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Research: 3D-Printing for Fundamental Science
Researchers at Uppsala University use additive manufacturing for innovative lab solutions.
Advancements in fundamental science originate in creative ideas and insightful postulates while they rest on reliable experimental verification. For the latter – as the word suggests – experimentation and trial in the lab is key. Often setting up challenging experiments to validate novel ideas pushes the boundaries of the state of the art leading to technological progress. Sometimes however, the interplay between basic science and applied technology can work in the opposite way, when new techniques become readily available at a relatively low cost on the consumer market, which lend themselves to be utilised in the lab.
Recently, Additive Manufacturing – commonly known as 3D-printing – became such a widespread technology empowering users at home, as well as in the lab, to fabricate plastic objects limited only by their drawing skills and imagination.
“Every experimental physics department should consider the use of 3D-printers. This emergent technology gives researchers easy access to rapid prototyping and very functional individual lab solutions.” says Merlin Pohlit, postdoctoral researcher in the Department of Physics and Astronomy at Uppsala University.
3D-printing allowed Merlin Pohlit to design and print prototypes of several parts, needed for a custom build sample stage of a cryostat, which will be used for electronic transport measurements at low temperatures (see Fig. 1). In such a system, the layout of the electric wiring has to be realised in a very limited space, while considering the heat conductance so that the sample can be cooled to temperatures below 10 Kelvin (approx. -263 oC). The ability to quickly build relatively cheap prototypes proved to be extremely helpful to get the design right before issuing an expensive fabrication in metal.
However, 3D-printing is not limited to prototyping parts before conventional machining as today’s printers are sufficiently accurate that printed parts can be used directly for experiments. Considering a slight shrinkage of the printed material, it is possible to print holes and working screw threads with the required dimensions. This allows for manufacturing functional components like the printed sample holder for resistivity measurements, with press fitted gold contacts, as shown in Fig. 2.
"I foresee 3D-printing widely used in academic education and research labs around the world, where single tailored parts are needed on a regular basis. However, for their scientific use often specific properties, like the vacuum compatibility of a material, are required. Therefore, I expect that 3D-printed parts will be used more and more frequently in experiments, as soon as more of these properties become better known. Yet, there is another factor despite being very useful: It is plain fun to arrive in the lab in the morning and see your sketched ideas became a printed reality, just while you were sleeping." says Merlin Pohlit.
Uppsala University was among the first research facilities incorporating 3D-printing into the lab environment and so far, it was a very fruitful endeavour. While a project successfully developed printable materials for application in neutron science, the scientific challenges resulted in the development of improved printer heads and more durable printer nozzles fabricated from ruby, which are now commercially available. Things have come full circle by improving the tool for the next generation of experiments to come.
For a thorough discussion of novel concepts regarding 3D printing using plastic materials and its scientific use, the interested reader is referred to the article by Anders et al., which is available as an Open Access article.
A. Olsson and A. R. Rennie, Boron carbide composite apertures for small-angle neutron scattering made by three-dimensional printing, Journal of Applied Crystallography 49, 696-699, 2016.
A. Olsson and M. S. Hellsing and A. R. Rennie, New possibilities using additive manufacturing with materials that are difficult to process and with complex structures, Physica Scripta 92, 053002, 2017. doi: 10.1088/1402-4896/aa694e<|endoftext|>
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Stonehenge’s layout holds clues.
Above the trilithons the pattern below, of four equally spaced parallel lines, is where four roof trusses were located, their weight resting on the capping stones. The trusses spanned a central void, a large hall used for gatherings.
The trilithons supported rof trusses – that why they exist!
Logic backs this assertion. Stonehenge’s stones tell us it’s builders were planning something big and magnificent. Why else go to such trouble to shape and shift massive stones, and cap them.
The roof trusses were formed at ground level and raised by pivoting. To do this solid stone was necessary – another reason why hard sarsen stone was chosen. Softer stone would crumble.
The diagram above explains why the trilithons are of different height. Raising four trusses becomes complicated if adjoining pairs of trilithons are the same height.
Though big the roof span at Stonehenge was not impossibly large. Westminster Hall is much wider!<|endoftext|>
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# NCERT Solutions Class 2
Students are introduced to new topics in almost every subject in Class 2. It is these topics that form the base of the student at an early age. Schools affiliated to CBSE recommend NCERT books for Class 2 because these have study material as per the guidelines prepared by the CBSE. When referring to CBSE Class 2 textbooks, students, teachers, and even parents look for accurate answers to the textbook questions. This is where NCERT Solutions for Class 2 by Extramarks can be of great help.
## NCERT Solutions for Class 2 English and Class 2 Mathematics
To solve the textbook problems and ensure that all the problems are correctly answered, NCERT class 2 solutions can be of great help. It lets the kids perform in a better way in their tests. All the solutions given are accurate, and kids can understand the concepts much better.
## NCERT Solutions for Class 2 Mathematics
Chapter 1 What is Long, What is Round?
Question: 1 What is Long? What is Round?
Question: 2 Do you also think so?
Answer: Yes, we also think that it is a pencil.
Chapter 2 Counting in Groups
Question 1: Number of pencils in your class
More than 45
Less than 45
Answer: Less than 45.
Question 2: Number of spokes in one cycle wheel
More than 20
Less than 20
Answer: Less than 20.
Chapter 3 How Much Can You Carry?
1. Guess and tell what Chhotu monkey used to help Chikky and Micky. Draw a picture of it in the monkey’s hand.
Answer: Chhotu monkey used a beam balance to help Chikky and Micky.
Chapter 4 Counting in Tens
1. In the morning, she counted her chickens.
(a) How many baskets of 10 chickens are there? ______
(b) How many chickens are there in all?
50 + 4 = _______
Answer: 50 + 4 = 54
Chapter 5 Patterns
1. Look at the pattern and fill up the boxes.
Answer: Fill the boxes with the given patterns.
Chapter 6 Footprints
1. Draw the missing things in the picture.
Answer: Draw tables, books, newspapers and chairs.
Chapter 7 Jugs and Mugs
(a) How many lemons will you need? ______
(b) How many spoons of sugar will you take? ______
Answer: 6 spoons of sugar
Lemon Drink Stall at a Village Fair.
Chapter 8 Tens and Ones
(a) How many students are left in all? __________
Answer:20 students are left in all.
(b) How many more teams can be made with all these students left? ___________
Answer:2 teams can be made with the students left.
Chapter 9 My Funday
1. Which day will come
(a) After Sunday? ______
(b) Before Sunday? ______
Chapter 10 Add Our Points
Q.Who won the game? ___________
Answer: The elephant won the game.
Q.Who lost the game? ___________
Answer: The tortoise lost the game.
The winner got bananas from Bunnoo.
Chapter 11 Lines and Lines
Q.Is there any number or letter that you cannot make with matchsticks?
1. What about writing letters using straight lines? Which ones are easy?
Answer: Some letters are elementary to write with matchsticks. The letters A, E, F, H, I, T, V, K, and W are easy to write with matchsticks.
Chapter 12 Give and Take
1. How many beads are taken by Razia and Sonu?
2. Razia and Sonu take 47 beads.
Chapter 13 The Longest Step
1. Can you tell why the elephant won?
Answer: The elephant took a big step while walking. So the elephant won.
Q.Who takes the most significant step?
A.The elephant takes the most significant step.
Chapter 14 Birds Come, Birds Go
1. Bunnu rabbits can eat 29 carrots in one week. Munnu rabbits can eat 42 carrots in one week. Who eats more in a week, and by how much?
___________ eats __________ more carrots.
A.Munnu eats 13 more carrots.
Q.Neha is 29 years old. Her mother is 58 years old. How many years older is Neha’s mother?
Mother is __________ years older than Neha.
A.Mother is 29 years older than Neha.
Chapter 15 How Many Ponytails?
Q.Find out what the different fruits are.
A.The different fruits are bananas, apples, oranges and mangoes.
## Class 2 English NCERT Solutions
Chapter 1 First Day at School
Q 1. Who was the first friend you made?
1. Do it yourself.
Q 2. What did you enjoy doing the most?
A.I enjoyed playing with my pet dog, Shanky.
Q 3. Do you have a pet at home who waits for you to return from school?
1. Yes, I have a pet dog at my home who waits for me to come back from school.
Chapter 2 Haldis Adventure
Summary In English
One day a girl named Haldi met a giraffe named Smiley on her way to school. The giraffe told her that he would feel happy whenever she stared at him. He asked her many questions about her school. Then Smiley dropped Haldi to her school. She saw beautiful scenes on her way to school while sitting on the back of the giraffe. On reaching the school, she came down from his back, and when she turned back to thank him, the giraffe had already gone away.
Haldi was surprised and happy
too. Then she remembered that she
would be late for school. So she said
to the giraffe, “I would love to talk
to you, but I must rush to school or I
will be late.”
Word- Meanings: Surprised ( सप्रेडि्ज्ड ) – struck with wonder, 3119 ad Arial Remembered ( रिमेंबरर्ड )–came in mind again, याद आया। Late ( लेट ) – not on time, विलंब, देर। Rush ( रस ) – to run fast, भागना।
Chapter 3- The Paddling-Pool
Question. Fill in the blanks with ‘before or “after”:
Answer: Before eating food, I wash my hands.
After reaching school, I sit in a class.
After eating food, I wash my mouth.
After reaching home, I do my homework.
Chapter 4 – I am Lucky
Q.If I were an octopus
I would be thankful
For my eight arms.
So I just think I am lucky to be “me”.
Not “you”, but “me”.
Word-Meanings: Octopus ( ऑक्टोपस )-a strange sea creature having eight arms, अष्टभुज।
Chapter 5-I Want
This is the story of a monkey who wants to become strong. One wise woman came to know about this. She gave one magic stick to the monkey to fulfil all his wishes. First of all, he wants to become a giraffe. Then he wants to become an elephant, and at last, he wants to become a zebra.
All his wishes are granted. At last, when he comes along the river, he finds himself looking so scary. He felt so bad after seeing his face. He told his mother that he was happy being himself. Then he wishes to become a monkey again. His wish was granted, and he became a monkey again. He threw away his magic stick.
Word-Meanings: Strong ( स्ट्रांग )-mighty, ताकतवर Wise ( वाइज़ )-Learned, बुद्धिमती। Magic wand (मैजिक वांड) -magic stick, जादू की छड़ी। wish (विश) -want, इच्छा।
The little monkey is happy.
An elephant comes down to the river.
Word-Meanings: Happy ( हैप्पी ) -in a joyful mood, खुश।
Chapter 6 A Smile
Question. Play the game with your partner. Close your eyes. Move your finger on the page while you sing.
Tic Tac Toe, Round I go, if I miss, I’ll take this. When you finish the poem, stop moving your finger. Open your eyes, see which letter you stop at. Use the clues to write the word beginning with the letter you find.
Chapter 7 The Wind and the Sun
Find a rhyming word from the story for each of these words:
bun hold boat
fan sunny pot
Answer. sun, cold, coat, man, funny, hot.
Question. Find three more ‘doing words in the story that ends in ‘ing’ and then write them here.
Answer. Putting wiping shining.
Chapter 8 Rain
Question. Where does the rain fall?
Answer. The rain falls everywhere. It falls on trees, fields, umbrellas and ships.
Chapter 9 Storm in the Garden
Question. Who was Sunu-sunu?
Ans. Sunu-sunu was a small snail.
Chapter 10 Zoo Manners
Make sentences using ‘this’ and ‘that’.
Answer: This is a pen.
That is my bicycle.
Chapter 11 Story Funny Bunny
Question 1: One day, _____________ (a nut/the sky) fell on Funny Bunny.
Answer: One day, a nut fell on Funny Bunny.
Question 2: Funny Bunny wanted to tell _______________ (the king/the cock/the sky) what he saw.
Answer: Funny Bunny wanted to tell the king what he saw.
Chapter 12 Poem Mr Nobody
Question 1:Who is Mr Nobody?
Answer: Mr Nobody is a funny little man who makes mischief in everybody’s house.
Chapter 13 Story Curlylocks and the Three Bears
Question 1:My sister’s hair is soft and curly.
Chapter 14 Poem On My Blackboard I Can Draw
Question 1:How many windows does the house have in this poem?
Answer: There are five little windows in the house.
Chapter 15 Story Make it Shorter
Question 1:What did Akbar order one day?
Answer: One day, Akbar drew a line on the floor and ordered his ministers to shorten the line without rubbing out any part of it.
Chapter 16 Poem I am the Music Man
Question 1:What are the two instruments that the music man can play?
Answer: The music man can play two instruments: piano and big drum.
Chapter 17 Story The Mumbai Musicians
Question 1:Why did the farmer tell Goopu to see the world?
Answer: The farmer told Goopu to see the world before he becomes too old to carry any more load on his back.
Chapter 18 Poem Granny Granny Please Comb My Hair
Question 1:What does the little girl want her Granny to do?
Answer: The little girl requests her Granny to comb her hair.
Chapter 19 Story The Magic Porridge Pot
Question 1:Where did Tara go one day?
Answer: One day, Tara went to a forest.
Chapter 20 Poem Strange Talk
Question 1:Are these sentences true or false?
A little green frog said, “Quack-quack.”
A little pig loved to make a row.
A duck only said, “Croak-croak.”
A pig cried aloud, “Wee-wee.”
Answer: A little green frog said, “Quack-quack.” False
A little pig loved to make a row. False
A duck only said, “Croak-croak.” False
A pig cried aloud, “Wee-wee.” True
Chapter 21 Story The Grasshopper and The Ant
Question. What would you hold if you were asked to store things for the winter? Name any three items.
Answer: I would like to store warm clothes like sweaters, jackets and socks for the winter season.
## How to help your child using NCERT Solutions for Class 2
You need to let your kid using the NCERT Solutions for class 2 understand that the knowledge they get is for learning and exploring different things and not merely to score grades.
Parents should correct the fundamental attributes error and motivate the kids at every step instead of being rigid with them. A reasonable timetable should be prepared so that the kids can work on their NCERT Solutions for every subject to learn without any confusion.
Important Tips For Students Related To NCERT Solutions for Class 2
• Students should start planning on how they intend to study the subject.
• Kids need to take interval and mid breaks.
The NCERT solutions can be an excellent reference for the kids to understand how and why the particular answer was solved in that way.
## FAQs (Frequently Asked Questions)
### 1. NCERT solutions are available for which subjects in class 2?
You can get the class 2 NCERT solution for English, Hindi and Mathematics.
### 2. What are the NCERT solutions for class 2?
This can be a significant learning step for the kids as they can get well versed with different chapters present in the class 2 subjects.<|endoftext|>
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# First Derivative Calculator(Solver) with Steps
Free derivatives calculator(solver) that gets the detailed solution of the first derivative of a function.
Function
1. Let $$u = - 3 x \sin^{2}{\left (x \right )} + 1$$.
2. The derivative of $$e^{u}$$ is itself.
3. Then, apply the chain rule. Multiply by $$\frac{d}{d x}\left(- 3 x \sin^{2}{\left (x \right )} + 1\right)$$:
1. Differentiate $$- 3 x \sin^{2}{\left (x \right )} + 1$$ term by term:
1. The derivative of the constant $$1$$ is zero.
2. The derivative of a constant times a function is the constant times the derivative of the function.
1. Apply the product rule:
$$f{\left (x \right )} = x$$; to find $$\frac{d}{d x} f{\left (x \right )}$$:
1. Apply the power rule: $$x$$ goes to $$1$$
$$g{\left (x \right )} = \sin^{2}{\left (x \right )}$$; to find $$\frac{d}{d x} g{\left (x \right )}$$:
1. Let $$u = \sin{\left (x \right )}$$.
2. Apply the power rule: $$u^{2}$$ goes to $$2 u$$
3. Then, apply the chain rule. Multiply by $$\frac{d}{d x} \sin{\left (x \right )}$$:
1. The derivative of sine is cosine:
The result of the chain rule is:
The result is: $$2 x \sin{\left (x \right )} \cos{\left (x \right )} + \sin^{2}{\left (x \right )}$$
So, the result is: $$- 6 x \sin{\left (x \right )} \cos{\left (x \right )} - 3 \sin^{2}{\left (x \right )}$$
The result is: $$- 6 x \sin{\left (x \right )} \cos{\left (x \right )} - 3 \sin^{2}{\left (x \right )}$$
The result of the chain rule is:
4. Now simplify:
Commands:
* is multiplication
oo is $\infty$
pi is $\pi$
x^2 is x2
sqrt(x) is $\sqrt{x}$
sqrt[3](x) is $\sqrt[3]{x}$
(a+b)/(c+d) is $\frac{a+b}{c+d}$
### The Most Important Derivatives - Basic Formulas/Rules
$\frac{d}{dx}a=0$ (a is a constant)
$\frac{d}{dx}x=1$
$\frac{d}{dx}x^n=nx^{n-1}$
$\frac{d}{dx}e^x=e^x$
$\frac{d}{dx}\log x=\frac1x$
$\frac{d}{dx}a^x=a^x\log x$
$(f\ g)' = f'g + fg'$ - Product Rule
$(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ - Quotient Rule
$\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$ - Chain Rule
$\frac{d}{dx}\sin(x)=\cos(x)$
$\frac{d}{dx}\cos(x)=-\sin(x)$
$\frac{d}{dx}\tan(x)=\sec^2(x)$
$\frac{d}{dx}\cot(x)=-csc^2(x)$
$\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\arccos(x)=-\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}$
$\frac{d}{dx}\text{arccot}(x)=-\frac{1}{1+x^2}$
$\frac{d}{dx}\text{arcsec}(x)=\frac{1}{x\sqrt{x^2-1}}$
$\frac{d}{dx}\text{arccsc}(x)=-\frac{1}{x\sqrt{x^2-1}}$
Contact email:<|endoftext|>
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by Lawrence and Lee | Grades
If the law is of such a nature that it requires you to be an agent of injustice to another, then I say, break the law." So wrote the young Henry David Thoreau in 1849. Three years earlier, Thoreau had put his belief into action and refused to pay taxes because of the United States government's involvement in the Mexican War, which Thoreau firmly believed was unjust. For his daring and unprecedented act of protest, he was thrown in jail. The Night Thoreau Spent in Jail is a celebrated dramatic presentation of this famous act of civil disobedience and its consequences. Its poignant, lively, and accessible scenes offer a compelling exploration of Thoreau's philosophy and life.<|endoftext|>
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The Tasmanian devil went through a serious decline due to a widespread infectious disease, but its numbers are now recovering — enough for it to be considered a pest by farmers because it preys on sheep and chickens.
Many researchers consider this mammal to be the marsupial equivalent of the hyena because of its body shape, short wide snout, and extremely strong jaws, which are capable of crushing bones.
Like the hyena, the Tasmanian devil is often described as intelligent and strong. In addition, it runs like a hyena and often feeds on carrion — usually small mammals and fish.
It also feeds on live invertebrates, such as beetle larvae.
A solitary nocturnal species, it spends the day resting in burrows or grassy nests, which it makes in hollow logs or thick brush.
At night it carefully sniffs out prey and waits patiently in ambush.
Name: Tasmanian Devil (Sarcophilus harrisii)
Family: Dasyuridae (Carnivorous Marsupials)
Range: Eastern Tasmania; extinct in Australia
Habitat: Dry eucalyptus forest
Diet: Carrion, insects, snakes, some vegetation, wallabies, and other small mammals
Head and Body Length: 21 to 30 inches (53 to 76 cm)
Tail Length: 9 to 12 inches (23 to 30 cm)
Weight: 9 to 20 pounds (4 to 9 kg)
Life Cycle: Mating March to May; gestation about 31 days, one to four young born
Description: Brownish black fur; long white patches on chest, sides and rump; pinkish snout; broad, massive head; powerful jaws with sharp, sturdy teeth; thickset, squat build; short, thick tail
Conservation Status: Not listed by the IUCN.
Related Content: Australia Zoo<|endoftext|>
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# Thread: Properties of Real Numbers
1. ## Properties of Real Numbers
This is a reference chart for anyone who needs to review algebra 1 real number properties or use in class.
Property (a, b and c are real numbers, variables or algebraic expressions)
Distributive Property
a • (b + c) = a • b + a • c
Sample
3 • (4 + 5) = 3 • 4 + 3 • 5
a + b = b + a
Sample
3 + 4 = 4 + 3
Commutative Property of Multiplication
a • b = b • a
Sample
3 • 4 = 4 • 3
a + (b + c) = (a + b) + c
Sample
3 + (4 + 5) = (3 + 4) + 5
Associative Property of Multiplication
a • (b • c) = (a • b) • c
Sample
3 • (4 • 5) = (3 • 4) • 5
a + 0 = a
Sample
4 + 0 = 4
Multiplicative Identity Property
a • 1 = a
Sample
4 • 1 = 4
a + (-a) = 0
Sample
4 + (-4) = 0
Multiplicative Inverse Property
a(1/a), where a cannot be zero.
Sample
2(1/2) = 1
Zero Property of Multiplication
a • 0 = 0
Sample
4 • 0 = 0
a + b is a real number
Sample
10 + 5 = 15 (a real number)
Closure Property of Multiplication
a • b is a real number
Sample
10 • 5 = 50 (a real number)
If a = b, then a + c = b + c.
