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# Binary — The Way Computers Read Numbers
Binary (/ˈbʌɪnəri/) usually refers to something that involves two things. Many people can look at binary code and instantly associate it with computers without a second thought, but what actually is binary?
# What is Binary?
Computers “read” all (or almost all) instructions in binary. Electricity flows through many transistors which can turn a switch on, or have no electricity run through the transistor leaving a switch off. This on/off structure is responsible for the binary nature of modern day computing.
# Counting Systems (Number bases)
If N-1 becomes larger than 9, we substitute the digits with letters. For example in base 16, the count goes 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. A number in base 16 would be 10AF (which would be 4271 in base 10!).
## Base 10: Our world
This is special because it means that in base 10, no other combination of numbers can have the same value of 7091. This fact is true for all numbers in any base as long as the two numbers you’re comparing are part of the same base.
## Base 2: Binary
There is no other number that you can write in base 2 that will give you the same value as 1101.
In short, binary is one way of representing a unique number. Just as we know 620 is unique in base 10, a number like 1010 is a representation of a unique number in base 2.
# So… How do we turn our numbers into binary (and vice versa)?
## Binary to Decimal
Which can be written as 1×8 + 1×4 +0×2 + 1×1. The final thing we need to do is to add it all up together. (8+4+0+1) = 13.
Therefore our final answer is 1101 = 13 in base 10.
## Decimal to Binary
1. Find the largest power of two that’s smaller than the number
2. Subtract the power of two from the number
3. Rinse and repeat with the remainder until you reach 0
Let’s do this with the number 175. The largest power of 2 that goes into 175 is 128. To test this, we would go through from the bottom. Does 2⁴ go into 175? Yes. Does 2⁵(32) go into 175? Yes… Does 2⁷(128) go into 175? Yes. Does 2⁸(256) go into 175? No. So the largest is 2⁷.
Our next step is to do 175–128 = 47.
Now we repeat the steps
• There is no 2⁶ term
• 47–32 = 15
• There is no 2⁴ term
• 15–8 = 7
• 7–4 = 3
• 3–2 = 1
• 1–1 = 0
So we can say that 175 = 1(2⁷) + 0(2⁶) + 1(2⁵)+ 0(2⁴) + 1(2³) + 1(2²)+1(2¹) + 1(2⁰), which can be written as 10101111.
And that’s about it! Computers use these numbers in all sorts of ways: Representing characters, adding numbers, executing commands, etc. Hopefully you were able to understand what binary is and how it works. Binary is another segment of maths that often gets overlooked as “computer language” and misunderstood by many people.
# Contact me if you have any other questions!📩
If you enjoyed this article, feel free to give me a follow. I plan on writing more articles to explain harder concepts in a simple fashion.
## If you have any more questions you can contact me at:
18-Year old Music, Programming and Computer Science enthusiast
18-Year old Music, Programming and Computer Science enthusiast
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# Math word problems cheat sheet
## REAL WORLD PROBLEMS:How to Write Equations Based on Algebra Word Problems
I know that you often sit in class and wonder, "Why am I forced to learn about equations, Algebra and variables?"
But... trust me, there are real situations where you will use your knowledge of Algebra and solving equations to solve a problem that is not school related. And... if you can't, you're going to wish that you remembered how.
It might be a time when you are trying to figure out how much you should get paid for a job, or even more important, if you were paid enough for a job that you've done. It could also be a time when you are trying to figure out if you were over charged for a bill.
This is important stuff - when it comes time to spend YOUR money - you are going to want to make sure that you are getting paid enough and not spending more than you have to.
Ok... let's put all this newly learned knowledge to work.
Click here if you need to review how to solve equations.
There are a few rules to remember when writing Algebra equations:
### Writing Equations For Word Problems
• First, you want to identify the unknown, which is your variable. What are you trying to solve for? Identify the variable: Use the statement, Let x = _____. You can replace the x with whatever variable you are using.
• Look for key words that will help you write the equation. Highlight the key words and write an equation to match the problem.
• The following key words will help you write equations for Algebra word problems:
altogether
increase
more
plus
sum
total
combine
difference
decrease
less
fewer
reduce
minus
per
times
product
double (2x)
triple (3x)
#### division
quotient
divided by
divided into
per
share
split
Let's look at an example of an algebra word problem.
### Example 1: Algebra Word Problems
Linda was selling tickets for the school play. She sold 10 more adult tickets than children tickets and she sold twice as many senior tickets as children tickets.
1. Let x represent the number of children's tickets sold.
2. Write an expression to represent the number of adult tickets sold.
3. Write an expression to represent the number of senior tickets sold.
4. Adult tickets cost \$5, children's tickets cost \$2, and senior tickets cost \$3. Linda made \$700. Write an equation to represent the total ticket sales.
5. How many children's tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?
As you can see, this problem is massive! There are 5 questions to answer with many expressions to write.
### Solution
1. In this problem, the variable was defined for you. Let x represent the number of children’s tickets sold tells what x stands for in this problem. If this had not been done for you, you might have written it like this:
Let x = the number of children’s tickets sold
2. For the first expression, I knew that 10 more adult tickets were sold. Since more means add, my expression was x +10. Since the direction asked for an expression, I don’t need an equal sign. An equation is written with an equal sign and an expression is without an equal sign. At this point we don’t know the total number of tickets.
3. For the second expression, I knew that my key words, twice as many meant two times as many. So my expression was 2x.
4. We know that to find the total price we have to multiply the price of each ticket by the number of tickets. Take note that since x + 10 is the quantity of adult tickets, you must put it in parentheses! So, when you multiply by the price of \$5 you have to distribute the 5.
5. Once I solve for x, I know the number of children’s tickets and I can take my expressions that I wrote for #1 and substitute 50 for x to figure out how many adult and senior tickets were sold.
### Where Can You Find More Algebra Word Problems to Practice?
Word problems are the most difficult type of problem to solve in math. So, where can you find quality word problems WITH a detailed solution?
The Algebra Class E-course provides a lot of practice with solving word problems for every unit! The best part is.... if you have trouble with these types of problems, you can always find a step-by-step solution to guide you through the process!
The next example shows how to identify a constant within a word problem.
### Example 2 - Identifying a Constant
A cell phone company charges a monthly rate of \$12.95 and \$0.25 a minute per call. The bill for m minutes is \$21.20.
1. Write an equation that models this situation.
2. How many minutes were charged on this bill?
### Solution
A cell phone company charges a monthly rate of \$12.95 and \$0.25 a minute per call. The bill for m minutes is \$21.20.
1. Write an equation that models this situation.
### Notes For Example 2
• \$12.95 is a monthly rate. Since this is a set fee for each month, I know that this is a constant. The rate does not change; therefore, it is not associated with a variable.
• \$0.25 per minute per call requires a variable because the total amount will change based on the number of minutes. Therefore, we use the expression 0.25m
• You must solve the equation to determine the value for m, which is the number of minutes charged.
The last example is a word problem that requires an equation with variables on both sides.
### Example 3 - Equations with Variables on Both Sides
You have \$60 and your sister has \$120. You are saving \$7 per week and your sister is saving \$5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve.
### Notes for Example 3
• \$60 and \$120 are constants because this is the amount of money that they each have to begin with. This amount does not change.
• \$7 per week and \$5 per week are rates. They key word "per" in this situation means to multiply.
• The key word "same" in this problem means that I am going to set my two expressions equal to each other.
• When we set the two expressions equal, we now have an equation with variables on both sides.
• After solving the equation, you find that x = 30, which means that after 30 weeks, you and your sister will have the same amount of money.
I'm hoping that these three examples will help you as you solve real world problems in Algebra!
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## Math Word Problem Keywords {Free Cheat Sheets}
Affiliate links may have been used in this post. FREE offers are often time-sensitive and may be limited time only.
Tackling word problems in math can be challenging for kids to learn and not any easier to teach. We called them story problems when I was in school. Check out this blog post for math word problem keywords with free cheat sheets to begin teaching this challenging math operations subject in your homeschool.
### Introducing word problems in math
There is no doubt that mathematical operations using words are difficult for kids. They go from counting numbers to doing math equations with numbers.
Then all of a sudden… there are words, just words. All of a sudden algebraic expressions and mathematical operations are POOF – words.
The lack of numbers and shift in mindset can completely throw off a lot of students. If kids have difficulty with reading, then that is yet another struggle for kids as they try to learn basic problems in math.
### Why do some kids struggle with word problems?
A key proponent in different operations in math is learning the keywords that prompt kids to understand which operation skill they need to use to solve the problem.
This means that they should master regular math problems first and be able to read with comprehension. You will shortly find that if these two skills aren’t somewhat mastered first, then word problems will become an issue.
Many times math is a subject best taught in sequential order. If one step is missed, then the future steps falter. This is much like how it is when teaching word problems.
The best thing for your children is for them to first:
• Be able to read well.
• Understand math concepts and phrases.
• Know to not rush, but focus on keywords, identify relevant information, and understand the text.
### What are keywords in math word problems?
Keywords in mathematical word operations are the words or phrases that will signal or show a student which type of math operation to choose in order to solve the math word problem.
The words used in operations are a strategy that helps the math problem make sense and draw connections to how it can be answered.
Basically, when using keywords, students must decipher whether they need to solve the math equation via addition, subtraction, multiplication, or division.
### What are the common keywords for math word problems?
Thankfully, there are keywords that our children can learn that help them work through their word problems. They are prompts that point them in the right direction.
Just like a different language needs words translated for comprehension, students translate the words… into math.
### Common special words that help with problem solving:
#### Keywords to identify addition in word problems
• total
• combine
• together
• altogether
• increased by
• increase
• more than
• plus
• sum
• together
• both
• in all
• larger
• larger than
• more
• in addition to
• how much in all
#### Keywords used in subtraction word problems
• take away
• left over
• decreased
• difference
• fewer than
• gave away
• how many more
• how many less
• less
• minus
• remain
• shorter than
• smaller than
#### Keywords for multiplication word problems
• cubed
• times
• each
• groups
• area
• product of
• per
• triple
• increased by
• multiplied by
• double/twice
#### Keywords used for division word problems
• divide
• equal group
• average
• each
• half of
• quarter
• ratio
• quotient
• share
• split by
• as much
• cut up
• how many in each
• percent
You can print off a free cheat sheet that has the above math keywords for word problems and add it to your homeschool binder.
### How can we help kids learn keywords for solving numberless word problems?
Teach kids steps for solving word problems until it becomes a habit or they get comfortable with the steps. First, they can look for the important information and write those down. (Read the problem carefully). Next, kids need to define or find the variables in the math equation.
From the keywords, kids can now determine what math operation to use. Translate the words to math. Then, kids can solve the math equation. This is where the skills of solving numbered equations are important.
Finally, students have to put their answers in the form of a word sentence. NOTE: Many times kids think after solving the equation they are done. However, the key to making sure they understand that word problems need word answers.
### Different Strategies to Familiarize Keywords in Word Operations
You can use some of these keywords as key vocabulary words in your homeschool.
Students can display subtraction, addition, multiplication, and multiplication handy reference posters on a bulletin board in your homeschool area. Students can also just list them on dry erase boards. These are perfect visual reminders for what keywords go with what math word problems.
Your students can also keep their keywords for math word problems with them as they study. They can place the list of keywords in a math folder or in an anchor chart – and then in their math folder.
Kids can keep the keyword poster sets in their math notebooks or keep them in a word problem journal.
Their strategy for learning word problem keywords all depends on how they best absorb information.
Students may do well using a combination of these methods. Either way, all of these different strategies can be used to get them comfortable in identifying the route to solve math word equations.
### Math Word Problem Keywords Cheat Sheets & Teaching Aids:
Get a Clue Free Download – Check out these word problem clue word handouts and posters to help your students with word problems. There 5 pages in all that will be handy for your kids in trying to find the correct answer while using the correct operation.
Subtraction Keywords/Addition Keywords – Until your kids memorize keywords and what they mean, this freebie can help. Grab these simple black and white printable signs. They will help kids look for keywords like larger numbers for subtraction word problems or addition keywords like in addition to
Words to Math – Keywords in math problems are essentially turning words into math. This graphic organizer printable is a quick reference for your students to use with numberless math word problems. Place them in a notebook chart or your homeschool classroom wall as a visual reminder.
Key Word Posters for Math Problems – Grab these word problem keyword handy reference posters for subtraction, addition, division, and multiplication. Each poster has its specific theme and specific words to solve all problem types. Kids will enjoy having practiced with these math key words posters.
Fun Key Word Sorting Activity – Your kids have now studied some keywords for problem solving in mathematical operations. Use this word problem sorting activity to test their knowledge in a fun engaging way. Add this fun activity to your test prep materials.
### Free Math Word Problems Keywords Cheat Sheet Download
You won’t want to miss our free cheat sheet download for different ways kids see keywords in various types of problems in mathematics. This math tool is everything your student needs and the perfect resource to reference keywords in math operations.
Math word problems are probably the first opportunity students get to understand how math relates to real world situations. The applications can be relevant in their real life experiences like going to the market.
#### However, the benefit to word problems doesn’t stop there…
With word problems, students develop their higher-order thinking and critical thinking skills.
Different types of word problems guide your students to applying math various math concepts at the same time. They have to know basic number sense, basic algebra skills, and even geometry when they attempt multiplication word problems.
If we do it the right way, kids won’t see word problems as a dreadful experience in math. Understanding word problems is a learning curve and doesn’t come easily to kids.
### Help For Math Struggles
Here are some resources to help you with a child struggling with math or if they seem to hate math.
Another important aspect of word problems is that they tell a parent/teacher if a child needs help in areas like reading comprehension or math number operations skills. This type of word math is a great evaluation of your student’s thinking processes.
We can, however, help make it a better experience for them by teaching it the right way.
Jeannette is a wife, mother and homeschooling mom. She has been mightily, saved by grace and is grateful for God’s sovereignty throughout her life’s journey. She has a Bachelor in English Education and her MBA. Jeannette is bi-lingual and currently lives in the Tongan Islands of the South Pacific. She posts daily freebies for homeschoolers!
Filed Under: Math Freebies for Homeschoolers
Sours: https://homeschoolgiveaways.com/2021/07/math-word-problem-keywords-cheat-sheets/
## Solving Word Problems in AlgebraInequality Word Problems
How are you with solving word problems in Algebra? Are you ready to dive into the "real world" of inequalities? I know that solving word problems in Algebra is probably not your favorite, but there's no point in learning the skill if you don't apply it.
I promise to make this as easy as possible. Pay close attention to the key words given below, as this will help you to write the inequality. Once the inequality is written, you can solve the inequality using the skills you learned in our past lessons.
I've tried to provide you with examples that could pertain to your life and come in handy one day. Think about others ways you might use inequalities in real world problems. I'd love to hear about them if you do!
Before we look at the examples let's go over some of the rules and key words for solving word problems in Algebra (or any math class).
### Word Problem Solving Strategies
• Read through the entire problem.
• Highlight the important information and key words that you need to solve the problem.
• Identify your variables.
• Write the equation or inequality.
• Solve.
• Write your answer in a complete sentence.
I know it always helps too, if you have key words that help you to write the equation or inequality. Here are a few key words that we associate with inequalities! Keep these handy as a reference.
### Inequality Key Words
• at least - means greater than or equal to
• no more than - means less than or equal to
• more than - means greater than
• less than - means less than
Ok... let's put it into action and look at our examples.
### Example 1: Inequality Word Problems
Keith has \$500 in a savings account at the beginning of the summer. He wants to have at least \$200 in the account by the end of the summer. He withdraws \$25 each week for food, clothes, and movie tickets.
• Write an inequality that represents Keith's situation.
• How many weeks can Keith withdraw money from his account? Justify your answer.
### Solution
Step 1: Highlight the important information in this problem.
Note: At least is a key word that notes that this problem must be written as an inequality.
Step 2: Identify your variable. What don't you know? The question verifies that you don't know how many weeks.
Let w = the number of weeks
Step 3: Write your inequality.
500 - 25w > 200
I know you are saying, "How did you get that inequality?"
I know the "at least" part is tricky. You would probably think that at least means less than.
But... he wants the amount in his account to be at least \$200 which means \$200 or greater. So, we must use the greater than or equal to symbol.
Step 4: Solve the inequality.
The number of weeks that Keith can withdraw money from his account is 12 weeks or less.
Step 5: Justify (prove your answer mathematically).
I'm going to prove that the largest number of weeks is 12 by substituting 12 into the inequality for w. You could also substitute any number less than 12.
Since 200 is equal to 200, my answer is correct. Any more than 12 weeks and his account balance would be less than \$200. Any number of weeks less than 12 and his account would stay above \$200.
That wasn't too bad, was it? Let's take a look at another example.
### Example 2: More Inequality Word Problems
Yellow Cab Taxi charges a \$1.75 flat rate in addition to \$0.65 per mile. Katie has no more than \$10 to spend on a ride.
• Write an inequality that represents Katie's situation.
• How many miles can Katie travel without exceeding her budget? Justify your answer.
### Solution
Step 1: Highlight the important information in this problem.
Note: No more than are key words that note that this problem must be written as an inequality.
Step 2: Identify your variable. What don't you know? The question verifies that you don't know the number of miles Katie can travel.
Let m = the number of miles
Step 3: Write the inequality.
0.65m + 1.75 < 10
Are you thinking, "How did you write that inequality?"
The "no more than" can also be tricky. "No more than" means that you can't have more than something, so that means you must have less than!
Step 4: Solve the inequality.
Since this is a real world problem and taxi's usually charge by the mile, we can say that Katie can travel 12 miles or less before reaching her limit of \$10.
Step 5: Justify (prove your answer mathematically).
Are you ready to try some on your own now? Yes... of course you are! Click here to move onto the word problem practice problems.
### Take a look at the questions that other students have submitted:
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Rapid Learning: Applied Math Word Problems - How to Solve Math Problems?
Begged Constance, but Nisha remained adamant. She silently began to style the girl's hair in a beautiful hairstyle, forced her to put on a tunic of the finest pale green silk, decorated with gold. Embroidery, tied her a thin belt and gave her elegant clogs made of morocco leather embroidered with emeralds.
Having dressed the girl, the slave girl suddenly ordered her to follow her. - Where are you taking me.
## Word problems sheet math cheat
She fiercely began to caress herself, experiencing an orgasm for the second time. - God, how good, you are like a magician. It never happened. You just guessed right with the moment when you started to cum on me. Twice in a row.
WHY 80% of STUDENTS struggle with MATH WORD PROBLEMS! + The BEST FREE Math Word Problem Generator!
The girl moaned softly when another rapist finished inside her and the unfortunate woman was carried back to the house, where they again threw. Her on the bed and left herself, to suffer and cry from powerlessness and pain, tearing apart not only the exhausted body, but also the girl's soul. Submission Lillian It was a new evening and Lillian was dreading the return of Jonathan, who had robbed her of her innocence last night by roughly taking her body.
At first, she herself was ready to surrender to him, but it seemed to her that it was only a dream. Of course, the leader of the robbers was handsome and strong, but he had no right to take away from her the most valuable thing that.
### Similar news:
As I expected, my charming interlocutor was really interested in this, she bent down to me and began to carefully look at. What I began to explain, and since we were sitting opposite each other, I could calmly move my gaze forward, as if trying to look into my face interlocutor. It worked.
6394 6395 6396 6397 6398
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1. Aggregate functions – Returns a single result from a group of rows.
Examples :- AVG, MAX, MIN, COUNT,SUM etc.
2. Numeric functions – Accepts numeric input and returns numeric result.
Examples :- SIN, COS, TAN, MOD, POWER, EXP etc.
3. Character functions – Returns character result from character input.
Examples :- CHR, LOWER, UPPER, TRIM etc.
4. Date functions – To manipulate date data.
Examples :- SYSDATE, ADD_MONTHS, TRUNC etc.
5. Conversion functions – To convert one data type to another data type
Examples :- TO_CHAR, TO_DATE, TO_NUMBER etc.
6. Analytical functions – It computes aggregate based on group of data
Examples :- RANK, DENSE_RANK, LAG etc.
Apart from these Oracle has few more functions like Collection functions, Large Object functions, XML functions, Data Mining functions, Hierarchical functions etc.
Some of the above listed functions which falls multiple categories. For example TRUNC is a Numeric as well as Date function. TRUNC(date) returns date value while TRUNC(number) returns numeric value.
Each individual SQL function explained in separate sections with examples.
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It is pretty odd that a calculator and a touch-tone telephone have exactly opposite layouts for their keypads, which have many identical components. The reasons behind the differences are not known for certain, but a few theories exist.
The first theory deals with the telephone's circuitry and tone-recognition hardware. When the touch-tone telephone was being designed in the late 1950s, the calculator and adding-machine designers had already established a layout that had 7, 8 and 9 across the top row. Data-entry professionals, and others who used calculators fairly regularly, were quite adept at navigating these keypads. They could hit the numbers extremely quickly, which was great for data entry, but not so great for dialing a touch-tone phone. The tone-recognition technology could not operate effectively at the speeds at which these specialists could dial the numbers. The telephone designers figured that if they reversed the layout, the dialing speeds would decrease and the tone-recognition would be able to do its job more reliably. This theory has little proof to substantiate it, but it does make sense.
A second theory refers to a study done by Bell Labs in 1960. This study involved testing several different telephone-keypad layouts to find out which was easiest to master. After testing several layouts, including one that used two rows with five numbers each and another that used a circular positioning, it was determined that the three-by-three matrix that had 1, 2 and 3 across the top was the easiest for people to use.
Another theory is based on the layout of a rotary telephone. On a rotary dial, 1 is at the top right and zero is on the bottom. When designing the new touch-tone keypad, putting the 1 on the top-right didn't make much sense, because Western writing is read from left to right. But putting 1 on the top-left, and the subsequent numbers to the right, did make sense. Using that formula, the resulting rows fell into place, with zero getting its own row at the bottom.
All of these theories attempt to explain why telephone and calculator keypads are exact opposites, yet no one theory can be pinpointed as the definitive reason. It is common practice today to use the telephone-keypad layout when designing new products that utilize a keypad, such as Automated Teller Machines.
Here are some interesting links:
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Subtracting Fractions
Multiplying Fractions
Parts of Fractions
Decimals
100
When adding fractions, they must have the same ______.
What is denominator?
100
When subtracting fractions, denominators must be ______.
What is common (the same).
100
2/4 x 3
1 2/4
100
the top digit in a fraction
What is the numerator?
100
What is 3/10 as a decimal?
0.3
200
When adding fractions with unlike denominators, you must find ______.
What is a common denominator?
200
The Carson family went fishing. They caught a 2 5/8 pound trout and a 1 4/8 pound bass. How many more pounds was the trout than the bass?
what is 1 1/8
200
To multiply fractions the denominators must be the same. True or False?
What is false?
200
The bottom digit of a fraction
What is the denominator?
200
What is 52/100 as a decimal?
0.52
300
When adding fractions with a common denominator, what do you do with the numerators?
300
When subtracting fractions with a common denominator, what do you do with the numerators?
What is subtract?
300
2 x 3/4
1 1/2
300
a fraction with the numerator greater than the denominator
What is an improper fraction?
300
Which decimal is bigger: 0.5 or 0.45?
0.5
400
2/10 +3/100
What is 23/100
400
What is simplest form?
400
1/2 and 2/4 are what kind of fractions?
What is equivalent fractions?
400
A number with a whole number and a fraction
What is a mixed fraction?
400
Order from least to greatest: 0.31, 0.3, 0.03
0.03, 0.3, 0.31
500
7/12 + 4/12= ?
What is 11/12
500
3/4 - 1/4
What is 2/4 (1/2)
500
2/5 x 6
2 2/5
500
A fraction with the same numerator and denominator.
What is an equivalent form of 1?
500
Order from greatest to least: 0.84, 0.54, 0.8, 0.9
0.9, 0.84, 0.8, 0.54
600
4/8 + 2/8
What is 6/8
600
9/12-6/12
What is 3/12
600
Ms. Stanbury made 3 pies. From each pie, she take a slice that is 1/8 of a pie so that she could do a porch drop off. How much pie is being dropped off?
What is 3/8?
600
1/4= x/12 what is x?
What is 3
600
Compare 0.7 and 0.70 using <,>, =
0.7 = 0.70
700
2/10 + 3/10
What is 1/2 (5/10)
700
14 2/5 - 9 1/5
What is 5 1/5
700
5 x 2/3
3 1/3
700
2/5=x/10
What is 4
700
What is the place called that the 6 is in in this decimal: 0.6
tenths place
800
1 2/4 + 2 3/4
What is 4 1/4
800
5/10 - 2/100
What is 48/100?
800
3/4 x 4
3
800
1/4 and 1/2. Are these equivalent fractions? how do you know?
No they are not equivalent.
800
What is the place called that the 2 is in, in this decimal? 0.02
hundreths place
900
3/5 + 1/4
What is 17/20
900
3 1/3 - 1 2/3
What is 1 2/3
900
2 1/2 x 3
7 1/2
900
2/3 and 4/6. Are these equivalent fractions and how do you know?
Yes they are equivalent because the shaded area of each fraction is the same.
900
How are fractions and decimals similar?
They are both parts of a whole
1000
1/3 + 1/4 + 2/6
What is 11/12
1000
64/100 - 4/10
24/100
1000
2 x 4/5
1 3/5
1000
3/4 and 2/3. Are these equivalent fractions, how do you know?
No they are not equivalent fractions. b
1000
What are three decimals between 0.4 and 0.52?
Any three answers are acceptable: 0.41, 0.42, 0.43, 0.44, 0.45, 0.46, 0.47, 0.48, 0.49, 0.5, 0.51
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### 3 4 systems of linear equations 2
1. 1. Systems of Linear Equations II Back to 123a-Home
2. 2. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons?
3. 3. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D.
4. 4. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D. 5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
5. 5. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D. 5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C. Exchanging the 2 coupons for pizzas and donuts, we would have 5P + 3D + 2C
6. 6. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D. 5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C. Exchanging the 2 coupons for pizzas and donuts, we would have 5P + 3D + 2C = 5P + 3D + 2(5P + 3D)
7. 7. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D. 5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C. Exchanging the 2 coupons for pizzas and donuts, we would have 5P + 3D + 2C = 5P + 3D + 2(5P + 3D) = 5P + 3D + 10P + 6D
8. 8. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D. 5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C. Exchanging the 2 coupons for pizzas and donuts, we would have 5P + 3D + 2C = 5P + 3D + 2(5P + 3D) = 5P + 3D + 10P + 6D = 15P + 9D or 15 slices of pizzas and 9 donuts.
9. 9. In math, the phrase "substitute (the expression) back into ...“ means to do the exchange, using the given coupon–expression, in the targeted equations, or expressions mentioned. Systems of Linear Equations II Example A. At Pizza Grande, one coupon C may be exchanged for three slices of pizza (P) and five donuts (D). We have 5 slices of pizzas, 3 donuts and 2 coupons. What do we have after exchanging the 2 coupons? The coupon value of C may be recorded as C = 5P + 3D. 5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C. Exchanging the 2 coupons for pizzas and donuts, we would have 5P + 3D + 2C = 5P + 3D + 2(5P + 3D) = 5P + 3D + 10P + 6D = 15P + 9D or 15 slices of pizzas and 9 donuts.
10. 10. There are two other methods to solve system of equations. Systems of Linear Equations II
11. 11. There are two other methods to solve system of equations. Substitution Method Systems of Linear Equations II
12. 12. There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. Systems of Linear Equations II
13. 13. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. Systems of Linear Equations II Example B. { There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation.
14. 14. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Systems of Linear Equations II Example B. { There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation.
15. 15. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Systems of Linear Equations II Example B. { There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. This is the coupon expression so we may exchange x = 5 – y
16. 16. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Then we replace x by (5 – y) in E1 and get 2(5 – y) + y = 7 Systems of Linear Equations II Example B. { There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. This is the coupon expression so we may exchange x = 5 – y
17. 17. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Then we replace x by (5 – y) in E1 and get 2(5 – y) + y = 7 10 – 2y + y = 7 Systems of Linear Equations II Example B. { There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. This is the coupon expression so we may exchange x = 5 – y
18. 18. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Then we replace x by (5 – y) in E1 and get 2(5 – y) + y = 7 10 – 2y + y = 7 10 – y = 7 3 = y Systems of Linear Equations II Example B. { There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. This is the coupon expression so we may exchange x = 5 – y
19. 19. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Then we replace x by (5 – y) in E1 and get 2(5 – y) + y = 7 10 – 2y + y = 7 10 – y = 7 3 = y Systems of Linear Equations II Example B. { Put 3 = y back to E2 to find x, we get x + 3 = 5 There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. This is the coupon expression so we may exchange x = 5 – y
20. 20. 2x + y = 7 E1 x + y = 5 E2 Solve by the substitution method. From E2, x + y = 5, we get x = 5 – y. Then we replace x by (5 – y) in E1 and get 2(5 – y) + y = 7 10 – 2y + y = 7 10 – y = 7 3 = y Systems of Linear Equations II Example B. { Put 3 = y back to E2 to find x, we get x + 3 = 5 x = 2 Therefore the solution is (2, 3) There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. This is the coupon expression so we may exchange x = 5 – y
21. 21. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Systems of Linear Equations II
22. 22. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. Systems of Linear Equations II
23. 23. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. Systems of Linear Equations II Example C. {
24. 24. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. Systems of Linear Equations II Example C. {
25. 25. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Systems of Linear Equations II Example C. {
26. 26. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7) in E2 and get 3x + 2(2x – 7) = 7 Systems of Linear Equations II Example C. {
27. 27. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7) in E2 and get 3x + 2(2x – 7) = 7 3x + 4x – 14 = 7 Systems of Linear Equations II Example C. {
28. 28. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7) in E2 and get 3x + 2(2x – 7) = 7 3x + 4x – 14 = 7 7x = 21 x = 3 Systems of Linear Equations II Example C. {
29. 29. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7) in E2 and get 3x + 2(2x – 7) = 7 3x + 4x – 14 = 7 7x = 21 x = 3 Systems of Linear Equations II Example C. { To find y, use the substitution equation, set x = 3 in y = 2x – 7
30. 30. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7) in E2 and get 3x + 2(2x – 7) = 7 3x + 4x – 14 = 7 7x = 21 x = 3 Systems of Linear Equations II Example C. { To find y, use the substitution equation, set x = 3 in y = 2x – 7 y = 2(3) – 7 y = -1
31. 31. We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. 2x – y = 7 E1 3x + 2y = 7 E2 Solve by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7) in E2 and get 3x + 2(2x – 7) = 7 3x + 4x – 14 = 7 7x = 21 x = 3 Systems of Linear Equations II Example C. { To find y, use the substitution equation, set x = 3 in y = 2x – 7 y = 2(3) – 7 y = -1 Hence the solution is (3, -1).
32. 32. Graphing Method Systems of Linear Equations II
33. 33. Graphing Method Given a system of linear equations the graph of each equation is a straight line. Systems of Linear Equations II
34. 34. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Systems of Linear Equations II
35. 35. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. Systems of Linear Equations II
36. 36. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II
37. 37. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D.
38. 38. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method.
39. 39. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 0 2x + y = 7
40. 40. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 0 2x + y = 7
41. 41. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 7/2 0 2x + y = 7 E2
42. 42. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 7/2 0 2x + y = 7 E2 (Check another point.)
43. 43. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 7/2 0 2x + y = 7 E1 E2 (0, 7) (7/2, 0) E1(Check another point.)
44. 44. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x + y = 5 x y 0 7 7/2 0 2x + y = 7 x y 0 0 (0, 7) (7/2, 0) E1(Check another point.)
45. 45. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 7/2 0 2x + y = 7 x + y = 5 x y 0 5 5 0 (0, 7) (7/2, 0) E1(Check another point.)
46. 46. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 7/2 0 2x + y = 7 (0, 7) (7/2, 0) (0, 5) (5, 0) E2 x + y = 5 x y 0 5 5 0 E1(Check another point.)
47. 47. Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. Systems of Linear Equations II 2x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example D. Use intercept method. x y 0 7 7/2 0 2x + y = 7 (0, 7) (7/2, 0) (0, 5) (5, 0) The intersection (2, 3) is the solution E1 E2 x + y = 5 x y 0 5 5 0 E1(Check another point.)
48. 48. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. Systems of Linear Equations II
49. 49. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. Systems of Linear Equations II x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example E.
50. 50. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. x + y = 7 x y 0 7 7 0 Systems of Linear Equations II x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example E.
51. 51. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. x + y = 7 x y 0 7 7 0 (0, 7) (7, 0) Systems of Linear Equations II x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example E.
52. 52. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. x + y = 7 x y 0 7 7 0 x y 0 5 5 0 (0, 7) (7, 0) Systems of Linear Equations II x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example E. x + y = 5
53. 53. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. x + y = 7 x y 0 7 7 0 x y 0 5 5 0 (0, 7) (7, 0) (0, 5) (5, 0) Systems of Linear Equations II x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example E. x + y = 5
54. 54. Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. x + y = 7 x y 0 7 7 0 x y 0 5 5 0 (0, 7) (7, 0) (0, 5) (5, 0) No intersection. No solution Systems of Linear Equations II x + y = 7 x + y = 5 Solve graphically.{ E1 E2 Example E. x + y = 5
55. 55. Graphs of dependent systems are identical lines, hence every point on the line is a solution. Systems of Linear Equations II
56. 56. Graphs of dependent systems are identical lines, hence every point on the line is a solution. Systems of Linear Equations II x + y = 5 2x + 2y = 10 Solve graphically.{ E1 E2Example F.
57. 57. Graphs of dependent systems are identical lines, hence every point on the line is a solution. 2x + 2y = 10 is the same as x + y = 5 Systems of Linear Equations II x + y = 5 2x + 2y = 10 Solve graphically.{ E1 E2Example F.
58. 58. Graphs of dependent systems are identical lines, hence every point on the line is a solution. 2x + 2y = 10 is the same as x + y = 5 x y 0 5 5 0 Systems of Linear Equations II x + y = 5 2x + 2y = 10 Solve graphically.{ E1 E2Example F.
59. 59. Graphs of dependent systems are identical lines, hence every point on the line is a solution. 2x + 2y = 10 is the same as x + y = 5 x y 0 5 5 0 (5, 0) (0, 5) Systems of Linear Equations II x + y = 5 2x + 2y = 10 Solve graphically.{ E1 E2Example F.
60. 60. Graphs of dependent systems are identical lines, hence every point on the line is a solution. 2x + 2y = 10 is the same as x + y = 5 x y 0 5 5 0 (5, 0) (0, 5) Every point is a solution, e.g.(0, 5), (2, 3), (5, 0)… (2, 3) Systems of Linear Equations II x + y = 5 2x + 2y = 10 Solve graphically.{ E1 E2Example F.
61. 61. Systems of Linear Equations II 4. {–x + 2y = –12 y = 4 – 2x Exercise. Solve by the substitution method. 1. {y = 3 – x 2x + y = 4 2. 3. {x = 3 – y 2x – y = 6 {x + y = 3 2x + 6 = y 5. {3x + 4y = 3 x = 6 + 2y 6. { x = 3 – 3y 2x – 9y = –4 10. Graph the inconsistent system { x + 3y = 4 2x + 6y = 8 {2x – y = 2 8x – 4y = 6 Problem 7, 8, 9: Solve problem 1, 2, and 3 by graphing. 11. Graph the dependent system
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# Daily Maths Revision – Week 11 Walkthrough
This week, we’re exploring:
### Question 1: Cones
You are given the formula for both the curved surface area and volume of a cone. So, finding these values is mostly about substitution. Just make sure you are using the radius for $$r$$, the slant height for $$l$$ and the perpendicular height for $$h$$.
$$\text{volume} = \frac{1}{3}\pi r^2h$$
$$\text{curved surface area} = \pi rl$$
Let’s find the volume and total surface area of this cone. We will give both answers in terms of $\boldsymbol{\pi}$.
The slant height is 13cm and the perpendicular height is 12cm. 10cm is the diameter but we want the radius, so we simply divide it by 2.
\begin{aligned} r &= 10 \div 2 \\ &= 5\text{cm} \\ h &= 12\text{cm} \\ l &= 13\text{cm} \end{aligned}
Volume
\begin{aligned} V &= \frac{1}{3}\pi r^2h \\ &= \frac{1}{3}\times pi \times 5^2 \times 12 \\ &= \boldsymbol{100\pi}\text{ cm}^2 \end{aligned}
Surface Area
We want to find the total surface area but the formula we are given only calculates the curved surface area. To find the total surface area, we need to add the base of the cone – a simple circle with the same radius as the cone.
\begin{aligned} \text{curved surface area} &= \pi r l \\ &= \pi \times 5 \times 13 \\ &= 65\pi \text{ cm}^2 \end{aligned}
\begin{aligned} \text{base of cone} &= \pi r^2 \\ &= \pi \times 5^2 \\ &= 25\pi \text { cm}^2\end{aligned}
\begin{aligned} \text{total surface area} &= 65\pi + 25\pi \\&= \boldsymbol{90\pi} \text{ cm}^2 \end{aligned}
### Question 2: Multiplying Algebraic Fractions
Multiplying algebraic fractions is exactly the same process as multiplying numerical fractions. We simply multiply the numerators together, multiply the denominators together and then simplify.
Write $$\frac{x + 3}{x + 7} \times\frac{x – 3}{2x + 1}$$ in the form $$\frac{x^2 – m}{ax^2 + bx + c}$$ where 𝒂, 𝒃, 𝒄 and 𝒎 are integers to be found.
Let’s start by multiplying the numerators and the denominators.
$$\frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1} = \frac{(x + 3)(x – 3)}{(x + 7)(2x + 1)}$$
The question asks us to write the fraction in a specific form. To do this, we simply expand the brackets. It can help to consider the numerator and denominator separately:
\begin{aligned} (x + 3)(x – 3) &= x^2 + 3x – 3x – 9 \\ &= x^2 – 9\end{aligned}
\begin{aligned} (x + 7)(2x + 1) &= 2x^2 + 14x + x + 7 \\ &= 2x^2 + 15x + 7\end{aligned}
Putting this back in the fraction:
$$\frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1} = \frac{x^2 – 9}{2x^2 + 15x + 7}$$
### Question 3: Translation
Translation keeps a shape the same size and orientation but simply changes its location. Usually, translations are given using vectors.
Translate the shape below by the vector $$\begin{pmatrix} -2 \\ 5 \end{pmatrix}$$.
The top number on the vector tells us how far to move the shape horizontally. If the number’s positive, we move the shape to the right and if it’s negative, we move the shape to the left. The bottom number tells us how far to move the shape vertically. If the number’s positive, we move the shape up and if it’s negative, we move the shape down. In this case, we move the shape 2 units left and 5 units up.
Pick a vertex (corner) and count the number of squares we need. This will give us the new location of the vertex.
Now, we can draw in the rest of the shape. Make sure it’s exactly the same size.
If asked to describe a transformation, we use a very similar method.
Describe the transformation that maps shape A onto shape B.
As before, pick a vertex on shape A and count the squares to get to the corresponding vertex on shape B.
So, the shape has been translated 3 units to the right and 4 units down. We can also write this as a vector: $$\begin{pmatrix} 3 \\ -4 \end{pmatrix}$$.
When asked to draw a quadratic graph, you will usually be given a table to complete. Unlike linear graphs, these graphs won’t be straight lines. Instead, they will be a smooth curve called a parabola.
Let’s use the table below to draw the graph with equation $$y = x^2 + 2x – 3$$.
We simply need to substitute each value of $$x$$ into the equation $$y = x^2 + 2x – 3$$. Be particularly careful with negative signs. If you are putting the values into a calculator, put brackets around any negative values.
\begin{aligned} x &= -3 & y&=(-3)^2+2\times(-3)-3 \\&&&=9-6-3\\&&&=0\end{aligned}
\begin{aligned} x &= 0 & y&=(0)^2+2\times(0)-3 \\&&&=0+0-3\\&&&=-3\end{aligned}
\begin{aligned} x &= 3 & y&=(3)^2+2\times(3)-3 \\&&&=9+6-3\\&&&=12\end{aligned}
Now, we can plot these points on a set of axes.
Finally, we can join the coordinates. Make sure you draw a smooth curve rather than lots of short, straight lines. This includes at the bottom of the graph – it continues to curve as it changes direction.
### Question 5: Completing the Square
Completing the square can be quite tricky – especially if the quadratic is in the form $$ax^2 + bx + c$$. Let’s start with a slightly easier expression.
Write $$x^2 + 5x – 2$$ in the form $$(x + a)^2 + b$$.
First, we halve the coefficient of $$x$$. In this case, that’s 5. This gives us the value of $$a$$: $$(x + \frac{5}{2})^2$$. Let’s expand this to see the expression:
\begin{aligned} \left(x + \frac{5}{2}\right)^2 &= \left(x + \frac{5}{2}\right)\left(x + \frac{5}{2}\right)\\ &= x^2 +\frac{5}{2}x+\frac{5}{2}x+ \frac{25}{4}\\ &= x^2 + 5x + \frac{25}{4}\end{aligned}
This is similar to the expression we started with. In fact, the first two terms are identical. The value of $$b$$ will adjust this expression to get the expression we started with. To find $$b$$, we will subtract $$\frac{25}{4}$$ from $$\left(x + \frac{5}{2}\right)^2$$. We should also include $$– 2$$ at this point.
\begin{aligned} x^2 + 5x – 2 &= \left(x + \frac{5}{2}\right)^2-\frac{25}{4} – 2 \\ &= \left(x + \frac{5}{2}\right)^2-\frac{33}{4}\end{aligned}
Write $$2x^2 – 20x + 7$$ in the form $$p(x + q)^2 + r$$ where $$p$$, $$q$$ and $$r$$ are integers to be found.
We will start by factorising out 2 from the expression before continuing as before:
\begin{aligned} 2x^2 – 20x + 7 &= 2\left[x^2-10x+\frac{7}{2}\right] \\ &= 2\left[\left(x-5\right)^2 -25 +\frac{7}{2} \right] \\&= 2\left[\left(x-5\right)^2 -\frac{43}{2} \right] \\ &= 2\left(x-5\right)^2 -43\end{aligned}
Don’t forget to read even more of our blogs here and you can find our main Daily Maths Revision Page here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.
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# sum to infinity
#### mathy1991
##### New Member
Hi, just wondering can anyone tell em how to sum this term from k=1 to infinity?
I used to know how to do it but I forgot, it's something to do with identifying a partial sum,
it is the sum of (e^(-0.02k))
#### Dason
Let $$X = \sum_{k = 1}^\infty \exp(-.02k)$$
Let's look at what this sum sort of looks like...
$$X = \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots$$
Now what happens if we pull that first term out?
$$X = \exp(-.02) * [\exp(-.02(1-1) + \exp(-.02(2-1)) +\exp(-.02(3-1)) + \ldots]$$
$$X = \exp(-.02)*[1 + \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots]$$
Then notice that inside that parenthesis we have the original series we're trying to figure out. This gives us an equation of the form X = K(1 + X) and what you're trying to solve for is X.
Hopefully that hint should get you there. We didn't really need to write it out what the sum actually looked like and could have instead just worked using summation notation but I think this way helps you visualize what is going on better.
#### mathy1991
##### New Member
Let $$X = \sum_{k = 1}^\infty \exp(-.02k)$$
Let's look at what this sum sort of looks like...
$$X = \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots$$
Now what happens if we pull that first term out?
$$X = \exp(-.02) * [\exp(-.02(1-1) + \exp(-.02(2-1)) +\exp(-.02(3-1)) + \ldots]$$
$$X = \exp(-.02)*[1 + \exp(-.02) + \exp(-.02*2) + \exp(-.02*3) + \ldots]$$
Then notice that inside that parenthesis we have the original series we're trying to figure out. This gives us an equation of the form X = K(1 + X) and what you're trying to solve for is X.
Hopefully that hint should get you there. We didn't really need to write it out what the sum actually looked like and could have instead just worked using summation notation but I think this way helps you visualize what is going on better.
Oh right thanks, so is the answer e^-0.02*(1/(1-e^-0.02))?
#### SmoothJohn
##### New Member
Now's a good time to investigate. Check the partial sum (maybe using excel) and compare it to your answer.
#### Dason
Looks fine to me. One way to check if this seems reasonable is just to have something compute the partial sum up to an arbitrarily large number for you. Using R this is trivial.
Code:
> k <- 1:10000
> sum(exp(-.02*k))
> # spits out an answer
> exp(-.02)/(1 - exp(-.02))
> # check if they're the same or at least really close.
#### mathy1991
##### New Member
Now's a good time to investigate. Check the partial sum (maybe using excel) and compare it to your answer.
Yeah I checked it with the website wolframalpha and it worked out the same
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The recent explosion in technology has led to the development of devices that would have been unfathomable just a few decades ago. Many of these technological advancements are enabled by the miniaturization of electronic components. Microscale devices can be used for a host of applications ranging from portable and implantable medical devices to wireless sensors.
Unfortunately, the development of functional microscale devices has been stalled by difficulties in miniaturizing energy storage to match. The high energy and high power density required for most applications is difficult to achieve in microbatteries due to their size and footprint restrictions. Though scientists have been researching a variety of possible workarounds, few functional microbatteries have been developed; the majority of the existing microbatteries designs simply cannot be manufactured easily.
A team of researchers has now fabricated microbatteries containing microelectromechanical and complementary metal-oxide-seminconductor (CMOS) devices using a futuristic 3D fabrication route. By combining 3D holographic lithography with conventional photolithography, the scientists demonstrated increased control of the electrode structure and spatial arrangement.
In general, the term "lithography" denotes a process that creates defined patterns from a variety of starting materials. Holographic lithography allows the construction of 3D lattices with a defined periodicity; the photolithography outlines a set of 2D solid structures that divide the 3D lattices, forming an interdigitated pattern.
For the new batteries, 3D holographic lithography was used to define a lattice with a typical periodicity of ~1 µm and ~40 percent porosity. This lattice was created on glass coated with indium tin oxide using four interfering laser beams that were arranged in an umbrella geometry. The 3D holographic structure was then infiltrated with a positive photoresist using conventional 2D photolithography, which defined the microbattery electrode distribution.
