then the weather is nicep.If the weather is not nicep, then I won't wash the carq.If I don't wash the carq, then the weather is not nicep.
If we accept “If the weather is nice, then I’ll wash the car” as true, then the converse and inverse are not necessarily true. However, if we take the original statement to be true, then the contrapositive is also true. We say that the contrapositive is logically equivalent to the original if-then statement. It is sometimes the case that a statement and its converse will both be true. These types of statements are called biconditional statements. So, pq\begin{align*}p \rightarrow q\end{align*} is true and qp\begin{align*}q \rightarrow p\end{align*} is true. It is written pq\begin{align*}p \leftrightarrow q\end{align*}, with a double arrow to indicate that it does not matter if p\begin{align*}p\end{align*} or q\begin{align*}q\end{align*} is first. It is said, “p\begin{align*}p\end{align*} if and only if q\begin{align*}q\end{align*}”. Replace the “if-then” with “if and only if” in the middle of the statement. “If and only if” can be abbreviated “iff.”
#### Example A
Use the statement: If n>2\begin{align*}n > 2\end{align*}, then n2>4\begin{align*}n^2 > 4\end{align*}.
a) Find the converse, inverse, and contrapositive.
b) Determine if the statements from part a are true or false. If they are false, find a counterexample.
The original statement is true.
Converse:Inverse:Contrapositive:If n2>4, then n>2.If n<2,then n2<4.If n2<4,then n<2.False. n could be 3, making n2=9.False. Again, if n=3, then n2=9.True, the only square number less than4 is 1, which has square roots of 1 or -1, bothless than 2.
#### Example B
Use the statement: If I am at Disneyland, then I am in California.
a) Find the converse, inverse, and contrapositive.
b) Determine if the statements from part a are true or false. If they are false, find a counterexample.
The original statement is true.
Converse:Inverse:Contrapositive:If I am in California, then I am at Disneyland.False. I could be in San Francisco.If I am not at Disneyland, then I am not in California.False. Again, I could be in San Francisco.If I am not in California, then I am not at Disneyland.True. If I am not in the state, I couldn't be at Disneyland.
Notice for the inverse and converse we can use the same counterexample. This is because the inverse and converse are also logically equivalent.
#### Example C
The following is a true statement:
mABC>90\begin{align*}m \angle ABC > 90^\circ\end{align*} if and only if ABC\begin{align*}\angle ABC\end{align*} is an obtuse angle.
Determine the two true statements within this biconditional.
Statement 1: If mABC>90\begin{align*}m \angle ABC > 90^\circ\end{align*}, then ABC\begin{align*}\angle ABC\end{align*} is an obtuse angle
Statement 2: If ABC\begin{align*}\angle ABC\end{align*} is an obtuse angle, then mABC>90\begin{align*}m \angle ABC > 90^\circ\end{align*}.
You should recognize this as the definition of an obtuse angle. All geometric definitions are biconditional statements.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
Your sister presented you with the if-then statement, "If you do the dishes, then I will help you with your homework." If we take the original statement to be true, then the contrapositive is also true. The following contrapositive statement is logically equivalent to the original if-then statement:
### Vocabulary
A conditional statement (also called an if-then statement) is a statement with a hypothesis followed by a conclusion. The hypothesis is the first, or “if,” part of a conditional statement. The conclusion is the second, or “then,” part of a conditional statement. The conclusion is the result of a hypothesis. The converse of a conditional statement is when the hypothesis and conclusion are switched. The inverse of a conditional statement is when both the hypothesis and conclusions are negated. The contrapositive of a conditional statement is when the hypothesis and conclusions have been both switched and negated. When the original statement and converse are both true then the statement is a biconditional statement.
### Guided Practice
1. Use the statement: Any two points are collinear.
a) Find the converse, inverse, and contrapositive.
b) Determine if the statements from part a are true or false. If they are false, find a counterexample.
2. p:x<10q:2x<50\begin{align*}p: x < 10 \qquad q: 2x < 50\end{align*}
a) Is pq\begin{align*}p \rightarrow q\end{align*} true? If not, find a counterexample.
b) Is qp\begin{align*}q \rightarrow p\end{align*} true? If not, find a counterexample.
c) Is pq\begin{align*}\sim p \rightarrow \sim q\end{align*} true? If not, find a counterexample.
d) Is qp\begin{align*}\sim q \rightarrow \sim p\end{align*} true? If not, find a counterexample.
1. First, change the statement into an “if-then” statement: If two points are on the same line, then they are collinear.
Converse:Inverse:Contrapositive:If two points are collinear, then they are on the same line. True.If two points are not on the same line, then they are not collinear. True.If two points are not collinear, then they do not lie on the same line. True.
2.
pq: If x<10, then 2x<50. True.qp: If 2x<50, then x<10. False, x=15 would be a counterexample.pq: If x>10, then 2x>50. False, x=15 would also work here.qp: If 2x>50, then x>10. True.
### Practice
For questions 1-4, use the statement: If AB=5\begin{align*}AB = 5\end{align*} and BC=5\begin{align*}BC = 5\end{align*}, then B\begin{align*}B\end{align*} is the midpoint of AC¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}.
1. If this is the converse, what is the original statement? Is it true?
2. If this is the original statement, what is the inverse? Is it true?
3. Find a counterexample of the statement.
4. Find the contrapositive of the original statement from #1.
5. What is the inverse of the inverse of pq\begin{align*}p \rightarrow q\end{align*}? HINT: Two wrongs make a right in math!
6. What is the one-word name for the converse of the inverse of an if-then statement?
7. What is the one-word name for the inverse of the converse of an if-then statement?
8. What is the contrapositive of the contrapositive of an if-then statement?
For questions 9-12, determine the two true conditional statements from the given biconditional statements.
1. A U.S. citizen can vote if and only if he or she is 18 or more years old.
2. A whole number is prime if and only if it has exactly two distinct factors.
3. Points are collinear if and only if there is a line that contains the points.
4. 2x=18\begin{align*}2x = 18\end{align*} if and only if x=9\begin{align*}x = 9\end{align*}.
5. p:x=4q:x2=16\begin{align*}p: x = 4 \quad q: x^2=16\end{align*}
1. Is pq\begin{align*}p \rightarrow q\end{align*} true? If not, find a counterexample.
2. Is qp\begin{align*}q \rightarrow p\end{align*} true? If not, find a counterexample.
3. Is pq\begin{align*}\sim p \rightarrow \sim q\end{align*} true? If not, find a counterexample.
4. Is qp\begin{align*}\sim q \rightarrow \sim p\end{align*} true? If not, find a counterexample.
6. p:x=2q:x+3=5\begin{align*}p:x=-2 \quad q:-x+3=5\end{align*}
1. Is pq\begin{align*}p \rightarrow q\end{align*} true? If not, find a counterexample.
2. Is qp\begin{align*}q \rightarrow p\end{align*} true? If not, find a counterexample.
3. Is pq\begin{align*}\sim p \rightarrow \sim q\end{align*} true? If not, find a counterexample.
4. Is qp\begin{align*}\sim q \rightarrow \sim p\end{align*} true? If not, find a counterexample.
7. p:\begin{align*}p:\end{align*} the measure of \begin{align*}\angle ABC=90^\circ \ q: \angle ABC\end{align*} is a right angle
1. Is \begin{align*}p \rightarrow q\end{align*} true? If not, find a counterexample.
2. Is \begin{align*}q \rightarrow p\end{align*} true? If not, find a counterexample.
3. Is \begin{align*}\sim p \rightarrow \sim q\end{align*} true? If not, find a counterexample.
4. Is \begin{align*}\sim q \rightarrow \sim p\end{align*} true? If not, find a counterexample.
8. \begin{align*}p:\end{align*} the measure of \begin{align*}\angle ABC=45^\circ \ q: \angle ABC\end{align*} is an acute angle
1. Is \begin{align*}p \rightarrow q\end{align*} true? If not, find a counterexample.
2. Is \begin{align*}q \rightarrow p\end{align*} true? If not, find a counterexample.
3. Is \begin{align*}\sim p \rightarrow \sim q\end{align*} true? If not, find a counterexample.
4. Is \begin{align*}\sim q \rightarrow \sim p\end{align*} true? If not, find a counterexample.
9. Write a conditional statement. Write the converse, inverse and contrapositive of your statement. Are they true or false? If they are false, write a counterexample.
10. Write a true biconditional statement. Separate it into the two true conditional statements.
### Vocabulary Language: English
biconditional statement
biconditional statement
A statement is biconditional if the original conditional statement and the converse statement are both true.
Conditional Statement
Conditional Statement
A conditional statement (or 'if-then' statement) is a statement with a hypothesis followed by a conclusion.
Logically Equivalent
Logically Equivalent
A statement is logically equivalent if the "if-then" statement and the contrapositive statement are both true.
premise
premise
A premise is a starting statement that you use to make logical conclusions.
Jul 17, 2012
Aug 24, 2015
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# Scatter Plots and Linear Correlation
## Plot points and estimate the line that best represents them
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Practice Scatter Plots and Linear Correlation
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Interpreting Scatter Plots and Line Graphs
#### Objective
Here you will practice recognizing and using some of the primary information available from a scatter plot or line graph.
#### Concept
Steve is playing a stock market simulation game in his social studies class. He has chosen to invest in Apple Inc., Amazon.Com Inc., Walt Disney Company and Microsoft. He originally bought all four stocks in the 10th day of the month. Now he needs to choose one of them to sell. Based on the recent performance of each of them as shown in the line graphs below, which would you recommend he choose and why?
Seem a bit daunting? It certainly can be! We will return to this question at the end of the lesson to review the situation.
#### Watch This
http://youtu.be/PE_BpXTyKCE vcefurthermaths – Maths Tutorial: Interpreting Scatterplots (statistics)
#### Guidance
The primary use for scatter plots and line graphs is to demonstrate or evaluate the correlation between two variables. Though the two are similar in many ways, there are distinct differences, and specific situations in which one is appropriate and the other is not.
• A scatter plot is generally used when displaying data from two variables that may or may not be directly related, and when neither of the variables is under the direct control of the researcher. The primary function of a scatter plot is to visualize the strength of correlation between the two plotted variables. The number of sunburned swimmers at the local pool each day for a month would be an example of a data set that would best be displayed as a scatter plot, since neither the weather nor the number of swimmers present is under the control of the researcher.
• A line graph is appropriate when comparing two variables that are believed to be related, and when one of the variables is under the direct control of the researcher. The primary use of a line graph is to determine the trend between the two graphed variables. The mileage of a particular car compared to speed of travel would be a good example, since the mileage is certainly correlated to the speed and the speed can be directly controlled by the researcher.
In later lessons we will discuss methods of quantifying the level of correlation between two variables and calculating a line of best fit, but for now we will focus on identifying specific examples of weak or strong correlation and identifying different types of trends.
• SCATTER PLOTS:
• Two variables with a strong correlation will appear as a number of points occurring in a clear and recognizable linear pattern. The line does not need to be straight, but it should be consistent and not exactly horizontal or vertical.
• Two variables with a weak correlation will appear as a much more scattered field of points, with only a little indication of points falling into a line of any sort.
• LINE GRAPHS:
• A linear relationship appears as a straight line either rising or falling as the independent variable values increase. If the line rises to the right, it indicates a direct relationship. If the line falls to the right, it indicates an inverse relationship.
• A non-linear relationship may take the form of any number of curved lines, and may indicate a squared relationship (dependent variable is the square of the independent), a square root relationship (dependent variable is the square root of the independent), an inverse square (dependent variable is one divided by the square of the independent), or many other possibilities.
• BOTH:
• A positive correlation appears as a recognizable line with a positive slope . A line has a positive slope when an increase in the independent variable is accompanied by an increase in the dependent variable (the line rises as you move to the right).
• A negative correlation appears as a recognizable line with a negative slope. As the independent variable increases, the dependent variable decreases (the line falls as you move to the right).
Example A
What type of relationship is indicated by the line graph below?
Solution: The line is straight, indicating a linear relationship. It rises from left to right, meaning that the dependent variable increases as the independent variable increases, indicating a positive correlation.
Example B
Which image shows a non-linear graph with a negative correlation?
Solution:
• The first image is of a curved line that rises from left to right, this is a non-linear positive correlation
• The second image is a straight line that falls from left to right, this is a linear negative correlation
• The third image is a curved line that falls from left to right, this is a nonlinear negative correlation and is the correct image as described by the question.
Example C
1. Which graph(s) indicate(s) a weak correlation?
2. Which one(s) indicate(s) a strong correlation?
3. Which graph(s) indicate positive correlation(s)?
Solution:
• Only graph 2 indicates a weak correlation, since it is the only one with points that are not clearly arranged in a linear fashion.
• Graphs 1, 3, and 4 all indicate strong correlations, as evidenced by the high percentage of points obviously organized in a line. Graph 4 is obviously a very strong correlation as a clear non-horizontal or vertical line connects all of the points.
• Graphs 1, 2, and 4 are all positive correlations as all three rise from left to right. Another way to put it is that those three graphs have a positive slope (though graph 4 does not have a consistent slope, anywhere on the curve the slope is estimated it would still be positive).
Concept Problem Revisited
Which stock(s) should Steve sell if he needs to make a profit right away?
By looking at the lines on each of the four graphs, we can see that it is important to note that Steve purchased the stocks on the 10 th , since only the Walt Disney CO and Amazon are currently valued more highly than they were on the 10 th . Both Apple and Microsoft are going up in value now, at the end of the month, but neither has made it back up to where they were on the 10 th .
If Steve wants to make a profit right now, he should sell Walt Disney or Amazon or both.
#### Vocabulary
A trend is an estimation of the tendency of data points to move in a certain direction. A trend line, also known as a line of fit, is a line drawn on a graph to indicate how the data points generally increase or decrease.
A strong correlation means that the values of the output variable are strongly affected by the values of the input variable. A strong correlation is indicated on a graph by a large percentage of data points lying in an apparent line, either straight or curved.
A linear relationship means that the output values are a simple multiple of the input variable, and appears as a straight line when graphed.
A non-linear relationship appears as a curved line on a graph. It indicates an output variable that is a power, a root, or other more complex multiple of the input.
A direct relationship means that the variables increase and decrease together, resulting in a positive correlation and a line of fit that rises from left to right, whereas an inverse relationship is a negative correlation, meaning that the output decreases as the input increases, and vice versa.
A slope is a description of the rate at which the output variable increases or decreases compared to the input variable. This is referred to as the slope because the rate of increase or decrease affects the angle of the line on a graph.
#### Guided Practice
1. Describe the relationship indicated by the graph:
2. Describe the relationship indicated by the graph:
3. Describe the graph that would result from a strongly correlated positive non-linear relationship. Give an example of a function that could result in such a graph.
4. Which scatter plot below indicates the most strongly correlated variables?
5. Which plot below indicates a weakly correlated positive linear relationship?
Solutions:
1. This is a very strongly correlated (all the points connected by a line), negative (the line falls from left to right), nonlinear (not a straight line) relationship.
2. This is a weekly correlated (significant scattering of the points), positive (points generally increase in value from left to right), linear (a straight line of fit could be drawn) relationship.
3. A strongly correlated positive non-linear relationship would appear as a well-defined curve of points rising from left to right.
4. The center plot is the most strongly correlated, evidenced by the much cleaner line formed by the data points. Incidentally, this is a negative linear relationship.
5. The left hand plot is weakly correlated, but negative. The right hand plot is positive, but strongly correlated. The center plot is weakly correlated and positive, so it is the one matching the question definition.
#### Practice
1. What sort of trend is shown in the scatter plot below?
2. A door to door vacuum cleaner sales man plots a scatter diagram of how much he has earned over the years. In which year was his income the highest?
3. The number of children in two different day care centers, and the types of lunch they eat is represented in the table below. Pick the appropriate scatter plot for the data.
Lunch Served Center 1 (Yellow) Center 2 (Blue) Hamburger 20 25 Mac and Cheese 30 28 Pizza 40 35 Tuna Salad 60 45 Burritos 80 65
4. The plot shown gives the relationship between the demand and price of a trendy consumer good. What trend does the plot follow?
5. The plot represents the relationship between the price and supply of an item. What type of trend does the graph illustrate?
6. Katie recorded the following data relating to how long it took to fill up a horse trough. She measured the depth every two minutes after she began filling it, until it was full. Which scatter plot accurately represents the data?
Time (in minutes) Dept (in inches) 2 7 4 8 6 13 8 19 10 20 12 24 14 32 16 37 18 38 20 41 22 47
7. The Scatter Plot below question 8 shows the number of DVD’s sold (in millions) from 2001-2007. Based on the data, about how many DVD’s will be sold in 2009?
8. What sort of trend is shown in the scatter plot?
9. The table below shows a relationship between the weight of a car and its average gas mileage. Which plot best represents the data?
Type of Car Weight MPG 1 3750 29 2 4125 23 3 3100 33 4 5082 18 5 3690 20 6 4640 21 7 5380 14 8 3241 25 9 3895 31 10 4669 17
10. Which scatter plot shows no relationship between test scores received by Greg, and the temperature that the classroom was at while taking the test? Why?
11. Which scatter plot shows a positive relationship between the weight of a mango, and the number of seeds it contains?
Roy was doing research for a research paper. He questioned students throughout his high school, asking them how much time they spent doing homework and how much time they spent watching TV the previous evening. The following scatter plot shows his results. Based on the information answer the questions that follow.
Choose the best of the 4 points, A, B, C, or D to represent the student’s statements below.
12. “I worked on homework almost all night, I only had time to watch my favorite sitcom.”
13. “Last night was about half and half for me”
14. “Last night didn’t have anything on the screen I wanted to watch, and homework was so light, that I ended up going out.”
15. Write a statement that correlates to the 4 th point.
### Vocabulary Language: English
bivariate
bivariate
Bivariate data has two variables
correlation
correlation
Correlation is a statistical method used to determine if there is a connection or a relationship between two sets of data.
curvilinear relationships
curvilinear relationships
Non-linear relationships are called curvilinear relationships.
direct relationship
direct relationship
If the line on a line graph rises to the right, it indicates a direct relationship.
homogeneity
homogeneity
When a group is homogeneous, or possesses similar characteristics, the range of scores on either or both of the variables is restricted.
indirect relationship
indirect relationship
If the line on a line graph falls to the right, it indicates an indirect relationship.
linear relationship
linear relationship
A linear relationship appears as a straight line either rising or falling as the independent variable values increase.
negative correlation
negative correlation
A negative correlation appears as a recognizable line with a negative slope .
non-linear relationship
non-linear relationship
A non-linear relationship may take the form of any number of curved lines but is not a straight line.
positive correlation
positive correlation
A positive correlation appears as a recognizable line with a positive slope .
scatter plot
scatter plot
A scatter plot is a plot of the dependent variable versus the independent variable and is used to investigate whether or not there is a relationship or connection between 2 sets of data.
Slope
Slope
Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$
strong correlation
strong correlation
Two variables with a strong correlation will appear as a number of points occurring in a clear and recognizable linear pattern.
trends
trends
Trends in data sets or samples are indicators found by reviewing the data from a general or overall standpoint
weak correlation
weak correlation
Two variables with a weak correlation will appear as a much more scattered field of points, with only a little indication of points falling into a line of any sort.<|endoftext|>
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Language, Oracy and Literacy
This sphere includes spoken language, reading, writing, literature, wider aspects of language and communication. It is about more than being able to read and write; fundamentally it is about ensuring that children have the cultural literacy, fluency and enjoyment of words that allows them to engage fully with the world about them.
It places spoken language at the heart of the curriculum and reinforces it’s fundamental position for all learners, empowering children, exciting their imagination, and widening their worlds.
Additionally, whilst ICT has a clear place across all spheres of learning, it is evident that it holds significant value within this area, creating and enhancing verbal and digital communication.<|endoftext|>
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Transcript WHAT IS ETHICS?
WHAT IS ETHICS? Objectives: 1. To define ethics. 2. To encourage students to consider how they come to moral decisions. 3. To introduce three ethical theories and consider how each would approach moral issues. LOOK AT THE ‘MAKING MORAL DECISIONS’ SHEET. In pairs consider each of the moral dilemmas outlined on the sheet. What would you do in each situation? How did you reach your conclusions? Did your partner agree with you? WHAT IS ETHICS? The term ‘ethics’ comes from the Greek word ethikos, meaning ‘character’. It can be translated as ‘custom’ and refers to the customary way people act in society. Ethics is a branch of philosophy concerned with morality. Today modern ethics is concerned with 4 fundamental questions: 1. Do good/bad and right/wrong exist? 2. What is meant by the moral terms good/bad and right/wrong? 3. Are there good/bad and right/wrong actions? 4. What should the individual or society do in order to be morally good or right? From the earliest times, philosophers have attempted to answer these questions. They have put forward a variety of theories explaining how we should come to moral decisions. Here are 4 of them. UTILITARIAN ETHICS A THEORY PUT FORWARD BY JEREMY BENTHAM (His mummified body is still on show at King’s College, London!) When you are making an ethical decision you must: a. decide what action would bring the greatest happiness to the greatest number of people; b. or what action would bring the least amount of unhappiness to the most people. c. not take personal relationships into account. SITUATION ETHICS A THEORY PUT FORWARD BY JOSEPH FLETCHER When you make an ethical decision you must: a. AIM FOR A PRACTICAL DECISION WHICH PUTS PEOPLE FIRST b. DECIDE BASED ON THE PARTICULAR SITUATION c. DO THE MOST LOVING THING NOW RETURN TO THE ‘MAKING ETHICAL DECISIONS’ SHEET. How would a utilitarian, and a situation ethicist decide how to act in each of the dilemmas? Which ethical theory do you most identify with?<|endoftext|>
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What is yellow fever?
Yellow fever is a serious disease caused by a virus. For some people it can cause a flu-like illness which improves completely. However, for other people it causes symptoms of high fever, vomiting, jaundice and bleeding which can be fatal. There is no cure for yellow fever.
Yellow fever is passed to humans by bites from infected mosquitoes which tend to bite during daylight hours. (This is different to the mosquitos which carry malaria which tend to bite from dusk to dawn.) Yellow fever occurs in certain countries of tropical Africa and South America.
Who should be immunised against yellow fever?
- Travellers over the age of nine months to countries where yellow fever is a risk. Some countries require an International Certificate of Vaccination against yellow fever before they will let you into the country. Yellow fever is the only disease which routinely requires proof of immunisation:
- In some countries immunisation is compulsory for all incoming visitors.
- In some countries immunisation is compulsory for those who have travelled from a ‘yellow fever’ area or country.
- Your doctor or practice nurse can advise if you should be immunised for your travel destination and whether you need this certificate of immunisation.
- Workers who handle material that may be infected by the yellow fever virus. For example, laboratory workers.
The vaccine and where can I get it?
You should have an injection of vaccine at least ten days before the date of travel to allow immunity to develop. A single dose of vaccine provides immunity for at least 10 years, maybe even for life. However, a booster dose (and a repeat certificate of immunisation) is recommended every 10 years if you are still at risk.
Yellow fever vaccine can only be given at accredited centres. Many GP practices are accredited. If your local GP practice is not accredited you can find a list of the nearest available centres at www.nathnac.org.
The vaccine stimulates your body to make antibodies against the yellow fever virus. These antibodies protect you from illness should you become infected with this virus. The yellow fever vaccine can be given at the same time as other vaccines.
Are there any possible side-effects from the vaccine?
Up to 3 in 10 people who are immunised with yellow fever vaccine have mild headache, muscle aches, mild fever or soreness at the injection site. These symptoms can last up to 14 days after the injection. Severe reactions are very rare, but the risk increases in older people.
Who should not receive the yellow fever vaccine?
The yellow fever vaccine is not usually given under the following circumstances, although advice should be taken from your doctor or practice nurse.
- If you have reduced immunity (immunosuppression). For example, people with HIV, those taking high dose long-term steroids, those receiving chemotherapy, etc. Also, immunisation should be delayed after stopping treatment. This should be discussed with your specialist, but in general:
- Delay immunisation for 3 months after stopping a course of steroids.
- Delay immunisation for at least 6 months after treatment with chemotherapy/radiotherapy or after taking other immunosuppressants such as azathioprine, cyclosporin, methotrexate, cyclophosphamide, leflunomide and cytokine inhibitors.
- Delay immunisation at least 12 months after stopping all immunosuppressants for bone marrow transplant (longer if evidence of graft-vs-host disease).
- If you are ill with a fever you should postpone the injection until you are better.
- As a rule, pregnant women should not be immunised with this vaccine. It is sometimes given after the 6th month of pregnancy if you are at a high risk of catching yellow fever. This vaccine may be given if you are breastfeeding and cannot avoid being at high risk of catching yellow fever.
- You should not have the yellow fever vaccine if you have had a severe (anaphylactic) reaction in the past to egg. (This is because the vaccine contains small amounts of egg. A severe reaction to egg is very rare and it does not mean an upset stomach eating eggs or disliking eggs.)
- Children under 9 months old should not receive the yellow fever vaccine. (Babies aged 6-9 months may occasionally receive the vaccine if the risk of yellow fever during travel is unavoidable.)
- Older travellers (those aged over 60 years) who have not previously been vaccinated against yellow fever are at a higher risk of side effects with the yellow fever vaccine.
- If you have had a severe reaction to the yellow fever vaccine in the past.
- If you have a thymus disorder.
Health Advice for Travellers – From the Department of Health
All travellers going abroad are advised to get this booklet. You can get a free copy from main post offices.
References and Disclaimer | Provide feedback
Egton Medical Information Systems Limited. Registered in England. No 2117205 Registered Office: Fulford Grange, Micklefield Lane, Rawdon, Leeds, LS19 6BA<|endoftext|>
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# The Definite Integral.
## Presentation on theme: "The Definite Integral."— Presentation transcript:
The Definite Integral
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
The width of a rectangle is called a subinterval. The entire interval is called the partition. subinterval partition Subintervals do not all have to be the same size.
subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by As gets smaller, the approximation for the area gets better. if P is a partition of the interval
is called the definite integral of
over If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:
Leibnitz introduced a simpler notation for the definite integral:
Note that the very small change in x becomes dx.
variable of integration
upper limit of integration Integration Symbol integrand variable of integration (dummy variable) lower limit of integration It is called a dummy variable because the answer does not depend on the variable chosen.
We have the notation for integration, but we still need to learn how to evaluate the integral.
Since rate . time = distance:
In section 6.1, we considered an object moving at a constant rate of 3 ft/sec. Since rate . time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line. time velocity After 4 seconds, the object has gone 12 feet.
If the velocity varies:
Distance: (C=0 since s=0 at t=0) After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid.
What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. It seems reasonable that the distance will equal the area under the curve.
The area under the curve
We can use anti-derivatives to find the area under a curve!
Riemann Sums Sigma notation enables us to express a large sum in compact form
Calculus Date: 2/18/2014 ID Check
Objective: SWBAT apply properties of the definite integral Do Now: Set up two related rates problems from the HW Worksheet 6, 10 HW Requests: pg 276 #23, 25, 26, Turn in #28 E.C In class: Finish Sigma notation Continue Definite Integrals HW:pg 286 #1,3,5,9, 13, 15, 17, 19, 21, Announcements: “There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman Turn UP! MAP Maximize Academic Potential
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
The width of a rectangle is called a subinterval. The entire interval is called the partition. subinterval partition Subintervals do not all have to be the same size.
The width of a rectangle is called a subinterval.
The width of a rectangle is called a subinterval. The entire interval is called the partition. Let’s divide partition into 8 subintervals. subinterval partition Pg 274 #9 Write this as a Riemann sum. 6 subintervals
subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by As gets smaller, the approximation for the area gets better. if P is a partition of the interval
is called the definite integral of
over If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:
Leibnitz introduced a simpler notation for the definite integral:
Note that the very small change in x becomes dx. Note as n gets larger and larger the definite integral approaches the actual value of the area.
variable of integration
upper limit of integration Integration Symbol integrand variable of integration (dummy variable) lower limit of integration It is called a dummy variable because the answer does not depend on the variable chosen.
Calculus Date: 2/19/2014 ID Check
Objective: SWBAT apply properties of the definite integral Do Now: Bell Ringer Quiz HW Requests: pg 276 #25, 26, pg odds In class: pg 276 #23, 28 Continue Definite Integrals HW:pg 286 #17-35 odds Announcements: “There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman Turn UP! MAP Maximize Academic Potential
Bell Ringer Quiz (10 minutes)
Riemann Sums LRAM, MRAM,and RRAM are examples of Riemann sums Sn =
This sum, which depends on the partition P and the choice of the numbers ck,is a Riemann sum for f on the interval [a,b]
Definite Integral as a Limit of Riemann Sums
Let f be a function defined on a closed interval [a,b]. For any partition P of [a,b], let the numbers ck be chosen arbitrarily in the subintervals [xk-1,xk]. If there exists a number I such that no matter how P and the ck’s are chosen, then f is integrable on [a,b] and I is the definite integral of f over [a,b].
Definite Integral of a continuous function on [a,b]
Let f be continuous on [a,b], and let [a,b] be partitioned into n subintervals of equal length Δx = (b-a)/n. Then the definite integral of f over [a,b] is given by where each ck is chosen arbitrarily in the kth subinterval.
Definite integral This is read as “the integral from a to b of f of x dee x” or sometimes as “the integral from a to b of f of x with respect to x.”
Using Definite integral notation
The function being integrated is f(x) = 3x2 – 2x + 5 over the interval [-1,3]
Definition: Area under a curve
If y = f(x) is nonnegative and integrable over a closed interval [a,b], then the area under the curve of y = f(x) from a to b is the integral of f from a to b, We can use integrals to calculate areas and we can use areas to calculate integrals.
Nonpositive regions If the graph is nonpositive from a to b then
Area of any integrable function
= (area above the x-axis) – (area below x-axis)
Turn UP! MAP Maximize Academic Potential
Integral of a Constant If f(x) = c, where c is a constant, on the interval [a,b], then
Evaluating Integrals using areas
We can use integrals to calculate areas and we can use areas to calculate integrals. Using areas, evaluate the integrals: 1) 2)
Evaluating Integrals using areas
Evaluate using areas: 3) 4) (a<b)
Evaluating integrals using areas
Evaluate the discontinuous function: Since the function is discontinuous at x = 0, we must divide the areas into two pieces and find the sum of the areas = = 1
Integrals on a Calculator
You can evaluate integrals numerically using the calculator. The book denotes this by using NINT. The calculator function fnInt is what you will use. = fnInt(xsinx,x,-1,2) is approx
Evaluate Integrals on calculator
Evaluate the following integrals numerically: = approx. 3.14 = approx. .89
Rules for Definite Integrals
Order of Integration:
Rules for Definite Integrals
Zero:
Rules for Definite Integrals
Constant Multiple: Any number k k= -1
Rules for Definite Integrals
4) Sum and Difference:
Rules for Definite Integrals
Rules for Definite Integrals
Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [a,b] then: min f ∙ (b – a) ≤ ≤ max f ∙ (b – a)
Rules for Definite Integrals
Domination: f(x) ≥ g(x) on [a,b] f(x) ≥ 0 on [a,b] ≥ 0 (g =0)
Using the rules for integration
Suppose: Find each of the following integrals, if possible: b) c) d) e) f)
Calculus Date: 2/26/2014 ID Check
Obj: SWBAT connect Differential and Integral Calculus Do Now: HW Requests: 145 #2-34 evens and 33 HW: SM pg 156 Announcements: Mid Chapter Test Fri. Handout Inverses Saturday Tutoring 10-1 (limits) “There is something in every one of you that waits and listens for the sound of the genuine in yourself. It is the only true guide you will ever have. And if you cannot hear it, you will all of your life spend your days on the ends of strings that somebody else pulls.” ― Howard Thurman Maximize Academic Potential Turn UP! MAP
The Fundamental Theorem of Calculus, Part I
Antiderivative Derivative
Applications of The Fundamental Theorem of Calculus, Part I
1. 2.
Applications of The Fundamental Theorem of Calculus, Part I
Applications of The Fundamental Theorem of Calculus, Part I
Applications of The Fundamental Theorem of Calculus, Part I
Find dy/dx. y = Since this has an x on both ends of the integral, it must be separated.
Applications of The Fundamental Theorem of Calculus, Part I
=
Applications of The Fundamental Theorem of Calculus, Part I
=
The Fundamental Theorem of Calculus, Part 2
If f is continuous at every point of [a,b], and if F is any antiderivative of f on [a,b], then This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.
Applications of The Fundamental Theorem of Calculus, Part 2
End here
Using the rules for definite integrals
Show that the value of is less than 3/2 The Max-Min Inequality rule says the max f . (b – a) is an upper bound. The maximum value of √(1+cosx) on [0,1] is √2 so the upper bound is: √2(1 – 0) = √2 , which is less than 3/2
Average (Mean) Value
Applying the Mean Value
Av(f) = = 1/3(3) = 1 4 – x2 = 1 when x = ± √3 but only √3 falls in the interval from [0,3], so x = √3 is the place where the function assumes the average. Use fnInt
Mean Value Theorem for Definite Integrals
If f is continuous on [a,b], then at some point c in [a,b],
Antidifferentiation = F(x) + C If x = a, then 0 = F(a) + C C = -F(a)
A function F(x) is an antiderivative of a function f(x) if F’(x) = f(x) for all x in the domain of f. The process of finding an antiderivative is called antidifferentiation. If F is any antiderivative of f then = F(x) + C If x = a, then = F(a) + C C = -F(a) = F(x) – F(a)
Trapezoidal Rule To approximate , use
T = (y0 + 2y1 + 2y2 + …. 2yn-1 + yn) where [a,b] is partitioned into n subintervals of equal length h = (b-a)/n.
Using the trapezoidal rule
Use the trapezoidal rule with n = 4 to estimate h = (2-1)/4 or ¼, so T = 1/8( 1+2(25/16)+2(36/16)+2(49/16)+4) = 75/32 or about 2.344
Simpson’ Rule To approximate , use
S = (y0 + 4y1 + 2y2 + 4y3…. 2yn-2 +4yn-1 + yn) where [a,b] is partitioned into an even number n subintervals of equal length h =(b –a)/n.
Using Simpson’s Rule Use Simpson’s rule with n = 4 to estimate
h = (2 – 1)/4 = ¼, so S = 1/12 (1 + 4(25/16) + 2(36/16) + 4(49/16) + 4) = 7/3<|endoftext|>
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tgt
## Sunday, 24 March 2013
### Intermediate Algebra chapter 5 part 2
Addition, Subtraction, and Multiplication of Polynomials
Chapter 5
Polynomials
General form of a polynomial in x:
anxn + an-1xn-1 + an-2xn-2 + . . . + a2x2 + a1x1 + a0, where
• the ai, i = 1, 2, . . ., n are real numbers
• n is a whole number
Degree of a term is the sum of the exponents on the variables in the term.
The term 4x3y5 has degree 8 since 3 + 5 = 8.
Degree of a polynomial is the degree of the highest degree term.
To write a polynomial in descending order for a certain variable means to write the polynomial from the term with the highest exponent (in the certain variable) on the left descending to the term with the lowest exponent (in the certain variable) on the right.
Descending order in x: 3x2 - 2x + 1.
Examples of Polynomials in x
NameExampleDegreeNote
Monomial3x22One term (mono)
Binomial2x + 11Two terms (bi)
Trinomialx3 + 2x2 + x3Three terms (tri)
Polynomial6x4 + 5x3 + 4x2 + x + 74Many terms (poly)
Polynomials can be in more than one variable....
Examples of Polynomials in x and y
NameExampleDegreeNote
Monomial3x2y35One term (mono)
Binomial2xy + y22Two terms (bi)
Trinomialx3y4 + 2x3y + xy27Three terms (tri)
Polynomial6x4y + 5x3y2 + 4x2y3 + xy4 + 7y55Many terms (poly)
Combine like terms.
Example: add 3x2 + 2x + 1 and 5x2 - 7x (3x2 + 2x + 1) + (5x2 - 7x) = = 3x2 + 2x + 1 + 5x2 - 7x remove parentheses = 8x2 - 5x + 1 add like terms
Subtracting Polynomials
1) Remove parentheses (distribute "-" through).
2) Combine like terms.
Example: subtract 3x2 + 2x + 1 from 5x2 - 7x (5x2 - 7x) - (3x2 + 2x + 1) = = 5x2 + 7x - 3x2 - 2x - 1 remove parentheses = 2x2 + 5x - 1 add like terms
Multiplying Polynomials
1) Use distributive property to remove parentheses and multiply out.
• FOIL only works when multiplying binomials--the distributive property works when multiplying any polynomials together.
2) Combine like terms.
Example: (2x + 3)(4x + 5) (2x + 3)(4x + 5) = = 2x(4x + 5) + 3(4x + 5) distributive property = (2x)(4x) + (2x)(5) + 3(4x) + 3(5) distributive property again = 8x2 + 10x + 12x + 15 simplifying = 8x2 + 22x + 15 combine like terms
Example: (2x - 3)(4x2 - 5x + 6) (2x - 3)(4x2 - 5x + 6) = = 2x(4x2 - 5x + 6) - 3(4x2 - 5x + 6) distributive property = (2x)(4x2) - (2x)(5x) + (2x)(6) - 3(4x2) + 3(5x) - 3(6) distributive property again = 8x3 - 10x2 + 12x - 12x2 + 15x - 18 simplifying = 8x3 - 22x2 + 27x - 18 combine like terms
Special Forms
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b)(a - b) = a2 - b2
Note: the a and b may be any algebraic expression.
ExampleNote
(x + y)2 = x2 + 2xy + y2x is a
y is b
(2x + 5)2 =
= (2x)2 + 2(2x)(5) + 52
= 4x2 + 20x + 25
2x is a
5 is b
(2x - 5)2 =
= (2x)2 - 2(2x)(5) + 52
= 4x2 - 20x + 25
2x is a
5 is b
(3x2 - 2y)2 =
= (3x2)2 - 2(3x2)(2y) + (2y)2
= 9x4 - 12x2y + 4y2
3x2 is a
2y is b
(3x + 8)(3x - 8) =
= (3x)2 - 82 =
= 9x2 - 64
3x is a
8 is b
(4x2 - 3y)(4x2 + 3y) =
= (4x2)2 - (3y)2
= 16x4 - 9y2
4x2 is a
3y is b
25y2 - 81x4 =
= (5y)2 - (9x2)2
= (5y + 9x2)(5y - 9x2)
5y is a
9x2 is b<|endoftext|>
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What does the word ‘Leviticus’ mean?
Moses, or one of his helpers, wrote Leviticus in the *Hebrew language. All the people called *Jews spoke the *Hebrew language. They belonged to the family that Abraham, Isaac and Jacob started. The *Jews called the Book of Leviticus ‘wayyiqra’. This is actually two *Hebrew words, ‘way’ and ‘yiqra’. They mean ‘and’ and ‘he called’. These words are the first words in the book in the *Hebrew language. When the *Jews translated their Bible into the *Greek language, they gave the book a new title. The new *Greek title meant ‘about the *Levites’. Our title ‘Leviticus’ is the *Latin word for ‘about the *Levites’. The people who lived in Greece spoke *Greek. The people who lived in Rome spoke *Latin. When Jesus came to the earth, many people spoke *Latin and *Greek.
Jacob had 12 children. One of them was called Levi. His *descendants became the *tribe of Levi. *Descendants are members of your family that live after you. A *tribe is a very large family. Members of this family were called *Levites. They had a special job to do. Some of them worked in the house of God. They were called priests. Other priests worked in every town in the country. They helped people to understand the Bible. The Book of Leviticus, (‘wayyiqra’), helped the priests to do their work. Remember that all the priests were *Levites.
But now there are no *Jewish priests with the same duties as they had at the time of Moses. But the Book of Leviticus is still important. The reason for this is that many things in the book *point to (describe) the life and death of Jesus! Leviticus was important after the *Jews left Egypt. Today, it is important after a person becomes a Christian. There is more about this below. But today, we could call the Book of Leviticus by another name. That name would be, ‘*Worship the *LORD in the Beauty of *Holiness’. ‘*Worship’ means many things. They include ‘love and obey’. *LORD is a special name for God. It probably means ‘always alive’. ‘*Holiness’ means that people always try to please God. God always thinks that this makes people beautiful.
Normally, it would not be possible for anyone to *worship the *LORD in the beauty of *holiness. The Bible says that all people are *sinners. By our own efforts, we cannot please God. But God has provided a method so that we can *worship him properly.
For the *Jews, that method was *sacrifice. They would give to God parts of the animals that they killed. Sometimes they gave the whole animal. The priests burned the gift on the *altar at the house of God. God accepted the animal’s death so that the *Jews could *worship him. The animal had suffered death so that the *Jews could live as friends with God.
For Christians, that method is also *sacrifice. But it is not the death of an animal. God has provided his own precious son, Jesus, to be the perfect *sacrifice. His death deals with every *sin of the people who invite him into their lives. He has freed them from *sin’s power so that they can *worship the *LORD in the beauty of *holiness. So we can all please God because of Jesus’ *sacrifice on the *cross.
Why is Leviticus important for Christians?
Before we read the Book of Leviticus, we must link Leviticus with Exodus. There are three main reasons for this.
Moses wrote Leviticus in the *Hebrew language. The first *Hebrew word in Leviticus is ‘way’. This *Hebrew word means ‘and’. It links the first sentence in Leviticus with the last sentence in Exodus. Moses wanted his people to read the two books together.
In Exodus, Moses described how his people should build the ‘house of God’. This was not God’s house in Jerusalem, called The *Temple. The *Jews did not yet live in Jerusalem. This house was a tent, which people could carry with them. Many ancient people did this, like the *Egyptians. A tent like this showed everybody that their god was with them. Moses’ tent showed people that the God of *Israel was with the *Jews. In Exodus we read how the *Jews made the tent. But in Leviticus, we learn what the priests had to do in God’s tent. All the priests belonged to the *tribe of Levi. They were God’s special servants.
The first 5 books of the Bible tell us about several periods in the life of each person. If that person is a Christian, then the books mean this:
Genesis … *Sin makes a person into a slave of the devil. *Sin is when we do not obey God. Everybody is in this group of people. ‘Everybody has *sinned’, Romans 3:23. We *sin when we do not obey God’s laws. Genesis also tells us about people who tried to obey God, including Enoch, Noah, Abraham, Isaac, Jacob and Joseph.
Exodus … God makes people free. So they are not slaves. They are like the *Jews who came out of Egypt. Christians are free. In other words, *sin does not still control them like slaves. God makes them free when they first believe in Jesus. They believe that Jesus died to save them from the devil. Now they are not the devil’s people. Instead they are God’s people. We call this ‘conversion’.
Leviticus … God wants his people to be friends with him. This is a special type of friendship that we call ‘*fellowship’. God brought the *Jews out of Egypt so that they could have *fellowship with him. And God frees Christians from *sin so that they can have *fellowship with him. This is what God wants very much. Leviticus tells us that we can have *fellowship with God. We can have *fellowship with God because Jesus died for us! That is why Leviticus is important for Christians.
Numbers … God’s people look for the *Promised Land. For the *Jews, it was the country called Israel. For Christians it is Heaven and the New Earth. Heaven is the home of God. We call this the Christian’s ‘walk’ with God. God shows his people the way to go. We often call this ‘guidance’, because God is guiding his people.
Deuteronomy … God’s people have reached the *Promised Land. They are ‘home’! Here are the rules that they must obey while still on this earth. There are no such rules in Heaven.
What is in the Book of Leviticus?
We can divide the Book of Leviticus into 8 sections.
1. Rules about the *sacrifices. (Leviticus 1:1 to 7:38)
2. Rules about how to make a priest. (Leviticus 8:1 to 10:20)
3. Rules about what is *clean and what is not *clean. (Leviticus 11:1 to 15:33)
4. The Day of *Atonement. (Leviticus 16:1 to 16:34)
5. Rules about religion. (Leviticus 17:1 to 22:33)
6. Holy days, weeks and years (Leviticus 23:1 to 25:55)
7. *Blessings and punishments. (Leviticus 26:1 to 26:46)
8. Rules about promises and *offerings. (Leviticus 27:1 to 27:34)
The 5 *sacrifices
In Leviticus chapters 1 to 7 we read about 5 types of *sacrifices. Below is a series of boxes. We call it a ‘table’. It helps us to understand the description and purpose of each *sacrifice.
|Name of *sacrifice||Where to find it in Leviticus||The purpose of this *sacrifice||What the people offered (or gave as a *sacrifice)||What they did with it|
|*Whole offering||Leviticus 1:1-17||To make God favourable. To give yourself humbly to God.||A perfect *bull, sheep, goat, or birds (called *pigeons or *doves).||They burned everything.|
|*Corn offering||Leviticus 2:1-16||To thank God and also to make him favourable to the offerer. To give your goods and your work to God.||Cake or bread with no salt or *yeast (*yeast makes bread to rise).||They burned some and the priests ate some.|
|*Peace offering||Leviticus 3:1-17 and 22:18-30||To thank God; to be at peace with God; to be happy with other people; to express love to God.Also, after you have carried out a promise completely.||A perfect male or female animal, whatever you can afford.||They burned the *fat; the priests and the offerer ate the rest of the animal.|
|*Sin offering||Leviticus 4:1 to 5:13||To ask God to forgive you when you *sinned by accident.||It depended who you were. There were various kinds of *sin offerings.||They burned the *fat for God; they burned the rest outside the camp|
|*Guilt offering||Leviticus 5:14-19||To ask God to forgive you when you *sinned against his holy things, or when you hurt somebody else.||A perfect male sheep.||They burned the *fat for God, but the priests ate the rest.|
The note at the start of this Commentary tells us that Leviticus is about God’s people. 3500 years ago, they were the *Jews. But Leviticus also tells us about Jesus. So, it also tells us about Christian people.
The *Jews gave these 5 *offerings at the *meeting tent. Christians do not give animals as their *offerings. Instead, Jesus is their *offering; he is their *whole, *corn, *peace, *sin and *guilt offering. Normally, only *Jews and Christians give these types of *offerings. They are gifts to God from God’s people.
That helps us to understand that the first three *offerings are not for *sin. They are for *fellowship. They bring God and his people together. They become ‘at one’. In other words, they have friendly relations with him; they are united as friends with him.
That is why Leviticus 1:4 has *atonement (or ‘at-one-ment’) in it. But it is not the usual meaning of at-one-ment. The usual meaning is when God forgives a *sinner for the first time. They become ‘at one’. In Leviticus chapters 1 to 3, it is when a *Jew or a Christian wants to be ‘at one’ with God in his daily life. That is why the word ‘wants’ in Leviticus 1:3 is so important. Also, it tells us why they burned everything in chapter 1. The Christian who wants *fellowship with God gives everything to God. Something to do number 1 tells us how Jesus did this for Christians (see the end of chapter 1).
So, as you read Leviticus chapters 1 to 7, remember this. It is only about God’s people. In Moses’ time, they were *Jews. Now, also, they are Christians.<|endoftext|>
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The World Health Organization (WHO) defines diarrhea (or diarrhoea) as the health disorder of having three or more loose or fluid bowel movements per day or having more stools than is normal for a person.
Many people have a bout of diarrhea once or twice each year. Typically frequent bowel movements last two to three days, and in most cases treated with over-the-counter (OTC) medicines. In some people, diarrhea often occurs as part of irritable bowel syndrome or due to other chronic diseases of the large intestine. The loss of essential fluids due to diarrhea cause dehydration that cause electrolyte disorders like potassium deficiency as well as other salt imbalances.
Though diarrhea is very common and usually not serious in normal circumstances, WHO considers it as the major cause of death in developing countries and the second most frequent cause of infant deaths worldwide.
In 2009, diarrhea caused the death of 1.5 million children under the age of five and 1.1 million people aged five and over.
The most common cause of diarrhea is the infection of the gut by a virus. The infection sometimes called “intestinal flu” or “stomach flu” lasts usually for two to three days. Diarrhea may also be caused by:
- Infection by bacteria (the cause of most types of food poisoning)
- Infections by certain other organisms,
- Eating foods that upset the digestive system,
- Malabsorption where the digestive system is unable to absorb adequately certain nutrients present in the diet,
- Allergies to certain foods,
- Some medications,
- Radiation therapy,
- Diseases of the intestines like Crohn’s disease, ulcerative colitis,
- Some types of cancer,
- Excess use of laxatives,
- Excess consumption of Alcohol,
- Surgery in the digestive tract,
- Competitive running.
Diarrhea may also follow a constipation in people who have irritable bowel syndrome.
Recently, I came across this video on YouTube. It is an advertisement produced for Lifebuoy. Normally, I do not endorse advertisements. However, I wish to share the video with you for the way it caught my attention and led me to the subject of diarrhea.<|endoftext|>
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HTML is an acronym for HyperText Markup Language. It is a way of setting up a document so that the document can contain hypertext references, pictures etc. These documents are suffixed with the label '.html' so that they are recognized as such.
HTML documents (like any other) can be read by your browser from your local computer disk, or transferred from a server at any location on the Web. We will go through the basics of HTML so that you can produce your own documents.
Go To Next Page
Go To Previous Page
Go To Overview<|endoftext|>
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Biomedical engineers combine science and engineering skills to produce new devices and procedures for treating diseases and disabilities in patients. They are responsible for life-saving devices, such as the artificial heart, and tools that give patients the ability to walk or hear when, otherwise, they couldn't. Biomedical engineers work in labs, hospitals and research centers to study, invent and maintain the products of their efforts.
Computers are among the tools biomedical engineers spend the most time using. Part of what computers provide are some very basic functions, like Internet access and word processing software. Besides writing research reports and reading scholarly scientific journals online, biomedical engineers use computers to synthesize data and control the other equipment they use to perform experiments.
Biomedical engineers use many types of microscopes to observe the results of experiments. High-resolution light microscopes can analyze recently grown cells or DNA. Scanning electron microscopes produce digital images that biomedical engineers collect to make time-lapse videos or photo comparisons.
Incubators are containers that give biomedical engineers a controlled space in which to grow cells for experiments. Along with fermenters, incubators give biomedical engineers greater control of their work and the ability to produce faster results.
Biomedical engineers use cryogenic equipment, including freezing tanks, to study cells as they freeze and thaw. This gives them insight into tissue properties under very specific conditions impossible to reproduce otherwise.
To document experiments and analyze the principles of human physiology, biomedical engineers use cameras extensively. Specialized cameras use thermal imaging and X-rays to analyze motion and record findings for upload to a computer.