Sample
If x = 10,
then x + 3 = 10 + 3
Subtraction Property of Equality
If a = b, then a - c = b - c.
Sample
If x = 10,
then x - 3 = 10 - 3
Multiplication Property of Equality
If a = b, then a • c = b • c.
Sample
If x = 10,
then x • 3 = 10 • 3
Division Property of Equality
If a = b, then a / c = b / c,
assuming c ≠ 0.
Sample
If x = 10,
then x / 3 = 10 / 3
Substitution Property
If a = b, then a may be substituted for b, or conversely.
Sample
If x = 5, and x + y = z,
then 5 + y = z.
Reflexive (or Identity) Property of Equality
a = a
Sample
12 = 12
Symmetric Property of Equality
If a = b, then b = a.
Sample
If 3 = 3*1, then 3*1 = 3
Transitive Property of Equality
If a = b and b = c,
then a = c.
Sample
If 2a = 10 and 10 = 4b,
then 2a = 4b.
Law of Trichotomy
Exactly ONE of the following holds:
a < b, a = b, a > b
Sample
If 8 > 6, then 8 6 and
8 is not < 6.
2. ## Re: Properties of Real Numbers
Is there a question in all of that?
3. ## Re: Properties of Real Numbers
Plato,
This is a reference chart for anyone who needs to review algebra 1 real number properties or use in class.
4. ## Re: Properties of Real Numbers
Those "properties" are all true of the rational numbers as well as the real numbers. To specifically talk about the real numbers you have to add a "completeness property": if a set of real numbers has an upper bound then it has a least upper bound.
5. ## Re: Properties of Real Numbers
Originally Posted by HallsofIvy
Those "properties" are all true of the rational numbers as well as the real numbers. To specifically talk about the real numbers you have to add a "completeness property": if a set of real numbers has an upper bound then it has a least upper bound.
This is a reference chart for anyone who wants or needs to quickly review algebra 1 properties. It is not my intention to tutor or teach properties of real numbers.
6. ## Re: Properties of Real Numbers
Hi,
I have anice video about these properties but on rational numbers. But no worries, it works with real numbers also. Check it out.<|endoftext|>
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Updated on April 16 at 4:19 p.m. ET
For millennia, the only planets we knew of were the ones in our own solar system. That changed in October 1995, when a pair of Swiss astrophysicists discovered a planet orbiting a sun-like star about 50 light-years from Earth, in the constellation Pegasus. For decades, scientists had suspected that other planets existed in the cosmos, and they finally had the proof.
The discovery of 51 Pegasi b, as it was called, was just the beginning. The astronomy community was witnessing “A Parade of New Planets,” declared a headline in Scientific American in 1996. In the months since the exoplanet discovery had been announced, the publication reported, astronomers had reported finding at least four more planets.
More than two decades later, the parade is still going. Today, there are 3,717 known exoplanets, and nearly 4,500 other suspected exoplanets waiting to be verified. More than 900 of them are thought to have a rocky surface like Earth’s.
On Wednesday, NASA will launch a new spacecraft designed to discover still more exoplanets. (The launch was scheduled for Monday but was postponed to conduct additional pre-flight analysis, SpaceX said.) The Transiting Exoplanet Survey Satellite, or TESS, will spend two years surveilling more than 200,000 stars, watching for evidence of planets around them.
TESS will employ a method different than the one that was used to discover 51 Pegasi b, but the spacecraft owes a great deal to the first known exoplanet. Without 51 Pegasi b, astrophysicists may not have seriously considered the technology that should allow TESS to find thousands of new planets in the Milky Way.
In 1995, Michel Mayor and Didier Queloz, both at the University of Geneva, were trying to find exoplanets with a technique called the radial-velocity method. Sometimes, when a planet orbits a star, the planet’s gravity causes the star to wobble ever so slightly. The wobbling motion produces shifts in the star’s light, which can be detected with special instruments from Earth. By studying these shifts, astrophysicists can figure out the mass of a planet and how long it takes to complete one orbit around its star.
Back then, some were skeptical that the radial-velocity method would work, and a number of searches using the technique had proved unsuccessful since the late 1980s. The discovery of 51 Peg, as Mayor and Queloz like to call the exoplanet, was the breakthrough everyone was waiting for.
But no one was prepared for the planet they found.
“When we discovered 51 Peg, it was quite an unusual object,” says Mayor, now a professor emeritus at the university. “It was absolutely not expected from theory.”
51 Pegasi b was about half the mass of Jupiter and orbited extremely close to its star. One trip all the way around took just four days. Astrophysicists didn’t think a planet that size could orbit so closely to a star. The innermost planet in our solar system, Mercury, is thousands of times less massive than Jupiter and takes 88 days to orbit the sun. When Mayor and Queloz announced their find at a scientific conference in Italy, some members of the field were skeptical, but a different team soon confirmed the discovery.
“The shock was so profound that 51 Peg completely changed our perspective of how we could look for planets,” says Queloz, now a physics professor at both Geneva and the University of Cambridge.
The discovery of 51 Pegasi b meant that astrophysicists could look much closer to a star to search for exoplanets. This led them to more seriously consider another technique for detecting planets, known as the transit method. Like the radial-velocity method, the transit method relies on a star’s light. Astronomers train their telescopes on a star and watch for any dimming in its brightness, which would occur when a planet passes in front of it and blocks out the light. The closer the planet is to its parent star, the easier it is for telescopes to spot it.
The first detection of an exoplanet using this method was announced in 2003 by astronomers from the Harvard-Smithsonian Center for Astrophysics. OGLE-TR-56b, located around 5,000 light-years away, is about the size of Jupiter and orbits 14 times closer to its star than Mercury does to the sun.
In 2009, the Kepler mission took the transit method to space. Kepler, a NASA spacecraft, launched into an orbit around the sun equipped with instruments to detect dips in the brightness of thousands of stars in its field of view. The mission has discovered thousands of confirmed and potential exoplanets since. NASA now announces the verification of new exoplanets so often—about every few months or so—that these discoveries are no longer headline-making news.
Today, the planets in our solar system seem like the weird ones. Kepler has found rocky planets 10 times the mass of Earth, gas giants the size of Jupiter with scorching temperatures, and even rogue planets floating around the galaxy without a star to call home. At least 30 exoplanets are about the size of Earth and orbit in the habitable zone of their star systems, that cosmic sweet spot where water exists as a liquid on the surface.
“I still think that today, without 51 Peg, Kepler would never have flown,” Queloz said. And if Kepler hadn’t flown, TESS probably wouldn’t have, either. Like Kepler, TESS will search for exoplanets using the transit method.
TESS’s timing couldn’t be better. Kepler is expected to run out of fuel and will cease operations sometime in the coming months. Engineers didn’t give Kepler a gas gauge, so they just have to watch for warning signs of low fuel and wait.
TESS will launch from Cape Canaveral in Florida on a SpaceX Falcon 9 rocket. Once in space, the spacecraft will fire its engines several times to position itself for an encounter with the moon’s gravity, which will push it into its final orbit.
From its vantage point in high-Earth orbit, above where satellites normally operate, TESS will have an unobstructed view of the sky. The telescope will spend two years staring into the cosmos, turning back only to beam home data, covering more area than Kepler did. Kepler could stare only at specific patches of sky at a time, but TESS will be able to see about 90 percent of it.
TESS will focus on exoplanets around the brightest, closest stars in the galaxy, the kind of target perfect for follow-up observations by other space observatories and ground-based telescopes. Astronomers predict TESS will discover more than 1,600 new exoplanets, including about 70 Earth-sized exoplanets.
Where Kepler and TESS leave off, powerful telescopes currently under construction pick up. In the next decade, observatories like the Extremely Large Telescope in Chile will take the study of exoplanets a step further. They will examine the atmospheres of other planets, looking for the molecules that we know can, under the right conditions, create a world suitable for life.
They may even tell us something about 51 Pegasi b. Last year, astrophysicists using an instrument on the Very Large Telescope in Chile said they detected traces of water in the atmosphere of the exoplanet. The object that kicked off the parade of planets more than two decades ago may hold more surprises for us still.
We want to hear what you think about this article. Submit a letter to the editor or write to [email protected].<|endoftext|>
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Thousands of years ago….
Across the Mediterranean Sea, there was a war being fought for centuries. A war within Greece,
a war fought between the Spartans and the Trojans. But along with that, they were also at war with the Persian empire.
Sparta and Troy, along with all of Greece, had been enemies, yet allies.
After the death of Darius II, king of the Persian Empire, his eldest son Artaxerxes had claimed the throne.
However, Artaxerxes’ younger brother Cyrus the Young had a power struggle against his older brother,
and this is what sparked the Battle of Cunaxa.
401 B.C., the Battle of Cunaxa
Over two milleniums ago,
an army of Greek soldiers found themselves isolated in the middle of the Persian Empire.
One thousand miles from safety, one thousand miles from the sea, one thousand miles,
with enemies on all sides.
Theirs’ was a story of desperate forced-march,
theirs’ was a story of courage…
Cyrus had a plan to unite the Persians and the Greeks in order to conquer the Persian Empire.
He had caught the imaginations of the Greeks,
and managed to recruit an army of Spartan mercenaries to aid him in his battle.
Cyrus was killed by Artaxerxes on the battlefield,
and Clearchus, general of the Spartans,
had taken charge and led the retreat back to the sea.
>The Battle of Cunaxa was fought in 401 BC between Cyrus the Younger
and his elder brother Arsaces, who had seized the Persian throne as Artaxerxes II in 404 BC.
Cyrus gathered an army of Greek mercenaries,
consisting of 10,400 hoplites and 2,500 peltasts, under the Spartan general Clearchus,
and met Artaxerxes at Cunaxa on the left bank of the Euphrates River, 70 kilometres North of Babylon.
Artaxerxes had about 150 scythed chariots compared to about 20 available to Cyrus.
Something like this same ratio probably applies to the ratio of non-Greek troops available to each side.
Artaxerxes certainly enjoyed a superiority in cavalry.
The tactical outcome of the battle is disputed but as Cyrus died in the battle it was a political and strategic victory for Arsaces.
On Cyrus’s death Clearchus assumed the chief command and conducted the retreat,
until, being treacherously seized with his fellow-generals by Tissaphernes,
he was handed over to Artaxerxes and executed. Stranded deep in enemy territory,
with most of their generals dead,
Xenophon played an instrumental role in encouraging
the “Ten Thousand” Greek army to march north to the Black Sea in an epic fighting retreat.
This story is recorded in Anabasis by Xenophon who accompanied the “expedition up country”.
see also History of Iran (Persia)
THE RETREAT OF THE TEN THOUSAND
XENOPHON’S advice pleased the Greeks. It was far better, they thought, to make the glorious attempt to return home, than basely to surrender their arms, and become the subjects of a foreign king.
They therefore said they would elect a leader, and all chose Xenophon to fill this difficult office. He, however, consented to accept it only upon condition that each soldier would pledge his word of honor to obey him; for he knew that the least disobedience would hinder success, and that in union alone lay strength. The soldiers understood this too, and not only swore to obey him, but even promised not to quarrel among themselves.
So the little army began its homeward march, tramping bravely over sandy wastes and along rocky pathways. When they came to a river too deep to be crossed by fording, they followed it up toward its source until they could find a suitable place to get over it; and, as they had neither money nor provisions, they were obliged to seize all their food on the way.
The Greeks not only had to overcome countless natural obstacles, but were also compelled to keep up a continual warfare with the Persians who pursued them. Every morning Xenophon had to draw up his little army in the form of a square, to keep the enemy at bay.
They would fight thus until nearly nightfall, when the Persians always retreated, to camp at a distance from the men they feared. Instead of allowing his weary soldiers to sit down and rest, Xenophon would then give orders to march onward. So they tramped in the twilight until it was too dark or they were too tired to proceed any farther.
After a hasty supper, the Greeks flung themselves down to rest on the hard ground, under the light of the stars; but even these slumbers were cut short by Xenophon’s call at early dawn. Long before the lazy Persians were awake, these men were again marching onward; and when the mounted enemy overtook them once more, and compelled them to halt and fight, they were several miles nearer home.
As the Greeks passed though the wild mountain gorges they were further hindered by the neighboring people, who tried to stop them by rolling trunks of trees and rocks down upon them. Although some were wounded and others killed, the little army pressed forward, and, after a march of about a thousand miles, they came at last within sight of the sea.
You may imagine what a joyful shout arose, and how lovingly they gazed upon the blue waters which washed the shores of their native land also.
But although Xenophon and his men had come to the sea, their troubles were not yet ended; for, as they had no money to pay their passage, none of the captains would take them on board.
Instead of embarking, therefore, and resting their weary limbs while the wind wafted them home, they were forced to tramp along the seashore. They were no longer in great danger, but were tired and discontented, and now for the first time they began to forget their promise to obey Xenophon.
To obtain money enough to pay their passage to Greece, they took several small towns along their way, and robbed them. Then, hearing that there was a new expedition on foot to free the Ionian cities from the Persian yoke, they suddenly decided not to return home, but to go and help them.
Xenophon therefore led them to Pergamus, where he gave them over to their new leader.
There were still ten thousand left out of the eleven thousand men that Cyrus had hired,
and Xenophon had cause to feel proud of having brought them across
the enemy’s territory with so little loss.
After bidding them farewell, Xenophon returned home,
and wrote down an account of this famous Retreat of the Ten Thousand in a book called the Anabasis.
This account is so interesting that people begin to read it as soon as they know a little Greek,and thus learn all about the fighting and marching of those brave men.<|endoftext|>
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In the Mediterranean and Near East, Carnelian was a common stone in ancient times. Between 4000 and 3000 B.C., Egypt had natural resources available for beading whereas Mesopotamia did not have such materials available. While Carnelian and Agate came from India and Afghanistan, Lapis Lazuli came from Afghanistan, and gold likely came from the Anatolia or Iran mountains (Dubin, 1995).
As a result of sophisticated goldsmithing techniques that were used for granulation and filigree to make beads, such craftsmanship was superior and they were crafted for Sumerian royalty in the city-state of Ur in southern Mesopotamia. “The products of Sumerian jewelers spread into the less-developed cultures of western Asia and into Anatolia, the probable source of the jewelers’ gold (Dubin, 1995:12). The jewelry techniques of the Sumerians also disseminated into southern Greece and Crete (Dubin, 1995).
Sumerian and Mesopotamian jewelry, and bead making were so important that it cannot be overemphasized. It had an indirect, if not a direct, impact on the ways subsequent cultures in western Asia and the Mediterranean adorned themselves! It also appears as though glass and glass beads originated from western Asia and possibly Sumeria. The earliest examples date back to the Akkad dynasty (2340-2180 B.C.) from Mesopotamia and from the Caucasus region in present-day Russia (Dubin, 1995).
These ancient cultures believed that Carnelian brought good luck, prevented misfortune, fought off diseases, dissipated fear, and warded off evil spirits and things; in essence, it was believed that Carnelian brought protection from harm and from envy. According to Eastern belief, thoughts of envy of another’s possessions or wealth would cause the other to lose that of which you were envious. (Hall, 2011).
Today, “Magnanimous Carnelian helps you be grateful for what you have and to give thanks for the good fortune of others, reinforcing universal abundance” (Hall, 2011:64).
In addition, it is believed that Carnelian has the ability to energize, enhance fertility, reduce inflammation, and removes/prevents depression (Hall, 2011).
Furthermore, it is believed that “Carnelian stimulates courage and action. It restores motivation, energizes the soul body, and helps turn dreams into realities” (Hall, 2011:64).
To harness the power of Carnelian, wear it on your body/person so that it is close to you at all times. Throughout the day that you are wearing it, touch it periodically and say, “‘I am grateful’” (Hall, 2011:65).
If you would like to give wearable art gifts that are natural, handmade, unique, and incorporate Carnelian, (and/or any other stones that are meaningful to you or the beautiful women in your life,) please contact me for a private jewelry consultation at [email protected] or (919) 649-2123.
Dubin, Lois Sherr. 1995. The History of Beads: From 30,000 B.C. to the Present, Concise Edition. New York: Harry N. Abrams, Incorporated.
Hall, Judy. 2011. 101 Power Crystals: The Ultimate Guide To Magical Crystals, Gems, And Stones For Healing And Transformation. Beverly, MA: Fair Winds Press, a member of Quarto Publishing Group, USA Inc.<|endoftext|>
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Every Question Helps You Learn
You'll need your Maths skills for this quiz.
# Letters for Numbers 1
Letters for Numbers involves a code that you have to work out.
In this series of 11-plus verbal reasoning quizzes letters stand for numbers. Work out the correct answer to each sum and choose the correct answer from the four choices available. There will always be a code letter for each of the four numbers, and the question will have three or four letters in it. Enjoy the first of this series and watch out for tricky answers!
Take a good look at the example given below and make sure you understand what is required. You'll need both your maths and reasoning skills for these quizzes, so get ready to use that amazing brain power!
Please note that these questions assume you have not yet learned BODMAS, so if you have, please don't use BODMAS!
Example: What is the answer to the sum below, written as a letter?
A = 5, B = 14, C = 12, D = 13, E = 3
A + C - E = ?
A
B
C
E
Answer: The correct answer is B. If we translate all the letters into numbers, our new sum is 5 + 12 - 3, which = 14. 14 = B.
1.
What is the answer to the sum, written as a letter?
D = 9, T = 18, N = 6, Z = 3
D + N - Z + N
D
T
N
Z
If we translate all the letters into numbers, our new sum is 9 + 6 - 3 + 6, which = 18. 18 = T. Interesting number alert! 9 is a square number
2.
What is the answer to the sum, written as a letter?
B = 6, C = 12, R = 18, V = 2
C x V - B
B
C
R
V
If we translate all the letters into numbers, our new sum is 12 x 2 - 6, which = 18. 18 = R
3.
What is the answer to the sum, written as a letter?
F = 7, W = 6, O = 12, R = 4
F - W + R + F
F
W
O
R
If we translate all the letters into numbers, our new sum is 7 - 6 + 4 + 7, which = 12. 12 = O. Interesting number alert! 7 is a funny number. If we look at the factors from 7, they add up to interesting numbers: 2 x 7 = 14 (adding up to 5) 3 x 7 = 21 (adding up to 3) 4 x 7 = 28 (adding up to 10 as well as 1 + 0 = 1) 5 x 7 = 35 (adding up to 8) 6 x 7 = 42 (adding up to 6). Can you guess what comes next? It always equals two less than the number before it
4.
What is the answer to the sum, written as a letter?
H = 16, S = 14, M = 12, P = 2
H + M ÷ P
H
S
M
P
If we translate all the letters into numbers, our new sum is 16 + 12 ÷ 2 , which = 14. 14 = S. Interesting number alert! 16 is a square number (4 x 4)
5.
What is the answer to the sum, written as a letter?
C = 31, G = 1, Q = 8, J = 4
C + G ÷ Q + J
C
G
J
Q
If we translate all the letters into numbers, our new sum is 31 + 1 ÷ 8 + 4, which = 8. 8 = Q. Interesting number alert! 31 is a prime number
6.
What is the answer to the sum, written as a letter?
E = 5, M = 12, D = 4, C = 1
M - E + D + C
E
M
D
C
If we translate all the letters into numbers, our new sum is 12 - 5 + 4 + 1, which = 12. 12 = M. Interesting number alert! 12 is a number where, if you add up the numbers it is made of, it equals 3. Double it and add those numbers (24 gives us 6) and it equals three more than the first two numbers added up to. No matter which factor you use, this works. Try it!
7.
What is the answer to the sum, written as a letter?
A = 30, D = 25, C = 65, B = 3
A x B - C
A
D
C
B
If we translate all the letters into numbers, our new sum is 30 x 3 - 65, which = 25. 25 = D
8.
What is the answer to the sum, written as a letter?
F = 9, X = 17, G = 4, K = 2
X - F ÷ G + K
X
F
G
K
If we translate all the letters into numbers, our new sum is 17 - 9 ÷ 4 + 2, which = 4. 4 = G. Interesting number alert! 17 is a prime number!
9.
What is the answer to the sum, written as a letter?
R = 81, S = 9, T = 10, U = 1
R ÷ S + U ÷ T
R
S
U
T
If we translate all the letters into numbers, our new sum is 81 ÷ 9 + 1 ÷ 10, which = 1. 1 = U. Interesting number alert! 81 is not only a square number (9 x 9) but its two numbers also add up to 9!
10.
What is the answer to the sum, written as a letter?
J= 13, Q = 9, D = 23, P = 117
J x Q - P + D
J
Q
P
D
If we translate all the letters into numbers, our new sum is 13 x 9 - 117 + 23, which = 23. 23 = D
Author: Stephen O'Hara<|endoftext|>
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In a research paper published in Physical Review Letters, Norman Yao, a UC Berkeley assistant physics professor has described how to create a new form of matter known as time crystals, and how to measure their various properties. The new form of matter can have various phases, such as the solid, liquid and gas states of water. Two independent groups following the blueprints provided by Yao have managed to create time crystals in labs.