This second resist of the electrodes consisted of solid straight walls inside the 3D lattice. Nickel was then deposited through the porous lattice. After that, oxygen was used to remove the photoresist template; finally, active materials for the cathode and anode were independently plated onto the battery construct.
These active materials are critical to the function of a battery; when a device is hooked up to both ends of a battery, a current is established within the device through the transfer of electrons from the anode to the cathode via a series of chemical reactions. Lithiated MnO2 was used as the cathode and Ni-Sn was used as the anode. The anode was independently cycled versus lithium metal six times before fresh electrolyte was cast on the micro battery, and it was capped using a silicon polymer. Using this method, the batteries were stable in air for several days, although they weren't tested for longer periods.
The battery itself was only 2 mm on a side and roughly 10 µm thick, and all its components were tiny. It contained a current collector consisting of a 4-mm2 area interdigitated 3D porous nickel scaffold. Individual electrodes had a rectangular cross-section that was about 10 µm high with a porosity of 60 percent. The electrodes were 35 µm wide with 15 µm spacing.
The packaged microbattery was tested between 3.2 and 1.4 volts. The microbattery was charged in an hour and discharged for various times. The batteries exhibited energy and power densities up to 6.5µWh cm-2 µm-1 and 3,600µWh cm-2 µm-1. (Given the radically different sizes, these numbers aren't really comparable to those of larger batteries currently on the market.)
Practical microbattery applications would require a cycle life of several hundred charge-discharge cycles. In this construct, the capacities of the cathode and anode are carefully matched. The careful control of the Ni-Sn cycling range, when added to the 3D Ni scaffold, improves the cyclability. The scientists think that the scaffold relieves the cycling-induced stress in the film. The microbatteries retained 80 percent or more of their initial capacity after cycling 100 times at varied rates; they also retained almost 90 percent of the initial capacity after driving an LED for 200 cycles.
These studies demonstrate that this new fabrication technique shows promise for the future development of functional microbatteries. Once put into practice, it could open the door for a new generation of electronics.
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# Suppose a differentiable function
Question:
Suppose a differentiable function $\mathrm{f}(\mathrm{x})$ satisfies the identity $f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y$,
for all real $x$ and $y$. If $\operatorname{Lim}_{x \rightarrow 0} \frac{f(x)}{x}=1$, then $f^{\prime}(3)$
is equal to
Solution:
Since, $\lim _{x \rightarrow 0} \frac{f(x)}{x}$ exist $\Rightarrow f(0)=0$
Now, $f^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{f(\mathrm{x}+\mathrm{h})-f(\mathrm{x})}{\mathrm{h}}$
$=\lim _{h \rightarrow 0} \frac{f(h)+x h^{2}+x^{2} h}{h}($ take $y=h)$
$=\lim _{h \rightarrow 0} \frac{f(h)}{h}+\lim _{h \rightarrow 0}(x h)+x^{2}$
$\Rightarrow f^{\prime}(x)=1+0+x^{2}$
$\Rightarrow f^{\prime}(3)=10$
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# Math in Focus Grade 2 Chapter 6 Practice 5 Answer Key Multiplying 10: Skip-counting and Using Dot Paper
Go through the Math in Focus Grade 2 Workbook Answer Key Chapter 6 Practice 5 Multiplying 10: Skip-counting and Using Dot Paper to finish your assignments.
## Math in Focus Grade 2 Chapter 6 Practice 5 Answer Key Multiplying 10: Skip-counting and Using Dot Paper
The numbers below the chart follow a pattern. Use the hundreds chart to find the missing numbers.
Use patterns to fill in the blanks.
Question 1.
Question 2.
2 × 1 = ___
2 × 10 = ____
2 × 1 = 2
2 × 10 = 20
Question 3.
3 × 1 = ___
3 × 10 = ___
3 × 1 = 3
3 × 10 = 30
Use patterns to fill in the blanks.
Question 4.
4 × 1 = ____
4 × 10 = ___
4 × 1 = 4
4 × 10 = 40
Question 5.
5 × 1 = _____
5 × 10 = ___
5 × 1 = 5
5 × 10 = 50
Question 6.
6 × 10 = ___
6 × 10 = 60
Question 7.
7 × 10 = ___
7 × 10 = 70
Solve.
Question 8.
There are 4 bundles of sticks. Each bundle has 10 sticks. How many sticks are there in all?
4 × 10 = ____
There are ___ sticks in all.
4 × 10 = 40
There are 40 sticks in all
Question 9.
Megan makes 6 bracelets. She needs 10 beads to make one bracelet. How many beads are needed to make the 6 bracelets?
___ × 10 = ___
____ beads are needed to make the 6 bracelets.
6 × 10 = 60
60 beads are needed to make the 6 bracelets.
Solve.
Question 10.
During sports day, 10 children form a group for a relay race. How many children are there in 8 groups?
____ × 10 = _____
There are ___ children in 8 groups.
8 × 10 = 80
There are 20 children in 8 groups.
Question 11.
The school band has 10 violins. Each violin has 4 strings on it. How many strings are on the 10 violins?
____ × ___ = ____
There are ___ strings on the 10 violins.
1 0 × 4 = 40
There are 40 strings on the 10 violins.
Use dot paper to multiply.
Question 12.
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The UN Statistics Division and the World Bank launched a new guide on Wednesday to help nations worldwide, especially in low- and middle-income countries, better manage data that contains geographical information – known officially as geospatial data.
The guide includes advice on how to collect, access and use geospatial information to develop effective policies, and more accurately support decision-makers in directing aid and development resources; ensuring that the most vulnerable are not left behind.
“Geospatial information is a critical component of national infrastructure and a blueprint of what happens where, and with proven societal and economic value,” said Stefan Schweinfest, Director of the Statistics Division, which is part of the UN Department of Economic and Social Affairs (DESA).
High-quality, timely geospatial information is often overlooked in policymaking - Anna Wellenstein (World Bank)
“Better understanding and management of digital location-based data and services, and good geospatial information integrated with urban planning and census data, can enable more efficient resource allocation for better service delivery,” he explained.
The guide, titled ‘Integrated Geospatial Information Framework’, makes concrete recommendations on establishing national geospatial information management processes and putting that information to use.
According to the UN Committee of Experts on Global Geospatial Information Management (UN-GGIM), with more reliable geospatial data, policy-makers, international organisations, civil society and others, will have better insights into the distribution of needs and ways to optimize development planning and investments.
“High-quality, timely geospatial information is often overlooked in policymaking, yet is fundamental to achieving inclusive growth and sustainable development,” said Anna Wellenstein, who leads land and geospatial activities at the World Bank.
Currently, all governments hold a considerable amount of geospatial information, including databases on who has access to education; communities most affected by poverty; areas at risk of disasters; as well as mobile data that can keep more people informed about disease outbreaks and weather patterns.
But the information, although critical to improve lives and livelihoods, is often not current, shared, or integrated with other necessary data.
“The Framework will help countries in building capacity for using geospatial technology to enhance informed government decision-making, facilitate private sector development, take practical actions to achieve a digital transformation, and bridge the geospatial-digital divide.”
The UN-GGIM stressed that this guide is also meant to help low- and middle-income countries move toward developing their “e-economies” to provide better social and economic services to citizens. For example, integrated geospatial information management can enable Small Island Developing States (SIDS) to better monitor climate change impacts, plan mitigation, and manage disaster risks.
Representatives from governments and geospatial information experts are attending the Eighth Session of the UN-GGIM in New York from 1 to 3 August to discuss efforts to enhance collaboration, coordination, and coherence in global geospatial information management. In November, high-level stakeholders will meet in Deqing, Zhejiang Province, China, at the UN World Geospatial Information Congress to ensure the widest and fullest use of geospatial information to advance social, economic, and environmental development.
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People - Ancient Greece: Agathon
(ca. 448–400 BC) An ancient Athenian tragic poet.
Agathon in Wikipedia
Agathon (Greek: Ἀγάθων) (ca. 448–400 BC) was an Athenian tragic poet whose works, up to the present moment, have been lost. He is best known for his appearance in Plato's Symposium, which describes the banquet given to celebrate his obtaining a prize for his first tragedy at the Lenaia in (416). He is also a prominent character in Aristophanes' comedy the Thesmophoriazusae. Agathon was the life-long companion of Pausanias, with whom he appears in both the Symposium and in Plato's Protagoras. Together with Pausanias, Agathon later moved to the court of Archelaus, king of Macedon, who was recruiting playwrights; it is here that he probably died around 401 BC. Agathon introduced certain innovations into the Greek theater: Aristotle tells us that the characters and plot of his Antheus were original and not, as was usual at the time, borrowed from mythological subjects. Agathon was also the first playwright to write choral parts which were apparently independent from the main plot of his plays.
Agathon is portrayed by Plato as a handsome young man, well dressed, of polished manners, courted by the fashion, wealth and wisdom of Athens, and dispensing hospitality with ease and refinement. The epideictic speech in praise of love which Agathon recites in the Symposium is full of beautiful but artificial rhetorical expressions, and has led some scholars to believe he may have been a student of Gorgias. In the Symposium, Agathon is presented as the friend of the comic poet Aristophanes, but this alleged friendship did not prevent Aristophanes from harshly criticizing Agathon in at least two of his comic plays: the Thesmophoriazousae and the (now lost) Gerytades. In a later play, the Frogs, Aristophanes softens his criticisms, but even so it may be only for the sake of punning on Agathon's name (ἁγαθός = "good") that he makes Dionysus call him a "good poet".
Agathon was also a friend of Euripides, another recruit to the court of Archelaus of Macedon.
Agathon's extraordinary physical beauty is brought up repeatedly in the sources; the historian W. Rhys Roberts observes that "ὁ καλός Ἀγάθων has become almost a stereotyped phrase." Our most detailed description of Agathon can be found in Aristophanes' comedy, the Thesmophoriazousae, in which Agathon appears as a pale, cleanshaven young man, dressed in women's clothes. Regrettably, it is hard to determine how much of Aristophanes' portrayal is fact and how much mere comic invention.
After a close reading of the Thesmophoriazousae, the historian Jane McIntosh Snyder observed that Agathon's costume was almost identical to that of the famous lyric poet Anacreon, as he is portrayed in early 5th-century vase-paintings. Snyder theorizes that Agathon might have made a deliberate effort to mimic the sumptuous attire of his famous fellow-poet, although by Agathon's time, such clothing, especially the κεκρύφαλος (an elaborate covering for the hair) had long fallen out of fashion for men. According to this interpretation, Agathon is mocked in the Thesmophoriazousae not only for his notorious effeminacy, but also for the pretentiousness of his dress: "he seems to think of himself, in all his elegant finery, as a rival to the old Ionian poets, perhaps even to Anacreon himself."
Agathon has been thought to be the subject of Lovers' Lips, attributed to the philosopher Plato:
Kissing Agathon, I had my soul upon my lips; for it rose, poor wretch, as though to cross over.
A looser translation reads:
Kissing Agathon, I found my soul at my lips. Poor thing! It went there, hoping--to slip across.
Although the authenticity of this epigram was accepted for many centuries, it was probably not composed for Agathon the tragedian; nor was it composed by Plato. Stylistic evidence suggests that the poem (with most of Plato's other alleged epigrams) was actually written some time after Plato had died: for its form is that of the Hellenistic erotic epigram, which did not become popular until after 300 B.C. According to 20th-century scholar Walther Ludwig, the poems were spuriously inserted into an early biography of Plato—sometime between 250 B.C. and 100 B.C.--and adopted by later writers from this source.
If you notice a broken link or any error PLEASE report it by clicking HERE
© 1995-2016 Bible History Online
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Playing a Pokémon-like card game about ecology and biodiversity can result in broader knowledge of species and a better understanding of ecosystems than traditional teaching methods, like slideshows, according to new research from the University of British Columbia.
An open-source project launched in 2010 by UBC biologist David Ng and collaborators, the Phylo Trading Card Game works similarly to Pokémon trading cards, but uses real organisms and natural events instead of imaginary characters. While the Phylo project has proven immensely popular around the world, this is the first study to have tested its efficacy as a teaching and learning tool.
Researchers examined how people who played the game retained information about species and ecosystems, and how it impacted their conservation behaviour. They compared the results to people who watched an educational slideshow, and those who played a different game that did not focus on ecosystems.
“Participants who played the Phylo game weren’t just remembering iconic species like the blue whale and sea otter, but things like phytoplankton, zooplankton and mycorrhizal fungi,” said lead author Meggie Callahan, a PhD candidate in the Institute for Resources, Environment and Sustainability. “They would say things like, ‘I really needed this card because it was the base of my ecosystem,’ or, ‘When my partner destroyed my phytoplankton it killed all of my chain of species.’ Obviously, the game is sending a strong message that is sticking with them.”
Participants in both the Phylo Game group and slideshow group improved their understanding of ecosystems and species knowledge, but those who played the Phylo Game were able to recall a greater number of species. They were also more motivated to donate the money they received to preventing negative environmental events, such as climate change and oil spills. (Study participants were rewarded with a toonie [$2] or a loonie [$1], and were given options to donate the money toward different causes.)
“The message for teachers is that we need to use all possible ways to engage the public and get them interested in and caring about the issues of species extinctions and ecosystem destructions,” said Callahan. “Something as simple as a card game can be adapted to any environment, from classrooms to field-based workshops, in any location. Our study shows that this can be a really beneficial way of learning about species, and their ecosystems and environments.”
Researchers used a deck created for the Beaty Biodiversity Museum that focused on British Columbia’s ecosystems, but there are many other versions of the Phylo cards circulating the world. A global community of artists, institutions, scientists and game enthusiasts have created numerous iterations of the game—including decks featuring west coast marine life, dinosaurs, microbes, and even a Women in Science version created by Westcoast Women in Engineering, Science and Technology.
“We have 20 to 30 decks and more coming every year,” said Ng. “Games have a way of enticing anybody.”
All Phylo decks are open-source and can be downloaded for free from the Phylo website. The Beaty deck, used in the study, is also available at the Beaty Biodiversity Museum gift shop.
The study, “Using the Phylo Card Game to advance biodiversity conservation in an era of Pokémon” appears in Palgrave Communications.
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Stem cells are mother cells that have the potential to become any type of cell in the body. One of the main characteristics of stem cells is their ability to self-renew or multiply while maintaining the potential to develop into other types of cells. Stem cells can become cells of the blood, heart, bones, skin, muscles, brain etc. There are different sources of stem cells but all types of stem cells have the same capacity to develop into multiple types of cells.
Stem cells (center ones) can develop into any cell type. They are valuable as research tools and might, in the future, be used to treat a wide range of diseases. Credit: Judith Stoffer
Types of stem cells
Pluripotent Stem Cells (PS cells)
These possess the capacity to divide for long periods and retain their ability to make all cell types within the organism. The best known type of pluripotent stem cell is the one present in embryos that helps babies grow within the womb. These are termed embryonic stem cells. These cells form at the blastocyst stage of development. A blastocyst is a hollow ball of cells that is smaller than a pinhead. The embryonic stem cells lie within this ball of cells. Recent research has enabled scientists to derive pluripotent cells from adult human skin cells. These are termed induced pluripotent stem cells or iPS cells.
Fetal stem cells
These are obtained from tissues of a developing human fetus. These cells have some characteristics of the tissues they are taken from. For example, those taken from fetal muscles can make only muscle cells. These are also called progenitor cells.
Adult stem cells
These are obtained from some tissues of the adult body. The most commonly used example is the bone marrow. Bone marrow is a rich source of stem cells that can be used to treat some blood diseases and cancers.
Discovery of stem cells
Scientists first studied the potential of stem cells in mouse embryos over two decades ago. Over years of research they discovered the properties of these stem cells in 1998. They found methods to isolate stem cells from human embryos and grow the cells in the laboratory.
Early studies utilized embryos created for infertility purposes through in vitro fertilization procedures and when they were no longer needed for that purpose. The use required voluntary donation of the embryos by the owners.
Potential for use
Stem cell research is improving by leaps and bounds. These may soon become the basis for treating diseases such as Parkinson's disease, diabetes, heart failure, cerebral palsy, heart disease and host of other chronic ailments.
Stem cells may also be used for screening new drugs and toxins and understanding birth defects without subjecting human volunteers to the toxins and drugs.
Reviewed by April Cashin-Garbutt, BA Hons (Cantab)
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# Graphical Methods of Solving Pair of Linear Equations in Two Variables
A system of linear equations is just a pair of two lines that may or may not intersect. Graph of a linear equation is a line. There are a few methods that can be used to solve two linear equations, for example, Substitution Method, Elimination Method, etc. They can also be solved using graphical methods in which we draw the graph of the equations and try to find that intersection point. First, let’s see some linear equation and their graphs.
### Linear Equations and Graphs
For placing the two linear equations on graph, draw them on the graph and see where and if they intersect. The lines intersecting on the graph give intersecting points known as the solution of the linear equations. Let’s look at a pair of linear equations and their graphs from real-life world problems.
Question 1: Roman went to a stationery shop and purchased 2 pencils and 3 erasers for Rs 9. His friend Sonam saw the new variety of pencils and erasers with Roman, and she also bought 4 pencils and 6 erasers of the same kind for Rs 18. Represent this situation algebraically and graphically.
Solution:
Let’s say cost of one pencil is Rs.x and cost of one eraser be Rs. y. Then the equations can be formulated as,
2x + 3y = 9,
4x + 6y = 18,
Let’s plot both of these on graph,
By putting them in intercept form,
The graph of both equations looks as follows,
Both of these lines have same equations, and thus they coincide.
Question 2: A coach of the cricket team bought 3 bats and 6 balls for Rs 3900. Later, he bought another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Solution:
Let’s say that cost of each bat is Rs. “x” and Balls is Rs. “y”.
3x + 6y = 3900
x + 3y = 1300
Let’s make graph of these equations, it can be made through any of the forms that were taught, but we will stick to intercept form
Now the graphs of both equations come out to be,
We can see that these lines intersect at one point.
### Graphical Method of Solution
In this method, we graph both the equations and then try to find the intersections. Now in graphing and finding the possible intersections, there can be three cases that can occur
1. Unique solution
2. No Solution
3. Infinitely Many Solutions
### Unique Solution
A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. In such cases, both the lines intersect each other.
For example:
Question 1: Solve the given pair of equations with the graphical method.
x + 3y = 6
2x – 3y = 12
Solution:
We need to first plot both of them on to find the solution. Let’s bring them in intercept form,
Both of these equations can now be plotted on graph.
We can see on graph that both these lines intersect on (6,0) so this is the solution to this pair of linear equations.
Question 2: Ragini went to a shop to buy some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Ragini bought.
Solution:
Let’s say there are “x” number of pants and “y” number of Skirts. Then following equations are formed.
y = 2x – 2
y = 4x – 4
Let’s draw graphs for both the equations,
From graph, we can see that these lines intersect at (1,0). So this is the solution for this system of equations
x = 1 and y = 0
That means that Ragini brought 1 pant and 0 skirts.
### No Solution
A pair of linear equations in two variables, which has no solution, is called an inconsistent pair of linear equations. In such cases both the lines are parallel, so they never intersect.
For example:
Question: Draw the linear equations on the graph for the given two equations,
7x- 9y = 4
7x- 9y = 12
Solution:
The graph for the given two equations will look like,
As it is clear that the lines are not intersecting each other at any point, they are parallel in nature. Hence, there is no solution for the given linear equations.
### Infinitely Many Solutions
A pair of linear equations in two variables, which have infinitely many solutions, is called a dependent pair of linear equations. In such cases, both the lines coincide with each other.
For example:
Let’s generalize the above cases,
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
These equations are,
1. Intersecting if,
2. Coincident if,
3. Parallel if,
Question: Draw the linear equations on the graph, also find the solution for the two linear equations.
8x+13y = 17
16x+26y = 34
Solution:
The two equations are not only parallel, they are overlapping each other, hence all the points are intersecting each other. Therefore, There exist infinitely many solutions for the given linear equations.
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By Robert J. Gieb
Freedom and dignity for all men. The idea inspired the founding fathers in the creation of a new government. "We hold these truths to be self-evident, that all men are created equal, that they are endowed by their Creator with certain unalienable Rights, that among these are Life, Liberty and the pursuit of Happiness." The application of the idea moved slowly in the ante bellum years following the revolution. Then that idea, that grand truth of equality and the unalienable right to life and liberty for all human beings, lurched forward amidst the pain and agony of the Civil War. This nation may have been formed in a revolutionary war waged to fulfill its citizens' declaration of their right to self-government, but the very heart and spirit of these United States of America came to life in the war between the states, in that bloody and painful conflict to bring equality and liberty to all in this great republic.
The subsequent collapse and failure of federal reconstruction efforts in the south, and the ensuing years of segregation and racial injustice that included in too many times and places terror and death, slowed but never stopped the progress toward freedom. Pushed on by the passage of the 13th, 14th, and 15th Amendments to the Constitution, the idea of full human dignity for all Americans moved closer to legal reality.
After World War II the drive for freedom and liberty for all began to pick up momentum. The civil rights movement, then led by Martin Luther King, marched doggedly through the streets of the Deep South. The content of one's character, and not the color of his skin should be the measurement, said Dr. King. Every human being has inherent human dignity, and the right to be treated as a human person. Not unimportantly, the federal courts were finally sympathetic to Dr. King's message.
One hundred years after the Civil War, the Supreme Court was in a new frame of mind about race. In sharp contrast to Scott v. Sandford in 1857 and Plessy v. Ferguson in 1896, the Supreme Court in post war America, beginning with Brown v. Board of Education in 1954, was ready to begin to make the founders' talk of equality and freedom a practical reality for black as well as white. By the late 1960s the Court's project was running along at a full gallop.
During this time the politics of civil rights changed. While historically the Republican Party was the party of Lincoln and emancipation, and the Democratic Party that of slavery and racial injustice, as the 20th century unfolded the latter became the party of civil rights. The old tradition of civil rights advocacy seemed to have faded in the Grand Old Party. In contrast, the civil rights wing of the Democratic Party, made up of the racially oppressed, but also of white liberals and assorted left wing groups with agendas of deconstruction, became ascendant. By the 1960s the segregationist wing of the party was being crushed by liberal federal courts, its own party leadership, and, not least, by public opinion informed and inflamed by a sympathetic media who had the weapon of television.
The great freedom train that had pulled out of the station so many years before, and which had been slowed to a crawl so many times, was now rolling down the tracks toward equality for all. "Four score and seven years ago our fathers brought forth on this continent, a new nation, conceived in Liberty, and dedicated to the proposition that all men are created equal," Abraham Lincoln had said on that November day in1863 at Gettysburg. By nine score and eight years after 1776 the idea of liberty for all was finally a reality, even if a very imperfect reality.
Then in 2008 the crowning glory came to the old civil rights movement. The freedom train finally pulled into the station. Barrack Obama, a black man, was elected the 44th president of the United States. This man, a member of a race that until the Civil War was considered by law to be not fully human, was, eleven score and twelve years from the founding, elected to the highest elected office in the land.
In a message to Congress on December 1, 1862 Lincoln told the assembled legislators: "In giving freedom to the slave, we assure freedom to the free -- honorable alike in what we give, and what we preserve. We shall nobly save, or meanly lose, the last best, hope of earth." Certainly, on matters of race the election of Obama in 2008 is proof that Americans did indeed "assure freedom to the slave," and thus did "assure freedom to the free." But even so, have we nobly saved or meanly lost the battle for human dignity and the struggle for legal protection for all human beings? Sadly the battle is still on, and there is a reason for this.
As the old segregationist strongholds and legal impediments fell to the Civil Rights Act of 1964, the Voting Rights Act of 1965, and a continuing procession of decisions from the Supreme Court and lower federal courts striking down the remaining vestiges of segregation and legal racism, the high drama of marches and confrontations, and of sweeping social and legal changes, began to wane. The murder of Martin Luther King in April of 1968 left the civil rights movement without strong leadership.
But at that same time there was new talk of equality and civil rights: gender equality. The freedom train took on a pink tint as the civil rights movement enthusiastically embraced the feminist movement, and the radical individualism that underpinned it. The hard political left, both black and white, was in charge of this union, and still is now. On the race side, the old familiar goal of racial equality was replaced by the whiney cries of victimization, and the extortive demands for preferential treatment. The immediate goal on the feminist side was the Equal Rights Amendment to the federal constitution. But the real prize on which their eyes were firmly affixed was abortion. That pink tint would soon enough turn to a bloody red.
Unlike its history on matters of race where it had surely not moved "with all due and deliberate speed," on gender and sex the Supreme Court was already in action when the feminist movement began to hit its full stride in the 1970s. With its 1965 decision in Griswold v. Connecticut, declaring that state's prohibition of the sale of contraceptives to be an unconstitutional violation of individual privacy, the court had laid the legal groundwork, and was primed to address the issue of abortion. It wasted no time in declaring to America that a majority of the high court had seen a Constitutional vision. In January of 1973 seven justices (the first black justice being one of the seven) solemnly pronounced in Roe v. Wade that they now detected from the gaseous vapors of the Constitution the absolute right of any mother to kill her child so long as the killing was done while at least an infinitesimal portion of the child's tiny body remained within the body of the mother. In the newspeak of our age, civil rights became the civil right to murder the innocent.
The traditional civil rights movement in America, transformed by radical feminism and infected by radical individualism, has now come full circle. What began as an honorable movement to bring liberty and legal recognition of human personhood to all human beings in this nation, and which achieved so much over so many years, has now ended in the denial of the very truth for which the battle was fought in the first place. What began as a noble cause has ended in the horror of now 50,000,000 innocent babies being cut and torn to pieces, burned to death with chemicals, and murdered by stabbing their tiny skulls and sucking out their brains. There are 4,000 unborn children murdered each day in abortion mills in the United States, and 1000 of those are black babies. One out of every three black children conceived in this county is killed in abortion.
It must be one of the great ironies of all history. From black slavery to black presidency, but with a black president who denies that all of his citizens are human and entitled to legal protection of their human personhood. The man who is the crowning glory of the civil rights movement does not believe in civil rights for all human beings.
In 1855 Lincoln wrote to Kentucky lawyer George Robertson complaining that those who were "pro choice" on slavery had, in their greed for power and money, come to the point where they now denied the fundamental truth of American democracy:
On the question of liberty, as a principle, we are not what we have been. When we were the political slaves of King George, and wanted to be free, we called the maxim that "all men are created equal" a self evident truth; but now when we have grown fat, and have lost all dread of being slaves ourselves, we have become so greedy to be masters that we call the same maxim "a self evident lie."
So, in spite of the years of struggle for civil rights for all human beings, our government leaders and our laws have come full circle back to ante bellum notions of inequality. If the pro slavery forces of 1855, in their greed for power and money, twisted the idea of equality and human dignity for personal benefit and gain, President Obama and his fellow pro abortionists do no less today. The "slave" has risen to power not to free all people, but to be a "slaveholder" himself, who denies the greatest truth of our national covenant.
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Huey P. Newton’s revolution may not have been televised, but his inspiration and message were certainly felt around the world.
When Newton and Bobby Seale created the Black Panther Party in 1966, six young Pacific Islander men took notice and in 1971 created The Polynesian Panther Party. The Polynesian Panthers were a revolutionary social justice movement dedicated to stopping racial inequalities carried out against indigenous Māori and Pacific Islanders in New Zealand.
Pacific Islanders is a term used to describe the Indigenous peoples of Oceania (Melanesia, Micronesia, and Polynesia).
In the 1950’s New Zealand’s economy struggled, which led to an influx of Pacific Islanders moving to the island. These people of color drastically changed the make of Auckland, New Zealand–the city was no longer all white.
White supremacy in New Zealand began to rear its ugly head and Pacific Islanders were subjected to redlining, racial profiling, disproportionate incarceration, and segregation in the sports world.
Fred Schmidt, Nooroa Teavae, Paul Dapp, Vaughan Sanft, Eddie Williams, and Will ‘Ilolahia founded the Polynesian Panther Party on June 16th, 1971, Samoa News reports.
The group began by working with the community through activism, education, and legal aid. Their organization was grassroots to its core. One of their first acts as an organization was helping out with cooking and ticket collecting at the NZ University Students’ Association Arts Council Rock Festival.
The Panthers eventually organized homework and tutoring centers for Pacific children, created education programs centered around teaching indigenous people their rights as New Zealand citizens, and even created free meal programs, as well as food banks for hundreds of families.
Another important aspect of the Polynesian Panther Party was its stance on gender equality. The organization challenged gender norms and saw all sexes as equal. A woman could hold any position, job, status, or rank. The group also hosted gender equality workshops.
The Polynesian Panthers also organized a prison-visit program, which provided families the opportunity to visit their loved ones in prison. They sent folks to speak with the people behind bars who didn’t have a family to visit as well. They believed that even people in prison deserved to experience socialization and empathy.
They also created a Police Investigation Group Patrol, or PIG Patrol, which monitored police activity in the community. Since police would often patrol areas where Pacific Islanders would hang out, the Polynesian Panthers kept an eye on the police to make sure they didn’t harass the residents.
Although times have changed in New Zealand, the legacy left by The Polynesian Panthers still lives on.
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# Common Problem Solving Strategies 1. 2. 3. 4. 5.
Guess (this includes guess and check, guess and improve) Act It Out (act it out and use equipment) Draw (this includes drawing pictures and diagrams) Make a List (this includes making a table) Think (this includes using skills you know already)
We have provided black line masters for these strategies so that you can make posters and display them in your classroom. There are two kinds of these. The first is just a list of strategies. You might find this useful for you and your children to refer to from time to time. The second consists of a page per strategy with space provided to insert the name of any problem that you come across that uses that particular strategy (Act it out, Draw, Guess, Make a List). We have found that this kind of poster provides good revision for children. It also establishes links across curriculum areas. Through these links, children can see that mathematics is not only connected by skills but also by processes. An In-Depth Look At Strategies Back to Top
We now look at each of the following strategies and discuss them in some depth. You will see that each strategy we have in our list is really only a summary of two or more others. 1 Guess This stands for two strategies, guess and check and guess and improve. Guess and check is one of the simplest strategies. Anyone can guess an answer. If they can also check that the guess fits the conditions of the problem, then they have mastered guess and check. This is a strategy that would certainly work on the Farmyard problem but it could take a lot of time and a lot of computation. Because it is such a simple strategy to use, you may have difficulty weaning some children away from guess and check. If you are not careful, they may try to use it all the time. As problems get more difficult, other strategies become more important and more effective. However, sometimes when children are completely stuck, guessing and checking will provide a useful way to start and explore a problem. Hopefully that exploration will lead to a more efficient strategy and then to a solution. Guess and improve is slightly more sophisticated than guess and check. The idea is that you use your first incorrect guess to make an improved next guess. You can see it in action in the Farmyard problem. In relatively straightforward problems like that, it is often fairly easy to see how to improve the last guess. In some problems though, where there are more variables, it may not be clear at first which way to change the guessing. 2 Act It Out We put two strategies together here because they are closely related. These are Act it Out and Use Equipment. Young children especially, enjoy using Act it Out. Children themselves take the role of things in the problem. In theFarmyard problem, the children might take the role of the animals though it is unlikely that you would have 87 children in your class! But if there are not enough children you might be able to press gang the odd teddy or two. There are pros and cons for this strategy. It is an effective strategy for demonstration purposes in front of the whole class. On the other hand, it can also be cumbersome when used by groups, especially if a largish number of students is involved. We have, however, found it a useful strategy when students have had trouble coming to grips with a problem.
The on-looking children may be more interested in acting it out because other children are involved. Sometimes, though, the children acting out the problem may get less out of the exercise than the children watching. This is because the participants are so engrossed in the mechanics of what they are doing that they dont see through to the underlying mathematics. However, because these children are concentrating on what they are doing, they may in fact get more out of it and remember it longer than the others, so there are pros and cons here. Use Equipment is a strategy related to Act it Out. Generally speaking, any object that can be used in some way to represent the situation the children are trying to solve, is equipment. This includes children themselves, hence the link between Act it Out and Use Equipment. One of the difficulties with using equipment is keeping track of the solution. Actually the same thing is true for acting it out. The children need to be encouraged to keep track of their working as they manipulate the equipment. In our experience, children need to be encouraged and helped to use equipment. Many children seem to prefer to draw. This may be because it gives them a better representation of the problem in hand. Also, if theyre a little older, they may feel that using equipment is only 'for babies'. Since there are problems where using equipment is a better strategy than drawing, you should encourage childrens use of equipment by modelling its use yourself from time to time. 3 Draw It is fairly clear that a picture has to be used in the strategy Draw a Picture. But the picture need not be too elaborate. It should only contain enough detail to solve the problem. Hence a rough circle with two marks is quite sufficient for chickens and a blob plus four marks will do for pigs. There is no need for elaborate drawings showing beak, feathers, curly tails, etc., in full colour. Some children will need to be encouraged not to over-elaborate their drawings (and so have time to attempt the problem). But all children should be encouraged to use this strategy at some point because it helps children see the problem and it can develop into quite a sophisticated strategy later. Its hard to know where Drawing a Picture ends and Drawing a Diagram begins. You might think of a diagram as anything that you can draw which isnt a picture. But where do you draw the line between a picture and a diagram? As you can see with the chickens and pigs, discussed above, regular picture drawing develops into drawing a diagram. Venn diagrams and tree diagrams are particular types of diagrams that we use so often they have been given names in their own right. Its probably worth saying at this point that acting it out, drawing a picture, drawing a diagram, and using equipment, may just be disguises for guessing and checking or even guessing and improving. Just watch children use these strategies and see if this is indeed the case. 4 Make a list Making Organised Lists and Tables are two aspects of working systematically. Most children start off recording their problem solving efforts in a very haphazard way. Often there is a little calculation or whatever in this corner, and another one over there, and another one just here. It helps children to bring a logical and systematic development to their mathematics if they begin to organise things systematically as they go. This even applies to their explorations. There are a number of ways of using Make a Table. These range from tables of numbers to help solve problems like the Farmyard, to the sort of tables with ticks and crosses that are often used in logic problems. Tables can also be an efficient way of finding number patterns. When an Organised List is being used, it should be arranged in such a way that there is some natural order implicit in its construction. For example, shopping lists are generally not organised. They usually grow
haphazardly as you think of each item. A little thought might make them organised. Putting all the meat together, all the vegetables together, and all the drinks together, could do this for you. Even more organisation could be forced by putting all the meat items in alphabetical order, and so on. Someone we know lists the items on her list in the order that they appear on her route through the supermarket. 5 Think In many ways we are using this strategy category as a catch-all. This is partly because these strategies are not usually used on their own but in combination with other strategies. The strategies that we want to mention here are Being Systematic, Keeping Track, Looking For Patterns, Use Symmetry and Working Backwards and Use Known Skills. Being Systematic, Keeping Track, Looking For Patterns and Using Symmetry are different from the strategies we have talked about above in that they are over-arching strategies. In all problem solving, and indeed in all mathematics, you need to keep these strategies in mind.
GEORGE POLYA ..
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Question
# The positive square root of 7+√48 is
A
7+23
B
7+3
C
2+3
D
3+2
Solution
## The correct option is C 2+√3 7+√48=7+√4×12 =7+2√12 =4+3+2√4×3 =(√4)2+(√3)2+2×(√4×√3) =(√4+√3)2 ∴ Square root=√4+√3 =2+√3
Suggest corrections
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crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00232.warc.gz
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A new theory may solve a longstanding puzzle surrounding the formation of rain from clouds. In the 10 January print issue of PRL, a team suggests that raindrop formation is a kind of nucleation phenomenon–like bubbles forming in soda or sugar crystals in honey. The theory seems to explain one effect of pollution on rain clouds over land and may eventually lead to better weather and climate forecasts.
Traditionally, atmospheric scientists have described rain development in two steps: Droplets with radii smaller than about 20 microns grow through condensation, acquiring water molecules from water vapor in the surrounding cloud. At around 20 microns gravity begins to take over, and the droplets fall slowly and erratically, growing through so-called collection by accreting smaller droplets. But according to this theory, a droplet should take longer than an hour to grow from 10 microns to 100 microns–a full-sized drizzle drop–whereas a typical precipitating cloud lasts only thirty minutes. Atmospheric scientists believe that other factors, such as cloud turbulence, allow droplets to form more quickly, but the details are not fully understood.
Now Robert McGraw and Yangang Liu of Brookhaven National Laboratory in Upton, NY, have proposed an entirely new way to approach the problem: they use so-called statistical methods, which describe the behavior of a large collection of elements, such as molecules in a gas. They treat drizzle drop formation as a kind of nucleation, like the process that forms sugar crystals in honey. Random fluctuations in honey at the molecular level cause sugar molecules to continually form small clumps, but below a certain size, they can’t grow. Once a clump reaches the critical size by random fluctuations, it becomes energetically favorable to add additional molecules and grow.
McGraw and Liu apply this theory at the larger scale of drizzle drops. Droplets constantly grow through condensation and evaporate at random, but once a droplet reaches a critical radius it can begin to grow more quickly through collection. In the team’s model, the rate at which droplets cross the size threshold can be calculated based on the cloud’s turbulence and the concentration and size distribution of the droplets. According to McGraw, enough of the droplets would cross the size threshold quickly to account for drizzle’s fast formation.
McGraw says the work also helps explain one effect that airborne pollutant particles have on the environment. These particles increase the concentration of droplets in clouds, which according to the theory, increases the critical radius they need to attain before they can grow. So polluted clouds are more stable and less likely to produce rain. This prediction agrees with observations: clouds over land–where there is more pollution–have longer lifetimes than do maritime clouds.
Marcia Baker, of the University of Washington in Seattle, says that McGraw and Liu’s theory is an interesting new formulation of an older idea–that cloud turbulence and fluctuations in water vapor content allow drizzle to form more quickly than it otherwise would. But she cautions that researchers still need to test the theory by comparing it with aircraft and remote-sensing data on drizzle formation. To make the theory more useful in practice, the team also hopes to better account for cloud turbulence in future versions. “That might allow weather forecasters and climate modelers to improve their predictions,” McGraw says.
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A pelvis MRI (magnetic resonance imaging) scan is a imaging test that uses a machine with powerful magnets and radio waves to create pictures of the area between the hip bones. This part of the body is called the pelvic area.
Structures inside and near the pelvis include the bladder, prostate and other male reproductive organs, female reproductive organs, lymph nodes, and pelvic bones.
An MRI does not use radiation. Single MRI images are called slices. The images are stored on a computer or printed on film. One exam produces dozens or sometimes hundreds of images.
MRI - pelvis; MRI - hips; Pelvic MRI with prostate probe; Magnetic resonance imaging - pelvis
How the test is performed
You may be asked to wear a hospital gown or clothing without metal fasteners. Certain types of metal can cause inaccurate images.
You lie on your back on a narrow table. The table slides into the middle of the MRI machine.
Small devices, called coils, may be placed around your hip area. These devices help send and receive the radio waves, and improve the quality of the images. If pictures of the prostate and rectum are needed, a small coil will be placed into your rectum. This coil must stay in place for about 30 minutes while the images are taken.
Some exams require a special dye called contrast media. The dye is usually given before the test through a vein (IV) in your hand or forearm. The dye helps the radiologist see certain areas more clearly.
During the MRI, the person who operates the machine will watch you from another room. The test most often lasts 30 to 60 minutes, but may take longer.
How to prepare for the test
You may be asked not to eat or drink anything for 4 to 6 hours before the scan.
Tell your doctor if you are afraid of close spaces (have claustrophobia). You may be given a medicine to help you relax and be less anxious. Or your doctor may suggest an open MRI in which the machine is not as close to the body.
Before the test, tell your health care provider if you have:
Brain aneurysm clips
Artificial heart valves
Heart defibrillator or pacemaker
Inner ear (cochlear) implants
Kidney disease or dialysis (you may not be able to receive contrast)
Worked with sheet metal in the past (you may need tests to check for metal pieces in your eyes)
Because the MRI contains strong magnets, metal objects are not allowed into the room with the MRI scanner:
Pens, pocketknives, and eyeglasses may fly across the room.
Items such as jewelry, watches, credit cards, and hearing aids can be damaged.
Pins, hairpins, metal zippers, and similar metallic items can distort the images.
Removable dental work should be taken out just before the scan.
How the test will feel
An MRI exam causes no pain. If you have difficulty lying still or are very nervous, you may be given a medicine to relax you. Too much movement can blur MRI images and cause errors.
The table may be hard or cold, but you can request a blanket or pillow. The machine produces loud thumping and humming noises when turned on. You can wear ear plugs to help reduce the noise.
An intercom in the room allows you to speak to someone at any time. Some MRIs have televisions and special headphones that you can use to help the time pass.
There is no recovery time, unless you were given a medicine to relax. After an MRI scan, you can resume your normal diet, activity, and medications.
Why the test is performed
This test may be done if a female has any of the following signs or symptoms:
Abnormal vaginal bleeding
A mass in the pelvis (felt during a pelvic exam or seenon another imaging test)
A pelvic mass that occurs during pregnancy
Endometriosis (usually only done after ultrasound)
Pain in the lower belly (abdominal) area
Unexplained infertility (usually only done after ultrasound)
Unexplained pelvic pain (usually only done after ultrasound)
This test may be done if a male has any of the following signs or symptoms:
Lumps or swelling in the testicles or scrotum
Undescended testicle (unable to be seen using ultrasound)
Unexplained pelvic or lower abdominal pain
Unexplained urination problems, including trouble starting or stopping urinating
A pelvic MRI may be done in both males and females who have:
Abnormal findings on an x-ray of the pelvis
Birth defects of the hips
Injury or trauma to the hip area
Unexplained hip pain
A pelvic MRI is also frequently done to see if certain cancers have spread to other areas of the body. This is called staging. Staging helps guide future treatment and follow-up and gives you some idea of what to expect in the future. A pelvic MRI may be used to help stage cervical, uterine, bladder, rectal, prostate and testicular cancers.
A normal result means your pelvic area appears normal.
Abnormal results in both males and females may be due to:
Birth defects of the hip joint
Consult your health care provider with any questions and concerns.
What the risks are
MRI contains no radiation. To date, no side effects from the magnetic fields and radio waves have been reported.
The most common type of contrast (dye) used is gadolinium. It is very safe. Allergic reactions to the substance rarely occur. But gadolinium can be harmful to patients with kidney problems who require dialysis. If you have kidney problems, tell your health care provider before the test
The strong magnetic fields created during an MRI can cause heart pacemakers and other implants to not work as well. It can also cause a piece of metal inside your body to move or shift.
Tests that may be done instead of a pelvic MRI include:
A CT scan may be done in emergency cases, since it is faster and usually available right in the emergency room.
Childs DC, Dalrymple NC. Female reproductive system. In: Dalrymple NC, Leyendecker JR, Oliphant M. Problem Solving in Abdominal Imaging. Philadelphia, PA: Elsevier Mosby; 2009:chap 21.
Dalrymple, NC. Ureters, bladder, and urethra. In: Dalrymple NC, Leyendecker JR, Oliphant M. Problem Solving in Abdominal Imaging. Philadelphia, PA: Elsevier Mosby; 2009:chap 19.
Gjelsteen AC. CT, MRI, PET, PET/CT, and ultrasound in the evaluation of obstetric and gynecologic patients. Surg Clin North Am. April 2008; 88(2): 361-90, vii.
Wilkinson ID, Paley MNJ. Magnetic resonance imaging: basic principles. In: Grainger RC, Allison D, Adam, Dixon AK, eds. Diagnostic Radiology: A Textbook of Medical Imaging. 5th ed. New York, NY: Churchill Livingstone; 2008:chap 5.
Zarka AI, Jung AJ, Dalrymple NC. Male reproductive system. In: Dalrymple NC, Leyendecker JR, Oliphant M. Problem Solving in Abdominal Imaging. Philadelphia, PA: Elsevier Mosby; 2009:chap 20.
Ken Levin, MD, private practice specializing in Radiology and Nuclear Medicine, Allentown, PA. Review provided by VeriMed Healthcare Network. Also reviewed by A.D.A.M. Health Solutions, Ebix, Inc., Editorial Team: David Zieve, MD, MHA, Bethanne Black, Stephanie Slon, and Nissi Wang.
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If 10 be added to four times a certain number,
Question:
If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.
Solution:
Let the number be $\mathrm{x}$.
$\therefore 10+4 x=5 x-5$
$\Rightarrow 10+5=5 x-4 x$
$\Rightarrow 15=x$
$\Rightarrow x=15$ (by transposition)
Therefore, the number is 15 .
|
crawl-data/CC-MAIN-2022-27/segments/1656103617931.31/warc/CC-MAIN-20220628203615-20220628233615-00530.warc.gz
| null |
# chapter2
```Ch. 2- Complex Numbers>Background
Chapter 2: Complex Numbers
I. Background
•
Complex number (z):
z = A + iB where
i
1
(note: engineers use
“real part”: Re{z} = A
“imaginary part”: Im{z} = B
(both A and B are real numbers)
•
Graphical Representation:
ex: 2+3i
3-4i
j
1
)
Ch. 2- Complex Numbers>Background
•
Polar vs. rectangular coordinates:
Rectangular coordinates: z = A + iB
Polar coordinates: z = reiθ
(note: |z| ≡ r , z = θ)
To move between the two:
A + iB = (r cosθ) + i (r sinθ) = reiθ
because eiθ = cosθ + sinθ (see section 7-9 for proof)
so:
2
2
A r cos
r A B
or
1 B
B
r
sin
tan A
θ is in radians, r ≥ 0
ex: z=3-i4 (write in polar coordinates)
Ch. 2- Complex Numbers>Background
•
Beware when using tan
1
( BA )
tan
1
( BA ) tan
1
( BA )
difference between θ and θ+π.
It should always find 2 0 2 .
You may need to add π to get to
ex: z = -3 + i4
In practice, polar form reiθ is often easier to use because it’s easier to
differentiate and integrate reiθ.
Ch. 2- Complex Numbers>Algebra>Complex Conjugate
II. Algebra with complex numbers
1) Complex conjugate:
Let z = A + iB
Then z (or z*) A iB is the complex conjugate.