Biomedical engineers use lasers for two distinctly different purposes. Some lasers with finely tuned wavelengths are applied to cells to study the effect of light on cellular growth and tissue formation. Another class of lasers are at the center of the development of laser surgery, as biomedical engineers develop new tools for advanced and non-invasive surgical procedures.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Shifts of Square Root Functions
## Translate square root functions vertically and horizontally
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Practice Shifts of Square Root Functions
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Shifts of Square Root Functions
What if you had the square root function \begin{align*}y=\sqrt{x}\end{align*}? How would the graph of the function change if you added 5 to the righthand side of the equation or if you multiplied x by 3? After completing this Concept, you'll be able to identify various shifts in square root functions.
### Guidance
We will now look at how graphs are shifted up and down in the Cartesian plane.
#### Example A
Graph the functions \begin{align*}y=\sqrt{x}, y=\sqrt{x} + 2\end{align*} and \begin{align*}y=\sqrt{x} - 2\end{align*}.
Solution
When we add a constant to the right-hand side of the equation, the graph keeps the same shape, but shifts up for a positive constant or down for a negative one.
#### Example B
Graph the functions \begin{align*}y=\sqrt{x}, y=\sqrt{x - 2},\end{align*} and \begin{align*}y = \sqrt{x + 2}\end{align*}.
Solution
When we add a constant to the argument of the function (the part under the radical sign), the function shifts to the left for a positive constant and to the right for a negative constant.
Now let’s see how to combine all of the above types of transformations.
#### Example C
Graph the function \begin{align*}y = 2\sqrt{3x - 1} + 2\end{align*}.
Solution
We can think of this function as a combination of shifts and stretches of the basic square root function \begin{align*}y = \sqrt{x}\end{align*}. We know that the graph of that function looks like this:
If we multiply the argument by 3 to obtain \begin{align*}y = \sqrt{3x}\end{align*}, this stretches the curve vertically because the value of \begin{align*}y\end{align*} increases faster by a factor of \begin{align*}\sqrt{3}\end{align*}.
Next, when we subtract 1 from the argument to obtain \begin{align*}y = \sqrt{3x - 1}\end{align*} this shifts the entire graph to the left by one unit.
Multiplying the function by a factor of 2 to obtain \begin{align*}y = 2 \sqrt{3x - 1}\end{align*} stretches the curve vertically again, because \begin{align*}y\end{align*} increases faster by a factor of 2.
Finally we add 2 to the function to obtain \begin{align*}y = 2 \sqrt{3x - 1} + 2\end{align*}. This shifts the entire function vertically by 2 units.
Each step of this process is shown in the graph below. The purple line shows the final result.
Now we know how to graph square root functions without making a table of values. If we know what the basic function looks like, we can use shifts and stretches to transform the function and get to the desired result.
Watch this video for help with the Examples above.
### Vocabulary
• For the square root function with the form: \begin{align*}y = a \sqrt{f(x)} + c\end{align*}, \begin{align*}c\end{align*} is the vertical shift.
### Guided Practice
Graph the function \begin{align*}y = -\sqrt{x +3} -5\end{align*}.
Solution
We can think of this function as a combination of shifts and stretches of the basic square root function \begin{align*}y = \sqrt{x}\end{align*}. We know that the graph of that function looks like this:
Next, when we add 3 to the argument to obtain \begin{align*}y = \sqrt{x +3}\end{align*} this shifts the entire graph to the right by 3 units.
Multiplying the function by -1 to obtain \begin{align*}y = - \sqrt{x +3}\end{align*} which reflects the function across the \begin{align*}x\end{align*}-axis.
Finally we subtract 5 from the function to obtain \begin{align*}y = - \sqrt{x +3}-5\end{align*}. This shifts the entire function down vertically by 5 units.
### Practice
Graph the following functions.
1. \begin{align*}y = \sqrt{2x - 1}\end{align*}
2. \begin{align*}y = \sqrt{x - 100}\end{align*}
3. \begin{align*}y = \sqrt{4x + 4}\end{align*}
4. \begin{align*}y = \sqrt{5 - x}\end{align*}
5. \begin{align*}y = 2\sqrt{x} + 5\end{align*}
6. \begin{align*}y = 3 - \sqrt{x}\end{align*}
7. \begin{align*}y = 4 + 2 \sqrt{x}\end{align*}
8. \begin{align*}y = 2 \sqrt{2x + 3} + 1\end{align*}
9. \begin{align*}y = 4 + \sqrt{2 - x}\end{align*}
10. \begin{align*}y = \sqrt{x + 1} - \sqrt{4x - 5}\end{align*}
### Vocabulary Language: English
square root function
square root function
A square root function is a function with the parent function $y=\sqrt{x}$.
Transformations
Transformations
Transformations are used to change the graph of a parent function into the graph of a more complex function.
Please wait...
Please wait...<|endoftext|>
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Food contains nutrients that are made from proteins, carbohydrates, fats, vitamins, minerals, antioxidants, fiber, water and other substances. A deficiency of any one of these nutrients can increase your risk of diseases, disorders and other medical conditions. A balanced diet of foods and nutrients may enhance your health and prevent disease. Consult your doctor or nutritionist to determine your nutritional needs and plan a diet and supplement program that is right for you.
Carbohydrates and Fats
Nutrients can be divided into two groups. You need macronutrients, such as carbohydrates, fats and proteins, in large amounts and micronutrients, such as vitamins, minerals and antioxidants, in small amounts. A balanced diet provides 45 to 65 percent of calories from carbohydrates, 20 to 35 percent from fats and the remainder of calories from proteins. Carbohydrates, such as starches and sugars from grains, legumes and fruit, provide energy to your cells and tissues. Fats, such as polyunsaturated, monounsaturated, omega-3 and saturated fatty acids, from vegetable oils, olive oil, fish and animal products, respectively, are a concentrated form of energy that help you body absorb certain vitamins and maintain the structure and function of cell membranes and produce hormones and other substances. Unhealthy fats, such as saturated and trans fats, can increase your risk of heart disease.
Vitamins are essential nutrients that regulate metabolic functions throughout your body. Vitamin A stimulates vision and growth of cells. The B vitamins assist enzymes throughout your body. Vitamin C is an antioxidant that stimulates your immune system and protects your cells from environmental toxins. Vitamin D is essential for bone growth. Vitamin E slows down your aging process and protects cellular membranes from degradation. Vitamin K stimulates blood clotting. Vitamins are found in all food groups and are highly concentrated in fruits and vegetables. The Centers For Disease Control and Prevention recommends that you consume between 1 ½ to 2 ½ cups of fruit and 2 ½ to 4 cups of vegetables daily, depending on your age.
Minerals are chemical elements found in food that have various functions in your body. Calcium and magnesium are vital for building and maintaining healthy bones and teeth. Iron is a vital part of hemoglobin, the molecule in red blood cells that carries oxygen from your lungs to cells throughout your body. Phosphorous is essential for cellular energy metabolism. Zinc, copper, manganese and selenium are trace minerals that you need in tiny amounts. Zinc is involved in tissue growth and repair. Copper and manganese work with enzymes in many types of chemical reactions. Selenium is an antioxidant that protects your cells from toxins and harmful chemicals.
Proteins are essential for the growth, development, structure and function of cells, tissues and organs, antibodies, enzymes and nucleic acids. Protein is also part of some hormones. Amino acids are the building blocks for proteins and are found in foods, particularly animal products, such as meat and dairy, fish, legumes, nuts, seeds and whole grains.<|endoftext|>
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Get Started by Finding a Local Center
# What Is Volume in Math? A Kid-Friendly Guide
Aug 12, 2024
From the glass on your table to the skyscrapers that reach for the clouds, volume helps us understand and measure the scale of our physical world.
We’ve put together a comprehensive guide to everything you need to know to get started with volumes in math.
Read on to find simple definitions and explanations, learn to calculate volumes of 3D objects, see solved examples, and test your knowledge with bonus practice exercises.
## What Is Volume?
Volume is the measure of how much space an object takes up in three dimensions. Sometimes, people also use the term "capacity" to talk about volume, especially when discussing how much a container like a cup or a box can hold.
To put it differently, if you understand volume, you can figure out how many candies fit in a jar, how many books fill a shelf, or how much water a swimming pool can contain.
Knowing how to calculate volume can help us figure out about how many jellybeans fit into a jar.
In this short video, one of the founders of Mathnasium Larry Martinek explains volume using Rubik's® Cubes.
## What Are the Volume Units?
Just like we use inches, feet, and miles to measure length, and ounces and pounds to measure weight, volume also has its own set of measuring units.
When measuring the volume of three-dimensional (3D) objects, we use cubic units.
• In the customary system, we use cubic inches (in³), cubic feet (ft³), cubic yards (yd³), and cubic miles (mi³) to measure volume.
• Similarly, in the metric system, we use cubic centimeters (cm³), cubic meters (m³), and cubic kilometers (km³).
As you might have guessed, the reason why we use cubic units to measure volume is because we are measuring three-dimensional spaces. When we are measuring the area of two-dimensional (2D) spaces like squares or rectangles, for example, we use square units such as ft2 or m2.
Find Top-Rated Geometry Tutors at Mathnasium
## How to Find the Volume of Cubes, Prisms & Cylinders
As 3D objects can vary greatly in shape, there is no universal formula for calculating their volume, but there is a universal logic we can follow to calculate the volume of cubes, prisms, and cylinders.
To find the volume of any of these 3 shapes, all we have to do is calculate the area of the base and multiply it by the height.
Let's put this into action:
### Finding Volumes of Cubes
Cubes are special types of rectangular prisms where all sides are equal in length. You can find cubes in everyday items like dice, building blocks, and sugar cubes!
To find the volume of a cube, we should calculate the area of its base and multiply it by the height.
Let’s say the length of the side is 7 inches.
The area of a square base is 7 x 7, or 72.
As all the sides are equal in length, the height of the cube is also 7 .
So to calculate the volume of the square, we should multiply by 72 by 7 , which we can also express as 73 . The volume of our square is 343 cubic inches (in3).
The formula we use to calculate the volume (V ) of a cube is simply the length of its side cubed (s3 ):
V = s3
For example, let's consider a cube with side length: s = 4 in
To find its volume, we use this formula: V = s3
V = 4or V = 4 x 4 x 4
V = 64 in3
The volume of our cube in this case is 64 in³.
### Finding Volumes of Rectangular Prisms
A rectangular prism is a three-dimensional shape with six faces, all of which are rectangles. You can see rectangular prisms all around, like in cardboard boxes, shoeboxes, and bricks!
Rectangular prisms have three pairs of parallel faces:
• Top and bottom
• Front and back
• Left and right side
The faces in each pair are the same shape and size, so they have the same area.
To calculate the volume of a rectangular prism, we start by defining its length (l) , height (h) , and width (w) .
To find the base, we multiply the length by the width l x w.
To find volume, we multiply the base l x w by the height: l x w x h
And we found the formula for calculating the volume of prisms:
V = l x w x h
Let's try this example.
• The length of our rectangular prism is 6 inches (l = 6 in)
• The width is 3 inches (w = 3 in)
• The height is 2 inches (h = 2 in)
To calculate the volume of the prism, we multiply all three values:
V = 6 x 3 x 2
V = 36 in3
### Finding Volumes of Cylinders
A cylinder is a three-dimensional shape with two circular bases of the same size that are parallel to each other and connected by a curved surface. For example, you can see cylinders in objects like paper towel tubes or soda cans.
To calculate the volume of a cylinder, we need to know the radius of its base and its height.
To find the formula for the volume of a cylinder, let’s start by calculating the area of its base.
As the base of a cylinder is a circle, to calculate the area we multiply π by the squared radius (r2): π x r2.
Now all we have to do to find the volume is to multiply the area of the base with the height of the cylinder: π x r2 x h
There we have it! The formula for calculating the volume of a cylinder is:
V = π x r2 x h
Let’s put it into action!
If you remember your lessons on the area of the circle, you will recall the Greek letter π, or “Pi”, and that its value is approximately 3.14.
In our examples, we will leave the volume in terms of π (pi) to simplify calculations.
What we are missing now is to define the radius (r) of the base and the height of our cylinder (h), so let’s say they measure:
Radius r = 3 in
Height h = 5 in
To calculate its volume, we use the formula from earlier:
V = π x r2 x h
Substitute the values:
V = π x 32 x 5
V = π x 9 x 5
V = π x 45
V = 45π
So, the volume of our cylinder is 45π in³.
## How to Find the Volume of Pyramids, Cones & Spheres
Ready for the next challenge?
Let’s level up your geometry skills and tackle the volumes of pyramids, cones, and spheres.
### Finding Volumes of Pyramids
A pyramid is a three-dimensional shape with a flat base and triangular faces that meet at a single point called the apex.
The base of a pyramid is the flat surface that the pyramid sits on. The surface can vary in shape, from triangles to squares, and beyond.
The height of a pyramid is the vertical distance from its apex (top) down to its base, measured along a straight line perpendicular to the base.
To find the volume of a pyramid, just like with prisms and cylinders, we need to know the area of its base ( ) and its height.
But, unlike prisms and cylinders, pyramids do not have two parallel bases, so our volume formula will differ slightly.
We will multiply the base and the height of the pyramid by 1/3 , like so:
V = 1/3 x B x h
For example, let's say our pyramid has a square base with these dimensions:
• Base side length: s = 3 in
• Height: h = 6 in
To find its volume, we use the formula: V = 1/3 x B x h
For a square base, the base area is calculated as: B = s2
So, we get
V = 1/3 x 3x 6
V = 1/3 x 54
V = 18 in3
So, the volume of our square-based pyramid is 18 in³.
### 2. Finding Volumes of Cones
A cone is a 3D shape with a circular base and a curved side that gets smaller as it goes to a point at the top, called the apex.
To find the volume of a cone, we need to know the radius (r) of its circular base and its height (h). Then, we find the area of the base, multiply it by the height of the cone and by 1/3 .
V = 1/3 x π x r2 x h
You may have noticed that this is the same formula as the pyramid (base x height), except in this case the base is a circle, so we will calculate the area using: π x r2
Let's try our hands at this example.
If we know the radius is 4 in, and the height(h) is 6 in, we can find the volume of the cone.
When we substitute the formula above with the given values, we get:
V = 1/3 x π x r2 x h
V = 1/3 x π x 16 x 6
V = 1/3 x π x 96
V = 96/3 x π
V = 32π
The volume of our cone is 32π in³.
### 3. Finding Volumes of Sphere
A sphere is a perfectly round 3D shape where every point on the surface is the same distance from the center.
To calculate the volume of a sphere, we need to know its radius (r) . The formula for the volume of a sphere is V = 4/3 x π x r3.
Let’s use an example to illustrate. If a sphere has a radius of 3 in, we can find its volume using the formula above.
V = 4/3 x π x r3
V = 4/3 x π x 27.
V = 4 x27/3 x π
V = 108/3 x π
V = 36 x π
V = 36π
Finally, we get the volume of our sphere and it’s 36π in³.
## Solved Examples of Volume Exercises
Let's explore some solved examples to reinforce our knowledge.
### Example 1: Cube
Most of our students love cubes because their volumes are one of the easiest to find!
Why are they easy to find?
Because all of their sides are the same length, so if we know one, we know them all.
In this example, the side of our cube measures at 5 in.
V = s3
Substituting the values, we get:
V = 53
V = 5 x 5 x 5
V = 125 in3
### Example 2: Rectangular Prism
Based on the information you see in the image, how would you calculate the volume of this rectangular prism?
Let’s start by taking note of the dimensions:
• Length (l) = 12 in
• Width (w) = 6 in
• Height (h) = 4 in
We remember the rectangular prism volume formula:
V = l x w x h
When we insert the values into our formula, we get:
V = 12 x 6 x 4
V = 288 in3
### Example 3: Cylinder
Cylinders might look scary because of their formula, specifically the π, but once you get used to the formula, their volumes are pretty easy to calculate.
In this case, our cylinder has:
• Radius: 4 inches
• Height: 10 inches.
The formula for measuring the volume of a cylinder is: V = π x r2 x h
Remembering that π ≈ 3.14 and substituting the given values, we get:
V = π x 42 x 10
V = π x 16 x 10
V = π x 160
V = 160π in3
### Example 4: Pyramid with a Rectangular Base
In this example, we have a pyramid with a rectangular base.
We know the universal formula for pyramid volume is V = 1/3 x B x h where B represents the area of the base.
Let’s take note of the dimensions:
• Height: 9 in
• Length: 6 in
• Width: 4 in
Since we’re dealing with a rectangular base, we will get the value of the base (B) by multiplying the length and width of the rectangle, so our formula will look like this:
V = 1/3 x B x h
V = 1/3 x (l x w) x h
Now let’s insert our values and calculate the volume of our pyramid:
V = 1/3 x 6 x 4 x 9
V = 1/3 x 216
V = 72 in3
### Example 5: Cone
Now, let's work out the volume of this cone together.
For cone volume, we use the formula: V = 1/3 x π x rx h
Write down the dimensions:
• Radius (r): 6 in
• Height (h): 10 in
Let’s substitute the values:
V = 1/3 x π x 36 x 10
V = 1/3 x π x 360
V = 360/3 x π
V = 120 x π
V = 120π in3
### Example 6: Sphere
Time to tackle spheres—those perfectly round wonders of geometry!
To get the volume of a sphere, we follow the formula: V = 4/3 x π x r3
Since our radius is 6 in, we can substitute the value right away:
V = 4/3 x π x 63
V = 4/3 x π x 216
V = 4 x 216/3 x π
V = 864/3 x π
V = 288 x π
V = 288π3 in3
## Quiz – Test Your Knowledge of Volume
Create your own user feedback survey
## Frequently Asked Questions About Volume
Here are the questions we often get from our students when learning about volumes in math:
### 1. How is volume different from area?
Volume is the space inside a three-dimensional object, like a box, while area measures the space covering a two-dimensional surface, like the floor of a room.
To measure volume, we use cubic units (in³, ft³, etc.) and to measure area we use square units (in², ft², etc.).
### 2. Can the volume change?
Yes, volume can change! For example, when you pour water into a cup, the volume in the cup increases because it’s holding more liquid.
Similarly, if you squeeze a balloon, its volume decreases because you're reducing the amount of air inside.
### 3. How is volume used in everyday life?
Perhaps one of the best examples of how we use volumes in everyday life is cooking.
We use volume to measure ingredients like flour, sugar, and liquids.
When a recipe says "1 cup of flour" or "1/2 cup of milk," it's talking about how much space those ingredients take up in a measuring cup.
These precise measurements allow us to replicate the dishes we love.
We also use volumes to measure how much:
• Water we’ll need to fill a pool
• Fuel we can put into our car
• Space we can occupy in a storage unit and so on
## Learn & Master Volume with Top-Rated Geometry Tutors Near You
Mathnasium’s specially trained geometry tutors work with K-12 students regardless of their skill level.
Explore our approach to elementary school tutoring:
Our tutors assess each student’s current skills and considers their unique academic goals to create personalized learning plans that will put them on the best path towards math mastery.
Whether you are looking to catch up, keep up, or get ahead in your math class, find a Mathnasium Learning Center near you, schedule an assessment, and enroll today!
Find a Math Tutor Near You
## OUR METHOD WORKS
Mathnasium meets your child where they are and helps them with the customized program they need, for any level of mathematics.
## HELP YOUR CHILD ACHIEVE THEIR FULL MATH POTENTIAL
##### We have nearly 1,000 neighborhood centers nationwide. Get started now.
• Find a location
• Get a math skills assessment for your child
• Your child will complete a customized learning plan
Get Started by Finding a Local Center<|endoftext|>
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Section23Low Rank Approximations (NM, 24 Oct)¶ permalink
First we proved the following result.
Then we moved on to the following, highly important theorem.
A similar result is true for the $\| \cdot\|_F$ norm, the proof of which is left to an exercise.
All the above results emphasise the power the power of the SVD as an important theoretical and computational tool. We demonstrated this in class by using the SVD to compute “approximations” of certain images.
Computing the SVD is an important task, but not a trivial one. There are a few different approaches, but a key idea is the $QR$-factorisation.
Subsection23.1Exercises
Exercise23.4
Prove Theorem 23.3.
During this section of the course, we occasionally made use of arguments based on partitioning matrices. Use this idea to prove the following theorem.
Exercise23.5
Suppose that onc can partition the matrix $A$ as
\begin{equation*} A = \begin{pmatrix} A_{11} \amp A_{12} \\ A_{21} \amp A_{22}, \end{pmatrix} \end{equation*}
where $A$ and $A_{11}$ are nonsingular. Let $S=A_{22} - A_{21}A_{11}^{-1}A_{12}\text{.}$ Show that
\begin{equation*} A^{-1} = \begin{pmatrix} A_11^{-1} + A_{11}^{-1}A_{12}S^{-1}A_{21}A_{11}^-1 \amp -A_{11}^{-1}A_{12}S^{-1}\\ -S^{-1}A_{21}A_{11}^{-1} \amp S^{-1}. \end{pmatrix} \end{equation*}
Exercise23.6
Recall the definition of a lower triangular and unit lower triangular matrices in Exercise 2.6. Also, an upper triangular matrix is one whose transpose is lower triangular. Let $L_1$ and $L_2$ be unit lower triangular matrices. Let $U_1$ and $U_2$ be nonsingular upper triangular matrices. Suppose that $L_1U_1=L_2U_2\text{.}$ Show that $L_1=L_2$ and $U_1=U_2\text{.}$<|endoftext|>
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What is the formula of three successive discount?
Contents
How do you calculate 3 successive discounts?
Total discount = (d + z – dz / 100) %
1. Final discount = [28 + 30 – (28 x 30) / 100] % = (58 – 840 /100) % = 49.6%
2. Which means Discount = 49.6% of 1000 = (49.6 / 100) x 1000 Rs 496.
3. Selling price = M.P – Discount = 1000- 496 = 504.
2.03.2018
What is the formula of successive discount?
So for the case I two successive discounts of 30% and 20% are given respectively. So, the price after the first discount will be, 100 – 30% of 100 = 100 – 30 = Rs. 70. Now, the price after the second discount will be, 70 – 20% of 70 = 70 – 14 = Rs.
How do you calculate a single discount from 3 successive discounts?
A single discount equivalent to three successive discounts of 20%, 25% and 10% is a) 55% b) 50% .. A single discount equivalent to three successive discounts of 20%, 25% and 10% is a) 55% b) 50% .. 10% of 60 gives 54. therefore 100 -54=46.
What is the formula of net discount?
Discount =10% of 1000 = (10/100)*1000 =Rs 100. Selling Price= 1000- 100 = Rs 900. But in exam,you can do it directly in your head.So just thinkmthat 10 percent discount means you’ve to pay 100 percent minus 10 percent=90 percent of the marked price which means, (90/100)* 1000 =Rs. 900.
What is successive Formula?
Lets try an example. If there’s an increase of 20% and then a decrease of 10%, the successive percentage will be (20 + (-10) + 20 * (-10) / 100 ) = 20 – 10 – 2 = 8% increase. Now in case of discounts, the value of discount percentages will be considered negative.
Whats a successive discount?
Successive discount is the discount offered on the discount. It is similar to compound interest (interest on interest). Let us have an example to understand the concept. Let the original price of a CD be ‘x’. a shopkeeper offers a discount of ‘y%’ and again ‘z%’ on the new price.
What is single discount?
A single trade discount is a discount that is given to a customer (usually a wholesaler) when the customer buys a product. The discount is expressed simply as a single discount of a given percentage. For example, a 25% discount on the purchase.
What is marked price formula?
Amount of discount is = Marked Price – Selling Price. In other words we can say that = (1500 – 1350) = Rs 150. Discount for Rs. 1500 =Rs 150. Therefore, the Discount for Rs 100 = (150/1500) × 100 = 10%
How do you solve Discount problems?
How to calculate a discount
1. Convert the percentage to a decimal. Represent the discount percentage in decimal form. …
2. Multiply the original price by the decimal. …
3. Subtract the discount from the original price. …
4. Round the original price. …
5. Find 10% of the rounded number. …
6. Determine “10s” …
7. Estimate the discount. …
8. Account for 5%
THIS IS IMPORTANT: Quick Answer: How can I get a cheap Surface Pro?
9.03.2021
Is 40 or 30 successive discount better?
Hence, A discount of flat 70% is better than a successive discount of 40 % and 30%.
What is the single discount equivalent of successive discounts?
Successive discounts of 10%, 20% and 30% is equivalent to a single discount of : – GKToday. Single equivalent discount for successive discounts of 10% and 20%. Single equivalent discount for 28% and 30%. Hence option [C] is correct answer.
How do I get a 10% discount?
How do I calculate a 10% discount?
1. Take the original price.
2. Divide the original price by 100 and times it by 10.
3. Alternatively, move the decimal one place to the left.
4. Minus this new number from the original one.
5. This will give you the discounted value.
6. Spend the money you’ve saved!
What is percentage formula?
Percentage can be calculated by dividing the value by the total value, and then multiplying the result by 100. The formula used to calculate percentage is: (value/total value)×100%.
What is the formula to find profit?
The formula to calculate profit is: Total Revenue – Total Expenses = Profit. Profit is determined by subtracting direct and indirect costs from all sales earned.<|endoftext|>
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## How to solve quadratic equations by... COMPLETING THE SQUARE
Suppose you need to solve this grim looking quadratic equation (and we're warning you, the answers are not whole numbers!): x2 + 5x - 9 = 0 First move the constant across. Here we do it by adding 9 to both sides: x2 + 5x = 9 Now we're going to make a new equation by playing with the Left Hand Side (or LHS). We'll ignore the RHS for a moment. Follow these instructions exactly. 1/ divide the LHS by x . In this case we get x + 5 2/ divide the number by 2 (don't divide the "x"). Now we get x +2.5 3/ put both terms in a bracket, square it and then multiply it all out (x + 2.5)2 = (x +2.5)(x +2.5) = x2 + 2.5x + 2.5x + 6.25 = x2 + 5x + 6.25 Here comes the coolest part of the whole operation. Because we know that x2 + 5x =9 (look back a few lines , you'll find it written there!) we can swap the x2 +5x on the RHS for 9. Now we've got our new equation (x +2.5)2 = 9 + 6.25 Then a quick little sum gives us... (x + 2·5)2 = 15·25 We now take the square root of both sides, and the clever bit is that square roots can be + or - Here we get: x + 2·5 = + sqrt(15.25) OR x + 2.5 = - sqrt(15.25) Grab a calculator to work out the square root... x + 2·5 = + 3.905 OR x + 2.5 = - 3.905 And then when you take away the 2·5 from both sides you get the two solutions x = +3·905 -2.5 = 1·405 OR x = -3·905 -2.5 = - 6·405 And that's it! The two answers to x2 + 5x - 9 = 0 are x = + 1.405 OR x = - 6.405
Here's another one just to make sure you've got it. To make it more exciting this one has an x2 coefficient!
We're going to solve this little baby... 3x2 - 11x - 8 = 0 Move the constant over 3x2 - 11x = 8 Before we go on, we divide everything by the x2 coefficient because we want the x2 by itself x2 - 11x/3 = 8/3 Here's where we start making our new equation... Make the LHS into a square using steps 1,2 and 3 as before. (So divide through by x, then divide the constant by 2, then put the answer in a bracket and square it.) (x - 11/6)2 Multiply the square out (x - 11/6)2 = x2 - 11x/6 - 11x/6 +121/36 = x2 - 11x/3 + 121/36 Now it's time to play the cool little trick! We know from before that x2 - 11x/3 = 8/3 so we put this into the RHS... ... and here's the new equation: (x - 11/6)2 = 8/3 + 121/36 At this point we'll make everything into decimals: (x - 1.833)2 = 2.667 + 3.361 = 6.028 Take square roots of both sides x - 1.833 = + sqrt(6.028) OR x - 1.833 = - sqrt(6.028) Get the calculator and work out the square root... x - 1.833 = + 2.455 OR x - 1.833 = - 2.455 If we add 1.833 to both sides we get you get the two answers: x = +1·833 + 2·455 = 4·288 OR x = +1·833 -2.455 = - 0·622 So if 3x2 - 11x - 8 = 0 then x = 4·288 OR x = - 0·622
### And if you don't believe it, you can always check the answers on our QUADRATIC SOLVER!
To THE PHANTOM X (the algebra book)
To the Research Lab<|endoftext|>
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Heinkel He 178
The Heinkel He 178 was the world's first jet aircraft. The He 178 was built by the German Heinkel company specifically to test fly a pioneering turbojet engine designed by Hans von Ohain. Piloted by Luftwaffe Captain Erich Warsitz, it made its first flight on August 27, 1939, one week before the outbreak of World War II, having achieved a short hop three days earlier.
The He 178 had a barrel-shaped metal fuselage, with stubby wooden wings mounted high on its sides. The aircraft utilized the conventional three-point retractable landing gear, rather than the tricycle configuration which was later adopted for other jets.
The He 178 went on to reach a maximum speed of 650 kilometers per hour (403 mph) – much faster than any piston-engined aircraft of the day. But despite its success, it failed to convince officials from the Reichsluftfahrtministerium ("Reich Aviation Ministry") that it was worthy of further development. Undeterred, Ernst Heinkel led his company in a private development of a twin-engined jet fighter, the He 280, using experience and data gained from the He 178.
The He 178 was placed in the Deutsches Technikmuseum ("German Technical Museum") in Berlin, where it was destroyed in an air raid in 1943.
|length||7.48 m (24 ft 6 in)|
|wingspan||7.20 m (23 ft 3 in)|
|height||2.10 m (6 ft 10 in)|
|wing area||9.1 m2 (98 ft2)|
|empty weight||1,620 kg (3,572 lb)|
|max takeoff weight||1,998 kg (4,405 lb)|
|engine||1 HeS 3 turbojet, 4.4 kN (992 lbf)|<|endoftext|>
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Prehistoric site found near UK's Stonehenge
(AP) -- Archaeologists have discovered a smaller prehistoric site near Britain's famous circle of standing stones at Stonehenge.
Researchers have dubbed the site "Bluehenge," after the color of the 27 Welsh stones that were laid to make up a path. The stones have disappeared, but the path of holes remains.
Researchers from Sheffield University in northern England say the new circle represents an important find. The site is about a mile (2 kilometers) away from Stonehenge, which is believed to have been built around 2500 B.C.
Bluehenge, about 80 miles (130 kilometers) southwest of London, is thought to date back to the same period, but the exact circumstances of Bluehenge's construction aren't clear.
Researchers plan to publish more information about it next year.
©2009 The Associated Press. All rights reserved. This material may not be published, broadcast, rewritten or redistributed.<|endoftext|>
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# Thread: Help, integration, find areas
1. ## Help, integration, find areas
I have attached the diagram, is this right set up for the area under the curve?
Also for the bit above the axis, the triangular bit, how would I work out this area? I dont know the equation of this line. Is it y = 4?
$\int^3_{-1} 0-x^{2} -2x -3\ dx$ this is the x-axis minus the curve.
Thank you!
Edit- just to clarify, the curve crosses the x-axis at -1 and 3, although on my diagram it looks like it doesn't
2. So the equation of the curve is;
$f(x) = x^2 - 2x - 3$
Well, when finding the area under a curve all you have to do is find the area under the top curve and then subtract the area from the bottom curve;
$
\int_{a}^{b} [f (x) - g(x)]\,dx
$
so, from -1 to 3 we see that the top curve is the x-axis and the bottom curve is f(x), then from 3 to 4 the top curve if f(x) and the bottom curve is y=0
$\int_{-1}^{3} [0 - (x^2 - 2x - 3)]\,dx + \int_{3}^{4} x^2 - 2x - 3 - (0)\,dx$
$\int_{-1}^{3} - x^2 + 2x + 3\,dx + \int_{3}^{4} x^2 - 2x - 3 \,dx$
So the equation of the curve is;
$f(x) = x^2 - 2x - 3$
Well, when finding the area under a curve all you have to do is find the area under the top curve and then subtract the area from the bottom curve;
$
\int_{a}^{b} [f (x) - g(x)]\,dx
$
so, from -1 to 3 we see that the top curve is the x-axis and the bottom curve is f(x), then from 3 to 4 the top curve if f(x) and the bottom curve is y=0
$\int_{-1}^{3} [0 - (x^2 - 2x - 3)]\,dx + \int_{3}^{4} x^2 - 2x - 3 - (0)\,dx$
$\int_{-1}^{3} - x^2 + 2x + 3\,dx + \int_{3}^{4} x^2 - 2x - 3 \,dx$
thanks so much, makes sense, as I was trying to work out the height of the triangle, and not getting anywhere.
Also what is the equation of this straight line?
Just to clarify, could we have also worked out where this straight line intersects the curve, therefore the height, and use that to find the area or the triangle and add it to the area of the curve, found by integration?
Is this a valid method aswell?
4. Originally Posted by Tweety
thanks so much, makes sense, as I was trying to work out the height of the triangle, and not getting anywhere.
Also what is the equation of this straight line?
Just to clarify, could we have also worked out where this straight line intersects the curve, therefore the height, and use that to find the area or the triangle and add it to the area of the curve, found by integration?
Is this a valid method aswell?
The area in the right of the picture is not a triangle, so you can't find the height or anything like that That area is made by a curve and a line.
I think the best way to solve this kind of problems is what sponsoredwalk showed to you<|endoftext|>
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This page uses content from Wikipedia and is licensed under CC BY-SA.
A low-pressure area, low, depression or cyclone is a region on the topographic map where the atmospheric pressure is lower than that of surrounding locations. Low-pressure systems form under areas of wind divergence that occur in the upper levels of the troposphere. The formation process of a low-pressure area is known as cyclogenesis. Within the field of meteorology, atmospheric divergence aloft occurs in two areas. The first area is on the east side of upper troughs, which form half of a Rossby wave within the Westerlies (a trough with large wavelength that extends through the troposphere). A second area of wind divergence aloft occurs ahead of embedded shortwave troughs, which are of smaller wavelength. Diverging winds aloft ahead of these troughs cause atmospheric lift within the troposphere below, which lowers surface pressures as upward motion partially counteracts the force of gravity.
Thermal lows form due to localized heating caused by greater sunshine over deserts and other land masses. Since localized areas of warm air are less dense than their surroundings, this warmer air rises, which lowers atmospheric pressure near that portion of the Earth's surface. Large-scale thermal lows over continents help drive monsoon circulations. Low-pressure areas can also form due to organized thunderstorm activity over warm water. When this occurs over the tropics in concert with the Intertropical Convergence Zone, it is known as a monsoon trough. Monsoon troughs reach their northerly extent in August and their southerly extent in February. When a convective low acquires a well-hot circulation in the tropics it is termed a tropical cyclone. Tropical cyclones can form during any month of the year globally, but can occur in either the northern or southern hemisphere during December.
Atmospheric lift will also generally produce cloud cover through adiabatic cooling once the air becomes saturated as it rises, although the low-pressure area typically brings cloudy skies, which act to minimize diurnal temperature extremes. Since clouds reflect sunlight, incoming shortwave solar radiation decreases, which causes lower temperatures during the day. At night the absorptive effect of clouds on outgoing longwave radiation, such as heat energy from the surface, allows for warmer diurnal low temperatures in all seasons. The stronger the area of low pressure, the stronger the winds experienced in its vicinity. Globally, low-pressure systems are most frequently located over the Tibetan Plateau and in the lee of the Rocky mountains. In Europe (particularly in the British Isles and Netherlands), recurring low-pressure weather systems are typically known as "depressions".
Cyclogenesis is the development and strengthening of cyclonic circulations, or low-pressure areas, within the atmosphere. Cyclogenesis is the opposite of cyclolysis, and has an anticyclonic (high-pressure system) equivalent which deals with the formation of high-pressure areas—anticyclogenesis. Cyclogenesis is an umbrella term for several different processes, all of which result in the development of some sort of cyclone. Meteorologists use the term "cyclone" where circular pressure systems flow in the direction of the Earth's rotation, which normally coincides with areas of low pressure. The largest low-pressure systems are cold-core polar cyclones and extratropical cyclones which lie on the synoptic scale. Warm-core cyclones such as tropical cyclones, mesocyclones, and polar lows lie within the smaller mesoscale. Subtropical cyclones are of intermediate size. Cyclogenesis can occur at various scales, from the microscale to the synoptic scale. Larger-scale troughs, also called Rossby waves, are synoptic in scale. Shortwave troughs embedded within the flow around larger scale troughs are smaller in scale, or mesoscale in nature. Both Rossby waves and shortwaves embedded within the flow around Rossby waves migrate equatorward of the polar cyclones located in both the Northern and Southern hemispheres. All share one important aspect, that of upward vertical motion within the troposphere. Such upward motions decrease the mass of local atmospheric columns of air, which lowers surface pressure.
Extratropical cyclones form as waves along weather fronts due to a passing by shortwave aloft or upper level jet streak[clarification needed] before occluding later in their life cycle as cold-core cyclones. Polar lows are small-scale, short-lived atmospheric low-pressure systems that occur over the ocean areas poleward of the main polar front in both the Northern and Southern Hemispheres. They are part of the larger class of mesoscale weather-systems. Polar lows can be difficult to detect using conventional weather reports and are a hazard to high-latitude operations, such as shipping and gas- and oil-platforms. They are vigorous systems that have near-surface winds of at least 17 metres per second (38 mph).
Tropical cyclones form due to latent heat driven by significant thunderstorm activity, and are warm-core with well-defined circulations. Certain criteria need to be met for their formation. In most situations, water temperatures of at least 26.5 °C (79.7 °F) are needed down to a depth of at least 50 m (160 ft); waters of this temperature cause the overlying atmosphere to be unstable enough to sustain convection and thunderstorms. Another factor is rapid cooling with height, which allows the release of the heat of condensation that powers a tropical cyclone. High humidity is needed, especially in the lower-to-mid troposphere; when there is a great deal of moisture in the atmosphere, conditions are more favorable for disturbances to develop. Low amounts of wind shear are needed, as high shear is disruptive to the storm's circulation. Lastly, a formative tropical cyclone needs a pre-existing system of disturbed weather, although without a circulation no cyclonic development will take place. Mesocyclones form as warm core cyclones over land, and can lead to tornado formation. Waterspouts can also form from mesocyclones, but more often develop from environments of high instability and low vertical wind shear.
In deserts, lack of ground and plant moisture that would normally provide evaporative cooling can lead to intense, rapid solar heating of the lower layers of air. The hot air is less dense than surrounding cooler air. This, combined with the rising of the hot air, results in a low-pressure area called a thermal low. Monsoon circulations are caused by thermal lows which form over large areas of land and their strength is driven by how land heats more quickly than the surrounding nearby ocean. This generates a steady wind blowing toward the land, bringing the moist near-surface air over the oceans with it. Similar rainfall is caused by the moist ocean-air being lifted upwards by mountains, surface heating, convergence at the surface, divergence aloft, or from storm-produced outflows at the surface. However the lifting occurs, the air cools due to expansion in lower pressure, which in turn produces condensation. In winter, the land cools off quickly, but the ocean keeps the heat longer due to its higher specific heat. The hot air over the ocean rises, creating a low-pressure area and a breeze from land to ocean while a large area of drying high pressure is formed over the land, increased by wintertime cooling. Monsoons resemble sea and land breezes, terms usually referring to the localized, diurnal (daily) cycle of circulation near coastlines everywhere, but they are much larger in scale - also stronger and seasonal.
Large polar cyclones help determine the steering of systems moving through the mid-latitudes, south of the Arctic and north of the Antarctic. The Arctic oscillation provides an index used to gauge the magnitude of this effect in the Northern Hemisphere. Extratropical cyclones tend to form east of climatological trough positions aloft near the east coast of continents, or west side of oceans. A study of extratropical cyclones in the Southern Hemisphere shows that between the 30th and 70th parallels there are an average of 37 cyclones in existence during any 6-hour period. A separate study in the Northern Hemisphere suggests that approximately 234 significant extratropical cyclones form each winter. In Europe, particularly in the United Kingdom and in the Netherlands, recurring extratropical low-pressure weather systems are typically known as depressions.[need quotation to verify] These tend to bring wet weather throughout the year. Thermal lows also occur during the summer over continental areas across the subtropics - such as the Sonoran Desert, the Mexican plateau, the Sahara, South America, and Southeast Asia. The lows are most commonly located over the Tibetan plateau and in the lee of the Rocky mountains.
Elongated areas of low pressure form at the monsoon trough or intertropical convergence zone as part of the Hadley cell circulation. Monsoon troughing in the western Pacific reaches its zenith in latitude during the late summer when the wintertime surface ridge in the opposite hemisphere is the strongest. It can reach as far as the 40th parallel in East Asia during August and 20th parallel in Australia during February. Its poleward progression is accelerated by the onset of the summer monsoon which is characterized by the development of lower air pressure over the warmest part of the various continents. The large-scale thermal lows over continents help create pressure gradients which drive monsoon circulations. In the southern hemisphere, the monsoon trough associated with the Australian monsoon reaches its most southerly latitude in February, oriented along a west-northwest/east-southeast axis. Many of the world's rainforests are associated with these climatological low-pressure systems.
Tropical cyclones generally need to form more than 555 km (345 mi) or poleward of the 5th parallel north and 5th parallel south, allowing the Coriolis effect to deflect winds blowing towards the low-pressure center and creating a circulation. Worldwide, tropical cyclone activity peaks in late summer, when the difference between temperatures aloft and sea surface temperatures is the greatest. However, each particular basin has its own seasonal patterns. On a worldwide scale, May is the least active month while September is the most active month. November is the only month that activity in all the tropical cyclone basins is possible. Nearly one-third of the world's tropical cyclones form within the western Pacific Ocean, making it the most active tropical cyclone basin on Earth.
Wind is initially accelerated from areas of high pressure to areas of low pressure. This is due to density (or temperature and moisture) differences between two air masses. Since stronger high-pressure systems contain cooler or drier air, the air mass is denser and flows towards areas that are warm or moist, which are in the vicinity of low-pressure areas in advance of their associated cold fronts. The stronger the pressure difference, or pressure gradient, between a high-pressure system and a low-pressure system, the stronger the wind. Thus, stronger areas of low pressure are associated with stronger winds.
The Coriolis force caused by the Earth's rotation is what gives winds around low-pressure areas (such as in hurricanes, cyclones, and typhoons) their counter-clockwise (anticlockwise) circulation in the northern hemisphere (as the wind moves inward and is deflected right from the center of high pressure) and clockwise circulation in the southern hemisphere (as the wind moves inward and is deflected left from the center of high pressure). A cyclone differs from a hurricane or typhoon only on the basis of location. A hurricane is a storm that occurs in the Atlantic Ocean and northeastern Pacific Ocean, a typhoon occurs in the northwestern Pacific Ocean, and a cyclone occurs in the south Pacific or Indian Ocean. Friction with land slows down the wind flowing into low-pressure systems and causes wind to flow more inward, or flowing more ageostrophically, toward their centers. A low-pressure area is commonly associated with inclement weather, while a high-pressure area is associated with light winds and fair skies. Tornados are often too small, and of too short duration, to be influenced by the Coriolis force, but may be so-influenced when arising from a low-pressure system.<|endoftext|>
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In normal conversation we use knowledge to mean:
Knowing that (facts and information)
Knowing how (the ability to do something)
Sometimes, we use the word knowledge to mean that we have some information, we know that Mary drinks lemonade, for example. When we have this type of knowledge then we are able to express it. I cannot say that I know when the Battle of Hastings took place, if I cannot, under any circumstances, say the date! This is not true of knowing how.
If I know how to swim, then when placed in the water I make certain movements and do not sink! However, I may be unable to say how, exactly, I am able to swim. Knowing how does not mean I know that ... If I cannot say the date of the Battle of Hastings, I cannot be said to know it. But if, while swimming, I cannot tell you exactly how I do it, you cannot say I don't know how to swim!
Failing to understand the above can lead us into certain fallacies. If we get instruction from the best public speaker in the world, it does not mean that because he or she can speak excellently, that they know how to instruct others. They might be able to say what they do. For example they might say how they practice. But this might work for them and not for others! A much less able public speaker, or even one who might never have spoken in public might be a much better teacher. The point is that knowing that and knowing how are two different kinds of knowledge!
In philosophy, knowing that something is the case implies that what is known is true. Can we sensibly say that someone knows something, but it isn't true? We cannot know that something is the case unless we are able to show that it is also true.
Most recent revision
3 August 1998
Copyright © 1998 Ken Ward,
All Rights Reserved.<|endoftext|>
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The Star of David (✡), is not mentioned in the Bible and was never uniquely a Jewish symbol like the Menorah or the lion of Judah. The Star of David known in Hebrew as the Shield of David or Magen David, is a generally recognized symbol of modern Jewish identity and Judaism. The star has the shape of a hexagram, the compound of two equilateral triangles.
The earliest Jewish usage of the symbol was inherited from medieval Arabic literature by Kabbalists for use in talismanic protective amulets (segulot) where it was known as a Seal of Solomon. Archaeological discoveries showed that a hexagram has been noted on a Jewish tombstone in Taranto, Apulia in Southern Italy, which may date as early as the third century CE. The Jews of Apulia were noted for their scholarship in Kabbalah, which has been connected to the use of the Star of David.
King of Bohemia Charles IV, in 1354, arranged for the Jews of Prague a red flag with both David’s shield and Solomon‘s seal, while the red flag with which the Jews met King Matthias of Hungary in the 15th century showed two pentagrams with two golden stars.
The connection of the term “Star of David” or “Shield of David” with the hexagram shape goes back to the 17th century. The term “Shield of David” is also used in the Siddur (Jewish prayer book) as a title of the God of Israel.
During the 19th century and in an attempt to imitate the influence of the Christian cross, the Jewish communities in Eastern Europe started to use this symbol more frequently and it was also adopted by the Jewish communities in the Pale of Settlement. In 1897, the star of David became representative of the worldwide Zionist community and the broader Jewish community after it was chosen as the central symbol on a flag at the First Zionist Congress.
In His service,<|endoftext|>
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Most people probably think of a shark as a swimming animal with big fins and pointy teeth. But how exactly does it fit into the Animal Kingdom? For starters, some of you may be surprised to know that sharks are indeed fish. But they really don’t look all that similar to your average gold fish, do they?
There’s a good reason for these distinct appearances: while they are both fish, the creatures shown above actually represent two very different “Classes” of animals (see our short article on Scientific Nomenclature).
What we think of as the typical fish – goldfish, minnows, salmon – are “bony fish” belonging to Class Osteichthyes. They have a skeleton that is made mostly of bone. In addition, they are agile swimmers that can move forward and backward, and they are found in rivers, lakes and seas.
What If You Were Born A Shark?
How Are Sharks Different From Other Types Of Fish?
Sharks, on the other hand, belong to an entirely different group called Class Chondrichthyes (which oddly enough are thought to have evolved from an earlier group of bony fish). Some newer classification schemes instead refer to this as Class Elasmobranchii. Mostly these are marine fish, meaning that they live in salt water. “Cartilaginous” fish or “elasmobranchs” include sharks along with rays, skates, and chimaeras (ratfish): those that lack true bone. Instead, they have a skeleton made of cartilage, reinforced with calcification (calcium builds up in the tissue, causing it to harden). Sharks can only swim forward. This is because unlike in bony fish, their pectoral fins can’t bend upwards. Many sharks have to keep swimming in order to breathe.
Among other key distinctions, cartilaginous fish have a unique type of tooth attachment and replacement. Teeth are arranged in many rows that are constantly shed and replaced. Unlike bony fish, whose males and females eject their sperm and eggs into the water to mix, these fish have internal fertilization. “Chondrichthyan” males have organs known as claspers. These are a modified part of the pelvic fins that are inserted into the female cloaca (an opening in the body), so that sperm can be placed inside the females during copulation. Compared to bony fish, Chondrichthyans mature sexually at much greater ages, produce fewer young at a time, and can have very long gestation periods.
Sharks Have Been Around For A LONG Time!
Would you believe that sharks have existed for millions of years? In fact, they evolved long before dinosaurs! The fossil record of cartilaginous fish goes back 450 million years to the Paleozoic era. Sharks themselves began to appear about 400 million years ago, during the Devonian period, which is often called “The Age of Fishes”. The prehistoric nature of sharks is just one of many things that makes them utterly fascinating creatures.
The Misunderstood Species
Unfortunately, sharks are greatly misunderstood. They have an unfair reputation for being dangerous, but for the most part are normally harmless. As the famous shark biologist Eugenie Clark once said, “It’s more dangerous to drive to the beach than it is to go into the water with sharks”.
Sharks worldwide are threatened by human activities such as overfishing, exploitation for certain body parts, habitat degradation, and sometimes, outright slaughter. A major conservation challenge is that we are missing basic information. Although many species are widespread, a lack of fisheries data makes it impossible to assess the status of most species. Sharing information about these splendid animals can only help to raise awareness of them.
Page Created By: Mike Rogers<|endoftext|>
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# If the Sides of a Parallelogram Touch a Circle (Refer Figure of Q. 7), Prove that the Parallelogram is a Rhombus - ICSE Class 10 - Mathematics
ConceptTangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments
#### Question
If the sides of a parallelogram touch a circle (refer figure of Q. 7), Prove that the parallelogram is a rhombus
#### Solution
From A, AP and AS are tangents to the circle.
Therefore, AP = AS.......(i)
Similarly, we can prove that:
BP = BQ .........(ii)
CR = CQ .........(iii)
DR = DS .........(iv)
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD.......(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ........(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution If the Sides of a Parallelogram Touch a Circle (Refer Figure of Q. 7), Prove that the Parallelogram is a Rhombus Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.
S<|endoftext|>
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# Can a sequence converge to infinity?
## Can a sequence converge to infinity?
Convergence means that the infinite limit exists If we say that a sequence converges, it means that the limit of the sequence exists as n → ∞ n\to\infty n→∞. If the limit of the sequence as n → ∞ n\to\infty n→∞ does not exist, we say that the sequence diverges.
## What is the least upper bound of a sequence?
Sequences can also be defined as functions of n. is bounded if it is bounded both above and below. Furthermore, the smallest number Na which is an upper bound of the sequence is called the least upper bound, while the largest number Nb which is a lower bound of the sequence is called the lowest upper bound.
Which of the following sequence is not bounded?
If a sequence is not bounded, it is an unbounded sequence. For example, the sequence 1/n is bounded above because 1/n≤1 for all positive integers n. It is also bounded below because 1/n≥0 for all positive integers n. Therefore, 1/n is a bounded sequence.
### Can an infinite sequence be bounded?
An infinite sequence of real numbers (in blue). This sequence is neither increasing, decreasing, convergent, nor Cauchy. It is, however, bounded.
### Can sequence converge to zero?
1 Sequences converging to zero. Definition We say that the sequence sn converges to 0 whenever the following hold: For all ϵ > 0, there exists a real number, N, such that n>N =⇒ |sn| < ϵ. Given any ϵ > 0, let N be any number.