Repeatedly tapping a chunk of jello introduces vibrations, that can repeat periodically after certain intervals of time. Scientists blasted two lasers at a one dimensional chain of ytterbium ions. One laser was to create a magnetic field, and another was to partially flip the spins of atoms. The atoms settled in a stable, repeating pattern of spin flipping, which is the definition of a time crystal. Nobel laureate Frank Wilczek first theorised the existence of time crystals in 2012, and other researchers independently proved that such a crystal could be made.
The UC Berkeley researchers bridged the gap between theory and implementation, and actually created one of the first examples of non-equilibrium materials. The discovery could have implications in quantum computing, and can theoretically be used to create perfect memories that do not degrade over time. “For the last half-century, we have been exploring equilibrium matter, like metals and insulators. We are just now starting to explore a whole new landscape of non-equilibrium matter.” Yao said.
If the atomic structures in known crystals can repeat in regular space intervals, why cannot the atomic structures repeat periodically in time? This was the question that Norman Yao investigated, resulting in the breakthrough implementation. There are more details about the experimental creation of time crystals on the UC Berkeley web site.<|endoftext|>
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# How Can Horizontally Stretching The Graph Of F (x)=x3 By A Factor Of B Can Be? (Solution found)
How to calculate horizontal and vertical graph stretches?
• Horizontal And Vertical Graph Stretches And Compressions (Part 1) The general formula is given as well as a few concrete examples. y = c f(x), vertical stretch, factor of c; y = (1/c)f(x), compress vertically, factor of c; y = f(cx), compress horizontally, factor of c; y = f(x/c), stretch horizontally, factor of c; y = – f(x), reflect at x-axis
## How do you stretch a graph horizontally by a factor of 3?
If g(x) = 3f (x): For any given input, the output iof g is three times the output of f, so the graph is stretched vertically by a factor of 3. If g(x) = f (3x): For any given output, the input of g is one-third the input of f, so the graph is shrunk horizontally by a factor of 3.
## How do you stretch a graph horizontally?
Key Takeaways
1. When by either f(x) or x is multiplied by a number, functions can “stretch” or “shrink” vertically or horizontally, respectively, when graphed.
2. In general, a vertical stretch is given by the equation y=bf(x) y = b f ( x ).
3. In general, a horizontal stretch is given by the equation y=f(cx) y = f ( c x ).
You might be interested: Uterus Stretching Pain When I Standing Up? (Best solution)
## How do you write a horizontal stretch by a factor of 2?
Thus, the equation of a function stretched vertically by a factor of 2 and then shifted 3 units up is y = 2f (x) + 3, and the equation of a function stretched horizontally by a factor of 2 and then shifted 3 units right is y = f ( (x – 3)) = f ( x – ).
## How do you find the factor of a horizontal stretch?
A horizontal stretch or shrink by a factor of 1/k means that the point (x, y) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). Consider the following base functions, (1) f (x) = x2 – 3, (2) g(x) = cos (x).
## Whats a horizontal stretch?
Horizontal stretches are among the most applied transformation techniques when graphing functions, so it’s best to understand its definition. Horizontal stretches happen when a base graph is widened along the x-axis and away from the y-axis. Understanding the common parent functions we might encounter.
## What is a horizontal stretch?
A horizontal stretching is the stretching of the graph away from the y-axis. A horizontal compression (or shrinking) is the squeezing of the graph toward the y-axis. • if k > 1, the graph of y = f (k•x) is the graph of f (x) horizontally shrunk (or compressed) by dividing each of its x-coordinates by k.
## How do you find the stretch factor of a graph?
1. Refer to: y=af(b(x−h))+k.
2. A vertical stretch is the stretching of a function on the x-axis.
3. A horizontal stretch is the stretching of a function on the y-axis.
4. For example:
5. b=12.
6. To vertically stretch we use this formula:
7. To horizontally stretch we use this formula:
8. x1=x12.
You might be interested: What Is Dynamic Stretching For Swimmers? (Question)
## How do you do a horizontal stretch and compression?
If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function. Given a function y=f(x) y = f ( x ), the form y=f(bx) y = f ( b x ) results in a horizontal stretch or compression.
## How do you translate a graph horizontally?
Horizontally translating a graph is equivalent to shifting the base graph left or right in the direction of the x-axis. A graph is translated k units horizontally by moving each point on the graph k units horizontally. g(x) = f (x – k), can be sketched by shifting f (x) k units horizontally.
## What is a horizontal shrink by 1 2?
The horizontal shrink means you shrink x by a factor of 1/2. Currently the slope on the right side of the V is 1, so to “shrink” it, you actually DIVIDE by 1/2, giving you a new slope of 2.
## Why are horizontal stretches opposite?
Why are horizontal translations opposite? While translating a graph horizontally, it might occur that the procedure is opposite or counter-intuitive. That means: For negative horizontal translation, we shift the graph towards the positive x-axis.<|endoftext|>
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# A cone a hemisphere and a cylinder stand on equal bases and have the same height. Find the ratio of their volumes
Manan Jain
12 Points
7 years ago
We know that all the three figures-cone,hemisphere and cylinder are on same base. It means that they will have the same base radius.
so, let the common radius of all figures be ‘r’.
also they have same height.
so, let the common heights of the three figures be ‘h’.
Now, volume of hemisphere=$\inline \frac{2}{3}\pi r^{3}$
volume of cylinder=$\inline \pi$r2h
volume of cone=$\inline \frac{1}{3}\pi r^{2}h$
so, volume of cone:volume of cylinder:volume of hemisphere=$\inline \frac{2}{3}\pi r^{3}$:$\inline \pi$r2h:$\inline \frac{1}{3}\pi r^{2}h$
{{(Cancel out the common values)}}
=$\inline \frac{2}{3}$r$\inline \pi$×r×r:$\inline \pi$r×r×h:$\inline \frac{1}{3}$$\inline \pi$r×r×h
=$\inline \frac{2}{3}$r:h:$\inline \frac{1}{3}$h
we know that in a hemisphere, radius=height of hemisphere,
so,
=$\inline \frac{2}{3}$h:h:$\inline \frac{1}{3}$h
=$\inline \frac{2}{3}$:1:$\inline \frac{1}{3}$
multiply the ratio by 3,we get
=2:3:1<|endoftext|>
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This article is designed to provide learners of all ages and skill levels with the basic tools and techniques needed to master multiplication. Whether you’re just starting with multiplication, looking to refresh your skills, or helping a young learner, this guide will offer valuable insights. It aims to make multiplication less intimidating and more approachable by breaking it down into its simplest elements. Continue reading the article to embark on this exciting numerical journey.
## Importance of Multiplication Skills
Multiplication is a fundamental arithmetic operation that is pivotal in many aspects of mathematics and daily living. It forms the basis for more complex mathematical concepts like fractions, algebra, and geometry. Beyond academics, multiplication skills are invaluable in everyday scenarios, including budgeting, cooking, and time management. Developing strong multiplication skills at an early age equips learners with a sturdy mathematical foundation, fostering logical thinking and problem-solving abilities.
## Traditional Multiplication Method: Step-by-Step Guide
The traditional multiplication method, long multiplication, is a fundamental process often taught in school to handle multi-digit numbers. Here’s a step-by-step guide to help you understand and employ this method:
1. Write the Numbers: Write down the two numbers you want to multiply, one above the other, aligning them on the right side. Draw a horizontal line under the two numbers.
1. Multiply by Units: Start by multiplying the digit in the unit’s place (rightmost digit) of the lower number with every digit of the upper number, from right to left. Write the result below the horizontal line.
1. Carry Forward: If the product is a two-digit number, write down the unit digit and carry forward the tens digit to the next multiplication.
1. Next Row: Move to the next digit in the lower number (tens place). Repeat the multiplication and carry forward process, but this time, write the result in the next row, starting one place to the left.
1. Continue Process: Repeat this process for every digit in the lower number.
1. Add Results: After completing all the rows, add them up. The total is the result of the multiplication.
Remember, practice is key to mastering the traditional multiplication method. Start with simple numbers and gradually progress to larger ones. With regular practice, this strategy will soon become second nature.
## What is Skip Counting, and How Can it Boost Your Multiplication Skills?
Skip counting is among the most effective multiplication strategies to enhance your multiplication abilities significantly. Fundamentally, skip counting involves counting forward by numbers other than one, for instance, counting by twos (2, 4, 6, 8, and so forth) or by fives (5, 10, 15, 20, and so on).
Skip counting lays the groundwork for understanding multiplication as it helps learners visualize and understand the concept of grouping, a cornerstone of multiplication. When children skip count, they add the same number repeatedly, paving the way toward understanding that multiplication is a more efficient way of adding the same number.
For example, five groups of three can be calculated by adding three, five times (3+3+3+3+3). Alternatively, it can also be calculated by skip counting by threes five times (3, 6, 9, 12, 15). Eventually, learners realize that this process can be further simplified by multiplying 5 x 3 = 15.
Skip counting is a powerful tool in the multiplication arsenal. By incorporating this strategy, learners can develop a more intuitive understanding of multiplication, making the transition to formal multiplication methods smoother and less daunting.
## Can Visual Representations Make Multiplication Fun and Easy?
Visual representations can indeed make multiplication engaging and approachable, particularly for visual learners who absorb information best when presented in a graphic or pictorial format. Arrays, number lines, and area models are a few visual strategies that can be employed to illustrate multiplication concepts.
Arrays are essentially grids with rows and columns representing multiplication equations. For instance, a 3×4 array represents the multiplication equation 3×4=12, with three rows of four squares each. This visual representation helps learners understand the concept of repeated addition and grouping inherent in multiplication.
On the other hand, number lines are fantastic tools for demonstrating skip counting or repeated addition. Multiplying 3×4, for example, can be shown as four jumps of three on a number line, reinforcing the connection between addition and multiplication.
Area models divide a rectangle into smaller squares or rectangles to represent a multiplication problem, where the length and width of the rectangle represent the two numbers being multiplied. This approach is useful for grasping more complex multiplication problems involving larger numbers.
By making multiplication visually tangible, these methods make the concept less abstract, aiding learners in grasping the essence of multiplication. Visual representations can be a lighthearted and engaging way of mastering multiplication, transforming learning into a fun-filled adventure.
## How to Make Multiplication Easier: Factoring and Distributive Property
The power of multiplication lies in its simplicity and efficiency, and understanding the principles of factoring and the distributive property can make it even easier. Factoring involves breaking down a number into its multiplication factors. For example, the number 12 can be factored into 3 and 4, as 3×4 equals 12. This becomes useful when dealing with larger numbers or when simplifying fractions.
The distributive property, on the other hand, is a multiplication strategy that allows you to break down complex multiplication problems into smaller, more manageable parts. Also known as the multiplication break-apart method, it is expressed as a(b + c) = ab + ac. This means multiplying a number by a group of numbers added together is the same as doing each multiplication separately.
Take the example of 6 x 14. This might seem like a tricky multiplication problem initially, but using the distributive property, you can break it down: 6 x (10 + 4) = (6 x 10) + (6 x 4) = 60 + 24 = 84. The distributive property allows splitting more challenging multiplication problems into ones that are easier to solve, making multiplication less intimidating, especially for young learners.
Understanding and utilizing factoring and the distributive property make multiplication easier and pave the way for understanding more complex mathematical operations. Armed with these strategies, you have additional tools to tackle multiplication more easily and efficiently.
## Struggling with Big Numbers? Grasping Multiplication for Two and Three-Digit Numbers
Multiplying two or three-digit numbers might seem daunting at first, but with the right approach, it can be simplified. Learners can confidently multiply larger numbers by building on the basic multiplication strategies and understanding place value. The process involves breaking down the problem into manageable parts and using basic multiplication and addition to find the solution.
Take, for example, the two-digit multiplication of 24 x 13. You can break down 13 into 10 and 3. Now, multiply 24 by 10 and 24 by 3 separately. The multiplication of any number by 10 is straightforward; simply add a zero to the end of the number. So, 24 x 10 = 240. For 24 x 3, use basic multiplication skills to get 72. Now add the two results together: 240 + 72 = 312. Thus, 24 x 13 = 312.
The same strategy can be applied to three-digit numbers. For instance, multiplying 123 x 24 breaks down 24 into 20 and 4. Multiply 123 by 20 and by 4 separately, then add the results.
This strategy, which builds on the distributive property, can make multiplying bigger numbers more manageable. With practice, this method can significantly enhance accuracy and speed in solving larger multiplication problems and lessen the fear of multiplying big numbers. Remember, mastery comes with practice, so keep practicing these multiplication strategies until they become second nature.
## Can You Multiply with Just Your Mind?
Yes, you can perform multiplication operations swiftly and accurately in your head by employing mental math techniques. Mental multiplication strategies primarily involve understanding number properties, leveraging place value, and utilizing basic multiplication facts.
For instance, when multiplying a number by 10, simply add a zero to the end of the number; for multiplying by 5, halve the number and multiply by 10. When multiplying by 9, multiply by 10 and subtract the original number.
To multiply two two-digit numbers mentally, break down the multiplication into smaller parts you can easily compute. For example, to multiply 21 by 13, you can calculate (20 x 10) + (20 x 3) + (1 x 10) + (1 x 3) = 200 + 60 + 10 + 3 = 273.
Mental multiplication might seem challenging at first, but you’ll become more adept with practice. It’s a powerful skill that can be extremely helpful in daily life, enabling quick estimations, improving number sense, and boosting confidence in mathematical abilities. Remember, the key to successful mental multiplication is practicing and understanding the underlying principles of the multiplication strategies outlined in this guide.
Tips for Faster Mental Multiplication
Mastering the art of mental multiplication can significantly speed up your math proficiency. Here are some valuable tips to help you multiply numbers faster mentally:
1. Understanding the Properties of Numbers: Familiarize yourself with the properties of numbers. For instance, any number multiplied by zero equals zero, and any number multiplied by one retains its value. These basic properties can save you time and effort when multiplying mentally.
1. Leveraging Simplification Techniques: Use simplification techniques to make multiplication more manageable. For example, if you’re multiplying a large number by a number ending in zero, simply perform the multiplication with the remaining digits and append the corresponding number of zeros at the end.
1. Rounding and Adjusting: Round complex numbers up or down to make the multiplication easier, then adjust the answer accordingly. For instance, if you want to multiply 99 by a number, you might find it easier to multiply 100 by the number and subtract the original number.
1. Breaking Down Larger Numbers: Break down larger numbers into smaller, more manageable parts. For example, for the multiplication 24 x 15, think of it as (20 x 15) + (4 x 15). This simplifies the problem into more manageable multiplications.
1. Practicing Regularly: Regular practice is key to improving mental multiplication speed. Start with simpler numbers and gradually move on to larger, more complex ones.
1. Using Skip Counting for Smaller Numbers: For smaller numbers skip counting can prove to be a faster technique. For instance, multiplying by 4 can be achieved by doubling the number twice.
1. Understanding the Distributive Property: The distributive property can be a valuable tool in mental multiplication. It allows you to break down complex problems into simpler ones, making mental calculations more manageable.
Remember, the goal of mental multiplication is not just speed but also accuracy. Therefore, take the time to ensure your calculations are correct even as you aim to improve your speed. With these tips and regular practice, your mental multiplication skills improve significantly.
## Strategies for Memorizing Multiplication Tables
One of the fundamental steps in mastering multiplication is familiarizing yourself with multiplication tables, also known as times tables. These tables provide a quick reference for multiplying small numbers and are vital for enhancing calculation speed and accuracy. Here are some strategies to make memorizing multiplication tables less daunting and more enjoyable:
1. Start Small: Begin with the easier times tables such as 1s, 2s, 5s, and 10s. These tables follow simple patterns that are easy to remember. For instance, any number multiplied by 1 remains the same, and for the 10s, you simply add a zero to the end of the number.
1. Use Visual Aids: Create a multiplication chart to refer to as you learn. Visual aids are highly effective when memorizing multiplication tables as they appeal to visual learners and offer a convenient reference point.
1. Leverage Patterns: Look for patterns in the tables. For example, in the 9s table, the tens digit increases by 1, and the ones digit decreases by 1 as you go down the table (9, 18, 27, 36, 45…).
1. Practice Regularly: Constant practice is key to memorizing multiplication tables. Make it a habit to review the tables you have learned daily, and gradually introduce new ones as you become more confident.
1. Use Flashcards: Flashcards with multiplication problems on one side and answers on the other provide an interactive and effective way to learn. They’re also portable, allowing you to practice anytime, anywhere.
1. Sing or Chant the Tables: Turn the tables into a song or a chant. The rhythm and melody can make the tables more memorable and fun to learn.
1. Use Educational Apps: Numerous educational applications are designed to make learning multiplication tables engaging through games and quizzes. These can be especially appealing to tech-savvy learners.
1. Apply Multiplication in Real-Life Situations: Apply the multiplication tables to real-life situations such as shopping or baking. This provides practice and demonstrates the relevance and utility of multiplication in everyday life.
Remember, mastering multiplication tables requires time and patience. It’s not a race, so learn at your own pace and move on to more complex tables only when you’re comfortable. You’ll have the multiplication tables memorized quickly with consistent practice and the right strategies.
## FAQs
Here are some frequently asked questions about different multiplication strategies to improve mathematical skills.
What are the strategies to teach multiplication?
Teaching multiplication can be made effective and enjoyable by employing various strategies. Here are a few approaches teachers can use:
1. Concrete Learning: Start with concrete learning using physical objects like counters, blocks, or beads to illustrate the concept of multiplication as repeated addition. This helps students visualize and better understand the process.
1. Skip Counting: Once students are comfortable with counting, introduce skip counting. It’s a precursor to multiplication, teaching children to count by twos, threes, or any number other than one.
1. Arrays: Use an array model for multiplication to represent problems visually. This helps students connect the numbers in a multiplication equation and the size of the array.
1. Number Line: Use a number line to illustrate multiplication. For example, for 3×4, make four jumps of three on a number line.
1. Multiplication Tables: Encourage students to learn multiplication tables. Make it interactive and fun by using songs, games, or flashcards.
1. Group Activities: Organize group activities that involve multiplication to promote collaborative learning.
1. Real-Life Examples: Incorporate real-life examples to demonstrate the practical application of multiplication. This helps students understand its relevance and importance.
1. Word Problems: Use word problems to practice multiplication. This enhances students’ problem-solving skills and allows them to apply their multiplication knowledge in various contexts.
1. Incorporate Technology: Use technology like educational apps or online games that can make learning multiplication fun and interactive.
1. Consistent Practice: Encourage regular and consistent practice. Like any other skill, multiplication becomes easier and more automatic with practice.
These multiplication strategies can contribute to a comprehensive and engaging approach to teaching multiplication. Remember, different students learn differently, so it’s beneficial to use a mix of these strategies to cater to diverse learning styles.
What is the standard multiplication strategy?
The standard multiplication strategy, often called the traditional method or long multiplication, involves breaking down the multiplication problem into simpler steps to obtain the final result.
Why teach multiplication strategies?
Teaching multiplication strategies is fundamental in developing a strong foundation in arithmetic, which is crucial in navigating other complex mathematical concepts. It enhances learners’ number sense, enabling them to understand and connect with numbers on a deeper level. These strategies also aid in improving computational skills, critical thinking, and problem-solving capacities.
What is the purpose of multiplication?
The purpose of multiplication is to simplify the process of repeated addition. Multiplication is a more efficient method when dealing with larger numbers or quantities. For example, instead of repetitively adding 5 ten times, it is quicker and easier to multiply 5 x 10.
How do students learn multiplication?
Students learn multiplication through a variety of methods, each serving to develop their understanding and fluency in the concept. Here are some key steps in the learning process:
1. Understanding the Concept: The first step is understanding multiplication as repeated addition. This foundational concept is often taught using concrete objects such as counters or blocks, which allow students to see and physically manipulate groups of items visually.
1. Learning Multiplication Facts: As students grasp the concept of multiplication, they begin to learn multiplication facts, often starting with the times tables of smaller numbers. This involves memorization and frequent practice.
1. Practicing with Worksheets and Problems: Worksheets with multiplication problems provide students with ample practice opportunities. They are also introduced to word problems that require multiplication, promoting the application of their skills in different contexts.
1. Utilizing Strategies: Students learn different strategies for multiplication, such as distributive, associative, and commutative properties of multiplication. These strategies and mental math techniques can help them solve more complex multiplication problems efficiently.
1. Applying in Real Life: Teachers often incorporate real-life situations that require multiplication in lessons. This helps students understand the practicality and relevance of multiplication in solving everyday problems.
1. Leveraging Technology: Educational apps and online games provide interactive and engaging ways for students to learn and practice multiplication. Resources like these often offer immediate feedback, allowing students to instantly correct mistakes and learn from them.
Learning multiplication is a step-by-step process that builds upon each previous step. It requires time, practice, and patience, but with the right approaches and resources, students can successfully master this important mathematical operation.<|endoftext|>
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Separation Of Liquid And Biosolids
After trash and bulk contamination are removed from waste water by screening, the next step is the removal of suspended matter. This can be accomplished by several methods, the simplest of which is gravity sedimentation. Wastewater is held in a tank or vessel until heavier particles have sunk to the bottom and light materials have floated to the top. The top of the tank can be skimmed to remove the floating material and the clarified liquid can be drained off. In batch mode sedimentation, several tanks of sewage will go through the settling process before the accumulated sludge is removed from the bottom of the tank.