In polar coordinates: z = reiθ z* = re-iθ
Why? Recall: z = reiθ = r (cosθ + isinθ)
z* = r (cosθ - isinθ)
= r (cos(-θ) + isin(-θ))
= re-iθ
ex:
1) z = 2 + i3
2) z = 3 – i4
3) z = 5ei(/3)
Note: z z* = |z|2 (Proof in a minute…)
For the remaining examples:
Let z1 = 2 + i3 r = 3.6, (θ) = 0.98 rad z1 = 3.6ei(0.98)
z2 = 4 – i2 r = 4.5, (θ) = -0.46 rad z1 = 4.5ei(-0.46)
(A + iB) + (C + iD) = (A+C) + i(B+D)
i
i
there is no easy way to do this in polar coordinates: r1 e r2 e ?
1
ex: z1 + z2 (in rectangular coordinates)
ex: z1 + z2 (in polar coordinates)
2
Ch. 2- Complex Numbers>Algebra>Multiplication
3) Multiplication: (A + iB) (C + iD) = AC + i(AB) + i(BC) – BD (term by term)
( r1e
ex: z1•z2 (rectangular)
ex: z1•z2 (polar)
i 1
)( r2 e
i 2
) ( r1 r2 ) e
i ( 1 2 )
Ch. 2- Complex Numbers>Algebra>Multiplying by Complex Conjugate
4) Multiplying by complex conjugate:
ex: z1z1*
|z1|2
This is true for any complex number: |z1|2 = z1z1*
Proof: Let z = A + iB
z z*= (A + iB) (A – iB)
= A2 – iAB + iBA + B2
= A2 + B2
= r2
= |z|2
Ch. 2- Complex Numbers>Algebra>Division
5) Division:
r1 e
r2 e
i 1
i 2
r1
r2
e
i ( 1 2 )
(more difficult in rectangular form)
ex:
ex:
z1
z2
z1
z2
(polar)
(rectangular)
Ch. 2- Complex Numbers>Algebra>More Complex Conjugates & Complex Equations
6) More complex conjugates:
Say I want the complex conjugate of a messy equation:
2 i 3 x ix
2
z3
9 x ix
2
3
4
Change all i -i
2 i 3 x ix
2
z3*
9 x ix
2
3
4
7) Complex equations: (A + iB) = (C + iD)
ex: z1 + z2 = x + i(3x + y)
iff A = C and B = D
Ch. 2- Complex Numbers>Algebra>Powers
8) Powers: do these in polar form: ( r1 e
i 1
n
) r1 e
n
rectangular form switch to polar first
ex: z12 (polar)
ex: z12 (rectangular)
in 1
9) Roots: Polar coordinates:
ex:
( re
i
1
n
1
n
) r e
i ( n )
z1
Check by changing back to rectangular coordinates.
Find another root (add 2 to ).
Convert back to rectangular coordinates:
Ch. 2- Complex Numbers>Algebra>Roots
In general: z
1
n
ex: z 4
ex: z 27
Ch. 2- Complex Numbers>Algebra>Roots
has n possible roots!
z 2
find
3
z
Ch. 2- Complex Numbers>Algebra>Complex Exponentials
10) Complex Exponentials:
let z = x + iy
then ez = ex+iy = exeiy = ex(cosy + isiny)
ex:
e
e
e
e
3 i
3 i 2
i
i 2
Ch. 2- Complex Numbers>Algebra>Trig Functions
11) Trig Functions:
e
e
e
e
(e
e
i
cos i sin
i
i
cos i sin
e
i
2 cos
cos
i
e
i
2
cos i sin
i
i
i
e
cos i sin )
e
i
2 i sin
sin
e
i
e
2i
i
Ch. 2- Complex Numbers>Algebra>Trig Functions
This is very useful for derivatives and integrals:
ex:
d
dz
(cos z )
Good Trick:
1
i
i
i
1
i
ex:
cos( 2 x ) cos( 3 x ) dx
Ch. 2- Complex Numbers>Algebra>Hyperbolic Functions
12) Hyperbolic Functions:
iz
iz
sin z
e e
2i
cos z
e e
2
iz
iz
Usual sin/cos functions
Likewise: tanh z
sinh z
cosh z
sech z
1
cosh z
etc…
↔
sinh z
cosh z
z
e e
2
z
z
e e
2
z
Hyperbolic functions
(entirely real)
Ch. 2- Complex Numbers>Algebra>Natural Logarithm
13) Natural logarithm:
i
i
ln z ln( re ) ln( r ) ln( e ) ln( r ) i
Note: i = i(+2) = i(+4) = …
So ln(z) has infinitely many solutions:
ln(z) = ln(r) + i = ln(r) + i(+2) = …
‘Principal solution’ has 0 , 2 .
Ch. 2- Complex Numbers>Example: RLC Circuit
Physics Example: RLC Circuit
applied emf
V V sin ωt
Then I I sin (ωω )
Find I0 & Φ and Z (the complex impedance).
From Physics 216:
V R IR RI 0 sin( t )
VL L
dI
dt
VC
1
C
LI 0 cos( t )
Idt
1
C
I 0 cos( t )
What is the impedance?
Z
V
I
V R VC V L
I
(yuck!)
Ch. 2- Complex Numbers>Example: RLC Circuit
It’s easier to do this:
V V0e
i t
Ch. 2- Complex Numbers>Example: RLC Circuit
What’s the physically real I?
i ( t )
I I 0e
where
V0
I0
R ( L
2
tan
1
1
C
1
C
)
2
L
R
Recall our actual driving voltage:
We wrote this as V 0 e
Only Im V 0 e
i t
i t
.
has any physical reality.
So, same goes for current:
The physically real current is Im I 0 e
i ( t )
I
0
sin( t ) with same I o & Φ as above.
Ch. 2- Complex Numbers>Example: RLC Circuit
Resonance
Defn: The frequency at which Z is entirely real:
Z R i L
1
C
So, at resonance ωR:
RL
1
RC
0 R
1
LC
(resonance frequency)
Note that for this circuit, I0 is max when ω=ωR:
I0
V0
R ( L
2
See plot of I0 vs. ω.
1
C
)
2
V0
R
(at resonance)
Ch. 2- Complex Numbers>Example: RLC Circuit
Complex Impedances
We found:
V R RI
XR R
V L i LI V L X L I
X L i L
VC
1
i C
I VC X C I
where
XC
1
i C
“reactance” or
“complex impedance”
Complex impedances behave just like resistors in series and parallel.
Ch. 2- Complex Numbers>Example: RLC Circuit
From Physics 216:
Wave : Asin( ω - kx φ)
Wavelength
: k
Period : ω
2π
T
2π
λ
λ
T
2π
k
2π
ω
Phase : φ
Velocity : v
Amplitude
ω
k
(to the right)
: A
Instead, it’s easier to use complex notation: Ae
i( ω( kx φ)
```
21 cards
19 cards
15 cards
14 cards
31 cards
|
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Throughout history, law has served as a mediator of the relationships between people. It is a set of rules that are enforceable through governmental and social institutions. In common use, law refers to the system of rules that govern action and procedure. The term also encompasses the people who work within the legal system.
There are three main types of legal systems: civil law, common law and international law. Each has its own unique characteristics, but they share several key characteristics. In general, civil law systems are less detailed than common law systems and require less judicial decision making. Common law systems are more formal and include explicit acknowledgment of decisions by the executive and legislative branches of government.
International law relates to matters of human rights and the international relations of nations. It is addressed in many multilateral treaties, which are signed by governments and deposited with the United Nations Secretary-General. In addition, many treaties are opened for ratification by the General Assembly. The International Law Commission promotes the progressive development of international law and the codification of international law.
Laws can be created by a single legislator, a group of legislators or an executive through decrees and other forms of state-enforced legislation. In addition, a nation’s constitution can influence the creation of laws. In countries that have a federal government, the creation of laws is governed by the federal government.
Laws can also be created by private individuals, through contract and arbitration agreements. In the United States, the US Uniform Commercial Code (UCC) and the antitrust statutes of the turn of the twentieth century have been used to codify the principles of common law commercial law.
In most countries, laws pertaining to business, property, and tax are regulated. There are also regulations governing utilities such as water and gas. Likewise, there are regulations on the provision of public services such as telecommunications.
Laws can also cover topics like personal, national, and foreign rights. Some of these topics include: personal property (movable objects and intangible rights), commercial law (rights to property), and immigration law (rights of foreigners to live and work in a nation-state).
The International Court of Justice, also known as the World Court, is the main United Nations dispute settlement organ. It was established in 1946 and has issued numerous judgments. It has also issued advisory opinions. Its current members include 34 of the world’s most important legal systems.
Law has shaped history, politics, and economics. It can be used to protect individuals and minorities against majorities, maintain the status quo, and to promote social justice. Some legal systems serve these purposes better than others. Law can also help maintain orderly social change and keep peace within a nation.
Law has also been described as the art of justice. In the United States, the legal profession plays an important role in ensuring that individuals have access to justice. Modern lawyers must earn a Bachelor of Laws or Master of Legal Studies and pass a qualifying examination before practicing law.
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English
How to Calculate The Probability of Combination
Tutorial on how to solve combination problems with examples.
Combination Formula
When order doesnt matter in arranging things, then its called as combination. Lets learn how to calculate the probability of combination problem with example using formula given below.
C(n,r)=
n! / r!(n−r)!
"!" - Symbol of Factorial
n - number of combinations
Steps To Follow
Step 1: Consider picking 3 people randomly from the group of 8 for playing an outdoor game. Find the possible number of combinations.
Step 2: Here you can choose people in any order, thus its a combination problem.
Substitute the given values in the combination formula.
n = 8
r = 3
C(n,r)= n! / r!(n−r)!
C(8, 3) = 8! / 3!(8-3)!
= 8! / 3!(5)!
=8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 (5 x 4 x 3 x 2 x 1)
=40320/6 (120)
=
40320/ 720
=56
The possible combinations are 56.
|
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Climate change models used to analyze changing conditions on Earth can be applied to at least one other planet in our solar system — Mars, a new study reveals.
Earth-based computer models can predict the age and location of glacial features on the Red Planet, scientists say. This extra-planetary generalization suggests that climate change models aren't so as inaccurate as they're sometimes portrayed, said study researcher William Hartmann, a senior scientist at the Planetary Science Institute in Tucson.
"Some public figures imply that modeling of global climate is 'junk science,' but if climate models can explain features observed on other planets, then the models must have at least some validity," Hartmann said in a statement.
Hartmann and his colleagues presented the new Mars climate research today (Oct. 16) at the annual meeting of the American Astronomical Society's Division for Planetary Sciences in Reno, Nev.
Today's Martian landscape is not known for its glaciers, but previous research has shown that during some periods of Mars' history, the planet's rotational axis has tilted more than 45 degrees. When this happens, climate models suggest that the polar ice cap tilted toward the sun evaporates, lingering in the atmosphere as water vapor and increasing the chances of snowfall in the hemisphere tilted away from the sun.
Planetary researchers have since used global climate models built for Earth but tweaked for Martian conditions to study Mars' glacial history. They've found indications of glacial features in a 40-mile-wide (64 kilometers) crater called "Greg" in the mid-latitude southern region of the planet. The surface layer of those features would have formed at the same time as the last extreme tilt, PSI scientists found. [ Photos of Mars: The Amazing Red Planet ]
Now, the researchers have combined four different measurements of Martian geology and climate, including climate predictions, the presence of glaciers, the ages of glacial surface layers and radar measurements of ice. The measurements and models are all consistent with one another, the researchers found.
"The bottom line is that the global climate models indicate that the last few intense deposits of ice occurred about 5 million to 15 million years ago, virtually centered on Greg crater, and that's just where the spacecraft data reveal glaciers whose surface layers date from that time," Hartmann said. "If global climate models indicate specific concentration of ice-rich features where and when we actually see them on a distant planet, then climate modeling should not be sarcastically dismissed. Our results provide an important, teachable refutation of the attacks on climate science on our home planet."
- The Reality of Climate Change: 10 Myths Busted
- Water Rushed On Mars, Curiosity Finds | Video
- The Search for Life on Mars (A Photo Timeline)
© 2013 Space.com. All rights reserved. More from Space.com.
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Native species are those that occur in the region in which they evolved. Plants evolve over geologic time in response to physical and biotic processes characteristic of a region: the climate, soils, timing of rainfall, drought, and frost; and interactions with the other species inhabiting the local community. Thus native plants possess certain traits that make them uniquely adapted to local conditions, providing a practical and ecologically valuable alternative for landscaping, conservation and restoration projects, and as livestock forage. In addition, native plants can match the finest cultivated plants in beauty, while often surpassing non-natives in ruggedness and resistance to drought, insects and disease.
The native plants list identifies native plant species currently recommended by the Virginia Native Plant Society, the Virginia Department of Conservation and Recreation, and other project cooperators for use in horticulture, land management, conservation, and restoration projects in Virginia. The list provides a selection of plant choices adapted to growing conditions in Virginia, focusing on those native species currently or potentially available in the nursery trade. For the most part, relatively common species have been included on the list, although a few less common species were also included due to their establishment in the trade and the general stability of their habitat in Virginia. Rare species were left off the list in order to protect the genetic integrity of naturally occurring populations of rare species and avoid the collection of rare plants.back to top of page
The use of native plants is on the rise across the country as more people discover their many benefits. An ever-widening selection of vigorous, nursery-propagated native plants is available from specialty growers and many larger nurseries as a result of this increased demand. This offers a much-needed alternative to wild collection or the purchase of wild-collected plants. Wild-collection threatens the existence of native species by causing net losses in population size and genetic diversity, and leaves the collector or purchaser with highly stressed plants that have a decreased likelihood of survival. These problems are multiplied when the plants are collected from a distant source population, then planted in a new location with different environmental conditions.
We encourage land managers, conservation professionals, restoration specialists, landscape designers, and private individuals to utilize local growers and nurseries that offer nursery-propagated native species, especially plants propagated from local populations. Once a good source of native plants has been located, the next step is choosing appropriate plants for a project. One of the greatest benefits of designing with native plants is their adaptation to local conditions. But it is important to select plants with growth requirements that best match the conditions in the area to be planted.
When planning projects utilizing native plant species, you can start with this list for information on which plants grow in each of the three major regions of Virginia. These three regions include: the Coastal Plain, from the Atlantic Ocean in the east to the Fall Line in the west; the Piedmont, from the Fall Line in the east to the Blue Ridge Mountains in the west; and the Mountains, including the Blue Ridge Mountain, the Ridge and Valley, and the Appalachian Plateau Provinces. Next study the minimum light and moisture requirements for each species, noting that some plants will grow under a variety of conditions. Recommended uses include wildlife benefits, horticulture and landscaping, conservation and restoration, or domestic livestock forage. Of course, many of these species are well suited to more than one of these categories.
For more information, refer to field guides and publications on local natural history for color, shape, height, bloom times, and specific wildlife value of the plants that grow in your region. For help in designing native plantings with combinations of species that would occur together naturally, visit a local natural area and observe common plant associations, spatial groupings, and habitat conditions. For specific recommendations and advice about project design, consult with a natural areas steward or a natural resource specialist.back to top of page
In North America, plant species are generally described as native if they occurred here prior to European settlement. This distinction is made because of the large-scale changes that have occurred since the arrival of the European settlers. The Europeans imported a variety of plants to this country, many are still the major component of traditional lawns and gardens. They also include many beneficial plants important in farming, such as vegetables and grains. Today, approximately 25% of flowering plants in North America are non-natives or alien species, most of Eurasian origin.
All plants are native to some region, and offer a variety of ecological, economic and aesthetic benefits. In fact, many alien species are beneficial, providing food and other valuable resources to society. It is only when a species is "out of place" that we become concerned. In these instances, invasive alien plants can pose a serious threat to biodiversity. Due to a lack of natural controls such as insect pests and competitors, some alien plants can easily become established in new areas. Once established, alien plant species can out compete and displace the native plant species, disrupting ecological processes and significantly degrading entire plant communities. Many invasive plants spread quickly and grow so densely that other species cannot get established in areas infested by these alien plant species. Common native plants can be crowded out, or their populations threatened due to hybridization with escaped ornamentals. Endangered species may be driven from their last habitats by invasive alien plant species. Aquatic invasive species can clog waterways, disrupt groundwater flows, degrade water quality, and lead to dramatic changes in native plant and animal communities.
Although a majority of invasive alien species come from other countries, they may also be introduced from a different region of the same country. For example, saltmarsh cordgrass is a dominant and important member of coastal saltmarsh communities along the Atlantic and Gulf Coasts. But when used for restoration projects in west coast marshes, it became invasive, out competing and replacing western species.
In contrast to invasive alien species, other non-native plants are unable to thrive without extra effort by land managers. For instance, they may originate in regions with abundant rainfall and soils rich in nutrients. If then introduced into a drier region with less fertile soils, they may require additional watering and fertilizer. The natural defenses plants evolve in their original habitats may not protect them in a new environment, requiring the application of pesticides to aid their growth.back to top of page
The benefit of growing plants within the region they evolved is they are more likely to thrive under the local conditions while being less likely to invade new habitats. Native plants are well adapted to local environmental conditions, maintain or improve soil fertility, reduce erosion, and often require less fertilizer and pesticides than many alien plants. These characteristics save time and money and reduce the amount of harmful run-off threatening the aquatic resources of our streams, rivers, and estuaries. In addition, functionally healthy and established natural communities are better able to resist invasions by alien plant species. So the use of native plants can help prevent the spread of alien species already present in a region and help avert future introductions. With the large variety of grasses, ferns, wildflowers, shrubs and trees from which to choose, native plants can fulfill any landscaping need, from simple container gardens to showy perennial borders to expansive public lawns and gardens.
Native plants provide familiar sources of food and shelter for wildlife. As natural habitats are replaced by urban and suburban development, the use of native plants in landscaping can provide essential shelter for displaced wildlife. Land managers can use native plants to maintain and restore wildlife habitat. Native wildlife species comprise a majority of the game and non-game animals we manage habitat for, and they evolved with native plant species. Although alien species are often promoted for their value as wildlife food plants, there is no evidence that alien plant materials are superior to native plants. For instance, on land managed for upland game animals, native warm season grasses (big and little bluestem, switch grass, Indian grass, coastal panic grass, gama grass), and other native forbs (butterfly weed, ironweed, Joe Pye weed) offer good sources of nutrition without the ecological threats associated with nonnative forage plants. Dramatic increases in nesting success of both game birds and songbirds have been observed in fields planted with native grasses, which also offer superior winter cover. In addition, warm season grasses provide productive and palatable livestock forage. (For more information on native warm season grasses contact the Virginia Department of Game and Inland Fisheries for the publication "Native Warm Season Grasses for Virginia and North Carolina: Benefits for Livestock and Wildlife.")
On a broader ecological scale, planting native species contributes to the overall health of natural communities. Disturbances of intact ecosystems that open and fragment habitat, such as land clearing activities, increase the potential of invasion by alien species. Native plants provide important alternatives to alien species for conservation and restoration projects in these disturbed areas. They can fill many land management needs currently occupied by nonnative species, and often with lower costs and maintenance requirements. Once established in an appropriate area, most native plant species are hardy and do not require watering, fertilizers, or pesticides.
In addition to ecological and land management benefits, the native flora of Virginia offers a surprising variety of color, form, and texture to gardeners and landscape designers. In fact, many familiar and popular landscaping plants such as black-eyed Susan, columbine, and bee balm are native to Virginia. Designing with natives allows the creation of distinctive natural landscapes including woodlands, meadows, and wetlands with unique regional character. In addition, native plants attract a greater variety of butterflies, hummingbirds, songbirds and other wildlife than traditional lawns. In fact, the greater the variety of native species included in a landscape, the more likely uncommon or rare species will be attracted to an area.back to top of page
Virginia is divided into several physiographic provinces based on their geologic history. Each province is unique in topography, soil pH, soil depth, elevation, availability of light, and hydrology. These characteristics all combine to influence the species of plants and animals found there. Virginia is unique, encompassing parts of five of these provinces, and thus a greater variety of natural landscapes than any other eastern state. For the purposes of this list, we have grouped the physiographic provinces into three regions: Coastal Plain, Piedmont, and Mountain.
Virginia's Coastal Plain is bordered by the Fall Line to the west and by the Atlantic Ocean, the Chesapeake Bay and its tributaries to the east. This is the youngest of the physiographic provinces, formed by sediments eroded from the Appalachian Highlands and deposited along the Atlantic shoreline. The Coastal Plain varies in topography from north to south. The northern Coastal Plain consists of the three peninsulas formed between the four major tributaries of the Chesapeake Bay; the Potomac, the Rappahannock, the York, and the James Rivers. In the north, the Northern Neck is somewhat hilly and well drained. As you move southward across the Middle Peninsula and Lower Peninsula the topography flattens until south of the James River the landscape is basically level. The Eastern Shore, separated from the mainland by the Chesapeake Bay, exhibits little topographic relief. These subtle differences in topography and the variety of fresh, brackish, and saltwater systems from ocean and inland bay to rivers, ponds, and bogs, have contributed to the great variety of natural communities found on the Coastal Plain.
Virginia's Piedmont Plateau province is a gently rolling upland bounded on the east by the Fall line and the west by the Blue Ridge Mountains. The western boundary of the Piedmont is characterized by distinct peaks and ridges, comprising the foothills of the Blue Ridge Mountains. To the east, the Piedmont continues to slope more gently toward the Fall Line. The Fall Line marks the zone of transition from the hard, resistant bedrock underlying the Piedmont to the softer sediments underlying the Coastal Plain. Streams are able to cut more easily through the sands, gravels, and clays of the Coastal Plain, and rivers widen as the topography flattens. In the northern part of the state this boundary is sharply delineated by falls and rapids. From foothills to rapids, these varying site conditions support a mosaic of plant communities.
The Mountain region of Virginia actually includes parts of three provinces; the Blue Ridge, the Ridge and Valley, and the Appalachian Plateau Physiographic Provinces. The Blue Ridge encompasses the Blue Ridge Mountains, a wedge of ancient rock that was uplifted over younger rocks when the Appalachian Mountains were formed. A narrow system of peaks in the north, the Blue Ridge widens south of Roanoke Gap into a broad plateau topped by the highest peaks in Virginia--Mount Rogers and Whitetop. The Ridge and Valley Province is characterized by long, even-crested, parallel ridges rising above intervening valleys of various size. The Valley of Virginia is included in this province, encompassing the large Shenandoah Valley, as well as the James, Roanoke, New River and the Clinch, Powell and Holston River valleys. The ridges of the Appalachian Plateau in far southwestern Virginia were not as folded and faulted as those of the Ridge and Valley, but formed from a high, unified plateau of nearly horizontal rock layers. The modern mountainous topography was created by streams cutting deeply through the plateau, forming an intricate network of narrow, steep valleys. The diversity in topography and geologic history of the Mountain region of Virginia gives rise to a rich array of natural communities and native species.back to top of page
This project is the result of a collaboration between the Virginia Department of Conservation and Recreation and the Virginia Native Plant Society, and was made possible by a grant from the National Fish and Wildlife Foundation and the Department of Environmental Quality's VA Coastal Program. Funds were also contributed by the Virginia Nurserymen's Association, the Virginia Chapter of the American Society of Landscape Architects, and the Lewis Ginter Botanical Garden.
In addition to those three organizations, the sponsors extend their considerable appreciation to the other collaborators who provided valuable advice and assistance throughout the life of the project:
Project participants share a commitment to protect native plant habitats, especially those that support rare, threatened, or endangered species. The use of native plant species--especially plants propagated from local populations--in land management, conservation, restoration, and horticultural projects will help maintain the ecological integrity of natural areas and preserve native biodiversity.
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Huge tornadoIn the United States have more tornadoes than any other country: about four times the estimated occurring in Europe, excluding waterspouts. This is mainly due to the unique geography of the Americas. North America is relatively large and extends from the tropics to the Arctic areas, and has no major mountain chain that goes from east to west and block air flow between these two areas. In the mid latitudes, where most of the tornadoes, the Rocky Mountains block moisture and atmospheric flow, allowing drier air exists at intermediate levels of the troposphere, and causing the formation of a low pressure area to east of those mountains. An increase in the flow of air from the Rockies favors the formation of a dry line if the flow is strong at higher levels while the Gulf of Mexico, east, provides abundant low level moisture from the atmosphere. This unique topography causes many collisions of warm air with cold air, which are the conditions that create strong and lasting storms. A large portion of these tornadoes form in that area in the center of the United States between the Rockies and the Gulf, known as Tornado Alley ("tornado alley"). This area also includes parts of Canada, mainly in Ontario and the Canadian prairies, although southeastern Quebec, the interior of British Columbia and western New Brunswick are also prone to tornadoes. Sometimes tornadoes are also strong in northeastern Mexico.
On average, in the United States about 1,200 tornadoes occur each year. The Netherlands reported the highest number of tornadoes per area of any country to register there more than 20 tornadoes, which equals 0.00048 km2 tornadoes per year, followed by the United Kingdom presented annually about 33, ie 0, 00,013 per km2; However, most are small and cause little damage. In absolute numbers, regardless of territorial extension, the UK experiences more tornadoes than any European country, excluding waterspouts.
Tornadoes kill an average of 179 people per year in Bangladesh, by far the largest amount in a country in the world. This is due to its high population density, poor quality of buildings, lack of knowledge about security measures to combat the tornadoes and other factors. Other countries in the world with frequent tornadoes include South Africa, Argentina, Uruguay, southern Brazil, Australia and New Zealand as well as portions of Europe and Asia.
Tornadoes are most common during the spring and less during the winter. Since the spring and autumn are transitional periods (warm to cold weather and vice versa) is more likely that the cold air is warm air , which causes for these stations are experienced peaks. However, the conditions for its formation can occur at any time of year. Tornadoes can also be generated from the eye of tropical cyclone landfall, which often happens in the fall and late summer.
The incidence of tornadoes is highly dependent on time of day, due to solar radiation. Worldwide, most tornadoes occur during the afternoon, between 3:00 pm and 7:00 pm time local, being the highest point at 5:00 pm. However, destructive tornadoes can occur at any time of day. The Gainesville tornado of 1936, one of the most devastating tornadoes in history, occurred at 8:30 am local time.
There Jola Mediterranean) in turn increases the amount of moisture in the atmosphere. The increased moisture can cause an increase in the occurrence of tornadoes, especially during the cold season.
Some evidence suggests that the phenomenon Southern Oscillation (ENSO, for its acronym in English) is slightly related to changes in the activity of tornadoes, this varies by season and region and depending on whether the phenomenon ENSO corresponds to that of El Niño or La Niña.
Climate change may affect tornadoes through teleconnections as when changing a jet stream and other major weather patterns. Although it is possible that global warming might affect the activity of the tornadoes, this effect can not yet be identifiable because of its complexity, the nature of storms and issues relating to the quality of databases. In addition, any effect would vary by region.
10 dni, Pogoda 16 dni, webtelly.org, 16 dni, Weather Travel
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ST. PAUL, MN (May 4, 1999) -- Six successive years of disease have taken their toll on many small grain farmers in the Red River Valley of North Dakota, Minnesota, and Manitoba. This extended episode is bringing ruin to many farmers in the region. Additional outbreaks in Midwestern and Eastern states of the USA as well as Central and Eastern Canada are leaving thousands of farmers searching for solutions. The culprit responsible for the vast devastation is Fusarium head blight, more commonly known as scab. This all consuming fungal disease shrivels the kernels of small grains such as wheat, rye and barley, significantly reducing yields.
"Moisture, at the time of flowering, is the main stimulus necessary for scab," says Robert W. Stack, plant pathologist at North Dakota State University and a member of The American Phytopathological Society. "If a wet environment exists for an extended period, even with low levels of the fungus in the field or temperatures that aren't usually favorable to disease development, severe scab disease can result."
During the first part of this century, scab was considered a major threat to wheat and barley and recently it has resurfaced worldwide increasing in intensity. A succession of "wet cycle" years beginning in 1993 are linked to the current scab epidemic. According to the United States Department of Agriculture (USDA), "From 1991 to 1997, American farmers lost 470 million bushels of wheat, worth $2.6 billion, because of the scab epidemic." These substantial losses recently provoked a national response resulting in the development of the "US Wheat and Barley Scab Initiative," a consortium of scientists and agribusiness leaders working together to solve the scab epidemic.
"Researchers are on the threshold of major breakthroughs using new methods and technologies to solve this disease problem," says Stack. "Breeding for disease resistance is underway worldwide and soon new cultivars with increased resistance to scab and developed by conventional breeding methods will be available."
For more information on scab, visit the APS May web feature story with photographs and links to additional sites at http://www.scisoc.org. The American Phytopathological Society (APS) is a professional scientific organization dedicated to the study and control of plant disease with 5,000 members worldwide.
The above post is reprinted from materials provided by American Phytopathological Society. Note: Materials may be edited for content and length.
Cite This Page:
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Each month, renowned astronomers share their latest research at Morrison Planetarium.
When we look to the skies, we see bright stars that lie far beyond the Solar System. Take the constellation Orion as an example—the brighter of Orion’s shoulders is the star Betelgeuse, originally named إبط الجوزاء (Ibt al-Jauza) by Arab astronomers, that lies nearly 650 light years away. As a red supergiant, Betelgeuse is much larger than the Sun. While we can’t determine its exact size with current technology, if Betelgeuse were in our Solar System, it would likely engulf Jupiter’s orbit!
But because Betelgeuse lies far away, it appears to the naked eye as a single point in the sky. As the ninth brightest star in Earth’s sky, we can usually make out its light even in large cities. But this is not true for all stars!
The photo attached to this article was taken in Indonesia, after the photographer’s neighbors had turned their lights out. The disk of our home galaxy, the Milky Way, appears as a hazy band that crosses through the center of the image, surrounded by thousands of stars. In comparison, if we visit the living roof of the California Academy of Sciences in Golden Gate Park during NightLife, we might see several hundred stars on a clear night—and we have very little chance of spotting the Milky Way.
The reason for the difference is the light generated by modern civilization. Stray light from street lighting, building or statue illumination, and other displays enters the atmosphere. While the total light escaping is faint compared to daylight, it is much brighter than the arriving light from most stars. This light washes out all but the brightest stars, like trying to read your iPad on a sunny day.
While the impact of light pollution on stargazing is of interest to urban astronomers, lighting the night has effects on terrestrial aspects of life as well. For example, many nocturnal species have evolved to rely on darkness to hide from predators. Similarly, some predators have novel advantages at night, such as the adapted eyes of owls, or the echolocation of bats. Human behaviors are also affected, as light peeking through a window can disrupt sleep patterns.
A recent study measured the rate of change of radiance and lit area across the globe, finding that both increased by 1.8 to 2.2 percent per year over the past four years. Growth is expected in developing nations, as increasing wealth enables villages to build infrastructure. But surprisingly, this increase in lighting was also seen in the most developed parts of the world.
What caused the increase in lighting in affluent countries? The satellite images showed a shift in color in many cities consistent with a shift from yellowish sodium vapor lamps to white LED lamps. LED lighting is significantly more efficient than older technologies and helps reduce our energy consumption. However, this benefit is only realized if we do not use those savings to install additional unnecessary lighting!
To help our community see the stars, and our nocturnal neighbors thrive in their ecosystems, we should be deliberate in how we use our new sources of lighting. If we are considering installing or upgrading lighting, here are some questions to consider: Will this light extend beyond my property? Does our lighting system focus all light downward? Does our light need to be on all night, or can it be motion activated or use a timer?
By continuing to improve our lighting habits, we can bring back our view of the classical cosmos! Betelgeuse is bright enough to be seen even by San Francisco stargazers, but the myriad details of Orion—the faint stars of his sword and shield, for example—require darker skies. Imagine if Orion could be seen in all his glory over the City by the Bay! New ways of thinking about lighting could restore the constellations to our night skies.
Image: Abdul Rahman
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3 Tutor System
Starting just at 265/hour
# 5. 2x – 1 = 14 – x
Given 2x – 1 = 14 – x
$$\Rightarrow$$ 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)
$$\Rightarrow$$ 3x = 15
$$\Rightarrow$$x = 15 ÷ 3 = 5
Hence x = 5 is the required solution.
Checking:
2x – 1 = 14 – x
Substituting x = 5 in the given equation,we have,
LHS= 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
RHS = 14 – x = 14 – 5 = 9
LHS = RHS
Hence x = 5 is the required solution.
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1. ## Combinatorics
How many different ten-digit number can be created using only 2s and 0s which is divisible by 4 and doesn't contain two neighbouring 0s?
I know that the last two digits as a number have to be a multiple of 4 so the number is: $\displaystyle \overline{\dots 20}$.
And how can I count the numbers not containing neighbouring 0s?
2. Originally Posted by james_bond
How many different ten-digit number can be created using only 2s and 0s which is divisible by 4 and doesn't contain two neighbouring 0s?
I know that the last two digits as a number have to be a multiple of 4 so the number is: $\displaystyle \overline{\dots 20}$.
And how can I count the numbers not containing neighbouring 0s?
I have not gone over it carefully since I have to go now, but here is a guess.
Generalise it for:$\displaystyle \mathop {a_n ...a_3 20}\limits^{\_\_\_\_\_\_\_\_\_\_\_\_}$ show that the number of possibilities is [tex] f_n [/Math] (nth fibonacci number) Try using induction (draw a diagram)
3. Originally Posted by james_bond
How many different ten-digit number can be created using only 2s and 0s which is divisible by 4 and doesn't contain two neighboring 0s? I know that the last two digits as a number have to be a multiple of 4 so the number is: $\displaystyle \overline{\dots 20}$.
I assume that 0202020220 will count as such a ten-digit number even though it begins with a zero.
$\displaystyle \sum\limits_{k = 0}^4 {\binom{9-k}{k}}=55$ is the answer.
In the first eight places in the number we can have anywhere from no 0’s(k=0) to at most four 0’s(k=4).
Then there will be 8-k 2’s creating 9-k to place and separate the 0’s.
4. Hello, James!
How many different ten-digit number can be created using only 2s and 0s
which is divisible by 4 and doesn't contain two neighbouring 0s?
I know that the last two digits as a number have to be a multiple of 4
. . so the number ends in: . $\displaystyle \hdots 20$
And how can I count the numbers not containing neighbouring 0s?
I assume that the leading digit cannot be zero.
So we have: .$\displaystyle 2\:\_\:\_\:\_\;\_\;\_\;\_\;\_\;2\;0$
. . Hence, we want 7-digit numbers with non-adjacent 0s.
I couldn't find a way to think through this task, so I drew a tree diagram.
. . And here's what I found . . .
. . $\displaystyle \begin{array}{c|c}\text{no. of digits} & \text{no. of ways} \\ \hline 1 & 2 \\ 2 & 3 \\ 3 & 5 \\ 4 & 8 \\ 5 & 13 \\ \vdots & \vdots \end{array}\quad\hdots\quad\text{These are Fibonacci numbers!}$
Hence, there are 34 seven-digit numbers with non-adjacent 0s.
Therefore, the answer is: .$\displaystyle {\bf{\color{blue}34}}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Plato allowed a leading zero, so his result is the next Fibonacci number, 55.
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# 75 percent of fifteen Dollars
### Percentage Calculator 2
is what percent of ? Answer: %
### Percentage Calculator 3
We think you reached us looking for answers to questions like: What is 75 percent of fifteen? Or maybe: 75 percent of fifteen Dollars
See the detailed solutions to these problems below.
## How to work out percentages explained step-by-step
Learn how to solve percentage problems through examples.
In all the following questions consider that:
• The percentage figure is represented by X%
• The whole amount is represented by W
• The portion amount or part is represented by P
### Solution for 'What is 75% of fifteen?'
#### Solution Steps
The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate".
Let's say that you need to find 75 percent of 15. What are the steps?
Step 1: first determine the value of the whole amount. We assume that the whole amount is 15.
Step 2: determine the percentage, which is 75.
Step 3: Convert the percentage 75% to its decimal form by dividing 75 into 100 to get the decimal number 0.75:
75100 = 0.75
Notice that dividing into 100 is the same as moving the decimal point two places to the left.
75.0 → 7.50 → 0.75
Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:
0.75 x 15 = 11.25 (answer).
The steps above are expressed by the formula:
P = W × X%100
This formula says that:
"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100".
The symbol % means the percentage expressed in a fraction or multiple of one hundred.
Replacing these values in the formula, we get:
P = 15 × 75100 = 15 × 0.75 = 11.25 (answer)
Therefore, the answer is 11.25 is 75 percent of 15.
### Solution for '75 is what percent of fifteen?'
The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount".
The following problem is of the type "calculating the percentage from a whole knowing the part".
#### Solution Steps
As in the previous example, here are the step-by-step solution:
Step 1: first determine the value of the whole amount. We assume that it is 15.
(notice that this corresponds to 100%).
Step 2: Remember that we are looking for the percentage 'percentage'.
To solve this question, use the following formula:
X% = 100 × PW
This formula says that:
"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100".
This formula is the same as the previous one shown in a different way in order to have percent (%) at left.
Step 3: replacing the values into the formula, we get:
X% = 100 × 7515
X% = 750015
So, the answer is 75 is 500.00 percent of 15
### Solution for '15 is 75 percent of what?'
The following problem is of the type "calculating the whole knowing the part and the percentage".
### Solution Steps:
Step 1: first determine the value of the part. We assume that the part is 15.
Step 2: identify the percent, which is 75.
Step 3: use the formula below:
W = 100 × PX%
This formula says that:
"To find the whole, divide the part by the percentage then multiply the result by 100".
This formula is the same as the above rearranged to show the whole at left.
Step 4: plug the values into the formula to get:
W = 100 × 1575
W = 100 × 0.2
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The parity
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Posted on: Thu Jun 09, 2011 8:37 am
Posts: 693
Joined: Fri May 13, 2011 6:51 pm
The parity
Basic strategies (perhaps trivial but may be useful to someone).
Second rule: The parity (a powerful tool for puzzles like the difficult 7x7 only subtractions on wednedays, etc.).
The addition of all numbers in any row or column can be even (10, 28, 36, 78 for the 4x4, 7x7, 8x8 and 12x12) or odd (15, 21, 45, 55, 153 for the 5x5, 6x6, 9x9, 10x10 and 17x17).
If we have a cage with n- the addition of all numbers inside must be even if n is even (we will call it “cage even”) and odd if n is odd (we will call it “cage odd”). For instance, in a cage of two cells with result 2-, both numbers inside must be even or both odd, and then the addition is even (7 and 5, 2 and 4, etc.). In a cage of two cells with result 3-, the two numbers inside must be of different parity, one odd and the other even, and the addition of both will be odd. Let’s now think in a three cells cage odd-, that is, a - b - c is odd; since a - b - c = a - (b+c), then “a” and “(b+c)” must be of different parity and consequently “a” + “(b+c)” = a + b + c is odd (for instance, 9-2-4 produces 3-, odd, 9+2+4=15, odd). Same demonstration applies to a three cells cage even-. Larger cages behave similarly since a - b – c – d… = a – (b+c+d…), etc.
Next step: The additon of all numbers inside two “cages odd” is even (and obviously if the two cages are even). Iterating this procedure and by comparing the result with the parity of the line or the “rectangle”, etc., we can determine the parity of individual cells (also knowing the parity of some subareas, like those with the x sign, we can eliminate combinations, etc.).
For instance, in the Jun 08 7x7 easy, the parity of the rectangle of the top 4 rows (excluding the cell d5) (observe that the 30x is an odd cage) is odd, so to complete the even parity (4x28) of the four rows, the cell d5 must be odd and “1” is the only possible value; e5 will then be 2. And now, checking the parity of the three bottom rows, the 12x in f1-g1 must be 4x3 (cage odd) and not 6x2 (cage even).
Posted on: Sat Jun 25, 2011 5:25 am
Posts: 8
Joined: Fri May 13, 2011 1:15 am
Re: The parity
Thanks for this tip. I have put it use and has been a definite help.
Posted on: Mon Jun 27, 2011 7:50 pm
Posts: 693
Joined: Fri May 13, 2011 6:51 pm
Re: The parity
larryb33 wrote:
Thanks for this tip. I have put it use and has been a definite help.
Welcome. I arrived to that conclusion when looking for some "error detection" applicable to the lines (rows or columns) (as in the computer's world) but I quickly observed that it was an all-purpose tool that could eliminate combinations in cages of the type "x" or the ":". For me many times is the key in the 9x9's. The parity rule can be combined with the "maximums and minimums in the sum of cages"; for instance, if you have s1 + s2 + s3 = 50 (where s1, s2 and s3 can be of any type of cage including "x" or ":") and you are analyzing a combination of s3 = 19, then s1 + s2 is odd and equals 31, and s1 and s2 have of course opposite parity between them, and also the only possibilities are those "pairs" of combinations for s1 and s2 that give that total of 31, and we can proceed analyzing the effect over the rest of the puzzle.
Posted on: Mon Sep 19, 2011 6:50 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Hi, clm.
In your explanation you refer to the puzzle (Jun 08 7x7 easy).
How can I obtain this historic puzzle to better understand your explanation?
Thank you. Cheers.
_________________
Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.
Posted on: Mon Sep 19, 2011 7:28 pm
Posts: 2209
Joined: Thu May 12, 2011 11:58 pm
Re: The parity
jomapil wrote:
In your explanation you refer to the puzzle (Jun 08 7x7 easy).
How can I obtain this historic puzzle to better understand your explanation?
clm wrote:
For instance, in the Jun 08 7x7 easy, the parity of the rectangle of the top 4 rows (excluding the cell d5) (observe that the 30x is an odd cage) is odd, so to complete the even parity (4x28) of the four rows, the cell d5 must be odd and “1” is the only possible value; e5 will then be 2. And now, checking the parity of the three bottom rows, the 12x in f1-g1 must be 4x3 (cage odd) and not 6x2 (cage even).
Here it is:
Last edited by pnm on Thu Sep 22, 2011 2:46 pm, edited 1 time in total.
Posted on: Mon Sep 19, 2011 7:31 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Thank you, pnm.
Cheers.
_________________
Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.
Posted on: Mon Sep 19, 2011 7:42 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Excuse me, pnm, but I think you made a mistake. If I'm right that puzzle refers to 7x7medium20110608.
Cheers.
_________________
Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.
Posted on: Mon Sep 19, 2011 9:16 pm
Posts: 693
Joined: Fri May 13, 2011 6:51 pm
Re: The parity
jomapil wrote:
Excuse me, pnm, but I think you made a mistake. If I'm right that puzzle refers to 7x7medium20110608.
Cheers.
Hi, jomapil. The graphic is the correct one, the one I was referring to. I cann't remember now if it was the 7x7 easy or medium for that date, but sure Patrick found the correct one. It is not possible for the moment to recall an old puzzle using the Puzzle id (identification introduced "recently" by Patrick), only the last week's puzzles can be directly seen. When I sent the post we had not defined yet the labelling for the cells (actually agreed as in Excel, after the discussion in the Forum, that is, letters for the columns and numbers for the rows, although this has not been implemented yet in the page, ... perhaps in the future upon Patrick's time availability), so I used letters for the rows and numbers for the columns in that post. And at that time I was not yet uploading graphics (sorry for that) because for me the procedure was not clear, and only after a few e-mails with Patrick, I was able to do it.
This was a very easy example for "the parity" rule. The parity of that cell must be odd, and since a cage "2:" has only (in a 7x7) the combinations 1-2, 2-4 and 3-6, that cell must contain a "1" (there is a 3 already in the column) and the roommate is a "2". Now that you know the even parity of this second cell you may determine the parity of the cage "12x" = odd (so 3-4 and not 2-6), considering the parity of the three bottom rows (even, 3 x 28 = 84; here we must be careful because, i.e., three rows will have odd parity in the case of a 5x5, 6x6, 9x9, 10x10, ..., respectively 3 x 15, 3 x 21, 3 x 45, 3 x 55, ...).
We can apply "the parity rule" to practically all puzzles, in many situations, eliminating possibilities and focusing the candidates for the cells and/or cages. The utility will be more evident with the use.
Posted on: Mon Sep 19, 2011 10:18 pm
Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: The parity
Thank you, clm, for your explanation and for the parity rule. At first sight it seems useful lowering the number of possibilities. I go to explore this concept in the next days and in the next puzzles.
Cheers and till the next.
_________________
Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.
Posted on: Mon Dec 02, 2013 9:51 pm
Posts: 693
Joined: Fri May 13, 2011 6:51 pm
Re: Some additional clarifications with respect to the parit
Some additional clarifications with respect to the parity rule.
Around two years and a half have elapsed since this concept was first introduced here in the Forum. Later, one year and a half ago, I sent another post to the section “Solving strategies and tips” named “5x5 **: The parity rule for children/beginners” with an example of how this rule could be applied to a 5x5 difficult. However, many new puzzlers have joined the site since then so my intention now is to refresh a little bit that initial idea and to help a little those who are not still using this concept either because they do not consider it useful or perhaps because they have not fully understood its meaning.
Ok, the idea is as follows: We start with a 2-cell cage “n-” where “n” is even, for instance, “4-”, in a 9x9 puzzle. To get an even difference you need both numbers to be simultaneously even or simultaneously odd, in this example [1,5], [2,6], [3,7], [4,8] or [5,9] (because if one is even and the other is odd the difference will be odd). Then if both have the same parity its sum will be even, either if they are both even or if they are both odd, in this example, for the cage “4-”, the sum would be 6, 8, 10, 12 or 14, respectively. And we say that this is an “even cage” (we define this as an “even cage”) because the sum of all numbers inside is even. We may not know the final numbers yet but we know that necessarily its sum will be even.
Now, lets consider a 2-cell cage “n-” where “n” is odd, for instance, “3-”, in a 9x9 puzzle. The numbers inside this cage must have different parity, one must be even and the other must be odd, because if both were even or both odd simultaneously the difference would be always even (but not odd). In this example the possibilities are [1,4], [2,5], [3,6], [4,7], [5,8] and [6,9], that is, one even number and one odd number. And being of different parity its sum will be odd, in our example, the sum would be 5, 7, 9, 11, 13 or 15, respectively. And we say that this is an “odd cage” because the sum of all numbers inside is odd. We may not know the final numbers yet but we can affirm that necessarily its sum will be odd.
This is the heart of the idea and soon we will see its utility. How is the sum of all numbers included in two “even” cages, for instance, “2-” and “4-”?: even plus even equals even. And the sum of all numbers contained in two “odd” cages, for instance, “1-” and “5-”?: it will be even since odd plus odd equals even. And if one of the cages is even and the other is odd, for instance, “6-” and “7-”?: the sum of all numbers within those two cages will be odd since even plus odd is odd.