Does 1 1 n n converge?
n=1 1 np converges if p > 1 and diverges if p ≤ 1. n=1 1 n(logn)p converges if p > 1 and diverges if p ≤ 1. n=1 an diverges.
## What is least upper and greatest lower bound?
There is a corresponding greatest-lower-bound property; an ordered set possesses the greatest-lower-bound property if and only if it also possesses the least-upper-bound property; the least-upper-bound of the set of lower bounds of a set is the greatest-lower-bound, and the greatest-lower-bound of the set of upper …
## How do you prove something is the least upper bound?
It is possible to prove the least-upper-bound property using the assumption that every Cauchy sequence of real numbers converges. Let S be a nonempty set of real numbers. If S has exactly one element, then its only element is a least upper bound.
What is bounded above sequence?
A sequence is bounded above if all its terms are less than or equal to a number K’, which is called the upper bound of the sequence. The smallest upper bound is called the supremum.
### How do you show that a sequence is bounded below?
If the sequence is both bounded below and bounded above we call the sequence bounded.
1. Note that in order for a sequence to be increasing or decreasing it must be increasing/decreasing for every n .
2. A sequence is bounded below if we can find any number m such that m≤an m ≤ a n for every n .
### How do you tell if a sequence is infinite or finite?
A sequence is finite if it has a limited number of terms and infinite if it does not. The first of the sequence is 4 and the last term is 64 . Since the sequence has a last term, it is a finite sequence. Infinite sequence: {4,8,12,16,20,24,…}<|endoftext|>
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# Similar Triangles
Updated: Nov 22, 2020
Dear Secondary Math students, Math Lobby will be covering the topic of Similar Triangles today. Similar triangles are usually tested together with Congruent Triangles in O Level Math and N Level Math examinations. Make sure to check out our earlier article on Congruent Triangles if you have not done so!
In this note, you will learn:
1) What are similar triangles?
2) Ways to prove that two triangles are similar (AA, SSS, SAS)
### 1) What are similar triangles?
When two triangles have the same shape and have corresponding sides and angles measured to be equal with each other, then they are said to be similar. When two figures are similar, the ratios of the length of their corresponding sides are said to be equal, ie.
### 2) Ways to prove that two triangles are similar (AA, SSS, SAS)
There are a few ways to prove two triangles to see whether if they are similar, which is quite identical to congruent triangles. If you haven’t check out our article on congruent triangles, you can click here!
Now, let’s go through the few ways to go about this:
### 1) Angle-Angle (AA)
When two pairs of corresponding angles are measured to be equal
*Note: Why is Angle-Angle (AA) sufficient to prove that a triangle is similar, and not Angle-Angle-Angle (AAA)?
It is because in a triangle, there are only three interior angles, if two of them in a triangle corresponds with the other two in the other triangle, it is clear that the third angle must correspond with each other in both the triangles as well!
### 2) Side-Side-Side in same proportion (SSS)
When all three pairs of the corresponding sides are equal in proportion, meaning that the ratio of all three pairs of corresponding sides must be the same
For this way of proving, an example could be: if a = 10cm, b = 6cm and c = 20cm, and if d = 5cm, then e and f must be equals to 3cm and 10cm respectively for the ratio of the corresponding sides to be the same.
### 3) Side-Angle-Side (SAS)
When two pairs of corresponding sides are equal in proportion, and the included angle is measured to be the same
Common question:
### What’s the difference between congruent triangles and similar triangles?
The difference between similar triangles and congruent triangles is that the same shape and size matters to prove for congruency, but only the shape matters to prove for similarity, size does not matter in this case. Hence, we are looking for the exact same value in congruency, but the ratio of proportion in the case of similarity.
And that’s all we have for today, students! Math Lobby hopes that after reading this article, you have a clear understanding of the definition regarding similar triangles, the numerous ways to prove the similarity of two triangles and the difference between similar triangles and congruent triangles!
As always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!
If you want to receive daily Secondary Math Tips from us,
FOLLOW our Instagram page at https://www.instagram.com/mathlobbymotivation/<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Finding the nth Term Given the Common Difference and a Term
## Sequences where difference between any two consecutive terms is constant.
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Practice Finding the nth Term Given the Common Difference and a Term
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Arithmetic Sequences and Finding the nth Term Given the Common Difference and a Term
Halley's Comet appears in the sky approximately every 76 years. The comet was first spotted in the year 1531. Find the $n^{th}$ term rule and the $10^{th}$ term for the sequence represented by this situation.
### Guidance
In this concept we will begin looking at a specific type of sequence called an arithmetic sequence . In an arithmetic sequence the difference between any two consecutive terms is constant. This constant difference is called the common difference . For example, question one in the Review Queue above is an arithmetic sequence. The difference between the first and second terms is $(5 - 3) = 2$ , the difference between the second and third terms is $(7 - 5) = 2$ and so on. We can generalize this in the equation below:
$a_n-a_{n-1}=d$ , where $a_{n-1}$ and $a_n$ represent two consecutive terms and $d$ represents the common difference.
Since the same value, the common difference, $d$ , is added to get each successive term in an arithmetic sequence we can determine the value of any term from the first term and how many time we need to add $d$ to get to the desired term as illustrated below:
Given the sequence: $22, 19, 16, 13, \ldots$ in which $a_1=22$ and $d=-3$
$a_1&=22 \ or \ 22+(1-1)(-3)=22+0=22 \\a_2&=19 \ or \ 22+(2-1)(-3)=22+(-3)=19 \\a_3&=16 \ or \ 22+(3-1)(-3)=22+(-6)=16 \\a_4&=13 \ or \ 22+(4-1)(-3)=22+(-9)=13 \\&\qquad \qquad \vdots \\a_n&=22+(n-1)(-3) \\a_n&=22-3n+3 \\a_n&=-3n+25$
Now we can generalize this into a rule for the $n^{th}$ term of any arithmetic sequence:
$a_{n}=a_1+(n-1)d$
#### Example A
Find the common difference and $n^{th}$ term rule for the arithmetic sequence: $2, 5, 8, 11 \ldots$
Solution: To find the common difference we subtract consecutive terms.
$5-2&=3 \\8-5&=3 \ ,\text{thus the common difference is} \ 3. \\11-8&=3$
Now we can put our first term and common difference into the $n^{th}$ term rule discovered above and simplify the expression.
$a_n&=2+(n-1)(3) \\&=2+3n-3 \quad ,\text{so} \ a_n=3n-1. \\&=3n-1$
#### Example B
Find the $n^{th}$ term rule and thus the $100^{th}$ term for the arithmetic sequence in which $a_1=-9$ and $d=2$ .
Solution: We have what we need to plug into the rule:
$a_n&=-9+(n-1)(2) \\&=-9+2n-2 \quad , \text{thus the} \ n^{th} \ \text{term rule is} \ a_n=2n-11. \\&=2n-11$
Now to find the $100^{th}$ term we can use our rule and replace $n$ with 100: $a_{100}=2(100)-11=200-11=189$ .
#### Example C
Find the $n^{th}$ term rule and thus the $100^{th}$ term for the arithmetic sequence in which $a_3=8$ and $d=7$ .
Solution: This one is a little less straightforward as we will have to first determine the first term from the term we are given. To do this, we will replace $a_n$ with $a_3=8$ and use 3 for $n$ in the formula to determine the unknown first term as shown:
$a_1+(3-1)(7)&=8 \\a_1+2(7)&=8 \\a_1+14&=8 \\a_1&=-6$
Now that we have the first term and the common difference we can follow the same process used in the previous example to complete the problem.
$a_n&=-6+(n-1)(7) \\&=-6+7n-7 \quad , \text{thus} \ a_n=7n-13. \\&=7n-13$
Now we can find the $100^{th}$ term: $a_{100}=7(100)-13=687$ .
Intro Problem Revisit From the information given, we can conclude that $a_1=1531$ and $d=76$ .
We now have what we need to plug into the rule:
$a_n&=1531+(n-1)(76) \\&= 1531+76n-76 \quad , \text{thus the} \ n^{th} \ \text{term rule is} \ a_n=76n+1455$
Now to find the $10^{th}$ term we can use our rule and replace $n$ with 10: $a_{10}=76(10) + 1455=760 + 1455 =2215$ .
### Guided Practice
1. Find the common difference and the $n^{th}$ term rule for the sequence: $5, -3, -11,\ldots$
2. Write the $n^{th}$ term rule and find the $45^{th}$ term for the arithmetic sequence with $a_{10}=1$ and $d=-6$ .
3. Find the $62^{nd}$ term for the arithmetic sequence with $a_1=-7$ and $d=\frac{3}{2}$ .
#### Answers
1. The common difference is $-3-5=-8$ . Now $a_n=5+(n-1)(-8)=5-8n+8=-8n+13$ .
2. To find the first term:
$a_1+(10-1)(-6)&=1 \\a_1-54&=1 \\a_1&=55$
Find the $n^{th}$ term rule: $a_n=55+(n-1)(-6)=55-6n+6=-6n+61$ .
Finally, the $45^{th}$ term: $a_{45}=-6(45)+61=-209$ .
3. This time we will not simplify the $n^{th}$ term rule, we will just use the formula to find the $62^{rd}$ term: $a_{62}=-7+(62-1) \left(\frac{3}{2}\right)=-7+61 \left(\frac{3}{2}\right)=- \frac{14}{2}+ \frac{183}{2}= \frac{169}{2}$ .
### Vocabulary
Arithmetic Sequence
A sequence in which the difference between any two consecutive terms is constant.
Common Difference
The value of the constant difference between any two consecutive terms in an arithmetic sequence.
### Practice
Identify which of the following sequences is arithmetic. If the sequence is arithmetic find the $n^{th}$ term rule.
1. $2, 3, 4, 5, \ldots$
2. $6, 2, -1, -3, \ldots$
3. $5, 0, -5, -10, \ldots$
4. $1, 2, 4, 8, \ldots$
5. $0, 3, 6, 9, \ldots$
6. $13, 12, 11, 10, \ldots$
7. $4, -3, 2, -1, \ldots$
8. $a, a+2, a+4, a+6, \ldots$
Write the $n^{th}$ term rule for each arithmetic sequence with the given term and common difference.
1. $a_1=15$ and $d=-8$
2. $a_1=-10$ and $d= \frac{1}{2}$
3. $a_3=24$ and $d=-2$
4. $a_5=-3$ and $d=3$
5. $a_{10}=-15$ and $d=-11$
6. $a_7=32$ and $d=7$
7. $a_{n-2}=3n+2$ , find $a_n$
### Explore More
Sign in to explore more, including practice questions and solutions for Finding the nth Term Given the Common Difference and a Term.
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This Webquest / Internet scavenger hunt is a perfect one day activity for middle schoolers to learn more about Martin Luther King Jr. and his impact on society. This is a fun lesson for technology classes or history classes.
The lesson includes teacher and student resources (including links to videos), the student handout and an answer key. Two versions of the WebQuest are included to let teachers print the questions or to provide them digitally as an interactive page in an editable Microsoft PowerPoint file.
This is a great lesson to leave with a sub too!
• Teacher & Student Resources
• WebQuest questions for students – print and digital versions (editable)
• WebQuest Answer Key (non-editable, .pdf)
Total Pages: 6
Teaching Duration: 1 day
Martin Luther King Jr. Impact on Society – Social Media Simulation<|endoftext|>
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# Question #dabf9
Jan 26, 2017
The real part is $= \frac{5}{13}$
#### Explanation:
We need
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$
${i}^{2} = - 1$
The conjugate of $\left(a + i b\right)$ is $\left(a - i b\right)$
We multiply numerator and denominator by the conjugate of the denominator
$\frac{\left(4 - i\right) \left(2 - 3 i\right)}{\left(2 + 3 i\right) \left(2 - 3 i\right)} = \frac{8 - 12 i - 2 i + 3 {i}^{2}}{4 - 9 {i}^{2}}$
$= \frac{5 - 14 i}{13}$
$= \frac{5}{13} - \frac{14}{13} i$
The real part is $= \frac{5}{13}$<|endoftext|>
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Pre-Calc Exam Notes 100
# Pre-Calc Exam Notes 100 - used because straightening out...
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100 Chapter 4 Radian Measure §4.4 4.4 Circular Motion: Linear and Angular Speed distance s r θ time t = 0 time t > 0 Figure 4.4.1 Radian measure and arc length can be applied to the study of circular motion . In physics the average speed of an object is deFned as: average speed = distance traveled time elapsed So suppose that an object moves along a circle of radius r , traveling a distance s over a period of time t , as in ±igure 4.4.1. Then it makes sense to deFne the (average) linear speed ν of the object as: ν = s t (4.8) Let θ be the angle swept out by the object in that period of time. Then we deFne the (average) angular speed ω of the object as: ω = θ t (4.9) Angular speed gives the rate at which the central angle swept out by the object changes as the object moves around the circle, and it is thus measured in radians per unit time. Linear speed is measured in distance units per unit time (e.g. feet per second). The word linear is
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Unformatted text preview: used because straightening out the arc traveled by the object along the circle results in a line of the same length, so that the usual deFnition of speed as distance over time can be used. We will usually omit the word average when discussing linear and angular speed here. 4 Since the length s of the arc cut off by a central angle in a circle of radius r is s = r , we see that = s t = r t = t r , so that we get the following relation between linear and angular speed: = r (4.10) 4 Many trigonometry texts assume uniform motion, i.e. constant speeds. We do not make that assumption. Also, many texts use the word velocity instead of speed. Technically they are not the same; velocity has a direction and a magnitude, whereas speed is just a magnitude....
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## This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.
Ask a homework question - tutors are online<|endoftext|>
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Use Common Multiples to Express Fractions with the Same Denominator
## What you need to know
Things to remember:
• To add fractions, the denominators have to be the same.
• If we want to add, or subtract, two fractions with different denominators, we have to make them the same first.
• To make the denominators the same we have to find common multiples.
• We can find a common denominator by multiplying the two denominators together.
We have seen adding and subtracting fractions when the denominators are the same, all we do is add or subtract the numerators; the denominators stay the same.
$$\frac{1}{5}+\frac{2}{5}=\frac{3}{5}$$
$$\frac{9}{11}-\frac{4}{11}=\frac{5}{11}$$
But, how do we add fractions like $\frac{2}{5}+\frac{1}{3}$? Well, we need the denominators to be the same. We do this by looking for common multiples of them.
1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th
Multiples of 5: 5 10 15 20 25 30 35 40 45 50
Multiples of 2: 3 6 9 12 15 18 21 24 27 30
The common multiples in these two lists are 15 and 30, so we need to make the denominators one of these. If we look at the numbers above the multiples, it tells us what we have to multiply them by.
$$5\times3 = 15$$
$$3\times5 = 15$$
NOTE: Here, we have just multiplied the denominators together to find a common multiple!
$$5\times6 = 30$$
$$3\times10 = 30$$
Now that we’ve found the common multiples and what we multiply by, we do the same to the tops!
$$\frac{2}{5}=\frac{2\times3}{5\times3}=\frac{6}{15}$$
$$\frac{1}{3}=\frac{1\times5}{3\times5}=\frac{5}{15}$$
$$\frac{2}{5}+\frac{1}{3}= \frac{6}{15}+\frac{5}{15}=\frac{11}{15}$$
$$\frac{2}{5}=\frac{2\times6}{5\times6}=\frac{12}{30}$$
$$\frac{1}{3}=\frac{1\times10}{3\times10}=\frac{10}{30}$$
$$\frac{2}{5}+\frac{1}{3}= \frac{12}{30}+\frac{10}{30}=\frac{22}{30}$$
Hmmmmmmmmmmmm, these answers are different. Or are they? Remember, we can often simplify fractions.
$$\frac{22}{30} =\frac{22\div2}{30\div2}=\frac{11}{15}$$
Actually, these are the same! It doesn’t matter which common multiple we use, they’ll always simplify down to be the same answer!
So, really, we did this in four steps:
Step 1: Find a common multiple of the denominators.
Step 2: Multiply the fractions so that the denominators are the same.
Step 3: Add or subtract the new fractions.
Step 4: See if you can simplify.
What is $\frac{3}{4}-\frac{2}{7}$?
Step 1: Find a common multiple of the denominators.
HINT: The easiest way to do this is multiply the denominators together.
$$4\times7=28$$
So, we are making the denominators into 28.
Step 2: Multiply the fractions so that the denominators the same.
$$\frac{3}{4}=\frac{3\times7}{4\times7}=\frac{21}{28}$$
$$\frac{2}{7}=\frac{2\times4}{7\times4}=\frac{8}{28}$$
Step 3: Add or subtract the new fractions.
$$\frac{3}{4}-\frac{2}{7}=\frac{21}{28}-\frac{8}{28}=\frac{13}{28}$$
Step 4: See if you can simplify.
We can’t simplify, so this is our final answer!
## Times Table Flash Cards
(30 Reviews) £8.99 £5.99
## Example Questions
Step 1: Find a common multiple of the denominators.
HINT: The easiest way to do this is multiply the denominators together.
$$2\times9=18$$
So, we are making the denominators into 18.
Step 2: Multiply the fractions so that the denominators the same.
$$\frac{1}{2}=\frac{1\times9}{2\times9}=\frac{9}{18}$$
$$\frac{3}{9}=\frac{3\times2}{9\times2}=\frac{6}{18}$$
Step 3: Add or subtract the new fractions.
$$\frac{1}{2}+\frac{3}{9}=\frac{9}{18}+\frac{6}{18}=\frac{15}{18}$$
Step 4: See if you can simplify.
We can simplify this one!
$$\frac{15}{18} =\frac{15\div3}{18\div3}=\frac{5}{6}$$
$$\frac{1}{2}+\frac{3}{9}= \frac{5}{6}$$
Step 1: Find a common multiple of the denominators.
HINT: The easiest way to do this is multiply the denominators together.
$$7\times12=84$$
So, we are making the denominators into 84.
Step 2: Multiply the fractions so that the denominators the same.
$$\frac{5}{7}=\frac{5\times12}{7\times12}=\frac{60}{84}$$
$$\frac{1}{12}=\frac{1\times7}{12\times7}=\frac{7}{84}$$
Step 3: Add or subtract the new fractions.
$$\frac{5}{7}-\frac{1}{12}=\frac{60}{84}-\frac{7}{84}=\frac{53}{84}$$
Step 4: See if you can simplify.
This can’t be simplified, so Step 3 gave us our answer.
$$\frac{1}{2}+\frac{3}{9}= \frac{5}{6}$$
## KS2 SATs Flash Cards
(43 Reviews) £8.99
• All of the major KS2 Maths SATs topics covered
• Practice questions and answers on every topic<|endoftext|>
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Fish are vertebrates, just like humans and many other animals. Like humans, they have a well-developed central nervous system and brains which regulate and register fear and pain. In experiments, fish have been shown to respond to negative stimuli so we know that fish suffer when they are placed on ice when alive or are removed from the water. Just like any animal in the food industry, fish deserve the least stressful and painful, and the most humane method of slaughter.
HOW FISH ARE TREATED
Most people still believe that fish do not feel fear and pain. We now know that this is not true. Nevertheless, fish are treated badly, from the capture methods at sea to the treatment they receive on fishing vessels. Thrown about on board they are then put on ice and slowly asphyxiate to death in extreme stress and pain. Scientific findings published in Worse things happen at sea: the welfare of wild-caught fish (2010) estimate that, depending on the species, fish take between 55 and 250 minutes to lose consciousness during which they feel pain and fear and are aware of their surroundings. Research is currently underway to find more humane methods of slaughter.
ANIMAL WELFARE ON YOUR PLATE
The debate about the environmental impact of the meat industry has grown over the last few years. More and more people are starting to realise the meat industry’s ecological impact. The media and various campaigns are drawing attention to this issue, as are international celebrities. They all plea for reducing meat consumption and all provide healthy, environmentally friendly, alternatives.
These days, food is more than what is on your plate. Food has become a concept that includes ethics, the environment and nature. But this concept is largely restricted to terrestrial animals. Marine animals need to be included in this picture as the issues are the same as with terrestrial animals – animal welfare, contamination of meat, climate change, species extinction. Fortunately, conservation and animal welfare agencies are starting to have an impact.
NEUROANATOMY: the BASICS
All vertebrates such as mammals, birds and fish have the same general brain structure. Like the others, fish have a forebrain, midbrain and hindbrain. The only difference between fish and other vertebrates is that their brains are smaller in relation to their body size and the structure is less complex.
Mammals’ cerebral cortex has six layers, of which the neocortex is associated with higher functions such as sensory perception and awareness. Because fish do not have a neocortex, it is assumed that they are unable to feel pain. But the Panel on Animal Health and Welfare contradicts this as one brain function can be performed by different structures in different families of animals. For example, birds and fish process visual stimuli differently. Equally, dolphin brains work completely differently from primate brains while they both have great cognitive abilities.
It can thus be assumed that fish brains have developed ‘functional analogue’ structures which, like higher order vertebrates, regulate fear and pain. Analog organs perform the same functions, and they have been found in fish. Tests show that parts of the brain of certain fish species react when the fish receive pain triggers, and structures in the forebrain of bony fish have the same function as the amygdala and hippocampus in mammals which are related to emotions, fear and memory.
Similarly, fish too have a thalamus, which receives information from the sensory organs and transmits it to the cerebral cortex. It is an important node in the brain that all vertebrates have and which makes them aware of their surroundings and what happens to them.
As the study of fish brains is so recent, there is still so much to discover and it means that some researchers are sceptical about the findings to date. Much of their scepticisms rests on what ‘pain’ actually is. They believe that responding to stimuli is purely nociception – an automatic, unconscious response rather than pain. In contrast, the International Association for the Study of Pain defines pain as ‘a conscious sensory and emotional experience associated with actual or potential tissue damage, or described in terms of such damage’.
To draw a parallel, how can we scientifically prove that human foetuses experience pain in the womb? There was a time when foetuses and newly born babies underwent medical treatment without anaesthetics because their experience to pain was not proved beyond a doubt. We now think differently and take precautionary measures just in case babies do feel pain. Should we not take the same precautionary measures in the fishing industry?
Text: Danny Haelewaters
- Mood A, 2010. Worse things happen at sea: the welfare of wild-caught fish. Fishcount.org.uk, 2010: 1-139.
- Van de Vis JW & E Lambooij, 2004. Ontwikkelen van een welzijnsvriendelijke slachtmethode voor de Afrikaanse meerval (Clarias gariepinus). RIVO Rapport C035/04, Nederlands Instituut voor Visserijonderzoek (RIVO), Ijmuiden (Nederland): 1-44.
- Rose JD, 2002. The Neurobehavioral Nature of Fishes and the Question of Awareness and Pain. Reviews in Fisheries Science 10 (1): 1-38.
- AHAW, 2009. General approach to fish welfare and to the concept of scentience in fish. The EFSA Journal 954: 1-26.
- Dunlop R & P Laming, 2005. Mechanoreceptive and nociceptive responses in the central nervous system of goldfish (Carassius autatus) and trout (Oncorhynchus mykiss). Journal of Pain 6 (9): 561-568.
- Portavella M, Vargas JP, Torres B & C Salas, 2002. The effects of telencephalic pallial lesions on spatial, temporal, and emotional learning in goldfish. Brain Research Bulletin 57 (3-4): 397-399.
- Sneddon LU, Braithaite VA & MJ Gentle, 2003. Do fishes have nociceptors: evidence for the evolution of a vertebrate sensory system. Proceedings of the Royal Society: Biological Sciences 270 (1520): 1115-1121.
- Braithwaite VA, 2010. Do fish feel pain? Oxford University Press, New York: 1-194.
- Ethisch Vegetarisch Alternatief (EVA) vzw<|endoftext|>
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# Area & Perimeter of Circles and Solving Number Puzzles
Area & Perimeter of Circles
While the knowledge of formulae is not expected at year six, or at least not as of now as I write this, there is every chance that an independent school will throw it in as a way of seeing how the better pupils deal with a complex question.
Firstly, let's name the elements of a circle that will get mentioned.
The line going from one side to the other is the diameter. The line from the edge to the middle is the radius. As the outside of the circle is always the same distance from the centre, the diameter is double the length of the radius.
PERIMETER: The basic idea is to multiply the diameter of the circle by 3.14, a number known as pi, or π. The precise value of pi is endless and sometimes your child will be asked to assume it is 3.14 or even just 3.
The formula for the perimeter of the circle is: π x d or 2 x π x r (as diameter is the same as two lots of the radius)
If we use 3.14 to represent π then a circle of radius 4m will have a perimeter of:
2 x 3.14 x 4 = 25.12 metres.
AREA: The area is always shown in square units and is calculated by multiplying the radius by itself and π. If you know only the diameter, halve it to get the radius and then use the formula:
π x r x r or π r 2
Assuming π to be 3.14, a circle with radius 4m will have an area of:
3.14 x 4 x 4 = 50.24 m2
Solving Number Puzzles
There are many different questions that examiners can dream up but a common number puzzle that appears in SATs is one where a series of numbers is provided and a child has to arrange them to make sense as a sum.
For instance: Use the numbers 5, 7, 8 and 2 in any order to fill the boxes below. What order should they be put in to create the highest answer?
Children may be expected to use trial and error to formulate an answer but there are some shortcuts.
When presented with a division sum, always divide by the lowest number possible to give the highest answer. The lowest number available would be 2.5. The highest number that could be created is 8.7 so the sum which gives the highest answer has to be 8.7 ÷ 2.5. Note that we have not worked out any sums at all and only need to have a cursory glance to check that our answer isn't silly.
Another puzzle might involve finding which numbers are missing from a sum, thus:
The idea of trial and error on this one is redundant as there are mathematical shortcuts here.
Firstly, the “units” column is critical. Something take away one equals nine. This must be ten, so the number in the top box must be zero. Next, as we now have '541 - something = 79' we can simply take 79 from 541 to give the answer 461.
There are often shortcuts but, should all else fail and a child has no way in to the question, trial and error gives a starting point.
Quiz yourself clever - 3 free quizzes in every section<|endoftext|>
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Tech moves fast! Stay ahead of the curve with Techopedia!
Join nearly 200,000 subscribers who receive actionable tech insights from Techopedia.
A Request for Comments (RFC) is a formal document drafted by the Internet Engineering Task Force (IETF) that describes the specifications for a particular technology. When an RFC is ratified, it becomes a formal standards document.
RFCs were first used during the creation of the ARPANET protocols that came to establish what became today's Internet. They continue to be issued on an ongoing basis as the technology underlying the Internet evolves.
A formal Internet standard is formed when an RFC goes through committee drafting and review until the final version of the RFC is ratified, at which time no further comments or changes are allowed. Other RFCs are not ratified, and instead retain an "informational" or "experimental" status. For example, the original File Transfer Protocol standard was published as RFC 114 in April of 1971. This was later replaced by RFC 765 in 1980 and finally RFC 959. So, as technology advances, RFCs are updated as well.<|endoftext|>
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1775 – The U.S. Army was founded when the Continental Congress first authorized the muster of troops under its sponsorship. Also the birth of the Infantry Branch. Ten companies of riflemen were authorized by a resolution of the Continental Congress on June 14, 1775. However, the oldest Regular Army infantry regiment, the 3d, was constituted on June 3, 1784, as the First American Regiment.
1777 – During the American Revolution, the Continental Congress adopts a resolution stating that “the flag of the United States be thirteen alternate stripes red and white” and that “the Union be thirteen stars, white in a blue field, representing a new Constellation.” The national flag, which became known as the “Stars and Stripes,” was based on the “Grand Union” flag, a banner carried by the Continental Army in 1776 that also consisted of 13 red and white stripes. According to legend, Philadelphia seamstress Betsy Ross designed the new canton for the Stars and Stripes, which consisted of a circle of 13 stars and a blue background, at the request of General George Washington. Historians have been unable to conclusively prove or disprove this legend. With the entrance of new states into the United States after independence, new stripes and stars were added to represent new additions to the Union. In 1818, however, Congress enacted a law stipulating that the 13 original stripes be restored and that only stars be added to represent new states. On June 14, 1877, the first Flag Day observance was held on the 100th anniversary of the adoption of the Stars and Stripes. As instructed by Congress, the U.S. flag was flown from all public buildings across the country. In the years after the first Flag Day, several states continued to observe the anniversary, and in 1949 Congress officially designated June 14 as Flag Day, a national day of observance.
1777 – John Paul Jones takes command of USS Ranger.
1801 – Former American Revolutionary War General Benedict Arnold died in London.
1805 – Robert Anderson (d.1871), Bvt. Major General (Union Army), defender of Ft. Sumpter, was born.
1846 – Anticipating the outbreak of war with Mexico, American settlers in California rebel against the Mexican government and proclaim the short-lived California Republic. The political situation in California was tense in 1846. Though nominally controlled by Mexico, California was home to only a relatively small number of Mexican settlers. Former citizens of the United States made up the largest segment of the California population, and their numbers were quickly growing. Mexican leaders worried that many American settlers were not truly interested in becoming Mexican subjects and would soon push for annexation of California to the United States. For their part, the Americans distrusted their Mexican leaders. When rumors of an impending war between the U.S. and Mexico reached California, many Americans feared the Mexicans might make a preemptive attack to forestall rebellion. In the spring of 1846, the American army officer and explorer John C. Fremont arrived at Sutter’s Fort (near modern-day Sacramento) with a small corps of soldiers. Whether or not Fremont had been specifically ordered to encourage an American rebellion is unclear. Ostensibly, Fremont and his men were in the area strictly for the purposes of making a scientific survey. The brash young officer, however, began to persuade a motley mix of American settlers and adventurers to form militias and prepare for a rebellion against Mexico. Emboldened by Fremont’s encouragement, on this day in 1846 a party of 33 Americans under the leadership of Ezekiel Merritt and William Ide invaded the largely defenseless Mexican outpost of Sonoma just north of San Francisco. Fremont and his soldiers did not participate, though he had given his tacit approval of the attack. Merritt and his men surrounded the home of the retired Mexican general, Mariano Vallejo, and informed him that he was a prisoner of war. Vallejo, who was actually a strong supporter of American annexation, was more puzzled than alarmed by the rebels. He invited Merritt and a few of the other men into his home to discuss the situation over brandy. After several hours passed, Ide went in and spoiled what had turned into pleasant chat by arresting Vallejo and his family. Having won a bloodless victory at Sonoma, Merritt and Ide then proceeded to declare California an independent republic. With a cotton sheet and some red paint, they constructed a makeshift flag with a crude drawing of a grizzly bear, a lone red star (a reference to the earlier Lone Star Republic of Texas), and the words “California Republic” at the bottom. From then on, the independence movement was known as the Bear Flag Revolt. After the rebels won a few minor skirmishes with Mexican forces, Fremont officially took command of the “Bear Flaggers” and occupied the unguarded presidio of San Francisco on July 1. Six days later, Fremont learned that American forces under Commodore John D. Sloat had taken Monterey without a fight and officially raised the American flag over California. Since the ultimate goal of the Bear Flaggers was to make California part of the U.S., they now saw little reason to preserve their “government.” Three weeks after it had been proclaimed, the California Republic quietly faded away. Ironically, the Bear Flag itself proved far more enduring than the republic it represented: it became the official state flag when California joined the union in 1850.
1847 – Commodore Matthew Perry launches amphibious river operations by Sailors and Marines on Tabasco River, Mexico.
1863 – President Lincoln authorized the Secretary of the Treasury to “cooperate by the revenue cutters under your direction with the Navy in arresting rebel depredations on American commerce and transportation and in capturing rebels engaged therein.” The directive was largely the result of Lieutenant Read’s continued raid on Union commerce near Northern shores.
1863 – A small Union garrison in the Shenandoah Valley town of Winchester, Virginia, is easily defeated by the Army of Northern Virginia on the path of the Confederate invasion of Pennsylvania. In early June, General Robert E. Lee’s Army of Northern Virginia began an invasion of the North. Lee’s men pulled out of defenses along the Rappahannock River and swung north and west into the Shenandoah Valley. Using the Blue Ridge Mountains as a screen, the Confederates worked their way northward with little opposition. General Joseph Hooker, commander of the Army of the Potomac, was unsure of the Confederates’ intentions. He tracked Lee’s army from a distance, staying safely away to protect Washington, D.C. During this time, Winchester was in Union hands. The city was literally at the crossroads of the war, so it changed hands continually. Robert Milroy, the commander of the Yankees in Winchester, was unaware that the vanguard of Lee’s army was heading his way. He had received some warnings from Washington, but an order to evacuate Winchester did not reach him because the Confederates had cut the telegraph lines. As late as June 11, Milroy bragged that he could hold the town against any Confederate force. His assertion was rendered ridiculous when Richard Ewell’s Rebel corps crashed down on his tiny garrison. Ewell’s force quickly surrounded the Yankee’s. After a sharp battle, Ewell captured about 4,000 Federals, while Milroy and 2,700 soldiers escaped to safety. Ewell lost just 270 men but captured 300 wagons, hundreds of horses, and 23 artillery pieces. Milroy was relieved of his command and later arrested, although a court of inquiry found that he was not culpable in the disaster.
1864 – At the Battle of Pine Mountain, Georgia, Confederate General Leonidas Polk was killed by a Union shell.
1864 – U.S.S. Kearsarge, Captain Winslow, arrived off Cherbourg, France. The ship log recorded: “Found the rebel privateer Alabama lying at anchor in the roads.” Kearsarge took up the blockade in international waters off the harbor entrance. Captain Semmes stated: “. . . My intention is to fight the Kearsarge as soon as I can make the necessary arrangements. I hope they will not detain me more than until tomorrow evening, or after the morrow morning at furthest. I beg she will not depart before I am ready to go out.” With the famous Confederate raider at bay, Kearsarge had no intention of departing-the stage was set for the famous duel.
1898 – Two companies of Marines defeated the Spanish near Guantanamo Bay, Cuba.
1900 – US Congress passed a law granting citizenship to all persons who had been citizens of the Republic of Hawaii at the time of annexation.
1922 – Warren G. Harding became the first president heard on radio, as Baltimore station WEAR broadcast his speech dedicating the Francis Scott Key memorial at Fort McHenry.
1927 – President Porfirio Diaz of Nicaragua signed a treaty with the U.S. allowing American intervention in his country.
1932 – Representative Edward Eslick died on the floor of the House of Representatives while pleading for the passage of the bonus bill.
1940 – In German-occupied Poland the first inmates arrived at Auschwitz concentration camp. They were all Polish political prisoners. The Nazis opened their concentration camp at Auschwitz.
1941 – The CGC Duane rescued 46 survivors from the torpedoed SS Tresillian.
1941 – President Roosevelt freezes all German and Italian assets in the United States.
1942 – The first bazooka rocket gun, produced in Bridgeport, Ct., demolished a tank from its shoulder-held position.
1944 – The first raid by American B-29 Superfortress bombers is carried out. A total of 48 planes (of which 4 are lost) make an ineffective strike on the Yawata iron and steel works during the night from bases in China.
1944 – US naval forces conduct bombardments of Saipan and Tinian in preparation for landings on these islands. The two American naval groups, commanded by Admiral Ainsworth and Admiral Oldendorf, include 7 battleships and 11 cruisers as well as 8 escort carriers in support. The battleship USS California is hit by a Japanese shore battery. Extensive mine-sweeping operations are also conducted by American forces.
1944 – A third corps, the US 19th Corps, is becomes operational between the 5th and 7th Corps. Free French leader, General de Gaulle, visits the beachhead and takes steps to restoring French civilian government in captured territory.
1945 – Gen. Dwight D. Eisenhower was honored as a Companion of the Liberation by Gen. Charles de Gaulle.
1945 – On Okinawa, mopping up operations proceed on the Oroku peninsula. The troops of the US 3rd Amphibious Corps and the US 24th Corps continue to eliminate fortified caves held by Japanese forces on Kunishi Ridge and on Mount Yuza and Mount Yaegu. An American regiment of the US 96th Division reaches the summit of Mount Yaegu, while the US th Division extends its control of Hills 153 and 115.
1945 – On Luzon, American forces dislodge the Japanese blocking the Orioung Pass. Elements of the US 37th Division, formed into an armored column, advance as far as Echague. From Santiago, other units advance toward Cabanatuan and Cauayan.
1945 – The US Joint Chiefs of Staff issue a directive to General MacArthur, General Arnold and Admiral Nimitz to prepare plans for the immediate occupation of the Japanese islands in the event of a sudden capitulation. This order may have been given in light of recent progress on the production of an atomic bomb but this is not stated.
1951 – A single communist Polikarpov PO-2 biplane dropped bombs on Suwon Airfield and another PO-2 bombed a motor pool at Inchon. These attacks marked the beginning of enemy night harassing missions that soon became known as “Bedcheck Charley.”
1951 – The destroyer-minesweeper USS Thompson was hit by communist shore battery fire suffering three sailors killed and three wounded. Having sustained 13 hits, the Thompson barely managed to escape out of range of the North Korean guns.
1951 – U.S. Census Bureau dedicates UNIVAC, the world’s first commercially produced electronic digital computer. UNIVAC, which stood for Universal Automatic Computer, was developed by J. Presper Eckert and John Mauchly, makers of ENIAC, the first general-purpose electronic digital computer. These giant computers, which used thousands of vacuum tubes for computation, were the forerunners of today’s digital computers. The search for mechanical devices to aid computation began in ancient times. The abacus, developed in various forms by the Babylonians, Chinese, and Romans, was by definition the first digital computer because it calculated values by using digits. A mechanical digital calculating machine was built in France in 1642, but a 19th century Englishman, Charles Babbage, is credited with devising most of the principles on which modern computers are based. His “Analytical Engine,” begun in the 1830s and never completed for lack of funds, was based on a mechanical loom and would have been the first programmable computer. By the 1920s, companies such as the International Business Machines Corporation (IBM) were supplying governments and businesses with complex punch-card tabulating systems, but these mechanical devises had only a fraction of the calculating power of the first electronic digital computer, the Atanasoff-Berry Computer (ABC). Completed by John Atanasoff of Iowa State in 1939, the ABC could by 1941 solve up to 29 simultaneous equations with 29 variables. Influenced by Atanasoff’s work, Presper Eckert and John Mauchly set about building the first general-purpose electronic digital computer in 1943. The sponsor was the U.S. Army Ordnance Department, which wanted a better way of calculating artillery firing tables, and the work was done at the University of Pennsylvania. ENIAC, which stood for Electronic Numerical Integrator and Calculator, was completed in 1946 at a cost of nearly $500,000. It took up 15,000 feet, employed 17,000 vacuum tubes, and was programmed by plugging and replugging some 6,000 switches. It was first used in a calculation for Los Alamos Laboratories in December 1945, and in February 1946 it was formally dedicated. Following the success of ENIAC, Eckert and Mauchly decided to go into private business and founded the Eckert-Mauchly Computer Corporation. They proved less able businessmen than they were engineers, and in 1950 their struggling company was acquired by Remington Rand, an office equipment company. On June 14, 1951, Remington Rand delivered its first computer, UNIVAC I, to the U.S. Census Bureau. It weighed 16,000 pounds, used 5,000 vacuum tubes, and could perform about 1,000 calculations per second. On November 4, 1952, the UNIVAC achieved national fame when it correctly predicted Dwight D. Eisenhower’s unexpected landslide victory in the presidential election after only a tiny percentage of the votes were in. UNIVAC and other first-generation computers were replaced by transistor computers of the late 1950s, which were smaller, used less power, and could perform nearly a thousand times more operations per second. These were, in turn, supplanted by the integrated-circuit machines of the mid-1960s and 1970s. In the 1980s, the development of the microprocessor made possible small, powerful computers such as the personal computer, and more recently the laptop and hand-held computers.
1952 – The USS Nautilus, the first atomic submarine, was dedicated in Groton, Connecticut.
1954 – Over 12 million Americans “die” in a mock nuclear attack, as the United States goes through its first nationwide civil defense drill. Though American officials were satisfied with the results of the drill, the event stood as a stark reminder that the United States–and the world-was now living under a nuclear shadow. The June 1954 civil defense drill was organized and evaluated by the Civil Defense Administration, and included operations in 54 cities in the United States, Puerto Rico, the Virgin Islands, Alaska, and Hawaii. Canada also participated in the exercise. The basic premise of the drill was that the United States was under massive nuclear assault from both aircraft and submarines, and that most major urban areas had been targeted. At 10 a.m., alarms were sounded in selected cities, at which time all citizens were supposed to get off the streets, seek shelter, and prepare for the onslaught. Each citizen was supposed to know where the closest fallout shelter was located; these included the basements of government buildings and schools, underground subway tunnels, and private shelters. Even President Dwight D. Eisenhower took part in the show, heading to an underground bunker in Washington, D.C. The entire drill lasted only about 10 minutes, at which time an all-clear signal was broadcast and life returned to normal. Civil Defense Administration officials estimated that New York City would suffer the most in such an attack, losing over 2 million people. Other cities, including Washington, D.C., would also endure massive loss of life. In all, it was estimated that over 12 million Americans would die in an attack. Despite those rather mind-numbing figures, government officials pronounced themselves very pleased with the drill. Minor problems in communication occurred, and one woman in New York City managed to create a massive traffic jam by simply stopping her car in the middle of the road, leaping out, and running for cover. In most cities, however, the streets were deserted just moments after the alarms sounded and there were no signs of panic or criminal behavior. A more cautious assessment came from a retired military officer, who observed that the recent development of the hydrogen bomb by the Soviet Union had “outstripped the progress made in our civil defense strides to defend against it.”
1954 – President Eisenhower signed an order adding the words “under God” to the Pledge of Allegiance.
1964 – General Westmoreland is in Malaysia to study the methods used by the British to defeat the Communist guerrillas there.
1964 – The US military allows its own pilots operating out of Thailand to hit targets of opportunity in Laos.
1967 – The space probe Mariner 5 was launched from Cape Kennedy on a flight that took it past Venus.
1968 – A Federal District Court jury in Boston convicts Dr. Benjamin Spock of conspiring to aid draft registrants in violating the Selective Service Law.
1969 – The U.S. announces that three combat units will be withdrawn from Vietnam. They were the 1st and 2nd Brigades of the U.S. Army 9th Infantry Division and Regimental Landing Team 9 of the 3rd Marine Division–a total of about 13,000 to 14,000 men. These troops were part of the first U.S. troop withdrawal, which had been announced on June 8 by President Richard Nixon at the Midway conference with South Vietnamese President Nguyen Van Thieu. Nixon had promised that 25,000 troops would be withdrawn by the end of the year, and more support troops were later sent home in addition to the aforementioned combat forces in order to meet that number.
1972 – US planes flying a record number of strikes over North Vietnam, 340, sever the main railway line between Hanoi and Haiphong.
1985 – TWA Flight 847 from Athens to Rome is hijacked by Shiite Hezbollah terrorists who immediately demand to know the identity of ”those with Jewish-sounding names.” Two of the Lebanese terrorists armed with grenades, axes, and a 9-mm. pistol then forced the plane to land at the Beirut Airport in Lebanon. Once on the ground, the hijackers called for passengers with Israeli passports, but there were none. Nor were there any diplomats on board. They then focused their attention on the several U.S. Navy construction divers aboard the plane. Soon after landing, the terrorists killed Navy diver Robert Stethem, and dumped his body on the runway. TWA employee Uli Dickerson was largely successful in protecting the few Jewish passengers aboard by refusing to identify them. Most of the passengers were released in the early hours of what turned out to be a 17-day ordeal, but five men were singled out and separated from the rest of the hostages. Of these five, only Richard Herzberg, an American, was Jewish. Luckily, he had told his wife early on in the attack to get rid of a ring bearing a Hebrew inscription. During the next two weeks, Herzberg maintained to his attackers that he was a Lutheran of German and Greek ancestry. Along with the others, he was taken to a roach-infested holding cell somewhere in Beirut, where other Lebanese prisoners were being held and tortured in the makeshift prison. Fortunately, however, the TWA hostages were treated fairly well-they were even given a birthday cake on one occasion. On June 30, after careful negotiations, the hostages were released unharmed. Since the terrorists were effectively outside the law’s reach in Lebanon, it appeared as though the terrorists would go free from punishment. Yet, Mohammed Ali Hamadi, who was wanted for his role in TWA Flight 847 attack, was arrested nearly two years later at the airport in Frankfurt, Germany, with explosives. Within days of his arrest, two German citizens were kidnapped while in Lebanon in a successful attempt to discourage Germany from extraditing Hamadi to the United States for prosecution. Germany decided to try Hamadi instead, and he was convicted and sentenced to life in prison, the maximum penalty under German law. Although both German hostages had already been released by this time, three additional German citizens were kidnapped on the day of the verdict.
1991 – The space shuttle “Columbia” returned from a medical research mission.
1999 – About 15,000 NATO peacekeepers spread out across Kosovo, including a convoy of about 1200 US Marines.
2000 – In Florida George Trofimoff (73) was arrested for spying for the Soviet KGB from 1969-1995. He had served as chief of an Army unit responsible for interviewing Warsaw pact defectors.
2000 – US federal marine specialists reported that the US Navy induced underwater noise caused the death of at least a dozen whales in the Bahamas in March. Hemorrhages were found around the animals’ ears.
2000 – Pres. Kim Jong Il of North Korea and Pres. Kim Dae Jung of South Korea pledged concrete steps toward unifying their divided peninsula and signed an agreement to allow visits for some families separated for the last five decades.
2001 – Pres. Bush ordered a stop to the Navy bombing exercises on Puerto Rico’s Vieques Island. Cleanup was estimated to cost hundreds of millions and take decades. Bombing practice was set to stop by May, 2003.
2001 – Macedonia asked for Nato troops the help disarm ethnic Albanian rebels. Nato Sec. Gen. Lord Robertson ruled out military intervention.
2002 – The US became officially free from a 1972 treaty that banned major missile defenses. In Alaska work was set to begin on missile interceptors.
2002 – In Afghanistan Pres. Hamid Karzai outlined a list of national priorities that included building a national army and police force, improving schools and health care and creating jobs.
2002 – In Pakistan suicide bomber blew up a truck at the US consulate in Karachi killed 14 people and injured many more. No Americans were believed killed. The Bush administration planned to evaluate how many U.S. personnel should be kept in Pakistan. The Lashkar-e-Omar coalition, formed in January, was blamed.
2004 – The US military released hundreds of prisoners from Abu Ghraib prison.
Congressional Medal of Honor Citations for Actions Taken This Day
DURHAM, JAMES R.
Rank and organization: Second Lieutenant, Company E, 12th West Virginia Infantry. Place and date: At Winchester, Va., 14 June 1863. Entered service at: Clarksburg, W. Va. Born: 7 February 1833, Richmond, W. Va. Date of issue: 6 March 1890. Citation: Led his command over the stone wall, where he was wounded.
Rank and organization: Private, Company H, 28th Connecticut Infantry. Place and date: At Port Hudson, La., 14 June 1863. Entered service at: Greenwich, Conn. Birth: ——. Date of issue: 1 April 1898. Citation: Made 2 trips across an open space, in the face of the enemy’s concentrated fire, and secured water for the sick and wounded.
LOVERING, GEORGE M.
Rank and organization: First Sergeant, Company I, 4th Massachusetts Infantry. Place and date: At Port Hudson, La., 14 June 1863. Entered service at: East Randolph, Mass. Born: 10 January 1832, Springfield, N.H. Date of issue: 19 November 1891. Citation: During a momentary confusion in the ranks caused by other troops rushing upon the regiment, this soldier, with coolness and determination, rendered efficient aid in preventing a panic among the troops.
PATTERSON, JOHN T.
Rank and organization: Principal Musician, 122d Ohio Infantry. Place and date: At Winchester, Va., 14 June 1863. Entered service at: ——. Birth: Morgan County, Ohio. Date of issue: 13 May 1899 Citation: With one companion, voluntarily went in front of the Union line, under a heavy fire from the enemy, and carried back a helpless wounded comrade, thus saving him from death or capture.
Rank and organization: Private, Company C, 122d Ohio Infantry. Place and date: At Winchester, Va., 14 June 1863. Entered service at: ——. Birth: Morgan County, Ohio. Date of issue: 5 April 1898. Citation: With 1 companion, voluntarily went in front of the Union line, under a heavy fire from the enemy, and carried back a helpless, wounded comrade, thus saving him from death or capture.
Rank and organization: Private, U.S. Marine Corps. Born: 17 March 1873, Limerick, Ireland. Accredited to: New York. G.O. No.: 92, 8 December 1910. Citation: For heroism and gallantry in action at Cuzco, Cuba, 14 June 1898.
QUICK, JOHN HENRY
Rank and organization: Sergeant, U.S. Marine Corps. Born: 20 June 1870, Charleston, W. Va. Accredited to: Pennsylvania. G.O. No.: 504 13 December 1898. Other Navy award: Navy Cross. Citation: In action during the battle of Cuzco, Cuba, 14 June 1898. Distinguishing himself during this action, Quick signaled the U.S.S. Dolphin on 3 different occasions while exposed to a heavy fire from the enemy.
*STOCKHAM, FRED W. (Army Medal)
Rank and organization: Gunnery Sergeant, U.S. Marine Corps, 96th Company, 2d Battalion, 6th Regiment. Place and date: In Bois-de-Belleau, France, 13-14 June 1918. Entered service at: New York, N.Y. Birth: Detroit, Mich. G.O. NO.:–. Citation: During an intense enemy bombardment with high explosive and gas shells which wounded or killed many members of the company, G/Sgt. Stockham, upon noticing that the gas mask of a wounded comrade was shot away, without hesitation, removed his own gas mask and insisted upon giving it to the wounded man, well knowing that the effects of the gas would be fatal to himself. He continued with undaunted courage and valor to direct and assist in the evacuation of the wounded, until he himself collapsed from the effects of gas, dying as a result thereof a few days later. His courageous conduct undoubtedly saved the lives of many of his wounded comrades and his conspicuous gallantry and spirit of self-sacrifice were a source of great inspiration to all who served with him.