The settling process can be hastened by use of chemical precipitants such as aluminum sulfate. Gentle stirring with rods, another method, encourages the aggregation of a number of fine, suspended materials. As the clots of material grow larger and heavier, they sink. Suspended matter can be encouraged to float by exposing it to fine bubbles, a method known as dissolved-air floatation. The bubbles adhere to the matter and cause it to float to the surface, where it can be removed by skimming.
Another method of filtration, generally used after gravity sedimentation, is deep bed filtration. Partially processed liquid from the sedimentation tanks, called effluent, flows over a bed of graded sand and crushed coal. This material not only strains the larger particles from the effluent, but further clarifies it by removing fine particles via adhesion. The filtering material attracts these small particles of sewage by electrostatic charge, pulling them out of the main flow and resulting in significantly clearer liquid. Alternately, effluent can be filtered by a fine mesh screen or cloth, in a method known as surface filtration, or solid material can be pulled out by centrifuge.
At this point the original raw sewage has been essentially separated into two parts: sludge, or biosolids; and clarified effluent. Both parts still contain disease carrying, oxygen consuming pathogens, and need further processing. Earlier, we discussed the biological decay of raw sewage. Theoretically, both biosolids and effluent can be processed using biological treatment methods, but at this point cost considerations come into play. Biological treatment of dense sludge is time-consuming, requiring large tanks to allow complete processing, whereas that of effluent is fairly efficient. Thus, biosolids are generally processed by different methods than effluent.
After settling out, biosolids can be removed from the bottom of the sedimentation tank. These tanks may have a conical shape to allow the sludge to be removed through a valve at the tip, or they may be flat bottomed. The sludge can be dried and incinerated at temperatures between 1,500–3,000°F (816–1,649°C), and the resulting ashes, if non-toxic, can be buried in a landfill. Composting is another method of sludge disposal. The biosolids can be mixed with wood chips to provide roughage and aeration during the decay process. The resulting material can be used as fertilizer in agriculture. Properly diluted, sludge can also be disposed of through land application. Purely municipal sludge, without chemicals or heavy metals, makes a great spray-on fertilizer for non-food plants. It is used in forestry, and on such commercial crops as cotton and tobacco. It must be monitored carefully, though, so that it does not contaminate ground water.<|endoftext|>
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### Course: Class 11 Physics (India)>Unit 4
Lesson 8: Basic differentiation rules
# Basic differentiation review
Review the basic differentiation rules and use them to solve problems.
## What are the basic differentiation rules?
Sum rule$\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}f\left(x\right)+\frac{d}{dx}g\left(x\right)$
Difference rule$\frac{d}{dx}\left[f\left(x\right)-g\left(x\right)\right]=\frac{d}{dx}f\left(x\right)-\frac{d}{dx}g\left(x\right)$
Constant multiple rule$\frac{d}{dx}\left[k\cdot f\left(x\right)\right]=k\cdot \frac{d}{dx}f\left(x\right)$
Constant rule$\frac{d}{dx}k=0$
The Sum rule says the derivative of a sum of functions is the sum of their derivatives.
The Difference rule says the derivative of a difference of functions is the difference of their derivatives.
The Constant multiple rule says the derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
The Constant rule says the derivative of any constant function is always $0$.
## What problems can I solve with basic differentiation rules?
You can find the derivatives of functions that are combinations of other, simpler, functions. For example, $H\left(x\right)$ is defined as $2f\left(x\right)-3g\left(x\right)+5$. We can find ${H}^{\prime }\left(x\right)$ as follows;
We used the basic differentiation rules to find that ${H}^{\prime }\left(x\right)=2{f}^{\prime }\left(x\right)-3{g}^{\prime }\left(x\right)$.
Now suppose we are also given that ${f}^{\prime }\left(3\right)=1$ and ${g}^{\prime }\left(3\right)=5$. We can find ${H}^{\prime }\left(3\right)$ as follows:
$\begin{array}{rl}{H}^{\prime }\left(3\right)& =2{f}^{\prime }\left(3\right)-3{g}^{\prime }\left(3\right)\\ \\ & =2\left(1\right)-3\left(5\right)\\ \\ & =-13\end{array}$
Problem 1
$x$
$1$$-1$$-18$$0$$4$
$G\left(x\right)=-4f\left(x\right)+3h\left(x\right)-2$
${G}^{\prime }\left(1\right)=$
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• what are some easy ways of remembering differentiation and integration rules for o'level students
• My Calculus professor in college said there is only one way you can actually memorize most of the rules and learn to quickly and effectively apply them: Do a ton of exercises. There's no two ways about it, I'm afraid.
• Isn't the difference rule exactly the same as the sum rule? f(x) - g(x) is the same thing as f(x) + (-g(x)), after all.
• Yes it is! Finding structure in these properties can only help you as you move through the course :)
• How can I discriminate all these different rule?
(1 vote)
• Practice, practice, practice. You can go back and watch the videos again, and re-work the exercises and quizzes until you feel comfortable with them. Best wishes!
• I am wondering if there are any prerequisites that I should be doing before I start AP Calculus. I am a fast learner and good at picking up new topics, but I am wondering if there are any holes in my education so far. (I am not doing this with school.)
(1 vote)
• You really need a good foundation in working with functions. You should be able to work with functions defined many different ways (tables, graphs, equations ect) and be able to recognize parent functions and apply transformations. You should also be able to use function notation very consistently with functions presented in any form. I have found that working hard with the students on this in their pre-calc courses made a huge difference to their success in AP calc.
• On a graph assume y = x^2.If we know how y is going to change with respect to x.Then,why do we need to differentiate?
(1 vote)
• Just because we know a function is changing with respect to x isn't enough. In most cases we like to know how it is changing.
(1 vote)
• do you find a derivittve of a function
(1 vote)
• Yes, you do need to find the derivative of the function that you're asked to find the derivative of! You can find the derivative of a function by applying the differentiation rules listed above.
(1 vote)
• f(y)=y/(y+x)..how do i find the derivative using the first principle?<|endoftext|>
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Informational (nonfiction), 118 words, Level G (Grade 1), Lexile 400L
The Spider's Web entertains readers as they learn about how spiders build and use webs. No matter how students feel about spiders, the amazing photographs of a spider building a web and feeding will hold their attention. Use this fascinating book to teach summarizing information and sequencing events.
Use of Kurzweil 3000® formatted books requires the purchase of Kurzweil 3000 software at www.kurzweiledu.com.
Guided Reading Lesson
Use of vocabulary lessons requires a subscription to VocabularyA-Z.com.
Visualize to understand text
Discriminate short vowel /i/ sound
Identify short vowel i
Grammar and Mechanics
Identify and use possessive nouns
Recognize and use the high-frequency word she
Think, Collaborate, Discuss
Promote higher-order thinking for small groups or whole class
You may unsubscribe at any time.<|endoftext|>
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As children enter the elementary school, they are guided by a Class Teacher in Grades 1- 8. Students and teachers develop a deep and enduring relationship through the shared experience of the main lesson curriculum, where one subject is taught for the first two hours of each day in blocks of three to four weeks.
During each Main Lesson block, the children make their own textbooks, full of illustrations, from subjects presented by the class teacher. It is the goal of each class teacher to bring the curriculum to their class in ways that will excite and enthuse, leading each student to question, learn, dream, and explore!
To learn more about the curriculum of each grade, visit the following pages:<|endoftext|>
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1. ## 4x^2-1/4x^2
4x^2-1/4x^2
its supposed to equal 15/16ths somehow?Or atleast thats what my teacher told me. Could you show me how this equals 15/16ths?
2. Ok well we see that $\displaystyle 4x^2 = 16$ from the bottom so now let's solve for x
$\displaystyle 4x^{2}=16$
$\displaystyle x^{2}=4$
$\displaystyle x =\sqrt{4}$
So we get $\displaystyle x = \pm 2$
(Remember that square roots always have 2 roots)
Now we can see if it satisfies the numerator.
$\displaystyle 4(2)^2-1 = 15$
$\displaystyle 4(4)-1 = 15$
$\displaystyle 16 - 1 = 15$
and of course $\displaystyle 15 = 15$
So we found that $\displaystyle x = \pm 2$
3. Originally Posted by bilbobaggins
4x^2-1/4x^2
its supposed to equal 15/16ths somehow?Or atleast thats what my teacher told me. Could you show me how this equals 15/16ths?
SnipedYou:
You can in this particular instance assume that the numerator is 15 and the denominator is 16, but you first have to show that $\displaystyle 4x^2 - 1$ and $\displaystyle 4x^2$ are relatively prime, a problem that I doubt bilbobaggins is going to want to do.
Here's the general method:
$\displaystyle \frac{4x^2 - 1}{4x^2} = \frac{15}{16}$
$\displaystyle 16(4x^2 - 1) = 15(4x^2)$
$\displaystyle 64x^2 - 16 = 60x^2$
$\displaystyle 4x^2 = 16$
$\displaystyle x^2 = 4$
Thus
$\displaystyle x = \pm 2$
-Dan
4. I wonder if he means $\displaystyle \frac{4x^2-1}{4x^2}=\frac{15}{16}$ for all x?
Well that is false. It cannot simplify down to $\displaystyle \frac{15}{16}$.
5. Is there a case where 2 consecutive numbers aren't relatively prime?
6. Originally Posted by SnipedYou
Is there a case where 2 consecutive numbers aren't relatively prime?
No, but I suspect if bilbobaggins didn't know how to solve the problem the long way (my way) I doubt he would have even considered the concept of relatively prime in solving this.
-Dan<|endoftext|>
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# A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by $F = -\hat{i}+2\hat{j}+3\hat{k}\;N$, where $\hat{i}$, $\hat{j}$, $\hat{k}$ are unit vectors along the x-, y- and z- axes of the system respectively. What is the work done by this force in moving the body a distance of $4\;m$ along the z-axis?
Last updated date: 20th Jun 2024
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Hint: Try to recall how you would express the work done as a dot product of the above two quantities. Also remember that the only contributing vector components are the non-zero components, both in magnitude and direction. In other words, focus only on the component of the force acting along the z-direction while computing your dot product.
Formula used:
Work done $W = \vec{F}\; .\vec{S}$ where F is the force vector and S is the displacement vector.
Let us first establish that a vector in 3-dimensions can be broken into 3 components:
The x-axis component $\hat{i}$,
The y-axis component $\hat{j}$, and
The z-axis component $\hat{k}$.
Each component of a vector depicts the magnitude of influence of that vector in a given direction. The $\hat{i}$, $\hat{j}$ and $\hat{k}$ represent unit vectors in the x-, y- and z-direction respectively, and the number that precedes them represent the magnitude of the vector in that direction.
Now, we have a body that can move only along the direction of the z-axis. This means that any distance that we take that this body covers will be in the z($\vec{k}$)-direction. Therefore, the distance that the body travels under the influence of the force can be represented by the displacement vector $\vec{S} = 0\hat{i}+0\hat{j}+4\hat{k}$.
The work done by the force $\vec{F} = -1\hat{i}+2\hat{j}+3\hat{k}$ to move the body by a distance $\vec{S} =4\hat{k}$ is given as the scalar product of the two, i.e.,:
$W = \vec{F}.\vec{S} = \left(-1\hat{i}+2\hat{j}+3\hat{k}\right). \left(4\hat{k}\right)$
$\Rightarrow W = \left(3\hat{k}\right). \left(4\hat{k}\right) = 12\;J$
Therefore, only the z-component of the force contributes to moving the body in the z-direction. Thus, the work done by the force in moving the body through a distance of $4\;m$ is $12\;J$
Note:
Remember that the dot product of two vectors results in a scalar quantity and hence is it not directional. Another form of expressing the dot product when instead of the individual components the angle $\theta$ between the two vectors is given is:
$W = \vec{F}.\vec{S} = |F||S|cos\theta$
In the above problem, we consider only $W = \left(3\hat{k}\right). \left(4\hat{k}\right)$, which means $W = 4 \times 3 \cos 0^{\circ} = 12\;J$ since $cos0^{\circ} =1$ . This is the same reason why we do not consider $\hat{i}.\hat{j}$ or $\hat{j}.\hat{k}$ or $\hat{i}.\hat{k}$ since for them, $\theta =90^{\circ} \Rightarrow cos 90^{\circ} = 0, \Rightarrow W=0$.
Therefore, the work done is numerically quantified only when the vectors are not perpendicular to each other and the vectors have non-zero components.<|endoftext|>
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# CASE STUDY QUESTION 3-Class X-Maths
CASE STUDY QUESTION 3-Class X-Maths
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the
surface of the water and the fly at the end of the string rests on the water 3.6 m
away and 2.4 m from a point directly under the tip of the rod. She is pulling the
string at the rate of 5 cm per second. Nazima’s friend observe her position and
draw a rough sketch by using A, B, C and D positions of tip, point directly under
the tip of the rod, fish and Nazima’s position (see the below figure). Assuming that
her string (from the tip of her rod to the fly) is taut, answer the following
questions
(a) What is the length AC?
(i) 2 m (ii) 3 m (iii) 4 m (iv) 5 m
(b) What is the length of string pulled in 12 seconds?
(i) 6 m (ii) 0.3 m (iii) 0.6 m (iv) 3 m
(c) What is the length of string after 12 seconds?
(i) 2.4 m (ii) 2.7 m (iii) 2 m (iv) 2.2 m
(d) What will be the horizontal distance of the fly from her after 12 seconds?
(i) 2.7 m (ii) 2.78 m (iii) 2.58 m (iv) 2.2 m
(e) The given problem is based on which concept?
(i) Triangles
(ii) Co-ordinate geometry
(iii) Height and Distance
(iv) None of these
Solutions:
a. AC2 = AB2 + BC2
[ By Pythagoras theorem ]
⇒ AC2 = (1.8)2 + (2.4)2
⇒ AC2 = 3.24 + 5.76
⇒ AC2 = 9
⇒ AC = 3m
b. She pulls the string at the rate of 5cm/s
∴ String pulled in 12 second = 12 × 5 = 60cm
= 0.6m
c. Length of string out after 12 second is AP.
⇒ AP = AC – String pulled by Nazima in 12 seconds.
⇒ AP = (3 − 0.6)m
=2.4m
d. In △ADB, AB2 + BP2=AP2 ⇒ (1.8)2 + BP2 = (2.4)2
⇒ BP2 = 5.76 − 3.24 ⇒ BP2 = 5.76 − 3.24
⇒ BP2 = 2.52 ⇒ BP=1.58 m
Horizontal distance of fly = BP + 1.2m
Horizontal distance of fly =1.58m + 1.2m
∴ Horizontal distance of fly = 2.78m
e. Triangles
error: Content is protected !!<|endoftext|>
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# Equations of Circles (x – a)2 + (y – b)2 = r2
## Presentation on theme: "Equations of Circles (x – a)2 + (y – b)2 = r2"— Presentation transcript:
Equations of Circles (x – a)2 + (y – b)2 = r2
(a, b) = center of the circle r = radius of the circle (x – 4)2 + (y + 3)2 = 36 What is the center and the radius of this circle? (4, -3) = center of the circle 6 = radius of the circle
Let’s Look at the Graph of (x – 4)2 + (y + 3)2 = 36
radius of the circle is 6 center of the circle is (4, -3)
Equations of an Ellipse
Standard Form Intercept Form The center is the origin The center is the origin The center is the origin The center is the origin
Let’s Look at the Graph of 16x2 + 36y2 = 576
center of the ellipse is the origin x-intercepts are 6 y-intercepts are 4
Rewriting the Equation of an Ellipse
Rewrite the following equation of an ellipse in intercept form. Rewrite the following equation of an ellipse in standard form. Just switch the coefficients of x and y. Since the equation must be equal to 1, divide by 400. Multiply by the common denominator. Simplify the fractions. Simplify the equation. I think I can do this without having to do too much work.
Graphing Circles and Ellipses
Graph the following circle. Write the equation of the circle whose graph looks like this. 8 6 4 2 -2 -4 -6 -8 8 6 4 2 -2 -4 -6 -8 8 6 4 2 -2 -4 -6 -8 Graph the following ellipse. Write the equations of the following two ellipses in both standard form and intercept form.
Equations of a Hyperbola
xy = k As x increases, y decreases so that the product of x and y is always k xy = 8 As x increases, y decreases so that the product of x and y is always 8 A hyperbola is a function The coordinate axes are asymptotes of the graph When k > 0, the graph is in quadrant I and quadrant III When k < 0, the graph is in quadrant II and quadrant IV Each branch of the hyperbola is the reflection of the other in the origin
Let’s Look at the Graph of xy = 8
Since k > 0, the graph is in quadrant I and quadrant III The x and y axes are asymptotes of the graph
Let’s Look at the Graph of xy = -8
Since k < 0, the graph is in quadrant II and quadrant IV The x and y axes are asymptotes of the graph
Graphing Hyperbolas Solve the equation for y. Make a table of values.
x y x y Since k is positive, the graph is in quadrants I and III Since k is negative, the graph is in quadrants II and IV
More Graphing Hyperbolas
Solve the equation for y. Make a table of values. x y x y Since k is positive, the graph is in quadrants I and III Since k is positive, the graph is in quadrants I and III
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# 11-7 Multiplying Integers Warm Up Find each product. 1. 8 42. 7 12 3. 3 94. 6 5 5. 80 66. 50 6 7. 40 908. 20 700 32 84 27 30 480 300 3,600 14,000.
## Presentation on theme: "11-7 Multiplying Integers Warm Up Find each product. 1. 8 42. 7 12 3. 3 94. 6 5 5. 80 66. 50 6 7. 40 908. 20 700 32 84 27 30 480 300 3,600 14,000."— Presentation transcript:
11-7 Multiplying Integers Warm Up Find each product. 1. 8 42. 7 12 3. 3 94. 6 5 5. 80 66. 50 6 7. 40 908. 20 700 32 84 27 30 480 300 3,600 14,000
11-7 Multiplying Integers Learn to multiply integers.
11-7 Multiplying Integers –66Product –(2 + 2 + 2) 2 + 2 + 2Addition the opposite of 3 groups of 2 Words –3 23 2Numbers
11-7 Multiplying Integers 6–6Product –[(–2) + (–2) + (–2)](–2) + (–2) + (–2) Addition the opposite of 3 groups of –2 Words –3 (–2) 3 (–2) Numbers
11-7 Multiplying Integers Additional Example 1: Multiplying Integers Find each product. A. 5 2 B. 4 (–5) 5 2 = 10Think: 5 groups of 2. 4 (–5) = –20 Think: 4 groups of –5. To find the opposite of a number, change the sign. The opposite of 6 is –6. The opposite of –4 is 4. Remember!
11-7 Multiplying Integers Additional Example 1: Multiplying Integers Find each product. C. –3 2 D. –2 (–4) –3 2 = –6 Think: the opposite of 3 groups of 2. –2 (–4) = 8 Think: the opposite of 2 groups of –4.
11-7 Multiplying Integers Check It Out: Example 1 Find each product. A. 3 4 B. 2 (–7) 3 4 = 12Think: 3 groups of 4. 2 (–7) = –14 Think: 2 groups of –7.
11-7 Multiplying Integers Check It Out: Example 1 Find each product. C. –5 3 D. –4 (–6) –5 3 = –15 Think: the opposite of 5 groups of 3. –4 (–6) = 24 Think: the opposite of 4 groups of –6.
11-7 Multiplying Integers MULTIPLYING INTEGERS If the signs are the same, the product is positive. 4 3 = 12 –6 (–3) = 18 If the signs are different, the product is negative. –2 5 = –10 7 (–8) = –56 The product of any number and 0 is 0. 0 9 = 0(–12) 0 = 0
11-7 Multiplying Integers Additional Example 2: Evaluating Integer Expressions Evaluate –7x for each value of x. A. x = –3 B. x = 5 –7xWrite the expression. –7 (–3) Substitute –3 for x. 21 The signs are the same, so the answer is positive. –7xWrite the expression. –7 5Substitute 5 for x. –35The signs are different, so the answer is negative. –7x means –7 x. Remember!
11-7 Multiplying Integers Check It Out: Example 2 Evaluate –4y for each value of y. A. y = – 2 B. y = 7 –4yWrite the expression. –4 (–2) Substitute –2 for y. 8The signs are the same, so the answer is positive. –4yWrite the expression. –4 7Substitute 7 for y. –28The signs are different, so the answer is negative.
11-7 Multiplying Integers Learn to divide integers.
11-7 Multiplying Integers Multiplication and division are inverse operations. To solve a division problem, think of the related multiplication. To find the mean of a list of numbers: 1. Add all the numbers together. 2. Divide by how many numbers are in the list. Remember!
11-7 Multiplying Integers Additional Example 1: Dividing Integers Find each quotient. A. –30 ÷ 6 B. –42 ÷ (–7) Think: What number times 6 equals –30? –5 6 = –30, so –30 ÷ 6 = –5. Think: What number times –7 equals –42? 6 (–7) = –42, so –42 ÷ (–7) = 6.
11-7 Multiplying Integers Check It Out: Example 1 Find each quotient. A. –15 ÷ 5 B. –36 ÷ (–6) Think: What number times 5 equals –15? –3 5 = –15, so –15 ÷ 5 = –3. Think: What number times –6 equals –36? 6 (–6) = –36, so –36 ÷ (–6) = 6.