Before applying this rule to any real situation we need to demonstrate something else: that the previous conclusions for the 2-cell cages can be extended to cages of any size. Really, this is quite simple: Suppose that we have a 4-cell cage “2-” in a 9x9 puzzle: a possibility would be, for instance, [1,2,4,9] because 9 - 4 - 2 - 1 = 2. Observe that we always depart from the higher number in the cage, 9 in this case, and we subtract the other numbers, but 9 - 4 - 2 - 1 = 9 - (4 + 2 + 1) (the introduction of the parenthesis is simply an arithmetic valid operation but in this way we convert the 4 numbers into 2 numbers by grouping all the small numbers in one), 9 and the value of the parenthesis must be of the same parity to get an even difference as exposed above (in this case both are odd, 9 and 7), consequently 9 and the value of the parenthesis have an even sum, that is, 9 + (4 + 2 + 1) must be even or, from another point of view, the sum of all numbers inside the cage is even (16 in this particular case) [since 9 + (4 + 2 + 1) = 9 + 4 + 2 + 1 removing now the parenthesis]; in summary, we say that “2-” is an “even” cage whichever is the number of cells; another possibility for that “2-” 4-cell cage, for instance, the combination [1,1,4,8]: 8 - 4 - 1 - 1 = 2 = 8 - (4 + 1 + 1), 8 and the value of the parenthesis must be of the same parity, both are even in this case (8 and 6) and, being of the same parity, its sum is even.
Lets go now with a 4-cell cage “n-”, where “n” is odd, to see that the sum of all numbers contained in the cage must be odd: a - b - c - d = a - (b + c + d), where “a” represents the higher number in the cage and “b”, “c” and “d”, the other numbers in the cage. Observe that “a” and “(b + c + d)” must necessarily be of different parity to produce an odd result and, consequently, a + (b + c + d) is odd, and this amount is equal to a + b + c + d (now removing the parenthesis) so the sum of all numbers within the cage is odd, and we say that we have an odd cage. An example with real numbers (just in case the use of letters may sound abstract): for a “3-” 4-cell cage, in a 10x10 puzzle, a possibility is the combination [2,2,3,10]: here 10 - 3 - 2 - 2 = 3, but 10 - 3 - 2 - 2 = 10 - (3 + 2 + 2) in which expression we see that 10 and the parenthesis (with a value of 7) have different parity what is a mandatory requirement to arrive to an odd difference, in this case the subtraction of 10 and 7 equals 3; consequently, if the parity of both things is different, when adding 10 and the parenthesis we will obtain an odd sum, 17 in this particular case. Another example, the combination [1,1,2,7] is another possibility for the cage “3-”: 7 - 1 - 1 - 2 = 3 = 7 - (1 + 1 + 2) where we see that 7 and the parenthesis (which value is 4) have different parity and its sum is odd: 7 + (1 + 1 + 2) = 7 + 4 = 7 + 1 + 1 + 2 = 11.
We have then arrived to the final conclusion: A cage “n-” is even if “n” is even and it is odd when “n” is odd (this happens always regardless of the numbers used to comply with the subtraction); or, expressed with different words, if “n” is even the sum of all numbers in the cage is even and if “n” is odd the sum of all numbers in the cage is odd.
Now the question is: What is the utility of all this?. Ok: because it can be applied to “innies”, “outies”, “multiplication” or “divison” cages, … , and to any number of rows or columns or areas (for instance, if the parity of the “inner area” is known, like in those 6x6’s when the corners are given, and we have a 4x4 "inner area", …), etc..
Applying the parity rule.
First example: Suppose, for instance, that in a 7x7 puzzle, like the subtraction only puzzle on Wednesdays, you have an “innie” cell in a line (row or column), you can determine the parity of this individual cell in this way: if you have in that line, for instance, three 2-cell cages, lets say “1-”, “2-” and “3-”, you know that the sum of the six numbers contained in these three cages is even, just by "adding" the parity of the three cages, even in this case, and, since the sum of all numbers in any line is 28 (addition rule in a 7x7 puzzle, 28 is the sum for evey row or column), the “alone” (innie) cell must be even = 28 - even. And this is very helpful, for instance, if the cage that contains this cell is “4-” you inmediately know that the cage must be [2,6] and you can write the candidates since [1,5] or [3,7] would always produce an odd number in that position and that‘s impossible.
Second example: Suppose that you have two columns (in a patterned 7x7) with individual cells and addition/subtraction cages (or, i.e., some well defined multiplication cages like a 3-cell “10x” or a 3-cell L-shape “25x”, “32x”, …) plus one “72x” 3-cell L-shape cage. Our purpose is to determine the parity of this last cage. You "iterate" the parity rule to the sum of the individual cells and those cages, consequently determining the parity of the sum of all the numbers involved and, since the parity of two complete columns is even in this case (in a 7x7 the sum of two columns is 2 x 28 = 56) you find the parity of the pending cage, the targeted multiplication cage “72x”: if that result is even the only valid combination is [2,6,6] and, if it is odd, the only valid combination is [3,4,6]. And now you can continue looking at or extending the procedure to the two rows (the perpendicular lines to the previous lines) containing that “72x” cage.
******
The parity rule will solve you many uncertainties in the day by day "Calcudoku job" and you will gain a lot of time not only when dealing with small puzzles but also with large puzzles. You will find its utility as soon as you start using it, specially in the 9x9's, the 7x7 subtraction only on Wednesday's, the actual 10x10 on Fridays to define, i.e., if "40x" is [4,10] or [5,8], that is, an even or an odd cage, the actual 8x8 medium on Sundays, most 5x5 difficult, etc..
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Bird populations in Finland have shifted northward by an average of 45 km between the 1970s and the 2010s due, in particular, to climate change. Before this, the effects of climate change have been studied through changes in the distribution of species. A new study method also takes into account regional shifts in population density.
A new study by the Finnish Environment Institute (SYKE) and the Finnish Museum of Natural History shows that northern species are retreating northward at greater speed than southern species are spreading to new areas.
Climate change has been predicted to affect the ranges of species as favourable climate conditions shift towards the poles. According to the new study, the mean weighted latitude of density of Finnish bird populations has shifted northward by an average of 45 km, as predicted.
The density of northern species has shifted more (73 km) than that of southern species (27 km). In as many as 23 species, such as black grouse, ruff, greenfinch and yellow wagtail, the mean weighted latitude of density has shifted northward by over a hundred kilometres. A southward shift of over a hundred kilometres was observed in only two species (common buzzard and common raven).
More accurate information through observation of density shifts
Unlike in previous studies, the study behind this new publication looked at the regional density shifts of species during different decades, starting from the 1970s. Density estimates provide significantly more information about species' response to climate change than only studying their ranges, which has been standard practice in these kinds of studies until now. Such studies have concluded that species' ranges shift northwards as the climate warms, but looking at density shifts provides more detailed information about changes in the abundance of species.
Changes in the mean weighted latitude of density were compared to distribution changes observed and recorded in bird atlases from 1974 to 2010. The comparison revealed that the range of southern species shifted more to the north on average than the mean centre of the weighted density of the species, while in northern species, density shifted more to the north than the range.
"This study shows that the populations of northern species are retreating northward at a faster rate than what studies looking at changes in range have shown in the past," says Leading Researcher Raimo Virkkala from the Finnish Environment Institute (SYKE). "These species are increasingly at great risk of disappearing from the Nordic countries, as the Arctic Ocean limits their shift to the north."
A million bird observations confirm the results
The study is based on the long-term monitoring of bird populations using the same line transect method. The observational data of the study consisted of nearly a million (990,301) observations of 94 different bird species, and the results have now been published as part of the international Global Change Biology publication series.
"Census data gathered by volunteer bird watchers is invaluable for the comprehensively study of long-term changes in bird populations. Each bird monitoring count is an important part of the national and international bird monitoring network," thanks Curator Aleksi Lehikoinen, the coordinator of the Finnish Museum of Natural History's bird census efforts.
- Raimo Virkkala, Aleksi Lehikoinen. Patterns of climate-induced density shifts of species: poleward shifts faster in northern boreal birds than in southern birds. Global Change Biology, 2014; DOI: 10.1111/gcb.12573
Cite This Page:
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Benefits of Being Bilingual
Encouraging your child to learn a second language isn’t just about giving them knowledge of vocabulary and grammar from another country. There’s a lot more to this gift than meets the eye.
Whether it’s boosting your child’s cognitive ability and overall test scores, or just helping to improve their brains, boost critical thinking and knowledge of international culture, the benefits of being bilingual are numerous and potent:
First of all, it increases cognitive ability and can even lead to higher test scores. While it is lamentable that our world is still so drive by standardized test scores, we should still take every advantage we can get. According to studies listed on the American Council on the Teaching of Foreign Languages (ACTFL) website, there is a large body of evidence to support the notion that learning a second language helps with academic achievement.
Secondly, it allows young people to think in a whole new way. When you ger really familiar with a new language, you start to think about concepts in a whole new way. A second language demands that we express ideas using different syntax and logic. It may require us to use metaphor or euphemism, or conversely to be more direct and literal. Either way, it lends a whole new perspective to a child’s thinking process, inspiring more of those “Eureka” moments along the way.
Next, and perhaps unexpectedly for some, a second language can greatly increase capacity for empathy. It should never be underestimated just how much understanding it takes to know when to use one language or another when communicating to another human being. What it takes is the speaker being able to comprehend which words will fall most effectively and kindly upon the ears of the interlocutor. If that’s not the perfect exercise in empathy, then we’re not sure what is.
A study in Canada found that being bilingual may have protective effects on your brain against problems like dementia and even Alzheimer’s. The study pointed to findings among Alzheimer’s patients in Canada, a group of bilingual adults, while showing greater brain atrophy than their monolingual counterparts, were nonetheless able to perform better in a series of cognitive tests carried out for the research. It appears that there may be a link between being bilingual and that kind of shielding effect that despite atrophy keeps your mind going for longer.
Being bilingual means being skilled, and skills bring confidence, which in turn drives happiness and contentment. Researchers Wayne Thomas and Virginia Collier extensively compared single-language classrooms to dual-lingual classrooms, finding that students not only performed better (as mentioned above) but also were happier in their school lives. Could it be that the confidence they gain, admiration and respect from peers, and the ability to comfortably communicate with more people around them adds up to greater happiness?
Becoming bilingual isn’t like flipping a switch for children, and there are drawbacks as well as benefits, of course. For many, however, seeing the difference in their child’s ability, personality and modes of thinking is enough to drive them to making that all-important decision to nurture them in a second language.
By Thomas Longrigg
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Mekong Delta provinces discuss the importance of maintaining ecosystem services in the 5th Mekong Delta Nature and Culture Conservation Forum, Vietnam
The Mekong river is one of the largest rivers in the world, second only to the Amazon in terms of biodiversity. It includes a vast array of ecosystems including mangrove forests, rivers and streams, sand dunes, seasonal mangrove forest-grasslands, and pond and inland ecosystems. Each provides Delta residents with different, essential benefits. For local economies, these ecosystems provide rich fisheries, fruits and/or alluvium for agriculture; in addition, they protect local residents and their surrounding environment with coastline protection, erosion control, flow regulation, microclimate conditioning and carbon absorption. They are home to many rare species, especially birds, like the red Sarus Crane (Grus antigone), and fish, like the Mekong giant catfish (Pangasianodon gigas).
Rapid, unplanned economic development, and unchecked population growth, as well as climate change, continually pressure these ecosystems and degrade the quality of their services . “The use of ecosystem services without proper planning has weakened or completely destroyed many [of] the Delta’s ecosystems. [They] are reduced in size, isolated, fragmented due to economic development (such as forest clearance for aquaculture or agriculture), infrastructure development, residential area enlargement, and contamination by production and wastewater. The construction of hydro dams upstream has also changed the natural flow of the river, leading to the loss of alluvium annually, and [making] the delta [even] more vulnerable to climate change. In the past, mangrove forest covered almost all coastal areas of the Delta but is disappearing rapidly. Now mangrove forest covers only Bac Lieu and Ca Mau provinces (about 77,00ha),” said Mr. Hoang Viet, Climate Change Co-ordinator of WWF-Vietnam.
Ecoystems and their services are nature’s gift to the Delta, and for many years, natural and economic development have been linked. To maintain ecosystems’ benefits to local people, an understanding of their natural cycles, as well as experience planning and protecting these habitats must be enhanced. WWF, along with Biodiversity Conservation Agency (BCA) raised this topic at the Forum for Provincial Delta Policy and Decision- Makers to allow learning from and sharing of best practices with experts from international organizations.
“Climate Change impacts on the region are already obvious; thus protecting, recovering and maintaining healthy ecosystems have become critical and must be strategic. The BCA, with technical support from WWF, other organizations and experts, has been drafting the first National Plan for Biodiversity Conservation, in which the importance of preserving and maintaining ecosystems are emphasized. Through this forum, we hope Mekong Delta provinces will have a better understanding about this issue, thus getting well prepared for developing and implementing their own plans when the National Plan is endorsed by the Government”, said Ms. Huynh Thi Mai, Deputy Director of BCA, Ministry of Natural Resources and Environment.
According to WWF experts, the recovery and maintenance of ecosystems and their services require synchronization of local policies to ensure sustainable livelihoods for those still dependent on rice planting, fishing and aquaculture. Increasing awareness of the need for environmental protection in local communities is also critical—along with the development of Mechanisms and tools to support ecosystem service payments (PES) .
Although there are challenges to maintaining the health and integrity of ecosystems, good news has just come to Mekong River Delta: Mui Ca Mau National Park, with support from the BCA and WWF, was officially recognized as the 5th RAMSAR site, an internationally significant wetland in Vietnam and 2088th in the world, by the RAMSAR Convention Secretariat. “We are very happy. After our tireless efforts and tremendous support from the MoNRE and other related departments, Mui Ca Mau National Park was recognized as the second RAMSAR in Mekong River Delta. This will be a chance for us to attract attention, support and investment from the community, agencies and organizations locally and globally in conservation activities of the Natural Park. We see this as positive pressure to maintain and upgrade wetland ecosystem services in the area, especially in the context of increasingly complicated climate change [impacts]. The certificate-handling ceremony will be held on 13th April in the Park by the MoNRE, the Ca Mau Provincial People Committee with support from WWF and IUCN, which will also be a chance to educate the surrounding communities on the importance of protecting this land,” said Mr. Tran Van Tuan, Director of Mui Ca Mau National Park.
After the forum, it is hoped that policy makers in the thirteen Mekong River Delta provinces will better understand the importance of maintaining ecosystem services for economic and social development; learn new approaches and gain experience in planning and maintenance, with particular regards to methods to evaluate and quantify ecosystem services. With hope, decision makers will be better informed and consider the health of ecosystem services when planning to use natural resources.
Nature and Culture Conservation for Sustainable Development of Mekong Delta Forum is a WWF’s initiative to facilitate information and experience exchange among policy makers, organizations, and experts on sustainable development and nature and culture conservation of the Mekong Delta. The first forum was hold in Can Tho in 2009 on ‘Rural Sustainable Development’; in Rach Gia, Kien Giang in 2010 on ‘Adapting to Climate Change in Mekong Delta’; in Ben Tre in 2011 on ‘Adaptive Management in the Context of Climate Change’ and in Cao Lanh, Dong Thap in 2012 on ‘Wise Use of Wetland’.
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Until now, scientists haven’t detected any planets outside of our own Milky Way Galaxy—it’s simply been too difficult to discern such small things from so far away. That should make you wonder: are there any planets outside the Milky Way? A new paper might be able to answer that question.
Researchers at the University of Oklahoma looking at a galaxy 3.8 billion light years away spotted evidence of planets. More specifically, they think that there should be at least 2,000 objects, ranging from moon- to Jupiter-sized, per main-sequence star in the galaxy, based on how the galaxy’s gravity warped the objects behind it. This is not direct evidence, mind you; no one has spotted any actual planets. But it’s evidence nonetheless.
“It is natural to hypothesize that planets are common in external [non-Milky Way] galaxies as well,” the authors write in the paper, published in Astrophysical Journal Letters. “However, we lack the observational techniques to test this hypothesis, because compared to their Galactic brethren, extragalactic planets are much farther away and much more difficult to separate from the host stars/galaxies.” So they came up with a more roundabout method.
Normally, researchers spot exoplanets through the way they dim the stars they orbit when they pass between the star and our telescopes. Sometimes they even identify distant planets through direct imaging, actually resolving the planet from its host star.
That doesn’t work if you’re looking at distant galaxies where you can barely tell the stars from one another. So instead, the researchers used a technique called “quasar microlensing.” What’s that? Well, according to Einstein’s theory of General Relativity, really massive things, like galaxies, stars, and even planets warp the shape of space itself. That means that they also bend the light traveling through the warped space. And if they bend a bunch of light rays the right way, they can actually act magnify the light behind it, acting like a huge magnifying glass.
In this case, a distant galaxy called RXJ1131−1231 has warped light in such a way that astronomers could see several magnified images of a galaxy behind it in the sky. As they looked closer and closer, the researchers noticed much tinier changes, like nicks or irregularities in a lens. In their new paper, the researchers demonstrated that a population of rogue planets (i.e. planets not orbiting stars) in RXJ1131−1231 could account for the lensing at the smallest scale.
Scientists have previously spotted this kind of microlensing event using the Andromeda galaxy as the magnifying glass, which some have proposed could be evidence for planets.
The paper doesn’t name any planets or anything like that, as the study’s author Eduardo Guerras says in a press release: “This galaxy is located 3.8 billion light years away, and there is not the slightest chance of observing these planets directly, not even with the best telescope one can imagine in a science fiction scenario.” But they were still able to use their calculations to estimate the number of planets and masses.
We tend to assume that other galaxies have planets—after all, why should the Milky Way be unique?—but this could be the first, albeit indirect, evidence for these distant worlds.
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Although many white Americans supported the ACS, free blacks were nearly unanimous in their condemnation of colonization, viewing it as little more than a wholesale deportation scheme. As a consequence, only about 4,000 free African Americans and 7,000 former slaves (freed on the condition that they immigrate) settled in Liberia between 1820 and 1864. Those who did go faced many challenges. The mortality rate from disease was extremely high, and relations between the colonists and Liberia's indigenous peoples were often strained.
Under the aegis of the ACS, a white "agent" governed the colony, but in 1839 the first steps toward self-rule were taken, and the colony became a Commonwealth. Under a new constitution, legislative powers were shared by the ACS-appointed governor and a popularly elected council. Although the first governor of the commonwealth was white, his successor was Joseph Jenkins Roberts, a freeborn native of Virginia who became Liberia's first black chief executive in 1841.
In 1847, with the blessing of the ACS, Liberia declared its independence and adopted a constitution modeled on that of the United States. Although Liberia was the first republic established on the African continent, it was denied diplomatic recognition by the United States until 1862, because of southern opposition to accrediting a black diplomat in Washington.
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This lesson is designed to integrate what the students have learned concerning the Articles of Confederation, The Constitutional Convention, The three branches of government and the approval of the Bill of Rights. Specifically, this lesson focuses on the standards by examining how the Constitution became the foundation of the government of the United States. They will also learn the key democratic principles in the Constitution.
Students will be able to:
Students describe the people and events associated with the development of the U.S. Constitution and analyze the Constitution’s significance as the foundation of the American republic.
1. List the shortcomings of the Articles of Confederation as set forth by their critics.
2. Explain the significance of the new Constitution of 1787, including the struggles over its ratification and the reasons for the addition of the Bill of Rights.
3. Understand the fundamental principles of American constitutional democracy, including how the government derives its power from the people and the primacy of individual liberty.
4. Understand how the Constitution is designed to secure our liberty by both empowering and limiting central government and compare the powers granted to citizens, Congress, the president, and the Supreme Court with those reserved to the states.
English Language Arts
Comprehension and Analysis of Grade-Level-Appropriate Text
2.3 Discern main ideas and concepts presented in texts, identifying and assessing evidence that supports those ideas.
2.4 Draw inferences, conclusions, or generalizations about text and support them with textual evidence and prior knowledge
1.2 Create multiple-paragraph expository compositions: a. Establish a topic, important ideas, or events in sequence or chronological order. b. Provide details and transitional expressions that link one paragraph to another in a clear line of thought. c. Offer a concluding paragraph that summarizes important ideas and details.
Research and Technology
1.3 Use organizational features of printed text (e.g., citations, end notes, bibliographic references) to locate relevant information.
1.4 Create simple documents by using electronic media and employing organizational features (e.g., passwords, entry and pull-down menus, word searches, a thesaurus, spell checks).
2.3 Write research reports about important ideas, issues, or events by using the following guidelines: a. Frame questions that direct the investigation. b. Establish a controlling idea or topic. c. Develop the topic with simple facts, details, examples, and explanations.
1.0 Listening and Speaking Strategies
Organization and Delivery of Oral Communication
1.4 Select a focus, organizational structure, and point of view for an oral presentation.
1.5 Clarify and support spoken ideas with evidence and examples.
1.6 Engage the audience with appropriate verbal cues, facial expressions, and gestures.
2.0 Speaking Applications (Genres and Their Characteristics)
2.2 Deliver informative presentations about an important idea, issue, or event by the follow-ing means: a. Frame questions to direct the investigation. b. Establish a controlling idea or topic. c. Develop the topic with simple facts, details, examples, and explanations.
2.3 Deliver oral responses to literature:
a. Summarize significant events and details.
b. Articulate an understanding of several ideas or images communicated by the literary work.
c. Use examples or textual evidence from the work to support conclusions
Visual and Performing Arts Content Standards
2.2 Demonstrate the use of blocking (stage areas, levels, and actor's position, such as full front, quarter, profile, and full back) in dramatizations.
3.1 Select or create appropriate props, sets, and costumes for a cultural celebration or pageant
Connections and Applications
5.1 Use theatrical skills to dramatize events and concepts from other curriculum areas, such as reenacting the signing of the Declaration of Independence in history social science.
Integration of Knowledge and Ideas
7. Integrate and evaluate content presented in diverse media and formats, including visually and quantitatively, as well as in words.*
College and Career Readiness Anchor Standards for Reading K-5
Key Ideas and Details
1. Read closely to determine what the text says explicitly and to make logical inferences from it; cite specific textual evidence when writing or speaking to support conclusions drawn from the text.
2. Determine central ideas or themes of a text and analyze their development; summarize the key supporting details and ideas.
3. Analyze how and why individuals, events, and ideas develop and interact over the course of a text.
Craft and Structure
4. Interpret words and phrases as they are used in a text, including determining technical, connotative, and figurative meanings, and analyze how specific word choices shape meaning or tone.
Range of Reading and Level of Text Complexity
10. Read and comprehend complex literary and informational texts independently and proficiently.
The political process involves participation; understanding requires student engagement.
The strength of a democracy is equal to the strength of its citizens. (We must understand, participate in, and further develop our system of government to ensure democracy).
E Pluribus Unum: out of many, one. (From a variety of sources and experiences, we have developed a successful government and legal system).
How is the Constitution a living document?
Is citizenship a right or a responsibility?
Higher Order Thinking Questions:
What were the shortcomings of the Articles of Confederation and why did some leaders want to change the articles? (Evaluation)
Why do you think the Constitution and the Bill of Rights were written? (Evaluation)
What kind of person wrote the Constitution and the Bill Of Rights? (Analysis, Evaluation)
Did the Constitution create a stronger central government? Explain (Evaluation)
How are the Constitution and The Bill of Rights living documents? (Analysis)
Students will be evaluated through informal checks for understanding, teacher observation, self-reflections, and performing an authentic task (GRASP) evaluated by a rubric.
Activity Steps: This lesson should be completed over a 2-week period per teacher discretion and student need.
|“Have you ever made up rules for a game and then found when playing, that the rules did not work well.
What kinds of problems did you have and what did you do to fix the problems
Did players disagree about how to fix or change the rules?
How were disagreements settled?"
|Attend to a formal introduction.
Take notes as needed
Review Articles of Confederation
|Review the Articles of Confederation.
Discuss the shortcomings of the Articles.
Discuss the Annapolis Convention and Shays’s Rebellion.
Discus the Constitutional Convention and the Participants
Discuss the Debates and the Compromises at the Convention
Discuss the purpose of the Constitution as stated in the Preamble
Compare the powers and functions of three branches of government
Discuss the struggle to get the Bill of Rights approved and explain the key rights that were guaranteed in the Bill of Rights
| Take notes as needed.
Students use history/social science text and/or other documents provided on the Articles and the Conventions, check and correct as a class for understanding
1- 2 weeks class time
|Simulation of Constitutional situations that require action by branch (es)
Apply knowledge about the Bill of Rights to a shield that depicts a few illustrations of these rights
Create a Hall of Fame that honors the constitution, the people and the events and ideas that were important to history.
|Student teams will use constitutional power cards that decide governmental actions. Some situations will require the students to place the power cards in a proper sequence that the government would act.
1 minute oral presentation of their character highlights
Come in Costume with props as needed and complete self reflection
Create Tableaus of an amendment in the Bill of Rights and perform in front of class
Reflect on the group process
Next steps include:
Lessons discussing a Constitutional democracy
Next steps include:
Compare the powers granted to citizens, to the three branches of the federal government and to the states
Special Needs of students are considered in this lesson:
Students are put into flexible grouping scenarios that will benefit learning for all types of learners and special needs. Hands-on learning with plenty of opportunities for movement, verbal, written, and nonverbal expression, and multiple learning modalities are available within the context of this lesson.
This lesson may be repeated with different amendments and or articles from the constitution. Students may want to investigate current events that could lead to amendments. Film the tableau’s and use music, (after practicing with the technology and guidance with the use of music and props). Skits could be performed in the place of tableau with structure provided.
Materials and Resources Needed:
History Social Science text, Internet access, wipe boards, project planner, additional classroom resources, paper, costumes, and props.
We The Kids: The Preamble to the Constitution of the United States, David Catrow
A Kids’ Guide to America’s Bill of Rights, Kathleen Krull
Shh! We're Writing the Constitution, Jean Fritz
. . .If You Were There When They Signed the Constitution , Elizabeth Levy
A More Perfect Union,The Story of Our Constitution, Betsy Maestro
DVD: Founding of our Federal Government
Context of the unit:
This unit is designed to focus on the history social science standards in 5.7.1 - 5.7. These focus on the constitution as the basis for our American republic. The students will learn how the United States Constitution came to be written and how it defines the national government. Students have learned the steps involved in the creation of “tableau” in previous studies.
Context of the lesson:
This lesson would be the conclusion of the unit. Students will have covered the Declaration of Independence and the end of the Revolutionary War and a review of how the new nation would organize government, which included writing state constitutions.
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# 6.3: Boxes and Boxes Extras
Difficulty Level: At Grade Created by: CK-12
Extras for Experts - Boxes and Boxes – Interpret pan balances to determine values of variables
Solutions
\begin{align*}1. \quad t = 2 \ \text{pounds}; \ u = 4 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ D, 3u = 12, \ \text{so} \ u = 4 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ C, 2t = 4, \ \text{so} \ t = 2 \ \text{pounds}.\end{align*}
\begin{align*}2. \quad v = 3 \ \text{pounds}; w = 1 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ H, v < 5 \ \text{or} \ 1, 2, 3 \ \text{or} \ 4 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ G, v = 3w, \ \text{so} \ v \ \text{must be a multiple of} \ 3.\!\\ {\;} \quad \ \text{Then} \ 3 = 3w \ \text{and} \ w \ \text{is} \ 1 \ \text{pound}.\end{align*}
\begin{align*}3. \quad y = 4, 8, \ \text{or} \ 12 \ \text{pounds}; z = 1, 2, \ \text{or} \ 3 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ J, y < 16 \ \text{or} \ 1, 2, 3, ..., 15 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ K, y = 4z, \ \text{so} \ y = \ \text{must be a multiple of} \ 4.\!\\ {\;} \quad \ \text{The possible multiples of} \ 4 \ \text{that are less than} \ 16 \ \text{are} \ 4, 8, \ \text{and} \ 12.\!\\ {\;} \quad \ \text{If} \ y = 4, 8, \ \text{or} \ 12, \ \text{then} \ z = 1, 2, \ \text{or} \ 3 \ \text{pounds}.\end{align*}
\begin{align*}4. \quad r = 1, 2 \ \text{or} \ 3 \ \text{pounds}; s = 10 \ \text{pounds}\!\\ {\;} \quad \ \text{From} \ B, 2s = 20, \ \text{so} \ s = 10 \ \text{pounds}.\!\\ {\;} \quad \ \text{From} \ A, 3r < 10, \ \text{or} \ 1, 2, 3, ..., 9 \ \text{pounds. So} \ r = 1, 2, \ \text{or} \ 3 \ \text{pounds}.\end{align*}
All weights are whole numbers of pounds.
What could be the weights? Tell how you figured it out.
All weights are whole numbers of pounds.
What could be the weights? Tell how you figured it out.
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Date Created:
Feb 23, 2012
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# rank of a matrix
First of all I am sorry because I have asked similar kind of question a few days ago.But I still have problem with row reductions when there are letters in a matrix.The question is asking the value of 'a' when the rank of matrix is 1 , 2 , 3 and 4. I am not good at row reductions.In each row operations the matrix became more confusing.If someone help me with the row reductions , I will be very happy. $$\left( \begin{array}{ccc} 1&1&2&0\\ 2&a+1&3&a-1\\ -3&a-2&a-5&a+1\\ a+2&2&a+4&-2a \end{array} \right)$$
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You just need to work the algebra, just like you do with regular algebra.
Subtract twice the first row from the second row to get $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ -3 & a-2 & a-5 & a+1\\ a+2 & 2 & a+4 & -2a \end{array}\right).$$ This is just algebra (e.g., the (2,2) entry is $a-1$ because $(a+1)-2 = a-1$).
Add three times the first row to the second row. The (3,2) entry will be $(a-2) + 3(1) = a+1$; the (3,3) entry will be $(a-5)+3(2) = a+1$. Etc. You get: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ 0 & a+1 & a+1 & a+1\\ a+2 & 2 & a+4 & -2a \end{array}\right).$$ Subtract $a+2$ times the first row from the fourth row; that means subtracting $1(a+2)$ from the first entry; $1(a+2)$ from the second entry; $2(a+2)$ from the third entry; and $0$ from the fourth entry. You get: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ 0 & a+1 & a+1 & a+1\\ 0 & -a & -a & -2a \end{array}\right).$$ Add the fourth row to the third row to simplify things: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & a-1 & -1 & a-1\\ 0 & 1 & 1 & -a+1\\ 0 & -a & -a & -2a \end{array}\right).$$ Exchange second and third row: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & 1 & 1 & -a+1\\ 0 & a-1 & -1 & a-1\\ 0 & -a & -a & -2a \end{array}\right).$$ Add $1-a$ times the second row to the third row: $$\left(\begin{array}{rrrr} 1 & 1 & 2 & 0\\ 0 & 1 & 1 & -a+1\\ 0 & 0 & -a & -1+2a-a^2\\ 0 & -a & -a & -2a \end{array}\right).$$
We are just doing algebra, only in several columns at the same time. Keep going until you get a matrix for which the rank will be easy to figure out, depending on the values of $a$. I've gotten you almost all of the way there.
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Okey.Thank you very much.I will keep on simplifying this matrix.thanks for your patience. – Piril Jun 21 '11 at 21:59
here is an answer for rank=1:
if rank=1, that means all the rows are dependent. The first row is given, so every other row should be a multiple of the first row. Let's start with the second row. since the first element on the second row is 2, and the first element on the first row is 1, then (a+1) should also be 2*1 and (a-1) should be 2*0. so $a$ should be equal to 1. However if you check the third row, substituting $a$ with 1, we have [-3, -1, -4, 2] which is not a multiple of the first row. hence rank=1 has no solution.
You can do row manipulation to make the matrix lower triangular. then the determinant would be the product of the elements on the diagonal.
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Thank you very much. – Piril Jun 21 '11 at 22:00
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# Kruskal Wallis H Test: Definition, Examples & Assumptions
Statistics Definitions > Kruskal Wallis H Statistic
## What is the Kruskal Wallis Test?
The Kruskal Wallis H test uses ranks instead of actual data.
The Kruskal Wallis test is the non parametric alternative to the One Way ANOVA. Non parametric means that the test doesn’t assume your data comes from a particular distribution. The H test is used when the assumptions for ANOVA aren’t met (like the assumption of normality). It is sometimes called the one-way ANOVA on ranks, as the ranks of the data values are used in the test rather than the actual data points.
The test determines whether the medians of two or more groups are different. Like most statistical tests, you calculate a test statistic and compare it to a distribution cut-off point. The test statistic used in this test is called the H statistic. The hypotheses for the test are:
• H0: population medians are equal.
• H1: population medians are not equal.
The Kruskal Wallis test will tell you if there is a significant difference between groups. However, it won’t tell you which groups are different. For that, you’ll need to run a Post Hoc test.
## Examples
1. You want to find out how test anxiety affects actual test scores. The independent variable “test anxiety” has three levels: no anxiety, low-medium anxiety and high anxiety. The dependent variable is the exam score, rated from 0 to 100%.
2. You want to find out how socioeconomic status affects attitude towards sales tax increases. Your independent variable is “socioeconomic status” with three levels: working class, middle class and wealthy. The dependent variable is measured on a 5-point Likert scale from strongly agree to strongly disagree.
## Running the H Test
Sample question: A shoe company wants to know if three groups of workers have different salaries:
Women: 23K, 41K, 54K, 66K, 78K.
Men: 45K, 55K, 60K, 70K, 72K
Minorities: 18K, 30K, 34K, 40K, 44K.
Step 1: Sort the data for all groups/samples into ascending order in one combined set.
20K
23K
30K
34K
40K
41K
44K
45K
54K
55K
60K
66K
70K
72K
90K
Step 2: Assign ranks to the sorted data points. Give tied values the average rank.
20K 1
23K 2
30K 3
34K 4
40K 5
41K 6
44K 7
45K 8
54K 9
55K 10
60K 11
66K 12
70K 13
72K 14
90K 15
Step 3: Add up the different ranks for each group/sample.
Women: 23K, 41K, 54K, 66K, 90K = 2 + 6 + 9 + 12 + 15 = 44.
Men: 45K, 55K, 60K, 70K, 72K = 8 + 10 + 11 + 13 + 14 = 56.
Minorities: 20K, 30K, 34K, 40K, 44K = 1 + 3 + 4 + 5 + 7 = 20.
Step 4: Calculate the H statistic:
Where:
• n = sum of sample sizes for all samples,
• c = number of samples,
• Tj = sum of ranks in the jth sample,
• nj = size of the jth sample.
H = 6.72
Step 5: Find the critical chi-square value, with c-1 degrees of freedom. For 3 – 1 degrees of freedom and an alpha level of .05, the critical chi square value is 5.9915.
Step 6: Compare the H value from Step 4 to the critical chi-square value from Step 5.
If the critical chi-square value is less than the H statistic, reject the null hypothesis that the medians are equal.
If the chi-square value is not less than the H statistic, there is not enough evidence to suggest that the medians are unequal.
In this case, 5.9915 is less than 6.72, so you can reject the null hypothesis.
------------------------------------------------------------------------------
Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!
Statistical concepts explained visually - Includes many concepts such as sample size, hypothesis tests, or logistic regression, explained by Stephanie Glen, founder of StatisticsHowTo.
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The Sun - power house
Live image: click to enlarge.
The Sun has existed for 5 billion (5,000,000,000, or 5×109) years. It is one of 150 billion stars in our galaxy, The Milky Way. Despite consuming 5 million tonnes of Hydrogen every second, there is enough material left for it to exist for another 5 billion years.
A Typical Star
Stars can exist with between 1/20 and 70× the mass of our Sun, making it somewhat average. Heavier stars have much shorter lifetimes, since their greater gravity increases the rate of fusion. This also makes them hotter and brighter than our Sun, whose surface temperature is about 6000K. More on stars.
On Earth, humans use Nuclear Fission to produce energy for some of our electricity supply. This is somewhat inefficient, and produces very dangerous nuclear waste. The Sun uses the technically more difficult Nuclear Fusion to produce its energy: 3×1026 joules per second! (3×1026 W).
Fig. 1: Fusion of Hydrogen into Helium
In studying the Sun, we humans have learned how to produce energy from Nuclear Fusion. This century, it may become our main source of energy.
Once it was learned that the Sun is just an ordinary star, scientists on earth realised that if we learned about the sun, we were learning about the stars. In the diagram below, the spectrum of the Sun is shown. As you can see, there are lots of unexpected black lines.
Fig. 2: The Solar Spectrum
The black lines seen in the solar spectrum are Fraunhofer lines. They indicate the presence of chemical elements in the Sun. Light from other stars is similar - from their spectra we learn about what they're made from. Click for more.
This photo was taken in 2002 by the Swedish 1-metre solar telescope on La Palma, and is the most detailed picture ever of the Sun's surface. Sunspots are caused by twists in the Sun's magnetic field. A new page will appear here shortly! Source: Institute for Solar Physics, Stokholm
Spectacular Auroras!The Sun's influence reaches out far beyond the earth. A "solar wind" of charged particles is continuously bathing all the planets. At earth, this causes the auroras - the so-called northern and southern lights. Read more. The Sun has been under study for years, but not until late 1995 did major discoveies start to be made with the launch of the the joint NASA/ESA probe, SOHO.
In April 1998, the British-built and operated CDS spectrometer on board discovered tornadoes wider than Africa! The Sun has even been seen to have regular quakes, as shown on the right. In June 1998, SOHO observed two comets crash into the Sun in quick succession, as shown top right. For the latest information, check the SOHO website.
Death of the Solar System
In 1 billion years, the Sun will be bigger and hotter than it is today. The Earth's oceans will have boiled away into space. Stars start to die when they stop burning hydrogen. The sun, a yellow, G-2 star, will (in 5 billion years) then become a red giant encompassing the orbit of Mercury. Mercury will disappear and Venus will lose its atmosphere and become a burnt out planet. The Earth will suffer the same fate, even though it is outside the red giant's immediate reach.
Life Cycles of Stars
An active tale. Check it out!
GCSE MaterialFind out much more about the sun on GCSE.com.
Sun | Mercury | Venus | Earth | Moon | Mars | Asteroids | Jupiter | Saturn | Uranus | Neptune | Pluto | X | Kuiper
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Aurora polaris (ger: Polarlicht; fr: aurore polaire)
Aurora is latin for 'dawn'. The Northern Lights and the aurora borealis are two names for the same thing. The term aurora borealis was introduced by Galileo Galilei in 1619 who wrongly thought that the aurora is caused by sunlight reflecting from the high atmosphere. However, from thereafter the name was used for Northern Lights. The proper name for the aurora of the southern hemisphere is the aurora australis. Together the aurora australis and the aurora borealis are known as the aurora polaris. Nowadays the simple name aurora is mostly used, as is the name Northern Lights.
In fact, the aurora is a luminous phenomenon occuring in the upper atmosphere at altitudes between about 100km and 1000km. Auroras are visible best from the auroral zone the region about 15-30 degrees from each magnetic pole with the greatest frequency of aurorae. The region within which auroral activity occurs is the auroral oval and is only visible from space. The center of the auroral oval lies at the geomagnetic poles - not to be confused with the geographic poles.
However, infrequently they might be seen as far down as the equator. Like giant curtains in the sky that slowly wave as if a gentle breeze were blowing, aurorae fill the entire sky with changing colours and motion and no two aurorae are ever alike. An auroral display may begin with a faint glow, an overall veil, or weak patches in the sky towards the pole. Eventually it will form an arc and may develop rays. Their intensity may vary, multiple bands are likely to form flickering and flaming in surges of brightness across the sky.
What Makes Northern Lights Happen? Click here.
Is Earth the only planet with aurorae?
Clearly no! Every rotating planet with a magnetosphere - or magnetic field should have aurorae. The proof has been given by the famous Hubble telescope in 1996 when it took breath-taking images of a Jupiter aurora. For it's lack of a magnetic field o ur Moon does not display aurorae - what a pity!.
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# Perfect Numbers
Math Lair Home > Topics > Perfect Numbers
Perfect numbers were given their name by the ancient Greek mathematicians, who mixed number theory with mysticism. A perfect number is a number that is equal to the sum of all of its (positive) divisors, excluding itself.
6 is the first perfect number because 6 = 1 + 2 + 3.
28 is also perfect, because 28 = 1 + 2 + 4 + 7 + 14.
The third perfect number is 496, and the fourth is 8128. All of these were found by the ancient Greeks. The fifth perfect number, 33,550,336, is much larger than the first four. It was first mentioned in a fifteenth-century manuscript.
Numbers that are not perfect are classified as either abundant or deficient. Abundant numbers are numbers whose sum of proper divisors is greater than the number itself. The smallest such number is 12, whose divisors sum to 1 + 2 + 3 + 4 + 6 = 16. Deficient numbers are numbers whose sum of proper divisors is less than the number itself.
The Greek mathematician Euclid discovered a pattern that allowed him to show that 496 and 8128 are perfect numbers. He proved (Elements, book IX, proposition 36) that, if 2 k − 1 (where k is a positive integer) is prime, then 2 k-1 × (2 k − 1) is perfect. Primes of the form 2 k − 1 are called Mersenne primes. For example, 8128 = 2 6 × (2 7 − 1). All of the perfect numbers generated by this method are even. In the 18th century, Leonhard Euler proved that all even perfect numbers are of Euclid's form.
On the other hand, odd perfect numbers are a mystery to mathematicians. No-one has found an odd perfect number, but no-one has been able to prove that all perfect numbers are even. There are many conditions that an odd perfect number must satisfy. Euler proved that an odd perfect number must be of the form p a × q b × r c × ..., where p, q, r, etc. are of the form 4n + 1, a is of the form 4n + 1, and b, c, etc. are all even. More recently, several other conditions were discovered. An odd perfect number must be a perfect square multiplied by an odd power of a single prime. It must have at least eight distinct prime factors. If 3 is not one of those factors, at least eleven distinct factors are required. It must also be divisible by a prime power greater than 10 20. The greatest prime factor must be greater than 300,000 and the second largest must be greater than 1000. Any odd perfect less than 10 9118 is divisible by the sixth power of some prime. In Richard Guy's Unsolved Problems in Number Theory, he states that the lower bound for an odd perfect number was (as of 1991) above 10 300, although he writes that there is some scepticism about the later proofs, possibly due to the sloppiness of the proofs. It is possible, though, that some number may just satisfy all of the conditions.
Currently, about 47 perfect numbers are known, although the Great Internet Mersenne Prime Search discovers new ones from time to time. Here are the first seven perfect numbers:
• 6
• 28
• 496
• 8,128
• 33,550,336
• 8,589,869,056
• 137,438,691,328
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## Algebra: A Combined Approach (4th Edition)
Published by Pearson
# Chapter 7 - Section 7.5 - Solving Equations Containing Rational Expressions - Exercise Set: 65
x = 6.667$^{\circ}$ or = 6$\frac{2}{3}$$^{\circ} Thus, \frac{450}{x} = 67.5^{\circ} and, \frac{150}{x} = 22.5^{\circ} #### Work Step by Step Given, that \frac{450}{x} and \frac{150}{x} are complementary angles. Thus, \frac{450}{x} + \frac{150}{x} = 90^{\circ} Since denominators are same (x), thus we can directly add the numerators ( 450 and 150) Thus, \frac{450}{x} + \frac{150}{x} = \frac{600}{x} Given \frac{600}{x} = 90^{\circ} , Thus x = \frac{600}{90}$$^{\circ}$ = $\frac{20}{3}$$^{\circ} = 6.667^{\circ} or = 6\frac{2}{3}$$^{\circ}$ Thus $\frac{450}{x}$ = 450 $\div$ $\frac{20}{3}$ = 450 $\times$ $\frac{3}{20}$ = 67.5$^{\circ}$ $\frac{150}{x}$ = 150 $\div$ $\frac{20}{3}$ = 150 $\times$ $\frac{3}{20}$ = 22.5$^{\circ}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Electrical Analogues of Simple Harmonic Motion in PCB Design
July 31, 2019
Serious students of mathematics, physics, and engineering know that there are a number of mathematical coincidences that pop up in different fields. The mathematical analysis applied to one type of problem in one field can be applied to a completely different problem in another field, typically because the equations that govern these problems are nearly identical.
Electronics and mechanics are not without their correspondence. Simple harmonic motion of a mechanical body and a harmonic response in a circuit can be described with the same equations of motion. Understanding this aids a systems designer in understanding how circuits respond to real oscillating drivers: a circuit may be driven with a periodic source, but that does not mean the resulting motion will be simple.
## Ideal Simple Harmonic Motion in Circuits
In the ideal case, and within the context of electronics design, simple harmonic motion refers to driving a circuit with a sinusoidal source. Linear circuits, meaning any circuit where the current and voltage (or derivatives of either) in different elements are directly proportional, will output a sinusoidal signal. This is a very important consideration as it allows the behavior of a circuit to be analyzed using Ohm’s law and Kirchoff’s laws.
All oscillators have a natural frequency, which refers to the frequency at which they would tend to oscillate if initially displaced from equilibrium. In the case of an undamped pendulum, the natural frequency depends on the length of the pendulum and the acceleration due to gravity. When the pendulum is initially displaced from equilibrium, it will naturally oscillate about the equilibrium position.
The electronic analog of a pendulum is an LC circuit, meaning a circuit that contains a capacitor and an inductor in series. In this case, when the circuit is initially displaced from equilibrium, meaning the capacitor initially holds some charge. Once the circuit is shorted, the current in the circuit will begin oscillating at its natural frequency, just as in the case of a mechanical oscillator.
Charging and discharging LC circuit, leading to simple harmonic motion during discharge
## Real Oscillations in a PCB Are Not Simple
Simple harmonic motion can be produced in a circuit when driven with an AC signal, and resonance can occur when the circuit is driven at a frequency that matches the resonance frequency. Mathematically, the current in the circuit will increase linearly over time and will eventually reach infinity. In a real circuit, even in a passive linear circuit, nonlinear effects will set in as the current grows. This will cause the current to saturate near some maximum value.