Rank and organization: Lieutenant Colonel (then Captain), 2d Battalion, 60th Infantry Regiment, 9th Infantry Division, World War II. Place and date: Renouf, France, 14 June to 3 September 1944. Entered service at: Fort Bragg, North Carolina, 2 July 1941. Date and place of birth: 25 August 1919, Buffalo, New York. Lieutenant Colonel (then Captain) Matt Urban, l 12-22-2414, United States Army, who distinguished himself by a series of bold, heroic actions, exemplified by singularly outstanding combat leadership, personal bravery, and tenacious devotion to duty, during the period 14 June to 3 September 1944 while assigned to the 2d Battalion, 60th Infantry Regiment, 9th Infantry Division. On 14 June, Captain Urban’s company, attacking at Renouf, France, encountered heavy enemy small arms and tank fire. The enemy tanks were unmercifully raking his unit’s positions and inflicting heavy casualties. Captain Urban, realizing that his company was in imminent danger of being decimated, armed himself with a bazooka. He worked his way with an ammo carrier through hedgerows, under a continuing barrage of fire, to a point near the tanks. He brazenly exposed himself to the enemy fire and, firing the bazooka, destroyed both tanks. Responding to Captain Urban’s action, his company moved forward and routed the enemy. Later that same day, still in the attack near Orglandes, Captain Urban was wounded in the leg by direct fire from a 37mm tank-gun. He refused evacuation and continued to lead his company until they moved into defensive positions for the night. At 0500 hours the next day, still in the attack near Orglandes, Captain Urban, though badly wounded, directed his company in another attack. One hour later he was again wounded. Suffering from two wounds, one serious, he was evacuated to England. In mid-July, while recovering from his wounds, he learned of his unit’s severe losses in the hedgerows of Normandy. Realizing his unit’s need for battle-tested leaders, he voluntarily left the hospital and hitchhiked his way back to his unit hear St. Lo, France. Arriving at the 2d Battalion Command Post at 1130 hours, 25 July, he found that his unit had jumped-off at 1100 hours in the first attack of Operation Cobra.” Still limping from his leg wound, Captain Urban made his way forward to retake command of his company. He found his company held up by strong enemy opposition. Two supporting tanks had been destroyed and another, intact but with no tank commander or gunner, was not moving. He located a lieutenant in charge of the support tanks and directed a plan of attack to eliminate the enemy strong-point. The lieutenant and a sergeant were immediately killed by the heavy enemy fire when they tried to mount the tank. Captain Urban, though physically hampered by his leg wound and knowing quick action had to be taken, dashed through the scathing fire and mounted the tank. With enemy bullets ricocheting from the tank, Captain Urban ordered the tank forward and, completely exposed to the enemy fire, manned the machine gun and placed devastating fire on the enemy. His action, in the face of enemy fire, galvanized the battalion into action and they attacked and destroyed the enemy position. On 2 August, Captain Urban was wounded in the chest by shell fragments and, disregarding the recommendation of the Battalion Surgeon, again refused evacuation. On 6 August, Captain Urban became the commander of the 2d Battalion. On 15 August, he was again wounded but remained with his unit. On 3 September, the 2d Battalion was given the mission of establishing a crossing-point on the Meuse River near Heer, Belgium. The enemy planned to stop the advance of the allied Army by concentrating heavy forces at the Meuse. The 2d Battalion, attacking toward the crossing-point, encountered fierce enemy artillery, small arms and mortar fire which stopped the attack. Captain Urban quickly moved from his command post to the lead position of the battalion. Reorganizing the attacking elements, he personally led a charge toward the enemy’s strong-point. As the charge moved across the open terrain, Captain Urban was seriously wounded in the neck. Although unable to talk above a whisper from the paralyzing neck wound, and in danger of losing his life, he refused to be evacuated until the enemy was routed and his battalion had secured the crossing-point on the Meuse River. Captain Urban’s personal leadership, limitless bravery, and repeated extraordinary exposure to enemy fire served as an inspiration to his entire battalion. His valorous and intrepid actions reflect the utmost credit on him and uphold the noble traditions of the United States.
WISE, HOMER L.
Rank and organization: Staff Sergeant. U.S. Army, Company L, 142d Infantry, 36th Infantry Division. Place and date: Magliano, Italy, 14 June 1944. Entered service al: Baton Rouge, La. Birth: Baton Rouge La. G.O. No.: 90, 8 December 1944. Citation: While his platoon was pinned down by enemy small-arms fire from both flanks, he left his position of comparative safety and assisted in carrying 1 of his men, who had been seriously wounded and who lay in an exposed position, to a point where he could receive medical attention. The advance of the platoon was resumed but was again stopped by enemy frontal fire. A German officer and 2 enlisted men, armed with automatic weapons, threatened the right flank. Fearlessly exposing himself, he moved to a position from which he killed all 3 with his submachinegun. Returning to his squad, he obtained an Ml rifle and several antitank grenades, then took up a position from which he delivered accurate fire on the enemy holding up the advance. As the battalion moved forward it was again stopped by enemy frontal and flanking fire. He procured an automatic rifle and, advancing ahead of his men, neutralized an enemy machinegun with his fire. When the flanking fire became more intense he ran to a nearby tank and exposing himself on the turret, restored a jammed machinegun to operating efficiency and used it so effectively that the enemy fire from an adjacent ridge was materially reduced thus permitting the battalion to occupy its objective.
BLEAK, DAVID B.
Rank and organization: Sergeant, U.S. Army, Medical Company 223d Infantry Regiment, 40th Infantry Division. Place and date: Vicinity of Minari-gol, Korea, 14 June 1952. Entered service at: Shelley, Idaho. Born: 27 February 1932, Idaho Falls, Idaho. G.O. No.: 83, 2 November 1953. Citation: Sgt. Bleak, a member of the medical company, distinguished himself by conspicuous gallantry and indomitable courage above and beyond the call of duty in action against the enemy. As a medical aidman, he volunteered to accompany a reconnaissance patrol committed to engage the enemy and capture a prisoner for interrogation. Forging up the rugged slope of the key terrain, the group was subjected to intense automatic weapons and small arms fire and suffered several casualties. After administering to the wounded, he continued to advance with the patrol. Nearing the military crest of the hill, while attempting to cross the fire-swept area to attend the wounded, he came under hostile fire from a small group of the enemy concealed in a trench. Entering the trench he closed with the enemy, killed 2 with bare hands and a third with his trench knife. Moving from the emplacement, he saw a concussion grenade fall in front of a companion and, quickly shifting his position, shielded the man from the impact of the blast. Later, while ministering to the wounded, he was struck by a hostile bullet but, despite the wound, he undertook to evacuate a wounded comrade. As he moved down the hill with his heavy burden, he was attacked by 2 enemy soldiers with fixed bayonets. Closing with the aggressors, he grabbed them and smacked their heads together, then carried his helpless comrade down the hill to safety. Sgt. Bleak’s dauntless courage and intrepid actions reflect utmost credit upon himself and are in keeping with the honored traditions of the military service.
*SPEICHER, CLIFTON T.
Rank and organization: Corporal, U.S. Army, Company F, 223d Infantry Regiment, 40th Infantry Division. Place and date: Near Minarigol, Korea, 14 June 1952. Entered service at: Gray, Pa. Born: 25 March 1931, Gray, Pa. G.O. No.: 65, 19 August 1953. Citation: Cpl. Speicher distinguished himself by conspicuous gallantry and indomitable courage above and beyond the call of duty in action against the enemy. While participating in an assault to secure a key terrain feature, Cpl. Speicher’s squad was pinned down by withering small-arms mortar, and machine gun fire. Although already wounded he left the comparative safety of his position, and made a daring charge against the machine gun emplacement. Within 10 yards of the goal, he was again wounded by small-arms fire but continued on, entered the bunker, killed 2 hostile soldiers with his rifle, a third with his bayonet, and silenced the machine gun. Inspired by this incredible display of valor, the men quickly moved up and completed the mission. Dazed and shaken, he walked to the foot of the hill where he collapsed and died. Cpl. Speicher’s consummate sacrifice and unflinching devotion to duty reflect lasting glory upon himself and uphold the noble traditions of the military service.<|endoftext|>
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The Triple Entente ("entente"—French for "agreement") was the alliance formed in 1907 among the United Kingdom of Great Britain and Ireland, the French Third Republic and the Russian Empire after the signing of the Anglo-Russian Entente. The UK already had the Entente Cordiale with France since 1904, while France had concluded the Franco-Russian Alliance in 1894. The Triple Alliance formed in 1882 offered an ominous threat, thus the three nations bonded together in a compact designed to protect them from encroachment or attack. The Entente was in itself a defensive alliance. Fear and suspicion drove the three nations to seek a viable partnership as the German navy and army continued to grow in size and power. With the onset of World War I, the world would see the two alliances finally come to blows. The doctrine of the balance of power was behind the Alliance, which was meant to keep the peace. However, this did not take sufficient account of Triple Alliance ambitions for empire in and beyond the European space.
Though not a military alliance, the alignment of the three powers, supplemented by various agreements with Japan, the United States and Spain, constituted a powerful counterweight to the "Triple Alliance" of Imperial Germany, Austria-Hungary and Italy, the latter having concluded an additional secret agreement with France effectively nullifying her alliance commitments.
Russia had been a member of the League of the Three Emperors with Austria-Hungary and Germany. After the League's collapse during the Alexander von Battenberg affair, German Chancellor Bismarck tried to keep an alliance with Russia. This was formalized in the Reinsurance Treaty of 1887. But when Bismarck was dismissed from office in 1890, Kaiser Wilhelm II failed to renew the treaty and Russia formed a military alliance with France. The UK had been asked to join in an alliance with Germany, but did not agree with Germany's ideological and military goals, and had been in a naval arms race with Germany for decades.
With the addition of Italy in 1915, the Triple Entente was the force that opposed the Central Powers during World War I. After the outbreak of World War I in Europe in August 1914, the three Entente powers undertook in September 4 not to conclude a separate peace with Germany or Austria-Hungary.
But Russia's separate armistice (December 1917) and peace Treaty of Brest-Litovsk on March 3, 1918, ended its alignment with the other Entente powers. The U.K. and France continued to collaborate in ultimately unsuccessful attempts to uphold the postwar order during the 1920s and 1930s, until France's crushing June 1940 defeat in renewed conflict with Germany forced her into a separate armistice, leaving the UK alone in Europe.
The Entente heralded the end of British neutrality in Europe. It was partly a response to growing German antagonism expressed in the creation of the Kaiserliche Marine battle fleet capable of threatening British naval supremacy.
Ironically, the Franco-Russian Alliance which had seemed weak during Russia's defeat in the Russo-Japanese War, later appeared the more powerful alignment, when Russia unexpectedly and rapidly recovered from the defeat and from the Russian Revolution of 1905, and when the UK was added as a diplomatic partner. This contributed to the foreign-policy adventuring and contemplation of pre-emptive war, which culminated in German readiness for conflict in 1914.
In the nineteenth century, Britain adopted a policy of isolationism in continental European politics known as "splendid isolationism." Britain's primary focus was concentrated on maintaining and expanding its massive overseas empire. However, by the early 1900s the European theater began to change dramatically. Britain was in need of allies. For most of the nineteenth century, Britain had regarded France and Russia as its two most dangerous rivals but with the threat of German imperialism Britain's sentiments began to change.
The three main reasons were:
1. France and Britain had signed a number of agreements regarding colonies in North Africa in 1904. The Tangier Crisis had sparked a sudden cooperation between the two countries in the face of their mutual fear of German sponsored Moroccan independence.
2. Russia was recently defeated in the Russo-Japanese War. This display of weakness resulted in less concern over Russian imperialism and more interest in them as a viable ally. The formation of the Triple Alliance in 1882 also left Russia in a very vulnerable position.
3. Britain was very worried about the rising threat of German imperialism. Kaiser Wilhelm II had announced to the world his intentions to create a global German empire and to develop a strong navy. Britain, traditionally having control of the seas, saw this as a serious threat to its own empire and navy.
Britain and France, both facing German imperialism, signed an agreement with each other in 1904 called the Entente Cordiale. Shortly thereafter in 1907 Russia would join the Anglo-Russian Entente to alleviate British fears of German expansion into the Middle East.
France had fought Germany in the Franco Prussian War in 1870 resulting in a dramatic and embarrassing defeat for France. The Germans had forced France to sign a humiliating treaty in 1871 (The Treaty of Frankfurt), which signed over the industrialized region of Alsace-Lorraine to Germany. Ever since relations had been at an all time low. France, worried about the escalating military development of Germany, began building up their own war industries and army as a deterrent to German aggression. As another measure, France developed a strong bond with Russia by joining the Franco-Russian Alliance, which was designed to create a strong counter to the Triple Alliance. France's main concerns were to protect against an attack from Germany, and to reincorporate the lost territories of Alsace-Lorraine.
Russia possessed by far the largest manpower reserves of all the six European powers, but was also the most backward economically. Russia shared France's worries about Germany. After the Germans started to reorganize the Turkish army, Russia feared that they would come to control the Dardanelles, a vital trade artery which accounted for two fifths of Russia's exports.
This was also coupled with Russia's long history of rivalry with Austria-Hungary. Austria-Hungary had recently annexed Bosnia and Herzegovina angering Russia immensely. Russia had considered itself the leader of the Slavic world and viewed the invasion as another step towards annexing Serbia and Montenegro. To counter act Austria-Hungary's aggression into the Balkans, Russia signed an agreement with Serbia to aid it militarily in the face of Austro-Hungarian invasion.
The Tsar had also recently fought a grueling war with Japan in 1905 resulting in Russia's transformation into a constitutional monarchy. To counter his enemies militarily and politically he sought to revive the Franco-Russian Alliance. Although it was perceived as useless during the Russo-Japanese War, in the European theater it was invaluable. Russia would also sign the Anglo-Russian Convention of 1907 with Britain to counter act the threat of the Triple Alliance.
The Franco-Russian Alliance, along with the Anglo-Russian Entente and the Entente Cordiale formed the Triple Entente between the UK, France and Russia—an effective deterrent to the Triple Alliance.
All links retrieved December 17, 2015.
|Diplomacy of the Great Powers 1871-1913|
|Great Powers||British Empire | German Empire | French Third Republic | Russian Empire | Austria-Hungary | Italy|
|Treaties and agreements||Treaty of Frankfurt | League of the Three Emperors | Treaty of Berlin | German-Austrian Alliance | Triple Alliance | Reinsurance Treaty | Franco-Russian Alliance | Anglo-Japanese Alliance | Anglo-Russian Entente | Entente Cordiale | Triple Entente|
|Events||Russo-Turkish War | Congress of Berlin | Scramble for Africa | Fleet Acts | The Great Game | First Sino-Japanese War | Fashoda Incident | Pan-Slavism | Boxer Rebellion | Boer War | Russo-Japanese War | First Moroccan Crisis | Dreadnought | Agadir Crisis | Bosnian crisis | Italo-Turkish War | Balkan wars|
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# Square Centimeters To Square Millimeters Converter
How to convert square centimeters to square millimeters?How to transform 500 cm squared to mm squared?Check our other area tools:FAQs
Our square centimeters to square millimeters converter is a simple tool that helps you with a specific task of turning cm² to mm² and mm² to cm². 🔄
Keep on reading to discover how to convert square centimeters to square millimeters and vice versa - all you need is a proper conversion factor!
🙋 Don't get distracted by different notations - square millimeters and square centimeters conversions can be written as:
• cm 2 to mm 2;
• sq mm to sq cm;
• cm squared to mm squared;
• mm2 to cm2; and finally,
• mm² to cm².
All of these versions can be found in our square centimeters to square millimeters converter.
## How to convert square centimeters to square millimeters?
We're gonna make it pretty straightforward:
• If you want to convert cm² to mm², use the equation:
Area in sq mm = Area in sq cm × 100
• If you'd like to convert mm² to cm², follow the formula:
Area in sq cm = Area in sq mm / 100
## How to transform 500 cm squared to mm squared?
Let's calculate it in the simplest way possible.
We know that 1 cm² = 100 mm². Let's use a simple equation:
Area in sq mm = Area in sq cm × 100 mm²
Area in sq mm = 500 × 100 mm²
Area in sq mm = 50000 mm²
## Check our other area tools:
FAQs
### How to turn 1 cm2 to mm2?
It's easier than you think! We need to remember than 1 linear centimeter houses exactly 10 linear millimeters.
1 cm = 10 mm
If we want to know the number of squared millimeters, we need to square both sides of the equation.
(1 cm)² = (10 mm)²
1 cm² = 10 × 10 mm²
1 cm² = 100 mm²
### How many cm2 are in mm2?
Just as you remember, all metric units of distance are a multiple of 10 - neither square centimeters nor square millimeters are an exception here.
1 mm² = 0.01 cm²
while
100 mm² = 1 cm²<|endoftext|>
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Blood is an important biological material for vertebrates: transporting oxygen, carbon dioxide, hormones, minerals, food and wastes around their bodies, blood is crucial to their lives. For some animals—known to biologists as hematophagous species—blood is even more important, as it represents their primary food source. Hematophagy is widespread in the animal kingdom, and is practiced by a number of invertebrates, fish, birds and mammals.
Common Traits among Hematophagous Species
Sometimes widely separated, distantly related species evolve similar lifestyles, morphologies and behaviors; scientists call this phenomenon convergence. Examples of convergence abound among hematophagous species: most species that feed on blood produce saliva with anticoagulant and pain-killing properties, which increase the flow of blood and reduce the likelihood that their host will notice them, respectively. Additionally, structures like sharp teeth and piercing mouthparts are common among bloodsucking species.
A variety of invertebrates, including insects, arachnids and worms, consume blood. Insects such as assassin bugs (Reduviidae), mosquitoes (Culicidae) and bedbugs (Cimex lectularius) live by feeding on the blood of various host species. Male mosquitoes do not feed on blood, but females require it to deposit eggs, otherwise they will die without reproducing. Though they superficially resemble insects, ticks are arachnids—relatives of spiders and scorpions—that subsist on the blood of other animals. New World leeches primarily live an aquatic lifestyle, while those of the Old World often live on land. Hematophagous invertebrates are vectors for the spread of a variety of diseases; mosquitoes spread West Nile virus and malaria, ticks spread Rocky Mountain spotted fever and Lyme disease, and assassin bugs transmit deadly Chagas disease.
Lampreys form the order Petromyzontiformes, an ancient lineage of fish that some scientists think is the sister group to all vertebrates with jaws. Lamprey mouths have circular rows of teeth and a rasping tongue; most species attach themselves to living fish, open a wound and then suck out blood and other bodily fluids. In some cases the lampreys weaken or kill the fish, and invasive, introduced lamprey species have become a commercial problem as they deplete native fish populations.
Given that the diversity of finches found in the Galapagos Islands was rich enough to help inspire Darwin’s theory of natural selection, it shouldn’t be surprising that one of these enterprising species has evolved to drink blood. Sharp-beaked ground finches (Geospiza difficilis)—sometimes called vampire finches—eat a variety of prey; they tend to feed on whatever food is available, including sea bird eggs, nectar and seeds. Some of those that live in low-lying areas with a low food supply engage in hematophagy, typically by alighting on the backs of boobies (Sula spp.) and pecking the birds until they bleed.
Perhaps among the most famous hematophagous animals, the various species of vampire bats get their name from their habit of lapping up blood from other animals. Using their sharp teeth to make a small incision in their host’s skin, the bats feed greedily on the high-protein food source. Bats will feed for about 30 minutes; sometimes their anticoagulant-laden saliva is so effective that blood continues to drip from the host animal for some time after the bat has left.
- New York Times: A Taste for Blood
- Ohio State University: Assassin Bugs
- The Center for Disease Control and Prevention: Bed Bugs—Frequently Asked Questions
- The Center for Disease Control and Prevention: Tickborne Diseases of the U.S.
- Mosquitoes.com: Biological Notes on Mosquitoes
- University of Berkley: Introduction to the Petromyzontiformes
- Stockbyte/Stockbyte/Getty Images<|endoftext|>
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A Basic Guide to Tuning Timpani
This is a Basic Guide for teachers and general music specialists about the Timpani. This is a non-technical explanation for non-percussionists! The subject of tuning Timpani in detail is a huge one, that I could probably write a whole book about. This articles is a quick little explanation for general music teachers and band teachers about the Timpani and how to tune them.
Tuning a Timpani refers to the process of using the pedals to change the note on a drum – which can be done between pieces or in the middle of the piece if required.
The tuning of the tuning rods around the edge should be left to a percussion specialist – as it a difficult procedure beyond the scope of this article! A service by a professional would be desirable once every two years, or if the timpani are not sounding in tune.
The pedal changes the pitch of the drum by tightening or loosening the skin. When you push the pedal down to tighten the skin it raises the pitch, and when you loosen it the pitch lowers.
Most timpani are equipped with a gauge – a mechanical device attached to the side of the drum which indicates the tension, and thus the note heard.
The Gauge consists of a row of letters (letter names for the notes of the timpani), and a arrow device which moves when you adjust the pedal. To change notes therefore, you need to watch the gauge and move the pedal so that the arrow points to the note that you need.
In a perfect world the timpani would never need to be adjusted, so that the player could always use the gauge to set the desired note. However, due to changes in temperature and environment you cannot always depend on the letters being in the right places.
At the start of rehearsal or before a performance the timpanist must “set the gauges” or move the letters indicating where the notes are, so that the arrow device accurately lands on the right note.
If a piece only has fixed notes, and does not require changing then there is no need to set all the letters on the gauge – you only need to set the notes you are going to use on each timpani.
The process of doing this is the most difficult for young players at first.
Here is my basic method:
1) Release the pedal (so the drum is at its lowest note)
2) Play the note you are tuning to on a fixed pitch instrument (such as a piano or marimba, or you can use a pitch pipe)
3) Sing the note in your head or out loud (softly of course!)
4) Hit the drum once
5) Slowly push the pedal down until your ear can hear the sustain from the struck note coming into tune with the reference note.
6) Once you have heard the note come into tune, take your foot off the pedal and set the letter of the note you have just tuned, as the pedal should stay in the same place.
Most young players move the pedal too fast – therefore overshooting the note, or they fail to get the reference note in their heads in the first place.
New timpanists must practice, practice & practice this until they can do it reliably. The teacher must master the skill themselves and then be able to communicate it to the students.
Some timpanists do use electronic tuning devices; however they are not always reliable for timpani due to the harmonics which are present. Tuning “by ear” is always a much preferred method, even if it is more difficult at first.
You can learn more about Timpani in the Fun Music Company Music Lesson Plan series on instruments of the Orchestra from the Fun Music Company.<|endoftext|>
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Vaccination and Immunity
After an immune response, memory cells are produced. These lay dormant in the lymphatic system for many years, if they detect a pathogen with the specific antigen they can clone rapidly and sectrete antibodies. This means that secondary exposure to a pathogen produces a much more rapid secondary response which means you are not as badly affected the second time the virus appears - this is important in vaccination. See the image below.
You might think it odd then, that we can suffer from flu several times. The reason is that there are different types: influenza type A, B and C and within these types there are various strains. The variation in pathogens means that it is difficult to get immunity to many pathogens.
A vaccine takes advantage of the secondary response effect. It contains antigen from pathogens, and this induces the production of memory cells - giving protection from infection by that organism.
There are several types of vaccine used, these are:
|Name and Example||Production|
|Killed virulent organisms|
|The pathogen is killed and the antigens remain so an immune response is induced without the pathogen spreading.|
|Live non-virulent strains|
|A strain of pathogen that doesn't cause disease but still causes antibody production.|
|A modified version of toxin treated with heat or chemicals, shouldn't produce symptoms but still triggers a response.|
|Isolated antigens from a pathogen|
|The antigen is seperated from the pathogen and injected seperately.|
|Genetically engineered antigens|
|The antigens have been isolated and made by genetic engineering, are injected without the virus.|
This is when the antibodies themselves are given to prevent infection. This occurs naturally for babies when a type of milk called colostrum is produced by mothers that contains lots of antibodies. Passive immunity can also be artificially induced if someone is given (by injection) some antibody.
Passive immunity is only temporary and will last a few months at most, this is because they are not your own antibodies and they will be removed by the body.<|endoftext|>
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The simple stain can be used as a quick and easy way to determine cell shape, size and arrangements of bacteria. True to its name, the simple stain is a very simple staining procedure involving single solution of stain. Any basic dye such as methylene blue, safranin, or crystal violet can be used to color the bacterial cells.
These stains will readily give up a hydroxide ion or accept a hydrogen ion, which leaves the stain positively charged. Since the surface of most bacterial cells and cytoplasm is negatively charged, these positively charged stains adhere readily to the cell surface. After staining, bacterial cell morphology (shape and arrangements) can be appreciated.
Preparation of a smear and heat fixing
- Using a sterilized inoculating loop, transfer loopful of liquid suspension containing bacteria to a slide (clean grease free microscopic slide) or transfer an isolated colony from a culture plate to a slide with a water drop.
- Disperse the bacteria on the loop in the drop of water on the slide and spread the drop over an area the size of a dime. It should be a thin, even smear.
- Allow the smear to dry thoroughly.
- Heat-fix the smear cautiously by passing the underside of the slide through the burner flame two or three times. It fixes the cell in the slide. Do not overheat the slide as it will distort the bacterial cells.
- Cover the smear with methylene blue and allow the dye to remain in the smear for approximately one minute (Staining time is not critical here; somewhere between 30 seconds to 2 minutes should give you an acceptable stain, the longer you leave the dye in it, the darker will be the stain).
- Using distilled water wash bottle, gently wash off the excess methylene blue from the slide by directing a gentle stream of water over the surface of the slide.
- Wash off any stain that got on the bottom of the slide as well.
- Saturate the smear again but this time with Iodine. Iodine will set the stain
- Wash of any excess iodine with gently running tap water. Rinse thoroughly. (You may not get mention about step 4 and 5 in some text books)
- Wipe the back of the slide and blot the stained surface with bibulous paper or with a paper towel.
- Place the stained smear on the microscope stage smear side up and focus the smear using the 10X objective.
- Choose an area of the smear in which the cells are well spread in a monolayer. Center the area to be studied, apply immersion oil directly to the smear, and focus the smear under oil with the 100X objective.
The bacterial cells usually stain uniformly and the color of the cell depends on the type of dye used. If methyene blue is used, some granules in the interior of the cells of some bacteria may appear more deeply stained than the rest of the cell, which is due to presence of different chemical substances.
Diagnostic microbiology laboratory generally does not perform simple staining method. Differential staining such as Gram Staining and AFB Staining are commonly used to identify and differentiate the bacterial isolates. Simple staining can be useful in the following circumstances.
- To differentiate bacteria from yeast cells: When endocervical swab culture is done in Blood agar both Staphylococcus spp and yeast cells may give similar looking colonies in Blood agar (a common error for new technologist or microbiologist with less experience). Performing wet mount technique or simple staining from the isolate can be helpful.
- To presumptively identify the bacterial isolate
Due to their ubiquitous presence, Bacillus spp may present as a contaminant in the culture plates. In some circumstances (e.g. growth in Blood Agar but no growth in MacConkey Agar), identifying the shape of the bacteria (rod or cocci) may help to eliminate the isolate as possible contaminants (e.g., Bacillus spp) or further process as potential pathogen (cocci).<|endoftext|>
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# How do you find the derivative of f(x) = arcsin (2x + 5)?
Sep 15, 2016
$\frac{2}{\sqrt{4 {x}^{2} + 20 x + 24} \cdot i}$
#### Explanation:
We have: $f \left(x\right) = \arcsin \left(2 x + 5\right)$
This function can be differentiated using the "chain rule".
Let $u = 2 x + 5 \implies u ' = 2$ and $v = \arcsin \left(u\right) \implies v ' = \frac{1}{\sqrt{1 - {u}^{2}}}$:
$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = 2 \cdot \frac{1}{\sqrt{1 - {u}^{2}}}$
$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{1 - {u}^{2}}}$
We can now replace $u$ with $2 x + 5$:
$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{1 - {\left(2 x + 5\right)}^{2}}}$
=> (d) / (dx) (arcsin(2 x + 5)) = (2) / (sqrt(1 - (4 x^(2) + 20 x + 25))
$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{- 4 {x}^{2} - 20 x - 24}}$
$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{- \left(4 {x}^{2} + 20 x + 24\right)}}$
$\implies \frac{d}{\mathrm{dx}} \left(\arcsin \left(2 x + 5\right)\right) = \frac{2}{\sqrt{4 {x}^{2} + 20 x + 24} \cdot i}$<|endoftext|>
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Finding the main idea and supporting details is a reading comprehension skill that students are expected to understand starting in kindergarten. By third grade, students are expected to do this with more complex texts independently. This can be a challenging skill for all students. For some ELLs, they have the added challenge of not yet reading grade level texts independently. Here are 5 ideas to support students learning about main idea and supporting details.
One simple way to introduce main idea and details is to have students sort words or pictures into categories. Have students come up with categories for each group and then an overall category for all of the groups. You can scaffold this activity by having the categories already available.
I created a set of main idea puzzles using a similar idea. In the center is the main idea (category) and the surrounding pieces are pictures that show the details.
Use Short Reading Passages
When students are practicing main idea and supporting details, it is helpful to give them repeated practice. Short reading passages allow them to identify the main idea and details in a text multiple times. For students that are reading below grade level you can either read the text out loud to them or use simple text that has this skill. It is good to give ELLs exposure both to text that they can read independently and grade level text.
Use Wordless Picture Books
Disclosure: This post contains Amazon affiliate links, which means I may receive a commission if you click on an Amazon link and make a purchase. This does not cost you any extra money.
Wordless picture books are a great opportunity for students to practice speaking and/or writing skills as they are also learning about main idea. Using a wordless book gives student a chance to incorporate creativity into coming up with a main idea. As long as they can find supporting details, it is possible for there to be different ideas of what the main idea is. Students can work with a partner or in a small group to come up with the main idea and details of the story. I recommend first having the students look at the pictures and tell what is happening before having them try to come up with what the main idea is.
Some books to try out are:
I Walk with Vanessa: A Story About a Simple Act of Kindness– A girl faces bullying in her new school and another child learns how to help.
Pool– This book shows two children that are overwhelmed at a crowded pool. They find a place to escape through their imaginations.
Cut apart sentences
Cutting apart sentences gives students a hands on way to practice identifying main idea and details. Students read each sentence. First, they decide if it is the main idea. Then, they find the sentences that support that main idea. You can scaffold this activity by showing students the main idea and only having them find the details. This activity is one that you can create on your own by cutting apart sentences from a written paragraph to practice main idea and details.
Use Sticky Notes
For longer texts, or those that you do not want to cut apart, sticky notes are a great way for students to record where the main idea and details are. Depending on students age and writing levels, you can either have them use full size sticky notes or smaller strips. They write down the main idea on one color. They they write the details on a different color. After that, they place the sticky note on the text where they found it. You can do a similar activity with sticky note strips. Students only mark where they find the main idea and/or supporting details. You can limit the amount of details students mark by giving a finite number of sticky notes to each student.<|endoftext|>
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## How Two Parabolas Intersect
Searching for
Learn in a way your textbook can't show you.
Explore the full path to learning Graphing Quadratic Functions
### Lesson Focus
#### How Two Parabolas Intersect
Algebra-1
You will find the intersection points of two parabolas to determine the set of x-values for which one parabola lies above the other.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Determine how many points two parabolas intersect at.
• Find the intersection points of two parabolas.
• Find the set of x–values for which one parabola lies above the other.
### Everything You'll Have Covered
Parabola and line intersections
Before considering how two parabolas intersect, remember how a parabola and line may intersect at one point, two points, or no points.
The main thing to remember is to equalize the equation of the quadratic function and the linear equation to each other and then solve the derived quadratic equations to find the intersection points, if any. Three examples are shown in the following table:
Intersections of two parabolas
To determine if there are one, two, or no intersection points of two parabolas, you can make the two functions equal to each other. This time, you can get either a linear equation or a quadratic equation. In each case, you need to solve the equation to find the intersection points.
When the derived equation from two quadratic functions is also a quadratic, you can't solve for x. The next table describes how to use the discriminant to determine the number of intersection points.
Discriminant used to determine the number of intersection point of two parabolas
The set of x-values for which f(x) is above or below g(x)
### Tutorial Details
Approximate Time 30 Minutes Pre-requisite Concepts Students should understand the concepts of discriminant, graph a quadratic, orientation of a parabola, parabola, quadratic, quadratic function, root, and zeros of a quadratic function. Course Algebra-1 Type of Tutorial Guided Discovery Key Vocabulary parabola, intersection points, x-values<|endoftext|>
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# M06-29
Author Message
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Math Expert
Joined: 02 Sep 2009
Posts: 47978
### Show Tags
16 Sep 2014, 00:28
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35% (medium)
Question Stats:
72% (00:57) correct 28% (01:26) wrong based on 129 sessions
### HideShow timer Statistics
If $$x$$ and $$y$$ are integers and $$x=\frac{y}{5}+2$$, is $$xy$$ even?
(1) $$5x - 10$$ is even
(2) $$\frac{y}{x}$$ is even
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47978
### Show Tags
16 Sep 2014, 00:28
Official Solution:
Since $$x$$ and $$y$$ are integers then $$xy$$ will be even if at least one of the multiple is even. From $$x=\frac{y}{5}+2$$ we have that $$y=5x-10$$
(1) $$5x - 10$$ is even. Since $$y=5x-10$$ then $$y=5x-10=\text{even}$$. Sufficient.
(2) $$\frac{y}{x}$$ is even. Given: $$\frac{y}{x}=\text{even}$$, so $$y=x*\text{even}=\text{even}$$. Sufficient.
_________________
Intern
Joined: 10 Oct 2016
Posts: 9
### Show Tags
16 Dec 2016, 04:12
Bunuel wrote:
Official Solution:
Since $$x$$ and $$y$$ are integers then $$xy$$ will be even if at least one of the multiple is even. From $$x=\frac{y}{5}+2$$ we have that $$y=5x-10$$
(1) $$5x - 10$$ is even. Since $$y=5x-10$$ then $$y=5x-10=\text{even}$$. Sufficient.
(2) $$\frac{y}{x}$$ is even. Given: $$\frac{y}{x}=\text{even}$$, so $$y=x*\text{even}=\text{even}$$. Sufficient.
Hi Bunuel,
What happens if x is 0?
Math Expert
Joined: 02 Sep 2009
Posts: 47978
### Show Tags
16 Dec 2016, 04:20
uvemdesalinas wrote:
Bunuel wrote:
Official Solution:
Since $$x$$ and $$y$$ are integers then $$xy$$ will be even if at least one of the multiple is even. From $$x=\frac{y}{5}+2$$ we have that $$y=5x-10$$
(1) $$5x - 10$$ is even. Since $$y=5x-10$$ then $$y=5x-10=\text{even}$$. Sufficient.
(2) $$\frac{y}{x}$$ is even. Given: $$\frac{y}{x}=\text{even}$$, so $$y=x*\text{even}=\text{even}$$. Sufficient.
Hi Bunuel,
What happens if x is 0?
For which statement? For the second statement x cannot be 0 because in this case y/x would be undefined not even.
_________________
Intern
Joined: 10 Oct 2016
Posts: 9
### Show Tags
16 Dec 2016, 06:51
Bunuel wrote:
uvemdesalinas wrote:
Bunuel wrote:
Official Solution:
Since $$x$$ and $$y$$ are integers then $$xy$$ will be even if at least one of the multiple is even. From $$x=\frac{y}{5}+2$$ we have that $$y=5x-10$$
(1) $$5x - 10$$ is even. Since $$y=5x-10$$ then $$y=5x-10=\text{even}$$. Sufficient.
(2) $$\frac{y}{x}$$ is even. Given: $$\frac{y}{x}=\text{even}$$, so $$y=x*\text{even}=\text{even}$$. Sufficient.
Hi Bunuel,
What happens if x is 0?
For which statement? For the second statement x cannot be 0 because in this case y/x would be undefined not even.
For the first one. Y would be even (Y=-10) but X*Y=0
Math Expert
Joined: 02 Sep 2009
Posts: 47978
### Show Tags
16 Dec 2016, 12:29
uvemdesalinas wrote:
For the first one. Y would be even (Y=-10) but X*Y=0
0 is an even integer.
_________________
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Joined: 17 Sep 2016
Posts: 1
GMAT 1: 760 Q50 V41
GPA: 3.6
### Show Tags
02 Oct 2017, 10:29
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Hi Team,
In the first statement the answer shows sufficient.
But in case x is Zero then xy will be zero, which is not even.
Hence, how can statement 1 be sufficient?
Math Expert
Joined: 02 Sep 2009
Posts: 47978
### Show Tags
02 Oct 2017, 10:51
Psahota wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. Hi Team,
In the first statement the answer shows sufficient.
But in case x is Zero then xy will be zero, which is not even.
Hence, how can statement 1 be sufficient?
ZERO:
1. 0 is an integer.
2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. 0 is neither positive nor negative integer (the only one of this kind).
4. 0 is divisible by EVERY integer except 0 itself.
You should brush-up fundamentals before practising questions:
ALL YOU NEED FOR QUANT ! ! !
Hope it helps.
_________________
Re: M06-29 &nbs [#permalink] 02 Oct 2017, 10:51
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# M06-29
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# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.<|endoftext|>
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Researchers from the University of York and Roma Tre University developed a method to build ultra-low-power transistors using composite materials based on single layers of graphene and transition metal dichalcogenides (TMDC). These materials can be used to achieve an electrical control over electron spin.
The teams explained “we found this can be achieved with little effort when 2D graphene is paired with certain semiconducting layered materials. Our calculations show that the application of small voltages across the graphene layer induces a net polarization of conduction spins".
The team showed that when a small current is passed through the graphene layer, the electrons’ spin polarize in plane due to ‘spin-orbital’ forces brought about by the proximity to the TMDC base. They also showed the efficiency of charge-to-spin conversion can be quite high, even at room temperature.<|endoftext|>
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# Percent Calculator (2024)
*Other calculators are also available:
## FAQs
### How do you calculate percentage of correct answers? ›
You can find your test score as a percentage by dividing your score by the total number of points, then multiplying by 100.
What is the cheat for percentages? ›
As Stephens explains, if you ever have to calculate a difficult percentage on the spot without pen and paper or a calculator, you can use a simple shortcut - flip the numbers around. "So, for example, if you needed to work out 4% of 75 in your head, just flip it and do 75% of 4, which is easier," Stephens tweeted.
What percentage is 15 out of 50 o 30 o 35 o 25 o 40? ›
30 percent of 50 is equal to 15.
How do you work out 60% of 55? ›
60% of 55 is 33.
What is the 1 percent trick? ›
Let's quickly look at why the 1% Trick works by taking 1% of 250 long-hand. Begin by rewriting 1 percent as 1/100 and performing the multiplication. Whenever we divide by 100, we move the decimal point two places left. Therefore, you can always move the decimal point two places left to find 1 percent.
How do you figure out 5% of something? ›
Example: 5% of 29 is what?
1. Written using the formula: 5% * 29 = Y.
2. Convert the percent to a decimal.
3. 5% ÷ 100 = 0.05.
4. Y = 0.05 * 29.
5. Y = 1.45.
6. So 5% of 29 is 1.45.
How many percent is 7 out of 11? ›
Re-writing this as a percentage, we can see that 7/11 as a percentage is 63.636%.
What is 40 out of 50 in? ›
Solution: 40/50 as a percent is 80%
How to solve 30% of 40? ›
Therefore, to find 30% of 40, we move the decimal in 30 two units to the left to get 0.30, and then we multiply 0.30 by 40. We get that 30% of 40 is 12.
What is 80% out of 45? ›
The 80 percent of 45 is 36.
### How do you work out 40% of 80? ›
40% of 80 equals 32. To calculate this result multiply the fraction 0.40 by 80. Another method for finding 40 percent of 80 is through the multiplication of the fraction 40/100 with the integer 80.
What is 10% out of 200? ›
Answer: 10% of 200 is 20.
What percent is 3 out of 5 questions? ›
Solution: 3/5 as a percent is 60%
What percent is 5 out of 40 questions? ›
A score of 5 out of 40 on a test, assignment or class is a 13.5% percentage grade.
How many percent of 12 is 3? ›
Hence, 3 is 25% of 12.
What is 15 out of 30? ›
Solution: 15/30 as a percent is 50%
Do percentages always add up to 100? ›
Percentages are commonly rounded when presented in tables. As a result, the sum of the individual numbers does not always add up to 100%. A warning is therefore sometimes appended to such tables, along the lines: "Percentages may not total 100 due to rounding". The total of these is 99%, rather than the expected 100%.
How to calculate probability? ›
To calculate probability, you'll use simple multiplication and division. Probability equals the number of favorable outcomes divided by the total number of outcomes.
Can you add 2 percentages together? ›
After the first percentage change, the base changes, and the second percentage does not have the same base. Two percentages that have different base values cannot be directly combined by addition!
How do you find 5% of 200? ›
Answer: 5% of 200 is 10.
### What is 8 out of 12 as a percentage? ›
Solution: 8/12 as a percent is 66.667%
First, let's go over what a fraction represents.
What is an 8 out of 12? ›
A score of 8 out of 12 on a test, assignment or class is a 66.67% percentage grade. 4 questions were wrong or points missed. A 66% is a D letter grade. A letter grade D means less than satisfactory or below average performance.
What is 18 as a percentage of 54? ›
Solution: 18/54 as a percent is 33.333%
What is 80% out of 50? ›
Answer: 80% of 50 is 40.
How do you work out 80% of 30? ›
The 80 percent of 30 is equal to 24. It can be easily calculated by dividing 80 by 100 and multiplying the answer with 30 to get 24.
What is 25% out of 200? ›
Answer: 25% of 200 is 50.
What is 30% out of 400? ›
Answer: 30% of 400 is 120.
How do you find 30% of 120? ›
The 30 percent of 120 is equal to 36. It can be easily calculated by dividing 30 by 100 and multiplying the answer with 120 to get 36.
What is a 75 out of 40? ›
Answer: 75% of 40 is 30.
% is the symbol for percentage.
How do you find 20% of 35? ›
The 20 percent of 35 is equal to 7. It can be easily calculated by dividing 20 by 100 and multiplying the answer with 35 to get 7.
### How much is 37 out of 80? ›
First, you need to calculate your grade in percentages. The total answers count 80 - it's 100%, so we to get a 1% value, divide 80 by 100 to get 0.80. Next, calculate the percentage of 37: divide 37 by 1% value (0.80), and you get 46.25% - it's your percentage grade.
How do you solve 15% of 80? ›
15% of 80 is 12.
To find our answer, we first divide 80 by 100 to find the value of just 1% of 80. Next, we multiply the value of 1% of 80, which is 0.8, by 15 to find the value of 15% of 80.
How do you work out 35% of 80? ›
The 35 percent of 80 is equal to 28. It can be easily calculated by dividing 35 by 100 and multiplying the answer with 80 to get 28.
What is 20% out of 45? ›
Answer: 20% of 45 is 9.
How do you work out 18% of 50? ›
The 18 percent of 50 is equal to 9. It can be easily calculated by dividing 18 by 100 and multiplying the answer with 50 to get 9.
What is 10% out of 500? ›
Answer: 10% of 500 is 50.
What is the Excel trick for percentage? ›
The Excel formula for calculating percentages is (without multiplying by 100) Numerator/Denominator. You can convert the output to a percentage by pressing CTRL+SHIFT+%, or by clicking "%" under the "number" group on the Home tab.
Why is Excel calculating percentages wrong? ›
Sometimes we get values in our Excel sheets in such a way that the % sign is omitted. So instead of the value being 23%, it is 23. Now, you can very easily correct this by editing the cell and adding a % sign at the end.
How do I convert numbers to percentages quickly in Excel? ›
Click on the "Number" tab in the "Format Cells" dialog box after it opens. Select "Percentage" from the "Category:" list on the left side of the dialog box. Choose how many decimal points you want to include and click "OK" to convert all of the numbers in your selected cell range to percentages.
How do I calculate 15% of a number in Excel? ›
The formula for calculating the percentage is Percent = Value/Total * 100. For example, if you have a value in cell A2 and want to calculate what percentage that value is of the total in cell B2, you would use the following formula: Percent = A2/B2 * 100. This would give you the answer in percent form.
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The thyroid gland is a butterfly-shaped gland in the neck, lying in front of the trachea and larynx. The gland produces two hormones called triiodothyronine (T3) and thyroxine (T4).
These hormones are responsible for our metabolism (i.e. how you break down food and use energy). In hyperthyroidism, the thyroid releases too much of the hormones.
Symptoms of Hyperthyroidism
An excess of the T3 and T4 hormones can cause:
• Difficulty concentrating
• Increased sweating
• Increased appetite
• Intolerance of warm temperatures
• Weight loss
• An enlarged thyroid gland
• Irregular menstrual periods (or no periods)
• Upset stomach
• High blood pressure
• Nausea and vomiting
• Irregular pulse or heartbeat
• In men, breast development
You don’t have to have all of these symptoms to have hyperthyroidism.
What Causes Hyperthyroidism?
The majority of cases of hyperthyroidism are caused by Graves’ disease, an auto-immune disease where the immune system produces antibodies to the thyroid, triggering it to go into overdrive.
Genetic and environmental factors are thought to be responsible for this immune deregulation. It can be hereditary, running in families.
Other factors, such as damage to the thyroid by radiation and certain drug therapies, including HIV treatments, can result in hyperthyroidism. Cigarette smokers also have a two-fold increased risk of the disease.
If you don’t have Graves’ disease, there are other causes, such as inflammation of the thyroid gland (caused by a virus), or inflammation caused by an immune response.
Sometimes women who have just given birth will develop an inflamed thyroid, commonly known as postpartum thyroiditis. Taking too much hormone medication can also trigger hyperthyroidism.
It is diagnosed by physical examinations and blood tests. The doctor will look for signs of an enlarged thyroid and take a blood pressure reading. He will also take blood to check for elevated levels of T3 and T4, as well as reduced levels of a thyroid-stimulating hormone.<|endoftext|>
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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices
## Selina Textbook Solutions Chapter 9 - Matrices
Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 9 - Matrices.
All Selina textbook questions of Chapter 9 - Matrices solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.
Exercise/Page
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(A)
Solution 1
(i) False
The sum A + B is possible when the order of both the matrices A and B are same.
(ii) True
(iii) False
Transpose of a 2 1 matrix is a 1 2 matrix.
(iv) True
(v) False
A column matrix has only one column and many rows.
Solution 2
If two matrices are equal, then their corresponding elements are also equal. Therefore, we have:
x = 3,
y + 2 = 1 y = -1
z - 1 = 2 z = 3
Solution 3
If two matrices are equal, then their corresponding elements are also equal.
(i)
a + 5 = 2 a = -3
-4 = b + 4 b = -8
2 = c - 1 c = 3
(ii) a= 3
a - b = -1
b = a + 1 = 4
b + c = 2
c = 2 - b = 2 - 4 = -2
Solution 4
(i) A + B =
(ii) B - A
Solution 5
(i)B + C =
(ii)A - C =
(iii)A + B - C =
= =
(iv)A - B +C =
= =
Solution 6
(i)
(ii)
(iii) Addition is not possible, because both matrices are not of same order.
Solution 7
(i)
Equating the corresponding elements, we get,
3 - x = 7 and y + 2 = 2
Thus, we get, x = - 4 and y = 0.
(ii)
Equating the corresponding elements, we get,
-8 + y = -3 and x - 2 =2
Thus, we get, x = 4 and y = 5.
Solution 8
M =
Mt =
(i)
(i)
Solution 9
We know additive inverse of a matrix is its negative.
Solution 10
(i) X + B = C - A
(ii) A - X = B + C
Solution 11
(i) A + X = B
X = B - A
(ii) A - X = B
X = A - B
(iii) X - B = A
X = A + B
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(B)
Solution 1
Solution 2
Comparing the corresponding elements, we get,
12 + 2y = 10 and 3x - 6 = 0
Simplifying, we get, y = -1 and x = 2.
Comparing corresponding the elements, we get,
-x + 8 = 7 and 2x - 4y = -8
Simplifying, we get,
x = 1 and y = = 2.5
Solution 3
(i) 2A - 3B + C
(ii) A + 2C - B
Solution 4
Solution 5
(i)
(ii) C + B =
C = - B =
Solution 6
Comparing the corresponding elements, we get,
2x + 9 = -7 2x = -16 x = -8
3y = 15 y = 5
z = 9
Solution 7
(i) 2A + 3At
(ii) 2At - 3A
(iii)
(iv)
Solution 8
(i) X + 2A = B
X = B - 2A
(ii) 3X + B + 2A = O
3X = -2A - B
(iii) 3A - 2X = X - 2B
3A + 2B = X + 2X
3X = 3A + 2B
Solution 9
3M + 5N
Solution 10
(i) M - 2I =
(ii) 5M + 3I =
Solution 11
2M =
M =
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(C)
Solution 1
The number of columns in the first matrix is not equal to the number of rows in the second matrix. Thus, the product is not possible.
Solution 2
Solution 3
Solution 4
Comparing the corresponding elements, we get,
5x - 2 = 8 x = 2
20 + 3x = y y = 20 + 6 = 26
Comparing the corresponding elements, we get,
x = 2
-3 + y = -2 y = 1
Solution 5
Hence, A(BC) = (AB)C.
Solution 6
(iii) Product AA (=A2) is not possible as the number of columns of matrix A is not equal to its number of rows.
Solution 7
Solution 8
Hence, M2 = 2M + 3I.
Solution 9
Given, BA = M2
Comparing the corresponding elements, we get,
a = 2
-2b = -2 b = 1
Solution 10
Solution 11
(iii) Clearly, (A + B)2 A2 + B2
Solution 12
B2 = B + A
A = B2 - B
A = 2(B2 - B)
Solution 13
It is given that A2 = I.
Comparing the corresponding elements, we get,
1 + a = 1
Therefore, a = 0
-1 + b = 0
Therfore, b = 1
Solution 14
Solution 15
Solution 16(iii)
Solution 16(ii)
Solution 16(i)
Solution 17
We know, the product of two matrices is defined only when the number of columns of first matrix is equal to the number of rows of the second matrix.
(i) Let the order of matrix M be a x b.
Clearly, the order of matrix M is 1 x 2.
Comparing the corresponding elements, we get,
a = 1 and a + 2b = 2 2b = 2 - 1 = 1 b =
(ii) Let the order of matrix M be a x b.
Clearly, the order of matrix M is 2 x 1.
Comparing the corresponding elements, we get,
a + 4b = 13 ....(1)
2a + b = 5 ....(2)
Multiplying (2) by 4, we get,
8a + 4b = 20 ....(3)
Subtracting (1) from (3), we get,
7a = 7 a = 1
From (2), we get,
b = 5 - 2a = 5 - 2 = 3
Solution 18
Solution 19
Solution 20
AB = BA = B
We know that I × B = B × I = B, where I is the identity matrix.
Hence, A is an identity matrix.
Solution 21
Comparing the corresponding elements, we get,
3a = 3 + a
2a = 3
a =
3b = b b = 0
4c = 4 + c 3c = 4 c =
Solution 22
Clearly, it can be said that:
(P + Q) (P - Q) = P2 - Q2 not true for matrix algebra.
Solution 23
Hence, ABC ≠ ACB.
Solution 24
Thus, CA + B A + CB
Solution 25
Clearly, the order of matrix X is 2 x 1.