11-7 Multiplying Integers Dividing Integers If the signs are the same, the product is positive. 24 ÷ 3 = 8 –6 ÷ (–3) = 2 If the signs are different, the quotient is negative. –20 ÷ 5 = –4 72 ÷ (–8) = –9 Zero divided by any integer equals 0. = 0 = 0 0 14 __ 0 –11 __ You cannot divide any integer by 0. Because division is the inverse of multiplication, the rules for dividing integers are the same as the rules for multiplying integers.
11-7 Multiplying Integers Additional Example 2A: Evaluating Integer Expressions Evaluate for each value of d. d = 16 Write the expression. = 16 ÷ 4 Substitute 16 for d. = 4 The signs are the same, so the answer is positive. d 4 __ d 4 16 4 __
11-7 Multiplying Integers Additional Example 2B: Evaluating Integer Expressions Evaluate for each value of d. d = –24 Write the expression. = –24 ÷ 4 Substitute –24 for d. = –6 The signs are different, so the answer is negative. d 4 __ d 4 -24 4 ___
11-7 Multiplying Integers Additional Example 2C: Evaluating Integer Expressions Evaluate for each value of d. d = –12 Write the expression. = –12 ÷ 4 Substitute –12 for d. = –3 The signs are different, so the answer is negative. d 4 __ d 4 -12 4 ___
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A real-time system includes hardware and software components that enable precise control over the execution of your code. You use a PC to develop code for a real-time system.
The following figure shows a basic setup for developing a real-time application.
- Development PC—The PC manages connections between devices in the system and provides a graphical environment to create and edit real-time code. When code runs on a real-time controller, you can use the PC as the user interface to modify VI panels and view data from the controller.
- Real-time controller—A real-time controller runs a Linux real-time operating system and software that allows you to set precise timing directives and deterministic execution for real-time code. The real-time controller can also provide precise timing when communicating with FPGAs and I/O hardware. For example, if you need a set of FPGAs to each perform a unique data processing task in a particular order, you can use a real-time controller to communicate directly with the FPGAs to guarantee that each FPGA in the sequence receives instructions at a precise time.
- Real-time system—A real-time system includes the chassis, real-time controller, and other modules in the chassis. You establish a connection between the real-time system and the development PC to access the real-time controller and other modules in the chassis.
- Network connection—A PC and a real-time system must connect to the same network so the PC and real-time controller can communicate with each other. Over this network connection, the PC deploys code to the real-time controller and acts as a live graphical user interface for the real-time controller.<|endoftext|>
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The history of Lutheranism in Europe is generally divided into several distinct periods. The first period, from 1520 to 1580, was one of doctrinal consolidation. Doctrinal disputes, especially that concerning antinomianism , began during Luther's lifetime, but became more heated after his death, when the controversy raised by Andreas Osiander over the meaning of Christ's death on the cross shook the whole German Evangelical Church. The opposing factions were the strict Lutherans, who refused any compromise with Rome or Calvinism, and the moderate wing, headed by Philip Melanchthon , who strove for reconciliation.
The period from 1580 to 1700 was called
the age of orthodoxy. Almost exclusive emphasis was put on right doctrine, and faith was understood as intellectual assent. During the early years of the 17th cent., Germany was racked by the Thirty Years War , and Lutheranism lost much of its territory. Religious boundaries were stabilized by the Peace of Westphalia (1648), which maintained that with slight exceptions the religion of the prince was to be the religion of his subjects. The latter part of the century saw a reaction against the prevailing orthodoxy in the form of Pietism .
In 1817, Frederick William III of Prussia sought to merge forcibly the Lutheran and Reformed churches of Prussia into a single organization called the Prussian Union. Some conservative Lutherans opposed this move and withdrew from the union to found the Evangelical Lutheran Church of Prussia as a free church. After World War I, the churches were no longer governed by state laws but still received state support.
In the unification of German culture under the Nazi regime, the church did not escape. In 1933 a national organization, the German Evangelical Church, was formed. Under the direction of the Nazi party it tried to develop a national racial church, with pure Aryan blood as a prerequisite for membership. A revolt against this movement, led by Martin Niemoeller , resulted in the founding of the Confessing Church and the formation of the Confessional Synod, which issued (1934) its declaration rejecting the Reich's interference with the church.
The end of the war saw the formation of the Evangelical Church in Germany (EKID), which is made up of members of both Lutheran and Reformed churches, and the United Evangelical Lutheran Church of Germany (VELKD), which functions as an expressly Lutheran constituency within the EKID. German churches have also cooperated wholeheartedly in the formation of the Lutheran World Federation (1947) and the World Council of Churches. The Lutheran Church is the established state church of Denmark, Iceland, Norway, and Finland; Sweden disestablished its Lutheran state church in 2000.
In North America
In North America, Lutherans from the Netherlands were among the settlers on Manhattan island in 1625. A congregation was formed there in 1648, but it was antedated by one established (1638) by Swedish settlers at Fort Christina (Wilmington) on the Delaware River. On nearby Tinicum Island the first Lutheran church building in the country was dedicated in 1646. Early in the 18th cent. exiles from the Palatinate established German Lutheran churches in New York, Pennsylvania, Delaware, and Maryland. The Salzburger migration to Georgia (1734) introduced Lutheranism in the South.
In the 18th cent., organization of the churches was begun by Heinrich Melchior Mühlenberg , who brought about the formation (1748) in Pennsylvania of the first synod in the country. The Synod of New York and adjoining states followed (1786); that of North Carolina was created in 1803. With the settlement of the Midwest, the West, and the Northwest, many small synods were formed by Norwegians, Danes, Finns, and other national groups.
Once there were about 150 distinct Lutheran bodies, but in 1918 many of the autonomous Lutheran bodies merged into the United Lutheran Church of America. The Evangelical Lutheran Synodical Conference of North America, formed in 1872, broke up in 1960, when the Wisconsin Evangelical Lutheran Synod (with almost 400,000 members, now the third largest Lutheran group in the United States) withdrew. The Lutheran Church–Missouri Synod, with some 2.5 million members, was also formerly part of the Evangelical Lutheran Synodical Conference of North America. It is now the second largest group of Lutherans. The American Lutheran Church, formed in 1961, and the Lutheran Church in America, formed in 1962, united to become the Evangelical Lutheran Church in America in 1988, now the largest Lutheran group, with nearly 4.8 million members. These groups comprise about 95% of North American Lutherans. In an ecumenical spirit, the Evangelical Lutheran's Churchwide Assembly agreed (1997) on a full communion with the Presbyterian Church (USA), the United Church of Christ, and the Reformed Church in America, and it reached a similar agreement with the Episcopal Church and the Moravian Church in 1999.
The Columbia Electronic Encyclopedia, 6th ed. Copyright © 2012, Columbia University Press. All rights reserved.
See more Encyclopedia articles on: Protestant Denominations<|endoftext|>
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(CN) – New findings that may explain why the North American ice sheet melted during one of the coldest periods of the last Ice Age also add to mounting evidence that climate change could lead to greater sea level rise than most existing models predict.
In a study published Thursday in the journal Nature, researchers from the University of Michigan show how small spikes in ocean temperature – rather than air temperature – likely fueled the eventual disintegration of the Laurentide Ice Sheet, which used to cover much of North America.
What caused the ice sheet to melt has puzzled scientists for decades, as the period in which it melted and splintered into the sea aligned with some of the coldest times in the last Ice Age, which ended about 10,000 years ago.
“We’ve shown that we don’t really need atmospheric warming to trigger large-scale disintegration events if the ocean warms up and starts tickling the edges of the ice sheets,” said lead author Jeremy Bassis.
The same mechanism is likely at work today on the Greenland ice sheet – and possibly Antarctica, according to the team.
“It is possible that modern-day glaciers, not just the parts that are floating but the parts that are just touching the ocean, are more sensitive to ocean warming than we previously thought.”
Scientists identified this potential mechanism in part due to a new, more accurate method to mathematically describe how ice breaks and flows, which Bassis developed several years ago. His model has enabled a deeper understanding of how ice could react to spikes in ocean or air temperature, and how such changes may affect sea level rise.
Other researchers used the model to predict in 2016 that melting Antarctic ice could increase sea levels by more than three feet. The previous estimate projected Antarctica would only contribute a few centimeters to sea level rise by 2100.
For the new study, Bassis and his team applied a version of this model to the climate of the last Ice Age, using ocean-floor sediment and ice-core records to estimate variations in water temperature. Their goal was to see if what’s happening today in Greenland could explain the behavior of Laurentide.
The collapse of the ancient ice sheet is an example of periods of rapid ice disintegration known as Heinrich events, which can be seen in sediment cores across the North Atlantic basin. The unusual sediment layers are what allowed scientists to initially identify Heinrich events.
“Decades of work looking at ocean sediment records has shown that these ice sheet collapse events happened periodically during the last Ice Age, but it has taken a lot longer to come up with a mechanism that can explain why the Laurentide Ice Sheet collapsed during the coldest periods only,” co-author Sierra Petersen said.
“This study has done that.”
The team set out to understand the size and timing of Heinrich events from the last Ice Age, creating simulations that were able to predict both. They even identified an additional Heinrich event that had previously been missed.
The model also factors in how the Earth’s surface reacts to the weight of the ice on top of it. Heavy ice depresses the planet’s surface, even pushing it below sea level – the point at which ice sheets are most vulnerable to warm ocean temperatures.
“There is currently large uncertainty about how much sea level will rise and much of this uncertainty is related to whether models incorporate the fact that ice sheets break,” Bassis said. “What we are showing is that the models we have of this process seem to work for Greenland as well as in the past, so we should be able to more confidently predict sea level rise.”
Bassis added that portions of Antarctica have similar geography to Laurentide.
“We’re seeing ocean warming in those regions and we’re seeing these regions start to change,” he said. “What we saw in our simulations is that just a small amount of ocean warming can destabilize a region if it’s in the right configuration, and even in the absence of atmospheric warming.”<|endoftext|>
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During the Cold War, the Warsaw Pact armies studied the disposition of rivers in western Europe. It concluded that they would have to cross water obstacles up to 100m wide every 35 to 60km. Every 100-150km, they would encounter a water obstacle between 100m and 300m wide. Every 250-300km they would encounter one that was wider still. In a war in western Europe, the Warsaw Pact armies expected to advance an average of 100km per day, leading to a significant number of river crossings. Since there was no guarantee of securing bridges intact, the Warsaw Pact put a great deal of emphasis on their ability to cross water obstacles. A range of equipment was created to bridge gaps or ferry vehicles over rivers. Echographs were developed, that could quickly measure water depth and river width. Many light armoured vehicles could swim. Main battle tanks carried snorkels that allowed them to wade through water up to 5m deep.
It seems likely that the armies of the Warsaw Pact preferred not to snorkel main battle tanks, and that ferries or pontoon bridges were preferred. None the less, a 1971 British Army intelligence report stated that the Soviet army considered it “a practical operation of war” (Army Technical Intelligence Review, April 1971). Every tank crew was fully trained in snorkelling. Training took place on purpose-built sites, with good facilities. Emphasis was placed on giving the crew confidence in their ability to snorkel well and safely.
Training was split into two phases. The first phase, lasting up to two months, concentrated on preparing the crews to operate tanks under water. Training covered swimming, diving, carrying out procedures underwater whilst wearing escape masks. There was a good deal of safety training, which helped with crew confidence and morale. Rescue operations were practised on simulators. Crews were not allowed to move onto the second phase until they had passed the first phase.
In the second phase of training, the crew got the chance to put their skills to the test. A five-meter deep lake would be used to practice driving underwater. Initially, drivers would drive 90m underwater, progressing to 150m as their skills improved. At least some sites also had facilities for blind driving, with the driver guided only by the tank’s gyro compass. After passing this second phase of training, crews would join their units.
Sealing and preparing a tank for snorkelling could take as little as 15 minutes. Older tanks took longer, up to half an hour. This would be done in a concealed area 3-5km from the river. When the tanks got to within 1-2km of the river, snorkels would be fitted. Tanks crossed at slow speed, in a column formation, with a 30m gap between vehicles. Drivers would not change gear or stop while in the water. Once across, the tank would have to stop while the crew removed the waterproofing. Until this was done, the turret could not be traversed or the gun fired. If a tank stalled in the river, the crew would flood it before escaping through the hatches.
An alternative method of snorkelling was to use winches to pull unmanned tanks across. A pair of armoured recovery vehicles would set up on the far bank with a pulley block and anchoring unit. Up to three tanks could be pulled across the river simultaneously. Using this method, 10-tank company could cross a 200m wide river in 35 minutes, assuming the tanks had already been sealed. The crews would cross separately, in amphibious vehicles such as APCs, or on boats.
Snorkelling was not to be carried out under fire, and in some cases, simply wouldn’t be possible. The entry bank had to be less than 25º, the exit bank less than 15º, and the current no more than three metres per second. In winter, drifting ice could damage the snorkel. The river bottom had to be reasonably firm, and free of boulders and craters.
Many lighter AFVs were amphibious, and could swim across water obstacles. All APCs from the BTR-50 onwards, and all IFVs, were fully amphibious. The BRDM series of reconnaissance vehicles, and even some self-propelled artillery and AA vehicles could swim. Most of these were propelled in the water by their tracks, although some used water jets, which gave a better performance in the water.
Water obstacles would only be crossed under fire as a last resort. In these cases, a great deal of artillery would be called upon to support the operation. If at all possible, helicopter troops would be landed on the far bank, and attack simultaneously with the crossing. Tanks would stay on the near bank to provide covering fire, while amphibious armoured vehicles swam across. Once a bridgehead was established, tanks and other vehicles would snorkel or be ferried across. These would then continue the advance, while engineers worked to build more permanent bridge crossings over the obstacle.
Soviet estimates found that two-thirds of the river obstacles they would encounter in Europe were less than 20m wide. This led to the development of vehicle-launched bridges capable of quickly crossing these gaps. The Polish army developed a tracked bridge, which was pushed into place by a tank. Small numbers of a T-34 based bridging tank were delivered to the Soviet army in 1957. This was soon superseded by the MTU-54, also sometimes referred to as the MTU or MTU-1.
In 1955, the MTU-54, a bridge layer based on a T-54 chassis, was introduced. The MTU-54 mounted a simple 12.3m bridge, carried horizontally on top of the vehicle. Unlike later vehicles, the bridge was not folded for transit. A chain drive mechanism moved the bridge forwards to launch, before being lowered into place. This method had the advantage of keeping the silhouette low whilst launching the bridge. The MTU-54 could bridge an 11m gap, and had a load capacity of 50 tonnes. Launch time was three to five minutes, and recovery can take place from either end.
Unusually, it was fitted with a DShKM machine gun, mounted in the centre of the vehicle. This had to be removed before launching the bridge. Later vehicles were fitted with a deep wading snorkel, NBC protection, and automatic fire supression system.
From 1967, the MTU-20 became the primary Soviet tank-launched bridge. This mounted a bridge on a T-55 chassis. In order to allow a longer span length whilst maintaining a low silhouette when launching, the ends of the bridge folded back when in transit. When launching the bridge, a stabiliser at the front was lowered. The ends of the bridge were then unfolded and the bridge rolled forward, before being lowered into place. The MTU-20 had a span length of 20m, with a load capacity of 60 tonnes. Launching the bridge took five minutes, recovery from either end took between five and seven minutes. Both launching and recovery could be carried out while the crew remained inside the vehicle. It was fitted with a deep wading snorkel, NBC protection, and automatic fire supression system.
The non-Soviet Warsaw Pact armies showed a preference for the more common scissor bridge design. Poland and East Germany jointly developed the BLG-60, which mounted a 50 tonne, 21.6m scissor bridge on a T-55 chassis. The bridge was launched by being lifted up to the vertical, then unfolded and lowered over the gap. This design gave a quicker launch time than the Soviet designs, at the expense of a very high silhouette during launch. The BLG-60 was fitted with NBC protection and a deep-wading snorkel. An improved version, the BLG-67, was introduced in the late 1970s.
Czechoslovakia also built its own bridge layer, the MT-55A. Like the BLG-60, this mounted a scissor bridge on a T-55 chassis. A front spade stopped the vehicle being tipped over by the weight of the bridge. Launch time was two to three minutes, recovery time five to six minutes. Both tasks could be carried out from inside the vehicle. It could span an obstacle of up to 18m, and load capacity was 50 tonnes. It had a gap measuring device and inclinometer to help with finding a suitable site for the bridge. Other equipment included infra-red night vision equipment, snorkel, automatic fire extinguisher, and NBC protection. Unusually, the Soviet army adopted the MT-55A, albeit in small numbers.
Initially, the scissor bridge carried by the MT-55A had circular holes in the sides of the bridge. Later models had solid sides. Multiple bridges could be combined to span larger gaps.
In 1974, a new bridge layer entered service, the MTU-72. These were made from existing T-72 tanks with the turret replaced with a bridge-launching mechanism. The bridge was of cantilever design, similar to that on the MTU-20, but made of an aliminium alloy. The 20m bridge had a load capacity of 50 tonnes and could span a gap of up to 18m. A second bridge could be launched from the first one to span a gap of up to 35m. Launching the bridge took three minutes, recovery eight minutes. Both launching and recovery could be carried out while the crew remained inside the vehicle.
A blade is fitted to the front of the hull, which is primarily intended to stabilise the vehicle during launching and recovery. It can also be used as a bulldozer blade. A deep wading snorkel, NBC protection system, fire suppression system, and thermal smoke generation unit are also fitted.
Amphibians & Ferries
The K-61 tracked amphibious ferry was introduced in 1950, a direct result of wartime experience with DUKW amphibious lorries supplied by the USA. It remained in Soviet service until the late 1960s. It was fully tracked, and based on a light AFV chassis, but was not armoured. The K-61 could carry eight wounded on stretchers, 40 fully-armed infantry, up to five tonnes of equipment, a lorry up to 2.5 tonnes, or an artillery piece.
Loading is done via ramps at the rear. Vehicles can be simply driven on, and a winch is provided for loading heavy equipment such as artillery pieces. It is driven through the water by a pair of propellers, at a speed of up to 10km/hour.
The PTS-M was introduced in 1966, as a replacement for the K-61. Larger than the K-61, it also had an amphibious trailer, although this was found to be of little use except in very calm waters. A more powerful engine meant that it could carry a greater load, 7.5 tonnes on land and 15 tonnes on water. Rear ramps are used to load cargo, with a winch for loading heavy non-motorised loads. An infra-red searchlight and infra-red driving lights were fitted.
The PTS-2 was a much improved and modernised version of the PTS-M, introduced in the late 1970s. It had a new suspension, derived from the MT-T artillery tractor. A pair of propellers drive the vehicle in the water. Load was the same as the PTS-M. The cab is fitted with NBC protection.
The GSP heavy amphibious ferry was introduced in 1959. A single ferry was made up of two distinct units, one left and one right. The two units were mirror images of each other, and not interchangeable. Before entering the water, a trim vane was erected at the front of the hull. The two units then entered the water separately and linked up. Once linked, the pontoons, which were inverted while in transit, were swung upright, and the treadways deployed. It was important that the pontoons were both unfolded together, to avoid overbalancing the whole. Assembly time was 6-10 minutes.
The ferry had hydraulically operated ramps at each end, allowing vehicles to be driven on at one end and off at the other. Maximum capacity was 52 tonnes, enough to carry a main battle tank. It was reported that tanks could fire their main armament whilst on the ferry, in good conditions.
The vehicle itself was tracked, with a suspension similar to the PT-76 light tank. It had infra-red driving lights, although these were only for use on land. The hull and pontoon was lightweight steel filled with plastic foam. The foam increased bouyancy and reduced vulnerability to enemy fire. Propulsion in the water was provided by four propellers (two per vehicle), mounted in tunnels under the hull. Maximum speed in the water was 8km/hour.
Initially known to NATO as the ABS(T), the PMM-2 was introduced in 1974 as a replacement for the GSP ferry, and possibly the PMP pontoon bridge. It is based on a BAZ-5937 chassis. Two aluminium folded pontoons are mounted on top, with entrance ramps. As the vehicle enters the water, the pontoons are hydraulically unfolded to either side of the vehicle, with the vehicle itself forming a centre section. It is propelled in the water by water jets.
As a ferry, units can be used individually (40 tonne capacity), in pairs (80 tonne capacity), or in threes (120 tonne capacity). A single vehicle can be used as a bridge, to span gaps of up to 17m. Up to ten vehicles can be combined to span larger gaps, with no need for bridging boats.
Most river crossings would have been assault crossings from the march, at sites that were only lightly defended, if at all. Reconnaissance patrols of up to platoon size, operating up to 50km ahead of the main body and equipped with specialised equipment, would find suitable sites. When a crossing site had been selected, a forward detachment, two to three hours ahead of the main body and avoiding contact with the enemy, would secure the site. A typical forward detachment would consist of a motor rifle battalion with an attached tank company and artillery battalion. Amphibians, ferries, air defence, anti-tank and chemical defence units would also be attached. Heliborne or occassionally airborne troops could also be used in this role.