In a PCB, every circuit will contain some capacitance and inductance due to the geometry of conductors in the circuit. These capacitance and inductance values are parasitic, but they should be included in circuit models in your board. The presence of these parasitic circuit elements means that each circuit in a PCB has some resonance frequency and can exhibit ringing when driven with digital pulses. This is problematic in transmission lines and illustrates the reason for using termination networks to suppress reflections and ringing.
### Damping in RLC Circuits
Unfortunately, real systems are not free of damping. In a mechanical oscillator, damping arises due to friction and drag. In an electrical oscillator, such as the LC circuit shown above, the wires have some resistance, which will dissipate power as the current oscillates. This causes the current in the circuit to drop to zero over time at an exponential rate, and the rate of damping is equal to half the total resistance in the circuit divided by the inductance. The presence of damping will cause the resonance frequency to shift away from the natural frequency, and it will limit the response of the circuit to some maximum value at resonance.
The typical method for examining the behavior of a circuit in the presence of parasitics is to model the circuit as an RLC network. The exact placement of the parasitic capacitor and inductor in the circuit will depend on the geometry of conductors and placement of components in the circuit.
It is important to note that even components themselves can have some parasitic resistance, capacitance, and inductance. For example, a real capacitor is not perfect and is actually modeled as its own RLC network. This means that components like capacitors actually exhibit a phenomenon called self-resonance. This self-resonance phenomenon can cause EMI problems with a capacitor is driven at its resonance frequency. It also means that a capacitor can exhibit ringing during discharge if the resistance in series with the capacitor is too low, meaning the response will be underdamped.
Effect of damping on the current in an RLC circuit
### Arbitrary Periodic Driving and Nonlinear Circuits
By “arbitrary” and “periodic”, one intends to mean “any signal that has repetitive shape and is not sinusoidal”. Real oscillators in electronic circuits can be driven with an arbitrary waveform. Examples include a stream of repeating digital pulses, a stream of Gaussian or Lorentzian pulses, sawtooth or triangle waves, or any other repetitive shape you can imagine.
Furthermore, not all circuit elements are linear. Instead, the relationship between current and voltage (or their derivatives) is not a straight line. A perfect example is a diode, where the output current is an exponential function of the input voltage. Another example is a transistor, where the output saturates once the input voltages become large. When these types of circuits are driven with a sinusoidal source, output from the circuit will not be sinusoidal, and there could be a phase shift that accumulates, depending on the capacitance and inductance in the circuit.
Arbitrary periodic (non-sinusoidal) sources will produce a periodic response in the circuit, but the response will not always have the same shape as the driver. There are a number of reasons for this. First, passives like inductors and capacitors exhibit a transient response that depends on the shape of an arbitrary driving pulse. Nonlinear circuit elements also affect the shape of the output pulse when the input driver is at a high signal level. As a useful example, consider the output from an inverting amplifier driven with a high voltage sinusoidal source. At high driving strength, the output will saturate as a square wave that is inverted compared to the input signal.
Any equivalent RLC network in a PCB that is driven with digital pulses can exhibit ringing if the circuit is underdamped. In general, any RLC circuit driven with a series of digital pulses will exhibit a transient response in the current in the circuit. This response can be either underdamped, perfectly damped, or overdamped, depending on the series resistance in the RLC network.
Working with the right PCB layout and design software can help you identify design your DC and AC networks to satisfy the maximum power transfer theorem. Allegro PCB Designer and Cadence’s full suite of design tools are designed with the layout and analysis tools you need to ensure power delivery throughout your DC networks and AC circuits.
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|Home | Articles | Forum | Glossary | Books|
<< cont. from part 1
3. THE CLASS C AMPLIFIER
Class C amplifiers are biased so that conduction occurs for much less than 180°.
Class C amplifiers are more efficient than either class A or push-pull class B and class AB, which means that more output power can be obtained from class C operation. The output amplitude is a nonlinear function of the input, so class C amplifiers are not used for linear amplification. They are generally used in radio frequency (RF) applications, including circuits, such as oscillators, that have a constant output amplitude, and modulators, where a high-frequency signal is controlled by a low-frequency signal.
-- Explain and analyze the operation of class C amplifiers
-- Describe basic class C operation
-- Discuss the bias of the transistor
-- Discuss class C power dissipation
-- Explain tuned operation
-- Determine maximum output power
-- Explain clamper bias for a class C amplifier
Basic Class C Operation
The basic concept of class C operation is illustrated in FIG. 21. A common-emitter class C amplifier with a resistive load is shown in FIG. 22(a). A class C amplifier is normally operated with a resonant circuit load, so the resistive load is used only for the purpose of illustrating the concept. It is biased below cutoff with the negative VBB supply.
The ac source voltage has a peak value that is slightly greater than |VBB| + VBE
so that the base voltage exceeds the barrier potential of the base-emitter junction for a short time near the positive peak of each cycle, as illustrated in FIG. 22(b). During this short interval, the transistor is turned on. When the entire ac load line is used, as shown in FIG. 22(c), the ideal maximum collector current is Ic(sat), and the ideal minimum collector voltage is Vce(sat).
The power dissipation of the transistor in a class C amplifier is low because it is on for only a small percentage of the input cycle. FIG. 23(a) shows the collector current pulses.
The time between the pulses is the period (T ) of the ac input voltage. The collector current and the collector voltage during the on time of the transistor are shown in FIG. 23(b).
To avoid complex mathematics, we will assume ideal pulse approximations. Using this simplification, if the output swings over the entire load, the maximum current amplitude is Ic(sat) and the minimum voltage amplitude is Vce(sat) during the time the transistor is on.
The power dissipation during the on time is, therefore,
PD(on) = Ic (sat)Vce (sat)
The transistor is on for a short time, ton, and off for the rest of the input cycle. Therefore, assuming the entire load line is used, the power dissipation averaged over the entire cycle is
Because the collector voltage (output) is not a replica of the input, the resistively loaded class C amplifier alone is of no value in linear applications. It is therefore necessary to use a class C amplifier with a parallel resonant circuit (tank), as shown in FIG. 24(a). The resonant frequency of the tank circuit is determined by the formula fr = 1/(2 pi radic.(LC)).
The short pulse of collector current on each cycle of the input initiates and sustains the oscillation of the tank circuit so that an output sinusoidal voltage is produced, as illustrated in FIG. 24(b). The tank circuit has high impedance only near the resonant frequency, so the gain is large only at this frequency.
The current pulse charges the capacitor to approximately +VCC, as shown in FIG. 25(a). After the pulse, the capacitor quickly discharges, thus charging the inductor.
Then, after the capacitor completely discharges, the inductor's magnetic field collapses and then quickly recharges C to near VCC in a direction opposite to the previous charge.
This completes one half-cycle of the oscillation, as shown in parts (b) and (c) of FIG. 25. Next, the capacitor discharges again, increasing the inductor's magnetic field.
The inductor then quickly recharges the capacitor back to a positive peak slightly less than the previous one, due to energy loss in the winding resistance. This completes one full cycle, as shown in parts (d) and (e) of FIG. 25. The peak-to-peak output voltage is therefore approximately equal to 2VCC.
The amplitude of each successive cycle of the oscillation will be less than that of the previous cycle because of energy loss in the resistance of the tank circuit, as shown in FIG. 26(a), and the oscillation will eventually die out. However, the regular recurrences of the collector current pulse re-energizes the resonant circuit and sustains the oscillations at a constant amplitude.
When the tank circuit is tuned to the frequency of the input signal (fundamental), re energizing occurs on each cycle of the tank voltage, Vr , as shown in FIG. 26(b). When the tank circuit is tuned to the second harmonic of the input signal, re-energizing occurs on alternate cycles as shown in FIG. 26(c). In this case, a class C amplifier operates as a frequency multiplier (x 2). By tuning the resonant tank circuit to higher harmonics, further frequency multiplication factors are achieved.
Maximum Output Power
Since the voltage developed across the tank circuit has a peak-to-peak value of approximately 2VCC, the maximum output power can be expressed as:
Rc is the equivalent parallel resistance of the collector tank circuit at resonance and represents the parallel combination of the coil resistance and the load resistance. It usually has a low value. The total power that must be supplied to the amplifier is:
Therefore, the efficiency is:
When the class C efficiency closely approaches 1 (100 percent).
[ EQN. 10]
Clamper Bias for a Class C Amplifier
FIG. 27 shows a class C amplifier with a base bias clamping circuit. The base-emitter junction functions as a diode.
When the input signal goes positive, capacitor C1 is charged to the peak value with the polarity shown in FIG. 28(a). This action produces an average voltage at the base of approximately -Vp.
This places the transistor in cutoff except at the positive peaks, when the transistor conducts for a short interval. For good clamping action, the R1C1 time constant of the clamping circuit must be much greater than the period of the input signal. Parts (b) through (f) of FIG. 28 illustrate the bias clamping action in more detail. During the time up to the positive peak of the input (t0 to t1), the capacitor charges to Vp - 0.7 V through the base-emitter diode, as shown in part (b). During the time from t1 to t2, as shown in part (c), the capacitor discharges very little because of the large RC time constant. The capacitor, therefore, maintains an average charge slightly less than Vp - 0.7 V.
Since the dc value of the input signal is zero (positive side of C1), the dc voltage at the base (negative side of C1) is slightly more positive than -(Vp - 0.7 V), as indicated in FIG. 28(d). As shown in FIG. 28(e), the capacitor couples the ac input signal through to the base so that the voltage at the transistor's base is the ac signal riding on a dc level slightly more positive than -(Vp - 0.7 V).
Near the positive peaks of the input volt age, the base voltage goes slightly above 0.7 V and causes the transistor to conduct for a short time, as shown in FIG. 28(f).
SECTION 3 CHECKUP
1. At what point is a class C amplifier normally biased?
2. What is the purpose of the tuned circuit in a class C amplifier?
3. A certain class C amplifier has a power dissipation of 100 mW and an output power of 1 W. What is its percent efficiency?
In this section, examples of isolating a component failure in a circuit are presented.
We will use a class A amplifier and a class AB amplifier with the output voltage monitored by an oscilloscope. Several incorrect output waveforms will be examined and the most likely faults will be discussed.
After completing this section, you should be able to
-- Troubleshoot power amplifiers
-- Troubleshoot a class A amplifier for various faults
-- Troubleshoot a class AB amplifier for various faults
Case 1: Class A
As shown in FIG. 30, the class A power amplifier should have a normal sinusoidal out put when a sinusoidal input signal is applied.
Now let's consider four incorrect output waveforms and the most likely causes in each case. In FIG. 31(a), the scope displays a dc level equal to the dc supply voltage, indicating that the transistor is in cutoff. The two most likely causes of this condition are (1) the transistor has an open pn junction, or (2) R4 is open, preventing collector and emitter current.
In FIG. 31(b), the scope displays a dc level at the collector approximately equal to the dc emitter voltage. The two probable causes of this indication are (1) the transistor is shorted from collector to emitter, or (2) R2 is open, causing the transistor to be biased in saturation. In the second case, a sufficiently large input signal can bring the transistor out of saturation on its negative peaks, resulting in short pulses on the output.
In FIG. 31(c), the scope displays an output waveform that indicates the transistor is in cutoff except during a small portion of the input cycle. Possible causes of this indication are (1) the Q-point has shifted down due to a drastic out-of-tolerance change in a resistor value, or (2) R1 is open, biasing the transistor in cutoff. The display shows that the input signal is sufficient to bring it out of cutoff for a small portion of the cycle.
In FIG. 31(d), the scope displays an output waveform that indicates the transistor is saturated except during a small portion of the input cycle. Again, it is possible that an incorrect resistance value has caused a drastic shift in the Q-point up toward saturation, or R2 is open, causing the transistor to be biased in saturation, and the input signal is bringing it out of saturation for a small portion of the cycle.
Case 2: ClassAB
As shown in FIG. 32, the class AB push-pull amplifier should have a sinusoidal output when a sinusoidal input signal is applied.
Two incorrect output waveforms are shown in FIG. 33. The waveform in part (a) shows that only the positive half of the input signal is present on the output. One possible cause is that diode D1 is open. If this is the fault, the positive half of the input signal forward biases D2 and causes transistor Q2 to conduct. Another possible cause is that the base-emitter junction of Q2 is open so only the positive half of the input signal appears on the output because Q1 is still working.
(a) D1 open or Q2 base-emitter open (b) D2 open or Q1 base-emitter open
The waveform in FIG. 33(b) shows that only the negative half of the input signal is present on the output. One possible cause is that diode D2 is open. If this is the fault, the negative half of the input signal forward-biases D1 and places the half-wave signal on the base of Q1. Another possible cause is that the base-emitter junction of Q1 is open so only the negative half of the input signal appears on the output because Q2 is still working.
SECTION 4 CHECKUP
1. What would you check for if you noticed clipping at both peaks of the output waveform?
2. A significant loss of gain in the amplifier of FIG. 30 would most likely be caused by what type of failure?
Application Activity: The Complete PA System
The class AB power amplifier follows the audio preamp and drives the speaker as shown in the PA system block diagram in FIG. 34. In this application, the power amplifier is developed and interfaced with the preamp that was developed in Section 6. The maximum signal power to the speaker should be approximately 6 W for a frequency range of 70 Hz to 5 kHz. The dynamic range for the input voltage is up to 40 mV. Finally, the complete PA system is put together.
The Power Amplifier Circuit
The schematic of the push-pull power amplifier is shown in FIG. 35. The circuit is a class AB amplifier implemented with Darlington configurations and diode current mirror bias. Both a traditional Darlington pair and a complementary Darlington (Sziklai) pair are used to provide sufficient current to an speaker load. The signal from the preamp is 8 Ω
capacitively coupled to the driver stage, Q5, which is used to prevent excessive loading on the preamp and provide additional gain. Notice that Q5 is biased with the dc output volt age (0 V) fed back through R1. Also, the signal voltage fed back to the base of Q5 is out of-phase with the signal from the preamp and has the effect of stabilizing the gain. This is called negative feedback. The amplifier will deliver up to 5 W to an speaker.
A partial datasheet for the BD135 power transistor is shown in FIG. 36.
1. Estimate the input resistance of the power amplifier in FIG. 35.
2. Calculate the approximate voltage gain of the power amplifier in FIG. 35? 8 Ω
The power amplifier is simulated using software with a 1 kHz input signal at near its maximum linear operation. The results are shown in FIG. 37 where an 8.2 ohm resistor is used to closely approximate the 8 ohm speaker.
3. Calculate the power to the load in FIG. 37.
4. What is the measured voltage gain? The input is a peak value.
5. Compare the measured gain to the calculated gain for the amplifier in FIG. 35.
The Complete Audio Amplifier
Both the preamp and the power amp have been simulated individually. Now, they must work together to produce the required signal power to the speaker. FIG. 38 is the simulation of the combined audio preamp and power amp. Components in the power amplifier are now numbered sequentially with the preamp components.
6. Calculate the power to the load in FIG. 38.
7. What is the measured voltage gain of the power amplifier?
8. What is the measured overall voltage gain?
Simulate the audio amplifier using software. Observe the operation with the virtual oscilloscope.
Prototyping and Testing
Now that the circuit has been simulated, the prototype circuit is constructed and tested.
After the circuit is successfully tested on a protoboard, it is ready to be finalized on a printed circuit board.
The power amplifier is implemented on a printed circuit board as shown in FIG. 39.
Heat sinks are used to provide additional heat dissipation from the power transistors.
9. Check the printed circuit board and verify that it agrees with the schematic in FIG. 35. The volume control potentiometer is mounted off the PC board for easy access.
10. Label each input and output pin according to function. Locate the single back side trace.
Troubleshooting the Power Amplifier Board A power amplifier circuit board has failed the production test. Test results are shown in FIG. 40.
11. Based on the scope displays, list possible faults for the circuit board.
Putting the System Together
The preamp circuit board and the power amplifier circuit board are interconnected and the dc power supply (battery pack), microphone, speaker, and volume control potentiometer are attached, as shown in FIG. 41.
12. Verify that the system interconnections are correct.
-- A class A power amplifier operates entirely in the linear region of the transistor's characteristic curves. The transistor conducts during the full 360 deg. of the input cycle.
-- The Q-point must be centered on the load line for maximum class A output signal swing.
-- The maximum efficiency of a class A power amplifier is 25 percent.
-- A class B amplifier operates in the linear region for half of the input cycle and it is in cutoff for the other half.
-- The Q-point is at cutoff for class B operation.
-- Class B amplifiers are normally operated in a push-pull configuration in order to produce an output that is a replica of the input.
-- The maximum efficiency of a class B amplifier is 79 percent.
-- A class AB amplifier is biased slightly above cutoff and operates in the linear region for slightly more than 180 deg. of the input cycle.
-- Class AB eliminates crossover distortion found in pure class B.
- - A class C amplifier operates in the linear region for only a small part of the input cycle.
-- The class C amplifier is biased below cutoff.
-- Class C amplifiers are normally operated as tuned amplifiers to produce a sinusoidal output.
-- The maximum efficiency of a class C amplifier is higher than that of either class A or class B amplifiers. Under conditions of low power dissipation and high output power, the efficiency can approach 100 percent.
Class A -- A type of amplifier that operates entirely in its linear (active) region.
Class AB -- A type of amplifier that is biased into slight conduction.
Class B -- A type of amplifier that operates in the linear region for 180 deg. of the input cycle because it is biased at cutoff.
Class C --- A type of amplifier that operates only for a small portion of the input cycle.
Efficiency--The ratio of the signal power delivered to a load to the power from the power supply of an amplifier.
Power gain -- The ratio of output power to input power of an amplifier.
Push-Pull -- A type of class B amplifier with two transistors in which one transistor conducts for one half-cycle and the other conducts for the other half-cycle.
1. Class A power amplifiers are a type of large-signal amplifier.
2. Ideally, the Q-point should be centered on the load line in a class A amplifier.
3. The quiescent power dissipation occurs when the maximum signal is applied.
4. Efficiency is the ratio of output signal power to total power.
5. Each transistor in a class B amplifier conducts for the entire input cycle.
6. Class AB operation overcomes the problem of crossover distortion.
7. Complementary symmetry transistors must be used in a class AB amplifier.
8. A current mirror is implemented with a laser diode.
9. Darlington transistors can be used to increase the input resistance of a class AB amplifier.
10. The transistor in a class C amplifier conducts for a small portion of the input cycle.
11. The output of a class C amplifier is a replica of the input signal.
12. A class C amplifier usually employs a tuned circuit.
1. If the value of R3 in FIG. 5 is decreased, the voltage gain of the first stage will (a) increase (b) decrease (c) not change
2. If the value of RE2 in FIG. 5 is increased, the voltage gain of the first stage will (a) increase (b) decrease (c) not change
3. If C2 in FIG. 5 opens, the dc voltage at the emitter of Q1 will (a) increase (b) decrease (c) not change
4. If the value of R4 in FIG. 5 is increased, the dc voltage at the base of Q3 will (a) increase (b) decrease (c) not change
5. If VCC in FIG. 18 is increased, the peak output voltage will (a) increase (b) decrease (c) not change
6. If the value of RL in FIG. 18 is increased, the ac output power will (a) increase (b) decrease (c) not change
7. If the value of RL in FIG. 19 is decreased, the voltage gain will (a) increase (b) decrease (c) not change
8. If the value of VCC in FIG. 19 is increased, the ac output power will (a) increase (b) decrease (c) not change
9. If the values of R1 and R2 in FIG. 19 are increased, the voltage gain will (a) increase (b) decrease (c) not change
10. If the value of C2 in FIG. 24 is decreased, the resonant frequency will (a) increase (b) decrease (c) not change
1. An amplifier that operates in the linear region at all times is (a) Class A (b) Class AB (c) Class B (d) Class C
2. A certain class A power amplifier delivers 5 W to a load with an input signal power of 100 mW. The power gain is (a) 100 (b) 50 (c) 250 (d) 5
3. The peak current a class A power amplifier can deliver to a load depends on the (a) maximum rating of the power supply (b) quiescent current (c) current in the bias resistors (d) size of the heat sink
4. For maximum output, a class A power amplifier must maintain a value of quiescent current that is (a) one-half the peak load current (b) twice the peak load current (c) at least as large as the peak load current (d) just above the cutoff value
5. A certain class A power amplifier has
The maximum signal power output is (a) 6 W (b) 12 W (c) 1 W (d) 0.707 W
6. The efficiency of a power amplifier is the ratio of the power delivered to the load to the (a) input signal power (b) power dissipated in the last stage (c) power from the dc power supply (d) none of these answers
7. The maximum efficiency of a class A power amplifier is (a) 25% (b) 50% (c) 79% (d) 98% Section 2 8. The transistors in a class B amplifier are biased (a) into cutoff (b) in saturation (c) at midpoint of the load line (d) right at cutoff
9. Crossover distortion is a problem for (a) class A amplifiers (b) class AB amplifiers (c) class B amplifiers (d) all of these amplifiers
10. A BJT class B push-pull amplifier with no transformer coupling uses (a) two npn transistors (b) two pnp transistors (c) complementary symmetry transistors (d) none of these
11. A current mirror in a push-pull amplifier should give an ICQ that is (a) equal to the current in the bias resistors and diodes (b) twice the current in the bias resistors and diodes (c) half the current in the bias resistors and diodes (d) zero
12. The maximum efficiency of a class B push-pull amplifier is (a) 25% (b) 50% (c) 79% (d) 98%
13. The output of a certain two-supply class B push-pull amplifier has a VCC of 20 V. If the load resistance is the value of Ic(sat) is (a) 5 mA (b) 0.4 A (c) 4 mA (d) 40 mA
14. The maximum efficiency of a class AB amplifier is (a) higher than a class B (b) the same as a class B (c) about the same as a class A (d) slightly less than a class B Section 3
15. The power dissipation of a class C amplifier is normally (a) very low (b) very high (c) the same as a class B (d) the same as a class A
16. The efficiency of a class C amplifier is (a) less than class A (b) less than class B (c) less than class AB (d) greater than classes A, B, or AB 50 Ω , VCEQ = 12 V and ICQ = 1 A.
17. The transistor in a class C amplifier conducts for (a) more than of the input cycle; (b) one-half of the input cycle (c) a very small percentage of the input cycle; (d) all of the input cycle 180°
Section 1 The Class A Power Amplifier
1. FIG. 42 shows a CE power amplifier in which the collector resistor serves also as the load resistor. Assume βDC = βac = 100.
(a) Determine the dc Q-point (ICQ and VCEQ).
(b) Determine the voltage gain and the power gain.
2. For the circuit in FIG. 42, determine the following:
(a) the power dissipated in the transistor with no load (b) the total power from the power supply with no load (c) the signal power in the load with a 500 mV input
3. Refer to the circuit in FIG. 42. What changes would be necessary to convert the circuit to a pnp transistor with a positive supply? What advantage would this have?
4. Assume a CC amplifier has an input resistance of 2.2 k Ω and drives an output load of 50 Ω .
What is the power gain?
5. Determine the Q-point for each amplifier in FIG. 43.
6. If the load resistor in FIG. 43(a) is changed to 50 Ω , how much does the Q-point change?
7. What is the maximum peak value of collector current that can be realized in each circuit of FIG. 43? What is the maximum peak value of output voltage in each circuit?
8. Find the power gain for each circuit in FIG. 43. Neglect
9. Determine the minimum power rating for the transistor in FIG. 44.
10. Find the maximum output signal power to the load and efficiency for the amplifier in FIG. 44 with a load resistor. 500 Ω r’ e.
Section 2 The Class B and Class AB Push-Pull Amplifiers
11. Refer to the class AB amplifier in FIG. 45.
(a) Determine the dc parameters VB(Q1), VB(Q2), VE, ICQ, VCEQ(Q1), VCEQ(Q2).
(b) For the 5 V rms input, determine the power delivered to the load resistor.
12. Draw the load line for the npn transistor in FIG. 45. Label the saturation current, Ic(sat), and show the Q-point.
13. Determine the approximate input resistance seen by the signal source for the amplifier of FIG. 45 if 50 Ω βac = 100.
14. If D2 has more voltage drop than D1, what effect does this have on the output?
15. Refer to the class AB amplifier in FIG. 46 operating with a single power supply.
(a) Determine the dc parameters VB(Q1), VB(Q2), VE, ICQ, VCEQ(Q1), VCEQ(Q2).
(b) Assuming the input voltage is 10 V pp, determine the power delivered to the load resistor.
16. Refer to the class AB amplifier in FIG. 46.
(a) What is the maximum power that could be delivered to the load resistor? (b) Assume the power supply voltage is raised to 24 V. What is the new maximum power that could be delivered to the load resistor?
17. Refer to the class AB amplifier in FIG. 46. What fault or faults could account for each of the following troubles? (a) a positive half-wave output signal (b) zero volts on both bases and the emitters (c) no output: emitter voltage = -15 V (d) crossover distortion observed on the output waveform
18. If a 1 V rms signal source with an internal resistance of is connected to the amplifier in FIG. 46, what is the actual rms signal applied to the amplifier input? Assume βac = 200.
Section 3 The Class C Amplifier 19. A certain class C amplifier transistor is on for 10 percent of the input cycle. If Vce(sat) = 0.18 V and Ic (sat) = 25 mA, what is the average power dissipation for maximum output?
20. What is the resonant frequency of a tank circuit with L = 10 mH and C = 0.001 mF?
21. What is the maximum peak-to-peak output voltage of a tuned class C amplifier with VCC = 12 V?
22. Determine the efficiency of the class C amplifier described in Problem 21 if VCC =15 V and the equivalent parallel resistance in the collector tank circuit is 50 Ω . Assume that the transistor is on for 10% of the period.
Section 4 Troubleshooting
23. Refer to FIG. 47. What would you expect to observe across RL if C1 opened?
24. Your oscilloscope displays a half-wave output when connected across RL in FIG. 47. What is the probable cause?
25. Determine the possible fault or faults, if any, for each circuit in FIG. 48 based on the indicated dc voltage measurements.
APPLICATION ACTIVITY PROBLEMS
26. Assume that the public address system represented by the block diagram in FIG. 34 has quit working. You find there is no signal output from the power amplifier or the preamplifier, but you have verified that the microphone is working. Which two blocks are the most likely to be the problem? How would you narrow the choice down to one block?
27. Describe the output that would be observed in the push-pull amplifier of FIG. 35 with a 2 V rms sinusoidal input voltage if the base-emitter junction of Q2 opened.
28. Describe the output that would be observed in FIG. 35 if the collector-emitter junction of Q5 opened for the same input as in Problem 27.
29. After visually inspecting the power amplifier circuit board in FIG. 49, describe any problems.
30. Referring to the datasheet in FIG. 50, determine the following:
(a) minimum for the BD135 and the conditions
(b) maximum collector-to-emitter voltage for the BD135
(c) maximum power dissipation for the BD135 at a case temperature of 25°C
(d) maximum continuous collector current for the BD135
31. Determine the maximum power dissipation for a BD135 at a case temperature of 50°C
32. Determine the maximum power dissipation for a BD135 at an ambient temperature of 50°C.
33. Describe what happens to the dc current gain as the collector current increases.
34. Determine the approximate hFE for the BD135 at IC = 20 mA.
35. Explain why the specified maximum power dissipation of a power transistor at an ambient temperature of 25°C is much less than maximum power dissipation at a case temperature of 25°C .
36. Draw the dc and the ac load lines for the amplifier in FIG. 51.
37. Design a swamped class A power amplifier that will operate from a dc supply of +15 V with an approximate voltage gain of 50. The quiescent collector current should be approximately 500 mA, and the total dc current from the supply should not exceed 750 mA. The output power must be at least 1 W.
38. The public address system in FIG. 34 is a portable unit that is independent of 115 V ac.
Determine the ampere-hour rating for the +15 V and the -15 V battery supply necessary for the system to operate for 4 hours on a continuous basis.
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Medium Chain Acyl-CoA Dehydrogenase Deficiency (MCADD)
What is MCADD?
MCADD stands for "medium chain acyl-CoA dehydrogenase deficiency". It is one type of fatty acid oxidation disorder. People with this type of disorder have problems breaking down fat into energy for the body. The enzyme "medium chain acyl-CoA dehydrogenase" is either missing or not working properly. This enzyme’s job is to break down certain fats in the food we eat into energy. It also breaks down fat already stored in the body. When the enzyme is missing or not working, the body cannot use fat for energy, and must rely only on glucose, a type of sugar. Although glucose is a good source of energy, there is a limited amount available. When all the glucose has been used up, the body tries to use fat for energy, but can't. This leads to low blood sugar and to the build up of harmful substances in the blood.
What are the symptoms?
Some of the first symptoms are extreme sleepiness, behavior changes, irritable mood, and poor appetite. Children with MCADD usually get symptoms for the first time between three months and two years of age. When children have symptoms, it's called a metabolic crisis. A crisis can often be set off by an infection, so illness may start with symptoms of fever, diarrhea, and vomiting. Low blood sugar then follows. If a metabolic crisis is not treated, a child with MCADD can develop breathing problems, seizures, and coma, sometimes leading to death.
What is the treatment?
The baby's primary doctor will work with a metabolic specialist to develop an individual treatment plan for the baby. A dietician familiar with MCADD may also help develop that treatment plan. Treatment can include:
Avoiding going a long time without food - Babies and young children with MCADD need to eat often to avoid low blood sugar or a metabolic crisis. They may need starchy snacks before going to bed and first thing in the morning. They may even need to be woken up during the night for another snack.
A special diet - Sometimes a low fat, high carbohydrate diet is recommended.
L-carnitine supplements - This is a safe and natural substance that helps the body make energy and helps the body get rid of harmful wastes.
Parents' Guide to MCADD (PDF)
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# Definition:Group
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## Definition
A group is a semigroup with an identity (that is, a monoid) in which every element has an inverse.
### Group Axioms
The properties that define a group are sufficiently important that they are often separated from their use in defining semigroups, monoids, and so on, and given recognition in their own right.
A group is an algebraic structure $\struct {G, \circ}$ which satisfies the following four conditions:
$(\text G 0)$ $:$ Closure $\ds \forall a, b \in G:$ $\ds a \circ b \in G$ $(\text G 1)$ $:$ Associativity $\ds \forall a, b, c \in G:$ $\ds a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(\text G 2)$ $:$ Identity $\ds \exists e \in G: \forall a \in G:$ $\ds e \circ a = a = a \circ e$ $(\text G 3)$ $:$ Inverse $\ds \forall a \in G: \exists b \in G:$ $\ds a \circ b = e = b \circ a$
These four stipulations are called the group axioms.
The notation $\struct {G, \circ}$ is used to represent a group whose underlying set is $G$ and whose operation is $\circ$.
### Group Law
The operation $\circ$ can be referred to as the group law.
### Product Element
Let $a, b \in G$ such that $= a \circ b$.
Then $g$ is known as the product of $a$ and $b$.
### Multiplicative Notation
When discussing a general group with a general group law, it is customary to dispense with a symbol for this operation and merely concatenate the elements to indicate the product element.
$x y$ is used to indicate the result of the operation on $x$ and $y$. There is no symbol used to define the operation itself.
$e$ or $1$ is used for the identity element.
$x^{-1}$ is used for the inverse element.
$x^n$ is used to indicate the $n$th power of $x$.
Compare with additive notation.
## Also denoted as
Some sources denote a group using the notation $\gen {G, \circ}$ for $\struct {G, \circ}$.
## Examples
### Example: $\dfrac {x + y} {1 + x y}$
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.
### Example: $x + y + 2$ over $\R$
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
$\forall x, y \in \R: x \circ y := x + y + 2$
Then $\struct {\R, \circ}$ is a group whose identity is $-2$.
### Example: $x + y + x y$ over $\R \setminus \set {-1}$
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
$\forall x, y \in \R: x \circ y := x + y + x y$
Let:
$\R' := \R \setminus \set {-1}$
that is, the set of real numbers without $-1$.
Then $\struct {\R', \circ}$ is a group whose identity is $0$.
### Example: $x^{-1} = 1 - x$
Let $S = \set {x \in \R: 0 < x < 1}$.
Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.
### Example: Group of Linear Functions
Let $G$ be the set of all real functions $\theta_{a, b}: \R \to \R$ defined as:
$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
where $a, b \in \R$ such that $a \ne 0$.
The algebraic structure $\struct {G, \circ}$, where $\circ$ denotes composition of mappings, is a group.
$\struct {G, \circ}$ is specifically non-abelian.
### Example: Operation Induced by Self-Inverse and Cancellable Elements
Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
$(1): \quad \exists e \in S: a \ast b = e \iff a = b$
$(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$
Then $\struct {S, \circ}$ is a group, where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.
## Examples of Structures which are not Groups
### Arbitrary Example: Order 4
Let $S = \set {1, 2, 3, 4}$.
Consider the algebraic structure $\struct {S, \circ}$ given by the Cayley table:
$\begin{array}{r|rrrr} \circ & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 3 & 1 \\ 3 & 3 & 2 & 4 & 3 \\ 4 & 4 & 3 & 1 & 2 \\ \end{array}$
Then $\struct {S, \circ}$ is not a group.
## Also see
• Results about groups can be found here.
## Historical Note
The term group was first used by Évariste Galois in $1832$, in the context of the solutions of polynomials in radicals. Augustin Louis Cauchy was also involved in this development.
The concept of the group as a purely abstract structure was introduced by Arthur Cayley in his $1854$ paper On the theory of groups.
The first one to formulate the set of axioms to define the structure of a group was Leopold Kronecker in $1870$.
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You are viewing an older version of this Concept. Go to the latest version.
# Definition of Variable
## A variable can be any unknown value, and will be a critical part of our curriculum beginning with algebra all the way up through calculus!
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Definition of Variable
What if you had a jar filled with dimes and quarters? You know that the total of the coins in the jar is $8.60. How could you write an equation to represent this situation? After completing this Concept, you'll be able to use variables to write equations like this one with unknown quantities. ### Watch This ### Guidance No one likes doing the same problem over and over again—that’s why mathematicians invented algebra. Algebra takes the basic principles of math and makes them more general, so we can solve a problem once and then use that solution to solve a group of similar problems. In arithmetic, you’ve dealt with numbers and their arithmetical operations (such as ). In algebra, we use symbols called variables (which are usually letters, such as ) to represent numbers and sometimes processes. For example, we might use the letter to represent some number we don’t know yet, which we might need to figure out in the course of a problem. Or we might use two letters, like and , to show a relationship between two numbers without needing to know what the actual numbers are. The same letters can represent a wide range of possible numbers, and the same letter may represent completely different numbers when used in two different problems. Using variables offers advantages over solving each problem “from scratch.” With variables, we can: • Formulate arithmetical laws such as for all real numbers and . • Refer to “unknown” numbers. For instance: find a number such that . • Write more compactly about functional relationships such as, “If you sell tickets, then your profit will be dollars, or “,” where “” is the profit function, and is the input (i.e. how many tickets you sell). #### Example A Write an algebraic equation for the perimeter and area of the rectangle below. To find the perimeter, we add the lengths of all 4 sides. We can still do this even if we don’t know the side lengths in numbers, because we can use variables like and to represent the unknown length and width. If we start at the top left and work clockwise, and if we use the letter to represent the perimeter, then we can say: We are adding ’s and 's, so we can say that: It's customary in algebra to omit multiplication symbols whenever possible. For example, means the same thing as or . We can therefore also write: Area is length multiplied by width. In algebraic terms we get: Note: by itself is an example of a variable expression; is an example of an equation. The main difference between expressions and equations is the presence of an equals sign (=). In the above example, we found the simplest possible ways to express the perimeter and area of a rectangle when we don’t yet know what its length and width actually are. Now, when we encounter a rectangle whose dimensions we do know, we can simply substitute (or plug in) those values in the above equations. In this chapter, we will encounter many expressions that we can evaluate by plugging in values for the variables involved. #### Example B Eric has some money in his savings account. How much more money does he need in order to buy a game that costs$98?
Solution:
Let be the money that Eric still needs and let be the money that Eric has in his savings account. Then, by subtracting the money he already has from the total money needed, we can figure out how much money he still needs:
#### Example C
Write an equation for the sum of 3 times some number and 5.
Solution:
Let be the total sum. Let be some number. Then 3 times some number is and then the sum of that and 5 is:
Watch this video for help with the Examples above.
### Vocabulary
• We use symbols called variables (which are usually letters, such as ) to represent numbers and sometimes processes.
• by itself is an example of a variable expression; is an example of an equation. The main difference between expressions and equations is the presence of an equals sign (=).
### Guided Practice
Alex has a certain amount of nickels and dimes in a jar. Write an algebraic equation for how much money she has, in terms of how many nickles and dimes she has.
Solution:
Let be the number of nickels and be the number of dimes that Alex has in the jar. Since each nickel is worth $0.05, the amount of money she has in nickels will be: Since each dime is worth$0.10, the amount of money she has in dimes will be:
This means that the total amount of money that Alex has will be:
Simplifying the expressions, we get:
### Practice
For 1-4, write the following in a more condensed form by leaving out a multiplication symbol.
For 5-10, write an equation for the following situations.
1. The amount of money Andrea has in a jar full of quarters and dimes.
2. The amount of money Michelle has in her coin purse if it only contains quarters, dimes and pennies.
3. The sum of 7 and 6 times some number.
4. 4 less than 20 times some number.
5. The amount of money you will earn if you are paid $10.25 an hour and spend$4.00 round trip to get too and from work.
6. A father earns a \$2000 dividend from an oil investment and distributes it equally amongst his children.
### Vocabulary Language: English
Equation
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
Variable Expression
Variable Expression
A variable expression is a mathematical phrase that contains at least one variable or unknown quantity.
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Albert Einstein’s general theory of relativity is one of the greatest achievements of 20th-century physics. Published in 1916, it says that what we perceive as the force of gravity in fact arises from the curvature of space and time. Einstein proposed that objects such as the sun and the Earth change this geometry. Most of the major points of his theory has been proven since, yet there are some – arguably the most exciting ones – which still need to be proved or disproved. Read more about them below the image.
Gravitational lens: a distribution of matter (such as a cluster of galaxies) between a distant light source and an observer that is capable of bending the light from the source as the light travels towards the observer. Unlike an optical lens, a gravitational lens produces a maximum deflection of light that passes closest to its center, and a minimum deflection of light that travels furthest from its center. Consequently, a gravitational lens has no single focal point, but a focal line.
Gravitational waves are disturbances in the curvature of spacetime, generated by accelerated masses, that propagate as waves outward from their source at the speed of light. Gravitational waves transport energy as gravitational radiation, a form of radiant energy similar to electromagnetic radiation.
Time dilation is a difference in the elapsed time measured by two clocks, either due to them having a velocity relative to each other, or by there being a gravitational potential difference between their locations. After compensating for varying signal delays due to the changing distance between an observer and a moving clock (i.e. Doppler effect), the observer will measure the moving clock as ticking slower than a clock that is at rest in the observer’s own reference frame. A clock that is close to a massive body (and which therefore is at lower gravitational potential) will record less elapsed time than a clock situated further from the said massive body (and which is at a higher gravitational potential).
Black holes are regions of spacetime exhibiting gravitational acceleration so strong that nothing – no particles or even electromagnetic radiation such as light – can escape from them. The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole. The boundary of the region from which no escape is possible is called the event horizon. Although the event horizon has an enormous effect on the fate and circumstances of an object crossing it, no locally detectable features appear to be observed. In many ways, a black hole acts like an ideal black body, as it reflects no light.
White holes: Physicists describe a white hole as a black hole’s ‘time reversal’, a footage of a black hole played backwards, much as a bouncing ball is the time reversal of a falling ball. While a black hole’s event horizon is a sphere of no return, a white hole’s event horizon is a boundary of no admission – space-time’s most exclusive club where no spacecraft can ever reach the region’s edge. Objects inside a white hole can leave and interact with the outside world, but since nothing can get in, the interior is cut off cut off from the universe’s past: No outside event will ever affect the inside. Although information and evidence regarding white holes remains inconclusive, the 2006 GRB 060614 has been proposed as the first documented observance of a white hole.
A wormhole is a speculative structure linking disparate points in spacetime which is based on a special solution of the Einstein field equations. A wormhole can be visualized as a tunnel with two ends at separate points in spacetime (i.e. different locations, or different points in time, or both). Wormholes are consistent with the general theory of relativity, but whether they actually exist remains to be seen. A wormhole could connect extremely long distances such as a billion light years or more, short distances such as a few meters, different universes, or different points in time.
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Ocular hypertension is a condition in which the measured eye pressure (intraocular pressure or IOP) is consistently greater than “normal." However, there is no obvious damage to the optic nerve as detected by an eye examination, optic nerve imaging, or evidence of visual field change. In other words, there is no evidence of glaucoma yet.
In considering what constitutes ocular hypertension, it is important to recognize that “normal” eye pressure likely depends on many different factors, including age and ethnicity, as well as factors such as corneal thickness (more on that below).
In the United States, the prevalence of ocular hypertension in non-Hispanic Whites who are 40 years of age and older is 4.5 percent, and increases up to 7.7 percent in 75 to 79 year olds. In Latinos, the prevalence across ages is quite similar. Because ocular hypertension occurs typically without symptoms, a majority of people with ocular hypertension remains undiagnosed. In the Los Angeles Latino Eye Study, 75 percent of Latinos with eye pressures greater than 21 mmHg were previously undiagnosed.
Video: Ocular Hypertension
Why is Ocular Hypertension Important?
Eye pressure is one of the major risk factors for the development of glaucoma. That being said, there are also plenty of individuals who never demonstrate elevated eye pressure but do have glaucoma. This is called “normal-tension” or “low-tension” glaucoma.
What are the Risk Factors?
It is estimated that three to six million persons in the United States have ocular hypertension. These are individuals whom an eye care provider will follow closely for the development of glaucoma. The likelihood of developing glaucoma increases with the number and relative strength of risk factors.
Risk factors of ocular hypertension include:
- higher eye pressure,
- older age,
- family history of glaucoma,
- lower ocular perfusion pressure (this is related to blood pressure and eye pressure),
- lower systemic blood pressure,
- thinner central cornea, and
- bleeding at the optic nerve head.
Central corneal thickness is an independent risk factor for the development of glaucoma. The cornea is the clear “window” of the eye and whenever eye pressure is measured, the biomechanical properties of the cornea play a role in the accuracy of the measurement.
When using applanation tonometry (one of the most common forms of eye pressure measurement), eye pressure may be overestimated with thicker corneas and underestimated with thinner corneas. There has been a lot of research trying to establish the exact relationship between eye pressure as measured by various techniques and central corneal thickness, but there is no widely accepted formula to account for an individual’s central corneal thickness. However, most eye care providers would take into consideration central corneal thickness. For example, a patient with thin corneas and high eye pressure would be considered higher risk than a patient with thick corneas and high eye pressure.
Low Blood Pressure
It is interesting to note that while higher blood pressure can indeed raise eye pressure by a very small amount, lower blood pressure is actually a risk factor for the development of glaucoma. This is likely due to the fact that lower blood pressure means lower perfusion pressure and decreased blood flow to the eye and optic nerve, which is a risk factor for the development of glaucoma.
How Quickly Can Ocular Hypertension Develop Into Glaucoma?
One well done and critical study that provided important information about ocular hypertension and the development of glaucoma is the Ocular Hypertension Treatment Study (OHTS). It demonstrated that the rate of untreated ocular hypertension patients in developing glaucoma was 9.5 percent in 5 years and 22 percent at 13 years, or about 2 percent per year. With treatment, the risk of developing glaucoma was reduced by about 50 percent.
When Should Treatment Begin?
So how do we decide whether a patient with ocular hypertension should be treated? This decision is complex, and should be undertaken together by both the patient and eye doctor. The eye doctor will consider the findings of the eye exam, risk factor assessment, testing such as optic nerve imaging and visual field testing and the patient’s preferences and ability to adhere to a treatment plan. The decision to begin treatment is particularly important in the case of ocular hypertension patients or glaucoma suspects, because treatment is long-term and exposes the patient to the side effects and costs of treatment. For some patients, the eye doctor and patient will reach the conclusion that the risk of developing glaucoma is high enough to justify treatment. The OHTS study provides some guidance on this topic as those with higher eye pressures and thinner corneas were most at risk for developing glaucoma.
There are also situations that are clearer with regard to when a patient with ocular hypertension should undergo treatment. Patients who demonstrate optic nerve damage or visual field changes over time should be treated because they have likely developed early glaucoma.
The bottom line is that a patient with ocular hypertension should have long-term follow-up with their eye doctor to monitor the potential development of glaucoma or to gauge the effectiveness of treatments that are implemented.
This content was last updated on: April 23, 2018
The information provided here is a public service of the BrightFocus Foundation and should not in any way substitute for personalized advice of a qualified healthcare professional; it is not intended to constitute medical advice. Please consult your physician for personalized medical advice. BrightFocus Foundation does not endorse any medical product, therapy, or resources mentioned or listed in this article. All medications and supplements should only be taken under medical supervision. Also, although we make every effort to keep the medical information on our website updated, we cannot guarantee that the posted information reflects the most up-to-date research.
These articles do not imply an endorsement of BrightFocus by the author or their institution, nor do they imply an endorsement of the institution or author by BrightFocus.
Some of the content may be adapted from other sources, which will be clearly identified within the article.
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a. Solid K has an electron configuration of [Ar]4s1, so it has an unpaired electron. This allows electric current to flow through the structure. KNO3 does not have an unpaired electron, so no electricity flow in.
b. SbCl3 has measurable dipole movement because the molecule is polar, so there is attraction between SbCl3 and other polar molecules. SbCl5 is non-polar, so there is no dipole interaction occurring.
|isn't chemistry fun!|
d. NaI is polar, so it is easy for it to bond with other molecules in a solution. I2 is non-polar and symmetrical, so it is harder to separate.
f. Sulfur has an expanded octet (12 electrons) because it is in the third energy level, so there are more options for bonding. Oxygen is only in the second energy level, and therefore is restricted to 8 electrons.
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POLITICIANS may have lost the plot on how to halt climate change. But technologists are forging ahead with a host of innovations that could halt the rise in greenhouse gas levels, says a UN panel of climate change experts in a report published this week.
The Intergovernmental Panel on Climate Change says that technical innovation has been faster than anticipated five years ago, when it made its last assessment. Wind turbines, hydrogen fuel cells, efficient car engines and the technology to bury carbon dioxide underground could become practical ways to cut greenhouse gas emissions.