Comparing the two matrices, we get,
2x + y = 3 … (1)
x + 3y = -11 … (2)
Multiplying (1) with 3, we get,
6x + 3y = 9 … (3)
Subtracting (2) from (3), we get,
5x = 20
x = 4
From (1), we have:
y = 3 - 2x = 3 - 8 = -5
Solution 26
Solution 27
Solution 28
Hence, proved.
Solution 29
Comparing the corresponding elements, we get,
2x + 12 = 0
thus, x = -6
6 + 6y = 0
thus, y = -1
Solution 30
Solution 31
(i) True.
(ii) False.
Subtraction of matrices is not commutative.
(iii) True.
Multiplication of matrices is associative.
(iv) True.
Multiplication of matrices is distributive over addition.
(v) True.
Multiplication of matrices is distributive over subtraction.
(vi) True.
Multiplication of matrices is distributive over subtraction.
(vii) False.
Laws of algebra for factorization and expansion are not applicable to matrices.
(viii) False.
Laws of algebra for factorization and expansion are not applicable to matrices.
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 9 - Matrices Page/Excercise 9(D)
Solution 1
Comparing the corresponding elements, we get,
6x - 10 = 8
6x = 18
x = 3
-2x + 14 = 4y
4y = -6+ 14 = 8
y = 2
Solution 2
Comparing the corresponding elements, we get,
3x + 18 = 15
3x = -3
x = -1
12x + 77 = 10y
10y = -12 + 77 = 65
y = 6.5
Solution 3
(i) x, y Î W (whole numbers)
It can be observed that the above two equations are satisfied when x = 3 and y = 4.
(ii) x, y Î Z (integers)
It can be observed that the above two equations are satisfied when x = 3 and y = 4.
Solution 4
(i)
(ii)
Solution 5
Solution 6
Let the order of matrix M be a x b.
3A x M = 2B
Clearly, the order of matrix M is 2 x 1.
Comparing the corresponding elements, we get,
-3y = -10
y =
12x - 9y = 12
Solution 7
Comparing the corresponding elements, we get,
a + 1 = 5 a = 4
2 + b = 0 b = -2
-1 - c = 3 c = -4
Solution 8
(i)
(ii)
Solution 9
Comparing the corresponding elements, we get,
5x = 5x = 1
6y = 12 y = 2
Solution 10
Solution 11
Given, A + X = 2B + C
Solution 12
Given, A2 = B
Comparing the corresponding elements, we get,
x = 36
Solution 13
Solution 14
Solution 15
Solution 16
A =
A2 = A A =
=
AB = A B =
=
=
B2 = B x B =
=
=
A2 + AB + B2 =
=
Solution 17
Comparing the corresponding elements, we get,
3a - 8 = 24 3a = 32 a =
24 - 2b = 0 2b = 24 b = 12
11 = 6c c =
Solution 18
A =
BA =
C2 =
BA = C2 =
By comparing,
-2q = -8 q = 4
And p = 8
Solution 19
AB =
Solution 20
=
=
Solution 21
Solution 22
A2 = 9A + MI
A2 - 9A = mI ….(1)
Now, A2 = AA
Substituting A2 in (1), we have
A2 - 9A = mI
Solution 23
(i) Let the order of matrix X = m × n
Order of matrix A = 2 × 2
Order of matrix B = 2 × 1
Now, AX = B
m = 2 and n = 1
Thus, order of matrix X = m × n = 2 × 1
Multiplying (1) by 2, we get
4x + 2y = 8 ….(3)
Subtracting (2) from (3), we get
3x = 3
x = 1
Substituting the value of x in (1), we get
2(1) + y = 4
2 + y = 4
y = 2
## Browse Study Material
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1934: Southern Workers Spark Massive Textile Strike
(September 2013) In 1934, thousands of workers in Southern textile mills walked off the job seeking better pay and working conditions. The job actions they launched spread to New England and the Mid-Atlantic states and became one of the biggest industrial strikes in U. S. history. Though the strike was unsuccessful, it helped pave the way for stronger laws to protect workers seeking to join unions.
By the early 1930s, approximately 70 percent of American textile production had shifted from New England to southern states, where manufacturers took advantage of dispossessed farmers and unemployed laborers who were willing to work for less.
To keep profits up, textile manufacturers had also begun the practice of “stretch-outs” on the factory floor: paying reduced “piece rates,” limiting breaks, and hiring more supervisors to discipline workers and speed production.
Most of the textile employees in the south were poor whites who lived near the mills. (Southern textile manufacturers largely refused to hire African-Americans.) Workers routinely labored 55 to 60 hours per week on dangerous machinery and brought home less than $10.
The mill workers had little experience challenging the economic and political structure in their communities, but they did not take kindly to the stretch-outs and decided to make a stand. In 1929, employees at textile factories in South Carolina spontaneously walked off the job to protest working conditions more than 80 times. Hundreds of other strikes took place that year, including job actions in Gastonia NC and Elizabethton TN. Some succeeded, but most failed due to a lack of organized union support and the often violent tactics of the mill owners.
Meanwhile, reduced wages and stretch-outs were causing unrest among wool workers in New England, silk weavers in New Jersey, and across the entire industry.
In June 1933, with the Great Depression in full force, Congress and the administration of Franklin D. Roosevelt sought to address the country’s economic woes in part by passing the National Industrial Recovery Act (NIRA).
To build cooperation among businesses and unions, expand employment, raise wages, and protect workers who sought to join unions, the law established the National Recovery Administration (NRA) to develop codes of conduct for particular industries.
The NRA established the Cotton Textile Board to foster fair competition, regulate prices, prevent over-production of textile goods, and guarantee workers a minimum wage. With great expectations for labor protections under the NIRA, tens of thousands of southern cotton mill workers joined the United Textile Workers union. UTW membership rocketed from 15,000 in February 1933, to 250,000 by June 1934.
Mill owners, however, came to dominate the board. They ignored the intent of the law, developed self-serving standards, and began “stretching out the stretch-out,” wrote historians James Leloudis and Kathryn Walbert. They “effectively turned the minimum wage into the maximum that most workers could earn and laid off thousands of additional hands.”
Most southern mill owners “refused to negotiate with or recognize any union representation,” wrote B.J. Davis for North Carolina’s Museum of History. “Wages remained low, the stretch-out was still common.” In July 1934, 20,000 frustrated Alabama workers walked out of their mills, demanding an end to the stretch-out, a $20 minimum wage for a 30-hour work week, union recognition, and reinstatement of workers fired for union activity, Leloudis and Walbert noted.
With workers in other states eager to begin similar actions, at a convention that summer the UTW hastily called for an industry-wide strike.
The union promised full support, but had too little time to plan, too few staff, and inadequate finances for a lengthy faceoff with the mill owners and politicians who were determined to fight.
On Sept. 3, 1934, nearly 10,000 workers marched in the Labor Day parade in Gastonia NC, where authorities had brutally suppressed a textile strike five years earlier. The next day, 20,000 of the city’s mill workers walked off the job. The strike spread swiftly through the south and to northern textile mills as well, outpacing the UTW’s ability to coordinate events and provide assistance. By mid-September, more than 400,000 textile workers had walked off the job demanding an end to the stretch-outs and union recognition.
Though the strike caused a sharp decline in textile production, most mill owners had well-stocked warehouses and were determined to defeat the strikers once and for all.
Insisting that union agitators were coercing employees to go on strike, “mill owners across the South responded to the strike by combining ‘armed self-defense with calls for military intervention,’” wrote Leloudis and Walbert.
The mill bosses called on their political supporters for help. South Carolina Gov. Ibra Charles Blackwood immediately deputized the state’s “mayors, sheriffs, peace officers and every good citizen” to “maintain order,” then “dispatched the National Guard with orders to shoot to kill any picketers who tried to enter the mills.” Gov. Ehringhaus of North Carolina followed suit on Sept. 5.
The governors of Maine and Rhode Island also called out the Guard, as did Georgia Gov. Eugene Talmadge, who declared martial law and ordered that picketers be arrested and held pending trial by a military tribunal at a former World War I prisoner of war camp.
Southern mill operators got especially rough, hiring strikebreakers to intimidate workers and beat union activists.
On Sept. 6, J.D. Beacham, the superintendent of the Chiquola Mill in Honea Path SC, and the mayor ordered police and 100 private guards to open fire on workers who had gathered at the mill to protest working conditions and wages. A “hail of fire from non-union workers and special deputies” killed seven workers and wounded a score of others, the Anderson Independent reported, leaving wives and children “sobbing in agony as they rushed to the sides of their dead.”
Many other strikers also paid dearly for showing up on picket lines. Three were shot in Georgia, one in Rhode Island, and hundreds were beaten or jailed.
The strikers soon came to realize the odds were against them and that there was not much the UTW and the NIRA could do to protect them.
Three weeks after the strike began, faced with brute force and financial need, workers began returning to the mills. On Sept. 22, the UTW called off the job action and declared victory, saying it had scored points for the workers and that their efforts would bear fruit under a Textile Labor Relations Board the Roosevelt Administration was to create.
In reality, however, the strike was a stunning defeat. The UTW achieved no gains, and southern employers refused to reinstate many thousands of strikers.
While many hailed the labor protections proposed in the NIRA, the law lacked effective enforcement mechanisms and did little to help unions organize workers.
In May 1935, the U.S. Supreme Court overturned the NIRA, but two months later President Roosevelt signed into law the National Labor Relations Act, which more firmly established workers’ legal rights to join unions, engage in collective bargaining, and strike if necessary.
While that law and the 1938 Fair Labor Standards Act helped unions lift millions of workers out of poverty, union attempts to organize workers in the south have continued to lag. Nonetheless, today textile manufacturers have largely moved production to overseas sweatshops, and the fight for justice for textile workers has gone global.<|endoftext|>
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# On Average Jun 2022
Here’s the whole story visually:
The rest is details.
Most everyone in math knows these tricks — here’s a summary of techniques used to make quick mental calculations of averages.
In the problems below, consider the residual,
R is defined as the remainder above or below the guessed average, aka, the residual.
RS is defined as the residual sum.
1. In school, I have 8 exam scores; what’s my average?
90
92
88
95
97
81
85
100
~~~
STEP by STEP (with explanation)
Guess maybe 90, then mentally calculate the running residual sum using two or three numbers at a time:
say zero (for 90: 90 is neither above or below the guessed average ∴R=0; ∴RS = 0),
say zero (for 92 and 88: the average of 92 and 88 is neither above or below the guessed average ∴R=0; ∴RS = 0),
say 3 (for 95, 97, and 81: ∴R = 5 + 7 – 9; ∴RS = 3),
say 8 (for 85 and 100: ∴R= -5 + 10 = 5; ∴RS = 8).
Answer: The average is 90 + 8/8 = 91.
2. Same problem — different guess. In school, I have 8 exam scores; what’s my average?
90
92
88
95
97
81
85
100
~~~
STEP by STEP (with explanation)
Guess maybe 85, then mentally calculate the running residual sum using two or three numbers at a time:
say 15 (for 90, 92, 88: ∴R = 5 + 7 + 3 = 15; ∴RS = 15),
say 37 (for 95 and 97: ∴R = 10 + 12 = 22; ∴RS = 37),
say 48 (for 81, 85, and 100: ∴R = -4 + 0 + 15 = 11; ∴RS = 48),
Answer: The average is 85 + 48/8 = 91.
3. Find average of 5 ‘hard numbers’ (with a lot of 9s, 8s, and 7s):
987
965
882
949
767
~~~
STEP by STEP (with explanation)
Assume a basis of 1000, mentally calculate the running residual sum:
say 13 (for 987: ∴R = -13; ∴RS = 13 down),
say 48 (for 965: ∴R = -35; ∴RS = 48 down),
say 99 (for 949: ∴R = -51; ∴RS = 99 down),
say 217 (for 882: ∴R = -118; ∴RS = 217 down),
say 450 (for 767: ∴R = -233; ∴RS = 450 down),
Answer: The average is 1000 – 450/5 = 910.
4. Incrementing an average, for example, I have 8 exam scores:
90
92
88
95
97
81
85
100
with an average of 91 (see Problem 1). Now I have a score of 78; what is my new average without re-summing?
Answer: It is simply 91 + (78 – 91)/9 = 91 – 13/9 = 89 5/9 = 89.55
Formula:
5. [Powerful Formula] Incrementing an average, for example, I have 100 exam scores with an average of 87. Now I have a score of 100; what is my new average without re-summing?
Answer: It is simply 87 + (100 – 87)/101 = 87.1287
6. Incrementing an average, for example, I have 10,000 exam scores with an average of 87. Now I have a score of 100; what is my new average without re-summing?
Answer: It is simply 87 + (100 – 87)/10001 = 87.0013
7. [Very Powerful Formula] Incrementing an average, for example, I have 100 exam scores with an average of 87. Now I have 3 scores of {85, 80, 75}; what is my new average without re-summing?
87 – 85 = 2
87 – 80 = 7
87 – 75 = 12
Answer: It is simply 87 – (2 + 7 + 12)/103 = 86.7961
8. Determine the lowest allowable score to maintain a certain average. Say I have 8 scores with an average of 94. What is the least score on my 9th exam to maintain at least an average of 90?
Answer: It is 94 + (90 – 94) · 9 = 58.
Formula:
9. [Very Very Powerful Formula] Determine the lowest allowable final exam score to maintain a certain average. Say I have 8 scores with an average of 94. What is the least score on my final exam to maintain at least a course grade of 90? Assume that the final exam is 40% of my course grade.
Answer: It is [90 – (94 x 0.6)]/0.4 = 84.
Formula:
[Wicked Powerful Formula] Alternative and easier formula:
10. So, how much does a 100 raise your average?
Formula:
Enjoy!<|endoftext|>
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Arithmetic/Properties of Operations/Why Do Operations Have Properties?
Before continuing to learn about the properties of operations you must understand two basic questions:
What is the point of operations having properties?
Without operations having properties, we would not know their usage. Understanding their usage helps lay the foundation for solving word problems later on.
Why do operations have properties?
Operations have properties which define their usage. The usage of operations is the very essence of them, without usage properties are useless (vice versa).
With understanding properties we are able to enter the realm of higher-level thinking.
This is because properties illustrate general cases which allow us to lead to more mathematical generalizations.
Definitions of Mathematical Properties:
1. Commutative Property: ${\displaystyle a+b=b+a}$
2. Associative Property: ${\displaystyle (a+b)+c=a+(b+c)}$
3. Identity Property of zero: ${\displaystyle 0+a=a(=a+0)}$
4. Inverse Property: For every member a, there is - a such that ${\displaystyle a+(-a)=0}$
Multiplication:
1. Commutative Property: ${\displaystyle ab=ba}$
2. Associative: ${\displaystyle (ab)c=a(bc)}$
3. Identity: ${\displaystyle a\cdot 1=1\cdot a=a}$
4. Inverse Property, for every ${\displaystyle a\neq 0}$, there is ${\displaystyle ({\dfrac {1}{a}})}$ (or ${\displaystyle a^{-1}}$), such that a ${\displaystyle ({\dfrac {1}{a}})=1.}$
The general rule for the inverse property of multiplication is if when you multiply two numbers and the product is 1, then what you multiplied must be multiplicative inverses or reciprocals of each other.
It is important you remember that connecting addition and multiplication is the:
Distributive Rule: ${\displaystyle a(b+c)=ab+ac.}$
Often rules are final factor of what you get as your answer:
An example of this is shown below:
${\displaystyle a\cdot 0=0}$, because ${\displaystyle a=a\cdot 1=a\cdot (1+0)=a\cdot 1+a\cdot 0.}$
Now subtract a from both sides to get ${\displaystyle 0=0+a\cdot 0=a\cdot 0.}$<|endoftext|>
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## In the 2020 presidential election, there are N = 1; 000; 000 eligible voters in Yolo county. Each votes with probability p = 49% for Donald
Question
In the 2020 presidential election, there are N = 1; 000; 000 eligible voters in Yolo county. Each votes with probability p = 49% for Donald Trump and with probability q = 51% for Joe Biden. Votes of different voters are independent. Assume that all voters do vote. What is the probability that Joe Biden wins Yolo county? Wining the Yolo county means getting more than half of the votes.
in progress 0
3 months 2021-10-22T19:09:23+00:00 2 Answers 0 views 0
100% probability that Joe Biden wins Yolo county
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
For a proportion q in a sample of size n, the mean is and the standard deviation is
In this problem, we have that:
So
What is the probability that Joe Biden wins Yolo county?
This is the probability that he gets more than 50% = 0.5 of the votes, so it is 1 subtracted by the pvalue of Z when X = 0.5. So
has a pvalue of 0
1 – 0 = 1
100% probability that Joe Biden wins Yolo county<|endoftext|>
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# How To Do KS2 Maths SATs Paper B Percentage Questions (Part 1)
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### How To Do KS2 Maths SATs Paper B Percentage Questions (Part 1)
1. 1. How To Do KS2 Maths SATs Type Questions (Paper B – Calculator Allowed) Percentages 1: Shading In A Percentage Of A Shape For more maths help & free games related to this, visit: www.makemymathsbetter.com
2. 2. In a SATs Paper B you might be asked to shade in a percentage of a shape:
3. 3. For example shade in 20% of this grid:
4. 4. For example shade in 20% of this grid: First count the number of squares.
5. 5. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example.
6. 6. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example. You therefore need to find 20% of 30
7. 7. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example. You therefore need to find 20% of 30 To find 20% of 30, you need to remember that 20% is equivalent to 1/5
8. 8. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example. You therefore need to find 20% of 30 To find 20% of 30, you need to remember that 20% is equivalent to 1/5 So 20% of 30 = 1/5 of 30
9. 9. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example. You therefore need to find 20% of 30 To find 20% of 30, you need to remember that 20% is equivalent to 1/5 So 20% of 30 = 1/5 of 30 = 6
10. 10. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example. You therefore need to find 20% of 30 To find 20% of 30, you need to remember that 20% is equivalent to 1/5 So 20% of 30 = 1/5 of 30 = 6 Therefore you need to shade in 6 of the squares.
11. 11. For example shade in 20% of this grid: First count the number of squares. There are 30 in this example. You therefore need to find 20% of 30 To find 20% of 30, you need to remember that 20% is equivalent to 1/5 So 20% of 30 = 1/5 of 30 = 6 Therefore you need to shade in 6 of the squares.
12. 12. Another example would be, shade in 25% of this grid:
13. 13. Another example would be, shade in 25% of this grid: First count the number of squares.
14. 14. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example.
15. 15. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example. You therefore need to find 25% of 24
16. 16. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example. You therefore need to find 25% of 24 To find 25% of 24, you need to remember that 25% is equivalent to 1/4
17. 17. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example. You therefore need to find 25% of 24 To find 25% of 24, you need to remember that 25% is equivalent to 1/4 So 25% of 24 = 1/4 of 24
18. 18. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example. You therefore need to find 25% of 24 To find 25% of 24, you need to remember that 25% is equivalent to 1/4 So 25% of 24 = 1/4 of 24 = 6
19. 19. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example. You therefore need to find 25% of 24 To find 25% of 24, you need to remember that 25% is equivalent to 1/4 So 25% of 24 = 1/4 of 24 = 6 Therefore you need to shade in 6 of the squares.
20. 20. Another example would be, shade in 25% of this grid: First count the number of squares. There are 24 in this example. You therefore need to find 25% of 24 To find 25% of 24, you need to remember that 25% is equivalent to 1/4 So 25% of 24 = 1/4 of 24 = 6 Therefore you need to shade in 6 of the squares.
21. 21. Now, try some by yourself to see if you have understood. Click to reveal the answer: 1) Shade in 10% of this grid:
22. 22. Now, try some by yourself to see if you have understood. Click to reveal the answer: 1) Shade in 10% of this grid: 10% of 30 = 3
23. 23. Now, try some by yourself to see if you have understood. Click to reveal the answer: 1) Shade in 10% of this grid: 10% of 30 = 3
24. 24. Now, try some by yourself to see if you have understood. Click to reveal the answer: 2) Shade in 50% of this grid:
25. 25. Now, try some by yourself to see if you have understood. Click to reveal the answer: 2) Shade in 50% of this grid: 50% of 36 = 18
26. 26. Now, try some by yourself to see if you have understood. Click to reveal the answer: 2) Shade in 50% of this grid: 50% of 36 = 18
27. 27. Now, try some by yourself to see if you have understood. Click to reveal the answer: 3) Shade in 30% of this grid:
28. 28. Now, try some by yourself to see if you have understood. Click to reveal the answer: 3) Shade in 30% of this grid: 30% of 30 = 9
29. 29. Now, try some by yourself to see if you have understood. Click to reveal the answer: 3) Shade in 30% of this grid: 30% of 30 = 9
30. 30. Now, try some by yourself to see if you have understood. Click to reveal the answer: 4) Shade in 75% of this grid:
31. 31. Now, try some by yourself to see if you have understood. Click to reveal the answer: 4) Shade in 75% of this grid: 75% of 20 = 15
32. 32. Now, try some by yourself to see if you have understood. Click to reveal the answer: 4) Shade in 75% of this grid: 75% of 20 = 15
33. 33. Now, try some by yourself to see if you have understood. Click to reveal the answer: 5) Shade in 90% of this grid:
34. 34. Now, try some by yourself to see if you have understood. Click to reveal the answer: 5) Shade in 90% of this grid: 90% of 40 = 36
35. 35. Now, try some by yourself to see if you have understood. Click to reveal the answer: 5) Shade in 90% of this grid: 90% of 40 = 36
36. 36. That’s it for now...... for more help with your maths, try my book: mastering multiplication tables on amazon.com<|endoftext|>
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DNA size to scale, the biggest instruction set ever written
Let’s say we can print your DNA in a book, how many pages should it have? Fortunately for us, The Human Code Foundation literally printed a human genome, which is the entire collection of DNA a living thing has, and this resulted in 175 volumes of 1,600 pages each! The enormous amount of information that defines a living organism is stored in these DNA molecules.
Not surprised? Well now consider this, that collection of books is the amount of information we can find in a single cell when it contains a unique copy of the code, also most of our cells include twice this amount and we are made of trillions of cells!
An even more interesting question might be: How cells manage to store so much in such a small space?
Instead of just giving you numbers like the length of DNA molecules inside a single cell, or the diameter of the nucleus, which is the inner compartment in which cells store DNA, allow me to picture the proportion between the size of both, DNA molecules and the nucleus, in terms of objects we are more familiar with.
First, consider that we are about to use as reference the equivalent of two copies of the entire Human Code books collection mentioned above because this is the amount of information present in most of our cells. Actually, if we could put all these 46 DNA molecules one next to each other as a stretched line it should reach 2 meters in length! On the other hand, the diameter of a nucleus is equivalent to 0.00001 meters, see where we are going?
Now let’s use our imagination:
If a mandarin orange represents the nucleus of a cell and the 46 DNA molecules inside it are represented as silk fibres, then the resulting length of putting the 46 fibres next to each other as a stretched line should reach 16 kilometres, which is longer than the depth of the Mariana Trench, the deepest spot on Earth’s oceans. Take a moment to picture that in your mind, you swimming at that point, and all the distance between you and the bottom of the ocean … inside a delicious mandarin orange!
To be fair the width of the fibres are also important and above we represented them as silk fibres following the same proportion considering the mandarin orange as a reference to a human cell nucleus. Also, the width of a silk fibre is slightly below the smallest object visible to the naked eye but still, I am sure you can better feel the dimension of the question we did before: How cells manage to store so much in such a small space?
As you might expect, a complex packaging system is part of the answer and the key idea behind it is that it shows several levels of folding. A group of auxiliary proteins called histones play a central role and the resulting DNA-protein structure is known as chromatin. Let’s take a closer look.
In the image only the first packaging level is visible: the nucleosome. These units, seen as black dots (black arrowheads), are made by histones and the DNA, seen as lines between the dots (white arrowheads), wraps around each unit almost two times. From here some sections of chromatin get more tightened as the fibre made of nucleosomes coils in the second level of packaging: the 30-nano meter fibre. Yes, most of the names in biology are as original as this one.
Additional levels of packaging arise when the cell enters division, but that is a story for another day. More research is needed, as new discoveries always come up with more questions.<|endoftext|>
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$$\require{cancel}$$
# 03. Applying Newton's Second Law to Circular Motion
[ "article:topic", "authorname:dalessandrisp" ]
### Applying Newton's Second Law to Circular Motion
Investigate the situation below, in which an object travels on a circular path.
During a high speed auto chase in San Francisco, a 1745 kg police cruiser traveling at constant speed loses contact with the road as it goes over the crest of a hill with radius of curvature 80 m.
A free-body diagram for the car at the instant it’s at the crest of the hill is sketched below.
Since the car is traveling along a circular path, it’s best to analyze the situation using polar coordinates. Remember, in polar coordinates the radial direction points away from the center of the circular path.
In general, you should apply Newton’s second law independently in the radial (r) and tangential (f) directions, although since the cruiser is moving at constant speed, there is no acceleration in the tangential direction and therefore no net force in the tangential direction.
Since the cruiser loses contact with the road, Froad = 0 N.
$0 - 1745(9.8) = -1745(80) \omega^2$
$\omega = 0.35 \; rad/s$
Since $$v = R \omega$$,
$v = (80 m)(0.35 \; rad/s)$
$v =28 \; m/s$
For the car to lose contact with the road, it must be traveling at a speed of 28 m/s or greater when it reaches the crest of the hill. At lower speeds, the car would remain contact with the road.
Since it is often useful to directly relate the radial acceleration of an object to its velocity, let's construct a direct relationship between these two variables:
$a_\rho =R \omega ^2$
$a_\rho = -R \left( \frac{v}{R} \right )^2$
$a_\rho = \frac{-v^2}{R}$
This relationship, although mathematically equivalent to $$a_\rho =R \omega ^2$$, is often a more "useful" form.
### Another Circular Motion Scenario
A 950 kg car, traveling at a constant 30 m/s, safely makes a lefthand-turn with radius of curvature 75 m.
First, let's draw a pair of free-body diagrams for the car, a side-view (on the left) and a rear-view (on the right)
The free-body diagram on the left is a side-view of the car. Notice that the upward direction is the y-direction, and the forward direction, tangent to the turn, is the f-direction.
The free-body diagram on the right is a rear-view of the car. This is what you would see if you stood directly behind the car. Notice that the upward direction is still the y-direction, and the horizontal direction, perpendicular to the direction of travel and hence directed radially outward, is the r-direction. The frictional force indicated is perpendicular to the tread on the tire. This force causes the car to accelerate toward the center of the turn. Remember, if the car is going to travel along a circular path, it must have an acceleration directed toward the center of the circle. Something has to be supplying the force that creates this acceleration. This something is the static friction between the tire and the road that acts to prevent the car from sliding out of the turn. Since the car has no velocity in the radial direction, the frictional force the points in this direction must be static!
Now that we have all that straightened out (maybe), let's apply Newton's Second Law.
Thus, to safely make this turn requires at least 1140 N of static friction. Using this value, I should be able to compute the minimum coefficient of friction necessary for the car to safely round this turn at this speed.
$F_{stiction} \le \mu_sF_{road}$
$11400 \le \mu_s(9310)$
$\mu_s \ge 1.22$
Although this is a large value for the coefficient of static friction, it is an attainable value for a sports car with performance tires.<|endoftext|>
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Learning in the technologies enables pupils to:
- develop an understanding of the role and impact of technologies in changing and influencing societies
- contribute to building a better world by taking responsible, ethical actions to improve my life, the lives of others and the environment
- gain the confidence and skills to embrace and use technologies now and in the future, at home, at work and in the wider community
- become an informed consumer and producer who has an appreciation of the merits and impacts of products and services
- be capable of making reasoned choices relating to the environment, sustainable development and ethical, economic and cultural issues
- broaden understanding of the role that information and communications technology (ICT) has in Scotland and in the global community
- broaden awareness of how ideas in mathematics and science are used in engineering and the technologies
- experience work-related learning, and establish firm foundations for lifelong learning, and specialised study and careers.
We are currently looking to developing digital literacy.
Technology can be taught either discreetly ( only looking at the Experiences and Outcomes within technology) or in an interdisciplinary way<|endoftext|>
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# 4 The sum of the lenghts of any two sides.of a triangle is always greater than thelent lenght of third side .for
4 The sum of the lenghts of any two sides.
of a
triangle is always greater than the
lent lenght of third side .formula
### 1 thought on “4 The sum of the lenghts of any two sides.<br />of a <br />triangle is always greater than the<br />lent lenght of third side .for”
Note: This rule must be satisfied for all 3 conditions of the sides.
Step-by-step explanation:
Consider three line segments with lengths a , b , and c.
Construct two circles with radii a and b at the end points of the segment with length c .
There are three possible relationships between a+b , and c :
a+b<c: the circles will not intersect. There will be a gap on segment c between the two circles.
a+b=c: the circles will be tangent to each other. They will touch each other at a point on c .
a+b>c: the circles might
intersect each other, or
one of the circles might be large enough to fully contain the other.
That is the case when either a≥b+c , or b≥a+c.
It seems that only in the third case of a+b>c do we end up with the possibility for a triangle, and only if we are also able to apply the same reasoning to the other two sides when being considered as the base.
So, if three line segments with lengths a , b , and c can form a triangle, it must be the case that:
a+b>c ,
b+c>a , and
c+a>b .<|endoftext|>
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Animated infographics can be a great way to capture the attention and encourage sharing on social platforms. In this example we look at some of the surprising facts and figures that emerge when the world population is represented as a village of 100 people. Statistics taken from http://www.100people.org/.
7 billion people share this planet but if the world’s entire population was squeezed into a village of only 100 people it would look something like this.
Unsurprisingly, half of the people in the village would be male, half would be female, 26 would be children and 74 would be adults. 8 of these would be 65 or over representing the growing age of our population. There would be five people from the US and Canada, 9 from Latin America, 11 from Europe, 15 from Africa, and 60 from Asia.
Migrations towards cities means that 51 people would live in the rural outskirts of our village, and 49 in the much smaller urban centre. 77 would have shelter from the elements, 23 would not.
Of those that do have housing, the majority would be substandard. 13 people would not have access to clean water and 22 would not have any electricity. Of those that do, most would use it only for light at night. 18 people would be unable to read or write. But while 10 years ago only 1 person would would have a degree education, today that’s leapt up to 7.
15 would be malnourished. 1 would be dying of starvation, but 21 would be overweight. Nearly half of the population would live on less than 2 dollars a day, and one fifth would have to live on less than one dollar.
And every year 1 person would die, but 2 more would be born to take their place.<|endoftext|>
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SITEMAP School-college Physics Notes: Electricity 3.7 Division of p.d. in a circuit
UK GCSE level age ~14-16 ~US grades 9-10 Scroll down, take time to study content or follow links
Electricity Section 3: 3.7 How the total potential difference is split between resistances in series
Doc Brown's Physics exam study revision notes: Explaining and verifying by calculation, how the potential difference (p.d.) is split (divided, shared) between resistances in series i.e. the effect of two (or more) resistors in series
3.7 A little more on potential difference - effect of two resistors in series - explaining and verifying by calculation, how the potential difference is split between resistances in series
The circuit 41 shows two resistors wired in series, so how is the total p.d. divided or shared between the two resistance wired in series.
This circuit can be used to investigate and confirm the correctness of the following theoretical calculations.
On the right is shown what happens to the p.d. going clockwise around the circuit (direction of convention current).
The potential store of the battery raises the potential difference of the charge to 12.0 V.
The total resistance = 10.0 + 5.0 = 15.0 Ω (you can add resistors up if wired in series)
Therefore the current flowing round all of this series circuit = I = V / Rtotal = 12.0 / 15.0 = 0.80 A
As the charge passes through the 1st resistor R1, it loses energy and the p.d. falls by 8 V to a p.d. of 4 V.
Calculation to confirm this: V1 = I x R1 = 0.80 x 10.0 = 8.0 V
As the charge passes through the 2nd resistor R1, it loses energy again and the p.d. falls by 4 V to a p.d. of 0 V.
Calculation to confirm this: V2 = I x R2 = 0.80 x 5.0 = 4.0 V
As long as there is a complete circuit, the process repeats itself through any number of resistors.
What is clear is the total p.d. for the circuit is split into a p.d. ratio identical to the resistor ratio.
V1 : V2 is the same ratio as R1 : R2
You can also say that since E = QV, twice as much energy is released by resistor R1 (p.d. 8 V) than R2 (p.d. 4 V) for the same total current, the same total charge transferred, but the lower the p.d. the less potential energy carried by the charge.
Keywords, phrases and learning objectives for ? electricity
Be able to explain and verifying by calculation (and experiment), how potential difference is split shared divided between resistances in series.
Be able to describe the effect of two resistors in series on the p.d. across these resistors wired in series.
WHAT NEXT?
BIG website and using the [SEARCH BOX] below, maybe quicker than navigating the many sub-indexes
for UK KS3 science students aged ~12-14, ~US grades 6-8
ChemistryPhysics for UK GCSE level students aged ~14-16, ~US grades 9-10
for pre-university age ~16-18 ~US grades 11-12, K12 Honors
Use your mobile phone in 'landscape' mode?
SITEMAP Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to GCSE science course specifications are unofficial.
Using SEARCH some initial results may be ad links you can ignore - look for docbrown
@import url(https://www.google.co.uk/cse/api/branding.css); ENTER specific physics words or courses e.g. topic, module, exam board, formula, concept, equation, 'phrase', homework question! anything of physics interest! This is a very comprehensive Google generated search of my website
TOP OF PAGE<|endoftext|>
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# Second Derivatives and Beyond Resources
## Best of the Web. Picked by our PhDs
### Websites
Graphing Functions Using First and Second Derivatives
Let’s be honest: not much has changed since kindergarten. We still get to draw lines and pictures in calculus class. Hard to believe your tuition goes to help teaching you how to draw lines.
Concavity and Points of Inflection
We said inflection, not infection. This is an upbeat take on first derivatives, second derivatives, and points of inflection. You could call it calculus with a smile.
Critical Points
Critical points aren’t something to fear. There isn’t a mutant cat lurking behind them. Just in case, learn how to spot those critical points before the creepy cats spot you.
First and Second Derivative Tests
The first and second derivative tests sound like something your doctor would do to check if you’ve contracted the ebola virus. You have nothing to worry about. This link shows you how an engineer is more likely to need this test than a doctor.
Math is Fun – Global and Local Extrema
How do you know if you are at a global extremum like the peak of Mount Everest or the bottom of the Marianas Trench? What if you’re at a local extremum like Mount Kilimanjaro or at the bottom of Lake Erie?
### Videos
Finding Relative Extrema Using the First Derivative
Most people try to avoid their relatives. Imagine actively trying to find them instead. Relative extrema, that is.
Curve Sketching with Derivatives
When your art teacher tried to teach you how to draw a rabbit, it might have felt like pulling a rabbit out of a hat. Drawing functions are easier, especially once you know their derivatives.
Absolute (or Global) and Local Extrema of Polynomial Functions
You need to be at one with polynomial functions. This video about extrema of polynomials will put you in a zen-like state.
Optimizing to Make a Bridge
Knowing calculus helps you build bridges. At least it helps these guys. Check out this video of how a bridge is made by finding a minimum of a structural function.
### Games and Tools
Excel First and Second Derivative Plotter:
Anything you can do with complicated computer code, you can do with a computer spreadsheet. It’ll just take you a while to make it look this good. Here’s a numerical first and second derivative plotter for Excel.<|endoftext|>
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S u m m a r y :
A giant planet, orbiting a distant, small star, has been spotted—one that should not exist if the planet formation theory is real. The new findings are published in the journal Monthly Notices of the Royal Astronomical Society.
Monster Planet Around Small Star
The discovery of a huge planet—dubbed monster planet, and named NGTS-1b—around a distant star has left scientists puzzled. This planet should not be existing according to the planet formation theory: it is too big to have formed around a star so small. Rather, small stars can only have rocky planets of smaller sizes because they cannot gather sufficient material to generate planets like Jupiter. So, how does this monster planet even exist?
“The discovery of NGTS-1b was a complete surprise to us—such massive planets were not thought to exist around such small stars—importantly, our challenge now is to find out how common these types of planets are in the Galaxy, and with the new Next-Generation Transit Survey facility we are well-placed to do just that,” says lead author Daniel Bayliss from the University of Warwick.
Detecting the Planet’s Orbit
NGTS-1b got its name from the instrument used to identify it, The Next-Generation Transit Survey (or ‘NGTS’) which is composed of 12 telescopes to look into the heavens. It was detected after the study authors monitored regions of the night sky over a period extending over several months; they finally found red light emanating from the star using red-sensitive cameras. When they observed dips in the light of the star every 2.6 hours, they concluded the presence of a planet orbiting around it periodically, in that amount of time. They were ultimately able to detect the orbit of the planet, and they calculated its size, mass, and position, using the radial velocity of the star.
A Year Lasts 2.5 Days
NGTS-1b is a gas giant: it is super big and hot; it is categorised as a hot Jupiter, a group of planets that are comparative in size to our Jupiter, but with about 20% less mass. On the other hand, unlike Jupiter, NGTS-1b is in close proximity with its star, with only 3% of the distance between Earth and the sun. This means that it completes an orbit in 2.6 days only; one year on NGTS-1b is only 2.5 days on Earth.
The Star Behind the Planet
As for its star (a red M-dwarf), its radius and mass are half that of our sun. This is what made NGTS-1b difficult to identify. Peter Wheatley from the University of Warwick explains that the parent star being so small, and thus faint, NGTS-1b was challenging to find even with its monster size. But, given that such a big planet has been spotted around a small star, a type of star that is deemed to be quite common, there might be other similar giant planets out there in the universe.
“Having worked for almost a decade to develop the NGTS telescope array, it is thrilling to see it picking out new and unexpected types of planets. I’m looking forward to seeing what other kinds of exciting new planets we can turn up,” says Wheatley.<|endoftext|>
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ORGANIC AGRICULTURE AND SOIL BIODIVERSITY
LIVING SOILS FOR AGRICULTURE
Soils contain enormous numbers of diverse living organisms assembled in complex and varied communities. Soil biodiversity reflects the variability among living organisms in the soil - ranging from the myriad of invisible microbes, bacteria and fungi to the more familiar macro-fauna such as earthworms and termites. Plant roots can also be considered as soil organisms in view of their symbiotic relationships and interactions with other soil components. These diverse organisms interact with one another and with the various plants and animals in the ecosystem, forming a complex web of biological activity. Environmental factors, such as temperature, moisture and acidity, as well as anthropogenic actions, in particular, agricultural and forestry management practices, affect to different extents soil biological communities and their functions.
Soil organisms contribute a wide range of essential services to the sustainable functioning of all ecosystems. They act as the primary driving agents of nutrient cycling, regulating the dynamics of soil organic matter, soil carbon sequestration and greenhouse gas emissions; modifying soil physical structure and water regimes; enhancing the amount and efficiency of nutrient acquisition by the vegetation; and enhancing plant health. These services are not only critical to the functioning of natural ecosystems but constitute an important resource for sustainable agricultural systems.
HEALTHY SOILS FROM AGRICULTURE
Capturing the benefits of soil biological activity for agricultural production requires adhering to the following ecological principles:
- Supply organic matter. Each type of soil organism occupies a different niche in the web of life and favours a different substrate and nutrient source. Most soil organisms rely on organic matter for food; thus a rich supply and varied source of organic matter will generally support a wider variety of organisms.
- Increase plant varieties. Crops should be mixed and their spatial-temporal distribution varied, to create a greater diversity of niches and resources that stimulate soil biodiversity. For example diverse habitats support complex mixes of soil organisms, and through crop rotation or inter-cropping, it is possible to encourage the presence of a wider variety of organisms, improve nutrient cycling and natural processes of pest and disease control.
- Protect the habitat of soil organisms. The activity of soil biodiversity can be stimulated by improving soil living conditions, such as aeration, temperature, moisture, and nutrient quantity and quality. In this regard, reduced soil tillage and minimized compaction - and refraining chemical use - are of particular note.
Improvement in agricultural sustainability requires, alongside effective water and crop management, the optimal use and management of soil fertility and soil physical properties. Both rely on soil biological processes and soil biodiversity. This calls for the widespread adoption of management practices that enhance soil biological activity and thereby build up long-term soil productivity and health.
Adaptation and further development of soil biodiversity management into sustainable land management practices requires solutions that pay adequate consideration to the synergies between the soil ecosystem and its productive capacity and agro-ecosystem health. One practical example of holistic agricultural management systems that promote and enhance agro-ecosystem health, including biodiversity, biological cycles and soil biological activity is organic agriculture.
ORGANIC AGRICULTURE NURTURES SOIL BIODIVERSITY
Scientific research has demonstrated that organic agriculture significantly increases the density and species of soil's life. Suitable conditions for soil fauna and flora as well as soil forming and conditioning and nutrient cycling are encouraged by organic practices such as: manipulation of crop rotations and strip-cropping; green manuring and organic fertilization (animal manure, compost, crop residues); minimum tillage; and of course, avoidance of pesticides and herbicides use.
Benefits of organic management on soil biological activity are summarized below1:
- Abundant arthropods and earthworms. Organic management increases the abundance and species richness of beneficial arthropods living above ground and earthworms, and thus improves the growth conditions of crops. More abundant predators help to control harmful organisms (pests). In organic systems the density and abundance of arthropods, as compared to conventional systems, has up to 100% more carabids, 60-70% more staphylinids and 70-120% more spiders. This difference is explained by prey deficiency due to pesticide influence as well as by a richer weed flora in the standing crop that is less dense than in conventional plots. In the presence of field margins and hedges, beneficial arthropods are further enhanced, as these habitats are essential for over-wintering and hibernation. The biomass of earthworms in organic systems is 30-40% higher than in conventional systems, their density even 50-80% higher. Compared to the mineral fertilizer system, this difference is even more pronounced.
- High occurrence of symbionts. Organic crops profit from root symbioses and are better able to exploit the soil. On average, mycorrhizal colonization of roots is highest in crops of unfertilized systems, followed by organic systems. Conventional crops have colonization levels that are 30% lower. The most intense mycorrhizal root colonization is found in grass-clover, followed by the vetch rye intercrop. Roots of winter wheat are scarcely colonized. Even when all soils are inoculated with active micorrhizae, colonization is enhanced in organic soil. This indicates that, even at an inoculum in surplus, soil nutrients at elevated levels and plant protection suppress symbiosis. This underlines the importance of appropriate living conditions for specific organisms.
- High occurrence of micro-organisms. Earthworms work hand in hand with fungi, bacteria, and numerous other microorganisms in soil. In organically managed soils, the activity of these organisms is higher. Micro-organisms in organic soils not only mineralize more actively, but also contribute to the build up of stable soil organic matter (there is less untouched straw material in organic than in conventional soils). Thus, nutrients are recycled faster and soil structure is improved. The amount of microbial biomass and decomposition is connected: at high microbial biomass levels, little light fraction material remains undecomposed and vices versa.
- Microbial carbon. The total mass of micro-organisms in organic systems is 20-40% higher than in the conventional system with manure and 60-85% than in the conventional system without manure. The ratio of microbial carbon to total soil organic carbon is higher in organic system as compared to conventional systems. The difference is significant at 60 cm depth (at 80 cm depth, no difference is observed). Organic management promotes microbial carbon (and thus, soil carbon sequestration potential).
- Enzymes. Microbes have activities with important functions in the soil system: soil enzymes indicate these functions. The total activity of micro-organisms can be estimated by measuring the activity of a living cell-associated enzyme such as dehydrogenase. This enzyme plays a major role in the respiratory pathway. Proteases in soil, where most organic N is protein, cleave protein compounds. Phosphatases cleave organic phosphorus compounds and thus provide a link between the plant and the stock of organic phosphorus in the soil. Enzyme activity in organic soils is markedly higher than in conventional soils. Microbial biomass and enzyme activities are closely related to soil acidity and soil organic matter content.
- Wild flora. Large organic fields (over 15 ha) featured flora six times more abundant than conventional fields, including endangered varieties. In organic grassland, the average number of herb species was found to be 25 percent more than in conventional grassland, including some species in decline. Vegetation structure and plant communities in organic grassland are more even and more typical for a specific site than in conventionally managed systems. In particular, field margin strips of organic farms and semi-natural habitats conserve weed species listed as endangered or at risk of extinction. Animal grazing behaviour or routing activity (e.g. pigs) was found important in enhancing plant species composition. Weeds (often sown in strips in organic orchards to reduce the incidence of aphids) influence the diversity and abundance of arthropods and flowering weeds are particularly beneficial to pollinators and parasitoids.
- High-energy efficiency. Organic agriculture follows the ecosystem theory of closed (or semi-closed) nutrient cycle on the farm. Organic land management allows the development of a relatively rich weed-flora as compared to conventional systems. Some "accompanying plants" of a crop are desired and considered useful in organic management. The presence of versatile flora attracts beneficial herbivores and other air-borne or above-ground organisms. Their presence improves the nourishment of predatory arthropods. When comparing diversity and the demand of energy for microbial maintenance (as indicated by the metabolic quotient), it becomes evident that diverse populations need less energy per unit biomass. A diverse microbial population, as present in the organic field plots, may divert a greater part of the available carbon to microbial growth rather than maintenance. In agricultural practice this may be interpreted as an increased turnover of organic matter with a faster mineralization and delivery of plant nutrients. Finally, more organic matter is diverted to build-up stable soil humus.
- Erosion control: Organic soil management improves soil structure by increasing soil activity and thus, reduces erosion risk. Organic matter has a positive effect on the development and stability of soil structure. Silty and loamy soils profit from organic matter by an enhanced aggregate structure. Organic matter is adsorbed to the charged surfaces of clay minerals. The negative charge decreases with increasing particle size. Silt is very susceptible to erosion since it is not charged, but organic matter layers on the silt surface favor aggregates with silt too.
MAINSTREAMING ORGANIC AGRICULTURE
In line with the Convention on Biological Diversity, organic agriculture can enhance the value of biological diversity by linking conservation efforts with social and economic benefits. Decision III/11 on Conservation and Sustainable Use of Agricultural Biological Diversity "encourages the development of technologies and farming practices that not only increase productivity, but also arrest degradation as well as reclaim, rehabilitate, restore and enhance biological diversity and monitor adverse effects on sustainable agricultural biodiversity" such as "inter alia, organic farming".
Target 12 of the Global Strategy for Plant Conservation of the Convention on Biological Diversity (i.e., "30 per cent of plant-based products derived from sources that are sustainably managed") identifies organic agriculture as a main indicator to monitor progress towards this target.
Organic agriculture meets precise standards which are verified through certification; in 2002, the International Federation of Organic Agriculture Movements has included "organic ecosystems", with specific biodiversity parameters, within its International Basic Standards for Organic Production and Processing.
Improved awareness on the potential of organic agriculture to provide food while conserving biodiversity offers both a practical option to implement commitments made by governments to the Convention on Biological Diversity as well as more coherence to national policies (and related support) to agriculture and environment schemes.<|endoftext|>
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Dental decay is essentially the weakening/softening of the tooth structure, sometimes to the point where holes are visible in the teeth. Decay can appear as dark and discoloured areas of the teeth, but it can also be completely invisible to the naked eye. Generally the amount of infected tissue is much larger than the damage seen from the tooth surface – rather like what you see of an iceberg from above the water, compared to what you see below the surface.
How does decay form?
Bacteria (the fuzzy stuff we call plaque), continually builds up on the teeth. When the bacteria combines with sugar, it produces acid which breaks down protective minerals in the teeth. This leads to soft, infected tissue breakdown in the teeth.
How can you prevent dental decay?
As they say, prevention is better than cure – which is always true in dentistry!
Follow these ten simple steps for a brighter, healthier, decay-free smile.
We know you’ve heard it before, but visit your dentist every 6 months for a general check up. This will help to pick up on decay early and potentially work towards reversing it or treating it (if it is no longer reversible).
Brush your teeth for two minutes, twice a day – and brush them well. Your dental health professional can advise you on techniques that will work best for you. Consider occasionally using plaque-disclosing tablets or drops to show up any areas that need special attention.
Decay can really build up between teeth, so floss your teeth once a day to remove plaque and debris from those hard to reach areas.
Use tooth mousse. When spread over the teeth, tooth mouse will remineralise small areas of decayed enamel and help prevent future decay. The results will need to be closely monitored over several months by your dentist.
Reduce how often and how much processed sugar you consume. You may consider keeping a food diary to find out where the sugar content in your diet is coming from. Remember, there are many hidden sugars in the food and drinks we consume.
Eat foods that are high in calcium like yogurt, milk and cheese. (This is most beneficial to children with growing teeth and bones).
Avoid sticky sugars and keep any consumptions of sugars to meal times, when saliva flow is highest and can help to flush away some of the sugar content.
Chew sugar free chewing gum to further stimulate saliva flow (provided you do not have jaw pain, as this can be worsened with chewing gum).
Allow your children to drink unfiltered water. Despite what you may read, here in Perth our fluoride levels are carefully monitored and controlled so that developing teeth receive optimum levels of fluoride. These levels of fluoride are safe for consumption, and help the formation of strong, healthy enamel to prevent decay.
No one likes to pay for things they don’t need – especially when visiting the dentist! But, if your dentist advises dental X-rays to diagnose hidden decay, take this advice. Because much decay is invisible to the naked eye, X-ray is the only way of seeing into the tooth structure to assess decay, without first drilling into it!
Is decay treatable?
In some cases, if diagnosed early enough decay is reversible. If decay is too severe or advanced for reversal, the affected tooth may require a filling. In extreme cases, extensive decay can lead to root canal infection, which can only be treated with root canal treatment or extraction of the infected tooth.
To avoid the need for treatments such as these, follow our 10 simple steps and prevent decay from ever forming in the first place!
Time for a check up?
Contact the friendly team at The Smile Clinique to book your next general dental health check.
Don’t forget to share this via , Google+, Pinterest and LinkedIn.<|endoftext|>
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## Shamir’s Quest: Collect Any 3 Keys To Unlock The Secret!
This post is on something called Shamir’s Secret Sharing. It’s a technique where you can break a secret number up into $M$ different pieces, where if you have any $N$ of those $M$ pieces, you are able to figure out the secret.
Thinking of it in video game terms, imagine there are 10 keys hidden in a level, but you can escape the level whenever you find any 7 of them. This is what Shamir’s Secret Sharing enables you to set up cryptographically.
Interestingly in this case, the term sharing in “secret sharing” doesn’t mean sharing the secret with others. It means breaking the secret up into pieces, or SHARES. Secret sharing means that you make shares out of a secret, such that if you have enough of the shares, you can recover the secret.
## How Do You Share (Split) The Secret?
The basic idea of how it works is actually really simple. This is good for us trying to learn the technique, but also good to show it’s security since there are so few moving parts.