If the crossing site was defended, the attack would be carried out with significant artillery and air support. River crossings got priority for air support, and were considered particularly vulnerable to enemy air attack. Air defence assets would be deployed close to the crossing site, and would cross the river as soon as feasible to extend their coverage.
The crossing itself would be carried out by APCs or IFVs swimming across the river, supported by tank and artillery fire from the near shore. A few tanks may have crossed in the first wave, but most would provide fire from the near bank and cross later. Artillery and anti-tank units would cross immediately after the infantry to provide support in holding the bridgehead. Tanks would cross using ferries, snorkelling, or bridges.
In the event of war in Western Europe, the Warsaw Pact expected their armies to advance quickly. The rivers of West Germany could not be allowed to slow the advance, and so considerable effort was put into developing equipment for their engineers. This equipment was simple, rugged, and supplied in significant quantities. Both peacetime exercises and experience in Afghanistan demonstrated that Soviet combat engineers were an effective force.<|endoftext|>
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The Tudor Rose
Although peace was still a bit unsettled, a beginning had been made. This window commemorates the start of that partnership, and the shift that was made towards the end of hostilities.
The War of the RosesThe War of the Roses had made enemies of the House of York and the House of Lancaster.
Upon the ending of their quarrel, the Tudor Rose became the symbol of unity.
The House of Lancaster, the ruling family, had a solid red rose which they rallied around.
The House of York, the challengers, surrounded themselves with a solid white rose.
For thirty years the conflict for the Throne of England threw good fortune towards one family, and then threw it towards the other.Eventually, the House of York won the last battle and the white rose was prominent.
The wheel of fortune spun just one more time, however, and with the death of Richard III, the House of Lancaster took the crown.
The red rose had come out on top.
Henry Tudor, House of Lancaster, had sworn that he would marry the eldest daughter of the House of York if he ascended the throne.
In 1485, just a few miles west of Leicester, Lancaster defeated York. One year later, Henry married Elizabeth.
The Tudor Rose
The Tudor dynasty had begun, and the single Tudor rose was born.
Taking the white rose of York as the center of the flower, and the red rose of Lancaster as the outside edging, the union of these two warring families had a beautiful symbol denoting unity and mutual regard.
Lovely examples of Tudor roses can be easily found today. Kings College Chapel, Cambridge and Hampton Court Palace have Tudor roses sprinkled everywhere. They can also be found in North Wales, on ancestral Tudor lands.
Article by "Tudor Rose"
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Unified modeling language
- “The Unified Modeling Language (UML) is a non-proprietary, object modeling and specification language used in software engineering. UML includes a standardized graphical notation that may be used to create an abstract model of a system: the UML model.” (Wikipedia:Unified Modeling Language).
- UML is not a method by itself, however it was designed to be compatible with any sort of object-oriented software development methods. As such it can be used to describe almost any sort of information processing architecture (including what learners do since learners can be modelled in terms of human information processing or what happens in an organization since an organization can be described in terms of information flows and procedures).
2 Modeling and diagram types
With UML you can model most every phase and object of the software development process. Technically speaking, a UML model consists of elements such as packages, classes, and associations. The corresponding UML diagrams are graphical representations of parts of the UML model. UML diagrams contain graphical elements (nodes connected by paths) that represent elements in the UML model.
According to Koper (2004), UML provides a collection of models and graphs to describe the structural and behavioural semantics of any complex information system.
There are two version of UML in use: UML 1.4 and UML 2. UML 2 models are more complex and their semantics may have changed. Some of the models provided in UML 1.4 are:
- UML use case models and scenario's to capture the user requirements and functionality of the system. Scenarios are instances of use cases.
- Class and object diagrams to specify the semantic information structure of a system. Object diagrams are instances of class diagrams.
- UML State diagrams to describe the dynamic behaviour of an object in a system.
- UML Interaction diagrams (sequence and collaboration diagrams) to model how groups of objects collaborate in some behaviour.
- UML Physical diagrams (deployment and component diagrams) to model the implementation structure of a system.
In UML 2.0 there are 13 types of diagrams. Some, like activity diagrams, are quite different from UML1.4, others (like use case) less. The specification provides the following taxonomy:
“Structure diagrams show the static structure of the objects in a system. That is, they depict those elements in a specification that are irrespective of time. The elements in a structure diagram represent the meaningful concepts of an application, and may include abstract, real-world and implementation concepts. For example, a structure diagram for an airline reservation system might include classifiers that represent seat assignment algorithms, tickets, and a credit authorization service. Structure diagrams do not show the details of dynamic behavior, which are illustrated by behavioral
diagrams. However, they may show relationships to the behaviors of the classifiers exhibited in the structure diagrams.
Behavior diagrams show the dynamic behavior of the objects in a system, including their methods, collaborations, activities, and state histories. The dynamic behavior of a system can be described as a series of changes to the system over time. (OMG-UML, 2005: 684).”
- UML behavioral modeling (behavior diagrams) allow to model what processes and activities must happen in the system being modeled
- UML structural modelling (structure diagrams) allow to model components of a system
- UML class diagram
- Composite structure diagram
- Component diagram
- Object diagram
- Package diagram
3 UML in education
- Design of educational software
Since UML is general formalism to describe information processing phenomena (like what people do, how systems are built, how programs interact etc.), UML can be used for educational software design, e.g. see Fle3's UML diagrams or Giesbers et al. (2007).
- Description of pedagogical scenarios
- Some researchers use UML to describe pedagogical scenarios, see also educational modelling languages
- Definition of pedagogical modeling languages
- Some educational modelling languages are also described as UML diagram, e.g. the semantic information model of IMS Learning Design (and former EML) has been expressed in UML. “The semantic, conceptual model has been expressed as a series of UML models, from which several bindings were generated automatically. E.g. for the IMS Learning Design specification a XML schema has been derived that keeps the semantics in the tag-names. However other bindings (RDF Schema/OWL, Topic Maps, SGML schema's, relational database schema's) could in principle be generated as well. This implies that the UML model is the dominant part of the specification; it captures the semantic structure and allows other representations to be generated from it. [.... ] It is expected that the semantic model underlying LD, as expressed in UML, is a critical component for the realisation of the Educational Semantic Web, because it provides a tested, generic and (within the IMS community) accepted semantic notation. Whether this model is implemented in XML, RDF-Schema, OWL,Topic Maps etc. depends which tools and technologies are used at any moment in time.” (Koper, 2004).
- coUML is a multi-purpose modeling language that can be used to design courses from global to detailed level.
- Definition of pedagogical use cases
- See use case
- Pedagogical workflow
4 Software tools
See Wikipedia's list of UML tools
- UML Drawing tools
- Violet - SourceForge site. Java-based UML editor. Tested: it works fine, but is not feature complete. It can be used for training and not too advanced diagrams. E.g. objects are missing from activity diagrams.
- ArgoUML ArgoUML is the leading open source UML modeling tool and includes support for all standard UML 1.4 diagrams. (Note: Some UML 2.0 languages are quite different)
- There exist some UML editor extensions to the Eclipse IDE, e.g. Violet UML Editor (not tested)
- Lucidchart is a SaaS that supports UML shape libraries
- UML Drawing plus code generators
- There exist many commercial tools
- Magic draw is popular and has an academic free version (for teaching, upon request only). There used to be community edition which is no longer available it seems.
- General-purpose drawing tools with UML support
- DIA - DIA (Wikipedia). Dia is free software/open source general-purpose diagramming software and has special objects to draw UML diagrams. Can export to various formats.
- UML drawings generators
- MetaUML, UML for LaTeX/MetaPost. Implements a subset of UML 2.0, i.e. curently (June 2008) subsets for Class, activity, use case, state machine and package diagrams. Important: Consult the manual since examples do not show the full potential of MetaUML.
- Extension:UML. Mediawiki extension (installed in this wiki).
- UML-based code generators
- BOUML (not tested)
5.1 Official / Standards
- Object Management Group (OMG)
- UML® Resource Page, a OMG web site.
- Unified Modeling Language (UML), version 2.1.2. There are two specifications that comprise the UML 2.1.2 specification: Superstructure and Infrastructure. There are also two specifications that relate to the UML2 specification (Diagram Interchange and Object Constraint Language).
5.2 Overview articles
- Unified Modeling Language (Wikipedia)
- Glossary of Unified Modeling Language terms
- List of UML tools (Wikipedia)
- UML tool (Wikipedia)
- Glossary of UML terms (Wikipedia)
- UML Quick Reference Card by Laurent Grégoire (2001).
- Practical UML: A Hands-On Introduction for Developers. Short tutorial by Randy Miller, CodeGear.
- Sparx UML Tutorials A series of introductions from SparxSystems.
- Introduction to the Diagrams of UML 2.0 by Scott W. Ambler, AgileModeling.com. A series of tutorials for the 13 UML 2 diagrams.
- Conrad Bock's articles on UML 2.0, e.g.
5.4 UML websites
- Uml bliki by Martin Fowler.
- UML® Resource Page, a OMG web site (Includes pointers to other web sites)
- Architecture and Design: Unified Modeling Language (UML) from Cetus-links.org. Good meta-index.
- Unified Modeling Language at IBM.
- Lucidchart UML resource page
- See all other UML-related articles. You will find examples in the articles and through links.
- UML diagrams of FLE3.
- Arlow, J., & Neustadt, I. (2002). UML and the Unified Process, London: Pearson Education.
- Booch, G., Jacobson, I., & Rumbaugh, J. (1998). Unified Modelling Language User Guide, Boston, MA: Addison-Wesley.
- Fowler, M. (2000). UML distilled (3rd ed.). Upper Saddle River, NJ: Addison-Wesley. ISBN 0321193687
- Giesbers, B., van Bruggen, J., Hermans, H., Joosten-ten Brinke, D., Burgers, J., Koper, R., & Latour, I. (2007). Towards a methodology for educational modelling: a case in educational assessment. Educational Technology & Society, 10 (1), 237-247. PDF
- Joosten-ten Brinke, D., van Bruggen, J., Hermans, H., Burgers, J., Giesbers, B., Koper, R., & Latour, I. (in press). Modeling assessment for re-use of traditional and new types of assessment. Computers in Human Behaviour.
- Koper, R. (2004). Use of the Semantic Web to Solve Some Basic Problems in Education: Increase Flexible, Distributed Lifelong Learning, Decrease Teacher's Workload. Journal of Interactive Media in Education, 2004 (6). Special Issue on the Educational Semantic Web. ISSN:1365-893X
- Randy Miller, Practical UML: A Hands-On Introduction for Developers, Borland Developer network. (This is a very short general UML tutorial).
- Donald Bell, UML basics: An introduction to the Unified Modeling Language, IBM Developper Works / Rational Rose. (IBM has a lot of UML and use case tutorials, needs some searching skills)
- OMG-UML (2003). UML Specification, version 1.4. Retrieved October 14, 2003, from http://www.omg.org/technology/documents/formal/uml.htm
- OMG-UML (2005). Unified Modeling Language (UML), version 2.1.2. retrieved 17:00, 5 June 2008 (UTC) from http://www.omg.org/technology/documents/formal/uml.htm
- There are two specifications that comprise the UML 2.1.2 specification: Superstructure and Infrastructure. There are also two specifications that relate to the UML2 specification (Diagram Interchange and Object Constraint Language).<|endoftext|>
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by Sylvia-Monique Thomas, University of Nevada Las Vegas
Brillouin scattering is named after Léon Nicolas Brillouin (1889-1969). The french physicist first predicted the inelastic scattering of light (photons) by thermally generated acoustic vibrations (phonons) in 1922. The soviet physicist Leonid Mandelstam (1879-1944) is believed to have discovered the scattering as early as 1918, but he published it only in 1926. Other commonly used names are Brillouin light scattering (BLS), Brillouin-Mandelstam scattering (BMS), and Brillouin-Mandelstam light scattering (BMLS).
In classical physics, Brillouin scattering can be described as reflection of the incident light from a three-dimensional diffraction grating produced by periodic density variations (refraction index changes) in a material due to propagating sound waves. Since the wave is traveling with the speed of sound, the scattered light frequency changes, too, it experiences a Doppler shift.
The distance between two waves is called wavelength and the amount of time between waves passing is called frequency, which are interrelated by the wave traveling speed. High frequency indicates short wavelength and low frequency indicates long wavelength.
During the approach of a wave at the observer the received frequency is higher than the emitted frequency, it is identical when the wave passes by, and it is lower during the decay of the wave.
When the source of the waves is moving toward the observer, each successive wave is emitted from a position closer to the observer than the previous wave; each wave takes less time to reach the observer. The time between arrivals of waves at the observer is reduced, causing a frequency increase.
While travelling, the distance between waves is reduced; the waves bundle. On the other hand, if the source of waves is moving away from the observer, each wave is emitted from a position farther from the observer; the arrival time between waves is increased, the frequency reduced. The distance between successive wave fronts is increased, the waves spread out.
For waves that propagate in a medium, such as sound waves, the velocity of the observer and of the source are relative to the medium in which the waves are transmitted. The magnitude of the Doppler shift depends on the motion of the source, the motion of the observer, and the motion of the medium.
From a quantum physics perspective, Brillouin scattering is an interaction between an electromagnetic wave and a density wave (photon-phonon scattering). Thermal motions of atoms in a material (e.g., solid, liquid) create acoustic vibrations, which lead to density variations and scattering of the incident light. The scattering is inelastic, which means that the kinetic energy of the incident light is not conserved (as in the case of elastic scattering) but the photon either loses energy to create a phonon (Stokes, kS= kI+ q, ωS = ωI+ ωq), or gains energy by absorbing a phonon (Anti-Stokes, kS= kI- q, ωS = ωI-ωq). Q and ωq are the wave vector and frequency of the phonon, and kS, ωS, kI and ωI are the wave vectors and frequencies of the scattered and incident photon, with 107 Hz < ωq < 1012 Hz.
Inelastic scattering schematic.There are two types of inelastic interaction between incident radiation and the vibrational modes in a crystal, as represented by phonons. In each case, a quantum of energy ho is exchanged. In a Stokes event the energy is transferred to the phonon, and the energy of the scattered radiation is reduced. In an anti-Stokes event, the energy is transferred from the phonon to the radiation, which is thus increased in energy. In the figure, h is Planck's constant; ω circular frequency, k photon wave vector, q phonon wave vector; y scattering angle; n refractive index; νth acoustic velocity. (--- Image courtesy H. Marquardt.
The frequency and path of the scattered light differ from those of the incident light. The magnitude of the photon frequency shift (ωB, Brillouin shift) depends on the wave length of the incident light (λ0), the refractive index (n) of the sample, the angle θ between incident and scattered light, and the phase velocity of the acoustic wave; it is equal to the energy of the phonon and can be used to measure the latter by analyzing a characteristic Brillouin spectrum, where only modes lying close to the Brillouin zone center are investigated (Q=0).
What is the difference to Raman spectroscopy?
Raman scattering also describes the inelastic scattering interaction of light with vibrations of matter. However, while Brillouin scattering involves the scattering of photons from low-frequency phonons, in Raman spectroscopy photons are scattered by interaction with vibrational and rotational transitions in molecules, and the frequency shift and material information are very different. While Brillouin spectroscopy measures the elastic behavior of a sample using an interferometer, Raman spectroscopy determines the chemical composition and molecular structure of a material using either an interferometer or, e.g., a dispersive (grating) spectrometer.
Brillouin spectroscopy is the most frequently used technique for measurements of single-crystal elastic properties of materials of geophysical interest. Elastic properties describe the temporary reversible volume and shape changes that occur when stresses are applied to a mineral and thus allow the calculation of the seismic wave velocities (seismic properties of the Earth) and the change in density that occurs when minerals are under extreme conditions (pressure and/or temperature). Seismic velocities (VP, VS) are related to the density (ρ), bulk (K) and shear (µ) moduli of a material.
The first Brillouin measurements of geological importance were carried out by Don Weidner and colleagues in 1975 [ref. 14]. In combination with the diamond anvil cell, phases that are stable at high pressure and high temperature can be investigated. The first experiments at very high pressure were performed by Bill Bassett and Edward Brody in 1977 [ref. 15]. If in addition combined with synchrotron radiation sound velocities and a material's volume (density) can be studied simultaneously.
The relationship of acoustic velocity (ν), phonon wave vector (q) and phonon frequency (ωq) is given by:
Acoustic velocities are related to elastic moduli and density of the sample, ν2 = c / ρ.
Advantages of Brillouin spectroscopy
- optical, non-contact technique (no coupling of transducers or other devices to the sample are needed to produce an acoustic excitation)
- suitable for measurements on polycrystalline samples and on small transparent or translucent samples of less than 100 µm in size
- can be performed on samples at high pressure and/or high temperature in a diamond anvil cells
Disadvantages of Brillouin spectroscopy
- only transparent samples can be measured (scattering efficiency of opaque samples is very low)
- low-temperature measurements are not feasible (scattering intensity is very low due to a decreasing number of vibrations with decreasing temperature)
What does a Brillouin system look like?
Interferometry is based on wave superposition. A Fabry-Pérot interferometer in particular is made of two parallel mirrors, where multiple light reflections between the two surfaces infer light wave interference. When two waves with the same frequency combine, the resulting pattern is determined by the phase difference between the two waves. Depending on the wavelength of the light, the angle the light travels between the reflecting surfaces, the thickness of the mirrors, and the refractive index of the material waves that are in phase will experience constructive interference (transmission peak maximum), while waves that are out of phase will experience destructive interference (transmission peak minimum). In an interferometer a single incident beam of coherent light is split into two beams by a grating or a partial mirror. The two beams will travel a different a path until they are recombined before arriving at a detector. The difference in the distance traveled by each beam creates a phase difference between them, which creates an interference pattern.
The intrinsic precision of Brillouin measurements is on the order of 0.1–0.5%. Typical frequencies of acoustic waves probed are in the range of tens of GHz. The accuracy of the elastic moduli determination can be in the order of 1%. In combination with a diamond anvil cell high-pressure Brillouin studies possible (to pressures > 100 GPa). High-temperature measurements can be performed using resistively heated DACs (to ~1700 K) or with CO2 laser heating (up to 3000K).
In the experiment monochromatic laser light is focused on a sample. The transparent sample can be either a well-sintered polycrystalline material or a single-crystal with highly-polished parallel (within a few hundredth of a degree) faces. A fraction of the incident light interacts with acoustic vibrations and induces Stokes and Anti-Stokes scattering. The frequency and intensity of the scattered light are measured by an interferometer.
What does a typical Brillouin spectrum look like?
How to determine elastic properties from a Brillouin experiment?
Elastic moduli of a material are related to the acoustic wave velocities and can be calculated using the following equation:
Angel, R.J., Jackson, J.M., Reichmann, H.J., Speziale, S. (2009) Elasticity Measurements on Minerals: a Review. European Journal of Mineralogy, 21, 525-550.
Bass, J.D. (2007) Theory and Practice – Techniques for Measuring High P/T Elasticity. In: Price, G.D. (Ed.) Treatise on Geophysics, 2, Elsevier BV, Amsterdam, 269-292.
Bass, J.D. & Parise, J.B. (2008) Deep Earth and Recent Developments in Mineral Physics. Elements, 4, 157-163.
Bass, J.D., Sinogeikin, S.V., and Li, B. (2008) Elastic Properties of Minerals: a key for Understanding the Composition and Temperature of Earth's Interior. Elements, 4, 165-170.
Bassett, W.A. (2001) Dynamic Measurements of Elastic Moduli of Samples at High Pressure and Temperature. In: Levy et al. (Eds.) Handbook of Elastic Properties of Solids, Liquids, and Gases, 1, 469-487.
Every, A.G. (2001) The Elastic Properties of Solids: Static and Dynamic Properties. In: Levy et al. (Eds.) Handbook of Elastic Properties of Solids, Liquids, and Gases, vol. 1, 3-36.
Nye, J.F. (1985) The Physical Properties of Crystals: Their Representation by Tensors and Matrices, Oxford, 329 pp.
Anderson O.L. & Isaak D.L. (1995) Elastic Constants of Mantle Minerals at High Temperature. In: Ahrens T.J. (Ed.) Mineral Physics and Crystallography: A Handbook of Physical Constants. American Geophysical Union, Washington, D.C., 64 -97.
Auld, B.A. (1990) Acoustic Fields and Waves in Solids, 432 pp.
Polian, A. (2003) Brillouin Scattering at High Pressure: An Overview, J. Raman spectroscopy, 34, 633.
Whitfield, C.H., Brody, E.M., and Bassett, W.A. (1976) Elastic moduli of NaCl by Brillouin scattering at high pressure in a diamond anvil cell. Review of Scientific Instruments, 47, 942-947.
Historical Papers Brillouin, L. (1922) Diffusion de la Lumiere et des Rayons X par un Corps Transparent Homogene: Influence de l'Agitation Thermique. Annales de Physique, 17, 88-122.
Mandelstam, L.I. (1926) Light Scattering by Inhomogeneous Media. Zhurnal Russii Fizicheskoi Khimii Ova., 58, 381.
Weidner, D., Swyler, K. and Carleton, H. (1975) Elasticity of Microcrystals. Geophysical Research Letters, 2, 189-192.