But critics believe that the IPCC has failed to give governments firm advice on how to make the new technologies work. They fear that the report, called Climate Change 2001: Mitigation will contribute to the political inaction that has followed last November's failed Kyoto Protocol talks on curbing climate change.
This is the third major report ...
To continue reading this article, subscribe to receive access to all of newscientist.com, including 20 years of archive content.
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+0
# if tan(a-b)=1/4, tan(a+b)=1/3 then tan2a=?
0
436
4
if tan(a-b)=1/4, tan(a+b)=1/3 then tan2a=?
Guest Oct 14, 2014
#2
+92225
+10
$$\\atan(\frac{1}{4})=A-B\qquad (1)\\\\ atan(\frac{1}{3})=A+B\qquad(2)\\\\ (1)+(2)\\\\ atan(\frac{1}{4})+atan(\frac{1}{3})=2A\\\\ tan[atan(\frac{1}{4})+atan(\frac{1}{3})]=tan[2A]\\\\$$
$$Now I am going to use the identity\\\\ \boxed{tan(\theta_1+\theta_2)=\frac{tan(\theta_1)+tan(\theta_2)}{1-tan(\theta_1)tan(\theta_2)}}\\\\\\ tan(2A)\\\\ =tan[atan(\frac{1}{4})+atan(\frac{1}{3})]\\\\ =\frac{\frac{1}{4}+\frac{1}{3}}{1-\frac{1}{4}\times\frac{1}{3}}\\\\ =\frac{\frac{7}{12}}{1-\frac{1}{12}}\\\\ =\frac{7}{12}\div \frac{11}{12}\\\\ =\frac{7}{12}\times \frac{12}{11}\\\\ =\frac{7}{11}\\\\ This answer is exact$$$Melody Oct 14, 2014 Sort: ### 4+0 Answers #1 +85819 +5 Using the tangent inverse tan-1(1/4) = (a - b) = about 14.04° ...... and tan-1(1/3) = (a +b) = about 18.43° So (a -b) + (a+b) = (14.04 + 18.43)° → 2a = about 32.47° So tan(2a) = tan(32.47°) = about .636363= 7/11 CPhill Oct 14, 2014 #2 +92225 +10 Best Answer $$\\atan(\frac{1}{4})=A-B\qquad (1)\\\\ atan(\frac{1}{3})=A+B\qquad(2)\\\\ (1)+(2)\\\\ atan(\frac{1}{4})+atan(\frac{1}{3})=2A\\\\ tan[atan(\frac{1}{4})+atan(\frac{1}{3})]=tan[2A]\\\\$$ $$Now I am going to use the identity\\\\ \boxed{tan(\theta_1+\theta_2)=\frac{tan(\theta_1)+tan(\theta_2)}{1-tan(\theta_1)tan(\theta_2)}}\\\\\\ tan(2A)\\\\ =tan[atan(\frac{1}{4})+atan(\frac{1}{3})]\\\\ =\frac{\frac{1}{4}+\frac{1}{3}}{1-\frac{1}{4}\times\frac{1}{3}}\\\\ =\frac{\frac{7}{12}}{1-\frac{1}{12}}\\\\ =\frac{7}{12}\div \frac{11}{12}\\\\ =\frac{7}{12}\times \frac{12}{11}\\\\ =\frac{7}{11}\\\\ This answer is exact$$$
Melody Oct 14, 2014
#3
+92225
+5
Thanks Chris
Melody Oct 14, 2014
#4
+85819
+5
I did mine "on the fly"....Melody's approach is actually "better"
CPhill Oct 14, 2014
### 13 Online Users
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Skip to 0 minutes and 15 secondsHello and welcome back to a step in practice-- we are dealing here with absolute values. In this video, we are going to solve just exercises from 1 to 4, and we leave exercise 5 to the PDF. In exercise 1, we are asked to prove the triangle inequality-- the absolute value of x plus y, less or equal than the sum of the absolute values. Now let us recall that the inequality, absolute value of x less or equal than a is equivalent to the fact that minus a is less than x, and at the same time, x is less than a.
Skip to 0 minutes and 55 secondsThis is the situation-- if a is positive, the elements whose absolute value is less than or equal than a are just those between minus a and a. This is the case a positive. And if a is negative, strictly negative, there are no x satisfying the first inequality, and there are no x satisfying both of these inequalities. So if we put a equal to the sum of the absolute values, it's enough to prove that x plus y lies between a and minus a. Now x is between its absolute value and the opposite of its absolute value. At the same time, y is between its absolute value and the opposite of the absolute value.
Skip to 1 minute and 45 secondsSo if we take the sum, term by term, we get minus the sum of the two absolute values less or equal than x plus y, less or equal than the sum of the absolute values. And thus, this implies that the triangular inequality is true.
Skip to 2 minutes and 7 secondsIn exercise 2, we've got the function, which is the difference of two absolute values-- precisely, the absolute value of x minus 1 and the absolute value of 4x plus 8. Now, x minus 1-- while the absolute value of x minus 1 depends on the sign of x minus 1. And analogously, the absolute value of 4x plus 8 depends on the sign of 4x plus 8. Now x minus 1 is greater or equal than 0, if and only if x is greater than 1. Whereas 4x plus 8 is greater or equal than 0 if and only if 4x is greater or equal than 8. That is, x greater or equal than minus 2.
Skip to 2 minutes and 59 secondsSo we consider three cases, depending whether x is less than minus 2, between minus 2 and 1, or greater than 1. So if x is strictly less than minus 2, then 4x plus 8 is negative. And also x minus 1 is negative. And therefore f of x equals-- well instead of the absolute values of x minus 1, we can write minus x minus 1, minus, minus, 4x plus 8. Which is minus x plus 1, plus 4x plus 8. That is 3x plus 9. If x belongs to the interval between minus 2 and 1, we take, for convenience, minus 2, and we do not take 1.
Skip to 3 minutes and 57 secondsThen well, x is greater than minus 2, so 4x plus 8 is greater or equal than 0. But x minus 1 is negative. And thus, f of x equals minus x minus 1, minus 4x plus 8.
Skip to 4 minutes and 22 secondsAnd this equals minus 5x plus 1, minus 8. So minus 7. Finally, if x is greater or equal than 1, then both terms are positive. 4x plus 8 is greater or equal than 0, x plus 1, x minus 1 is greater or equal than 0. And thus, we get f of x equal to x minus 1, minus 4x plus 8, which is exactly minus 3x, minus 9.
Skip to 5 minutes and 8 secondsIn exercise 3, we are asked to find the formula for the maximum between two real numbers-- a and b.
Skip to 5 minutes and 20 secondsNow notice that if we draw the segment between a and b, well, we can have b be on the right hand side or a on the right hand side. But in any case, there are some quantities that do not depend on the position of a with respect to b, like the midpoint within between a and b, which is always equal to a plus b over 2. And also the distance between a and b, which is always the absolute value of b minus a. We get the maximum between a and b by adding to the midpoint, half of the distance between a and b.
Skip to 6 minutes and 10 secondsTherefore, the maximum between a and b is the midpoint plus the absolute value of b minus a over 2. And this is a formula involving just absolute values and sums and quotients. Now in exercise 4, we've three sets. Let us write them in a more compact way. The set A is the set of numbers from minus 3 to plus infinity, whereas the set B is the set of numbers that are strictly less than 6-- so minus infinity, 6. And C is the set of numbers whose distance to 5 is less than or equal to 2. And so we get the closed interval 5 minus 2, 5 plus 2. That is, the interval 3, 7-- with 3, and 7 on it.
Skip to 7 minutes and 14 secondsNow it is convenient in order to find the intersection of the three sets to draw each set as a subset of the real line. So we draw 3 real lines, and on each of the real lines, we represent the sets A, B, and C. So the values that count here are minus 3, 3, 6, and 7. So we consider minus 3, 3, 6, and 7.
Skip to 7 minutes and 49 secondsAnd in the first real line, we consider the set A, in the second set B, in the third the set C. And we draw one more line here with the intersection of A with B and C. So let us draw the set A is the set from minus 3 to plus infinity. And we take minus 3--
Skip to 8 minutes and 18 secondsthe set B goes from minus infinity to 6, but we do not take 6, so this is excluded.
Skip to 8 minutes and 32 secondsAnd also the set C goes from 3 to 7, so we take a set from 3 to 7. And we take 7. So be careful. We do not take 6, but we take 7, 3 here, and minus 3 here. So the intersection is what is in common between the three sets. So we look in each column. Here, we do not take this column, we do not take this column, but we take this one. For sure we'll take from 3 to 6, but be careful we do not take 6, and we take 3.
Skip to 9 minutes and 20 secondsWe do not take the column between 3, 6, and 7, and we don't take the column between 7 plus infinity. So the intersection of the three sets is the set from 3-- the interval from 3 to 6. We take 3, and we do not take 6. And this ends exercise 4. Exercise 5 is solved just in the PDF. So see you on the next step.
Absolute value in practice
The following exercises are solved in this step.
We invite you to try to solve them before watching the video.
In any case, you will find below a PDF file with the solutions.
Prove the triangle inequality \(|x+y|\le |x|+|y|\) for all \(x,y\in\mathbb R\)
Write \(f(x) = |x-1| - |4x + 8|\) as an expression where there are no absolute values.
Express the maximum \(\max\lbrace a,b\rbrace\) between the real numbers \(a\) and \(b\) in a formula involving just sums, products and the absolute value of real numbers. (Hint: what is the distance from \(\max\lbrace a,b\rbrace\) to the midpoint of the segment joining \(a\) to \(b\)?)
Let \(A=\lbrace x\in\mathbb R:\, x\ge -3\rbrace, B=\lbrace x\in\mathbb R:\, x<6\rbrace,\) \( C=\lbrace x\in \mathbb R:\, |x-5|\le 2\rbrace\). Find the set \(A\cap B\cap C\).
Exercise 5. [Solved only in the PDF file]
Knowing that \(\sqrt 3=1.732050808…\) give an estimate of \(|\sqrt 3-1.73205|\).
© Università degli Studi di Padova
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# GDP Growth Rate
GDP growth rate or simply growth rate of an economy is the percentage by which the real GDP of an economy increases in a period. If the growth rate of an economy is g, its output doubles in 70/g periods.
When an economy’s growth rate is positive, the economy’s output is increasing, and it is said to be in recovery or in economic boom. But when the growth rate is negative, the economy is in a recession i.e. output is contracting.
The growth rate of GDP differs from the growth rate of GDP per capita simply because GDP per capita also depends on the population of the country which grows independently of the output. Growth rate of GDP per capita is a better measure of improvement in standard of life of an average person in the economy.
You must be wondering why we use the rate of change in real GDP as a measure of an economy’s growth rate instead of the rate of change of nominal GDP. It is because we are interested in finding out the increase in productive capacity of the economy. It means that we need to exclude the increase in GDP solely due to increase in price level.
The percentage change in nominal GDP broadly equals the growth rate (g) plus inflation rate (π).
$$\text{%\ Change in Nominal GDP}=\text{g}\ +\ \pi$$
## Calculation
The following formula can be used to calculate growth rate of an economy for a single period:
$$\text{g}=\frac{{\text{GDP}} _ \text{n}-{\text{GDP}} _ {\text{n}-\text{1}}}{{\text{GDP}} _ {\text{n}-\text{1}}}$$
Where GDPn is the real GDP in current year and GDPn-1 is the real GDP in the previous period.
If we want to calculate the average compound growth rate over multiple periods, we need to use the following formula:
$$\text{g}=\left(\frac{{\text{GDP}} _ \text{t}}{{\text{GDP}} _ \text{0}}\right)^\frac{\text{1}}{\text{t}}-\text{1}$$
Where GDPt is the latest real GDP, GDP0 is the earlier GDP and t is the number of periods.
The equation for compound average growth rate can be modified to find out the formula for future value of GDP given a constant growth rate g over t periods:
$${\text{GDP}} _ \text{t}={\text{GDP}} _ \text{0}\times{(\text{1}+\text{g})}^\text{t}$$
## Example
The following table shows US GDP data in trillions of US Dollars.
Year 2015 2016 2017
Nominal GDP 18.12 18.62 19.39
Real GDP (2012 Dollars) 17.37 17.66 18.05
The GDP growth rate for 2016 can be worked out as follows:
$$\text{g} _ {\text{2016}}=\frac{{\rm \text{GDP}} _ {\text{2016}}-{\rm \text{GDP}} _ {\text{2015}}}{{\rm \text{GDP}} _ {\text{2015}}}=\frac{\text{17.66}-\text{17.37}}{\text{17.37}}=\text{1.67%}$$
Please note that the growth rate of 1.67% worked out above is lower than the percentage change in nominal GDP in 2016 of 2.76%. The difference of 1.09% is attributable to change in price level. This is precisely the reason that the percentage change in nominal GDP curve is above the real GDP percentage growth curve in the graph below (obtained from FRED):
The average growth rate from 2015 to 2017 can be worked out using the formula for CAGR:
$$\text{g} _ {\text{2017}-\text{2015}}=\left(\frac{\text{18.05}}{\text{17.37}}\right)^\frac{\text{1}}{\text{2}}-\text{1}=\text{1.94%}$$
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# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter 2 –
Class XI Chapter 2 – Relations and Functions Maths ______________________________________________________________________________ 1,2,3,4 and B 1,5,9,11,15,16 A A B { 1,1 , 1,5 , 1,9 , 1,11 , 1,15 , 1,16 , 2,1 , 2,5 , 2,9 , 2,11 , 2,15 , 2,16 3,1 , 3,5 , 3,9 , 3,11 , 3,15 , 3,16 , 4,1 , 4,5 , 4,9 , 4,11 , 4,15 , 4,16 } It is given that f 1,5 , 2,9 , 3,1 , 4,5 , 2,11 (i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product . Thus, f is a relation from A to B. (ii) Since the same first element i.e, 2 corresponds to two different images i.e., 9 and 11, relation f is not a function. AB Question 11: Let f be the subset of Justify your answer. Solution 11: ZZ The relation f is defined as defined by f ab, a b : a, b Z f ab, a b : a, b Z . If f a function from Z to Z: We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B. Since 2,6, 2, 6 Z, 26, 2 6 , 2 6, 2 6 f 12,8 , 12, 8 f i.e., It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and 8. Thus, relation f is not a function. Question 12: Let A 9,10,11,12,13 Find the range of f. Solution 12: A 9,10,11,12,13 Prime factor of 9 = 3 Prime factors of 10 = 2, 5 Prime factor of 11 = 11 Prime factor of 12 = 2, 3 Prime factor of 13 = 13 and let f : A N be defined by f n the highest prime factor of n. and let f : A N is defined as f n The highest prime factor of n Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 2 – Relations and Functions Maths ______________________________________________________________________________ 9 f The highest prime factor of 9 = 3 f 10 The highest prime factor of 10 = 5 f 11 The highest prime factor of 11 = 11 f 12 The highest prime factor 12 = 3 f 13 The highest prime factor of 13 = 13 The range of f is the set of all Range of f 3,5,11,13 f n , where n A . Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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Home » 5th Roots » 5th Root of 17
# 5th Root of 17
The 5th root of 17 is the number, which multiplied by itself 5 times, is 17. In other words, this number to the power of 5 equals 17.
Besides the real value of
along with an explanation, on this page you can also find what the elements of the 5th root of 17 are called.
In addition to the terminology, we have a calculator you don’t want to miss:
Reset
⁵√17 = 1.7623403478
If you have been looking for the 5th root of seventeen, then you are right here, too.
## Fifth Root of 17
In this section we provide you with important additional information about the topic of this post:
The term can be written as ⁵√17 or 17^1/5.
As the index 5 is odd, 17 has only one 5th real root: ⁵√17 sometimes called principal 5th root of 17.
If you want to know how to find the value of this root, then read our article 5th Root located in the header menu.
There, we also discuss the properties for index n = 5 by means of examples: multiplication, division, exponentiation etc.
Next, we have a look at the inverse function.
### Inverse of 5th Root of 17
Extracting the 5th root is the inverse operation of ^5:
In the following paragraph, we are going to name the elements of this √.
## What is the 5th Root of 17?
You already have the answer to that question, and you also know about the inverse operation of 17 5th root.
Keep reading to learn what the parts are called.
• ⁵√17 is the 5th root of 17 symbol
• 5 is the index
• 5th root = 1.7623403478
5th root of 17 = 1.7623403478
As a sidenote: All values on this page have been rounded to ten decimal places.
Now you really know all about ⁵√17, including its values, parts and the inverse.
If you need to extract the 5th root of any other real number use our calculator above.
Simply insert the number of which you want to find the 5th root (e.g. 17); the calculation is done automatically.
If you like our information about ⁵√17, then a similar 5th root you may be interested in is, for example: 5th root of 19.
In the following table you can find the n-th root of 17 for n = 2,3,4,5,6,7,8,9,10.
## Table
The aim of this table is to provide you with an overview of the nth roots of 17.
217Square Root of 17²√17±4.1231056256
317Cube Root of 17³√172.5712815907
417Forth Root of 17⁴√17±2.0305431849
517Fifth Root of 17⁵√171.7623403478
617Sixth Root of 17⁶√17±1.6035216215
717Seventh Root of 17⁷√171.4989198721
817Eight Root of 17⁸√17±1.4249712926
917Nineth Root of 17⁹√171.3699873177
1017Tenth Root of 17¹⁰√17±1.3275316749
A few lines down from here we review the FAQs.
## Fifth Root of Seventeen
If you have been searching for what’s the fifth root of seventeen or 5-th root of 17, then you are reading the right post as well.
The same is true if you typed 5 root of 17 or 17 5 root in the search engine of your preference, just to name a few similar terms.
Right below you can find the frequently asked questions in the context.
## FAQs About the 5th Root of 17
Click on the question which is of interest to you to see the collapsible content answer.
### How Many Real Fifth Roots Does 17 Have?
17 has one real fifth root, because the index 5 is odd.
### What to the 5th Power Equals 17?
The 5th root of 17 to the power of 5 equals 17.
### How Do You Find the Fifth Root of 17?
Start with an initial guess such that 5 times that value equals approximately 17, then keep improving the guess until you have the required precision.
### What is 17 to the 5th Root?
17 to the 5th root = 17^1/5 = 1.7623403478.
### What Number is the 5th Root of 17?
The 5th root of 17 = 1.7623403478.
### How Do You Solve the 5-th Root of 17?
To compute the 5-th root of 17 use a calculator, and/or employ the Newton–Raphson method.
### What is the Value of 5 Root 17?
The value of 5 root 17 is 1.7623403478.
If something remains unclear do not hesitate getting in touch with us.
We are constantly trying to improve our site, and truly appreciate your feedback.
Ahead is the wrap-up of our information.
## Summary
To sum up, the fifth root of 17 is 1.7623403478.
Finding the 5th root of the number 17 is the inverse operation of rising the ⁵√17 to the power of 5. That is to say, (1.7623403478)5 = 17.
Further information related to ⁵√17 can be found on our page 5th Root.
Note that you can also locate roots like ⁵√17 by means of the search form in the menu and in the sidebar of this site.
If our article about the fifth √ 17 has been useful to you , then press some of the share buttons located at the bottom of this page.
If you have any questions about the 5-th root of 17, fill in the comment form below.
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## 1. 1 (8 points). Calculate the area of the quadrilateral with vertices at (0,0), (4,2), (2,2), and (4,6) An explicit line integral or sum of line integrals equal to this area is:_________________________________ The area is: _______________________________
Solution. This can be done with Green's formula:
Area = Integral[(-y) dx} counterwise along the perimeter of the figure. The
given points are listed in counterclockwise order,
On the first leg, we can take x=4 t, y = 2 t, 0 <= t <= 1:
In[1]:=
` Int1 = Integrate[(-2 t)*4, {t,0,1}]`
Out[1]=
``` -4
```
On the second leg, we can take x=4 - 2 t, y = 2, 0 <= t <= 1:
In[2]:=
` Int2 = Integrate[(-2)*(-2), {t,0,1}]`
Out[2]=
``` 4
```
On the third leg, we can take x=2 t + 2, y = 2+ 4 t, 0 <= t <= 1:
In[3]:=
` Int3 = Integrate[(-2 - 4 t)*(2), {t,0,1}]`
Out[3]=
``` -8
```
On the fourth leg, we return to the origin and can take x=4 - 4 t, y = 6 - 6 t, 0 <= t <= 1:
In[4]:=
` Int4 = Integrate[(-6 + 6 t)*(-4), {t,0,1}]`
Out[4]=
``` 12
```
In[5]:=
` Area = Int1 + Int2 + Int3 + Int4`
Out[5]=
``` 4
```
The explicit expression asked for is
Area = Integrate[(-2 t)*4, {t,0,1}]
+ Integrate[(-2)*(-2), {t,0,1}]
+ Integrate[(-2 - 4 t)*(2), {t,0,1}]
+Integrate[(-6 + 6 t)*(-4), {t,0,1}],
and its value is 4. You could equally well calculate the integral of x dy.
Up to Solutions to Test 4
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Courses
# NCERT Solutions - Chapter 14: Statistics, Class 10, Maths(Part 3) Notes - Class 10
## Class 10: NCERT Solutions - Chapter 14: Statistics, Class 10, Maths(Part 3) Notes - Class 10
The document NCERT Solutions - Chapter 14: Statistics, Class 10, Maths(Part 3) Notes - Class 10 is a part of Class 10 category.
All you need of Class 10 at this link: Class 10
Exercise 14.3
Q1 :
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
To find the class marks, the following relation is used.
Class Mark= (upper class limit + lower class limit)/2
Taking 135 as assumed mean (a), di, ui,fiuare calculated according to step deviation method as follows.
From the table, we obtain
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 - 145.
Modal class = 125 - 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal cla
Q2 :
If the median of the distribution is given below is 28.5, find the values of x and y.
The cumulative frequency for the given data is calculated as follows.
From the table, it can be observed that n = 60
45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 - 30.
Therefore, median class = 20 - 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
From equation (1),
8 + y = 15
y = 7
Hence, the values of x and y are 8 and 7 respectively.
Q3 :
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.
From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than
is 78, belonging to interval 35 - 40.
Therefore, median class = 35 - 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
Therefore, median age is 35.76 years.
Q4 :
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table
:
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula
assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5… 171.5 - 180.5)
The given data does not have continuous class intervals. It can be observed that the difference between two class
intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower class limits
respectively.
Continuous class intervals with respective cumulative frequencies can be represented as follows.
From the table, it can be observed that the cumulative frequency just greater than
is 29,
belonging to class interval 144.5 - 153.5.
Median class = 144.5 - 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
Therefore, median length of leaves is 146.75 mm.
Q5 :
Find the following table gives the distribution of the life time of 400 neon lamps:
Find the median life time of a lamp.
Thecumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than is 216, belonging to class
interval 3000 - 3500.
Median class = 3000 - 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
Median
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.
Q6 :
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames?
Also, find the modal size of the surnames.
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than
is 76, belonging to class
interval 7 - 10.
Median class = 7 - 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
To find the class marks of the given class intervals, the following relation is used.
Class Mark= (upper class limit + lower class limit)/2
Taking 11.5 as assumed mean (a), di, ui,and fiui are calculated according to step deviation method as follows.
Q7 :
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
The cumulative frequencies with their respective class intervals are as follows.
Cumulative frequency just greater than
is 19, belonging to class interval 55 - 60.
Median class = 55 - 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
= 56.67
Therefore, median weight is 56.67 kg.
Exercise 14.4 :
Q1 :
The following distribution gives the daily income of 50 workers of a factory.
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
The frequency distribution table of less than type is as follows.
Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.
Q2 :
During the medical check-up of 35 students of a class, their weights were recorded as follows:
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
The given cumulative frequency distributions of less than type are
Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.
Here, n = 35
So,n/2 = 17.5
Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5.
It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below.
The cumulative frequency just greater than
is 28, belonging to class interval 46 - 48.
Median class = 46 - 48
Lower class limit (l) of median class = 46
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2
Q3 :
The following table gives production yield per hectare of wheat of 100 farms of a village
Change the distribution to a more than type distribution and draw ogive.
The cumulative frequency distribution of more than type can be obtained as follows.
Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be
obtained as follows.
The document NCERT Solutions - Chapter 14: Statistics, Class 10, Maths(Part 3) Notes - Class 10 is a part of Class 10 category.
All you need of Class 10 at this link: Class 10
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Bed Bugs are heat sensitive, wingless blood-feasting bugs that are found everywhere, including the United States. Initially they start life like a small but visible egg, about the dimension of a poppy seed. From there they feed on blood and kill bed bugs become juvenile or “nymph” stage bugs. As they develop into adults, brown or red (fed) bed bugs (Here you can get data about what is bed bug) get about the dimensions of an apple seed. Bed bugs can live up to 20 months and might go without a blood meal for 1 ½ years. The female bed bug can lay over 500 eggs in her life cycle, which hatch in about 10 days, given the best temperatures.
It is believed that bed bugs don’t transmit disease. However, they can cause an allergic skin response and bacterial infection from scratching.
If found on time, bed bugs can be effectively managed.
Bites are normally the first warning sign of a bed bug problem, but not all bug bites are bedbugs. Here you can get data about what is bed bug. Since 1998, Amherst School has had only one bed bug incident, which was confined to a single room. That isolated incident occurred, like most bed bug problems, because they were shifted to one site from another in personal belongings such as back packs, bedding, clothing and suitcases.
Bed Bug Bites
Bed bug bites, unlike bites which you might get from a gnat, mosquito, no-see-um or similar pest generally present with a number of attacks, normally in a directly row on the arms, legs, neck or torso. Bed bugs tend to gather together in hidden, undisturbed places where a person sits or sleeps.
* Bed bugs are typically found on the bed, along the seams and sides of the bed mattress and box springs, on the headboard, and the bed frame.
* When assessing a bed or furniture for the possibility of bed bugs, we find for clusters of live bugs, shedding skins, dark colored fecal spots and the eggs. We look for blood spotting on the bed linens, where the bed bugs bite the host * Bed bugs are also known to cover in cracks, such as in baseboards along the floor at the wall.
* Bed bugs are as well known to conceal in cracks, for example in baseboards along the floor on the wall.
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noun, plural: amyloplasts
(botany) A type of leucoplast found in the cytoplasm of a plant cell, and serves as storage organelle of amylopectin
Plastids are organelles involved in the synthesis and storage of food. They are found within the cells of photosynthetic eukaryotes. In plants, plastids may develop into these forms: (1) chloroplasts, (2) chromoplasts, (3) gerontoplasts, and (4) leucoplasts. Leucoplasts are colourless plastids because they lack pigments. Their role is primarily for storage. Depending on the content of the leucoplasts, they may be amyloplasts, elaioplasts, proteinoplasts, or tannosomes.
Amyloplast is a leucoplast that is primarily involved in storing starch and detecting gravity. As for storing starch, the amyloplasts transform glucose into starch by polymerization of glucose and store the starch grains in the stroma. Most of the amyloplasts can be found in underground storage tissues of plants, such as potato. The amyloplasts, though, can turn into chloroplasts, such as seen in potato tubers that are exposed to light. They become green as the amyloplasts convert into chloroplasts.
As for detecting gravity, the amyloplasts are able to perceive gravity (gravitropism). These amyloplasts involved in gravitropism are referred to as statoliths.
Word origin: Gk ámylon (starch) + Gk plastós (formed, molded)
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# Data Structures - Merge Sort Algorithm
Merge sort is a sorting technique based on divide and conquer technique. With worst-case time complexity being Ο(n log n), it is one of the most respected algorithms.
Merge sort first divides the array into equal halves and then combines them in a sorted manner.
## How Merge Sort Works?
To understand merge sort, we take an unsorted array as the following −
We know that merge sort first divides the whole array iteratively into equal halves unless the atomic values are achieved. We see here that an array of 8 items is divided into two arrays of size 4.
This does not change the sequence of appearance of items in the original. Now we divide these two arrays into halves.
We further divide these arrays and we achieve atomic value which can no more be divided.
Now, we combine them in exactly the same manner as they were broken down. Please note the color codes given to these lists.
We first compare the element for each list and then combine them into another list in a sorted manner. We see that 14 and 33 are in sorted positions. We compare 27 and 10 and in the target list of 2 values we put 10 first, followed by 27. We change the order of 19 and 35 whereas 42 and 44 are placed sequentially.
In the next iteration of the combining phase, we compare lists of two data values, and merge them into a list of found data values placing all in a sorted order.
After the final merging, the list should look like this −
Now we should learn some programming aspects of merge sorting.
### Algorithm
Merge sort keeps on dividing the list into equal halves until it can no more be divided. By definition, if it is only one element in the list, it is sorted. Then, merge sort combines the smaller sorted lists keeping the new list sorted too.
```Step 1 − if it is only one element in the list it is already sorted, return.
Step 2 − divide the list recursively into two halves until it can no more be divided.
Step 3 − merge the smaller lists into new list in sorted order.
```
### Pseudocode
We shall now see the pseudocodes for merge sort functions. As our algorithms point out two main functions − divide & merge.
Merge sort works with recursion and we shall see our implementation in the same way.
```procedure mergesort( var a as array )
if ( n == 1 ) return a
var l1 as array = a[0] ... a[n/2]
var l2 as array = a[n/2+1] ... a[n]
l1 = mergesort( l1 )
l2 = mergesort( l2 )
return merge( l1, l2 )
end procedure
procedure merge( var a as array, var b as array )
var c as array
while ( a and b have elements )
if ( a[0] > b[0] )
add b[0] to the end of c
remove b[0] from b
else
add a[0] to the end of c
remove a[0] from a
end if
end while
while ( a has elements )
add a[0] to the end of c
remove a[0] from a
end while
while ( b has elements )
add b[0] to the end of c
remove b[0] from b
end while
return c
end procedure
```
To know about merge sort implementation in C programming language, please click here.
Advertisements
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# 3 Act Math: Butter Believe It
This task best fits 3.NF.A.3.a-b as an introduction to equivalent fractions.
## Act 1
What did you notice? What do you wonder?
Focus Question: How many will go on each plate so that everyone gets the same amount?
## Act 2
How many full bars were there?
## Extension:
What if the bars were cut in fourths and eighths?
# 3 Act Math: Pretzel Cut
This task is a little different than most, but I really think it has some good exploration built in and it lends itself to more of an open task. The standards that I think best fit this task are Math Practice 1, 3, and 4. If you think there is a good fit with a content standard, please comment below. There is definitely some partitioning of the whole which could be a good application of 3.GA.2.
## Act 1
What did you notice? What do you wonder?
Focus Question: How will the bread be cut?
## Act 2
I made 7 cuts. Number of Pretzels
# 3 Act: Broken Heart
This task is meant to emphasize partitioning on a line of symmetry. Common Core Standard 4.G.A.3.
## Act 1
What did you notice? What do you wonder?
Focus Question: Where will I need to cut the heart for it to be symmetrical (or to cover itself)?
## Act 2
Use this recording sheet for students.
# 3 Act: Ho Ho Ho
I happened upon some Little Debbie snacks at the grocery store a few days ago and got some ideas for some new tasks. This task is meant to show fractions as division by dividing the number of Ho Ho’s by the number of plates (or people). The common core standard this best fits is 5.NF.B.3.
## Act 1
What do you notice? What do you wonder?
Focus Question: How many Ho Ho snacks will each person (or plate) get?
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Language development consists of two important components:
- oral language and vocabulary development
- literacy development (phonemic, phonological, and print awareness)
In this article, we’ll focus on exploring oral language and vocabulary development.
Vocabulary and oral language development are the foundation of literacy skills, such as reading and writing. Research shows that exposing children to unfamiliar words in conversation has a positive impact on vocabulary development. Using complex, spicy words with young children benefits their reading comprehension well into the fourth grade!
The best way to support vocabulary and oral language development is simply to talk to your child and read with them every day. Here are some additional ways for you to promote oral language and vocabulary development at home:
- Hold eye contact as you talk with your baby. This helps establish an understanding of the norms of communication.
- Use gestures as you talk to your infant. For example, tap your lips with your fingers when you say eat. You can tell your baby, “It’s time to eat. We are going to eat a banana.” Whenever you say “eat” repeat the motion. With consistency, your child will use these gestures to talk to you.
- As your baby chats, listen and repeat their babbles. Allow time for them to respond. This is the start of oral language and will show the flow and form of a conversation.
- Read your child’s favorite books and sing their favorite songs repeatedly. Encourage them to tell parts of the stories and sing the songs with you.
- When you are talking to your child about something, point to it. Similarly, if your child points to something, point to it and tell them what it is.
- Build on your child’s comments to make bigger phrases or sentences. For example, if your child says, “big dog,” you can add to their statement by saying, “Yes, there is a big dog by the tree.”
- As you read to your child, ask them questions about what’s happening in the story and how they feel about it. This will help them to practice listening, retelling, and verbalizing thoughts. Ask questions that require your child to talk about the events of the story sequentially.
- As your child chats, listen and ask questions. The best questions are open-ended, meaning they don’t have a right or wrong answer. Allow time for them to respond. In addition to asking follow-up questions, ask your child to explain what they’ve just said or to give more detail.
- Have frequent conversations with your child about things of interest to them. Use spicy, unfamiliar words and talk about what those words mean.
- Encourage your child to make comparisons of objects or events using descriptive words. For example, if you have two different leaves, ask your child how they are the same or different.
- Have your child tell a story about something they recently experienced. Ask them what happened first, next, and last. Encourage them to tell the story sequentially.
- When shopping, talk with your child about what you plan to buy. Encourage them to describe the items using precise words. Have them practice giving directions by explaining how to find the item as you walk through the store.
Here are some additional online resources to continue exploring language development at home from the National Association for the Education of Young Children (NAEYC):
12 Ways to Support Language Development for Infants and Toddlers
Reinforcing Language Skills for our Youngest Learners
Supporting the Development of Creativity
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The ocean maintained a relatively steady temperature throughout most of the 20th century before rising steeply, new research suggests.
In estimations of ocean heat content—important when assessing and predicting the effects of climate change—calculations have often presented the rate of warming as a gradual rise from the mid-20th century to today.
“The lag is catching up and the ocean is warming more strongly now.”
The new findings, which could overturn that assumption, may have significant implications for what we might expect in the future.
“There wasn’t an onset of an imbalance until about 1990, which is later than most estimates,” says Timothy DeVries, an associate professor in the geography department at the University of California, Santa Barbara, and a coauthor of the new paper in Nature Communications.
According to the study, the period from 1950 to 1990 saw temperature fluctuations in the water column but no net warming. After 1990, the entire water column switched from cooling to warming.
These findings are the result of the addition of a largely underexplored factor in ocean heat content (OHC): deep ocean temperatures.
“Prior studies didn’t consider the deep ocean,” says Aaron Bagnell, a graduate scholar in DeVries’ laboratory and the paper’s lead author.
Because of the challenges involved in getting temperature measurements in the deep ocean (below 2,000 meters) that region has gone largely unaccounted for, and data has been sparse. “There is some existing data, from research cruises and autonomous floats,” Bagnell says.
The researchers used an autoregressive artificial neural network (ARANN) and machine learning methods to connect the dots between data points and “produce a single consistent estimate of the top-to-bottom OHC change for 1946 to 2019.” The result was a trend that delays warming by decades over previous models.
There are two main possibilities for why the effects of global warming took so long to reach the ocean, De Vries says.
“One is that anthropogenic warming might have been weaker than previously thought during the 20th century, perhaps due to the cooling effects of aerosol pollution,” he says. The other is that the deep ocean may still be exhibiting the effects of climate events long past.
“The ocean remembers.”
“It can take centuries for climate signals to propagate from the surface to the interior,” he says. Thus, the effects of a cooling event such as the Little Ice Age might be deep history to us on the surface, but the echoes of the event may have continued to resonate in the deep ocean into the 20th century, providing a buffer to the warming Earth.
The delayed cooling effect ended in 1990, after which ocean temperatures, according to the study, have been accelerating upward.
“The lag is catching up and the ocean is warming more strongly now,” Bagnell says. The Atlantic Ocean and Southern Ocean are currently where most of the warming is, with the Pacific Ocean and Indian Ocean not far behind.
Ocean warming is a concern on many levels, as it can cause changes in circulation, reduce its ability to absorb carbon, and fuel more intense storms, in addition to causing sea level rise and creating inhospitable environments for undersea life.
If the trend continues, the effects might last centuries, thanks to the same lag that kept the oceans cool until the last 30 years.
“The ocean remembers,” DeVries says.
Source: UC Santa Barbara
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Search
# Classroom Case Studies, Grades 6-8 Part E: Critiquing Student Lessons (20 minutes)
## Session 10: 6-8, Part E
In the Grades 6-8 curriculum, students are frequently asked to think about patterns, but often their “pattern sniffing” skills end with simply finding the next object.
For each of the problems below, answer the following questions:
Note 8
a. What algebraic ideas are in this problem? b. How are patterns used in this problem? c. What mathematics do you think students would learn from this problem? d. How would you modify the problem, or what additional questions might you ask, to incorporate the framework for analyzing patterns?
### Problems
Problem E1
How many tiles will be in the eighth figure of this pattern? How do you know?
Problem E2
a. Make a drawing of the next V-patterns. b. Is it possible for a V-pattern to have 84 dots? Why or why not? c. How many pairs are there in each V-pattern? d. How many dots will be in the sixth V-pattern? e. Make a V-pattern with 19 dots. f. Fill in the missing values in the table below. V-number means the number of pairs. Describe any patterns you see.
V-Number Number of Dots 1 3 2 5 3 7 4 ? 5 ? 6 ?
### Notes
Note 8
The two lessons in this section take a qualitatively different approach to students’ thinking about patterns. The types of questions asked in the Math in Context patterns problem allow students to use the physical representation to make conjectures about the pattern’s rule. In the first problem from Gateways, students are simply asked to supply a specific term in the sequence. In the framework developed on patterns, we were careful to include habits of mind that extend beyond finding a specific term. How do you know there are a specific number of dots or blocks in the sequence? If you know one term in the sequence, how could you get the next? Can you find a rule that will give you the number of dots or blocks for any term? These kinds of questions extend students’ thinking about predicting with patterns.
### Solutions
Problem E1
a. This problem shows the first three figures of a growing pattern, and requires students to find the number of tiles needed to make the 8th figure in the pattern. b. This problem uses a growing pattern, and the nth term uses n more tiles than the previous term. c. This problem can be used to help students learn to write a general rule for the nth term of a function. d. This problem could be extended to find the number of tiles needed for the 10th, 100th, and then the nth term. Ask students to describe how to build each term. Look for answers that relate each term to the previous term, and then relate the figure to its place in the pattern. For example, students might notice that the 1st term uses 1 tile, the 2nd term uses 1 + 2, or 3 tiles, the 3rd term uses 1 + 2 + 3, or 6 tiles, and so forth. The nth term uses 1 + 2 + 3 + … + n tiles.
Problem E2
a. This problem also shows the first three figures of a growing pattern. This problem not only requires students to find the number of dots needed to make the 6th figure in the pattern, however; it also requires students to determine whether it is possible for one of the terms to use 84 dots. b. This problem uses a growing pattern, and the nth term uses 2 more dots than the previous term. c. This problem can be used to help students learn to write a general rule for the nth term of a function. d. This problem asks students to find the number of dots needed for the 100th term, but could be further extended to ask for the number of dots needed for the nth term. Again ask students to describe how to build each term. Look for answers that relate each figure to its place in the pattern. For example, students might notice that the 1st term uses 1 dot + 1 pair of dots, the 2nd term uses 1 dot + 2 pairs of dots, the 3rd term uses 1 dot + 3 pairs of dots, and so forth. The nth term uses 1 dot + n pairs of dots, for a total of 2n + 1 dots.
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Online Calculators
# Quartile Calculator
Quartile Calculator
Quartile Statistics
First Quartile
Q1 =
Second Quartile
Q2 =
Third Quartile
Q3 =
Interquartile Range
IQR =
Median = Q2
$$\widetilde{x}$$ =
Minimum
Min =
Maximum
Max =
Range
R =
For more detailed statistics use the
Descriptive Statistics Calculator
## Calculator Use
This quartile calculator and interquartile range calculator finds first quartile Q1, second quartile Q2 and third quartile Q3 of a data set. It also finds median, minimum, maximum, and interquartile range.
Enter data separated by commas or spaces.
You can also copy and paste lines of data from spreadsheets or text documents. See all allowable formats in the table below.
## Quartiles
Quartiles mark each 25% of a set of data:
• The first quartile Q1 is the 25th percentile
• The second quartile Q2 is the 50th percentile
• The third quartile Q3 is the 75th percentile
The second quartile Q2 is easy to find. It is the median of any data set and it divides an ordered data set into upper and lower halves.
The first quartile Q1 is the median of the lower half not including the value of Q2. The third quartile Q3 is the median of the upper half not including the value of Q2.
## How to Calculate Quartiles
1. Order your data set from lowest to highest values
2. Find the median. This is the second quartile Q2.
3. At Q2 split the ordered data set into two halves.
4. The lower quartile Q1 is the median of the lower half of the data.
5. The upper quartile Q3 is the median of the upper half of the data.
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q2.
## How to Find Interquartile Range
The interquartile range IQR is the range in values from the first quartile Q1 to the third quartile Q3. Find the IQR by subtracting Q1 from Q3.
• IQR = Q3 - Q1
## How to Find the Minimum
The minimum is the smallest value in a sample data set.
Ordering a data set from lowest to highest value, x1 ≤ x2 ≤ x3 ≤ ... ≤ xn, the minimum is the smallest value x1. The formula for minimum is:
$\text{Min} = x_1 = \text{min}(x_i)_{i=1}^{n}$
## How to Find the Maximum
The maximum is the largest value in a sample data set.
Ordering a data set from lowest to highest value, x1 ≤ x2 ≤ x3 ≤ ... ≤ xn, the maximum is the largest value xn. The formula for maximum is:
$\text{Max} = x_n = \text{max}(x_i)_{i=1}^{n}$
## How to Find the Range of a Set of Data
The range of a data set is the difference between the minimum and maximum. To find the range, calculate xn minus x1.
$R = x_n - x_1$
Acceptable Data Formats
Type
Unit
Options
Actual Input Processed
Column (New Lines)
42
54
65
47
59
40
53
42, 54, 65, 47, 59, 40, 53
Comma Separated (CSV)
42,
54,
65,
47,
59,
40,
53,
or
42, 54, 65, 47, 59, 40, 53
42, 54, 65, 47, 59, 40, 53
Spaces
42 54
65 47
59 40
53
or
42 54 65 47 59 40 53
42, 54, 65, 47, 59, 40, 53
Mixed Delimiters
42
54 65,,, 47,,59,
40 53
42, 54, 65, 47, 59, 40, 53
### References
[1] Wikipedia contributors. "Quartile." Wikipedia, The Free Encyclopedia. Last visited 10 April, 2020.
Cite this content, page or calculator as:
Furey, Edward "Quartile Calculator | Interquartile Range Calculator" at https://www.calculatorsoup.com/calculators/statistics/quartile-calculator.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators
Last updated: September 19, 2023
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Teenage depression: Prevention begins with parental support
Teenage depression can affect nearly every aspect of your child's life. Understand what you can do to help prevent teenage depression, including possible mental health therapy.By Mayo Clinic Staff
Teenage depression is a serious health concern that causes a persistent feeling of sadness and loss of interest in activities. Although there's no sure way to prevent teenage depression, these strategies might help.
Offer unconditional support
A strong parent-child relationship can help prevent depression. To build — or maintain — a positive relationship with your child:
- Set aside time each day to talk
- Encourage your child to express his or her feelings
- Praise his or her strengths, whether it's in academics, music, athletics, relationships or other areas
- Offer positive feedback when you notice positive behavior
- Respond to your child's anger with calm reassurance rather than aggression
If your child is reluctant to talk, spend time in the same room. Even if you're not talking, a caring attitude can speak volumes.
Foster friendship and social support
Positive peer experiences and strong friendships can help prevent depression. Encourage your child to spend time with supportive friends. Playing team sports or taking part in other organized activities might help by boosting your child's self-esteem and increasing his or her social support network.
At the same time, be alert to the possible issues associated with early dating. Even typical romantic experiences, such as flirting and dating, can be challenging for teens — and might contribute to symptoms of depression.
Encourage physical activity
Regular physical activity — regardless of the level of intensity — might play a role in reducing teenage depression and anxiety.
For adolescents, the Department of Health and Human Services recommends one hour or more of physical activity a day. This includes aerobic activities — such as running, swimming, walking and jumping rope — and muscle-strengthening activities, such as climbing a rock wall or lifting weights.
Sept. 03, 2015
See more In-depth
- Merry SN, et al. Psychological and educational interventions for preventing depression in children and adolescents. Cochrane Database of Systematic Reviews. http://onlinelibrary.wiley.com/doi/10.1002/14651858.CD003380.pub3/abstract. Accessed July 17, 2015.
- Biddle SJH, et al. Physical activity and mental health in children and adolescents: A review of reviews. British Journal of Sports Medicine. 2011;45:886.
- Gladstone TRG, et al. The prevention of adolescent depression. Psychiatric Clinics of North America. 2011;34:35.
- Primack BA, et al. Using ecological momentary assessment to determine media use by individuals with and without major depressive disorder. Archives of Pediatric and Adolescent Medicine. 2011;165:360.
- Gangwisch JE, et al. Earlier parental set bedtimes as a protective factor against depression and suicidal ideation. Sleep. 2010;33:97.
- Davila J, et al. Romantic and sexual activities, parent-adolescent stress, and depressive symptoms among early adolescent girls. Journal of Adolescence. 2009;32:909.
- 2008 Physical Activity Guidelines for Americans. U.S. Department of Health and Human Services. http://www.health.gov/paguidelines/guidelines/default.aspx. Accessed July 17, 2015.
- Mason MJ, et al. Adolescents' social environment and depression: Social networks, extracurricular activity and family relationship influences. Journal of Clinical Psychology in Medical Settings. 2009;16:346.
- Schwartz OS, et al. Emotion socialization within the family environment and adolescent depression. Clinical Psychology Review. 2010;32:447.
- Maras D, et al. Screen time is associated with depression and anxiety in Canadian youth. Preventive Medicine. 2015;73:133.
- Hoecker JL. Adolescent sleep deprivation. Indian Pediatrics. 2008;45:181.