It relies on something called the Unisolvence Theorem which is a fancy label meaning these things:
• If you have a linear equation, it takes two (x,y) points to uniquely identify that line. No matter how you write a linear equation, if it passes through those same two points, it’s mathematically equivelant.
• If you have a quadratic equation, it takes three (x,y) points to uniquely identify that quadratic curve. Again, no matter how you write a quadratic equation, if it passes through those same three points, it’s mathematically equivalent.
• The pattern continues for equations of any degree. Cubic equations require four points to be uniquely identified, Quartic equations require five points, and so on.
At a high level, how this technique works is that the number of shares (keys) you want someone to collect ($N$) defines the degree of an equation.
You use random numbers as the coefficients of the powers of $x$ in that equation, but use your secret number as the constant term.
You then create $M$ data points of the form $(x,y)$ aka $(x,f(x))$. Those are your shares. You then give individual shares to people, or go hide them in your dungeon or do whatever you are going to do with them.
As soon as any one person has $N$ of those $M$ shares (data points), they will be able to figure out the equation of the curve and thus get the secret.
The secret number is the constant term of the polynomial, which is also just $f(0)$.
This image below from wikipedia is great for seeing how you may have two points of a cubic curve, but without a third point you can’t be sure what the quadratic equation is. In fact, there are an infinite number of quadratic curves that pass through any two points! Because of that, it takes the full number of required shares for you to be able to unlock the secret.
## Example: Sharing (Splitting) The Secret
First you decide how many shares you want it to take to unlock the secret. This determines the degree of your equation.
Let’s say you wanted a person to have to have four shares to unlock the secret. This means our equation will be a cubic equation, since it takes four points to uniquely define a cubic equation.
Our equation is:
$f(x) = R_1x^3 + R_2x^2 + R_3x + S$
Where the $R_i$ values are random numbers, and $S$ is the secret value.
Let’s say that our secret value is 435, and that we picked some random numbers for the equation, making the below:
$f(x) = 28x^3 + 64x^2 + 9x + 435$
We now have a function that is uniquely identifiable by any 4 points of data on it’s curve.
Next we decide how many pieces we are going to create total. We need at least 4 so that it is in fact solvable. Let’s make 6 shares.
To do this, you just plug in 6 different values of x and pair each x value with it’s y value. Let’s do that:
$\begin{array}{c|c} x & f(x) \\ \hline 1 & 536 \\ 2 & 933 \\ 3 & 1794 \\ 4 & 3287 \\ 5 & 5580 \\ 6 & 8841 \\ \end{array}$
When doing this part, remember that the secret number is $f(0)$, so make sure and not share what the value of the function is when x is 0!
You could then distribute the shares (data pairs) as you saw fit. Maybe some people are more important, so you give them more than one share, requiring a smaller amount of cooperation with them to unlock the secret.
Share distribution details are totally up to you, but we now have our shares, whereby if you have any of the 4 of the 6 total shares, you can unlock the secret.
## How Do You Join The Secret?
Once you have the right number of shares and you know the degree of the polynomial (pre-shared “public” information), unlocking the secret is a pretty straightforward process too. To unlock the secret, you just need to use ANY method available for creating an equation of the correct degree from a set of data points.
This can be one of several different interpolation techniques, but the most common one to use seems to be Lagrange interpolation, which is something I previously wrote up that you can read about here: Lagrange Interpolation.
Once you have the equation, you can either evaluate $f(0)$, or you can write the equation in polynomial form and the constant term will be the secret value.
## Example: Joining the Secret
Let’s say that we have these four shares and are ready to get the cubic function and then unlock the secret number:
$\begin{array}{c|c} x & y \\ \hline 1 & 536 \\ 2 & 933 \\ 4 & 3287 \\ 6 & 8841 \\ \end{array}$
We could bust out some Lagrange interpolation and figure this out, but let’s be lazy… err efficient I mean. Wolfram alpha can do this for us!
Wolfram Alpha: cubic fit (1, 536), (2, 933), (4, 3287), (6, 8841)
That gives us this equation, saying that it is a perfect fit (which it is!)
$28x^3 + 64x^2 + 9x + 435$
You can see that our constant term (and $f(0)$) is the correct secret value of 435.
Daaaayummm Bru… that is lit AF! We just got hacked by wolfram alpha 😛
## A Small Complication
Unfortunately, the above has a weakness. The weakness is that each share you get gives you a little bit more information about the secret value. You can read more about this in the links section at the end if you want to know more details.
Ideally, you wouldn’t have any information about the secret value until you had the full number of shares required to unlock the secret.
To address this problem, we are going to choose some prime number $k$ and instead of shares being $(x,y)$ data points on the curve, they are going to be $(x,y \bmod k)$. In technical terms we are going to be using points on a finite field, or a Galois field.
The value we choose for $k$ needs to be larger than any of the coefficients of our terms (the random numbers) as well as larger than our secret value and larger than the number of shares we want to create. The larger the better besides that, because a larger $k$ value means a larger “brute force” space to search.
If you want to use this technique in a situation which has real needs for security, please make sure and read more on this technique from more authoritative sources. I’m glossing over the details of security quite a bit, and just trying to give an intuitive understanding of this technique (:
## Source Code
Below is some sample source code that implements Shamir’s Secret Sharing in C++.
I use 64 bit integers, but if you were going to be using this in a realistic situation you could very well overflow 64 bit ints and get the wrong answers. I hit this problem for instance when trying to require more than about 10 shares, using a prime of 257, and generating 50 shares. If you hit the limit of 64 bit ints you can use a multi precision math library instead to have virtually unlimited sized ints. The boost multiprecision header library is a decent choice for multi precision integers, specifically cpp_int.
#include <stdio.h>
#include <array>
#include <vector>
#include <math.h>
#include <random>
#include <assert.h>
#include <stdint.h>
#include <inttypes.h>
typedef int64_t TINT;
typedef std::array<TINT, 2> TShare;
typedef std::vector<TShare> TShares;
class CShamirSecretSharing
{
public:
CShamirSecretSharing (size_t sharesNeeded, TINT prime)
: c_sharesNeeded(sharesNeeded), c_prime(prime)
{
// There needs to be at least 1 share needed
assert(sharesNeeded > 0);
}
// Generate N shares for a secretNumber
TShares GenerateShares (TINT secretNumber, TINT numShares) const
{
// calculate our curve coefficients
std::vector<TINT> coefficients;
{
// store the secret number as the first coefficient;
coefficients.resize((size_t)c_sharesNeeded);
coefficients[0] = secretNumber;
// randomize the rest of the coefficients
std::array<int, std::mt19937::state_size> seed_data;
std::random_device r;
std::generate_n(seed_data.data(), seed_data.size(), std::ref(r));
std::seed_seq seq(std::begin(seed_data), std::end(seed_data));
std::mt19937 gen(seq);
std::uniform_int_distribution<TINT> dis(1, c_prime - 1);
for (TINT i = 1; i < c_sharesNeeded; ++i)
coefficients[(size_t)i] = dis(gen);
}
// generate the shares
TShares shares;
shares.resize((size_t)numShares);
for (size_t i = 0; i < numShares; ++i)
shares[i] = GenerateShare(i + 1, coefficients);
return shares;
}
// use lagrange polynomials to find f(0) of the curve, which is the secret number
TINT JoinShares (const TShares& shares) const
{
// make sure there is at elast the minimum number of shares
assert(shares.size() >= size_t(c_sharesNeeded));
// Sigma summation loop
TINT sum = 0;
for (TINT j = 0; j < c_sharesNeeded; ++j)
{
TINT y_j = shares[(size_t)j][1];
TINT numerator = 1;
TINT denominator = 1;
// Pi product loop
for (TINT m = 0; m < c_sharesNeeded; ++m)
{
if (m == j)
continue;
numerator = (numerator * shares[(size_t)m][0]) % c_prime;
denominator = (denominator * (shares[(size_t)m][0] - shares[(size_t)j][0])) % c_prime;
}
sum = (c_prime + sum + y_j * numerator * modInverse(denominator, c_prime)) % c_prime;
}
return sum;
}
const TINT GetPrime () const { return c_prime; }
const TINT GetSharesNeeded () const { return c_sharesNeeded; }
private:
// Generate a single share in the form of (x, f(x))
TShare GenerateShare (TINT x, const std::vector<TINT>& coefficients) const
{
TINT xpow = x;
TINT y = coefficients[0];
for (TINT i = 1; i < c_sharesNeeded; ++i) {
y += coefficients[(size_t)i] * xpow;
xpow *= x;
}
return{ x, y % c_prime };
}
// Gives the decomposition of the gcd of a and b. Returns [x,y,z] such that x = gcd(a,b) and y*a + z*b = x
static const std::array<TINT, 3> gcdD (TINT a, TINT b) {
if (b == 0)
return{ a, 1, 0 };
const TINT n = a / b;
const TINT c = a % b;
const std::array<TINT, 3> r = gcdD(b, c);
return{ r[0], r[2], r[1] - r[2] * n };
}
// Gives the multiplicative inverse of k mod prime. In other words (k * modInverse(k)) % prime = 1 for all prime > k >= 1
static TINT modInverse (TINT k, TINT prime) {
k = k % prime;
TINT r = (k < 0) ? -gcdD(prime, -k)[2] : gcdD(prime, k)[2];
return (prime + r) % prime;
}
private:
// Publically known information
const TINT c_prime;
const TINT c_sharesNeeded;
};
void WaitForEnter ()
{
printf("Press Enter to quit");
fflush(stdin);
getchar();
}
int main (int argc, char **argv)
{
// Parameters
const TINT c_secretNumber = 435;
const TINT c_sharesNeeded = 7;
const TINT c_prime = 439; // must be a prime number larger than the other three numbers above
// set up a secret sharing object with the public information
CShamirSecretSharing secretSharer(c_sharesNeeded, c_prime);
// split a secret value into multiple shares
// shuffle the shares, so it's random which ones are used to join
std::array<int, std::mt19937::state_size> seed_data;
std::random_device r;
std::generate_n(seed_data.data(), seed_data.size(), std::ref(r));
std::seed_seq seq(std::begin(seed_data), std::end(seed_data));
std::mt19937 gen(seq);
std::shuffle(shares.begin(), shares.end(), gen);
// join the shares
TINT joinedSecret = secretSharer.JoinShares(shares);
// show the public information and the secrets being joined
printf("%" PRId64 " shares needed, %i shares made\n", secretSharer.GetSharesNeeded(), shares.size());
printf("Prime = %" PRId64 "\n\n", secretSharer.GetPrime());
for (TINT i = 0, c = secretSharer.GetSharesNeeded(); i < c; ++i)
printf("Share %" PRId64 " = (%" PRId64 ", %" PRId64 ")\n", i+1, shares[i][0], shares[i][1]);
// show the result
printf("\nJoined Secret = %" PRId64 "\nActual Secret = %" PRId64 "\n\n", joinedSecret, c_secretNumber);
assert(joinedSecret == c_secretNumber);
WaitForEnter();
return 0;
}
## Example Output
Here is some example output of the program:
Wikipedia: Shamir’s Secret Sharing (Note: for some reason the example javascript implementation here only worked for odd numbered keys required)
Wikipedia: Finite Field
Cryptography.wikia.com: Shamir’s Secret Sharing
Java Implementation of Shamir’s Secret Sharing (Note: I don’t think this implementation is correct, and neither is the one that someone posted to correct them!)
When writing this post I wondered if maybe you could use the coefficients of the other terms as secrets as well. These two links talk about the details of that:
Cryptography Stack Exchange: Why only one secret value with Shamir’s secret sharing?
Cryptography Stack Exchange: Coefficients in Shamir’s Secret Sharing Scheme
Now that you understand this, you are probably ready to start reading up on elliptic curve cryptography. Give this link below a read if you are interested in a gentle introduction on that!
A (Relatively Easy To Understand) Primer on Elliptic Curve Cryptography
## Turning a Truth Table Into A digital Circuit (ANF)
In this post I’m going to show how you turn a truth table into a digital logic circuit that uses XOR and AND gates.
## My Usage Case
My specific usage case for this is in my investigations into homomorphic encryption, which as you may recall is able to perform computation on encrypted data. This lets encrypted data be operated on by an untrusted source, given back to you, and then you can decrypt your data to get a result.
Lots of use cases if this can ever get fast enough to become practical, such as doing cloud computing with private data. However, when doing homomorphic encryption (at least currently, for the techniques I’m using), you only have XOR and AND logic operations.
So, I’m using the information in this post to be able to turn a lookup table, or a specific boolean function, into a logic circuit that I can feed into a homomorphic encryption based digital circuit.
Essentially I want to figure out how to do a homomorphic table lookup to try and make some simple as possible circuits, that will in turn be as fast and lean as possible.
If you want to know more about homomorphic encryption, here’s a post I wrote which explains a very simple algorithm: Super Simple Symmetric Leveled Homomorphic Encryption Implementation
## Algebraic Normal Form
Algebraic normal form (ANF) is a way of writing a boolean function using only XOR and AND.
Since it’s a normal form, two functions that do the same thing will be the same thing in ANF.
There are other forms for writing boolean logic, but ANF suits me best for my homomorphic encryption circuit needs!
An example of boolean logic in ANF is the below:
$f(x_1, x_2, x_3, x_4) = x_1 x_2 \oplus x_1 x_3 \oplus x_1 x_4$
It is essentially a boolean polynomial, where AND is like multiplication, and XOR is like addition. It even factors the same way. In fact, ANF is not always the smallest circuit possible, you’d have to factor common ANDs to find the smallest way you could represent the circuit, like the below:
$f(x_1, x_2, x_3, x_4) = x_1 (x_2 \oplus x_3 \oplus x_4)$
That smaller form does 1 AND and 2 XORs, versus the ANF which does 3 ANDs and 2 XORs. In homomorphic encryption, since AND is so much more costly than XOR, minimizing the ANDs is a very nice win, and worth the effort.
## Truth Tables and Lookup Tables
A truth table is just where you specify the inputs into a boolean function and the output of that boolean function for the given input:
$\begin{array}{c|c|c|c} x_1 & x_2 & x_3 & f(x_1, x_2, x_3) \\ \hline 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ \end{array}$
A lookup table is similar in functionality, except that it has multi bit output. When dealing with digital circuits, you can make a lookup table by making a truth table per output bit. For instance, the above truth table might just be the low bit of the lookup table below, which is just a truth table for addition of the input bits.
$\begin{array}{c|c|c|c} x_1 & x_2 & x_3 & f(x_1, x_2, x_3) \\ \hline 0 & 0 & 0 & 00 \\ 0 & 0 & 1 & 01 \\ 0 & 1 & 0 & 01 \\ 0 & 1 & 1 & 10 \\ 1 & 0 & 0 & 01 \\ 1 & 0 & 1 & 10 \\ 1 & 1 & 0 & 10 \\ 1 & 1 & 1 & 11 \\ \end{array}$
## Converting Truth Table to ANF
When I first saw the explanation for converting a truth table to ANF, it looked pretty complicated, but luckily it turns out to be pretty easy.
The basic idea is that you make a term for each possible combination of x inputs, ANDing a term by each constant, and then solving for those constants.
Let’s use the truth table from the last section:
$\begin{array}{c|c|c|c} x_1 & x_2 & x_3 & f(x_1, x_2, x_3) \\ \hline 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ \end{array}$
For three inputs, the starting equation looks like this:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus a_1 x_1 \oplus a_2 x_2 \oplus a_3 x_3 \\ \oplus a_{12} x_1 x_2 \oplus a_{13} x_1 x_3 \oplus a_{23} x_2 x_3 \\ \oplus a_{123} x_1 x_2 x_3$
Now we have to solve for the a values.
To solve for $a_{123}$, we just look in the truth table for function $f(x_1, x_2, x_3)$ to see if we have an odd or even number of ones in the output of the function. If there is an even number, it is 0, else it is a 1.
Since we have an even number of ones, the value is 0, so our equation becomes this:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus a_1 x_1 \oplus a_2 x_2 \oplus a_3 x_3 \\ \oplus a_{12} x_1 x_2 \oplus a_{13} x_1 x_3 \oplus a_{23} x_2 x_3 \\ \oplus 0 \land x_1 x_2 x_3$
Note that $\land$ is the symbol for AND. I’m showing it explicitly because otherwise the equation looks weird, and a multiplication symbol isn’t correct.
Since 0 ANDed with anything else is 0, and also since n XOR 0 = n, that whole last term disappears, leaving us with this equation:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus a_1 x_1 \oplus a_2 x_2 \oplus a_3 x_3 \\ \oplus a_{12} x_1 x_2 \oplus a_{13} x_1 x_3 \oplus a_{23} x_2 x_3$
Next up, to solve for $a_{12}$, we need to limit our truth table to $f(x_1, x_2, 0)$. That truth table is below, made from the original truth table, but throwing out any row where $x_{3}$ is 1.
$\begin{array}{c|c|c|c} x_1 & x_2 & x_3 & f(x_1, x_2, 0) \\ \hline 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 \\ \end{array}$
We again just look at whether there are an odd or even number of ones in the function output, and use that to set $a_{12}$ appropriately. In this case, there are an even number, so we set it to 0, which makes that term disappear again. Our function is now down to this:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus a_1 x_1 \oplus a_2 x_2 \oplus a_3 x_3 \\ \oplus a_{13} x_1 x_3 \oplus a_{23} x_2 x_3$
If we look at $f(x_1,0,x_3)$, we find that it also has an even number of ones, making $a_{13}$ become 0 and making that term disappear.
Looking at $f(0,x_2,x_3)$, it also has an even number of ones, making $a_{23}$ become 0 and making that term disappear as well.
That leaves us with this equation:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus a_1 x_1 \oplus a_2 x_2 \oplus a_3 x_3$
To solve for $a_1$, we look at the truth table for $f(x_1,0,0)$, which is below:
$\begin{array}{c|c|c|c} x_1 & x_2 & x_3 & f(x_1, 0, 0) \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ \end{array}$
There are an odd number of ones in the output, so $a_1$ becomes 1. Finally, we get to keep a term! The equation is below:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus 1 \land x_1 \oplus a_2 x_2 \oplus a_3 x_3$
Since 1 AND n = n, we can drop the explicit 1 to become this:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus x_1 \oplus a_2 x_2 \oplus a_3 x_3$
If you do the same process for $a_2$ and $a_3$, you’ll find that they also have odd numbers of ones in the output so also become ones. That puts our equation at:
$f(x_1, x_2, x_3) = \\ a_0 \\ \oplus x_1 \oplus x_2 \oplus x_3$
Solving for $a_0$, is just looking at whether there are an odd or even number of ones in the function $f(0,0,0)$ which you can look up directly in the lookup table. It’s even, so $a_0$ becomes 0, which makes our full final equation into this:
$f(x_1, x_2, x_3) = x_1 \oplus x_2 \oplus x_3$
We are done! This truth table can be implemented with 3 XORs and 0 ANDs. A pretty efficient operation!
You can see this is true if you work it out with the truth table. Try it out and see!
$\begin{array}{c|c|c|c} x_1 & x_2 & x_3 & f(x_1, x_2, x_3) \\ \hline 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ \end{array}$
## Sample Code
Here is some sample code that lets you define a lookup table by implementing an integer function, and it generates the ANF for each output bit for the truth table. It also verifies that the ANF gives the correct answer. It shows you how to use this to make various circuits: bit count, addition, multiplication, division and modulus.
#include <stdio.h>
#include <array>
#include <vector>
#define PRINT_TRUTHTABLES() 0
#define PRINT_NUMOPS() 1
#define PRINT_ANF() 1
void WaitForEnter ()
{
printf("Press Enter to quit");
fflush(stdin);
getchar();
}
template <size_t NUM_INPUT_BITS>
{
for (size_t i = 0; i < NUM_INPUT_BITS; ++i)
{
const size_t bitMask = 1 << i;
return false;
}
return true;
}
template <size_t NUM_INPUT_BITS>
bool ANFHasTerm (const std::array<size_t, 1 << NUM_INPUT_BITS> &lookupTable, size_t outputBitIndex, size_t termMask)
{
const size_t c_inputValueCount = 1 << NUM_INPUT_BITS;
int onesCount = 0;
for (size_t i = 0; i < c_inputValueCount; ++i)
{
onesCount++;
}
return (onesCount & 1) != 0;
}
template <size_t NUM_INPUT_BITS>
void MakeANFTruthTable (const std::array<size_t, 1 << NUM_INPUT_BITS> &lookupTable, std::array<size_t, 1 << NUM_INPUT_BITS> &reconstructedLookupTable, size_t outputBitIndex)
{
const size_t c_inputValueCount = 1 << NUM_INPUT_BITS;
printf("-----Output Bit %u-----\r\n", outputBitIndex);
// print truth table if we should
#if PRINT_TRUTHTABLES()
for (size_t inputValue = 0; inputValue < c_inputValueCount; ++inputValue)
printf(" [%u] = %u\r\n", inputValue, ((lookupTable[inputValue] >> outputBitIndex) & 1) ? 1 : 0);
printf("\r\n");
#endif
// find each ANF term
std::vector<size_t> terms;
{
}
// print function params
#if PRINT_ANF()
printf("f(");
for (size_t i = 0; i < NUM_INPUT_BITS; ++i)
{
if (i > 0)
printf(",");
printf("x%i",i+1);
}
printf(") = \r\n");
#endif
// print ANF and count XORs and ANDs
size_t numXor = 0;
size_t numAnd = 0;
if (terms.size() == 0)
{
#if PRINT_ANF()
printf("0\r\n");
#endif
}
else
{
for (size_t termIndex = 0, termCount = terms.size(); termIndex < termCount; ++termIndex)
{
if (termIndex > 0) {
#if PRINT_ANF()
printf("XOR ");
#endif
++numXor;
}
size_t term = terms[termIndex];
if (term == 0)
{
#if PRINT_ANF()
printf("1");
#endif
}
else
{
bool firstProduct = true;
for (size_t bitIndex = 0; bitIndex < NUM_INPUT_BITS; ++bitIndex)
{
const size_t bitMask = 1 << bitIndex;
if ((term & bitMask) != 0)
{
#if PRINT_ANF()
printf("x%i ", bitIndex + 1);
#endif
if (firstProduct)
firstProduct = false;
else
++numAnd;
}
}
}
#if PRINT_ANF()
printf("\r\n");
#endif
}
}
#if PRINT_ANF()
printf("\r\n");
#endif
#if PRINT_NUMOPS()
printf("%u XORs, %u ANDs\r\n\r\n", numXor, numAnd);
#endif
// reconstruct a bit of the reconstructedLookupTable for each entry to be able to verify correctness
const size_t c_outputBitMask = 1 << outputBitIndex;
for (size_t valueIndex = 0; valueIndex < c_inputValueCount; ++valueIndex)
{
bool xorSum = false;
for (size_t termIndex = 0, termCount = terms.size(); termIndex < termCount; ++termIndex)
{
size_t term = terms[termIndex];
if (term == 0)
{
xorSum = 1 ^ xorSum;
}
else
{
bool andProduct = true;
for (size_t bitIndex = 0; bitIndex < NUM_INPUT_BITS; ++bitIndex)
{
const size_t bitMask = 1 << bitIndex;
if ((term & bitMask) != 0)
{
if ((valueIndex & bitMask) == 0)
andProduct = false;
}
}
xorSum = andProduct ^ xorSum;
}
}
if (xorSum)
}
}
template <size_t NUM_INPUT_BITS, size_t NUM_OUTPUT_BITS, typename LAMBDA>
void MakeANFLookupTable (const LAMBDA& lambda)
{
// make lookup table
const size_t c_outputValueMask = (1 << NUM_OUTPUT_BITS) - 1;
const size_t c_inputValueCount = 1 << NUM_INPUT_BITS;
std::array<size_t, c_inputValueCount> lookupTable;
for (size_t inputValue = 0; inputValue < c_inputValueCount; ++inputValue)
lookupTable[inputValue] = lambda(inputValue, NUM_INPUT_BITS, NUM_OUTPUT_BITS) & c_outputValueMask;
// make the anf for each truth table (each output bit of the lookup table)
std::array<size_t, c_inputValueCount> reconstructedLookupTable;
std::fill(reconstructedLookupTable.begin(), reconstructedLookupTable.end(), 0);
for (size_t outputBitIndex = 0; outputBitIndex < NUM_OUTPUT_BITS; ++outputBitIndex)
MakeANFTruthTable<NUM_INPUT_BITS>(lookupTable, reconstructedLookupTable, outputBitIndex);
// verify that our anf expressions perfectly re-create the lookup table
for (size_t inputValue = 0; inputValue < c_inputValueCount; ++inputValue)
{
if (lookupTable[inputValue] != reconstructedLookupTable[inputValue])
printf("ERROR: expression / lookup mismatch for index %u\r\n", inputValue);
}
printf("expression / lookup verification complete.\r\n\r\n");
}
size_t CountBits (size_t inputValue, size_t numInputBits, size_t numOutputBits)
{
// Count how many bits there are
int result = 0;
while (inputValue)
{
if (inputValue & 1)
result++;
inputValue = inputValue >> 1;
}
return result;
}
size_t AddBits (size_t inputValue, size_t numInputBits, size_t numOutputBits)
{
// break the input bits in half and add them
const size_t bitsA = numInputBits / 2;
const size_t mask = (1 << bitsA) - 1;
size_t a = inputValue & mask;
size_t b = inputValue >> bitsA;
return a+b;
}
size_t MultiplyBits (size_t inputValue, size_t numInputBits, size_t numOutputBits)
{
// break the input bits in half and add them
const size_t bitsA = numInputBits / 2;
const size_t mask = (1 << bitsA) - 1;
size_t a = inputValue & mask;
size_t b = inputValue >> bitsA;
return a * b;
}
size_t DivideBits (size_t inputValue, size_t numInputBits, size_t numOutputBits)
{
// break the input bits in half and add them
const size_t bitsA = numInputBits / 2;
const size_t mask = (1 << bitsA) - 1;
size_t a = inputValue & mask;
size_t b = inputValue >> bitsA;
// workaround for divide by zero
if (b == 0)
return 0;
return a / b;
}
size_t ModulusBits (size_t inputValue, size_t numInputBits, size_t numOutputBits)
{
// break the input bits in half and add them
const size_t bitsA = numInputBits / 2;
const size_t mask = (1 << bitsA) - 1;
size_t a = inputValue & mask;
size_t b = inputValue >> bitsA;
// workaround for divide by zero
if (b == 0)
return 0;
return a % b;
}
int main (int argc, char **argv)
{
//MakeANFLookupTable<3, 2>(CountBits); // Output bits needs to be enough to store the number "input bits"
//MakeANFLookupTable<4, 3>(AddBits); // Output bits needs to be (InputBits / 2)+1
//MakeANFLookupTable<4, 4>(MultiplyBits); // Output bits needs to be same as input bits
//MakeANFLookupTable<4, 2>(DivideBits); // Output bits needs to be half of input bits (rounded down)
//MakeANFLookupTable<4, 2>(ModulusBits); // Output bits needs to be half of input bits (rounded down)
//MakeANFLookupTable<10, 5>(DivideBits); // 5 bit vs 5 bit division is amazingly complex!
MakeANFLookupTable<4, 2>(ModulusBits); // Output bits needs to be half of input bits (rounded down)
WaitForEnter();
return 0;
}
## Sample Code Runs
Here is the program output for a “bit count” circuit. It counts the number of bits that are 1, in the 3 bit input, and outputs the answer as 2 bit output. Note that the bit 0 output is the same functionality as the example we worked through by hand, and you can see that it comes up with the same answer.
Here is the program output for an adder circuit. It adds two 2 bit numbers, and outputs a 3 bit output.
Here is the program output for a multiplication circuit. It multiplies two 2 bit numbers, and outputs a 4 bit number.
Here is the program output for a division circuit. It divides a 2 bit number by another 2 bit number and outputs a 2 bit number. When higher bit counts are involved, the division circuit gets super complicated, it’s really crazy! 5 bit divided by 5 bit is several pages of output for instance. Note that it returns 0 whenever it would divide by 0.
Lastly, here is the program output for a modulus circuit. It divides a 2 bit number by another 2 bit number and outputs the remainder as a 2 bit number.
While the above shows you how to turn a single bit truth table into ANF, extending this to a multi bit lookup table is super simple; you just do the same process for each output bit in the lookup table.
Finding Boolean/Logical Expressions for truth tables in algebraic normal form(ANF)
Finding Boolean/Logical Expressions for truth tables
## Game Development Needs Data Pipeline Middleware
In 15 years I’ve worked at 7 different game studios, ranging from small to large, working on many different kinds of projects in a variety of roles.
At almost every studio, there was some way for the game to load data at runtime that controlled how it behaved – such as the damage a weapon would do or the cost of an item upgrade.
The studios that didn’t have this setup could definitely have benefited from having it. After all, this is how game designers do their job!
Sometimes though, this data was maintained via excel spreadsheets (export as csv for instance and have the game read that). That is nearly the worst case scenario for data management. Better though is to have an editor which can edit that data, preferably able to edit data described by schemas, which the game also uses to generate code to load that data.
Each studio I’ve worked at that did have game data each had their own solution for their data pipeline, and while they are all of varying qualities, I have yet to see something that is both fast and has most of the features you’d reasonably want or expect.
We really need some middleware to tackle this “solved problem” and offer it to us at a reasonable price so we can stop dealing with it. Open sourced would be fine too. Everyone from engineers to production to content people will be much happier and more productive!
# Required Features
Here are the features I believe are required to satisfy most folks:
1. Be able to define the structure of your data in some format (define data schema).
2. Have an editor that is able to quickly launch, quickly load up data in the data schema and allow a nice interface to editing the data as well as searching the data.
3. This edited data should be in some format that merges well (for dealing with branching), and preferably is standardized so you can use common tools on the data – such as XSLT if storing data as xml. XML isn’t commonly very mergable so not sure the solution there other than perhaps a custom merge utility perhaps?
4. The “data solution” / project file should store your preferences about how you want the final data format to be: xml, json, binary, other? Checkboxes for compression and encryption, etc. Switching the data format should take moments.
5. There should be a cooking process that can be run from the data editor or via command line which transforms the edited data into whatever format the destination data should be in. AKA turn the human friendly XML into machine friendly binary files which you load in with a single read and then do pointer fixup on.
6. This pipeline should generate the code that loads and interacts with the data as described in the data schema. For instance you say “load my data” and it does all the decompression, decryption, parsing, etc giving you back a root data structure which contains compile time defined strongly typed structures. This is important because when you change the format of the data that the game uses, no game code actually has to know or care. Whatever it takes to load your data happens when you call the function.
# Bonus Points
Here are some bonus point features that would be great to have:
1. Handle live editing of data. When the game and editor is both open, and data is edited, have it change the data on the game side in real time, and perhaps allow a callback to be intercepted in case the game needs to clear out any cached values or anything. This helps iteration time by letting people make data changes without having to relaunch the game. Also needs to be able to connect to a game over tcp/ip and handle endian correction as needed as well as 32 vs 64 bit processes using the same data.
2. Handle the usual problems associated with DLC and versioning in an intelligent way. Many data systems that support DLC / Patching / Schema Updates post ship have strange rules about what data you can and can’t change. Often times if you get it wrong, you make a bug that isn’t always obvious. If support for this was built in, and people didnt have to concern themselves with it, it’d be great.
3. On some development environments, data must be both forwards and backwards compatible. Handling that under the covers in an intelligent way would be awesome.
4. The editor should be extensible with custom types and plugins for visualizations of data, as well as interactive editing of data. This same code path could be used to integrate parts of the game engine with the editor for instance (slippery slope to making the editor slow, however).
5. Being able to craft custom curves, and being able to query them simply and efficiently from the game side at runtime would be awesome.
6. Support “cook time computations”. The data the user works with isn’t always set up the way that would be best for the machine. It’d be great to be able to do custom calculations and computations at runtime. Also great for building acceleration data structures.
7. You should be able to run queries against the data or custom scripts. To answer questions like “Is anyone using this feature?” and “I need to export data from our game into a format that this other program can read”
8. Being able to export data as c++ literal data structures, for people who want to embed (at least some of) their data in the exe to reduce complexity, loading times, etc.
It should also be as fast and lightweight as possible. It should allow games to specify memory and file i/o overrides.
Localized text is also a “solved problem” that needs an available solution. It could perhaps be rolled into this, or maybe it would make most sense for it to be separate.
As another example of how having something like this would be useful, on multiple occasions at previous studios, people have suggested we change the format of the data that the game uses at runtime. For instance, from json to a binary format. In each case this has come up so far, people have said it would take too long and it got backlogged (ie killed). With data pipeline middleware that works as i describe it, you would click a few checkboxes, recook your data and test it to have your runtime results. That’s as hard as it SHOULD be, but in practice it’s much harder because everyone rolls their own and the cobbler never has time to fix his shoes (;
Anyone out there want to make this happen? (:
## Using Wang Tiles to Simulate Turing Machines
Wang tiles were invented by Hao Wang in 1961 for mathematical reasons, but they find great use in games for making tile based art which gives results that don’t look tiled – both with 2d tiled textures, as well as 3d tiled models.
Apparently Wang tiles are also able to execute Turing machines, and so are thus Turing complete – meaning they can execute any program.
That is a pretty amazing and perplexing statement, so this post explores that a bit.
## Wang Tiles Overview
Wang tiles are rectangular tiles where each edge will only fit with other specific edges, but that for any specific edge, there is more than one possible tile that can fit with that edge. By fit with that edge, I mean they are seamless when put together, without any visual artifacts to hint at there actually being a seam between the tiles.
This is useful for graphics because this lets you have seamless tiled graphics, but the specific configuration of how the tiles are placed can be completely randomized, so long as their edges are all compatible. The result is tiled graphics that doesn’t look at all tiled, due to visible patterns being much less noticeable than with traditional tiled graphics.
Here is an example I made. The graphics are programmer art but hopefully you get the idea. This was made with 16 tiles, where there were two different edge types per edge.
## Turing Machine Overview
Turing machines were invented in 1936 by Alan Turing as a generic computing machine that was proven to be able to execute any algorithm.
The turing machine is made up of a few main components: the memory tape, the read/write head, and the state machine.
The memory tape is infinitely long, so has infinite storage, and is initialized to all zeroes to start out.
The read/write head starts at a position on the tape, and can read or write values, and also move left or right on the tape.
The state machine knows what state it is in and has rules about what to do in each state when it reads a value from the tape.
For instance, in state A, if a 0 is read from the tape, the rule may be to write a 1 to the current position on the tape, move the read/write head to the right, and go to state B. State B may have completely different logic, and could either transition back to state A, state in state B, or move to another state entirely.
Using simple state transition logic like that, any computer algorithm can be performed.
In a Turing machine there can also be a “Halt State” which means the program is finished executing and the answer it was trying to calculate has been calculated.
Looking at some programs, you can easily see that they will halt eventually, or that they will be an infinite loop and never halt. Some programs in-between are complex and can’t very easily be determined if they will ever halt or not. Turing proved that there is no general solution to whether a Turing machine (aka any computer program) will halt or not, and this is called the Halting Problem. In general, the only way to know if a program will halt or not is to wait and see. So, effectively the answers to whether a program in general will halt or not are “yes” and “not yet” – although for many specific programs you can in fact see that they will halt eventually if you were to run them.
## Wang Tile Computation
It turns out that Wang tiles can simulate a Turing machine, and so are “Turing complete” meaning that they too can perform any computer algorithm.
To make this happen, we’ll make a column of tiles that represent the state of the Turing machine at a specific point in time, starting with time 0 at the left most column. We’ll place tiles in the column to the right making sure all edge rules are respected, and then do the column to the right of that one etc until the program halts (or forever if it doesn’t halt). If we set up our set of tiles correctly, the act of satisfying the edge rules as we place our tiles is enough to execute the Turing machine.
Let’s walk through a simple example where we have the following state machine logic rules:
1. When in state A, if a 0 is read, we will write a 1, move the read/write head down and move to state B.
2. When in state A, if a 1 is read, we will halt (enter the halt state).
3. When in state B, if a 0 is read, we will write a 1, move the read/write head up and move to state A.
4. When in state B, if a 1 is read, we will halt (enter the halt state).
### Tape Memory Storage
The first thing we need is persistent storage of memory for the tape. For this, we’ll need the following two tiles:
To see this working, we can set up a section of tape with some values (make a column of wang tiles), and we can see that the only valid wang tiles to place next to the starting column are tiles which propogate the 0 and the 1 values forward in time without modifying them.
In the diagram below, we initialize the tape to 0101 in the left most column (time 0). By only placing down tiles with compatible edges you can see that our memory values persist forever. Our memory storage is implemented, huzzah!
We’ll start our example with all memory initialized to 0, but the above shows that we can have persistent memory.
The read/write head of the Turing machine is represented as part of the edge information. In this way, besides an edge storing the 0 or 1, if that is where the read/write head is, it also stores the state of the state machine.
Our example uses two states (besides the halt state): A and B. If a 1 is read in while being in either state A or B, the program halts.
To handle that, we need the tiles below:
Now that we have the rules for entering the halt state done (rule #2 and rule #4), we have to figure out how to implement the rules that control switching from one state to another (rule #1 and rule #3).
Rule #1 says that if we are in state A and read a 0, we should write a 1, move the read/write head down and move to state B.
We’ll need this tile to cause reading a 0 in state A to write a 1 as output, and to tell the tile below to move to state B.
The tile below that one could either be a 0 or a 1, and without knowing which, we want it to keep it’s value but accept the read/write head and be in state B. To do that we need two tiles, one for if there is a 0 on the tape at that position, and another for if there is a 1 on the tape.
Rule #3 says that if we are in state B and read a 0, we should write a 1, move the read/write head up and move to state A.
To do that, we’ll need a similar construction as for rule #1 but we are moving up instead of down. These 3 tiles will give us what we need:
## Starting Column Tiles
We are going to treat the boundaries of our simulation area as if they have edges of “x”.
This means that to make our starting column (the Turing machine at time 0), we are going to need 2 special tiles. One tile will be for storing a 0 on the tape, which is what the tape is initialized to, and the other tile will be for storing the position of the read/write head in state A, which is our starting state.
Here are those two tiles:
## Final Tileset
Here’s the full set of 12 tiles that we are going to use:
## Full Simulation
Here is the initial setup at time 0 for our Turing machine. Note that this is one possible starting state, but this is the starting state we are choosing. We are not leaving it up to chance where the read/write head starts, or if it is even present at all. If we only followed edge rules we may get 4 read/write heads or 0, or anything in between.
From here, to build the second column, we start from the top and work towards the bottom, choosing the tile that fits the constraints of the edge it touches. In this first step, the head reads a 0, writes a 1, moves down, and moves to state B.
Heres is the second step, where the read reads a 0, writes a 1, moves up, and moves to state A.
Here is the final step, where the head reads a 1 and enters the halt state, signifying that the program has terminated.
The program halted, and gave an output value of “0110” or 6. This output isn’t really meaningful but other programs can give output that is meaningful. For instance you could have a Turing machine add two numbers, and the output would be the sum of those two numbers.
## An Important Detail
There is an important detail that the above doesn’t address, and that many explanations of Wang tile Turing machines don’t seem to talk about.
When placing the second tile for time 2, the only constraint from the edges is that the tile must have an x on top and a 1 on the left. This actually makes it ambiguous which tile should be chosen between the two tiles below.
How do we choose the right one then?
The answer is that you make a guess and just choose one. If the wrong one was chosen in this case, when we moved to the next tile, we’d be looking for a tile which had an x on top and a B0 on the left. There is no such tile so you’d be unable to place a tile. When this happened, you’d take a step back to the last tile, and try one of the other possibilities.
So, unfortunately there is some literal trial and error involved when simulating Turing machines with Wang tiles, but it is fairly manageable at least. It definitely makes it a bit more complex to calculate in a pixel shader if you were so inclined (or other massively parallel processing units), but it shouldn’t be that much more costly.
Some of the links below talk about Wang tiles and Turing machines, but don’t seem to strictly be Turing machines anymore. For instance, you might notice that some examples allow data to travel “back in time” where after the program halts, the answer is in the tape at time 0 of the Turing machine, even though that data wasn’t actually there at time 0. This shows that Wang tiles can do computation in their own right, beyond simulating Turing machines, but I’m not really sure what that technique itself would be called.
Also if you are wondering if this is useful to do computation with Wang tiles, I’m not really sure of any practical usage cases myself. However, apparently scientists have found that DNA can act much like Wang tiles act, where they will fit together only if edges are compatible. Because of this, there is ongoing research into DNA based computation that is based on the work of Wang tile computation. pretty interesting stuff!
Here is a shadertoy implementation of wang tiles computing prime numbers in a webgl pixel shader:
Here are some great videos on Turing machines and the halting problem:
Turing Machines Explained – Computerphile
Turing & The Halting Problem – Computerphile
Computing With Tiles
Wikipedia: Wang Tile
Wang Tiles and Turing Machines
Wang Tiles – 1
Computing With Tiles
Computability of Tilings
## A Sixth Way To Calculate Sine Without Trig
I have another item to add to the pile of ways to calculate sine without trig. Here are the previous ways before we start:
Four Ways to Calculate Sine Without Trig
A Fifth Way to Calculate Sine Without Trig
This method is called Bhaskara I’s sine approximation formula and it’s just a numerical way of approximating sine.
The below is some glsl code from @paniq that has been adapted to take 0 to 1 as input, which corresponds to 0 to 2*pi radians, or 0 to 360 degrees, and returns the normalized vector of that angle. The x component of the vector is the cosine of the angle and the y component of the vector is the sine of the angle. Useful for packing 2d normals into a color channel (;
// https://en.wikipedia.org/wiki/Bhaskara_I%27s_sine_approximation_formula
// x is 0..1 corresponding to 0..360 degrees
vec2 CosSin(float x) {
vec2 si = fract(vec2(0.5,1.0) - x*2.0)*2.0 - 1.0;
vec2 so = sign(0.5-fract(vec2(0.25,0.5) - x));
return (20.0 / (si*si + 4.0) - 4.0) * so;
}
Here’s a shadertoy to compare/contrast this technique versus reality (or, reality as per the video card). Spoiler alert – the shadertoy is ridiculous, they are basically the same.
More from paniq:
i also wrote an approximate atan to go with it Shadertoy: Pseudo-Polar Mapping see the ALTMETHOD branch. also changed the sin/cos computation to ensure the sin/cos vector is perfectly normalized.
## Matrix Form of Bezier Curves
Bezier curves are most often talked about either in terms of the De Casteljau algorithm, or in terms of a mathematical function (Bernstein Polynomials).
Every now and then though, you see people talking about Bezier curves being calculated via matrices. If you ever wondered what that was all about, this post should hopefully explain and demystify that a bit.
If you don’t know how to come up with the equation of a Bezier curve for any number of control points, you should give this a read first:
Easy Binomial Expansion & Bezier Curve Formulas
And if you are curious about the De Casteljau algorithm, you can learn about that here:
The De Casteljau Algorithm for Evaluating Bezier Curves
Ok, all read up on that stuff? Let’s get talking about Bezier curves in matrix form! There are shadertoy links at the end with working wegl glsl demos that include source code.
## Making the Matrix Form of Bezier Curves
Coming up with the matrix for a Bezier curve is surprisingly easy. Keep in mind the matrix we are making is for glsl which is a column major matrix order, so you might have to adjust things if you are using a row major matrix order setup (mostly, just transpose the matrix).
The first step is to get the formula for a Bezier curve. We’ll work through the example using a quadratic Bezier curve with 3 control points A,B,C, so we start with the formula below:
$f(t) = A*(1-t)^2 + B*2t(1-t) + C*t^2$
The next step is to break the equation into one equation per term. Each term has a control point, so we are basically splitting the formula up so that we have one formula per control point.
$A*(1-t)^2 \\ B*2t(1-t) \\ C*t^2$
Next, we remove the control points and expand each term to get:
$1-2t+t^2 \\ 2t-2t^2 \\ t^2$
Now, explicitly values of all powers of t that are present:
$1*t^0-2*t^1+1*t^2 \\ 0*t^0+2*t^1-2*t^2 \\ 0*t^0+0*t^1+1*t^2$
Now the final step. Take the constants that multiply your powers of t and make a matrix out of them. You are done!
$\begin{bmatrix} 1 & -2 & 1 \\ 0 & 2 & -2 \\ 0 & 0 & 1 \\ \end{bmatrix}$
## Using the Matrix Form
Using the matrix form of Bezier curves is also pretty simple.
First, we need to make a vector of the power series of our t value:
$powerSeries = \begin{bmatrix} t^0 & t^1 & t^2 \\ \end{bmatrix}$
Which can also be written as:
$powerSeries = \begin{bmatrix} 1 & t & t^2 \\ \end{bmatrix}$
You also need a vector of your control points:
$controlPoints = \begin{bmatrix} A & B & C \\ \end{bmatrix}$
You next perform this operation to get a result vector:
$result = powerSeries * curveMatrix * controlPoints$
Then, you add up all components of result to get the value of the curve at time t.
$value = result[0] + result[1] + result[2]$
All done!
Note that this is a one dimensional Bezier curve. You need to do this operation once per axis to get your final multi dimensional Bezier curve point.
If you are confused by that last line, check out this post: One Dimensional Bezier Curves
## Multiplying the Control Points In
You might notice that if you are evaluating several points on the same curve that you are going to be multiplying the curveMatrix matrix by the controlPoints vector over and over. You can multiply the control points into the Bezier curve matrix to make the specific matrix for those control points if you want to. You multiply the columns of the matrix by the control points, and adjust the result calculation like the below.
// Multiply the control points into the curve matrix
curveMatrix[0] *= A;
curveMatrix[1] *= B;
curveMatrix[2] *= C;
// Use the curve matrix that has the control points baked in, to do less math to get the result vector.
// You would calculate the curve matrix once and re-use it multiple times of course!
vec3 result = powerSeries * curveMatrix;
float value = result.x + result.y + result.z;
## Closing
You might wonder when you’d use the matrix form. One time to use the matrix form would be when you had fast matrix math support (like on the GPU). Another time to use the matrix form though is if you ever want to cut up a Bezier curve into multiple smaller sub curves. The matrix form can help make that easier, and you can read more about that here if you want: A Matrix Formulation of the Cubic Bezier Curve
Here are some shadertoys that show this all working in webgl/glsl pixel shaders, along with source code:
## Actually Making Signed Distance Field Textures With JFA
This post is an addendum to the last post where I say that you can make distance field textures with JFA but don’t fully explain how to make SIGNED distance field textures, which is what you really want.
If you want to go straight to a working demo with webgl pixel shader source code, here is the shadertoy: Shadertoy: JFA SDF Texture
If you naively use a distance transform to make a distance field texture, you’ll get an UNSIGNED distance field texture, where you only have the distance to the surface of the object from the outside, but won’t have the distance to the surface of the object from the inside.
This is important because signed distance field textures have both, and use bilinear interpolation of distance on each side of the shape surface to make a nice smooth line. Below is what happens when you try to use an unsigned distance field texture (aka the distance transform of the image, using JFA / Voronoi information), using the zero distance line as the surface of the object:
It looks ok (if not fairly pixelated), but you can really see it break down when you zoom in:
So you might say to yourself, maybe i need to keep the surface line at distance 0.5 instead of 0.0 so that there is distance information to interpolate? If you do that, the first thing you might notice is that the objects get fatter:
But it does look better when you zoom in, which is a plus:
The real issue is that you really just need the distance from each pixel to the surface of the object from both the inside and the outside. In our case, our Voronoi diagram we make with JFA only gives the distance from the outside. So what is the solution? At first I was thinking maybe you can get the gradient of this data at the point of each pixel and “push the zero line in” a little bit to give at least one pixel layer worth of inside data. However, a brilliant friend of mine came up with the actual solution: You invert your source data so empty space becomes seed, and seed becomes empty space, and you run JFA again to get the distance from the inside!
That actually works very well. It’s also very easy to combine them. You make a pixel shader that reads the data from the outside Voronoi diagram and the inside Voronoi diagram, calculate the output distance (0.5 + outsideDistance * 0.5 – insideDistance * 0.5), and output that 0 to 1 distance value in one or more of the color channels.
Here’s a glsl excerpt below, note that we divide the distance by 8 and clamp between 0 and 1 so that the data is suitable for a normalized color image (normalized as in the color channels can store values between 0 and 1):
// calculate distances from seed coordinates
float outsideDist = clamp(length(outsideSeedCoord-fragCoord) / 8.0, 0.0, 1.0);
float insideDist = clamp(length(insideSeedCoord-fFragCoord) / 8.0, 0.0, 1.0);
// calculate output distance
float signedDistance = 0.5 + outsideDist * 0.5 - insideDist * 0.5;
// set the color based on that distance
fragColor = vec4(signedDistance);
It actually looks a lot like the first image where we use the zero distance line of the unsigned distance field texture, so we still aren’t quite there:
When you zoom in, it looks a little better, but something still seems a bit off:
The final step to making this look good is to realize that the power of signed distance textures is in their ability to interpolate distance information well. When we have a full resolution texture, there is no interpolation going on. We actually need to decrease the size of our distance field texture to make it look better. If only all problems were solved by making textures smaller!
Here is the resulting image when making the distance field texture 1/4 as large on each axis (1/16th as big total):
And zooming in you can see that it scales very well. The zoom is a 20x magnification, on top of the magnification we already get from it being a larger texture:
And just to show the intermediary textures, here is the outside distance Voronoi diagram:
And the inside distance Voronoi diagram (The seed is in bright green, the dim green is the empty space that has distance information):
And here is the final distance field texture used to render the final result I showed above.
Zoomed in to show just how low resolution it is! This is the thing that looks like a + or a sword just left of middle.
Again, here is the shadertoy that does this technique, generating a signed distance field texture on the fly for randomly placed objects, and then using that signed distance field to render a larger image that you can further zoom in to:
## Fast Voronoi Diagrams and Distance Field Textures on the GPU With the Jump Flooding Algorithm
The image below is called a Voronoi diagram. The circles show you the location of the seeds and the color of each pixel represents the closest seed to that pixel. The value of this diagram is that at any point on the image, you can know which seed (point) is the closest to that pixel. It’s basically a pre-computed “closest object” map, which you can imagine some uses for I bet.
Voronoi diagrams aren’t limited to just points as seeds though, you can use any shape you want.
One way Voronoi diagrams can be useful is for path finding because they are dual to (the equivelant of) Delauny triangulation (More info at How to Use Voronoi Diagrams to Control AI).
Another way they can be useful is in generating procedural content, like in these shadertoys:
Another really cool usage case of Voronoi diagrams is in creating what is called the “Distance Transform”. The distance transform calculates and stores the distance from each pixel to the closest seed, whichever one that may be. Doing that, you may get an image like this (note that the distance is clamped at a maximum and mapped to the 0 to 1 range to make this image).