Bassett, W.A. & Brody, E.M (1977) Brillouin Scattering: A new way to Measure Elastic Moduli at High Pressures. In: Manghnani, M.H. & Akimoto, S. (Eds.) High Pressure Research - Applications in Geophysics, New York, 519-532.
Recent Review Papers
Speciale, S., Marquardt, H. & Duffy, T.S. (2014) Brillouin scattering and its application in Geosciences. Reviews in Mineralogy and Geochemistry, 78, 543-603.
- A short biography of L.N. Brillouin
- Tutorial on Brillouin zones
- Free software to visualize Brillouin zones
- Java applet to visualize Brillouin zones
These materials are being developed with the support of COMPRES, the Consortium for Materials Properties Research in Earth Sciences, under NSF Cooperative Agreement EAR 10-43050 and is partially supported by UNLV's High Pressure Science and Engineering Center, a DOE NNSA Center of Excellence supported under DOE NNSA Cooperative Agreement No. DE FC52-06NA26274.<|endoftext|>
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# How do you find the x and y intercepts for f(x) = (x-3)^2 + 17?
Aug 2, 2017
$\text{y-intercept "=26" no x-intercepts}$
#### Explanation:
$\text{to find the intercepts}$
• " let x = 0, in the equation for y-intercept"
• " let y = 0, in the equation for x-intercepts"
$x = 0 \to y = {\left(- 3\right)}^{2} + 17 = 26 \leftarrow \textcolor{red}{\text{ y-intercept}}$
$y = 0 \to {\left(x - 3\right)}^{2} + 17 = 0$
$\Rightarrow {\left(x - 3\right)}^{2} = - 17$
$\forall x \in \mathbb{R} \textcolor{w h i t e}{x} {\left(x - 3\right)}^{2} \ge 0$
$\Rightarrow \text{ there are no x-intercepts}$
graph{(x-3)^2+17 [-80, 80, -40, 40]}<|endoftext|>
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Rhode Island Standards
E 1. Individuals and societies make choices to address the challenges and opportunities of scarcity and abundance.
E 1 (5-6)-1. Students demonstrate an understanding of basic economic concepts by…
E 1 (5-6)-1.c. Identifying and differentiating between surplus, subsistence, and scarcity.
E 1 (5-6)-3. Students demonstrate an understanding that societies develop different ways to deal with scarcity and abundance by…
E 1 (5-6)-3.b. Identifying how scarcity impacts the movement of people and goods.
E 2. Producers and consumers locally, nationally, and internationally engage in the exchange of goods and services.
E 2 (5-6)-1. Students demonstrate an understanding of the variety of ways producers and consumers exchange goods and services by…
E 2 (5-6)-1.b. Identifying and explaining how supply, demand, and incentives affect consumer and producer decision making (e.g., division of labor/specialization).
RI.HP. Historical Perspectives/Rhode Island History
HP 4. Historical events and human/natural phenomena impact and are influenced by ideas and beliefs.
HP 4 (5-6)-2. Students demonstrate an understanding that innovations, inventions, change, and expansion cause increased interaction among people (e.g., cooperation or conflict) by…
HP 4 (5-6)-2.a. Citing examples of how science and technology have had positive or negative impacts upon individuals, societies and the environment in the past and present.
HP 4 (5-6)-2.c. Describing important technologies and advancements, including writing systems, developed by a particular civilization/ country/ nation.<|endoftext|>
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CMMAP - Studying Clouds and Climate
Clouds are an important part of Earth's weather and climate. Scientists use computer models to study our planet's climate. These computer models include models of clouds. It's hard to model clouds because these models need to include both large and small things. Some parts of cloud models need to explain very big things like hurricanes that can be more than a hundred miles across. Other parts of cloud models need to explain very small things like raindrops and snowflakes.
CMMAP (pronounced "see-map") stands for the Center for Multi-Scale Modeling of Atmospheric Processes. CMMAP scientists are working on a new way to model the climate that will help us to better understand the roles clouds play today and in the future as our climate changes. Climate model grid cells are usually about the size of the state of Delaware. The problem with modeling clouds is that they are much smaller than this, and there could be thousands of clouds in an area the size of Delaware at any given time.
One solution that CMMAP scientists are working on is to make the grid cells small enough to simulate clouds. This has been done using cells the size of city blocks. Although these models do a good job simulating clouds that are realistic, they are very slow. Even the fastest computers in the world are too slow to simulate global climate with city-block-sized cells. Someday this will be possible, but that is many years away.
We need better climate models today, so CMMAP is developing a radical new approach that bridges the gap between the city-block-sized grid cells and the Delaware-sized grid cells. This "middle ground" involves adding small grid cells to a tiny part of each Delaware-sized cell, which provides a representative sample of the grid cell rather than including all of the information involved.
The links below have more information about the CMMAP project, how CMMAP is helping people teach and learn about climate, and the science of clouds and climate modeling.
- Science Education page on the main CMMAP web site at CSU
- Little Shop of Physics (LSOP)
- CMMAP project home page at CSU
- Movies about Clouds, Weather, and Climate - from CMMAP and LSOP<|endoftext|>
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## Monday, August 19, 2024
### June 2024 Algebra 2 Regents Part IV
This exam was adminstered in January 2024 .
### June 2024 Algebra 2, Part IV
A correct answer is worth up to 6 credits. Partial credit can be given. Work must be shown or explained.
37. Megan is performing an experiment in a lab where the air temperature is a constant 73°F and the liquid is 237°F. One and a half hours later, the temperature of the liquid is 112°F. Newton’s law of cooling states T(t) = Ta + (T0 - Ta)e-kt where:
T(t): temperature, °F, of the liquid at t hours
Ta: air temperature
T0: initial temperature of the liquid
k: constant
Determine the value of k, to the nearest thousandth, for this liquid.
Determine the temperature of the liquid using your value for k, to the nearest degree, after two and a half hours.
Megan needs the temperature of the liquid to be 80°F to perform the next step in her experiment. Use your value for k to determine, to the nearest tenth of an hour, how much time she must wait since she first began the experiment.
In the first part, to solve for k, you have to substitute all the other variables and then isolate k.
T(t) = Ta + (T0 - Ta)e-kt
112 = 73 + (237 - 73)e-1.5k
39 = 164e-1.5k
39/164 = e-1.5k
ln(39/164) = -1.5k
-1.4363 = -1.5k
0.9575 = k
k = 0.958 to the nearest one-thousandth of an hour.
For the second part, substitute t = 2.5 hours.
T(t) = Ta + (T0 - Ta)e-kt
T(2.5) = 73 + (237 - 73)e-2.5(.958)
T(2.5) = 87.9523 = 88 degrees.
For the final part, set T(t) = 80 and solve for t.
T(t) = Ta + (T0 - Ta)e-kt
80 = 73 + (237 - 73)e-.958t
7 = 164e-.958t
7/164 = e-.958t
ln(7/164) = -.958t
-3.1539 = -.958t
3.2922 = t
It would take about 3.3 hours.
End of Exam
How did you do?
Questions, comments and corrections welcome.
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You can now order my newest book Burke's Lore, Briefs: I See What You'll Do There, written by Christopher J. Burke, which contains two humorous fantasy stories
Order the softcover or ebook at Amazon.
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Also, check out Devilish & Divine, an anthology filled with stories of angels and devils by 13 different authors, and In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides.
Available in softcover or ebook at Amazon.
If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.<|endoftext|>
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How to Simplify Radicals. Video Tutorial with interactive practice problems showing all the work
$\sqrt{45} = \color{red}{\sqrt{9}} \sqrt{5} = \color{red}{?}$
Video Tutorial with practice problems
Some Necessary Vocabulary The radicand refers to the number under the radical sign. In the radical below, the radicand is the number '5'.
### Some Necessary Background Knowledge
#### I. Know your Perfect Squares!
Before you learn how to simplify radicals,you need to be familiar with what a perfect square is. Also, you should be able to create a list of the first several perfect squares. This is easy to do by just multiplying numbers by themselves as shown in the table below.
List Perfect Squares
2*2 4
3*3 9
4*4 16
5*5 25
6*6 36
7*7 49
8*8 64
9*9 81
10*10 100
11*11 121
12*12 144
13*13 169
#### II. You can rewrite a radical as the product of two radical factors of its radicand !
That's a very fancy way of saying that you can rewrite radicals as shown in the table below.
### How to Simplify Radicals Steps
Let's look at to help us understand the steps involving in simplifying radicals.
Step 1
Find the largest perfect square that is a factor of the radicand.
4 is the largest perfect square that is a factor of 8.
Step 2
Rewrite the radical as a product of the square root of 4 (found in last step) and its matching factor(2).
Step 3
Simplify.
##### Problem 1
Step 1
Find the largest perfect square that is a factor of the radicand (72).
36 is the largest perfect square that is a factor of 72.
Step 2
Rewrite the radical as a product of the square root of 36 (found in last step) and its matching factor (2).
Step 3
Simplify.
##### Problem 2
Step 1
Find the largest perfect square that is a factor of the radicand (50).
25 is the largest perfect square that is a factor of 50.
Step 2
Rewrite the radical as a product of the square root of 25 (found in last step) and its matching factor (2).
Step 3
Simplify.
##### Problem 3
You know the deal. Just follow the steps.
Step 1
Find the largest perfect square that is a factor of the radicand (75).
25 is the largest perfect square that is a factor of 75.
Step 2
Rewrite the radical as a product of the square root of 25 (found in last step) and its matching factor (3).
Step 3
Simplify.
##### Problem 4
Step 1
Find the largest perfect square that is a factor of the radicand (32).
16 is the largest perfect square that is a factor of 32.
Step 2
Rewrite the radical as a product of the square root of 16 (found in last step) and its matching factor (2).
Step 3
Simplify.
##### Problem 5
Hopefully, by know you know how to simplify radicals.
Step 1
Find the largest perfect square that is a factor of the radicand (200).
100 is the largest perfect square that is a factor of 200.
Step 2
Rewrite the radical as a product of the square root of 100 (found in last step) and its matching factor (2).
Step 3
Simplify.
##### Problem 6
Step 1
Find the largest perfect square that is a factor of the radicand (108).
36 is the largest perfect square that is a factor of 108.
Step 2
Rewrite the radical as a product of the square root of 108 (found in last step) and its matching factor (3).
Step 3
Simplify.
$${\sqrt{36}} \sqrt{3} = 6 \sqrt{3}$$
##### Problem 7
Ok, this question is a trick one to see if you really understand step 1 of how to simplify radicals.
cannot be simplified because this radicand (26) does not have any perfect square factors. Therefore, you cannot simplify it.
### How to Simplify Radicals with Coefficients
Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root.
Step 1
Find the largest perfect square that is a factor of the radicand (just like before).
4 is the largest perfect square that is a factor of 8.
Step 2
Rewrite the radical as a product of the square root of 4 (found in last step) and its matching factor(2).
Step 3
Simplify.
Step 4
Multiply original coefficient (3) by the 'number that got out of the square root' (2).
### Practice Simplifying Radicals with Coefficients
##### Problem 8
Step 1
Find the largest perfect square that is a factor of the radicand (just like before).
4 is the largest perfect square that is a factor of 20.
Step 2
Rewrite the radical as a product of the square root of 4 (found in last step) and its matching factor(5).
Step 3
Simplify.
Step 4
Multiply original coefficient (6) by the 'number that got out of the square root ' (2).
##### Problem 9
Step 1
Find the largest perfect square that is a factor of the radicand (just like before).
16 is the largest perfect square that is a factor of 80.
Step 2
Rewrite the radical as a product of the square root of 16 (found in last step) and its matching factor(5).
Step 3
Simplify.
Step 4
Multiply original coefficient (2) by the 'number that got out of the square root ' (2).
##### Problem 10
Step 1
Find the largest perfect square that is a factor of the radicand (just like before).
25 is the largest perfect square that is a factor of 125.
Step 2
Rewrite the radical as a product of the square root of 25 (found in last step) and its matching factor(5).
Step 3
Simplify.
Step 4
Multiply original coefficient (4) by the 'number that got out of the square root ' (5).<|endoftext|>
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Permutations – Example and Practice Problems
With permutations, we can count the number of different ways of choosing objects from a set if the order of the objects does matter. This is different from combinations, where the order of the objects does not matter.
Here, we will start with a summary of permutations and look at their formula. Then, we will see several examples with answers to understand the application of the permutations formula.
ALGEBRA
Relevant for
Learning about permutations with solved examples.
See examples
ALGEBRA
Relevant for
Learning about permutations with solved examples.
See examples
Summary of permutations
A permutation is a list of objects, in which the order is important. Permutations are used when we are counting without replacing objects and order does matter. If the order doesn’t matter, we use combinations.
In general P(n, k) means the number of permutations of n objects from which we take k objects. Alternatively, the permutations formula is expressed as follows:
where:
• n represents the total number of elements in a set
• k represents the number of selected objects
• ! is the factorial symbol
To solve permutations problems, we have to remember that the factorial (denoted as “!”) is equal to the product of all positive integers less than or equal to the number preceding the factorial. For example, .
In the following examples, we will see the application of the permutations formula. Each example has its respective detailed solution, which can be used to understand the reasoning in the answer to each exercise.
EXAMPLE 1
Find the result of the permutation .
We have to use the permutations formula , and substitute and :
We can simplify this by writing to 7! as :
EXAMPLE 2
Find the result of the permutation .
We use the permutations formula using the values and :
In this case, we have nothing to simplify, so we have to calculate 8!:
EXAMPLE 3
Find the number of permutations .
In this case, we have the values and :
We recognize that we can write 9! as .Therefore, we simplify the 3!:
EXAMPLE 4
How many permutations are there with 4 objects and 2 places?
We can recognize the values and . Therefore, we use the formula substituting these values:
We rewrite the 4! as and we simplify:
EXAMPLE 5
In how many ways can a president, treasurer, and secretary be chosen from 7 candidates?
The problem involves 7 candidates, of which we chose 3. Therefore, we have the values and :
Now, we write 7! as . Then, we simplify the 4!:
EXAMPLE 6
In how many ways can the first 3 places be awarded in a race with 5 participants?
We recognize the values and :
We rewrite the factorial 5! as . Then, we simplify the 2!:
Permutations – Practice problems
Practice and test your knowledge of permutations. Select an answer and check it to see if you chose the correct answer. If you need help, you can look at the solved examples above.
In how many ways can the positions of president and vice president be assigned from a group of 8 people?
Escoge una respuesta<|endoftext|>
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5-E Model for Teaching Inquiry Science
Elementary Inquiry Science Is Important
Elementary science lays a foundation to help students develop investigation skills which they will use later on in more advanced science. It also fosters a background in communication and critical thinking, both useful life skills. By incorporating these different steps of scientific inquiry, students will develop a better understanding of how to investigate questions, present evidence, use different tools, and explore different ethical issues in science.
Your slide tray is being processed.<|endoftext|>
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Date of Award
Languages & Literature
The Elizabethan drama as characterized by Shakespeare amid his contemporaries was the culmination of several forces which had long been exerting their influence on the English playwrights. Prominent among these forces was that of the classical drama, whose tendencies affected both comedy and tragedy. It was from the tragedies of Seneca that the Elizabethans developed their tragedies of sheer horror and blood. The comedies of Plautus and Terrance taught English dramatists "how to build a plot along structural lines of a five act pattern and how to develop a complicated intrigue based on the motif of mistaken identity; it supplied them with new types of comic characters and it brought home to them the difference between crude slap-stick comedy and the more intellectual species of wit.’’ 1 This influence is clearly shown in the following native English plays: Ralph Roister Boister (1534- 1541), Gammer Gurton's Needle (1575), and The Supposes (1566).
Ryan, Joseph, "Shakespeare's Use Of Source Material In The Comedy Of Errors" (1933). Languages and Literature Undergraduate Theses. 117.<|endoftext|>
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Parametric Equations of a Circle
Parametric Equations of a Circle
If P (x, y) is a point on the circle with center C (h, k) and radius r, then x = h + r sin θ, y = k = r cos θ where 0 ≤ θ < 2π.
Proof: Let θ be the angle made by the line $$\overleftrightarrow{CP}$$ with x – axis in the positive direction let D, M be the projections of C, P on x – axis respectively.
Let Q be the projection of C on PM.
Now ∠PCQ = θ
cos θ = CQ/ CP,
sin θ = PQ/ CP
Therefore CQ = CP Cosθ
= r Cosθ
PQ = CP sinθ
= r sinθ
x = OM = OD + DM
=OD + CQ
= h + r cosθ
Y = PM = PQ + QM
= PQ + CD
= k + r sinθ
A point on the circle x² + y² = r² is taken in the form (r cosθ, r sinθ). The point (r cosθ, r sinθ). is simply denoted as point θ.
Example 1: The parametric equation of the circle (x-3) ² + (y-6) ² = 8².
Solution: Given that
(x-3) ² + (y-6) ² = 8²
Above equation h = 3, k = 6 and r = 8
The parametric equations are
x = h + r cosθ
y = k + r sinθ
x = 3 + 8 cosθ
y = 6 + 8 sinθ
where 0 ≤ θ < 2π
Example 2: Find the parametric equation of the circle 4(x² + y²) = 9.
Solution: Given circle equation is 4(x² + y²) = 9
(x² + y²) = 9/4
x² + y² = (3/2)² … (1)
center (h, k) = (0, 0)<|endoftext|>
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# Find the general solution of the differential equation $x\large\frac{dy}{dx}$$+2y=x^2\log x ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• (i) Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • (ii) Find the integrating factor (I.F) = e^{\int Pdx}. • (iii) Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: The given problem is a first order linear equation. To write it in the form, let us divide throughout by x \large\frac{dy}{dx} +\frac{2y}{x} =$$ x \log x$
The I.F = $\int \large\frac{2}{x }$$= 2\log x or \log x^2 The I.F = e^{\log x^2} = x^2 The solution is yx^2 = \int x^2\log x . x^2 + C yx^2 =\int x^3\log x.dx + C \int x^3 \log x dx can be done by integration by parts. Step 2: Let u = \log x;du = \large\frac{dx}{x} and dv = x^3dx or v =\large\frac{x^4}{4} Hence integration of x^3\log xdx = \log x(\large\frac{x^4}{4}) - \int \large\frac{1}{x}(\large\frac{x^4}{4})$$.dx$
$\qquad\qquad\qquad\qquad\qquad\quad=\log x.(\large\frac{x^4}{4}) - (\large\frac{1}{4})\frac{x^4}{4}$
Hence the required solution is $yx^2 = \log x(\large\frac{x^4}{4}) -\frac{ x^4}{16}$<|endoftext|>
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# Most Misunderstood Math Standards in Grade 4
Are these instructional mistakes happening in your classroom?
In my last post we explored the Most Misunderstood Mathematics Standards in Grade 3. It has been thrilling to see the reactions on both Twitter and Facebook. I would love to continue to engage with math educators and can’t wait to hear your thoughts! Leave me a comment below!
In this post, we will dive right into the most misunderstood math standards in Grade 4. I want to say again this is not an evaluation of teacher instruction in Grade 4; rather, it is a place where we can learn together about the current state of instruction in fourth grade and name concrete ways to get better. Together!
Two of my favorite standards in Grade 4 require us develop conceptual understanding in order to reach procedural skill: 4.NBT.B.5 and 4.NBT.B.6. Let’s dig in!
4.NBT.B.5
Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
In Grade 4, students are introduced to an area model for multi-digit multiplication, which connects to the 3rd grade understanding that the area of a rectangle is the product of its width and length. For this blog post, we will examine the connections between the area model, partial products, and the distributive property for two-digit by two-digit numbers.
Illustrating partial products with area models (Examples from the Progressions Documents found here.)
The example on the left is an area model illustrating partial products. Partial products uses place value concepts to solve multi-digit multiplication problems. It’s based on the distributive (grouping) property of multiplication. “The distributive property allows numbers to be decomposed into base-ten units, products of the units to be computed, then combined. By decomposing the factors into base-ten units and applying the distributive property, multiplication computations are reduced to single-digit multiplications and products of numbers with multiples of 10, of 100, and of 1000. Students can connect diagrams of areas or arrays to numerical work to develop understanding of general base-ten multiplication methods.” The area model provides us with a pictorial representation of the concept we want students to understand and makes the important connections to place value explicit for students.
4.NBT.B.6
Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
Similar to multiplication, we will examine division using an area model to illustrate partial quotient.
In partial-quotient division, place value and an understanding of multiplication are used to find partial quotients.
Historically, multi-digit division and multiplication have been solved by memorizing steps, mnemonics, or acronyms without connections to place value, area, and the properties. How many of you have heard Does McDonald’s Serve Cheese Burgers when discussing fourth grade division instruction? Be honest! I certainly will. Before I knew how important understanding the math concepts underpinning multiplication and division were, I thought a catchy phrase was just the thing to help my students remember steps to solving a problem. However, the saying, Does (division) McDonalds (multiplication) Serve (subtraction) Cheese (check) Burgers, (bring down) misses the mark when we think about deeply understanding division concepts.
Let’s make a commitment to one another. We commit to teaching the concepts of multiplication and division. We will not use acronyms or mnemonics to teach important math concepts. We can do it!
Thank you for your commitment to excellent daily math instruction for all students. See you next month when we dive into Grade 5.