- A family guide: What families need to know about adolescent depression. http://www.nami.org/Template.cfm?Section=Child_and_Adolescent_Action_Center&template=/ContentManagement/ContentDisplay.cfm&ContentID=24806. Accessed July 17, 2015.
- The depressed child. American Academy of Child and Adolescent Psychiatry. https://www.aacap.org/aacap/Families_and_Youth/Facts_for_Families/Facts_for_Families_Pages/The_Depressed_Child_04.aspx. Accessed July 17, 2015.
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A few weeks ago I left a note for nullification deniers regarding some of their more frequent errors. A quick perusal of Wikipedia clears most of these up, so there’s no excuse for always being so ignorant of the subject matter. Still, some opponents of nullification seem to know plenty of its history, and yet remain plagued with misunderstanding.
For those intellectually honest enough to address the popular origins of nullification in the United States – the Virginia and Kentucky Resolutions – a common rebuttal is to reference those states which opposed the “Principles of ‘98.” This fact seems to be used in order to marginalize the use of nullification and offers a convenient means of confirmation bias for those addressing this point. However, this argument is fraught with a number of problems. It ignores certain relevant historical facts that help explain the motive behind rejecting nullification at the time, it establishes a faulty chain of reasoning, and begs an important ethical question related to politics.
Understanding the manner in which officials were elected to federal office, and the political context of nullification, is essential to unraveling this argument. Events in world history also played a major role in this episode. In brief, here is how the process worked in the early part of American history.
Since George Washington’s first term, presidents were selected much in the same way as they are now, by the electoral college. Vice presidents however were not chosen by the presidential nominee and his political party prior to the election, as they are now. Until ratification of the 12th Amendment in 1804, vice presidents were chosen by the electoral college as runner-up to the presidency. This was meant to prevent deadlocks in the election process and it tended to serve as at least a partial check on the executive branch as well. With the advent of political parties the two offices were sometimes filled by opposing partisans. While a vice president couldn’t unilaterally scuttle a president’s agenda, close ties in the senate might be swayed by the subordinate executive. In 1796, when Federalist John Adams was selected over Democratic-Republican Thomas Jefferson, there was such a division.
With the Quasi-War with France underway, and with government officials in various countries growing increasingly uneasy with the ongoing French Revolution, congress passed the Alien and Sedition Acts in June of 1798. Among the provisions of these new laws, criticism of the Federalist-controlled government, namely John Adams himself, was prohibited. Conveniently left out of this proscription on free speech was Thomas Jefferson, the president’s chief political rival, who happened to be somewhat sympathetic to the revolutionary efforts in France. Because the Federalists largely controlled the federal government at the time, this omission should not come as a surprise.
This also helps to explain why Virginia and Kentucky stood alone in opposition to the Adams administration. In all, seven states passed formal denunciations of nullification. They were the New England States of Connecticut, Massachusetts, New Hampshire, Rhode Island, and Vermont; and the middle States of Delaware and New York. The former had long been controlled by the Federalist Party, and the latter had recently been taken over by Federalists in the previous congress. Several of these states included in their resolutions rather sharp condemnations of nullification. For example, Vermont’s legislature passed a resolution which read in part:
the General Assembly of the state of Vermont do highly disapprove of the resolutions of the General Assembly of Virginia, as being unconstitutional in their nature, and dangerous in their tendency. It belongs not to state legislatures to decide on the constitutionality of laws made by the general government….
In February of 1799 the legislature in Massachusetts, Adams’ home state, wrote we “explicitly declare […] the acts of Congress, commonly called ‘the Alien and Sedition Acts,’ not only constitutional, but expedient and necessary.” Given how contemporary politics is so deeply partisan, it should not surprise us to find that political allegiances may have had something to do with this episode in history.
Now, there appears a certain dilemma for opponents of nullification who buttress their argument by pointing to the New England States’ rejection of the principle. The very foundation of nullification is that the states are rightly charged with interpreting the constitution. Necessarily, the act of nullifying an unconstitutional federal law may only be done after a state has determined the act in violation of the constitution. Opponents of nullification argue the supreme court is the sole arbiter in deciding the validity of federal laws, and thus deny this role to the states.
But in pointing to the interpretations of the constitution by other states – which is what Vermont, Massachusetts, and the others did – someone denying nullification has contradicted his own thesis. In appealing to Vermont and saying “see, Vermont said Kentucky can’t nullify federal law,” an opponent of nullification is implicitly admitting that the states indeed have a role in deciding the constitutionality of a federal law. It’s irrelevant whether a state endorses nullification or opposes it, state interpretation is really at the heart of this argument.
The Massachusetts legislature did in fact acknowledge this, writing that “they do not themselves claim the right, nor admit the authority of any of the state governments, to decide upon the constitutionality of the acts of the federal government….” So they claimed no right to do so, but offered their opinion on the matter. It seems obvious that cognitive dissonance isn’t a feature unique in contemporary political debate.
Setting aside these two previous objections, let’s assume there was no political loyalty between the federal government and the states opposing the Virginia and Kentucky Resolutions. Let us also imagine that state interpretation of the constitution is recognized as a legitimate function by approved opinion. Does a critic of nullification have a valid point in bringing up the majority of states’ opposition to the “rightful remedy?”
The answer is still no.
Majority opinion does not make an action or policy right or wrong. Whether a single despotic ruler declares an action virtuous, or all but one voter in a pure democracy decide to implement a new policy, such opinions hold absolutely no moral weight. In fact, here’s a good rule of thumb: on questions of moral legitimacy relating to government, almost invariably the majority holds an incorrect position. So often the majority is willing to sacrifice the rights and liberties of the minority for their own gain that this is practically axiomatic. Review court case after court case, or browse through reams of congressional voting rolls; the majority opinion almost always goes in favor of usurpation and illegitimate force being applied to otherwise free and peaceful individuals.
For our purposes here it is not my intention to make a complete argument for individual liberty. However, the reader should at least take away from this piece that positive rights are illegitimate constructs of the state, and adherence to the non-aggression principle is the morally appropriate default for human relations. In deciding what is right and what is wrong, rather than look to the majority for guidance, the question that must be answered is this: will this law be compatible with Natural Law; are the negative rights of any one person bolstered or degraded by this action? For further reading, see Judge Andrew Napolitano, here.
While the comparison between the rights of the individual and the rights of states isn’t perfect, some parallels exist related to the constitution.
Despite being parties to the federal constitution, each state retains its own sovereignty. The 9th and 10th Amendments express this idea, reserving all rights not delegated to the federal government. If one state or a group of states decides to delegate more authority to the federal government, it does not ipso facto strip the relevant authority from dissenting states. The fact that some states, or all but one state, find a federal statute acceptable does not make it so for the remainder. In part, this is where the basis for nullification rests – the other states are free to resist through non-compliance.
To reemphasize my point above: the opinion of other states has no legal or moral standing regarding nullification. None.
In the same way that political gamesmanship drives contemporary politics, it did so in the earliest days of the republic. This, and not some idealized view of early America helps to explain the opposition to Kentucky and Virginia’s handling of the Alien and Sedition Acts. Relying on any state’s interpretation of the constitution necessarily requires one to accept the role states have in deciding the legal boundaries assigned to the federal government. Whether a state accepts a law or chooses to render it void is not important here; the crucial point is that they have the prerogative to decide which course to take. Majority opinion notwithstanding, the states aren’t bound by the federal government’s extra-constitutional legislation. It only takes one to say “no.”
Resisting the tendency toward absolute deference to the federal government is critical to preserving liberty. Nullification is the proper means for this. Indeed it is the only vehicle left with which to adequately restrain the federal government.
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# Algberic question
• Jun 17th 2013, 08:22 AM
sscgeek
Algberic question
If x^a*x^b*x^c =1 then the value of a^3+b^3+c^3 is
a)9
b)abc
c) a+b+c
d) 3 abc
• Jun 17th 2013, 09:55 AM
ebaines
Re: Algberic question
Given:
$\displaystyle x^ax^bx^c = x^{(a+b+c)} = 1$
then $\displaystyle a+b+c = 0$, and so $\displaystyle c= -(a+b)$. Hence:
$\displaystyle a^3 + b^3 + c^3 = a^3 + b^3 - (a+b)^3 = a^3 + b^3 -(a^3 + 3a^2b + 3b^2c + b^3) = -3ab( a+b) = 3abc$
• Jun 17th 2013, 10:33 AM
Soroban
Re: Algberic question
Hello, sscgeek!
Quote:
$\displaystyle \text{Given: }\:x^ax^bx^c \:=\:1$
$\displaystyle \text{Find the value of: }\:a^3+b^3+c^3$
. . $\displaystyle (a)\;9 \qquad (b)\;abc \qquad (c)\;a+b+c \qquad (d)\;3 abc$
We have: .$\displaystyle x^{a+b+c} \:=\:1 \quad\Rightarrow\quad a+b+c \:=\:0$
Cube both sides: .$\displaystyle (a+b+c)^3 \:=\:0^3$
. . $\displaystyle a^3 + 3a^2b + 3ab^2 + b^3 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + c^3 + 6abc \:=\:0$
. . $\displaystyle a^3 + b^3 + c^3 + (3a^2b+2ab^2 + 3abc)$
. . . . . . . . . . . . . . $\displaystyle + (3b^2c+3bc^2 + 3abc) + (3a^2c+3ac^2 {\color{blue}+\: 3abc}) {\color{blue}-\: 3abc} \:=\:0$
. . $\displaystyle a^3+b^3+c^3 + 3ab\underbrace{(a+b+c)}_{\text{This is 0}} + \:3bc\underbrace{(a+b+c)}_{\text{This is 0}} + \:3ac\underbrace{(a+b+c)}_{\text{This is 0}} -\:3abc\:=\:0$
. . . . . . . $\displaystyle a^3+b^3+c^3 - 3abc \;=\;0$
. . . . . . . . . $\displaystyle a^3+b^3+c^3 \;=\;3abc\;\;\hdots\;\;\text{answer (d)}$
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As the center of the colonies at the time of the American Revolution, the Hudson River Valley provided a nexus for the conflict and hosted many key figures, battles, and political events throughout the eight years of war. The Sons of Liberty, as active in New York as they were in Massachusetts, printed broadsides, encouraged boycotts, rallied, rioted, and dumped British tea into the New York Harbor even as Patriot housewives throughout the Valley threw their own “tea parties” at the expense of merchants and Loyalist neighbors. The social fabric was ripped apart by the struggle first between the powerful coalition of the DeLanceys and Livingstons and then the Loyalists and Whigs or Patriots.
The New York Provincial Congress established itself at the White Plains Courthouse in July 1776, creating the State of New York with its acceptance of the Declaration of Independence on July 9, 1776. New York adopted its Constitution in Kingston on April 20, 1777 and on February 6, 1778, ratified the Articles of Confederation, tying its fate to the rest of the United States of America.
Throughout the War, battles raged from Manhattan through the Mid-Hudson, including White Plains (1776), Forts Clinton and Montgomery (1777), Kingston (1777), and Stony Point (1779). In October 1777 Patriots watched helplessly as the British burned sites as far north as Clermont before turning back toward New York City. Major General Horatio Gates would right some of the wrongs when he accepted the surrender of Lieutenant General John Burgoyne’s British army at Saratoga on October 17, 1777, marking the turning point in the war. Starting in January 1778 the Americans would follow up on this victory by turning their attention from the ruins of Fort Constitution to building Fortress West Point with its famous chain across the Hudson, a complex that General George Washington called the “key of America.” In the summer of 1781, the French commander, the comte de Rochambeau, marched his 5,000-man army from Rhode Island to Philipsburg, New York, to join the Continental Army, first in the siege of New York and then in the pivotal Yorktown Campaign in Virginia.
Conflict would rage on the frontiers as Native American tribes allied with the British raided settlements and an American army under Major General John Sullivan and New York Brigadier General James Clinton exacted retribution. General Washington and his Main Army would encamp at Newburgh and New Windsor Cantonment in 1782 and 1783 as an Army of Observation for the British army in New York City. Finally on November 25, 1783, Governor George Clinton would lead the Americans into that city after the British evacuation. A War that started in Massachusetts had shifted and remained centered and then ended in New York.
In addition to the prominent roles played by the likes of New York’s first Governor, George Clinton, unsung heroes of the Hudson Valley did their duty as well. Sybil Ludington, New York’s own sixteen-year-old female Paul Revere, rode out of Carmel to raise the militia in defense of the burning Danbury, Connecticut. Chief Daniel Nimham of the Wappingers, a Native American member of the Sons of Liberty and a captain in the American militia lost his life in battle for the cause of liberty.
The American Revolution played out along the Hudson’s banks –from the first riots protesting the British Quartering Act on Golden Hill in Lower Manhattan, to the chaining of the Hudson and Benedict Arnold’s attempted betrayal of West Point in the Highlands in 1780, to the Battle of Saratoga along its northern shores where Arnold played the role not of traitor, but of hero. The Hudson River Valley was essential to the birth of our Nation.
The Great Chain
Major General Alexander McDougall, commander of the Hudson Highlands, wrote cryptically in his diary on April 30, 1778: “chain put aCross.” He was of course referring to the Great Chain at Fortress West Point. As of April 1778 Captain-Lieutenant Thomas Machin with his chain had finally done what Colonel James Clinton and Christopher Tappen had envisioned almost three years before: he had blocked the Hudson River from Fort Constitution to West Point. From spring until fall for the rest of the Revolution, the Great Chain was now a feature of the Highlands. Until at least 1782, soldiers removed the chain each winter and reinstalled it each spring to avoid destruction by the freezing river. General William Heath described the chain as he supervised its removal on November 14, 1780: “The links of this chain were probably 12 inches wide, and 18 inches long; the iron about 2 inches [ 2 ¼] square. This heavy chain was buoyed up by very large logs, of perhaps 16 or more feet long, a little pointed at the ends, to lessen their opposition to the force of the water on flood and ebb. The logs were placed at short distances from each other, the chain carried over them, and made fast to each by staples, to prevent their shifting; and there were a number of anchors dropped at distances, with cables made fast to the chain, to give it a greater stability.” This formidable obstacle of chain and rafts, which may have weighed as much as 100 tons, became the centerpiece of the fortifications at West Point—Washington’s “key of America”–as the supporting batteries, forts, and redoubts rose above it. In that spirit today, the United States Military Academy at West Point each year graduates leaders of character who are keys to the national security of the United States.
A link and swivel bolt from the Second Chain are displayed in Boscobel’s Carriage House visitor center.
General George Washington understood that the Hudson River was the nexus of population, industry, agriculture, commerce, communications, and logistics. As a strategist, he recognized that the Hudson was at once an avenue and a barrier, particularly in the Highlands. It was an invasion route to and from Canada at the one end and New York City on the other. Command of the Hudson influenced the economy and affected the movement of manpower and supplies. In his “Sentiments on a Peace Establishment” in 1783, Washington argued that the defense of the fortifications at West Point on the Hudson River–his major pivot point throughout the war–had been the “key of America.” He probably had first thought about the significance of the Hudson in the French and Indian War, because as early as May 1775 he had joined the other members of a committee of the Continental Congress to recommend fortifying both of its sides in the Hudson Highlands. The Continental Congress accepted the committee’s findings on May 25th and directed the erection of batteries there “in such manner as will most effectually prevent any vessels passing that may be sent to harass the inhabitants on the borders of said river.” The results would be first Fort Constitution on Constitution Island across the Hudson from West Point in 1775 and then Fortress West Point itself beginning in 1778. Washington would never waver in his commitment to the Hudson throughout the war and ventured away from it only when he felt that he had no other choice. He personally spent about one-third of the war in or near the Hudson Highlands. West Point remains the oldest continuously occupied post of the United States Army.
Boscobel possesses a panoramic view across Constitution Marsh and Constitution Island to West Point.
*Special thanks to Col. Jim Johnson for contributing these texts.
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# Percent and Ratios: Practice Questions RRB JE
Get top class preparation for SSC-CGL right from your home: fully solved questions with step-by-step explanation- practice your way to success.
1. What is 35% of a number if 12 is 15% of a number?
1. 5
2. 12
3. 28
4. 33
5. 62
2. A computer is on sale for \$1600, which is a 20% discount off the regular price. What is the regular price?
1. \$1800
2. \$1900
3. \$2000
4. \$2100
5. \$2200
3. A car dealer sells a SUV for \$39, 000, which represents a 25% profit over the cost. What was the cost of the SUV to the dealer?
1. \$29, 250
2. \$31, 200
3. \$32, 500
4. \$33, 800
5. \$33, 999
4. After having to pay increased income taxes this year, Edmond has to sell his BMW. Edmond bought the car for \$49, 000, but he sold it for a 20% loss. What did Edmond sell the car for
1. \$24, 200
2. \$28, 900
3. \$35, 600
4. \$37, 300
5. \$39, 200
5. At a company fish fry, ½ in attendance are employees. Employees'spouses are ⅓ of the attendance. What is the percentage of the people in attendance who are not employees or employee spouses?
1. 10.5%
2. 16.7%
3. 25%
4. 32.3%
5. 38%
6. If 6 is 24% of a number, what is 40% of the same number
1. 8
2. 10
3. 15
4. 20
5. 25
7. 25% of 400 =
1. 100
2. 200
3. 800
4. 10, 000
5. 12, 000
8. 22% of \$900 =
1. 90
2. 198
3. 250
4. 325
5. 375
9. Which of the following percentages is equal to 0.45?
1. 0.045%
2. 0.45%
3. 4.5%
4. 45%
5. 0.0045%
10. Which of these percentages equals 1.25?
1. 0.125%
2. 12.5%
3. 125%
4. 1250%
5. 1250.5%
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|Previous Lesson||Next Lesson|
|C1 part 1||C1 Lesson 10 Carbonates||C1 Lesson 12 Cement, Mortar and Concrete|
Learning outcomes and Specification referenceEdit
|C1.2.1 Calcium carbonate||C1 Lesson 11 Quarries|
|Know that limestone is needed for buildings and that the positive benefits of using this material should be considered against the negative aspects of quarrying.|
Full Specification TextEdit
C1.2.1 Additional guidance:
Candidates should know that limestone is needed for buildings and that the positive benefits of using this
material should be considered against the negative aspects of quarrying.
- Students are to write the chemical formula for calcium carbonate (CaCO3) and state the 3 elements are in this compound.
- Students to write down any formations of calcium carbonate they can think of (chalk, limestone, marble..). Higher ability students to write down what type of rock formation calcium carbonate is (Sedimentary) and how they are formed.
Activities to introduce new ideasEdit
- No activities have been added to this lesson plan in this section
Activities to practice applying new knowledgeEdit
- File:Lime cycle questions.doc
- File:Limestone exam question2.doc
- File:Limestone Quarrying Debate.doc
- File:Limestone exam question markscheme.doc
- No demonstrations have been added to this lesson plan
- No links have been added to this lesson plan
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All decisions we make in design and materials affect the health of ourselves,
our children, and the planet.
SIMPLE, NATURAL, SITE-SPECIFIC DESIGN
- COLLABORATION WITH THE SITE AND NATURAL SYSTEMS
- Imagine what the site would design for itself.
- Leave the surrounding area more natural than before construction.
- NATURAL STRUCTURES AND INTERIORS
- Respect sun, earth, plants, and water.
- Design for full advantage of the sun.
- Minimize electromagnetic radiation within buildings.
- Use more natural materials and fewer chemicals.
- APPROPRIATE TECHNOLOGY AND MATERIALS
- Take advantage of local resources and craftsmanship.
- Use energy-saving products.
- Find out how, where, and with what each product is made.
- Know how the product is installed, how long it lasts, and how it can be reused and recycled.
- Know how much energy it takes to transport, maintain, and use each product.
- Be aware of the effect each product will make on nearby living beings.
- PASSIVE SOLAR DESIGN
- Place spaces where the sun can warm them.
- Use house elements as thermal storage to balance heating.
- NATURAL, LOW-ENERGY LIGHTING
- Use natural light to decrease electricity and add dimension to all rooms.
- Light spaces with minimum wattage.
- Decrease energy usage with warm, natural-colored fluorescent lighting fixtures.
- Use task lighting in work areas.
- NATURAL HEATING AND COOLING
- Provide natural, even heat, and lower thermostat settings, with radiant hydronic slab heating.
- Construct for low heat loss.
- Use natural breezes to build thermal mass and passive cooling.
- Design outdoor rooms.
- Invest in high-quality windows and doors.
HEALTHY INTERIOR ENVIRONMENTS
- THE BREATHING HOUSE
- Construct houses that both insulate and breathe.
- Use heat recovery ventilation systems to save heat and moisture.
- NATURAL MATERIALS AND HEALTHY INTERIORS
- Make sure interior construction materials and finish systems are alive and breathing, instead of plastic and dead.
- Use wood, stone, earth, brick, grasses, cork, recycled tile and vegetable oils. Do not use plastic, chemicals, or toxins.
- Instead of wall-to-wall carpet and vinyl flooring, use natural flooring and area rugs.
- Use natural, non-toxic, and low VOC (volatile organic compound) paints and sealers.
- Avoid non-renewable and environmentally damaging materials, like CFC foams and particle board.
- NATURAL INDOOR SETTINGS
- Use interior plants to remove toxins from the air and provide fresh oxygen.
- Add interior water environments, which provide relaxing sounds and essential moisture.
- ELECTRONIC MAGNETIC FIELDS
- Explore the effects of electrical layouts on personal health.
- Eliminate high electric loads in relaxing and sleeping areas.
- Control electric wiring by switching off sleeping areas at night.
- ENGINEERED WOOD
- Use structural elements manufactured with
a minimum amount of high-grade wood.
- SUSTAINABLE LUMBER
- Use certified sustainable lumber.
- Use fast-growing materials like
- RE-USED WOOD
- Use re-cut lumber from recently dismantled buildings.
- RECYCLED MANUFACTURED MATERIALS
- Use materials manufactured with a high recycled content.
- LONG-LIFE AND LOW-MAINTENANCE MATERIALS
- Use materials that last, with minimum or no maintenance costs.
- LOW-ENERGY MATERIALS
- Use materials that require minimal energy to manufacture, transport, and grow.
- RECYCLED AND ALTERNATIVE CONSTRUCTION SYSTEMS
- Straw-bale construction.
- Cob (straw and earth) construction.
- Recycled ICF wall construction.
- Pre-fabricated panel wall & roof systems.
- Post and beam construction.
- Construction methods that the owner can use to build the home themselves.
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Set of inflammatory medical conditions, affecting the tissues surrounding the teeth is periodontal disease, otherwise called gum disease. As the name suggests, the gums become swollen, red and may also bleed in the early stage of the disease. This stage is called as gingivitis.
It can also happen due to pulling away of gum from the tooth, loss of bone and falling out of teeth; This usually happens in its acute form called periodontitis. Thus all the supporting tissues: the cementum that covers the root, the periodontal ligament and the alveolar bone are affected.
According to the latest research, this gum disease also causes other health problems like,
- Increasing risk of clogged arteries due to gum disease is believed to worsen existing heart disease and also increases the risk of stroke.
- Pregnant ladies having gum disease may deliver the baby too early. This premature baby may be of low birth weight.
- Diabetic patients also suffering from the periodontal disease may face difficulties in controlling their sugar level.
- And last but not the least it also causes lung infections, as the bacteria from the mouth may find their way to lungs and may cause severe pneumonia.
What Causes Periodontal Disease?
- The primary cause is inadequate oral hygiene which causes accumulation of a mycotic and bacterial matrix at the gum line, called dental plaque.
- Other causes include microbic plaque accumulation, such as restoration overhangs and root proximity.
- Another factor which contributes to other health hazards is smoking. Smokers have more bone loss, attachment loss and tooth loss in comparison to non-smokers because smoking decreases wound healing and suppresses antibody production.
- Many diseases such as Down syndrome, diabetes, etc. that decreases patient’s resistance to infection, increase the chances for periodontitis.
- Economic status also plays its cards in the game as periodontitis may also be related to higher stress.
- Genetics also play a role; they could explain why some patients with good plaque control have advanced periodontitis, while some others with not so good or poor plaque control are free from the disease.
- Typically, the risk of periodontitis in diabetics is estimated to be between 2-3 times higher. It’s known to involve aspects of inflammation, immune functioning, neutrophil activity and cytokine biology.
Signs And Symptoms Of Periodontal Disease
In many individuals, the disease would have increased before they opt for treatment because there are only a few signs in the initial stages of periodontitis. Some of the symptoms are:
- Redness, bleeding of gums while brushing teeth.
- Swelling of gums.
- Bad breath along with a persistent metallic taste in the mouth.
- Apparent lengthening of teeth due to gingival recession or by rough brushing or with a stiff toothbrush.
- Deep gaps or pockets between the teeth and the gums.
- Loose teeth, in the later stages.
- As gingival inflammation and bone destruction are largely painless people should not treat it as insignificant.
Prevention Of Periodontal Disease
Addition of some of the habits of daily oral hygiene may help you prevent from periodontitis. Below is the list of measures you can take to prevent periodontal diseases:
- Regular and proper brushing of teeth, i.e., at least twice a day.
- Directing the toothbrush bristles under the gum line while brushing for removal of bacterial-mycotic growth and disrupt the formation of subgingival plaque.
- Using antiseptic mouthwash to flush away bacterial matrix and cure gingivitis.
- Using periodontal trays that allow the medication to stay in place.
Treatments for periodontal disease
- Professional cleaning at the dentist’s office.
- Shadi Heidarian, a Palo Alto dentist, highly recommends opting for a special periodontal cleaning called scaling and root planing. This procedure involves removal of tartar, plaque and toxins from above and below the gum line, called scaling. The removal of spots on root surfaces is made smooth, known as planing.
- Pocket reduction procedure: By folding back the gum tissue dentist can remove infectious bacteria and smooth areas of damaged bone, allowing the gum tissue reattach to healthy bone.
- Gum Grafts: Roots that are exposed to gum recession can be covered with gum grafts which reduces sensitivity and protect the roots from decay.
- Regenerative procedures: the surgical procedure that helps the growth of bone in an area where periodontal disease has destroyed the bone.
Oral health and hygiene should not be neglected as it is the cause of many other diseases. You should keep your oral health in check so that you can carry that beautiful smile wherever you go!
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For at least 100 million years, there have been poppy seed-sized blood-sucking creatures called ticks carrying bacteria that cause various illnesses in humans.
“Yeah, they’re ancient,” says Timothy Driscoll, an associate professor of biology at West Virginia University who studies ticks at the university’s Vector-borne Infectious Disease Laboratory.
Ticks go through four life stages: egg, larva, nymph and adult. After an egg hatches, a tick must have a blood-meal at each stage in order to survive. Each stage also requires a new host. After completing their two- to three-year lifespan, female ticks detach from a host and lay eggs, typically in the springtime. These eggs hatch in summer and larval ticks emerge. Larvae are active but less likely to transmit tick-borne pathogens during this stage. Between fall and spring, as the weather warms, larvae molt and enter a more active stage as nymphs.
In their nymphal stage during spring and summer, ticks begin to seek out human hosts. They stay attached to their host for days before dropping off and transitioning into the adult stage. Ticks are most difficult to detect in this stage because they are so small, measuring less than 2mm.As an adult, ticks are questing for their final host. Adult ticks become inactive during the winter months, resuming their quest for a blood-meal in late February to early March.
A changing climate has both shortened and warmed the winter season in some regions, contributing to the ideal environmental conditions for ticks. Consequently, tick populations are growing and incidences of tick-borne illnesses like Lyme disease have been on the rise.
Before a tick spreads disease, however, they first need to find a meal. They do so by waiting in leaf litter and low-lying brush. This waiting is called questing. Ticks use their third and fourth pair of legs to rest on the tips of leaves and blades of grass. And because they can’t fly or jump, they sit, with their first pair of legs outstretched.
When someone — or something — brushes along the edge of the forest, ticks latch on with their two front legs.
A tick can attach to any part of the human body. Often ticks are found in hard-to-see areas such as the groin, armpits or scalp, according to the Centers for Disease Control and Prevention. Cam Heustess, a Western North Carolina construction worker, found a tick on his hip last summer while completing a project on the New River.
“I came in from building a garden shed off the New River, and one had attached on my hip,” he says.
Heustess submitted his tick to a tick-ID program in North Carolina and discovered it was a blacklegged tick.
These are one of the most common ticks among the Southeastern United States that bite humans, according to the CDC .
“In West Virginia and Appalachia in general, if you pick up a tick it’s probably going to be either the blacklegged tick or it’s going to be the American dog tick,” says WVU Associate Professor Timothy Driscoll.
A third tick, the lone star tick, is also present in the region but not found quite as often. These are the “three main flavors,” he says.
So why did that blacklegged tick end up on Heustess’ hip? Ticks find their hosts by detecting things like breath and body odors, or by sensing body heat.
“One of the big things they can sense is carbon dioxide in warm-blooded animals,” Driscoll says.
“Mammals give off CO2 as they perspire and ticks can use that as a cue that says ‘Hey, there’s some warm-blooded thing walking by,” he adds.
Lyme disease is the most common vector-borne disease in the United States. It is caused by the bacterium Borrelia burgdorferi and is transmitted to humans through the bite of infected blacklegged ticks. Most cases of Lyme disease can be cured within 2 to 4 weeks, with a course of oral antibiotics; however, patients can sometimes experience symptoms for more than 6 months following treatment.
Rocky Mountain Spotted Fever
Rocky Mountain spotted fever is a potentially fatal bacterial disease spread through the bite of an infected American dog tick, Rocky Mountain wood tick and the brown dog tick. It most commonly occurs in North Carolina, Tennessee, Missouri, Arkansas and Oklahoma.
Alpha-gal Syndrome, or red meat allergy, is an allergic reaction to consuming red meat. It is commonly reported among people living in the Southeastern United States and believed to be triggered by the bite of a lone star tick or blacklegged tick.
Driscoll states that ticks can also detect or respond to different things in a host animal’s hair, dander or skin.
A (sometimes barbed) feeding tube and a cement-like secretion allow a tick to stay firmly attached to its host. If undetected, a tick will stay latched onto its host long enough to get engorged before falling off. Sometimes this can take up to 36 hours.
It’s highly unlikely that a human is the first blood-meal for a tick.
“Humans are more of an incidental host,” says Alexis Barbarin, North Carolina’s state public health entomologist.
In the nymphal stage, she says, a tick will predominantly feed on small rodents and mammals, like mice or deer, and some birds. The blacklegged tick, for example, feeds mostly on white-footed mice, scientists have found.
While feeding, a tick may ingest a bloodborne infection from the host. The tick will pass on this infection to its next host. Sometimes, these infections carry the pathogens that cause vector-borne diseases.
The tick that latched on to Heustess was carrying Lyme disease, an incident which researchers say becomes more common every year. In the hospital, Heustess said his doctors told him he could be managing his illness for the rest of his life, and so far he has. Heustess says there are times when his illness flares up; seemingly out of nowhere he gets feverish and his joints ache.
“We always encourage people, regardless of what time of year it is, that they should take some measures to reduce their exposure to ticks and prevent vector-borne disease,” Barbarin says.
Ticks are tiny creatures who pack a powerful but infectious punch. Simple wardrobe changes can make all the difference when trying to avoid them.
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The 20 enjoyable, interactive classroom activities that are included will help your students understand the text in amusing ways. Fun Classroom Activities include group projects, games, critical thinking activities, brainstorming sessions, writing poems, drawing or sketching, and more that will allow your students to interact with each other, be creative, and ultimately grasp key concepts from the text by "doing" rather than simply studying.
1. Vietnam War
Research an aspect of the Vietnam War and present it to the class in a five-minute presentation.
2. History of Vietnam
Research an aspect of Vietnam's history. Present your findings to the class in a five-minute presentation.
Research communism, including when it began and how it has evolved. Write a one-page paper on your findings.
4. The Cold War
Research an aspect of the Cold War, including when it began, why it began, and how it came to an end. Write a one-page paper...
This section contains 583 words
(approx. 2 pages at 300 words per page)
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Sudden infant death syndrome (SIDS) refers to the unexpected, unexplained death of a child less than one year old. The occurrence of SIDS is rare during the first month of life, peaks at 2 to 4 months of age, then gradually decreases.
Experts do not know why some infants suddenly die. Research hopes to find a preventable cause. Many theories exist and it is likely that research will reveal that among babies who die of SIDS there are several different, but important causes. Some suggested causes include:
Area of the brain involved in regulation of breathing.
Copyright © 2005 Nucleus Communications, Inc. All rights reserved.
A risk factor is something that increases your chance of getting a disease or condition.
Risk factors include:
A baby that dies of SIDS typically appears healthy. He or she may have had a
or gastrointestinal infection in the two weeks prior to death. There usually are no warning signs that a baby is about to die of SIDS.
All possible illnesses and causes of death must be ruled out before a diagnosis of SIDS is made. A complete investigation will take place including:
Emergency medical personnel should be called as soon as the infant is discovered not breathing. Infant, not adult, cardiopulmonary resuscitation (CPR) should be started immediately (both types of CPR are taught at basic life support courses which are widely available in the US). Seek immediate medical care even if the baby starts breathing again. The cause of the incident should be fully evaluated. Families may need grief counseling after the death. Some parents find support groups helpful.
There is no way to predict which infants will die of SIDS. Several actions may help you lower your child's risk of SIDS:
Make sure anyone else caring for your child is also aware of these recommendations, especially that the baby should be placed on his or her back for sleeping.
National SIDS/Infant Death Resource Centerhttp://www.sidscenter.org
Sudden Infant Death Syndrome Alliancehttp://www.sidsalliance.org
SIDS Society Calgarywww.sidscalgary.ca/
American Academy of Pediatrics SIDS Task Force. The changing concept of sudden infant death syndrome: diagnostic coding shifts, controversies regarding the sleeping environment, and new variables to consider in reducing risk.
Current Pediatrics. 17th
edition. McGraw-Hill; 2005.
Galland BC et al. Prone versus supine sleep position: a review of the physiological studies in SIDS research.
J Paediatr Child Health. 2002;38(4):332-338.
Hunt CE, Hauck FR. Sudden infant death syndrome.
National Institute of Child Health and Human Development website. Available at:
Nelson Textbook of Pediatrics. 18th
edition. WB Saunders; 2007.
Sudden Infant Death Syndrome Alliance website. Available at:
The #1 daily resource for health and lifestyle news!
Your daily resource for losing weight and staying fit.
We could all use some encouragement now and then - we're human!
Explore your destiny as you discover what's written in your stars.
The latest news, tips and recipes for people with diabetes.
Healthy food that tastes delicious too? No kidding.
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If you know what a semicolon is then use it when writing in English
Semicolons are used similarly to commas and not at all like colons. This is the general rule of thumb that is easy to remember. Now to make matters a little complex, semicolons can be compared to commas, which are used when a sentence is put together to form a whole and complete one. However, when a sentence proves to be too long, yet carrying the necessary components and those components cannot be separated into different sentences, a semicolon can be used in order to not break a sentence down and destroy the essence of its meaning. Semicolons are also used in instances when ideas, two or more, are make to be part of a sentence, while they are given equal worth or rank. Semicolons are also used for brining two independent clauses together, which then eliminates the pause that exists if a period exists and needs to be filled with either and, but, nor, or yet as well as anything that brings two separate and independent statements together. Don’t forget about the fact that semicolons are closer to commas than they seem to be so use them instead of commas wherever a comma would prove inappropriate.
When you want to use the semicolon, make sure you are working with two sentences that can stand their own ground independently; meaning they are complete sentences. This way you are able to form a bond between them and strengthen the relationship between them by using a semicolon. Make sure not to use capital letters on the first word that follows a semicolon; they are designed to informally connect and strengthen each other that way. Below are some examples of the correct form of semicolon usage:
The tiger looked dead into my eyes; that is what tigers do right before they pounce on you for the kill.
Basketball players high five a lot on the court when they score; getting and giving high fives is a part of basketball.
The objective for world economic development is hidden; governments make sure some secrets are not exposed; this way, our secrets are not used against us by our enemies.
Polar bears are extremely powerful and aggressive animals; beneath that rage to eat people lies a gentle creature who only wants to protect her cubs.
You can work your way through college all you want, but there is no guarantee to make it through in one piece; and the reality is much harder than one can imagine, so work to save money, put yourself in a financially comfortable position and then push for academic success.
Learn and improve your English with ENG45.com
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Prove summation formula for binomial coefficients [duplicate]
Possible Duplicate:
simple binomial theorem proof
Prove that:
$$\sum_{k=0}^n \binom{k+a}{k}=\frac{(n+a+1)!}{n! (a+1)!},$$
where $a$ is a constant, without using induction. A probabilistic proof would be nice.
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marked as duplicate by Brian M. Scott, Mike Spivey, Hagen von Eitzen, Chris Eagle, Brandon CarterDec 10 '12 at 22:19
You’ll find both a combinatorial argument and a proof by induction at the earlier question cited above. – Brian M. Scott Dec 10 '12 at 21:15
@BrianM.Scott: And that was just yesterday! I'm not sure how I missed that. – Mike Spivey Dec 10 '12 at 21:18
@Mike: Well, I did have a slight advantage! (Good to see you back, by the way.) – Brian M. Scott Dec 10 '12 at 21:20
Presumably $n$ and $a$ are nonnegative integers. The right side is ${n+a+1} \choose n$. Consider $n+a+1$ items, of which you want to choose $n$. At least one of the first $n+1$ must be left out. If the first item to be left out is number $n+1-k$ (where $0 \le k \le n$), then you already have chosen the first $n-k$ items, you are skipping number $n+1-k$ and you need to choose $k$ out of the remaining $n+a+1-(n+1-k) = a+k$. The number of ways to do that is ${a+k} \choose k$.
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A similar result can be found here. The corollary is the result you seek. The proof of the corollary is on the same page. Hope this helps.
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carotenoid(redirected from Carotinoid)
Also found in: Dictionary, Thesaurus, Medical.
Any of a class of yellow, orange, red, and purple pigments that are widely distributed in nature. Carotenoids are generally fat-soluble unless they are complexed with proteins. In plants, carotenoids are usually located in quantity in the grana of chloroplasts in the form of carotenoprotein complexes. Carotenoprotein complexes give blue, green, purple, red, or other colors to crustaceans, echinoderms, nudibranch mollusks, and other invertebrate animals. Some coral coelenterates exhibit purple, pink, orange, or other colors due to carotenoids in the calcareous skeletal material. Cooked or denatured lobster, crab, and shrimp show the modified colors of their carotenoproteins.
The general structure of carotenoids is that of aliphatic and aliphatic-alicyclic polyenes, with a few aromatic-type polyenes. Most carotenoid pigments are tetraterpenes with a 40-carbon (C40) skeleton. More than 300 carotenoids of known structure are recognized, and the number is still on the rise.
There are several biochemical functions in which the role of carotenoids is well understood. These include carotenoids in the photosynthetic apparatus of green plants, algae, and photosynthetic bacteria, where carotenoids function as a blue light-harvesting pigment (antenna or accessory pigment) for photosynthesis. Thus carotenoids make it possible for photosynthetic organisms more fully to utilize the solar energy in the visible spectral region. See Chlorophyll, Photosynthesis
Another function of carotenoids is to protect biological systems such as the photosynthetic apparatus from photodynamic damage. This is done by quenching the powerful photodynamic oxidizing agent, singlet oxygen, produced as an undesirable by-product of the exposure of pigmented organisms to light.
Perhaps the most important industrial application of carotenoids is in safe coloration of foods, as exemplified in the coloring and fortification of margarine and poultry feedstuff.
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From 1915, social injustice, poverty and fear drove half a million African Americans from the South to the North, in what was called The Great Migration. In the North, African Americans had to live in ghettos and still faced segregation, but new economic opportunities were available to them. The release of the film 'The Birth of a Nation' led to increased racism and the growth of the Ku Klux Klan.
Produce two lists detailing living conditions for African Americans in the South and the North and relate these to push and pull factors that led to the great migration. Look in more detail at the attitude and activities of the Ku Klux Klan. This could be split into the three main periods of Klan activity, 1860s and 70s, 1915-1944 and 1950s and 60s. In each case look at reasons why people joined, based on the negative factors they wished to prevent and the positive factors they wanted to benefit from. The example of 'Birth of a Nation' could link to a study of media influence. Explore how the iconography of the Klan can be traced through the film and the books it was based on back to the romanticised views of old Scotland and England portrayed by the works of Sir Walter Scott.
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The Collapse of France, Dunkirk Retreat, and Battle of Britian
- Grades: 6–8, 9–12
Collapse of France. During the winter of 1939-40 the French army and the German Wehrmacht faced one another in what was regarded satirically as the sitzkrieg, or sit-down war. The world waited in anticipation of a major conflict between two powerful forces. On May 13, 1940, a bridgehead was established at Sedan, considered the gateway to France, and then suddenly, on May 16, a day after the Dutch capitulation, the German blitzkrieg was released on northern France. German mechanized forces outflanked the Maginot Line, surprised the Allies by attacking through the wooded Ardennes rather than the Belgian plain, and drove the British Expeditionary Force (BEF) from the Continent at Dunkirk ( Dunkerque). On June 5 the Germans launched another offensive southward from the Somme. They entered Paris unopposed on June 14 and forced France to sign an armistice at Compiègne on June 22, 1940.
The fall of France was an extraordinary victory for Hitler. The supposedly unbeatable French army had melted away before the onslaught of his mobile units, with their convincing display of mechanized power. Nazi Germany then occupied most of France and permitted the establishment of a friendly government at Vichy, in central France on the Allier River.
The Vichy Government was headed by Marshal Henri Pétain, hero of World War I, and Pierre Laval, a collaborationist. Disgruntled French patriots rallied around Gen. Charles de Gaulle, who pronounced himself leader of the Free French.
During the early months of the war Benito Mussolini maintained Italy's neutrality. When France was about to fall he decided to join the Nazis. Declaring war on the Allies on June 10, 1940, he invaded southern France in what U.S. president Franklin D. Roosevelt described as a "stab-in-the-back."
Dunkirk Retreat. During the Belgian campaign the Germans drove rapidly across southeastern Belgium and turned toward Abbeville on the French coast, thereby isolating Allied troops. The BEF and its French comrades appeared to be doomed. While some of the troops of the French First Army sold their lives in a fierce rearguard action, from British ports sailed one of the strangest armadas in history composed of destroyers, motor launches, private yachts, old ferries, steamers, even fishing smacks, about 850 vessels in all. While planes of the Royal Air Force (RAF) provided an umbrella over the scene to drive off German bombers, the fleet of British vessels moved to Dunkirk and proceeded to evacuate about 338,000 British, French, and Belgian troops from May 26 to June 4, 1940. Not only was a military disaster turned into a propaganda victory, but several hundred thousand experienced troops were saved for future action against the Axis.
Battle of Britain. Hitler, anticipating further eastern conquests, hoped that Britain would accept German control of the Continent and seek peace. But Britain shunned the chancellor's overtures of July 1940, and, in August, Hermann Goering's Luftwaffe began an all-out attack on British ports, airfields, and industrial centers and, finally, on London. The goal was to crush British morale and wipe out the RAF in preparation for Operation Sea Lion, an invasion of England.
The Battle of Britain was the first great air battle in history. For 57 nights London was attacked by an average force of 160 bombers. The outnumbered RAF, employing the effective Spitfire fighter and aided by radar, destroyed 1,733 aircraft while losing 915 fighters. German air power could not continue sustaining such heavy losses, and in October, Operation Sea Lion was postponed indefinitely.
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November 14, 2018
Carbon dioxide (CO2) is a major cause of global warming. Researchers use complex computer models to calculate the global circulation of this greenhouse gas. The oceans have a major influence on climate regulation. New research now helps to calculate this influence more precisely. These new findings are the result of a research project by scientists from Jacobs University and the Max Planck Institute for Marine Microbiology in Bremen in collaboration with colleagues from the Alfred Wegener Institute Helmholtz Center for Polar and Marine Research in Bremerhaven, the Marum Centre for Marine Environmental Sciences at the University of Bremen and the University of Gothenburg.
When the porous aggregates sink faster towards the seabed, the generated faster fluid flow can provide more oxygen to the aggregate. Illustration: Science Advances
Oceans contain about 50 times more carbon than the atmosphere, and about 20 times more carbon than land. Algae and organic particles in the upper, sunlit water layer bind the CO2, sink to the seabed and deposit there. These so-called aggregates are the main actors in the transport of organic carbon from the surface into the deep sea. By absorbing CO2 from the atmosphere, they play an important role in climate regulation.
In the project funded by the German Research Foundation (DFG), the scientists have now taken a close look at the individual aggregates and calculated the rate of their oxygen consumption. For example, they found that previous assumptions about oxygen flux into the aggregates were far too high. "We have achieved very accurate results that can be extended to other aggregate types and provide better estimate of oxygen consumption by sinking aggregates," says Dr. Arzhang Khalili, Professor of Computer Science at Jacobs University Bremen and at the Max Planck Institute for Marine Microbiology. Marine processes can now be integrated into existing climate and carbon cycle models and thus make them more realistic. The results of the study were recently published in the journal "Science Advances".
Link to article:
Questions will be answered by:
Prof. Dr. Arzhang Khalili | Professor of Computer Science
a.khalili [at] jacobs-university.de | +49 421 200 3256
akhalili [at] mpi-bremen.de | +49 421 2028 63
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The Domestication of the Slave Trade in the United States
Paper to be delivered at the Gilder Lehrman Center for the Study of Slavery and Abolition
Yale University, October 1999
Between 1770 and 1820, the United States withdrew from the international slave trade, During the same period, a substantial slave trade within the United States developed, uprooting slaves from the Atlantic seaboard and transplanting them in the western interior and Deep South. This domestication of the slave trade was one of the most important social, economic and political transformations in the United States from the Revolution to Civil War. It greatly contributed to the consolidation and expansion of slavery during an era fraught with dangers for slaveowners in North America.