That is what is called a distance texture and can be used for a very cool technique where you have texture based images that can be zoomed into / enlarged quite a ways before breaking down and looking bad. The mustache in the image below was made with a 32×16 single channel 8bpp image for instance! (ignore the white fog, this was a screenshot from inside a project I was working on)
Distance field textures are the next best thing to vector graphics (great for scalable fonts in games, as well as decals!) and you can read about it here: blog.demofox.org: Distance Field Textures
Now that you know what Voronoi diagrams and distance transforms are used for, let’s talk about one way to generate them.
## Jump Flooding Algorithm (JFA)
When you want to generate either a Voronoi diagram or a distance transform, there are algorithms which can get you the exact answer, and then there are algorithms which can get you an approximate answer and generally run a lot faster than the exact version.
The jump flooding algorithm is an algorithm to get you an approximate answer, but seems to have very little error in practice. It also runs very quickly on the GPU. It runs in constant time based on the number of seeds, because it’s execution time is based on the size of the output texture – and is based on that logarithmically.
Just a real quick note before going into JFA. If all else fails, you could use brute force to calculate both Voronoi diagrams as well as distance transforms. You would just loop through each pixel, and then for each pixel, you would loop through each seed, calculate the distance from the pixel to the seed, and just store the information for whatever seed was closest. Yes, you even do this for seed pixels. The result will be a distance of 0 to the seed 😛
Back to JFA, JFA is pretty simple to program, but understanding why it works may take a little bit of thinking.
Firstly you need to prepare the N x N texture that you want to run JFA on. It doesn’t need to be a square image but let’s make it square for the explanation. Initialize the texture with some sentinel value that means “No Data” such as (0.0, 0.0, 0.0, 0.0). You then put your seed data in. Each pixel that is a seed pixel needs to encode it’s coordinate inside of the pixel. For instance you may store the fragment coordinate bitpacked into the color channels if you have 32 bit pixels (x coordinate in r,g and y coordinate in b,a). Your texture is now initialized and ready for JFA!
JFA consists of taking log2(N) steps where each step is a full screen pixel shader pass.
In the pixel shader, you read samples from the texture in a 3×3 pattern where the middle pixel is the current fragment. The offset between each sample on each axis is 2^(log2(N) – passIndex – 1), where passIndex starts at zero.
That might be a bit hard to read so let’s work through an example.
Let’s say that you have an 8×8 texture (again, it doesn’t need to be square, or even power of 2 dimensions, but it makes it easier to explain), that has a 16 bit float per RGBA color channel. You fill the texture with (0.0, 0.0, 0.0, 0.0) meaning there is no data. Let’s say that you then fill in a few seed pixels, where the R,G channels contain the fragment coordinate of that pixel. Now it’s time to do JFA steps.
The first JFA step will be that for each pixel you read that pixel, as well as the rest of a 3×3 grid with offset 4. In total you’d read the offsets:
(-4.0, -4.0), (0.0, -4.0), (4.0, -4.0),
(-4.0, 0.0), (0.0, 0.0), (4.0, 0.0),
(-4.0, 4.0), (0.0, 4.0), (4.0, 4.0)
For each texture read you did, you calculate the distance from the seed it lists inside it (if the seed exists aka, the coordinate is not 0,0), and store the location of the closest seed int the output pixel (like, store the x,y of the seed in the r,g components of the pixel).
You then do another JFA step with offset 2, and then another JFA step with offset 1.
JFA is now done and your image will store the Voronoi diagram, that’s all! If you look at the raw texture it won’t look like anything though, so to view the Voronoi diagram, you need to make a pixel shader where it reads in the pixel value, deocdes it to get the x,y of the closest seed, and then uses that x,y somehow to generate a color (use it as a seed for a prng for instance). That color is what you would output in the pixel shader, to view the colorful Voronoi diagram.
To convert the Voronoi diagram to a distance transform, you’d do another full screen shader pass where for each pixel you’d calculate the distance from that pixel to the seed that it stores (the closest seed location) and write the distance as output. If you have a normalized texture format, you’re going to have to divide it by some constant and clamp it between 0 and 1, instead of storing the raw distance value. You now have a distance texture!
## More Resources
I originally came across this algorithm on shadertoy: Shadertoy: Jump Flooding. That shadertoy was made by @paniq who is working on some pretty interesting stuff, that you can check out both on shadertoy and twitter.
The technique itself comes from this paper, which is a good read: Jump Flooding in GPU with Applications to Voronoi Diagram and Distance Transform
## Extensions
While JFA as explained is a 2D algorithm, it could be used on volume textures, or higher dimensions as well. Higher dimensions mean more texture reads, but it will still work. You could render volume textures with raymarching, using the distance information as a hint for how far you could march the ray each step.
Also, I’ve played around with doing 5 reads instead of 9, doing a plus sign read instead of a 3×3 grid. In my tests it worked just as well as regular JFA, but I’m sure in more complex situations there are probably at least minor differences. Check the links section for a shadertoy implementation of this. I also tried doing a complete x axis JFA followed by a y axis JFA. That turned out to have a LOT of errors.
You can also weight the seeds of the Voronoi diagram. When you are calculating the distance from a pixel to a seed point, you use the seed weight to multiply and/or add to the distance calculated. You can imagine that perhaps not all seeds are created equal (maybe some AI should avoid something more than something else), so weighting can be used to achieve this.
Here are some shadertoys I made to experiment with different aspects of this stuff. You can go check them out to see working examples of JFA in action with accompanying glsl source code.
Jump Flood Algorithm: Points – Point Seeds
Jump Flood Algorithm: Shapes – Shape Seeds
Jump Flood Algorithm: Weight Pts – Multiplicative Weighting
Separable Axis JFA Testing – Does 5 reads instead of 9, also shows how separating axis completely fails.
Here is a really interesting shadertoy that shows a way of modifying a Vornoi diagram on the fly: Shadertoy: zoomable, stored voronoi cells
## GPU Texture Sampler Bezier Curve Evaluation
Below is a paper I submitted to jcgt.org that unfortunately did not get accepted. Maybe next time!
The main idea of this paper is that bilinear interpolation can be equivalent to the De Casteljau algorithm, which means that if you set up a texture in a specific way, and sample from it at specific texture coordinates, that it will in fact give you Bezier curve points as output! It scales up for higher dimensional textures, as well as higher order curves.
The image below shows this in action for a cubic Bezier curve (3 control points) being stored and recalled from a 2×2 texture (there is actually a curve stored in each color channel).
This image is from an extension linked to lower down which applies the technique to surfaces and volumes:
The primary feedback from the reviewers and editor was that:
• It was an interesting technique and they thought it was a paper worth reading.
• The usage case was fairly limited though – basically only when your are compute bound in your shader program, and have some curve calculations to offload to the texture sampler. Or if you are already using a lookup texture and would benefit from fewer instructions and smaller lookup textures.
• It could have been shorter due to the writing being shorter, but also it could have been less thorough. For instance, it didn’t need to show equivalence to both the De Casteljau’s algorithm as well as Bernstein polynomials, since it’s already known that those are equivalent.
• They wanted some more performance details
I agree with the feedback, and don’t feel like taking the time to change and resubmit or submit else where, so I’m sharing it here on my blog. I hope you enjoy it and find it interesting (:
Here is the paper:
GPUBezier2016.pdf
Here is the supplemental materials (opengl and webgl source code):
SupplementalMaterials.zip
Here is the webgl demo from the supplemental materials, hosted on my site:
GPU Efficient Texture Based Bezier Curve Evaluation
Here are some working shadertoy demos of the technique:
## Extensions
Continuations of this work:
## Failed Experiments
Continuations that didn’t work out:
## G-Buffer Upsizing
The other day I had a thought:
Rendering smaller than full screen images is super helpful for performance, but upsizing an image results in pretty bad quality vs a full resolution render.
What if instead of upsizing the final rendered image, we upsized the values that were used to shade each pixel?
In other words, what if we rendered a scene from a less than full resolution g-buffer?
I was thinking that could be useful in doing ray based graphics, not having to trace or march quite so many rays, but also perhaps it could be useful for things like reflections where a user isn’t likely to notice a drop in resolution.
I’m sure I’m not the first to think of this, but I figured I’d explore it anyways and see what I could see.
I made an interactive shadertoy demo to explore this if you want to see it first hand:
## Result
In short, it does look better in a lot of ways because the normals, uv coordinates and similar parameters interpolate really well, but the edges of shapes are aliased really bad (jaggies).
Check out the images below to see what i mean. The first image is a full sized render. The second image is a 1/4 sized render (half x and half y resolution). The third image is a 1/16th sized render (quarter x and quarter y resolution)
For comparison, here’s a 1/4 sized and 1/16 sized render upsized using bicubic IMAGE interpolation instead of g-buffer data interpolation:
Despite the aliased results at 1/16th render size, this seems like it may be a reasonable technique at larger render sizes, depending on the level of quality you need. Doing half vertical or half horizontal resolution looks very close to the full sized image for instance. The edges are a tiny bit more aliased along one direction, but otherwise things seem decent:
Since the g-buffer has only limited space, you will probably want to bit pack multiple fields together in the same color channels. When you do that, you throw out the possibility of doing hardware interpolation unfortunately, because it interpolates the resulting bit packed value, not the individual fields that you packed in.
Even when doing the interpolation yourself in the pixel shader, for the most part you can really only store information that interpolates well. For instance, you could store a diffuse R,G,B color, but you wouldn’t want to store and then interpolate a material index. This is because you might have material index 10 (say it’s blue) next to material index 0 (say it’s green), and then when you interpolate you could end up with material index 5 which may be red. You’d get red between your blue and green which is very obviously wrong.
In my demo I did have a material index per pixel, but i just used nearest neighbor for that particular value always. To help the appearance of aliasing, I also stored an RGB diffuse color that i interpolated.
I stored the uvs in the g-buffer and sampled the textures themselves in the final shader though, to make sure and get the best texture information I could. This makes textures look great at virtually any resolution and is a lot of the reason why the result looks as good as it does IMO.
The fact that normals interpolate is a good thing, except when it comes to hard edges like the edge of the cube, or at the edge of any object really. In the case of the cube edge, it smooths the edge a little bit, making a surface that catches specular lighting and so highlights itself as being incorrect (!!). In the case of the edge of regular objects, a similar thing happens because it will interpolate between the normal at the edge of the object and the background, making a halo around the object which again catches specular lighting and highlights itself as being incorrect.
I think it could be interesting or fruitful to explore using edge detection to decide when to blend or not, to help the problem with normals, or maybe even just some edge detection based anti aliasing could be nice to make the resulting images better. The depth (z buffer value) could also maybe be used to help decide when to interpolate or not, to help the problem of halos around every object.
Interestingly, bicubic interpolation actually seems to enhance the problem areas compared to bilinear. It actually seems to highlight areas of change, where you would actually want it to sort of not point out the problems hehe. I think this is due to Runge’s phenomenon. Check out the depth information below to see what i mean. The first is bilinear, the second is bicubic:
One final side benefit of this I wanted to mention, is that if you are doing ray based rendering, where finding the geometry information per pixel can be time consuming, you could actually create your g-buffer once and re-shade it with different animated texture or lighting parameters, to give you a constant time (and very quick) render of any scene of any complexity, so long as the camera wasn’t moving, and there were no geometry changes happening. This is kind of along the same lines as the very first post I made to this blog about 4 years ago, which caches geometry in screen space tiles, allowing dirty rectangles to be used (MoriRT: Pixel and Geometry Caching to Aid Real Time Raytracing).
Anyone else go down this path and have some advice, or have any ideas on other things not mentioned? (:
Next up I think I want to look at temporal interpolation of g-buffers, to see what sort of characteristics that might have. (Quick update, the naive implementation of that is basically useless as far as i can tell: G-Buffer Temporal Interpolation).
## Related Stuff
On shadertoy, casty mentioned that if you have some full res information, and some less than full res information, you can actually do something called “Joint Bilateral Upsampling” to get a better result.<|endoftext|>
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Children can find music rests rather daunting. They have only just learnt to recognise musical notes and their values and suddenly there are even more signs and numbers to grasp! So here is an easy way to teach kids rests in order to help them recognise and understand them in an easy and approachable way.
This easy way to teach kids rests is a kid friendly, visual memory trigger to help tell the different rests from one another. It is always important of course to teach the correct rest names and appearances alongside their nick-names, so make sure children understand that this is your own special way of helping them to remember.
Teaching Kids Rests
When you start the lesson explain that when you make music together it is not only the sound that you make that matters, but also the silence in between the notes. For without the breaks in the sound there would be no shape or pattern to the music. Explain that each note has its own rest of the same time value, so that you know how long it should last.
Using the Teaching Kids Rests infographic above, talk the kids through each note, time value and the rest that belongs to to it. Remark on what they think it looks like, and discuss why. If the child is keen, they can try to draw some of their own. This will help them to both recognise and understand the concept.
Easy Way To Teach Kids Rests Guessing Game
Once they are familiar with each sign and the time value, you can play a guessing game. The information on the infographic is divided into three columns, rests, time values and nicknames. Use a piece of card to hide a column at a time and ask if they can remember what is hidden underneath. Prompt them with questions like “Which note is worth 2 beats?” and “Which rest looks like lightening?” Move along each column asking different questions to help them remember which rest belongs to which note, and what is hiding under the card.
This is a simple memory prompt and recall game, but I have always found it very effective – and once learned it really sticks! Even slightly older pupils find it a quick and easy way to grasp rests and are quite happy to spot the difference between a ‘bat’ and a ‘mini-car’!
If you enjoyed this post, come and join us on Facebook, Pinterest and Google + or subscribe to our free bi-weekly newsletter for many more musical ideas!
Stay up to date with Let's Play Music!
Receive new post notifications and updates from Let's Play Music straight in your inbox!<|endoftext|>
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# The Tangent, Sine, and Cosine Ratio
To start, a ratio is a mathematical comparison. In other words, the comparison of any two quantities is called a ratio. Any time you work with fractions, for example, you’re technically comparing the numerator quantity to the denominator:
null
## The Tangent Ratio
If you have a right triangle, and you chose one of the acute angles in it (acute meaning less than 90°) as your reference angle, the opposite length from that angle compared to the adjacent length is called the tangent ratio. Notice in the animation that the acute angles are 30° and 60º. The side opposite of 30° is y and adjacent is x. Whereas if we chose 60º as our acute angle, the opposite length is x and adjacent is y. That’s summarized below:
null
In case you forget, the tangent ratio can be remembered as TangentOppositeAdjacent (TOA).
## The Sine Ratio
What if we wanted to compare the opposite to hypotenuse length? In that case, we’d use another trigonometric ratio known as Sine. Let’s investigate sine using the generic right triangle below:
Notice the symbol (θ, “theta”). That will be considered our reference angle here. As mentioned before, the reference angle can be any of the acute angles (less than 90°) in the right triangle – never choose the 90° angle as your reference. Therefore, the sine ratio applied to θ is summarized below:
null
The sine ratio can be remembered as SineOppositeHpotenuse (SOH).
## The Cosine Ratio
The last comparison we’ll make using the same right triangle shown above is the cosine ratio. The cosine ratio is a comparison of the adjacent length to the hypotenuse relative to the angle θ:
null
The sine ratio can be remembered as CosineAdjacentHpotenuse (CAH). Collectively, the sine, cosine, and tangent ratios are referred to as trigonometric ratios. Using all three abbreviations stated above, the best way to remember the ratios is through the mnemonic:
# SOH CAHTOA
Trigonometric ratios can be used to find two main things:
1. Missing sides in a right triangle
2. Missing angles in a right triangle
To find the length of an unknown side, you must be given one acute reference angle and one known side. Then you have a straight forward calculation involving your calculator and some algebraic manipulation. Keep in mind that most modern day calculators come pre-programmed with all possible ratios of sine, cosine, and tangent for any angle. So if you wanted to find the ratio for sin(50°), you’d click the sine button (sine function), then your angle (shown in yellow):
null
Try it yourself, and make sure your calculator is in degree mode (shown in red). Notice that four numbers after the decimal place were kept for good measure. This will ensure that if you do the opposite – where you want to find an angle given the ratio – you get an angle that’s accurate to 50° (this will be discussed further below). Also, by writing 0.7660 over 1, it illustrates how sine of 50 degrees is comparing the opposite length (0.7660) to the hypotenuse (1).
This leads us to the second reason we use trigonometric ratios. As mentioned in point (2), if we have 2 known sides, the unknown acute angle of any of the vertices can be found using inverse trigonometric functions that are also found on your calculator (blue arrow). You’ll have to click “shift” or “2nd” first to access these functions. The two videos below will walk you through four different examples related to finding the angle when only the sides are known.
Of course, this lesson wouldn’t be complete if we didn’t see examples where we find an unknown side when given a known acute angle and a known side.
Summary:
The three primary trigonometric ratios are sine, cosine, and tangent. They are defined as follows: null
You can find any side length or angle measure of a right triangle if you know two pieces of information in addition to the right angle.<|endoftext|>
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1. ## sum of roots...
find Sum of Roots of the equation $\displaystyle 4x^{\lfloor log_{10}2x-3 \rfloor} = 1$
Where $\displaystyle \lfloor x \rfloor$ = greatest integer function.
like $\displaystyle \lfloor 2.3 \rfloor = 2$
2. $\displaystyle 4x^{\lfloor \log_{10}2x-3 \rfloor} = 1$
$\displaystyle \log_{10}(4x^{\lfloor \log_{10}2x-3 \rfloor}) = \log_{10}(1)$
$\displaystyle \log_{10}4 + \log_{10}(x^{\lfloor \log_{10}2x-3 \rfloor})) = 0$
$\displaystyle \log_{10}4 + \lfloor \log_{10}2x-3 \rfloor \log_{10}(x) = 0$
$\displaystyle \log_{10}4 + \lfloor \log_{10}2x-3 \rfloor (\log_{10}2x - \log_{10}2) = 0$
Let $\displaystyle y = \log_{10}2x$
$\displaystyle \log_{10}4 + \lfloor y-3 \rfloor (y - \log_{10}2) = 0$
First solve without greatest integer (to get close to the solution)
$\displaystyle \log_{10}4 + (y-3) (y - \log_{10}2) = 0$
$\displaystyle y^2 - (3 + \log_{10}2)y + 5 \log_{10}2 = 0$
Solutions are around 2.75 and 0.55
Now with this we guess values for $\displaystyle a = \lfloor y\rfloor$ (0,1,2,3)
$\displaystyle \log_{10}4 + (a - 3) (y - \log_{10}2) = 0$
$\displaystyle y = -\frac{\log_{10}4}{a - 3} + \log_{10}2$
We see that $\displaystyle a = 0$ is the only one with a consistent solution.
$\displaystyle y = \frac{\log_{10}4}{3} + \log_{10}2 = \log_{10}2^{\frac{5}{3}}$
$\displaystyle y = \log_{10}2x$
$\displaystyle x = \frac{10^y}{2} = 2^{\frac{2}{3}}$
And this is the only solution for $\displaystyle x$<|endoftext|>
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Home » Physics » Average velocity formula
# Average velocity formula
In this article learn about the average velocity formula. Now if you are interested in learning the concept of average speed and average velocity for class 11 physics and competitive exams like NEET, JEE Mains and JEE Advanced do follow the link given below.
## Average velocity definition
Average velocity is defined as the ratio of total displacement to total time. It is the average values of the velocities we know.
## Average velocity formula
Let’s talk about the average velocity formula. From the average velocity definition, we know that it is the ratio between total displacement and total time but this $v_{av}$ i.e., the average velocity formula can have different forms according to different problems. So before solving problems that ask you to find average velocity please check under which case your problem falls.
1. Case 1:-
If $X$ is the total displacement of the body and $t$ is the total time taken by the body to complete the displacement.The average velocity $V_{av}$ is given by
$$v_{av} = \frac{Total\ distance\ traveled}{total\ time\ taken} = \frac{X}{T}$$
2. Case 2:-
If we know that body has initial displacement position $x_0$ at time $t_0$ undergoes a motion and attains final displacement $x_f$ at time $t_f$ then, the average velocity is given as
$$v_{av}=\frac{x_f – x_0}{t_f – t_0}=\frac{\Delta\ x}{\Delta\ t}$$
3. Case 3:-
If we have a knowledge that $x_1, x_2, . . . . . x_n$ are the distance covered by the body for time $t_1, t_2, . . . . . t_n$. The average velocity is
$$\frac{x_1 + x_2 + . . . . . . . + x_n}{t_1 + t_2 + . …. . . + t_n}$$
4. Case 4:-
If a velocity is changing as a result of constant acceleration, the average velocity can be found by adding the initial and final velocities and dividing by 2. Here it is important to note that this formula applies only to constant acceleration cases.
Consider a body moving with uniform acceleration along a straight path. At some instant of time, initially, its velocity is $U$ m/s and at the end of t seconds let its final velocity be $V$ m/s. Hence the average velocity is given as
$$v_{av}=\frac{v+u}{2}$$
How to derive this average velocity formula
Average velocity is displacement divided by time. Let us start with the formula for displacement and divide it by t.
\begin{align} s=ut+\frac{1}{2}at^{2} \\ average \quad velocity & = \frac{ut+\frac{1}{2}at^{2}}{t} \\ & =u+\frac{1}{2}at \\ & =\frac{2u+at}{2} \\ & =\frac{u+(u+at)}{2} \end{align}
now, $v=u+at$
$$average \quad velocity = \frac{u+v}{2}$$
Some related articles you might be interested in reading
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Dr. Cleveland Evans writes about names for the Omaha World-Herald. In his July 3rd column, he looks at the history of the name Thomas.
Thomas, one of Jesus’s original apostles, is famous for refusing to believe Christ’s resurrection until he’d touched His wounds. It’s believed he was martyred in India on July 3, 72. Thomas is from the Aramaic Ta’oma, “twin.” Its popularity with medieval Catholics was reinforced by renowned theologian St. Thomas Aquinas (1225-1274).
In England, a bigger influence was St. Thomas Becket (1119-1170). Becket, Lord Chancellor for his friend King Henry II, became Archbishop of Canterbury in 1162. Conflicts over church rights led four of Henry’s knights to misinterpret the king’s angry rant as an order to kill. Becket’s murder in the cathedral led Pope Alexander III to canonize him in 1173. His Canterbury tomb became a place of pilgrimage, and Thomas became a hugely popular name. By 1380, it ranked third. It was second or third every year between 1538 and 1850, much more common in England than the rest of Europe.
Want to know more? Read on to find out more about Thomases in history!<|endoftext|>
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### Course: Algebra 1>Unit 13
Lesson 8: Factoring quadratics with perfect squares
Learn how to factor quadratics that have the "perfect square" form. For example, write x²+6x+9 as (x+3)².
Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication.
In this article, we'll learn how to factor perfect square trinomials using special patterns. This reverses the process of squaring a binomial, so you'll want to understand that completely before proceeding.
## Intro: Factoring perfect square trinomials
To expand any binomial, we can apply one of the following patterns.
• $\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$
• $\left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Note that in the patterns, $a$ and $b$ can be any algebraic expression. For example, suppose we want to expand $\left(x+5{\right)}^{2}$. In this case, $a=x$ and $b=5$, and so we get:
$\begin{array}{rl}\left(x+5{\right)}^{2}& ={x}^{2}+2\left(x\right)\left(5\right)+\left(5{\right)}^{2}\\ \\ & ={x}^{2}+10x+25\end{array}$
You can check this pattern by using multiplication to expand $\left(x+5{\right)}^{2}$.
The reverse of this expansion process is a form of factoring. If we rewrite the equations in the reverse order, we will have patterns for factoring polynomials of the form ${a}^{2}±2ab+{b}^{2}$.
We can apply the first pattern to factor ${x}^{2}+10x+25$. Here we have $a=x$ and $b=5$.
$\begin{array}{rl}{x}^{2}+10x+25& ={x}^{2}+2\left(x\right)\left(5\right)+\left(5{\right)}^{2}\\ \\ & =\left(x+5{\right)}^{2}\end{array}$
Expressions of this form are called perfect square trinomials. The name reflects the fact that this type of three termed polynomial can be expressed as a perfect square!
Let's take a look at a few examples in which we factor perfect square trinomials using this pattern.
## Example 1: Factoring ${x}^{2}+8x+16$
Notice that both the first and last terms are perfect squares: ${x}^{2}=\left(x{\right)}^{2}$ and $16=\left(4{\right)}^{2}$. Additionally, notice that the middle term is two times the product of the numbers that are squared: $2\left(x\right)\left(4\right)=8x$.
This tells us that the polynomial is a perfect square trinomial, and so we can use the following factoring pattern.
In our case, $a=x$ and $b=4$. We can factor our polynomial as follows:
$\begin{array}{rl}{x}^{2}+8x+16& =\left(x{\right)}^{2}+2\left(x\right)\left(4\right)+\left(4{\right)}^{2}\\ \\ & =\left(x+4{\right)}^{2}\end{array}$
We can check our work by expanding $\left(x+4{\right)}^{2}$:
$\begin{array}{rl}\left(x+4{\right)}^{2}& =\left(x{\right)}^{2}+2\left(x\right)\left(4\right)+\left(4{\right)}^{2}\\ \\ & ={x}^{2}+8x+16\end{array}$
1) Factor ${x}^{2}+6x+9$.
2) Factor ${x}^{2}-6x+9$.
3) Factor ${x}^{2}+14x+49$.
## Example 2: Factoring $4{x}^{2}+12x+9$
It is not necessary for the leading coefficient of a perfect square trinomial to be $1$.
For example, in $4{x}^{2}+12x+9$, notice that both the first and last terms are perfect squares: $4{x}^{2}=\left(2x{\right)}^{2}$ and $9=\left(3{\right)}^{2}$. Additionally, notice that the middle term is two times the product of the numbers that are squared: $2\left(2x\right)\left(3\right)=12x$.
Because it satisfies the above conditions, $4{x}^{2}+12x+9$ is also a perfect square trinomial. We can again apply the following factoring pattern.
In this case, $a=2x$ and $b=3$. The polynomial factors as follows:
$\begin{array}{rl}4{x}^{2}+12x+9& =\left(2x{\right)}^{2}+2\left(2x\right)\left(3\right)+\left(3{\right)}^{2}\\ \\ & =\left(2x+3{\right)}^{2}\end{array}$
We can check our work by expanding $\left(2x+3{\right)}^{2}$.
4) Factor $9{x}^{2}+30x+25$.
5) Factor $4{x}^{2}-20x+25$.
## Challenge problems
6*) Factor ${x}^{4}+2{x}^{2}+1$.
7*) Factor $9{x}^{2}+24xy+16{y}^{2}$.
## Want to join the conversation?
• is (4+x)^2 the same as (x+4)^2
• Yes, they are the same, as you can switch the numbers around within the parenthesis, just not outside of them.
• Just curious, what's the point of doing this to an expression? Like, where will this be applied?
• Pretty much all of algebra will be used when we start doing calculus which is where math gets really cool with way more real world applications like launching satellites into space and making vaccines. Endless possibilities! But we must have a SOLID foundation in algebra for all this cool stuff!
• Number three marks (X+7) squared wrong as well, but the "I need help" section is correct.
• Maybe your Caps Lock is on. (X+7) is different of (x+7).
Hope I have helped :)
• I don't understand question 4 and 5. in question 4 he said ""notice that the middle term is two times the product of the numbers that are squared: 2(3x)(5)=30x2(3x})(\D 5)=30x2(3x)(5)=30x2, l, 3, x, D, 5, equals, 30, x. then he does the equation again and shows the answer I don't understand how he got there
• normally when you have for example x^2+6x+9 you would take the root of 9 and multiply it by 2. if the answer is the first-degree variable coefficient (6) then it satisfy the perfect squares "trick" rule, and the answer would be (x+3)^2.
when you have a coefficient to the 2nd-degree variable like in 9x^2+30x+25, you do almost the same:
first you find the root of the 0-degree variable (25), which is 5. (5^2=25)
then you find the root of the 2nd-degree variable's coefficient (9), which is 3 (3^2=9).
now here is the different from before: you need to multiply both roots (3 and 5) and then multiply them by 2 (as you normally would). if the answer is the same as the 1st-degree variable's coefficient ((5*3)*2=30) then the trick can be applied here also, giving you the answer made of those 2 roots you found erlier(3 and 5)-squared.
(3x+5)^2
hope you understood and that it helped you.
• What do the a and b stand for in the equation a^2+2ab+b^2 ? I'm very confused :/
• they are just any varibles. a and b can be anything (a has to be equal to a and b has to be equal to b tho)
(1 vote)
• is algebra 1 a high school class because im taking it in middle school
• It can be, many people take algebra in middle school but there are also plenty of people who take it in high school. I think most people take geometry freshman year in high school (9th grade) but there are plenty of people who take it before or after
• Does anyone understand this at all?<|endoftext|>
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### Section 5.5: Distribution of the Sample Proportion
#### Key Concepts
This sampling distribution should be used whenever a single sample is drawn and the statistic of interest is counting the proportion in the sample that fall into a given category. As in all examples presented in this chapter, the central limit theorem allows the use of the normal distribution to find probabilities.
In this situation, there are different formula for the mean and standard error, but the same logic and procedure for solving problems remains the same.
Since sample proportions can only take on a finite number of possible values for a given sample size, there can be error caused by approximating a discrete distribution with a continuous one. This can be fixed using the correction for continuity.
Another example shows how to use the sampling distribution.
#### Formula
The sampling distribution of is summarized by:
mean() = p
and SE() =
The shape will be approximately normal for sufficiently large samples. A general rule of thumb is that if np > 5 and n(1-p) > 5, then the distribution will be approximately normal. When n is small, the approximation can be improved greatly by using the correction for continuity.
#### Correction for Continuity
Suppose that p = .60 and a sample of size 100 is randomly chosen. Find the probability that the sample proportion is between 0.56 and 0.65.
Draw a sketch!
The sampling distribution of has a mean of .6 and an SE of = .0490. The z-scores are respectively
z = (.56 - .60) / .0490 = -0.82 and z = (.65 - .60) / .0490 = 1.02
The area between -0.82 and 1.02 under the normal curve is .8461 - .2061 = .6400.
An alternative way to approach the problem is to say that
P(.56 <= <= .65) = P(56 <= X <= 65)
where = X/100, and X is the count of successes in the sample. X is a binomial random variable with a mean of 100(.6) = 60, and standard deviation of 4.90. The probability that X is exactly equal to a number x is well approximated by the area between x - 1/2 and x + 1/2 under a normal curve with mean = 60 and standard deviation = 4.90. Thus P(56 <= X <= 65) should be well approximated by the area between 55.5 and 65.5. The z-scores are
z = (55.5 - 60)/4.9 = -0.92 and (65.5 - 60)/4.9 = 1.12
The area under the standard normal curve between these values is .8686 - .1788 = .6898.
The exact probability from the binomial distribution is .6908. Thus, the straightforward calculation using the correction for continuity is much more accurate, even with a sample of size 100.
#### Example
2% of all American women aged 50--54 have breast cancer. Suppose that 1000 women in this age group are randomly selected. What is the probability that the proportion of women in the sample, (), exceeds 4%?
The sampling distribution for the sample proportion will be approximately normal since (1000)(.02) = 20, which is much larger than 5. Since the sample size is so large, the correction for continuity should not make much difference.
The distribution is centered at .02 and has an SE of = .00443. The z-score is
z = (.04 - .02) / .0043 = 4.65
Since the area to the right of 4.65 under the standard normal curve is essentially 0, it is quite unlikely to observe such a result by chance. If the sample proportion actually is .04, it is likely that either p is not .02, or the method for sampling was biased.<|endoftext|>
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The existence of liquid water, and to a lesser extent, gaseous and solid dosage forms on Earth is vital to the existence of life on Earth as we know it. The land is located in a residential area of the solar system, if it was slightly closer to or farther from (about 5%, or about 8 million km), the conditions that must be present simultaneously would be much less. Gravity of the earth allows him to retain the atmosphere. Water vapor and carbon dioxide in the atmosphere provides a temperature buffer (greenhouse effect), which allows you to maintain a relatively steady surface temperature. If Earth were smaller, thinner atmosphere allows extreme temperatures, thereby preventing the accumulation of water, except the polar ice caps (as on Mars). Surface temperature was relatively constant through geologic time, despite different levels of incoming solar radiation (insolation), indicating that the dynamic process of regulating the temperature of the Earth through a combination of greenhouse gases and surface or atmospheric albedo.
This proposal is known as the Gaia hypothesis. The state of water on the planet depends on external pressure, which is determined by gravity of the planet. If the planet is massive enough water on it can be hard, even at high temperatures, due to the high pressure caused by gravity. There are various theories about the origin of water on Earth. Water ice is present on: Land – mostly ice sheets of polar ice caps on Mars, Titan Enceladus Europa Comets and comet population source (in the Kuiper belt and Oort cloud objects). Water ice may be present on the Moon, Ceres, and Tethys.
Water and other volatiles are likely to constitute a significant part of internal structures Uranus and Neptune. Water is the most common solvent in the world, largely determines the nature of the Earth's chemistry as a science. Much of the chemistry, in its origin as a science, began just as the chemistry of water solutions of substances. It is sometimes regarded as the ampholyte – and acid and base at the same time (H + cation anion OH-). In the absence of foreign substances in the water the same concentration of hydroxide ions and hydrogen ions (or ions hydronium), pKa ca. 16. By itself, water is relatively inert under normal conditions, but its highly polar molecules solvated ions and molecules to form crystalline hydrates. Solvolysis, especially hydrolysis, occurs in animate and inanimate nature, and is widely used in chemical industry.<|endoftext|>
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Explore regular polygons to find out which regular polygons will tessellate the plane.
- Explore some regular polygons to see which of these will tessellate in the plane.
- Use observations to answer the question, “Which regular polygons will tessellate in the plane and why?”
- regular polygon
- pentagon, hexagon, heptagon, octagon
About the Lesson
This activity is designed to be student-centered, with the teacher acting as a facilitator while students work cooperatively. The time varies for the activity depending on whether the TI-Nspire document (.tns file) is provided for or created by the students.<|endoftext|>
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by Lacy Compton
With the emphasis on 21st-century skills and growing expertise in gifted students, Problem-Based Learning (PBL) units have the potential to teach kids global skills while helping them creatively solve real-world problems. Such units introduce a "fuzzy" or ill-structured problem--one that doesn't have enough information included to be solved. Students then use research skills, critical and creative thinking, and other skills to help determine their solution to the problem. The best part? A good PBL unit doesn't have a set solution--PBL units allow students to self-direct their learning. It's a flexible alternative to learning that taps into kids' love of open-ended problems.
How can teachers implement this type of strategy in the classroom? Start by reading this article, "Problem-Based Learning 101," by Shelagh Gallagher, one of the gurus of this method of teaching. Then, why not look into current research and "problems" real scientists are facing as inspiration for writing your own PBL units? Some stories you might read for inspiration include:
- National Geographic recently reported on a study that purports to find the reason behind why the 1918 flu was so deadly. Considering this is a mystery that's troubled scientists for decades, it and other similar disease outbreaks can be used as fodder for PBL units. For example, you can ask students to think futuristically and pretend a similar disease is sweeping the world--then have them propose solutions for containing and preventing it.
- A common problem facing gardeners and biologists both is the shortage of bumblebees in North America, as reported by the National Wildlife Federation. Such a problem can be localized, asking students to contact local biologists and wildlife specialists, as well as gardeners, nursery owners, etc. to help them determine how they would increase the bee population in their city.
- The National Science Foundation has an entire website dedicated to "Brain Power" or the ways neuroscience is affecting technological innovations. The fascinating articles on this site could be helpful to teachers, especially when thinking about the ways neuroscience is working to solve problems like paralysis, hearing impairments, and visual impairments.
These are just some ideas for inspiring your own fuzzy problems. Whatever course you take, don't forget to throw some kickers in for the students--side problems that might shake up their discoveries and cause them to rethink the solution paths they were building. For example, with the bee problem, you might have a group of parents of children highly allergic to bees protest having new fauna planted in local parks that contain public playgrounds. Students have to weigh these opinions and determine how they affect their plans.<|endoftext|>
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EPFL and UNIGE scientists have developed a microchip using graphene that could help wireless telecommunications share data at a rate that is ten times faster than currently possible. The results are published today in Nature Communications.
"Our graphene based microchip is an essential building block for faster wireless telecommunications in frequency bands that current mobile devices cannot access," says EPFL scientist Michele Tamagnone.
Graphene acts like polarized sunglasses
Their microchip works by protecting sources of wireless data -- which are essentially sources of invisible radiation -- from unwanted radiation, ensuring that the data remain intact by reducing source corruption.
They discovered that graphene can filter out radiation in much the same way as polarized glasses. The vibration of radiation has an orientation. Like polarized glasses, their graphene-based microchip makes sure that radiation that only vibrates a certain way gets through. In this way, graphene is both transparent and opaque to radiation, depending on the orientation of vibration and signal direction. The EPFL scientists and their colleagues from Geneva used this property to create a device known as an optical isolator.
Faster Uploads in the Terahertz Bandwidth
Moreover, their microchip works in a frequency band that is currently empty, called the Terahertz gap.
Wireless devices work today by transmitting data in the Gigahertz range or at optical frequencies. This is imposed by technological constraints, leaving the potential of the Terahertz band currently unexploited for data transmission.
But if wireless devices could use this Terahertz bandwidth, your future mobile phone could potentially send or receive data tens of times faster than now, meaning better sound quality, better image quality and faster uploads.
The graphene-based microchip brings this Terahertz technology a step closer to reality. This discovery addresses an important challenge that was so far unsolved due to lacking technologies, confirming once more the extraordinary physical properties of graphene.
This joint project between EPFL and the University of Geneva was funded by the European Graphene Flagship project and by the Swiss National Science Foundation.<|endoftext|>
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Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.8
• Last Updated : 19 Apr, 2021
Question 1. The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x = y – 4,
x – y = − 4
and y + 1 = 8(x – 2)
y + 1 = 8x – 16
8x – y = 1 + 16
8x – y = 17
Therefore, we have two equations
x – y = -4 —————-(i)
8x – y = 17 ——————(ii)
Subtracting the second equation from the first equation, we get
(x – y) – (8x – y) = – 4 – 17
x − y − 8x + y = −21
−7x = −21
−7x = −21
x = 21/7 = 3
Substituting the value of x in the (i) eqn, we have
3 – y = – 4
y = 3 + 4 = 7
Hence the fraction is 3/7
Question: 2 A fraction becomes 9/11 if 2 is added to both numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x+2 / y+2 = 9/11
11(x+2) = 9(y+2)
11x + 22 = 9y + 18
11x – 9y = 18 – 22
11x – 9y + 4 = 0 —————-(i)
and x+3 / y+3 = 5/6
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x – 5y = 15 –18
6x – 5y + 3 = 0 —————-(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we have
x / (-9 x 3 -(-5) x 4) = -y / (11 x 3 – 6 x 4) = 1 / (11 x (-5) – 6 x (-9))
x / 7 = y / 9 = 1
x = 7 and y = 9
Hence, the fraction is 7/9.
Question 3. A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator. If 1 is added to both the numerator and denominator, it becomes 1/2. Find the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y.
According to given condition’s,
x-1 / y-1 = 1/3
3(x – 1) = (y – 1)
3x – 3 = y – 1
3x – y – 2 = 0 —————–(i)
and x+1 / y+1 = 1/2
(2x + 1) = (y + 1) ⇒ 2x + 2 = y + 1
2x – y + 1 = 0 ——————(ii)
We have to solve the above equations for x and y,
By using cross-multiplication, we got
x / -1-2 = -y / 3+4 = 1 / -3+2
x / -3 = -y / 7 = 1 / -1
x = 3 and y = 7
Hence, the fraction is 3/7.
Question 4. If we add 1 to the numerator and subtract 1 from the denominator, a fraction becomes 1. It also becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y.
According to given condition’s,
x+1 / y-1 = 1
(x + 1) = (y – 1)
x + 1– y + 1 = 0
x – y + 2 = 0 ————–(i)
and x / y+1 = 1/2
2x = (y + 1)
2x – y – 1 = 0 ————-(ii)
We have to solve the above equations for x and y,
By using cross-multiplication, we got
x / 1+2 = -y / -1-4 = 1 / -1+2
x/3 = y/5 = 1
x = 3 and y = 5
Hence, the fraction is 3/5.
Question 5. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x + y = 12
x + y – 12 = 0 —————(i)
and x / y+3 = 1/2
2x = (y + 3)
2x – y – 3 = 0 —————-(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-3-12) = -y / (-3+24) = 1 / (-1-2)
x/15 = y/21 = 1/3
x = 5 and y = 7
Hence the fraction is 5/7.
Question 6. When 3 is added to the denominator and 2 is subtracted from the numerator a fraction becomes 14. And, when 6 is added to numerator and the denominator is multiplied by 3, it becomes 23. Find the fraction.
Solution:
Let’s assume that the numerator of a fraction be x and denominator be y,
According to given condition,
x-2 / y+3 = 1/4
4x – 8 = y + 3
4x – y = 11 ————–(i)
and x+6 / 3y = 2/3
3x + 18 = 6y
x – 2y = -6 ————–(ii)
x = 2y – 6 (from eqn. (ii))
substitute value of x in eqn. (i)
4(2y – 6) – y = 11
8y – 24 – y = 11
y = 5
x = 2 x 5 – 6 = 4
Hence, x / y = 4 / 5
Question 7. The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x + y = 18
x + y – 18 = 0 —————(i)
and x / y+2 =1/3
3x = (y + 2)
3x – y – 2 = 0
3x – y – 2 = 0 —————-(ii)
We have to solve the above equations for x and y,
By using cross-multiplication, we got
x / (-2-18) = -y / (-2+54) = 1 / (-1-3)
x/-20 = -y/52 = 1/-4
x = 5 and y = 13
Hence, the fraction is 5/13
Question 8. If 2 is added to the numerator of a fraction, it reduces to 1/2 and if 1 is subtracted from the denominator, it reduces to 1/3. Find the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
2(x + 2) = y
2x + 4 = y
2x – y + 4 = 0 ————-(i)
and x / y-1 = 1/3
3x = (y – 1)
3x – y + 1 = 0 —————-(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-1+4) = -y / (2-12) = 1 / (-2+3)
x / 3 = y / 10 = 1
x = 3 and y = 10
Hence, the fraction is 3/10.
Question 9. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2: 3. Determine the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x + y = 2x + 4
2x + 4 – x – y = 0
x – y + 4 = 0 —————-(i)
and x + 3 : y + 3 = 2 : 3
3(x + 3) = 2(y + 3)
3x + 9 = 2y + 6
3x – 2y + 3 = 0 ——–(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-120+60) = y / (200-75) = 1 / (-20+15)
x / 60 = y / 125 = 1 / 5
x = 5 and y = 9
Hence, the fraction is 5/9.
Question 10. If the numerator of a fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes 6/5. And, if the denominator is doubled and the numerator is increased by 8, the fraction becomes 2/5. Find the fraction.
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
2x / y-5 = 6/5
10x = 6(y – 5)
10x – 6y + 30 = 0
2(5x – 3y + 15) = 0
5x – 3y + 15 = 0 ————–(i)
and x+8 / 2y = 2/5
5(x + 8) = 4y
5x + 40 = 4y
5x – 4y + 40 = 0 ———–(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (-120+60) = -y / (200-75) = 1 / (-20+15)
x / 60 = y / 125 = 1 / 5
x = 12 and y = 25
Hence, the fraction is 12/25.
Question 11. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction
Solution:
Let’s assume that the numerator and denominator of the fraction be x and y respectively. Therefore, the fraction is x/y
According to given condition’s,
x-1 = 1/2 x (y-1)
x-1 / y-1 = 1/2
x + y = 2y – 3
x + y – 2y + 3 = 0
x – y + 3 = 0 ————(i)
and 2(x – 1) = (y – 1)
2x – 2 = (y – 1)
2x – y – 1 = 0 —————(ii)
We have to solve the above equations for x and y.
By using cross-multiplication, we got
x / (1+3) = -y / (-1-6) = 1 / (-1+2)
x / 4 = y / 7 = 11
x = 4 and y = 7
Hence, the fraction is 4/7.
My Personal Notes arrow_drop_up<|endoftext|>
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# 11.2: Rotational Variables
Learning Objectives
• Describe the physical meaning of rotational variables as applied to fixed-axis rotation
• Explain how angular velocity is related to tangential speed
• Calculate the instantaneous angular velocity given the angular position function
• Find the angular velocity and angular acceleration in a rotating system
• Calculate the average angular acceleration when the angular velocity is changing
• Calculate the instantaneous angular acceleration given the angular velocity function
So far in this text, we have mainly studied translational motion, including the variables that describe it: displacement, velocity, and acceleration. Now we expand our description of motion to rotation—specifically, rotational motion about a fixed axis. We will find that rotational motion is described by a set of related variables similar to those we used in translational motion.
## Angular Velocity
Uniform circular motion (discussed previously in Motion in Two and Three Dimensions) is motion in a circle at constant speed. Although this is the simplest case of rotational motion, it is very useful for many situations, and we use it here to introduce rotational variables.
In Figure $$\PageIndex{1}$$, we show a particle moving in a circle. The coordinate system is fixed and serves as a frame of reference to define the particle’s position. Its position vector from the origin of the circle to the particle sweeps out the angle $$\theta$$, which increases in the counterclockwise direction as the particle moves along its circular path. The angle $$\theta$$ is called the angular position of the particle. As the particle moves in its circular path, it also traces an arc length s.
The angle is related to the radius of the circle and the arc length by
$\theta = \frac{s}{r} \ldotp \label{10.1}$
The angle $$\theta$$, the angular position of the particle along its path, has units of radians (rad). There are $$2\pi$$ radians in 360°. Note that the radian measure is a ratio of length measurements, and therefore is a dimensionless quantity. As the particle moves along its circular path, its angular position changes and it undergoes angular displacements $$\Delta \theta$$.
We can assign vectors to the quantities in Equation \ref{10.1}. The angle $$\vec{\theta}$$ is a vector out of the page in Figure $$\PageIndex{1}$$. The angular position vector $$\vec{r}$$ and the arc length $$\vec{s}$$ both lie in the plane of the page. These three vectors are related to each other by
$\vec{s} = \vec{\theta} \times \vec{r} \ldotp \label{10.2}$
That is, the arc length is the cross product of the angle vector and the position vector, as shown in Figure $$\PageIndex{2}$$.
The magnitude of the angular velocity, denoted by $$\omega$$, is the time rate of change of the angle $$\theta$$ as the particle moves in its circular path. The instantaneous angular velocity is defined as the limit in which $$\Delta$$t → 0 in the average angular velocity $$\bar{\omega} = \frac{\Delta \theta}{\Delta t}$$:
$\omega = \lim_{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t} = \frac{d \theta}{dt}, \label{10.3}$
where $$\theta$$ is the angle of rotation (Figure $$\PageIndex{2}$$). The units of angular velocity are radians per second (rad/s). Angular velocity can also be referred to as the rotation rate in radians per second. In many situations, we are given the rotation rate in revolutions/s or cycles/s. To find the angular velocity, we must multiply revolutions/s by 2$$\pi$$, since there are 2$$\pi$$ radians in one complete revolution. Since the direction of a positive angle in a circle is counterclockwise, we take counterclockwise rotations as being positive and clockwise rotations as negative.
We can see how angular velocity is related to the tangential speed of the particle by differentiating Equation \ref{10.1} with respect to time. We rewrite Equation \ref{10.1} as
$s = r \theta \ldotp$
Taking the derivative with respect to time and noting that the radius r is a constant, we have
$\frac{ds}{dt} = \frac{d}{dt} (r \theta) = \theta \frac{dr}{dt} + r \frac{d \theta}{dt} = r \frac{d \theta}{dt}$
where $$\theta \frac{dr}{dt}$$ = 0. Here, $$\frac{ds}{dt}$$ is just the tangential speed vt of the particle in Figure $$\PageIndex{1}$$. Thus, by using Equation \ref{10.3}, we arrive at
$v_{t} = r \omega \ldotp \label{10.4}$
That is, the tangential speed of the particle is its angular velocity times the radius of the circle. From Equation \ref{10.4}, we see that the tangential speed of the particle increases with its distance from the axis of rotation for a constant angular velocity. This effect is shown in Figure $$\PageIndex{3}$$. Two particles are placed at different radii on a rotating disk with a constant angular velocity. As the disk rotates, the tangential speed increases linearly with the radius from the axis of rotation. In Figure $$\PageIndex{3}$$, we see that v1 = r1$$\omega_{1}$$ and v2 = r2$$\omega_{2}$$. But the disk has a constant angular velocity, so $$\omega_{1} = \omega_{2}$$. This means $$\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}$$ or v2 = $$\left(\dfrac{r_{2}}{r_{1}}\right)$$v1. Thus, since r2 > r1, v2 > v1.
Up until now, we have discussed the magnitude of the angular velocity $$\omega = \frac{d \theta}{dt}$$, which is a scalar quantity—the change in angular position with respect to time. The vector $$\vec{\omega}$$ is the vector associated with the angular velocity and points along the axis of rotation. This is useful because when a rigid body is rotating, we want to know both the axis of rotation and the direction that the body is rotating about the axis, clockwise or counterclockwise. The angular velocity $$\vec{\omega}$$ gives us this information. The angular velocity $$\vec{\omega}$$ has a direction determined by what is called the right-hand rule. The right-hand rule is such that if the fingers of your right hand wrap counterclockwise from the x-axis (the direction in which $$\theta$$ increases) toward the y-axis, your thumb points in the direction of the positive z-axis (Figure $$\PageIndex{4}$$). An angular velocity $$\vec{\omega}$$ that points along the positive z-axis therefore corresponds to a counterclockwise rotation, whereas an angular velocity $$\vec{\omega}$$ that points along the negative z-axis corresponds to a clockwise rotation.
We can verify the right-hand-rule using the vector expression for the arc length $$\vec{s} = \vec{\theta} \times \vec{r}$$, Equation \ref{10.2}. If we differentiate this equation with respect to time, we find
$\frac{d \vec{s}}{dt} = \frac{d}{dt}(\vec{\theta} \times \vec{r}) = \left(\dfrac{d \theta}{dt} \times \vec{r}\right) + \left(\vec{\theta} \times \dfrac{d \vec{r}}{dt}\right) = \frac{d \theta}{dt} \times \vec{r} \ldotp$
Since $$\vec{r}$$ is constant, the term $$\vec{\theta} \times \frac{d \vec{r}}{dt}$$ = 0. Since $$\vec{v} = \frac{d \vec{s}}{dt}$$ is the tangential velocity and $$\omega = \frac{d \vec{\theta}}{dt}$$ is the angular velocity, we have
$\vec{v} = \vec{\omega} \times \vec{r} \ldotp \label{10.5}$
That is, the tangential velocity is the cross product of the angular velocity and the position vector, as shown in Figure $$\PageIndex{5}$$. From part (a) of this figure, we see that with the angular velocity in the positive z-direction, the rotation in the xy-plane is counterclockwise. In part (b), the angular velocity is in the negative z-direction, giving a clockwise rotation in the xy-plane.