The domestication of the slave trade grew out of the tension between economic opportunity and political crisis in the late eighteenth century. Rising demand for cotton and sugar in the world market and the restructuring of the Caribbean plantation economy created new opportunities for planters in the United States to profit from the use of slave labor, while territorial expansion in the southern United States incorporated a vast new region into the orbit of the plantation system. However, revolutionary stirrings among slaves throughout the Americas intensified fears of slave resistance and rebellion within the United States. Many slaveholders joined opponents of slavery in concluding that the importation of slaves from Africa and the Caribbean endangered the security of the country. Southern slaveholders hoped that the combination of prohibiting the importation of foreign slaves and dispersing the slave population through the southwest would strengthen the security of slavery in the United States.
Through the domestication of the slave trade, slaveholders and their allies fashioned a distinction between slaveholding and slavetrading which became integral to proslavery doctrine. Southern politicians and intellectuals applied the distinction between slaveholding and slavetrading to the expansion of slavery into the new territories of the southwest, but a substantial internal slave trade developed anyway. For the most part, the task of regulating the internal movement of slaves was left up to the southern states. All the southern states experimented with constitutional and legislative restrictions on the domestic slave trade in the early nineteenth century. These measures did not retard the growth of the internal slave trade, but they did express the emerging logic of proslavery doctrine. Their failure demonstrated the contradiction between the ethical outlook of proslavery doctrine and the commercial reality of slavery in the American South.
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#### MAT-09.AR.F.02
9th Grade (MAT) Targeted Standard (AR) Algebraic Reasoning (F) Functions Learners will develop a foundational knowledge of functions and use them to model relationships between quantities.
## Progressions
Functional Relationships
• MAT-08.AR.F.01 Defend whether a relation is a function from various representations using appropriate function language.
• MAT-08.AR.F.02 Compare and contrast properties of two linear functions, each represented in a different way (algebraically, graphically, numerically in tables, and/or by descriptions).
• MAT-08.AR.F.03 Compare and contrast linear and non-linear functions represented in different ways (algebraically, graphically, numerically in tables, and/or by descriptions).
• MAT-08.AR.F.04 Model a linear relationship between two quantities by creating a table, graph, and equation. Interpret the rate of change and initial value of a linear function in terms of the situation it models.
• MAT-08.AR.F.05 Describe qualitatively the functional relationship between two quantities by analyzing a graph, including where the function is constant, increasing, or decreasing; linear or nonlinear; and discrete or continuous. Create a graph that exhibits the qualitative features of a function described.
• MAT-09.AR.F.01 Determine whether a relationship is a function given a table, graph, or words, identifying x as an element of the domain and f(x) as an element in the range. Determine the domain and range of a function in context.
• MAT-09.AR.F.02 Use function notation, evaluate functions for inputs in their domains and interpret statements that use function notation in terms of context.
• MAT-09.AR.F.03 Sketch key features (to include intercepts, maximums, minimums, and lines of symmetry, where applicable) of linear, exponential, and quadratic functions modeling the relationship between two quantities using tables, graphs, written descriptions, and equations.
• MAT-09.AR.F.04 Relate the domain of a linear, quadratic, or exponential function to its graph and, where applicable, to the quantitative relationship it describes.
• MAT-09.AR.F.05 Calculate and interpret the average rate of change of a linear, quadratic, or exponential function (presented algebraically or as a table) over a specified interval. Estimate the rate of change from a graph.
• MAT-09.AR.F.06 Write a function defined by an expression in different but equivalent forms to reveal and explain the different properties of the function.
• MAT-09.AR.F.07 Compare key features of two linear, exponential, or quadratic functions, each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions).
• MAT-09.AR.F.08 Identify situations that can be modeled with linear, quadratic, and exponential functions.
• MAT-09.AR.F.10 Find the inverse of a linear function and describe the relationship between the domain, range, and graph of the function and its inverse. Graph the inverse of a linear function.
• MAT-09.AR.F.11 Interpret the parameters of a linear, quadratic, or exponential function in terms of context.
• MAT-10.GM.02 Represent transformations in the plane. Describe transformations as functions that take points in the plane as inputs and give other points as outputs. Compare transformations that preserve distance and angle to those that do not (i.e., rigid versus non-rigid motion).
• MAT-10.GM.26 Recognize that the radian measure of an angle is the ratio of the length of the arc to the length of the radius of a circle.
• MAT-12.AR.F.01 Write a function that describes a relationship between two quantities.
• MAT-12.AR.F.02 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.
• MAT-12.AR.F.03 Write a function defined by an expression in different but equivalent forms to reveal and explain the different properties of the function.
• MAT-12.AR.F.04 Identify the effect of transformations on the graph of a function by replacing f(x) with af(x), f(bx), f(x - h), and f(x) + k, for specific values of a, h, and k (both positive and negative). Find the value of a, b, h, and k given the graph of the function. Recognize even and odd functions from their graphs and equations.
• MAT-12.AR.F.05 Find inverse functions.
• MAT-12.AR.F.06 Apply the inverse relationship between exponents and logarithms to solve problems.
• MAT-12.AR.F.07 Compare key features of two functions, each represented in a different way (algebraically, graphically, numerically, in tables, or by verbal descriptions).
• MAT-12.AR.F.08 Use tables, graphs, verbal discussions, and equations to interpret and sketch the key features of a function modeling the relationship between two quantities.
• MAT-12.AR.F.09 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes.
• MAT-12.AR.F.11 Analyze and graph functions expressed symbolically (by hand in simple cases and using technology for more complicated cases), identifying key features of the graph.
• MAT-12.AR.F.12 Compare the end behavior of linear, quadratic, and exponential functions using graphs and/or tables to show that a quantity.
• MAT-12.AR.F.13 Determine whether a linear, quadratic, polynomial, exponential, logarithmic, or trigonometric model fits the situation. Determine an appropriate mathematical model in context (with or without technology).
• MAT-12.AR.F.14 Write arithmetic and geometric sequences both recursively and with an explicit formula and convert between the two forms. Use sequences to model situations.
• MAT-12.AR.F.16 Extend right triangle trigonometry and apply knowledge of the unit circle to determine values of sine, cosine, and tangent for multiples of π /3, π/4 and π/6.
• MAT-12.AR.F.17 Use the Pythagorean Identity sin²(θ) + cos²(θ) = 1 to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle.
• MAT-12.AR.F.18 Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle.
• MAT-12.AR.F.19 Use the unit circle to express the values of sine, cosine, and tangent for π - x, π + x, and 2π - x in terms of their values for x, where x is any real number.
• MAT-12.AR.F.20 Use the unit circle to explain the symmetry (odd and even) and the periodicity of trigonometric functions.
• MAT-12.AR.F.21 Create a trigonometric function to model periodic phenomena.
• MAT-12.AR.F.22 Restrict the domain of a trigonometric function to construct its inverse.
• MAT-12.AR.F.23 Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions and interpret them in context.
• MAT-12.AR.F.24 Know and apply the addition and subtraction formulas for sine, cosine, and tangent to solve problems.
|
crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00799.warc.gz
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# It takes a boat 2 hrs to travel 16 miles down stream. The same boat takes 4 hrs for the return trip.?
Whats the speed of the water?
Relevance
• Julian
Lv 6
8 years ago
v = speed of the boat
c = speed ot the water (current)
(v+c)*2 = 16 => v+c = 8
(v-c)*4 = 16 => v-c = 4
-------------
2c = 8-4 = 4
c = 2
answer : speed of the water = 2mph
• Anonymous
8 years ago
The speed of the boat + the river is its speed downstream, so
speed = distance/time
= 16/2
= 8 m/hr
The speed of the boat - the river is its speed upstream, as the water is travelling the opposite direction to the boat, so
speed = 16/4
= 4 m/hr
let the speed of the boat = b, and the speed of the water = w
b + w = 8
b - w = 4
add the two equations together to get
2b = 12
b = 6
Substitute into one of the above equations
6 + w = 8
w = 2 miles/hour
• 8 years ago
S=vxt so v=s/t.
S= 16 miles
t= 4 hrs
v= ?
v= 16/4 = 4 miles per hour.
That should be the correct answer.
v = speed
S= distance travelled
t= time
Whoops, used the Dutch formula, I believe the English formula is D=Rxt or something like that.
So S= D, v=R and t stays t. Sorry!
Source(s): Physics
• 4 years ago
formula of velocity: v=d/t, the place V is velocity, D is distance and T is time. appearing some rearranging we get that t=d/v In equation under v is velocity of the boat and r is the fee of the flow 40 8/(v+r)=3 40 8/(v-r)=12 fixing that equation we get that r=6
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crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00422.warc.gz
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• Jan 9th 2009, 05:54 PM
cnmath16
Question:
There are 8 opposite-sex married couples at a party. Two people are chosen at random to win a door prize.
a) What is the probability that the 2 people will be married to each other?
My answer: n(S) = (16 2) n(E) = ( 8 1)
P(married couple chosen) = (8 1) ÷ (16 2) = 8/20 = 1/15
Therefore, the probability is 1/15.
c) If 6 people are chosen, what is the probability that they are 3 married couples?
My answer: n(S) = (16 6)
n (E) = (8 3)
Therefore, P(3 married couples chosen) = (8 3)÷ (16 6) = 56/8008 = 7/1001
Thus, the probability that the 6 people chosen will be 3 married couples is 7/1001.
• Jan 9th 2009, 11:55 PM
Probability
Hello cnmath16
Looks good to me, apart from your typo in (a): should be 8/120 = 1/15; and in (c) the answer cancels to 1/143.
• Jan 10th 2009, 09:44 AM
Soroban
Hello, cnmath16!
And thank you for showing us your work!
As Grandad pointed out: .$\displaystyle \frac{7}{1001} \:=\:\frac{1}{143}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
A bit of trivia to keep in mind: .$\displaystyle 1001 \:=\:7\times11\times13$
This is the basis for a mathematical trick.
Decide in advance whether you want your volunteer to get a Lucky or Unlucky number.
Lend him a calculator and stand some distance from him.
Instruct him to select any three-digit number mentally.
Have him enter his number twice, forming a six-digit nunber.
From across the room, you "concentrate" on his number.
Then you announce that he selected a very lucky (or unlucky) number.
You add that his number just happens to be divisible by eleven.
Have him divide by 11.
You "concentrate" on his new number.
If he is to be Lucky, say that his number just happens to be divisible by the unlucky 13.
. . To get rid of it, have him divide by 13.
If he is to be Unlucky, say "Too bad, your number is divisible by the lucky 7.
. . We have to get rid of it." .Have him divide by 7.
"Concentrating", you say that it is divisible by his original three-digit number.
Have him divide by it.
Ask him for his final answer.. . It will be 7 or 13.
And you can say, "See? I told you that it was a Lucky (or Unlucky) number."
I assume you can figure out why this works.
.
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crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00499.warc.gz
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# Derivation of S = ut + 1/2at^2
1. Jul 26, 2008
### vijay_singh
Hi
I tried to derive the distance traveled by a body at contact acceleration from the definition of acceleration (increase in speed every sec), but the ended with a different result. Can you see what I am doing wrong.
u = initial speed
t = time taken
S = {distance in 1st sec} + {distance in 2nd sec } + {distance in 3rd sec) + ............... + {distance in t sec}
S = {u } + {u + a} + {u + 2a } + ................... + {u + (t - 1)a}
S = u * t + {a + 2a + .........+ (t - 1)a }
S = ut + a( 1 + 2 + 3 ........+ (t-1))
S = ut + a * t * (t -1) / 2
Vijay
2. Jul 26, 2008
### rcgldr
The distance moved is based on average velocity during each period, not the final velocity at the end of each time period:
S = {u + (1/2)a } + {u + (3/2)a} + {u + (5/2)a } + ... + {u + ((2t-1)/2)a}
To calculate the sum of the coefficients for a:
Code (Text):
c = ( 1)/2 + ( 3)/2 + ( 5)/2 + ... + (2t-5)/2 + (2t-3)/2 + (2t-1)/2
2c = ( 1)/2 + ( 3)/2 + ( 5)/2 + ... + (2t-5)/2 + (2t-3)/2 + (2t-1)/2
+ (2t-1)/2 + (2t-3)/2 + (2t-5)/2 + ... + ( 5)/2 + ( 3)/2 + ( 1)/2
-------------------------------------------------
(2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2 + (2t )/2
= t^2
c = 1/2 t^2
S = u * t + 1/2 * a * t^2
Last edited: Jul 26, 2008
3. Jul 27, 2008
### Hootenanny
Staff Emeritus
A much more straight forward method would be to solve the second order ODE:
$$\frac{d^2s}{dt^2} = \text{const.}$$
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crawl-data/CC-MAIN-2018-47/segments/1542039741491.47/warc/CC-MAIN-20181113194622-20181113220622-00107.warc.gz
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While experimenting, a group of scientists noticed that a A virus, VP882, was able to intercept and read the chemical messages between the bacteria to determine when was the best time to strike. Cholera bacteria communicate through molecular signals, a phenomenon known as quorum sensing, to check their population number. The signal in question is called DPO. VP 882, a subcategory of bacteria’s natural predator, the bacteriophage, waits for the bacteria to multiply and is able to check for the DPO. Once there is enough bacteria, in the experiment’s case they observed cholera, the virus multiples and consumes the bacteria like an all-you-can-eat buffet. The scientists tested this by introducing DPO to a mixture of the virus and bacteria not producing DPO and found that that the bacteria was in fact being killed.
The great part about VP 882 is it’s shared characteristic with a plasmid, a ring of DNA that floats around the cell. This makes it easier to possibly genetically engineer the virus so that it will consume other types of bacteria. This entails it can be genetically altered to defeat other harmful bacterial infections, such as salmonella.
Current phage therapy is flawed because phages can only target a single type of bacteria, but infections can contain several types of different bacteria. Patients then need a “cocktail” with a variety of phages, which is a difficult due to the amount of needed testing in order to get approved for usage. With the engineering capability of using a single type of bacteria killer and the ability to turn it to kill bacteria, phage therapy might be able to advance leaps and bounds.
As humans’ storage of effective antibiotics depletes, time is ticking to find new ways to fight bacterial infections. Are bacteriophages and bacteria-killing viruses like VP 882, the answers?
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Protein secondary structure
In biochemistry and structural biology, protein secondary structure is the general three-dimensional form of local segments of proteins. Secondary structure can be formally defined by the pattern of hydrogen bonds of the protein (such as alpha helices and beta sheets) that are observed in an atomic-resolution structure. More specifically, the secondary structure is defined by the patterns of hydrogen bonds formed between amine hydrogen and carbonyl oxygen atoms contained in the backbone peptide bonds of the protein. The secondary structure may alternatively defined based on the regular pattern of backbone dihedral angles in a particular region of the Ramachandran plot; thus, a segment of residues with such dihedral angles may be called a helix, regardless of whether it has the correct hydrogen bonds. The secondary structure may be provided by crystallographers in the corresponding PDB file.
Secondary structure does not describe the specific identity of amino acids in the protein which are defined as the primary structure, nor the global atomic positions in three-dimensional space, which are considered to be tertiary structure. Other types of biopolymers such as nucleic acids also possess characteristic secondary structures.
|Geometry attribute||α-helix||310 helix||π-helix|
|Residues per turn||3.6||3.0||4.4|
|Translation per residue||1.5 Å (0.15 nm)||2.0 Å (0.20 nm)||1.1 Å (0.11 nm)|
|Radius of helix||2.3 Å (0.23 nm)||1.9 Å (0.19 nm)||2.8 Å (0.28 nm)|
|Pitch||5.4 Å (0.54 nm)||6.0 Å (0.60 nm)||4.8 Å (0.48 nm)|
The most common secondary structures are alpha helices and beta sheets. Other helices, such as the 310 helix and π helix, are calculated to have energetically favorable hydrogen-bonding patterns but are rarely observed in natural proteins except at the ends of α helices due to unfavorable backbone packing in the center of the helix. Other extended structures such as the polyproline helix and alpha sheet are rare in native state proteins but are often hypothesized as important protein folding intermediates. Tight turns and loose, flexible loops link the more "regular" secondary structure elements. The random coil is not a true secondary structure, but is the class of conformations that indicate an absence of regular secondary structure.
Amino acids vary in their ability to form the various secondary structure elements. Proline and glycine are sometimes known as "helix breakers" because they disrupt the regularity of the α helical backbone conformation; however, both have unusual conformational abilities and are commonly found in turns. Amino acids that prefer to adopt helical conformations in proteins include methionine, alanine, leucine, glutamate and lysine ("MALEK" in amino-acid 1-letter codes); by contrast, the large aromatic residues (tryptophan, tyrosine and phenylalanine) and Cβ-branched amino acids (isoleucine, valine, and threonine) prefer to adopt β-strand conformations. However, these preferences are not strong enough to produce a reliable method of predicting secondary structure from sequence alone.
Low frequency collective vibrations are thought to be sensitive to local rigidity within proteins, revealing beta structures to be generically more rigid than alpha or disordered proteins. Neutron scattering measurements have directly connected the spectral feature at ~1 THz to collective motions of the secondary structure of beta-barrel protein GFP.
Hydrogen bonding patterns in secondary structures may be significantly distorted, which makes an automatic determination of secondary structure difficult. There are several methods for formally defining protein secondary structure (e.g., DEFINE, DSSP, STRIDE, SST).
The Dictionary of Protein Secondary Structure, in short DSSP, is commonly used to describe the protein secondary structure with single letter codes. The secondary structure is assigned based on hydrogen bonding patterns as those initially proposed by Pauling et al. in 1951 (before any protein structure had ever been experimentally determined). There are eight types of secondary structure that DSSP defines:
- G = 3-turn helix (310 helix). Min length 3 residues.
- H = 4-turn helix (α helix). Min length 4 residues.
- I = 5-turn helix (π helix). Min length 5 residues.
- T = hydrogen bonded turn (3, 4 or 5 turn)
- E = extended strand in parallel and/or anti-parallel β-sheet conformation. Min length 2 residues.
- B = residue in isolated β-bridge (single pair β-sheet hydrogen bond formation)
- S = bend (the only non-hydrogen-bond based assignment).
- C = coil (residues which are not in any of the above conformations).
'Coil' is often codified as ' ' (space), C (coil) or '-' (dash). The helices (G,H and I) and sheet conformations are all required to have a reasonable length. This means that 2 adjacent residues in the primary structure must form the same hydrogen bonding pattern. If the helix or sheet hydrogen bonding pattern is too short they are designated as T or B, respectively. Other protein secondary structure assignment categories exist (sharp turns, Omega loops etc.), but they are less frequently used.
Secondary structure is defined by hydrogen bonding, so the exact definition of a hydrogen bond is critical. The standard H-bond definition for secondary structure is that of DSSP, which is a purely electrostatic model. It assigns charges of to the carbonyl carbon and oxygen, respectively, and charges of to the amide nitrogen and hydrogen, respectively. The electrostatic energy is
According to DSSP, an H-bond exists if and only if is less than −0.5 kcal/mol. Although the DSSP formula is a relatively crude approximation of the physical H-bond energy, it is generally accepted as a tool for defining secondary structure.
The rough secondary-structure content of a biopolymer (e.g., "this protein is 40% α-helix and 20% β-sheet.") can be estimated spectroscopically. For proteins, a common method is far-ultraviolet (far-UV, 170–250 nm) circular dichroism. A pronounced double minimum at 208 and 222 nm indicate α-helical structure, whereas a single minimum at 204 nm or 217 nm reflects random-coil or β-sheet structure, respectively. A less common method is infrared spectroscopy, which detects differences in the bond oscillations of amide groups due to hydrogen-bonding. Finally, secondary-structure contents may be estimated accurately using the chemical shifts of an initially unassigned NMR spectrum.
Predicting protein tertiary structure from only its amino acid sequence is a very challenging problem (see protein structure prediction), but using the simpler secondary structure definitions is more tractable.
Early methods of secondary-structure prediction were restricted to predicting the three predominate states: helix, sheet, or random coil. These methods were based on the helix- or sheet-forming propensities of individual amino acids, sometimes coupled with rules for estimating the free energy of forming secondary structure elements. Such methods were typically ~60% accurate in predicting which of the three states (helix/sheet/coil) a residue adopts. The first widely used technique to predict protein secondary structure from the amino acid sequence was the Chou–Fasman method.
A significant increase in accuracy (to nearly ~80%) was made by exploiting multiple sequence alignment; knowing the full distribution of amino acids that occur at a position (and in its vicinity, typically ~7 residues on either side) throughout evolution provides a much better picture of the structural tendencies near that position. For illustration, a given protein might have a glycine at a given position, which by itself might suggest a random coil there. However, multiple sequence alignment might reveal that helix-favoring amino acids occur at that position (and nearby positions) in 95% of homologous proteins spanning nearly a billion years of evolution. Moreover, by examining the average hydrophobicity at that and nearby positions, the same alignment might also suggest a pattern of residue solvent accessibility consistent with an α-helix. Taken together, these factors would suggest that the glycine of the original protein adopts α-helical structure, rather than random coil. Several types of methods are used to combine all the available data to form a 3-state prediction, including neural networks, hidden Markov models and support vector machines. Modern prediction methods also provide a confidence score for their predictions at every position.
Secondary-structure prediction methods are continuously benchmarked, e.g., EVA (benchmark). Based on these tests, the most accurate methods at present are Psipred, SAM, PORTER, PROF, and SABLE. The chief area for improvement appears to be the prediction of β-strands; residues confidently predicted as β-strand are likely to be so, but the methods are apt to overlook some β-strand segments (false negatives). There is likely an upper limit of ~90% prediction accuracy overall, due to the idiosyncrasies of the standard method (DSSP) for assigning secondary-structure classes (helix/strand/coil) to PDB structures, against which the predictions are benchmarked.
Accurate secondary-structure prediction is a key element in the prediction of tertiary structure, in all but the simplest (homology modeling) cases. For example, a confidently predicted pattern of six secondary structure elements βαββαβ is the signature of a ferredoxin fold.
Both protein and nucleic acid secondary structures can be used to aid in multiple sequence alignment. These alignments can be made more accurate by the inclusion of secondary structure information in addition to simple sequence information. This is sometimes less useful in RNA because base pairing is much more highly conserved than sequence. Distant relationships between proteins whose primary structures are unalignable can sometimes be found by secondary structure.
- Linderstrøm-Lang KU (1952). Lane Medical Lectures: Proteins and Enzymes. Stanford University Press. p. 115. ASIN B0007J31SC.
- Schellman JA, Schellman CG (1997). "Kaj Ulrik Linderstrøm-Lang (1896-1959)". Protein Sci. 6 (5): 1092–100. doi:10.1002/pro.5560060516. PMC 2143695. PMID 9144781.
He had already introduced the concepts of the primary, secondary, and tertiary structure of proteins in the third Lane Lecture (Linderstram-Lang, 1952)
- Steven Bottomley (2004). "Interactive Protein Structure Tutorial". Retrieved January 9, 2011.
- Perticaroli S, Nickels JD, Ehlers G, O'Neill H, Zhang Q, Sokolov AP (2013). "Secondary structure and rigidity in model proteins". Soft Matter 9 (40): 9548. doi:10.1039/C3SM50807B.
- Perticaroli S, Nickels JD, Ehlers G, Sokolov AP (2014). "Rigidity, secondary structure, and the universality of the boson peak in proteins". Biophys. J. 106 (12): 2667–74. doi:10.1016/j.bpj.2014.05.009. PMID 24940784.
- Nickels JD, Perticaroli S, O'Neill H, Zhang Q, Ehlers G, Sokolov AP (2013). "Coherent neutron scattering and collective dynamics in the protein, GFP". Biophys. J. 105 (9): 2182–7. doi:10.1016/j.bpj.2013.09.029. PMC 3824694. PMID 24209864.
- Richards FM, Kundrot CE (1988). "Identification of structural motifs from protein coordinate data: secondary structure and first-level supersecondary structure". Proteins 3 (2): 71–84. doi:10.1002/prot.340030202. PMID 3399495.
- Kabsch W, Sander C (Dec 1983). "Dictionary of protein secondary structure: pattern recognition of hydrogen-bonded and geometrical features". Biopolymers 22 (12): 2577–2637. doi:10.1002/bip.360221211. PMID 6667333.
- Frishman D, Argos P (Dec 1995). "Knowledge-based protein secondary structure assignment". Proteins 23 (4): 566–579. doi:10.1002/prot.340230412. PMID 8749853.
- Konagurthu AS, Lesk AM, Allison L (Jun 2012). "Minimum message length inference of secondary structure from protein coordinate data". Bioinformatics (Oxford, England) 28 (12): i97–i105. doi:10.1093/bioinformatics/bts223. PMC 3371855. PMID 22689785.
- Pelton JT, McLean LR (2000). "Spectroscopic methods for analysis of protein secondary structure". Anal. Biochem. 277 (2): 167–76. doi:10.1006/abio.1999.4320. PMID 10625503.
- Meiler J, Baker D (2003). "Rapid protein fold determination using unassigned NMR data". Proc. Natl. Acad. Sci. U.S.A. 100 (26): 15404–9. doi:10.1073/pnas.2434121100. PMC 307580. PMID 14668443.
- Chou PY, Fasman GD (Jan 1974). "Prediction of protein conformation". Biochemistry 13 (2): 222–245. doi:10.1021/bi00699a002. PMID 4358940.
- Chou PY, Fasman GD (1978). "Empirical predictions of protein conformation". Annual Review of Biochemistry 47: 251–276. doi:10.1146/annurev.bi.47.070178.001343. PMID 354496.
- Chou PY, Fasman GD (1978). "Prediction of the secondary structure of proteins from their amino acid sequence". Advances in Enzymology and Related Areas of Molecular Biology 47: 45–148. PMID 364941.
- Simossis VA, Heringa J (Aug 2004). "Integrating protein secondary structure prediction and multiple sequence alignment". Current Protein & Peptide Science 5 (4): 249–66. doi:10.2174/1389203043379675. PMID 15320732.
- Pirovano W, Heringa J (2010). "Protein secondary structure prediction". Methods Mol. Biol. 609: 327–48. doi:10.1007/978-1-60327-241-4_19. PMID 20221928.
- Karplus K (2009). "SAM-T08, HMM-based protein structure prediction". Nucleic Acids Res. 37 (Web Server issue): W492–7. doi:10.1093/nar/gkp403. PMC 2703928. PMID 19483096.
- Pollastri G, McLysaght A (2005). "Porter: a new, accurate server for protein secondary structure prediction". Bioinformatics 21 (8): 1719–20. doi:10.1093/bioinformatics/bti203. PMID 15585524.
- Yachdav G, Kloppmann E, Kajan L, Hecht M, Goldberg T, Hamp T et al. (2014). "PredictProtein--an open resource for online prediction of protein structural and functional features". Nucleic Acids Res. 42 (Web Server issue): W337–43. doi:10.1093/nar/gku366. PMC 4086098. PMID 24799431.
- Adamczak R, Porollo A, Meller J (2005). "Combining prediction of secondary structure and solvent accessibility in proteins". Proteins 59 (3): 467–75. doi:10.1002/prot.20441. PMID 15768403.
- Kihara D (Aug 2005). "The effect of long-range interactions on the secondary structure formation of proteins". Protein Science : A Publication of the Protein Society 14 (8): 1955–1963. doi:10.1110/ps.051479505. PMC 2279307. PMID 15987894.
- Qi Y, Grishin NV (2005). "Structural classification of thioredoxin-like fold proteins". Proteins 58 (2): 376–88. doi:10.1002/prot.20329. PMID 15558583.
Since the fold definition should include only the core secondary structural elements that are present in the majority of homologs, we define the thioredoxin-like fold as a two-layer / sandwich with the βαββαβ secondary-structure pattern.
- Branden C, Author J (1999). Introduction to protein structure (2nd ed.). New York: Garland Science. ISBN 978-0815323051.
- Pauling L, Corey RB (1951). "Configurations of Polypeptide Chains With Favored Orientations Around Single Bonds: Two New Pleated Sheets". Proc. Natl. Acad. Sci. U.S.A. 37 (11): 729–40. PMC 1063460. PMID 16578412. (The original beta-sheet conformation article.)
- Pauling L, Corey RB, Branson HR (1951). "The structure of proteins; two hydrogen-bonded helical configurations of the polypeptide chain". Proc. Natl. Acad. Sci. U.S.A. 37 (4): 205–11. PMC 1063337. PMID 14816373. (alpha- and pi-helix conformations, since they predicted that helices would not be possible.)
- NetSurfP – Secondary Structure and Surface Accessibility predictor
- PSSpred A multiple neural network training program for protein secondary structure prediction
- Genesilico metaserver Metaserver which allows to run over 20 different secondary structure predictors by one click
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WHAT IS OSTEOPOROSIS?
The currently accepted definition of osteoporosis is “systemic skeletal disease characterized by low bone mass and micro architectural deterioration of bone tissue, with a consequent increase in bone fragility and susceptibility to fracture risk”.
Bone mineral density can be measured with reasonable accuracy. These tests form the basis for diagnosis of osteoporosis and the prediction of fracture risk. Bone mass refers to the amount of bone tissue contained in the skeleton. Bone mass can be expressed in terms of bone mineral content (the totalgrams of bone mineral within a given area of bone) or in terms of bone mineral density (the bone mineral content normalized for the projected
Normally bone density peaks between the ages of 30 to 40 and in subsequent years the bone density decreases.
If the decrease is significant enough, the so called “fracture threshold” is reached. At this level of bone density the patient is at significant risk of fracture. These thresholds are reached at different ages and the extend of bone loss
varies depending on the peak bone mass and generic and environmental factors, including activity level and diet.
The decrease of bone density in post-menopausal women is significantly greater than in pre-menopausal women or men. The average annual bone loss in post-menopausal women is 1% – 2% and in men 0.2% – 0.5%. Although the rate of bone loss in women is the highest in the years after the menopause, it continues in many patients for many years and
increases again after the age of 70. Annual losses can even reach 3% -5% during the first years following the menopause. In women the major cause of bone loss and osteoporosis is estrogen withdrawal, most commonly associated with the menopause and declining ovarian function, but any cause of estrogen deficiency can cause bone loss.
“Electromagnetic field treatment for Osteoporosis”. © Ben Philipson Curatronic Ltd.
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NCERT Solutions: Exercise Miscellaneous- Straight Lines
# Exercise Miscellaneous- Straight Lines NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download
Question 1: Find the values of k for which the line is
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.
ANSWER : - The given equation of line is
(k – 3) x – (4 – k2yk2 – 7k + 6 = 0 … (1)
(a) If the given line is parallel to the x-axis, then
Slope of the given line = Slope of the x-axis
The given line can be written as
(4 – k2y = (k – 3) xk2 – 7k 6 = 0
,
which is of the form ymx + c.
∴Slope of the given line =
Slope of the x-axis = 0
Thus, if the given line is parallel to the x-axis, then the value of k is 3.
(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is .
Now, is undefined at k2 = 4
k2 = 4
⇒ k = ±2
Thus, if the given line is parallel to the y-axis, then the value of k is ±2.
(c) If the given line is passing through the origin, then point (0, 0) satisfies the
given equation of line.
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.
Question 2: Find the values of θ and p, if the equation is the normal form of the line .
ANSWER : - The equation of the given line is .
This equation can be reduced as
On dividing both sides by , we obtain
On comparing equation (1) to , we obtain
Since the values of sin θ and cos θ are negative,
Thus, the respective values of θ and p are and 1
Question 3: Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.
ANSWER : - Let the intercepts cut by the given lines on the axes be a and b.
It is given that
ab = 1 … (1)
ab = –6 … (2)
On solving equations (1) and (2), we obtain
a = 3 and b = –2 or a = –2 and b = 3
It is known that the equation of the line whose intercepts on the axes are a and b is
Case I: a = 3 and b = –2
In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6.
Case II: a = –2 and b = 3
In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6.
Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.
Question 4: What are the points on the y-axis whose distance from the line is 4 units.
ANSWER : - Let (0, b) be the point on the y-axis whose distance from line is 4 units.
The given line can be written as 4x + 3y – 12 = 0 … (1)
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = –12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1y1) is given by .
Therefore, if (0, b) is the point on the y-axis whose distance from line is 4 units, then:
Thus, the required points are and .
Question 5: Find the perpendicular distance from the origin to the line joining the points
ANSWER : - The equation of the line joining the points is given by
It is known that the perpendicular distance (d) of a line AxByC = 0 from a point (x1y1) is given by .
Therefore, the perpendicular distance (d) of the given line from point (x1y1) = (0, 0) is
Question 6: Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.
ANSWER : - The equation of any line parallel to the y-axis is of the form
xa … (1)
The two given lines are
x – 7y + 5 = 0 … (2)
3xy = 0 … (3)
On solving equations (2) and (3), we obtain .
Therefore, is the point of intersection of lines (2) and (3).
Since line xa passes through point , .
Thus, the required equation of the line is .
Question 7: Find the equation of a line drawn perpendicular to the line through the point, where it meets the y-axis.
ANSWER : - The equation of the given line is .
This equation can also be written as 3x + 2y – 12 = 0
, which is of the form ymxc
∴Slope of the given line
∴Slope of line perpendicular to the given line
Let the given line intersect the y-axis at (0, y).
On substituting x with 0 in the equation of the given line, we obtain
∴The given line intersects the y-axis at (0, 6).
The equation of the line that has a slope of and passes through point (0, 6) is
Thus, the required equation of the line is .
Question 8: Find the area of the triangle formed by the lines yx = 0, x + y = 0 and xk = 0.
ANSWER : - The equations of the given lines are
y – x = 0 … (1)
xy = 0 … (2)
x – k = 0 … (3)
The point of intersection of lines (1) and (2) is given by
x = 0 and y = 0
The point of intersection of lines (2) and (3) is given by
xk and y = –k
The point of intersection of lines (3) and (1) is given by
xk and yk
Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (kk).
We know that the area of a triangle whose vertices are (x1y1), (x2y2), and (x3y3) is .
Therefore, area of the triangle formed by the three given lines
Question 9: Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2xy – 3 = 0 may intersect at one point.
ANSWER : - The equations of the given lines are
3x + y – 2 = 0 … (1)
px + 2y – 3 = 0 … (2)
2xy – 3 = 0 … (3)
On solving equations (1) and (3), we obtain
x = 1 and y = –1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).
p (1) + 2 (–1) – 3 = 0
p – 2 – 3 = 0
p = 5
Thus, the required value of p is 5.
Question 10: If three lines whose equations are concurrent, then show that
ANSWER : - The equations of the given lines are
y = m1x + c1 … (1)
y = m2x + c2 … (2)
y = m3x + c3 … (3)
On subtracting equation (1) from (2), we obtain
On substituting this value of x in (1), we obtain
is the point of intersection of lines (1) and (2).
It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).
Hence,
Question 11: Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.
ANSWER : - Let the slope of the required line be m1.
The given line can be written as , which is of the form y = mx + c
∴Slope of the given line =
It is given that the angle between the required line and line x – 2y = 3 is 45°.
We know that if θ is the acute angle between lines l1 and l2 with slopes m1 and m2 respectively, then .
Case I: m1 = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = 3 (x – 3)
y – 2 = 3x – 9
3xy = 7
Case II: m1 =
The equation of the line passing through (3, 2) and having a slope of is:
Thus, the equations of the lines are 3xy = 7 and x + 3y = 9.
Question 12: Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
ANSWER : - Let the equation of the line having equal intercepts on the axes be
On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain .
is the point of intersection of the two given lines.
Since equation (1) passes through point ,
∴ Equation (1) becomes
Thus, the required equation of the line is .
Question 13: Show that the equation of the line passing through the origin and making an angle θ with the line .
ANSWER : - Let the equation of the line passing through the origin be y = m1x.
If this line makes an angle of θ with line y = mx + c, then angle θ is given by
Case I:
Case II:
Therefore, the required line is given by .
Question 14: In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?
ANSWER : - The equation of the line joining the points (–1, 1) and (5, 7) is given by
The equation of the given line is
x + y – 4 = 0 … (2)
The point of intersection of lines (1) and (2) is given by
x = 1 and y = 3
Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k. Accordingly, by section formula,
Thus, the line joining the points (–1, 1) and (5, 7) is divided by line
x + y = 4 in the ratio 1:2.
Question 15: Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2xy= 0.
ANSWER : - The given lines are
2xy = 0 … (1)
4x + 7y + 5 = 0 … (2)
A (1, 2) is a point on line (1).
Let B be the point of intersection of lines (1) and (2).
On solving equations (1) and (2), we obtain .
∴Coordinates of point B are .
By using distance formula, the distance between points A and B can be obtained as
Thus, the required distance is .
Question 16: Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x y = 4 may be at a distance of 3 units from this point.
ANSWER : - Let y = mx + c be the line through point (–1, 2).
Accordingly, 2 = m (–1) + c.
⇒ 2 = –m + c
c = m + 2
y = mx + m + 2 … (1)
The given line is
x + y = 4 … (2)
On solving equations (1) and (2), we obtain
is the point of intersection of lines (1) and (2).
Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,
Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.
Question 17: Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.
ANSWER : - The equation of the given line is
x + 3y = 7 … (1)
Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.
Since line (1) is perpendicular to AB,
The mid-point of line segment AB will also satisfy line (1).
Hence, from equation (1), we have
On solving equations (2) and (3), we obtain a = –1 and b = –4.
Thus, the image of the given point with respect to the given line is (–1, –4).
Question 18: If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
ANSWER : - The equations of the given lines are
y = 3x + 1 … (1)
2y = x + 3 … (2)
y = mx + 4 … (3)
Slope of line (1), m1 = 3
Slope of line (2),
Slope of line (3), m3 = m
It is given that lines (1) and (2) are equally inclined to line (3). This means that
the angle between lines (1) and (3) equals the angle between lines (2) and (3).
Thus, the required value of m is .
Question19: If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.
ANSWER : - The equations of the given lines are
x + y – 5 = 0 … (1)
3x – 2y + 7 = 0 … (2)
The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by
It is given that .
, which is the equation of a line.
Similarly, we can obtain the equation of line for any signs of .
Thus, point P must move on a line.
Question 20: Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
ANSWER : - The equations of the given lines are
9x + 6y – 7 = 0 … (1)
3x + 2y + 6 = 0 … (2)
Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by
The perpendicular distance of P (h, k) from line (2) is given by
Since P (h, k) is equidistant from lines (1) and (2),
9h + 6k – 7 = – 9h – 6k – 18
⇒ 18h + 12k + 11 = 0
Thus, the required equation of the line is 18x + 12y + 11 = 0.
Question 21: A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
ANSWER : - Let the coordinates of point A be (a, 0).
Draw a line (AL) perpendicular to the x-axis.
We know that angle of incidence is equal to angle of reflection. Hence, let
∠BAL = ∠CAL = Φ
Let ∠CAX = θ
∴∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]
= 180° – θ – 180°+ 2θ
= θ
∴∠BAX = 180° – θ
From equations (1) and (2), we obtain
Thus, the coordinates of point A are .
Question 22: Prove that the product of the lengths of the perpendiculars drawn from the points
ANSWER : - The equation of the given line is
Length of the perpendicular from point to line (1) is
Length of the perpendicular from point to line (2) is
On multiplying equations (2) and (3), we obtain
Hence, proved.
Question 23: A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.
ANSWER : - The equations of the given lines are
2x – 3y + 4 = 0 … (1)
3x + 4y – 5 = 0 … (2)
6x – 7y + 8 = 0 … (3)
The person is standing at the junction of the paths represented by lines (1) and (2).
On solving equations (1) and (2), we obtain .
Thus, the person is standing at point .
The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point .
∴Slope of the line perpendicular to line (3)
The equation of the line passing through and having a slope of is given by
Hence, the path that the person should follow is .
The document Exercise Miscellaneous- Straight Lines NCERT Solutions | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
All you need of Commerce at this link: Commerce
## Mathematics (Maths) Class 11
75 videos|238 docs|91 tests
## FAQs on Exercise Miscellaneous- Straight Lines NCERT Solutions - Mathematics (Maths) Class 11 - Commerce
1. How do I solve miscellaneous exercises on straight lines for the JEE exam?
Ans. To solve miscellaneous exercises on straight lines for the JEE exam, you can follow these steps: 1. Read the given problem statement carefully and identify the information provided. 2. Use the equation of a straight line, y = mx + c, to form an equation based on the given conditions. 3. Simplify the equation and bring it to the standard form, Ax + By + C = 0, if required. 4. Solve the equation by applying the appropriate concepts of straight lines, such as finding the slope, distance between points, or intersection of lines. 5. Verify your solution and check if it satisfies all the given conditions in the problem statement. 6. If required, draw a figure or graph to visualize the problem and better understand the solution.
2. What are the important concepts and formulas to remember while solving straight lines for the JEE exam?
Ans. Some important concepts and formulas to remember while solving straight lines for the JEE exam are: 1. Slope of a line: The slope of a line passing through two points (x1, y1) and (x2, y2) is given by m = (y2 - y1) / (x2 - x1). 2. Point-slope form: The equation of a line passing through a point (x1, y1) with slope m is given by y - y1 = m(x - x1). 3. Slope-intercept form: The equation of a line with slope m and y-intercept c is given by y = mx + c. 4. Distance between two points: The distance between two points (x1, y1) and (x2, y2) is given by d = sqrt((x2 - x1)^2 + (y2 - y1)^2). 5. Parallel and perpendicular lines: Two lines are parallel if their slopes are equal, and they are perpendicular if the product of their slopes is -1. 6. Intersection of lines: The intersection point of two lines can be found by solving their equations simultaneously.
3. Are there any shortcut techniques or tricks to solve straight line problems quickly for the JEE exam?
Ans. While there are no specific shortcut techniques or tricks to solve straight line problems quickly, you can follow some strategies to improve your speed and accuracy: 1. Practice regularly to enhance your problem-solving skills and speed. 2. Understand and memorize the important concepts and formulas related to straight lines. 3. Break down complex problems into smaller, manageable parts to solve them step by step. 4. Visualize the problem by drawing figures or graphs to gain better insights into the solution. 5. Use logical reasoning and common sense to eliminate incorrect options in multiple-choice questions. 6. Solve previous year JEE question papers and mock tests to familiarize yourself with the exam pattern and time management.
4. Can I score well in the JEE exam by focusing only on the miscellaneous exercises of straight lines?
Ans. While solving miscellaneous exercises on straight lines is essential for a comprehensive understanding of the topic, it is not sufficient to score well in the JEE exam. The JEE syllabus covers various other subjects and topics, and it is important to have a balanced preparation strategy. You should allocate sufficient time to all subjects, including physics, chemistry, and mathematics, and practice a wide range of problem-solving techniques. Additionally, it is crucial to solve previous year question papers, take mock tests, and seek guidance from experienced teachers or mentors to enhance your overall performance in the JEE exam.
5. Is it necessary to draw graphs while solving miscellaneous exercises on straight lines for the JEE exam?
Ans. Drawing graphs can be a helpful visual aid while solving miscellaneous exercises on straight lines for the JEE exam, especially when dealing with geometric interpretations or understanding the behavior of lines. However, it is not always necessary to draw graphs. If the problem statement provides sufficient information and you can visualize the scenario mentally, you can proceed with solving the equations and finding the solution algebraically. Drawing graphs can be a personal preference or a technique to enhance understanding, but it is not a mandatory step in every problem-solving process.
## Mathematics (Maths) Class 11
75 videos|238 docs|91 tests
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Chimpanzees and Their Toolsets
Unlike chimpanzees in West and East Africa, who use a single tool to extract termites, chimpanzees in Congo Basin, Central America, use toolsets: perforating twigs, puncturing sticks, and fishing probes to harvest insects from towering earthen mounds scattered across lowland forests and underground nests.
It turns out that chimpanzees living in this region have the most sophisticated arsenal of tool-using skills documented in the animal kingdom. Aside from their specialized toolsets for harvesting termites, honey, and termites, they also customize the implements with different modifications in order to improve their efficiency.
Trying to untangle how chimpanzees in this area acquire these complex tool tasks, Stephanie Musgrave who is a biological anthropologist at the University of Miami screened thousands of hours of video footage that record visits to termite nests, including those by leopards, gorillas, forest elephants, in the Republic of Congo’s Goualougo Triangle. Her reward was identifying more than 660 hours of periodic visits by 25 young chimpanzees belonging to a notoriously elusive subspecies of chimpanzees. The recorded 15 years of footage captured the development of their tool-using skills from birth until maturity.
Key Features In The Human Culture
In the study, chimpanzees learn to use and make their unique termite-extracting toolsets, Musgrave and her colleagues with the Goualougo Triangle Ape Project provide novel insights into how cultures of chimpanzee persist over generations and maybe how technology came to be a defining aspect of the evolution of humanity.
Musgrave shares that one of the key features of human culture is its remarkable complexity. It is what we call cumulative. However, the expansion and continuation of such research depend on the long-term preservation of wild chimpanzees and their cultures. Both are increasingly endangered by human activities.#
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DISCOVER
# How to calculate braking distance
Updated February 21, 2017
Drivers must pay attention to their surroundings. This includes calculating the braking distance from the cars and objects around a driver. Knowing your car's the braking distance to avoid accidents can save lives and money.
Determine the speed the car is traveling--the higher the speed, the greater the braking distance. For this example, a speed of 70 miles per hour will be used.
Establish the stopping distance. This is the distance from your car to the object in front of it. To estimate stopping distance when driving, remember that the average car length is 15 feet. So four car lengths is roughly equal to 60 feet. When driving 70 mph, the stopping distance is 102.7 feet per second (fps = 1.467 x mph).
Institute a deceleration rate. This is the rate that applying the brakes slows the vehicle. This rate is typically 20 feet per second.
Establish the stopping time. When traveling at 70 mph, which equals 102.7 feet per second if the deceleration rate is 20 fps, the stopping time equals 102.7/20 = 5.135 seconds.
Ascertain the thinking distance, which is reaction time as it relates to distance. Reaction time can take an average of 2 seconds to set in and realize that there is a problem. The reaction time can vary given the age of the driver, condition of the car, distraction in the vehicle and condition of the road. The formula to make a precise calculation 102.7 fps x 2 (reaction time in seconds) = 205.4.
Calculate the total braking distance. This formula is 1/2 the initial velocity in feet per second multiplied by the time required to stop, which is 0.5 x 102.7 x 5.135 = 263.68. The calculated thinking distance is 2 x 102.7 = 205.4. Add the two numbers together. 469.08 feet is the total braking distance.
#### Tip
To estimate stopping time, pick a fixed object (a bridge, building or tree) and start counting the seconds when the car in front passes the fixed object. Stop counting when the driven car passes the fixed object.
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