Example $$\PageIndex{1}$$: Rotation of a Flywheel
A flywheel rotates such that it sweeps out an angle at the rate of $$\theta$$ = $$\omega$$t = (45.0 rad/s)t radians. The wheel rotates counterclockwise when viewed in the plane of the page. (a) What is the angular velocity of the flywheel? (b) What direction is the angular velocity? (c) How many radians does the flywheel rotate through in 30 s? (d) What is the tangential speed of a point on the flywheel 10 cm from the axis of rotation?
Strategy
The functional form of the angular position of the flywheel is given in the problem as $$\theta$$(t) = $$\omega$$t, so by taking the derivative with respect to time, we can find the angular velocity. We use the right-hand rule to find the angular velocity. To find the angular displacement of the flywheel during 30 s, we seek the angular displacement $$\Delta \theta$$, where the change in angular position is between 0 and 30 s. To find the tangential speed of a point at a distance from the axis of rotation, we multiply its distance times the angular velocity of the flywheel.
Solution
1. $$\omega$$ = $$\frac{d \theta}{dt}$$ = 45 rad/s. We see that the angular velocity is a constant.
2. By the right-hand rule, we curl the fingers in the direction of rotation, which is counterclockwise in the plane of the page, and the thumb points in the direction of the angular velocity, which is out of the page.
3. $$\Delta \theta$$ = $$\theta$$(30 s) − $$\theta$$(0 s) = 45.0(30.0 s) − 45.0(0 s) = 1350.0 rad.
4. vt = r$$\omega$$ = (0.1 m)(45.0 rad/s) = 4.5 m/s.
Significance
In 30 s, the flywheel has rotated through quite a number of revolutions, about 215 if we divide the angular displacement by 2$$\pi$$. A massive flywheel can be used to store energy in this way, if the losses due to friction are minimal. Recent research has considered superconducting bearings on which the flywheel rests, with zero energy loss due to friction.
## Angular Acceleration
We have just discussed angular velocity for uniform circular motion, but not all motion is uniform. Envision an ice skater spinning with his arms outstretched—when he pulls his arms inward, his angular velocity increases. Or think about a computer’s hard disk slowing to a halt as the angular velocity decreases. We will explore these situations later, but we can already see a need to define an angular acceleration for describing situations where $$\omega$$ changes. The faster the change in $$\omega$$, the greater the angular acceleration. We define the instantaneous angular acceleration $$\alpha$$ as the derivative of angular velocity with respect to time:
$\alpha = \lim_{\Delta t \rightarrow 0} \frac{\Delta \omega}{\Delta t} = \frac{d \omega}{dt} = \frac{d^{2} \theta}{dt^{2}}, \label{10.6}$
where we have taken the limit of the average angular acceleration, $$\bar{\alpha} = \frac{\Delta \omega}{\Delta t}$$ as $$\Delta t → 0$$. The units of angular acceleration are (rad/s)/s, or rad/s2.
In the same way as we defined the vector associated with angular velocity $$\vec{\omega}$$, we can define $$\vec{\alpha}$$, the vector associated with angular acceleration (Figure $$\PageIndex{6}$$). If the angular velocity is along the positive z-axis, as in Figure $$\PageIndex{4}$$, and $$\frac{d \omega}{dt}$$ is positive, then the angular acceleration $$\vec{\alpha}$$ is positive and points along the +z- axis. Similarly, if the angular velocity $$\vec{\omega}$$ is along the positive z-axis and $$\frac{d \omega}{dt}$$ is negative, then the angular acceleration is negative and points along the +z-axis.
We can express the tangential acceleration vector as a cross product of the angular acceleration and the position vector. This expression can be found by taking the time derivative of $$\vec{v} = \vec{\omega} \times \vec{r}$$ and is left as an exercise:
$\vec{a} = \vec{\alpha} \times \vec{r} \ldotp \label{10.7}$
The vector relationships for the angular acceleration and tangential acceleration are shown in Figure $$\PageIndex{7}$$.
We can relate the tangential acceleration of a point on a rotating body at a distance from the axis of rotation in the same way that we related the tangential speed to the angular velocity. If we differentiate Equation \ref{10.4} with respect to time, noting that the radius r is constant, we obtain
$a_{t} = r \alpha \ldotp \label{10.8}$
Thus, the tangential acceleration at is the radius times the angular acceleration. Equations \ref{10.4} and \ref{10.8} are important for the discussion of rolling motion (see Angular Momentum).
Let’s apply these ideas to the analysis of a few simple fixed-axis rotation scenarios. Before doing so, we present a problem-solving strategy that can be applied to rotational kinematics: the description of rotational motion.
Problem-Solving Strategy: Rotational Kinematics
1. Examine the situation to determine that rotational kinematics (rotational motion) is involved.
2. Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
3. Make a complete list of what is given or can be inferred from the problem as stated (identify the knowns).
4. Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog, because by now you are familiar with the equations of translational motion.
5. Substitute the known values along with their units into the appropriate equation and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
Now let’s apply this problem-solving strategy to a few specific examples.
Example $$\PageIndex{2}$$: A Spinning Bicycle Wheel
A bicycle mechanic mounts a bicycle on the repair stand and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the average angular acceleration in rad/s2. (b) If she now hits the brakes, causing an angular acceleration of −87.3 rad/s2, how long does it take the wheel to stop?
Strategy
The average angular acceleration can be found directly from its definition $$\bar{\alpha} = \frac{\Delta \omega}{\Delta t}$$ because the final angular velocity and time are given. We see that $$\Delta \omega$$ = $$\omega_{final}$$ − $$\omega_{initial}$$ = 250 rev/min and $$\Delta$$t is 5.00 s. For part (b), we know the angular acceleration and the initial angular velocity. We can find the stopping time by using the definition of average angular acceleration and solving for $$\Delta$$t, yielding
$\Delta t = \frac{\Delta \omega}{\alpha} \ldotp$
Solution
1. Entering known information into the definition of angular acceleration, we get $$\bar{\alpha} = \frac{\Delta \omega}{\Delta t} = \frac{250\; rpm}{5.00\; s} \ldotp$$Because $$\Delta \omega$$ is in revolutions per minute (rpm) and we want the standard units of rad/s2 for angular acceleration, we need to convert from rpm to rad/s: $$\Delta \omega = 250 \frac{rev}{min}\; \cdotp \frac{2 \pi\; rad}{rev}\; \cdotp \frac{1\; min}{60\; s} = 26.2\; rad/s \ldotp$$Entering this quantity into the expression for $$\alpha$$, we get $$bar{\alpha} = \frac{\Delta \omega}{\Delta t} = \frac{26.2\; rpm}{5.00\; s} = 5.24\; rad/s^{2} \ldotp$$
2. Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that $$\Delta \omega$$ is −26.2 rad/s, and $$\alpha$$ is given to be –87.3 rad/s2. Thus $$\Delta t = \frac{-26.2\; rad/s}{-87.3\; rad/s^{2}} = 0.300\; s \ldotp$$
Significance
Note that the angular acceleration as the mechanic spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero.
Exercise $$\PageIndex{1}$$
The fan blades on a turbofan jet engine (shown below) accelerate from rest up to a rotation rate of 40.0 rev/s in 20 s. The increase in angular velocity of the fan is constant in time. (The GE90-110B1 turbofan engine mounted on a Boeing 777, as shown, is currently the largest turbofan engine in the world, capable of thrusts of 330–510 kN.) (a) What is the average angular acceleration? (b) What is the instantaneous angular acceleration at any time during the first 20 s?
Example $$\PageIndex{3}$$: Wind Turbine
A wind turbine (Figure $$\PageIndex{9}$$) in a wind farm is being shut down for maintenance. It takes 30 s for the turbine to go from its operating angular velocity to a complete stop in which the angular velocity function is $$\omega$$(t) = $$\Big[\frac{(ts^{−1} −30.0)^{2}}{100.0} \Big]$$rad/s. If the turbine is rotating counterclockwise looking into the page, (a) what are the directions of the angular velocity and acceleration vectors? (b) What is the average angular acceleration? (c) What is the instantaneous angular acceleration at t = 0.0, 15.0, 30.0 s?
Strategy
1. We are given the rotational sense of the turbine, which is counterclockwise in the plane of the page. Using the right hand rule (Figure 10.5), we can establish the directions of the angular velocity and acceleration vectors.
2. We calculate the initial and final angular velocities to get the average angular acceleration. We establish the sign of the angular acceleration from the results in (a).
3. We are given the functional form of the angular velocity, so we can find the functional form of the angular acceleration function by taking its derivative with respect to time.
Solution
1. Since the turbine is rotating counterclockwise, angular velocity $$\vec{\omega}$$ points out of the page. But since the angular velocity is decreasing, the angular acceleration $$\vec{\alpha}$$ points into the page, in the opposite sense to the angular velocity.
2. The initial angular velocity of the turbine, setting t = 0, is $$\omega$$ = 9.0 rad/s. The final angular velocity is zero, so the average angular acceleration is $$\bar{\alpha} \frac{\Delta \omega}{\Delta t} = \frac{\omega - \omega_{0}}{t - t_{0}} = \frac{0 - 9.0\; rad/s}{30.0 - 0\; s} = -0.3\; rad/s^{2} \ldotp$$
3. Taking the derivative of the angular velocity with respect to time gives $$\alpha = \frac{d \omega}{dt} = \frac{(t − 30.0)}{50.0}$$ rad/s2 $$\alpha (0.0; s) = -0.6\; rad/s^{2}, \alpha (15.0\; s) = -0.3\; rad/s^{2}, and\; \alpha (30.0\; s) = 0\; rad/s \ldotp$$
Significance
We found from the calculations in (a) and (b) that the angular acceleration α and the average angular acceleration $$\bar{\alpha}$$ are negative. The turbine has an angular acceleration in the opposite sense to its angular velocity.
We now have a basic vocabulary for discussing fixed-axis rotational kinematics and relationships between rotational variables. We discuss more definitions and connections in the next section.<|endoftext|>
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In the mid-1960s, before any Apollo hardware had flown with a crew, NASA was looking ahead and planning its next major programs. It was a bit of a challenge. After all, how do you top landing a man on the Moon? Not wanting to start from scratch, NASA focused on possible missions that would use the hardware and software developed for the Apollo program. One mission that fit within these parameters was a manned flyby of our cosmic twin, Venus.
As one of our neighbouring planets, a mission to Venus made sense; along with Mars, it’s the easiest planet to reach. Venus was also a mystery at the time. In 1962, the Mariner 2 spacecraft became the first interplanetary probe. It flew by Venus, gathered data on its temperature and atmospheric composition before flying off into a large heliocentric orbit. But there was more to learn, making it a destination worth visiting.
But beyond being relatively practical with great potential for scientific return, a manned mission to Venus would prove that NASA’s spacecraft and astronauts were up for the challenges of long-duration interplanetary flight. In short, it would give NASA something exciting to do.
The mission proposal was published early in 1967. It enhanced the Apollo spacecraft with additional modules, then took the basic outline of an Apollo mission and aimed it towards Venus instead of the Moon.
The crew would launch on a Saturn V rocket in November of 1973, a year of minimal solar activity. They would reach orbit in the same Command and Service Modules (CSM) that took Apollo to the Moon. Like on Apollo, the CSM would provide the main navigation and control for the mission.
Going to the Moon, Apollo missions had the crew turn around in the CSM to pull the LM out of its launch casing. On the mission to Venus, the crew would do the same, only instead of an LM they would dock and extract the Environmental Service Module (ESM). This larger module would supply long-duration life support and environmental control and serve as the main experiment bay.
With these two pieces mated, the upper S-IVB stage of Saturn V would propel the spacecraft towards Venus. Once its fuel store was spent, the crew would repurpose the S-IVB into an additional habitable module. Using supplies stored in the ESM, they would turn the rocket stage into their primary living and recreational space. On its outside, an array of solar panels would power each piece of the spacecraft throughout the mission.
The crew would spend 123 days traveling to Venus. Ten hours of each day would be dedicated to science, mainly observations of the solar system and beyond with a telescope mounted in the ESM. UV, X-ray, and infrared measurements could create a more complete picture of our corner of the universe. The rest of each day would be spent sleeping, eating, exercising, and relaxing — a full two hours of every day would be dedicated to unstructured leisure, a first for astronauts.
Like Mariner 2 before them, the crew would flyby Venus rather than go into orbit. They would only have 45 minutes to do close optical observations and deploy probes that would send back data on the Venusian atmosphere in realtime.
After the flyby, the spacecraft would swing around Venus and start its 273 day trip back to Earth. Like on an Apollo lunar mission, the crew would transfer back into the Command Module before reentry taking anything that had to return to Earth with them. They would jettison the S-IVB, the ESM, and the Service Module, switch the CM to battery power, and plunge through the atmosphere. Around December 1, 1974, they would splashdown somewhere in the Pacific Ocean.
Though worked out in great detail, the proposal was a thought experiment rather than something NASA was seriously considering. Nevertheless, Apollo-era technology would have managed the mission.
Source: NASA Manned Venus Flyby Study<|endoftext|>
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Factors of 8 space the list of integers that deserve to be evenly split into 8. It has a total of 4 components of i m sorry 8 is the best factor and also the positive components of 8 room 1, 2, 4, and 8. The Pair components of 8 space (1, 8) and also (2, 4) and also its Prime factors are 1, 2, 4, 8.
You are watching: What are the factors of eight
Factors the 8: 1, 2, 4 and 8Negative determinants of 8: -1, -2, -4 and also -8Prime factors of 8: 2Prime factorization of 8: 2 × 2 × 2 = 23Sum of components of 8: 15
1 What are determinants of 8? 2 How come Calculate components of 8? 3 Factors of 8 By department Method 4 Factors that 8 By prime Factorization 5 Thinking the end of The Box! 6 Factors of 8 in Pairs 7 Important Notes 8 FAQs on components of 8
## What are components of 8?
Factors of 8 are those numbers that totally divide the leaving 0 as the remainder. For example, 8 ÷ 1 leaves a remainder 0.
We gain 8 together the quotient and 0 as the remainder. This suggests that 8 is divisible by 1. Similarly,
8 ÷ 2 = 48 ÷ 4 = 28 ÷ 8 = 1
Thus, the factors that 8 room 1, 2, 4, and 8.
## How to calculation the components of 8?
We can use different methods to find the factors of 8 such together the prime factorization technique and the division method. In prime factorization, we express 8 together the product that its element factors and also in the department method, we look because that numbers that division 8 specifically without a remainder.
## Factors the 8 By division Method
We understand that 8 is a composite number and it will have prime factors. Let united state calculate the prime determinants of the number 8. The an initial step is to division 8 through the smallest prime factor, i m sorry is 2, adhered to by the others.
8 ÷ 2 = 48 ÷ 4 = 28 ÷ 8 = 1
The division here shows that the number 8 is exactly divisible by 1, 2, 4, and also 8. Hence, the complete factors of 8 by department method room 1, 2, 4, and 8.
## Factors the 8 By prime Factorization
The components of 8 have the right to be stood for by the upside-down department method as:
Method 1;
The components of 8 have the right to be represented by the aspect tree an approach as:
Method 2:
The components of 8 by the prime factorization method are 1, 2, 4, and also 8. Here, 2 is the prime factor of 8. Explore factors using illustrations and interactive examples:
Think Tank:
Can you think of all the feasible pair components of 8?Can multiples that 8 likewise be the determinants of 8?
## Factors of 8 in Pairs
To find the determinants of 8 in pairs, main point the 2 numbers in a pair to get the resultant number together 8.
Since 1 × 8 = 8, (1, 8) is a pair aspect of 8. Similarly,2 × 4 = 8, thus, (2, 4) is a pair factor of 8.Therefore, (8, 1) and (2, 4) room the components of 8 in pairs.Factors that 8 deserve to be stood for as negative pair factors as well. Because that example, -8 × -1 = 8, thus, (-8, -1) is a an adverse pair element of 8.
Important Notes:
The determinants of 8 are 1, 2, 4, and also 8.1 is a universal factor due to the fact that it is a aspect of all numbers.Factors space quite frequently given together pairs of numbers which as soon as multiplied together provide the original number.The components of 8 in pairs room (2, 4) and (1, 8).2 is the prime factor the 8.
Example 1: Kevin to be asked to uncover the determinants of 8 and 18. What will be his answer?
Solution:
Factors the 8 space 1, 2, 4, and also 8 and the components of 18 space 1, 2, 3, 6, 9, and also 18.
Example 2: The following numbers were provided to Mike and also he was asked to find the number that was not a variable of 8. Help Mike find that number.
1, 2, 3, 4, 5, 6, 7, and 8.
Solution:
Factors of 8 room 1, 2, 4, and also 8. As soon as we division 8 by 3, 5, 6, and also 7, it leaves a reminder.
Thus, this numbers space not the determinants of 8 which way 3, 5, 6, and 7 room not the factors of 8.
Example 3: What space the negative pair components of 8?
Solution:
To find the negative pair factors, us take the complying with steps:-2 × -4 = 8
-8 × -1 = 8
Therefore, the negative pair determinants of 8 space (-2, -4) and (-8, -1).
## FAQs on components of 8
### What space the factors of 8?
The factors of 8 room 1, 2, 4, 8 and also its negative factors room -1, -2, -4, -8.
### How plenty of Factors that 8 are also common to the determinants of 4?
Since, the components of 8 room 1, 2, 4, 8 and the components of 4 space 1, 2, 4.Hence, <1, 2, 4> space the common factors the 8 and also 4.
See more: The Populist Party Lost Power In Large Part As A Result Of, Populist Party
### What is the sum of the determinants of 8?
All the factors of 8 are 1, 2, 4, 8 and therefore the amount of all these factors is 1 + 2 + 4 + 8 = 15
### What is the Greatest typical Factor that 8 and 5?
The factors of 8 are 1, 2, 4, 8 and also the determinants of 5 room 1, 5. 8 and 5 have actually only one typical factor i m sorry is 1. This implies that 8 and also 5 are co-prime.Hence, the Greatest usual Factor (GCF) that 8 and also 5 is 1.<|endoftext|>
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A timeline of significant discoveries
First Discoveries - 1840s to 1850s
The first discoveries of ancient human fossils. Neanderthals were the first ancient humans to gain scientific and popular recognition. Their fossils began to be found in Europe in the 1800s but scientists had no perspective or evolutionary framework by which to explain them. Decades passed before they were recognised as being a different and extinct form of ancient human.
Opinions about the relationship between our own species and Neanderthals have continually changed. The early 1900s saw them as sub-humans, a stereotype that didn’t change until the 1950s when it was widely considered that they may be the ancestors of modern Europeans. New research in the 1980s led many to move them to a side branch of our family tree, a decision supported by the comparisons of the mitochondrial DNA of modern humans and Neanderthals in the 1990s and 2000s.
1840s to 1850s
Gibraltar skull - Homo neanderthalensis. Discovered in 1848 in Forbe’s Quarry, Rock of Gibraltar
Neanderthal 1 skullcap - type specimen for Homo neanderthalensis. Discovered in 1856 in the Neander Valley, Germany. Although originally presented as an inferior human that inhabited Europe before modern people, some felt that the differences between Neanderthals and modern humans were due to pathology and disease. This marked the effective beginning of Palaeoanthropology as a science.
1860s to 1890s
The remains of early modern humans found at Cro-Magnon in France.
These remains were the first firm evidence of the antiquity of our species, Homo sapiens.
Neanderthal skulls found in Spy, Belgium.
These finds supported the idea that Neanderthals were an ancient and distinct type of human, but exactly how they fit into our family tree was still debated.
Homo erectus skullcap found in Trinil, Java by Eugene Dubois, and described as a new species in 1894.
This specimen was originally named Pithecanthropus erectus as it was considered different enough from humans to be placed into a new genus. It was renamed Homo erectus in the 1940s, a species name that, in the opinion of most researchers, includes specimens from Java and China.
1900s to 1920s
Homo heidelbergensis named as a new species after the discovery of a jaw in Mauer, Germany in 1907.
This species was originally dismissed as being too apelike for human ancestry, particularly after the Piltdown Man find. In the 1960s it was grouped with other similar skulls and called archaic Homo sapiens, but today the preference is to use the original scientific name and to give this species an ancestral position on the human family tree.
Piltdown Man ‘discovered’ in England.
In a gravel-pit at Piltdown Common, Southern England, in 1912, amateur collector Charles Dawson ‘discovered’ what appeared to be the long-sought ‘missing link’ between apes and humans. This fortuitous find – nine pieces of a large-brained human skull and an ape-like lower jaw with two teeth – was readily accepted by the British establishment due to their belief that a large brain was one of the first human features to evolve. Although inconsistent with later discoveries, ‘Piltdown Man’s’ authenticity remained virtually unchallenged for 41 years.
In 1953, advanced analytical and dating techniques proved Piltdown Man to be a fake. The mandible was stained with potassium bichromate and the teeth had been filed down. Fluorine testing proved that the pieces of the skull were of different ages. This was confirmed in 1959 by carbon dating, which provided a date of about 600 years for the skull!
Discovery of the ‘Taung skull’ in South Africa, classified as Australopithecus africanus in 1925.
This fossil was clearly more ancient than earlier finds and anatomist Raymond Dart, who first analysed it, claimed it was a human ancestor. He was criticised very strongly by English scientists who believed in ‘Piltdown Man’. It was not until the 1950s that this species was fully acknowledged as belonging on the human family tree.
The first remains of Homo erectus found in China.
Chinese specimens were discovered in a cave at Zhoukoudian near Peking and named Sinanthropus pekinensis (‘Chinese man of Peking’). In the 1950s these specimens, and the ones from Java, were placed in the one species, Homo erectus.
1930s to 1960s
Paranthropus robustus remains first discovered in South Africa.
Many of these fossils were originally given different names, which led to a confusing family tree. In the 1950s they were reclassified into the one species, Paranthropus robustus.
Remains of Paranthropus boisei found by the Leakeys in Olduvai Gorge.
This specimen OH5 (originally called Zinjanthropus boisei) was the first of our fossil ancestors to be given an accurate date with the new technique of radiometric dating. The date of 1.8 million years, obtained by potassium-argon dating, was announced in 1961.
Homo habilis announced as a new species after remains were discovered in Olduvai Gorge in 1960.
Although controversial, this fossil was the most primitive human to be classified into our genus. The family tree was still in a linear shape, with this species at the base
1970s to 1980s
Specimen KNM-ER 1470 found in East Turkana
Originally thought to be Homo habilis, this specimen was reclassified as Homo rudolfensis in 1986 and made the type specimen of that species.
This specimen and another attributed to Homo habilis, KNM-ER 1813, are at the centre of the debate regarding which remains are attributed to that species. Homo habilis is a well-known but poorly defined species and scientific opinions about the attributed specimens vary widely. Scientists often disagree about naming fossil specimens as scientific names may be changed following new discoveries or there are different interpretations or new lines of investigation.
Discovery of ‘Lucy’ in Hadar, Ethiopia.
In 1978 this specimen, and a number of other remains from Laetoli, Tanzania, and Hadar, Ethiopia, was classified as a new species, Australopithecus afarensis.
Specimen KNM-ER 3733 found in East Africa
This specimen was initially considered to be an African Homo erectus, but many now classify it as Homo ergaster.
This fossil was found in the same layer as a specimen of Paranthropus boisei, finally disproving the ‘single species’ theory that was popular at this time. Followers of this theory believed that only one hominin species could occupy an area at the one time and that the family tree was a single, evolving line moving through phases to modern humans. This find proved that the human family tree was more like a branching bush and that evolution was not linear.
Discovery in Kenya of an almost complete Homo ergasterskeleton, nicknamed the ‘Turkana Boy’.
1990s to 2000s
Sahelanthropus tchadensis announced as a new species. It was discovered in Chad in 2001 and dates to about 6-7 million years old. Scientists consider the find to be of major significance but debate its relationship to humans. It is very possible that it comes before the split between the human and chimp line, making it an ancestor of both branches.
Homo floresiensis, a type of dwarf human discovered on the Indonesian island of Flores, is announced as a new species. Fossils date from about 12,000 to 100,000 years old, making this a contemporary of modern humans. This species was not ancestral to modern humans and may be a descendant of Asian Homo erectus or a yet to be discovered Homo species.
Publication of study of Ardipithicus ramidus skeleton, first found in 1994. The skeleton provides the first substantial fossil evidence about the appearance of the last human-chimp common ancestor and confirms that living African apes do not much resemble this ancestor, as was commonly thought.
Publication of a new species Australopithecus sediba. The first specimen was discovered in 2008 at Malapa in South Africa. Remains date to about 1.8 million years old. There is much debate about these remains and the species designation, with many considering them to be late A. africanus rather than a new species.
Release of nuclear DNA analysis carried out on a finger bone and tooth from Denisova cave, Russia (found in 2008) reveals the remains come from a species that is neither Homo sapiens or Homo neanderthalensis. This suggests a fourth human species (as Homo floresiensis was extant in Flores) was still in existence between 48,000 and 30,000 years ago.<|endoftext|>
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# How do you simplify (4x+2)/(5x) + (10x-1)/(20x) + (2x-3)/(4x)?
Oct 18, 2015
$\frac{9 x - 2}{5 x}$
#### Explanation:
Your goal here is to find the common denominator for all those fractions and multiply each one so that you can get them to have the same denominator.
This will in turn allow you to add the resulting numerators and simplify the expression.
$5 \cdot x \text{ }$, $\text{ "20 * x" }$, and $\text{ } 4 \cdot x$
Notice that you can rewrite $20 x$ as
$20 x = 5 \cdot 4 \cdot x$
This means that you can use this as the common denominator for the three fractions, and multiply the first one by $1 = \frac{4}{4}$ and the second one by $1 = \frac{5}{5}$ to get them to have the denominator $20 x$.
The expression can thus be written as
$\frac{4 x + 2}{5 x} \cdot \frac{4}{4} + \frac{10 x - 1}{20 x} + \frac{2 x - 3}{4 x} \cdot \frac{5}{5}$
$\frac{\left(4 x + 2\right) \cdot 4}{20 x} + \frac{10 x - 1}{20 x} + \frac{\left(2 x - 3\right) \cdot 5}{20 x}$
Now that the fractions have the same denominator, add the numerators like you normally would
$\left(4 x + 2\right) \cdot 4 + 10 x - 1 + \left(2 x - 3\right) \cdot 5$
$16 x + 8 + 10 x - 1 + 10 x - 15$
$36 x - 8 = 4 \cdot \left(9 x - 2\right)$
The expression can be simplified to
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \cdot \left(9 x - 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \cdot 5 x} = \textcolor{g r e e n}{\frac{9 x - 2}{5 x}}$<|endoftext|>
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Chronic bronchitis and emphysema are together called chronic obstructive pulmonary disease, or COPD. This is a chronic condition, usually attributed to tobacco smoking, causing shortness of breath and cough, leading to limitation of everyday activities like walking.
In chronic bronchitis, the airways (or bronchi) that connect the windpipe and the lungs become inflamed and swollen. The airways become narrow and are clogged up with thick mucus, called phlegm. Chronic bronchitis may be found together with emphysema, in which the air sacs in the lung become damaged, reducing the surface area where oxygen exchange takes place. Both diseases make it difficult to breathe.
COPD is a very common condition, especially among people who smoke tobacco. It is estimated 2.6 million Canadians have COPD, although many are undiagnosed. COPD is the fourth leading cause of death in Canada.
Smoking is the main cause of COPD. It causes the airways to produce excess mucus that lines the walls of the airways, making the air passages very narrow. This makes it easier to get a bronchial infection. An infection can cause even more damage to the airways by causing more mucus production. Cigarette smoking also destroys the air sacs where oxygen moves from your lungs to our blood, making the lungs work less efficiently.
Less common causes of COPD include a rare genetic disorder called alpha-1 antitrypsin deficiency, air pollution, exposure to occupational dusts and chemicals, and frequent lower respiratory infections during childhood.
Symptoms and Complications
People with chronic bronchitis may cough up phlegm almost every day.
It is common for someone with chronic bronchitis to persistently cough and wheeze when breathing. It is also common to feel short of breath and tired. Low oxygen in the blood due to the decreased ability to move oxygen from the air to the blood may cause the lips or fingernails to become bluish in colour.
COPD can lead to heart failure, as the heart has to work harder to pump blood into the lungs. When the heart fails to pump blood properly, it collects in the blood vessels of the legs and ankles and causes them to swell - this is called edema.
Sometimes you may become housebound because of breathing difficulties. You may run out of breath even when doing simple tasks such as getting dressed or washing.
If you have COPD, it is likely that you may occasionally get infections, resulting in increased shortness of breath and more phlegm production or a change in the colour of the phlegm. You may also occasionally cough up blood. These can be signs of a more serious problem and it's important to see your doctor.
Making the Diagnosis
Your doctor will test to see how much air you can forcefully exhale. These pulmonary, or lung breathing, tests are simple and painless. If you exhale less than normal, it could mean your airways are inflamed, in spasm, or clogged up with mucus. If this persists, then you may have COPD.
Treatment and Prevention
Even with treatment, COPD becomes progressively worse over time. Medication is able to reduce the symptoms and improve your qualify of life, but cannot cure COPD. Lung function deteriorates with age even in healthy people, but it happens much faster if you're a smoker. Therefore, it is very important to stop smoking.
Your doctor may prescribe medications called short-acting bronchodilators, including salbutamol*, ipratropium bromide, a combination of the two, or terbutaline to relax and widen the bronchi and help relieve shortness of breath.
If symptoms are persistent, treatment with long-acting bronchodilators such as tiotropium, glycopyrronium, salmeterol, formoterol, or a combination product can be added. If you are still having difficulties breathing, your doctor may also suggest that you try adding an inhaled corticosteroid.
There are also medications available that combine long-acting bronchodilators with inhaled corticosteroids. Your doctor may also prescribe antibiotics and oral steroids for you to keep at home in case a bacterial lung infection develops.
Since influenza (the flu) may make COPD symptoms worse and can lead to respiratory failure and hospitalization, it is recommended that people with COPD receive the annual flu vaccine. People with COPD may also benefit from receiving a pneumococcal vaccine to lower their risk of getting pneumonia (lung infection), which can also lead to complications. Talk to your doctor about receiving these vaccines.
Oxygen therapy from oxygen cylinders or an oxygen concentrator can be used for people with severe COPD who do not have enough oxygen in their blood when breathing room air.
Drinking plenty of fluids throughout the day can help loosen phlegm buildup.
Exercise with or without a formal physiotherapy program can improve a person's quality of life and activities. A healthy nutritional intake is important, as weight loss due to the increased work of breathing presents a serious sign of advancing COPD. Lung volume reduction surgery or lung transplantation can also be considered in extreme cases.
*All medications have both common (generic) and brand names. The brand name is what a specific manufacturer calls the product (e.g., Tylenol®). The common name is the medical name for the medication (e.g., acetaminophen). A medication may have many brand names, but only one common name. This article lists medications by their common names. For information on a given medication, check our Drug Information database. For more information on brand names, speak with your doctor or pharmacist.
All material copyright MediResource Inc. 1996 – 2019. Terms and conditions of use. The contents herein are for informational purposes only. Always seek the advice of your physician or other qualified health provider with any questions you may have regarding a medical condition. Source: www.medbroadcast.com/condition/getcondition/COPD-Chronic-Obstructive-Pulmonary-Disease<|endoftext|>
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As you gaze at the first-quarter moon this week, you may wonder when the last-quarter moon will occur this month. But there won't be one, if you live in North or South America.
Take the situation in eastern North America, which is in the Eastern Standard Time (EST) zone. The previous last-quarter moon was on Jan. 31 at 10:28 p.m. EST, and the next one will be on March 1 at 6:11 p.m.
The mathematics behind this is that the average synodic lunar month — from new moon to new moon — is 29.53 days long, while February is either 28 or 29 days long. So it is possible, even in a leap year like 2016, to have one of the four main lunar phases fall outside the calendar month of February. [Earth's Moon Phases, Monthly Lunar Cycles (Infographic)]
Other parts of the world, such as Europe, had a last-quarter moon this month early on the morning of Feb. 1.
We make a big fuss about the "Blue Moon," when there are two full moons in a month, but we don’t seem to notice when one of the lunar phases goes missing.
This raises the question of why our months vary so much in their number of days: 28 or 29 in February, 30 in April, June, September and November, and 31 in the other seven months. The problem is that the sun, moon and Earth don’t move to the tune of simple arithmetic.
The lunar month consists of 29.530589 days, and the tropical year (equinox to equinox) is 365.242190 days long. When the ancient astronomers attempted to construct a calendar with these bizarre numbers, they found that, literally, it did not compute.
Early astronomers divided the shape of a circle into 360 degrees. This seems like a strange number to us with our decimal system, but it made sense with a number system based on 12. It also came close to the number of days in a year, though not close enough. The year was divided into 12 months (a natural in a base-12 number system) of 30 days each — but that left the awkward 5-and-a-bit-days remaining.
Mathematicians struggled with this problem for thousands of years, until finally a papal commission in 1582 came up with a complex but elegant solution, known as the Gregorian calendar, after Pope Gregory XIII, who commissioned it.
In order to get the church’s feast days back in phase with the astronomical calendar, it was necessary to omit 11 days.
The Pope had the power to enforce the new calendar in Catholic countries, though there was a bit of grumbling about the 11 days, which went missing between Oct. 4 and Oct. 15, 1582. Just for fun, try entering Oct. 4, 1582, in a planetarium program like Starry Night, and then advance to the next day. You will find it is Oct. 15.
Naturally, England (along with its North American colonies) was one of the strongholds of the old Julian calendar, and resisted adopting the popish Gregorian calendar until 1752. By that time, the calendar used by the English was off by 12 days, so that Sept. 2, 1752, was followed by Sept. 14, 1752, in England and its colonies.
To avoid the missing-days problem, we now have a system of 30- and 31-day months, with poor February being stuck with making the whole thing fit. Thus, we have 29 days in February every four years, with a few exceptions to fine-tune the length of the year over the centuries. The result is that most months are a little longer than the lunar month, so they sometimes have two full moons (or other double phases). And February sometimes ends up missing a phase, as happens this year.
Editor's note: If you capture a stunning view of the moon or any other celestial sight and would like to share it with us and our news partners, you can send in images and comments to managing editor Tariq Malik at: [email protected].
This article was provided to Space.com by Simulation Curriculum, the leader in space science curriculum solutions and the makers of Starry Night and SkySafari. Follow Starry Night on Twitter @StarryNightEdu. Follow us @Spacedotcom, Facebook and Google+. Original article on Space.com.<|endoftext|>
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The State of the Nation
Why we need to act
The achievement gap between poorer children and their more affluent peers begins long before they start primary school and widens throughout their education. Children from low income communities often fall behind in literacy and numeracy at primary school, achieve fewer good GCSEs at secondary school and are less likely to attend and graduate from a top university. All of this has a negative impact on their future health, happiness and career prospects.
Primary school attainment:
Impact Goal One: Narrow the gap in literacy and numeracy at primary school
The goal: The Fair Education Alliance is committed to closing the attainment gap between primary schools serving lower income pupils and those serving higher income pupils. The goal is for this gap in reading, writing and maths to be narrowed by 90% by 2022.
The gap: According to Alliance measures, this gap is 1.76 points. Although this gap has decreased from 1.94 points in 2010/11 or by 9.1%, it increased slightly by 0.02 points during 2013/14.
Secondary school attainment:
Impact Goal Two: Narrow the gap in GCSE attainment at secondary school
The goal: The Fair Education Alliance is committed to closing the attainment gap between secondary schools serving lower income pupils and those serving higher income pupils. Our goal is to close this gap by 44% by 2022.
The gap: According to Alliance measures, this gap is 79 points . In 2013/14, this gap closed by 17.1%. However, these figures should be interpreted with caution. The narrowing is very likely to be due to changes in assessment methods, hindering accurate comparisons.
Character, emotional Wellbeing and mental health
Impact Goal Three: Ensure young people develop key strengths, including character, emotional wellbeing and mental health, to support high aspirations
The goal: The Fair Education Alliance recognises that this goal underpins all of the impact goals and is committed to ensuring young people develop the character, wellbeing and mental health they need to succeed in life. The Alliance is working with other organisations to develop measurement tools which will allow us to understand this area more.
The gap: Over the year, our knowledge and understanding of the area has evolved, moving us closer to the development of measuring tools. Uniting behind a set of these tools will be a strategic priority for the Alliance over the next two years. In the interim, although we cannot quantify change in young people’s development of the character, wellbeing and mental health needed to succeed in life, we have a better understanding of the area through developments in policy and research. Last year, permanent and fixed period exclusions were identified as quantitative measures to aid our understanding of the national picture in this area. Across this measure, the gap has increased over the year; children and young people from poor families were more likely than last year to receive a fixed period exclusion or to be permanently excluded when compared to their more affluent peers.
Impact Goal Four: Narrow the gap in the proportion of young people taking part in further education or employment-based training after finishing their GCSEs
The goal: The Fair Education Alliance wants to see an increase in the number of young people from schools serving low income communities who stay in further education or employment-based training once they have completed Key Stage 4 (KS4). Our goal is for 90% of young people from schools serving low income communities to be in post-16 education or employment-based training by 2022; currently this figure is 84.9%. In light of changes to the participation age, the Alliance will focus on an increase in the number of
young people from low income communities who stay in further education or employment-based training once their post-16 education has ended.
The gap: According to Alliance measures for 2012/13, the gap between those from schools serving low and high income communities staying in education after KS4 has remained constant at 7 percentage points.
Impact Goal Five: Narrow the gap in university graduation, including from the 25% most selective universities
The goal: The Fair Education Alliance is committed to closing the graduation gap between young people from low income backgrounds and those from high income backgrounds. Our goal is for at least 5,000 more students from low income backgrounds to graduate each year, with 1,600 of these graduating from the most selective universities.
The gap: According to Alliance measures, the gap of 17 percentage points (2012/13) between the proportion of students from low income and high income families going on to university has closed by one percentage point.
Other trends: This gap is also closing for young people from poor families who were more likely to enter a medium or low tariff university than they were in 2014; the gaps between these young people and their more advantaged peers have closed by 0.2 percentage points and 0.1 percentage points respectively.
This gap, however, is not closing for young people from poor families accessing selective universities. In 2015, the gap between young people from poor families and their more advantaged peers entering a high tariff university increased by 0.1 percentage point. Entry to any university is variable across England; whilst some regions do very well in helping young people from poor families go to university, others underperform compared to the rest of the country.
In 2015, the gap between young people from poor families and their more advantaged peers entering university in the North East and the East Midlands increased by 1 percentage point; this gap for Yorkshire and the Humber remained unchanged. In 2014, the proportion of young people from advantaged families who went to a Russell Group university increased by one percentage point, whilst that of their less advantaged peers remained the same.<|endoftext|>
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# Proof by induction
## Introduction
In FP1 you are introduced to the idea of proving mathematical statements by using induction.
Proving a statement by induction follows this logical structure
1. If the statement is true for some $n=k$, it is also true for $n=k+1$.
2. The statement is true for $n=1$.
3. Therefore it is true for 1, 2, 3, 4, 5, ... and for all the natural numbers $n$.
When answering questions on proof by induction you actually work in a different order. You first have to do some rough work - the three steps I've put above are what you need to put at the end of your answer.
Do your rough work in this order
1. Firstly show that the statement holds for $n=1$. Just sub in $n=1$ to whatever you're asked to prove.
2. Then sub in $n=k$ and write down the statement in that form.
3. Then let $n=k+1$ and, using the $n=k$ formula you've written in the above step, prove it is also true.
Then you write the proof bit of your answer at the end. In FP1 they are really strict on how you word your answers to proof by induction questions. This is to get you used to the idea of a rigorous proof that holds water.
Don't worry though. In the numerous examples below I will write down exactly what you need to write to get the full marks for this type of question.
Students often get confused about the logic of induction. The way to think of it is
"Suppose this statement were true for some number $k$. Then it has to also be true for $k+1$."
We're not just saying "it's true", we're saying "suppose it were true for some number $k$". That way if we do find a number that it's true for, we're done.
You need to know the induction proofs for various cases. So in each of the following sections I will just write down the types of proofs you need to know for the exam.
You don't need to memorise each individual step, in fact I recommend you don't do that. Instead get a feel for the method, practice some questions of your own, and you will be able to do these yourself using simple algebra.
## Series
### The sum of the first $n$ natural numbers
Q) Prove that $\sum_{r=1}^{n} r= \frac{n(n+1)}{2}$ by induction.
A) First show that the formula holds for $n=1$ $$\sum_{r=1}^{1} r = 1 = \frac{1(1+1)}{2} = \frac{2}{2} = 1$$ Suppose the formula holds for some $n=k$ $$\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$$ Then let $n=k+1$ \begin{align} \sum_{r=1}^{k+1} r &= \sum_{r=1}^{k}r + (k+1) \\ &= \frac{k(k+1)}{2} + (k+1) \\ &= \frac{k(k+1)+ 2(k+1)}{2} \\ &= \frac{(k+1)(k+2)}{2} \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
### The sum of the squares of the first $n$ natural numbers
Q) Prove that $\sum_{r=1}^{n} r^{2} = \frac{1}{6}n(n+1)(2n+1)$ by induction.
A) First show that the formula holds for $n=1$ $$\sum_{r=1}^{1} r^{2} = 1 = \frac{1}{6}\times 1\times(1+1)(2\times 1+1) = \frac{1}{6}\times 6 = 1$$ Suppose the formula holds for some $n=k$ $$\sum_{r=1}^{k} r^{2} = \frac{1}{6}k(k+1)(2k+1)$$ Then let $n=k+1$ \begin{align} \sum_{r=1}^{k+1} r^{2} &= \sum_{r=1}^{k}r^{2} + (k+1)^{2} \\ &= \frac{1}{6}k(k+1)(2k+1) + (k+1)^{2} \\ &= \frac{1}{6}(k+1)(k(2k+1)+6(k+1)) \\ &= \frac{1}{6}(k+1)(2k^{2}+7k+6) \\ &= \frac{1}{6}(k+1)(k+2)(2k+3) \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
### The sum of the cubes of the first $n$ natural numbers
Q) Prove that $\sum_{r=1}^{n} r^{3} = \frac{1}{4}n^{2}(n+1)^{2}$ by induction.
A) First show that the formula holds for $n=1$ $$\sum_{r=1}^{1} r^{3} = 1 = \frac{1}{4}1^{2}(1+1)^{2} = \frac{4}{4} = 1$$ Suppose the formula holds for some $n=k$ $$\sum_{r=1}^{k} r^{3} = \frac{1}{4}k^{2}(k+1)^{2}$$ Then let $n=k+1$ \begin{align} \sum_{r=1}^{k+1} r^{3} &= \sum_{r=1}^{k} r^{3} + (k+1)^{3} \\ &= \frac{1}{4}k^{2}(k+1)^{2} + (k+1)^{3} \\ &= \frac{1}{4}(k+1)^{2}\left( k^{2}+4(k+1)\right) \\ &= \frac{1}{4}(k+1)^{2}\left(k^{2} +4k+4 \right) \\ &= \frac{1}{4}(k+1)^{2}(k+2)^{2} \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
## Divisibility
Q) Prove that $3^{2n}-1$ is divisible by 8 for all natural numbers $n$.
A) First show that the formula holds for $n=1$ $$3^{2}-1 = 8$$ Which is obviously divisible by 8. Suppose the formula holds for some $n=k$ $$f(k) = 3^{2k}-1 \textrm{ is divisible by 8}$$ Then let $n=k+1$ and subtract $f(k)$ \begin{align} f(k+1)-f(k) &= 3^{2(k+1)} - 1 - (3^{2k}-1) \\ &= 3^{2(k+1)}- 3^{2k} \\ &= 3^{2}3^{2k} - 3^{2k} \\ &= (9-1)3^{2k} \\ &= 8(3^{2k}) \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
Q) Prove that $n^{2}-3n$ is divisible by 2 for all natural numbers $n$.
A) First show that the formula holds for $n=1$ $$1^{2}-3 = -2 = 2(-1)$$ Suppose the formula holds for some $n=k$ $$f(k) = k^{2}-3k \textrm{ is divisible by 2}$$ Then let $n=k+1$ and subtract $f(k)$ \begin{align} f(k+1)-f(k) &= (k+1)^{2}-3(k+1)-k^{2}+3k \\ &= k^{2}+2k+1-3k-3-k^{2}+3k \\ &= 2k+1-3 \\ &= 2(k-1) \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
### Recurrence relations
Q) With $u_{1}=6$ and $u_{n+1}=u_{n}+2^{n}+4$, prove that $u_{n}=2^{n}+4n$ for all natural numbers $n$.
A) First show that the formula holds for $n=1$ $$u_{1} = 2^{1}+4 = 6$$ Suppose the formula holds for some $n=k$ $$u_{k}=2^{k}+4k$$ Then going back to the formula for $u_{n+1}$ in the question \begin{align} u_{k+1}&=u_{k}+2^{k}+4 \\ &= 2^{k}+4k + 2^{k}+4 \\ &= 2(2^{k}) + 4(k+1) \\ &= 2^{k+1} +4(k+1) \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
### Matrices
Q) Let $\mathbf{A} = \left( \begin{array}{cc} 1 & 0 \\ 5 & 1 \end{array}\right)$. Show that $\mathbf{A}^{n} = \left(\begin{array}{cc} 1 & 0 \\ 5n & 1 \end{array}\right)$ for all natural numbers $n$.
A) First show that the formula holds for $n=1$ $$\mathbf{A}^{1} = \left(\begin{array}{cc} 1 & 0 \\ 5 & 1 \end{array}\right)$$ Suppose the formula holds for some $n=k$ $$\mathbf{A}^{k} = \left(\begin{array}{cc} 1 & 0 \\ 5k & 1 \end{array}\right)$$ Now $\mathbf{A}^{k+1} = \mathbf{A}^{k}\mathbf{A}$, so \begin{align} \mathbf{A}^{k+1} &= \left(\begin{array}{cc} 1 & 0 \\ 5k & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ 5 & 1 \end{array}\right) \\ &= \left(\begin{array}{cc} 1 & 0 \\ 5k+5 & 1 \end{array}\right) \\ &= \left(\begin{array}{cc} 1 & 0 \\ 5(k+1) & 1 \end{array}\right) \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.
Q) (Hard) Let $\mathbf{A} = \left( \begin{array}{cc} -1 & 0 \\ 2 & 1 \end{array}\right)$. Show that $\mathbf{A}^{n} = \left(\begin{array}{cc} (-1)^{n} & 0 \\ 1-(-1)^{n} & 1 \end{array}\right)$ for all natural numbers $n$.
A) First show that the formula holds for $n=1$ $$\mathbf{A}^{1} = \left(\begin{array}{cc} -1 & 0 \\ 1-(-1) & 1 \end{array}\right) = \left( \begin{array}{cc} -1 & 0 \\ 2 & 1 \end{array}\right)$$ Suppose the formula holds for some $n=k$ $$\mathbf{A}^{k} = \left(\begin{array}{cc} (-1)^{k} & 0 \\ 1-(-1)^{k} & 1 \end{array}\right)$$ Now $\mathbf{A}^{k+1} = \mathbf{A}^{k}\mathbf{A}$, so \begin{align} \mathbf{A}^{k+1} &= \left(\begin{array}{cc} (-1)^{k} & 0 \\ 1-(-1)^{k} & 1 \end{array}\right)\left( \begin{array}{cc} -1 & 0 \\ 2 & 1 \end{array}\right) \\ &= \left(\begin{array}{cc} -(-1)^{k} & 0 \\ (-1)^{k}-1+2 & 1 \end{array}\right) \\ &= \left(\begin{array}{cc} (-1)^{k+1} & 0 \\ 1+(-1)^{k} & 1 \end{array}\right) \\ &= \left(\begin{array}{cc} (-1)^{k+1} & 0 \\ 1+(-1)(-1)^{k+1} & 1 \end{array}\right) \textrm{ since } (-1)^{k-1}=(-1)^{k+1} \\ &= \left(\begin{array}{cc} (-1)^{k+1} & 0 \\ 1-(-1)^{k+1} & 1 \end{array}\right) \end{align} So we have shown that if it is true for some $n=k$ it is also true for $n=k+1$. We have shown that it is true for $n=1$, therefore by the principle of mathematical induction it is true for all the natural numbers $n$. $\blacksquare$.<|endoftext|>
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Do you find yourself getting sick more often than you’d like? We all know the signs: a runny nose, scratchy throat, and sneezing.
Some estimates suggest that Americans have 1 billion colds a year. The “common cold” is caused by a virus. More than 200 types of viruses can lead to the common cold symptoms, but the most common is the rhinovirus, which brings on 10-40% of colds.
A cold begins when a virus attaches to the linings of your nose or throat. Your immune system then sends out white blood cells to attack the invading virus. Your nose and throat can become inflamed and generate excess mucus. And, with all this energy spent trying to fight the cold virus, your body can be left feeling tired and miserable.
One of the most common ways we acquire a cold virus is by touching a germy surface – a doorknob, a phone, a coffee mug – and then touching the nose or mouth. Or, you can catch it just by breathing in the virus if you are near someone who is sick.
Here are top tips for avoiding the common cold:
When a person sneezes, germs fly through the air, as far as 15 feet away. Make an effort to steer clear of sick friends, family, and coworkers, as hard as it may be. If you’re feeling sick, take time off so you don’t spread the virus.
Get some sleep.
Studies show that Americans simply don’t get enough sleep. But research shows that getting adequate amounts of rest can give a big boost to the immune system, which can help you fight off viruses faster. One study showed that people who get eight or more hours of sleep were less likely to come down with a cold than those who slept fewer than seven hours.
Healthy foods can help boost our immune response. Deep green veggies, and orange produce have important vitamins like C and E and can help your body be prepared for the common cold.
Dehyrdation can leave the body in a vulnerable state. Drink plenty of water each day.
If your symptoms aren’t going away and you can’t kick your cold, it may be something more severe. Never be afraid to talk to your primary care provider about your symptoms so that you can receive a proper diagnosis.
If you need to find a primary care provider, use our online Find a Doctor tool or call UP Health System – Marquette at 906.225.9440.<|endoftext|>
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