## 9 thoughts on “Most Misunderstood Math Standards in Grade 4”
1. Sarah says:
Thank you so much for your posts! I find them quite helpful.
Do you mind sharing a link to the wiring diagram mentioned in the table at towards the beginning of the article? Since it’s an image, the hyperlink doesn’t actually work. Thank you!
2. C. Senegal says:
Thanks for your post. I can definitely understand how decomposing during division could be confusing.
3. Stephanie says:
These misundersoond standards posts are fantastic. By chance is there one for 5th grade?
Yes! It’s been drafted! Posting later this week!
4. Bridget Trymbulak says:
I enjoyed the acronyms for division!
5. Lisette says:
ALL Math Master Teachers in my parish needs to read this article. Wait!! HSE Math Master Teacher needs to read this article. As an educator, I try my best to keep up with research. I attended a summer program for educator. While attending, the professor talked about math misconceptions. She also provided us with books.
Well, I read the books and try my best to teach so my students would not have misconceptions about the standard I would teach. The MMT would enter my classroom sit for a while than jump up and say you are not teaching the student correctly. You are teaching them tricks.
I was teaching standard 4.NBT.B.5 using area model, distributive property, etc. My MMT stated that’s NOT the “Most Efficient” way to solve that problem.
Here comes the Principal. If your MMT said don’t teach it that way than you better not teach it that way. I said “okay” and continued teaching it using area model, distributive property, etc.
Well, I am no longer there because of the MMT, but I was not going to continue to teach our future her way.<|endoftext|>
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Data Visualization Workshop
- What is Data Visualization
- Types of Visual Diagrams
- D3 API Docs
- List of D3 Modules
- Selections, Data, and Demo
- Enter and Append
- Scales and Axis
- First Challenge
- Second Challenge
- Enter and Update
- Exit and Merge
- Third Challenge
- Force Layout
- D3 and Vuejs
- Data Visualization Best Practices
- General Data Visualization Resources
- Data Visualization Experts and Companies
- GitBook Link
What is Data Visualization
- Data visualization is the presentation of data in a pictorial or graphical format.
- It enables decision makers to see analytics presented visually
- Decision makers can then grasp difficult concepts or identify new patterns.
- With interactive visualization, you can take concepts further and teach concepts better.
- You can drill down into charts and graphs for more detail,
- and interactively change what data you see and how it’s processed.
Types of Visual Diagrams
A diagram is a symbolic representation of information according to some visualization technique.
- A bar chart or bar graph is a chart or graph that presents categorical data with rectangular bars with heights or lengths proportional to the values that they represent.
- The bars can be plotted vertically or horizontally.
- A vertical bar chart is sometimes called a line graph.
- A line chart or line graph is a type of chart which displays information as a series of data points called 'markers' connected by straight line segments.
- It is a basic type of chart common in many fields.
- It is similar to a scatter plot except that the measurement points are ordered (typically by their x-axis value) and joined with straight line segments.
- A line chart is often used to visualize a trend in data over intervals of time – a time series – thus the line is often drawn chronologically.
- A scatter plot (also called a scatter graph, scatter chart, scattergram, or scatter diagram) is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data.
- If the points are color-coded, one additional variable can be displayed.
- The data is displayed as a collection of points, each having the value of one variable determining the position on the horizontal axis and the value of the other variable determining the position on the vertical axis
D3 API Docs
D3 4.0 is a collection of modules that are designed to work together
- You can use the modules independently, or you can use them together as part of the default build.
- The source and documentation for each module is available in its repository.
Follow the links below to learn more:
- API Reference
- Release Notes
- d3.js on Stack Overflow
- d3-js Google Group
- d3-js Slack Channel (Invite)
- d3-js Gitter Channel
- d3-js IRC Channel => #d3.js on irc.freenode.net
List of D3 Modules
dynamic geometry computation
visual variables mapping
visual variables mapping (selection, data-binding, attributes)
high level set of visual variables mapping (component)
user events and geometry computation
Data Visualization Best Practices
There is a wonderful illustration in this Tableau White Paper on Data Visualization Best Practices
General Data Visualization Resources
- Dashing D3js
- Visualizing Data - a fantastic blog about data visualization.
- Flowing Data - another data visualization blog.
- Information Aesthetics - Another good blog.
- Shirley Wu - Excellent blog on Data Visualization.
Data Visualization Experts and Companies
- Jer Thorp - Data artist.
- Nicolas Feltron - Data artist.
- Stamen - Data visualization company, excellent blog, heavy focus on maps.
- Fathom - Another interesting data visualization company.<|endoftext|>
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As the cold weather rolls in we cozy up inside, pour a cup of hot chocolate and light up the fireplace, ahh the great advancements of civilization! But what about all the animals? How do they survive the frigid weather of the northeast? Adaptation, migration and hibernation- three key terms to winter survival. Shorter days (photoperiod) as well as a change in temperature trigger a response in animals to prepare for winter; this is hardwired into them as their internal biological calendar. Some responses are obligatory, while others are consequential, but all are quite amazing. Let’s explore a few.
Although hibernation is the common term, it is lesser known that there are few “true hibernators.” This is because while many animals will spend their winter sleeping, not many sleep the whole way through like our groundhog friend “Staten Island Chuck”; instead, they experience torpor or brumation (dormancy in cold-blooded animals). Torpor is a state of inactivity where there is a reduction in the metabolic rate, a lower body temperature and slower heart beat.
When an animal experiences hibernation, torpor, or brumation, they must have additional adaptations to survive these long stretches of inactivity. Many mammals survive the cold season by growing in thick coats of fur, sometimes even slightly changing color to camouflage better in their environments, like the white tailed deer. A white tailed deer’s winter coat is also made up of hollow hairs that trap in air which is warmed by the body. Shrews, voles, chipmunks and white-footed mice will spend much of their time in underground runways and tunnels. Though insects are still readily available for insectivorous creatures, like the shrews, herbivores and omnivores would likely not survive without their caches of food that they have laboriously stored away.
Not all animals live off their larders and caches for the winter. So how do they survive on the small availability of food? Well, many fast for long periods of time; but surviving without food for long expanses is no easy feat. Cold-blooded animals are especially adept at utilizing very little energy, partly because they are not producing their own heat. Garter snakes will gather in the thousands in suitable hibernacula. Other terrestrial reptiles and amphibians will typically seek out deep cracks in logs or rocks or dig below the frost line or into the leaf litter. The wood frog is a spectacular example of adaptation. They produce antifreeze in their system (in this case a fancy term for sugar or glucose), which helps to keep its organs from freezing. The frogs can survive being frozen solid for months on end, during which they have no brain activity and no heartbeat.
Aquatic frogs in a reduced activity state must stay in rich oxygenated conditions, as opposed to aquatic turtles that nestle down into the mud. Snapping turtles and painted turtles can live in anaerobic environments during the winter months. They reduce their metabolism by about 90% and get the little energy they need from stored fat. The lack of oxygen will build up lactic acid in their systems; but their shell will release chemical buffers to break down and store the lactic acid until they have the opportunity to bask in the sun releasing the buildup altogether.
Waterfowl like ducks and geese must forage for food to survive. In freezing water they use their uropygial/preen gland to produce an oil to waterproof their feathers as well as a system called countercurrent heat exchange. Ever wonder how a duck can walk through snow and ice and swim through frigid water without getting frostbite on their feet? The arteries containing warm blood flowing from the heart lay close to the veins moving the colder blood from their feet. This warms up the venous blood and cools down the arterial blood so the temperature difference between the ice and their feet is lessened, reducing heat loss.
Many native animal species, from reptiles to birds to mammals, possess remarkable adaptations that allow them to perform great feats of survival.
By Renee Allen<|endoftext|>
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2008 IMO Problems/Problem 1
Problem
An acute-angled triangle $ABC$ has orthocentre $H$. The circle passing through $H$ with centre the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with centre the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with centre the midpoint of $AB$ intersects the line $AB$ at $C_1$ and $C_2$. Show that $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ lie on a circle.
Solution
Let $M_A$, $M_B$, and $M_C$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. It's not hard to see that $M_BM_C\parallel BC$. We also have that $AH\perp BC$, so $AH \perp M_BM_C$. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that $H$ is on the radical axis of the circles centered at $M_B$ and $M_C$, so $A$ is too. We then have $AC_1\cdot AC_2=AB_2\cdot AB_1\Rightarrow \frac{AB_2}{AC_1}=\frac{AC_2}{AB_1}$. This implies that $\triangle AB_2C_1\sim \triangle AC_2B_1$, so $\angle AB_2C_1=\angle AC_2B_1$. Therefore $\angle C_1B_2B_1=180^{\circ}-\angle AB_2C_1=180^{\circ}-\angle AC_2B_1$. This shows that quadrilateral $C_1C_2B_1B_2$ is cyclic. Note that the center of its circumcircle is at the intersection of the perpendicular bisectors of the segments $C_1C_2$ and $B_1B_2$. However, these are just the perpendicular bisectors of $AB$ and $CA$, which meet at the circumcenter of $ABC$, so the circumcenter of $C_1C_2B_1B_2$ is the circumcenter of triangle $ABC$. Similarly, the circumcenters of $A_1A_2B_1B_2$ and $C_1C_2A_1A_2$ are coincident with the circumcenter of $ABC$. The desired result follows.<|endoftext|>
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# Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3
You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3
9th Maths Exercise 5.3 Question 1.
Find the mid-points of the line segment joining the points
(i) (-2, 3) and (-6, -5)
(ii) (8, -2) and (-8, 0)
(iii) (a, b) and (a + 2b, 2a – b)
(iv) $$\left(\frac{1}{2},-\frac{3}{7}\right)$$ and $$\left(\frac{3}{2}, \frac{-11}{7}\right)$$
Solution:
9th Maths Exercise 5.3 Solution Question 2.
The centre of a circle is (-4, 2). If one end of the diameter of the circle is (-3, 7) then find the other end.
Solution:
∴ The other end is (-5, -3)
9th Maths Exercise 5.3 Samacheer Kalvi Question 3.
If the mid-point (x, y) of the line joining (3, 4) and (p, 7) lies on 2x + 2y + 1 = 0, then
what will be the value of p?
Solution:
Class 9 Maths Exercise 5.3 Solution Question 4.
The midpoint of the sides of a triangle are (2, 4), (-2, 3) and (5, 2). Find the coordinates of the vertices of the triangle.
Solution:
Class 9th Maths Chapter 5 Exercise 5.3 Question 5.
O(0, 0) is the centre of a circle whose one chord is AB, where the points A and B are (8, 6) and (10, 0) respectively. OD is the perpendicular from the centre to the chord AB. Find the coordinates of the mid-point of OD.
Solution:
Exercise 5.3 Class 10 Samacheer Kalvi Question 6.
The points A (-5, 4), B (-1, -2) and C(5, 2) are the vertices of an isosceles right¬angled triangle where the right angle is at B. Find the coordinates of D so that ABCD is a square.
Solution:
Ex 5.3 Class 9 Maths Solutions Question 7.
The points A(-3, 6), B(0, 7) and C(1, 9) are the mid points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallelogram.
Solution:
In a parallelogram diagonals bisect each other and diagonals are not equal.
10th Maths Exercise 5.3 Samacheer Kalvi Question 8.
A (-3, 2), B (3, 2) and C(-3, -2)are the vertices of the right triangle, right angled at A. Show that the mid point of the hypotenuse is equidistant from the vertices.
Solution:
Hence proved<|endoftext|>
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# Math
A rectangle has a width that is 1/8 its length. If the perimeter of the rectangle is 126 inches, what is the width of the rectangle?
1. 👍 0
2. 👎 0
3. 👁 61
1. l=8w
126=2l+2w
126=2l+2*1/8 l
126=2l+ l/4 l=2.25 l
solve for l, then w.
1. 👍 0
2. 👎 0
2. You were given:
A rectangle has a width that is 1/8 its length. If the perimeter of the rectangle is 126 inches, what is the width of the rectangle?
Length = we don't know and so I'll call it x.
Width = 1/8(x), which can be written x/8.
The perimeter given is 126 inches.
We use the formula for finding the perimeter of a rectangle.
P = 2L + 2W, where P = perimeter, L = length and W = width.
Is this clear so far?
126 inches = 2(x) + 2(x/8)
I will not use the word inches again for P because you know we are talking about a perimeter of 126 inches, right?
126 = 2x + x/4
We now have a fractional equation.
How do we remove the fraction part?
We multiply each term by the LCD, which so happens to be 4.
126(4) = 2x(4) + (x/4)(4)
504 = 8x + x
504 = 9x
To find x, we divide BOTH sides of the equation by 9.
504 divided by 9 = x
56 = x
We are looking for the width, right?
The width is (1/8)(x), which can also be written x/8.
We just found x, right?
To find the width, replace x with 56 in the fraction x/8 and then divide the numerator by the denominator.
So, x/8 becomes 56/8 = 7
What is the width of this rectangle?
That's it!
1. 👍 0
2. 👎 0
posted by Guido
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Can someone help me set up the equations thanks. Directions: Solve each of the following applications. Give all answers to the nearest thousandth. Problem: Geometry. The length of a rectangle is 1 cm longer than its width. If the
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How Digestion Works
Wellness and Prevention
The human body uses the process of digestion to break down food into a form that can be absorbed and used for fuel. The organs of the digestive system are the mouth, esophagus, stomach, pancreas, liver, gallbladder, small intestine, large intestine and anus. Recognizing how these organs work together to digest food is key to understanding how digestion works.
The digestive process begins in the mouth. Even before eating begins, the anticipation of eating stimulates glands in the mouth to produce saliva. The digestive system carries out three primary processes: mixing food, moving food through the digestive tract (peristalsis) and using chemicals to break down food into smaller molecules.
Consult a health care provider about specific questions regarding how digestion works and any related digestive health problems.<|endoftext|>
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# Using Prime Factorization to Reduce Fractions
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
Having to reduce fractions is a common occurrence in mathematics and even in our daily lives. This lesson will define prime factorization and show how to use it to reduce fractions.
## Reducing Fractions
Imagine you are talking to a friend. Your friend is telling you about a really yummy fruit juice they make and they tell you that the juice is 8/10 fresh pineapple juice. Then, they pause and say, 'Well, actually the juice is 4/5 fresh pineapple juice.' What just happened there? How did the pineapple juice content go from 8/10 to 4/5? What happened is your friend reduced the fraction 8/10 to an equivalent fraction 4/5.
Reducing a fraction involves removing a common factor that both the numerator and denominator have in common, where a common factor is a number that divides into both the numerator and the denominator evenly. In the case of your friend, they mentally reduced 8/10 by recognizing that both the numerator and denominator have a factor of 2. On paper, the work would be as follows.
When we reduce a fraction, we create an equivalent fraction, or a fraction that has the same value. In other words, 8/10 and 4/5 have the same value. There are a lot of different ways to go about reducing fractions. We are going to look at a method that uses prime factorization.
## Prime Factorization
Prime factorization of a number involves factoring the number completely until all the factors are prime, where a prime number is a number that is only divisible by 1 and itself. For example, consider the number 56. To find the prime factorization of 56, we just start factoring and keep going until all the factors are prime numbers.
56 = 7*8 7 is prime, factor 8 further 56 = 7*2*4 7 and 2 are prime, factor 4 further 56 = 7*2*2*2 All the factors are prime
We see the prime factorization of 56 is 7*2*2*2. What's really neat about prime factorization is that it doesn't matter what two factors you start with, you will still get the same prime factorization of that number. To illustrate this, suppose we started with 56 = 2*28 instead of 56 = 7*8.
56 = 2*28 2 is prime, factor 28 further 56 = 2*7*4 2 and 7 are prime, factor 4 further 56 = 2*7*2*2 All the factors are prime.
The factors are in a different order, but that doesn't matter. We still get that the prime factorization of 56 is 7*2*2*2. Now let's look at how to use prime factorization to reduce fractions.
## Using Prime Factorization to Reduce Fractions
To use prime factorization to reduce fractions, we follow these steps.
1. Find the prime factorization of the numerator and the denominator, and rewrite the fraction with those prime factorizations.
2. Cancel out any common factors in the numerator and the denominator.
3. Multiply any factors that are left in the numerator together and any factors that are left in the denominator together. This is your fraction in reduced form.
Let's take a look at your friend's juice again. Your friend said that the juice is 8/10 pineapple juice. To reduce this fraction using prime factorization, we first find the prime factorization of the numerator and the denominator and rewrite the fraction.
8 = 2*4 2 is prime, factor 4 further 8 = 2*2*2 All the factors are prime
The prime factorization of 8 is 2*2*2. As far as 10 goes, 10 = 2*5, and 2 and 5 are both prime, so the prime factorization of 10 is 2*5. Now we replace the 8 and 10 with their prime factorizations in the fraction.
Next, we cancel out any common factors in the numerator and denominator.
Lastly, we multiply any leftover factors in the numerator together and any leftover factors in the denominator together.
We see that 8/10 reduced is 4/5, which we already knew because your friend told us so! However, now we know how they did it.
## Another Example
Let's look at another fraction that is a little more involved. Suppose we want to reduce the fraction 68/112. The first thing we want to do is find the prime factorization of the numerator and the denominator.
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This article reports that exposure to systemic insecticides, known as neonicotinoids, and pesticides has been one of the main reasons behind the disappearance of bees.
Systemic insecticides known as neonicotinoids have become the fastest growing insecticides in the world. Two prominent examples, imidacloprid and clothianidin, are used as seed treatments in hundreds of crops. Virtually all of today’s genetically engineered Bt corn, for instance, is treated with neonicotinoid insecticides.
Bee colonies began disappearing in the United States shortly after the U.S. Environmental Protection Agency (EPA) allowed these new insecticides on the market, and a debate has since been raging over whether or not these chemicals are indeed contributing to the serious honeybee die-offs that have been occurring around the world.
Now the European Food Safety Authority (EFSA) has released a report that may put the debate to rest, as they’ve ruled neonicotinoid insecticides are essentially “unacceptable” for many crops.
Is This the “Death Knell” for Neonicotinoids?
The European Commission asked EFSA to assess the risks associated with the use of three common neonicotinoids – clothianidin, imidacloprid and thiamethoxam – with particular focus on:
Their acute and chronic effects on bee colony survival and development
Their effects on bee larvae and bee behavior
The risks posed by sub-lethal doses of the three chemicals
One of the glaring issues that EFSA came across was a widespread lack of information, with scientists noting that in some cases gaps in data made it impossible to conduct an accurate risk assessment. Still, what they did find was “a number of risks posed to bees” by the three neonicotinoid insecticides.
The Authority found that when it comes to neonicotinoid exposure from residues in nectar and pollen in the flowers of treated plants:1
“…only uses on crops not attractive to honeybees were considered acceptable.”
As for exposure from dust produced during the sowing of treated seeds, the Authority ruled “a risk to honeybees was indicated or could not be excluded…” According to certain environmental groups, the ruling could be the “death knell” for neonicotinoid pesticides.2
Pesticides Also Linked to Honeybee Colony Failures
Exposure to pesticides has been associated with changes in bee behavior and reductions in colony queen production, both of which could have detrimental impacts on the life of the colony. Last year, the impact of pesticides on individual bee behavior, and its subsequent impact on the colony as a whole, was also revealed. Bees given access to two commonly used agricultural pesticides (neonicotinoid and pyrethroid) were adversely affected in numerous ways, including:3
Fewer adult worker bees emerged from larvae
A higher proportion of foragers failed to return to the nest
A higher death rate among worker bees
An increased likelihood of colony failure
The researchers said:
“Here we show that chronic exposure of bumble bees to two pesticides (neonicotinoid and pyrethroid) at concentrations that could approximate field-level exposure impairs natural foraging behavior and increases worker mortality leading to significant reductions in brood development and colony success.
We found that worker foraging performance, particularly pollen collecting efficiency, was significantly reduced with observed knock-on effects for forager recruitment, worker losses and overall worker productivity. Moreover, we provide evidence that combinatorial exposure to pesticides increases the propensity of colonies to fail.”
What Makes Neonicotinoid Pesticides so Toxic?
Neonicotinoid insecticides are known as systemic chemicals because they disrupt the central nervous system of insects, leading to paralysis and death. It’s been suggested that even sub-lethal doses of the insecticides may be negatively impacting bees.
Because neonicotinoids are water soluble and very pervasive, they get into the soil and groundwater where they can accumulate and remain for many years and generate long-term toxicity to the hive. They enter the vascular system of the plant and are carried to all parts of it, as well as to the pollen and nectar. Neonicotinoids affect insects’ central nervous systems in ways that are cumulative and irreversible. Even minute amounts can have profound effects over time.
One of the observed effects of these insecticides is weakening of the bee’s immune system. Forager bees bring pesticide-laden pollen back to the hive, where it’s consumed by all of the bees. Six months later, their immune systems fail, and they fall prey to secondary, seemingly “natural” bee infections, such as parasites, mites, viruses, fungi and bacteria. Pathogens such as Varroa mites, Nosema, fungal and bacterial infections, and Israeli Acute Paralysis Virus (IAPV) are found in large amounts in honeybee hives on the verge of collapse.
In addition to immune dysfunction and opportunistic diseases, the honeybees also appear to suffer from neurological problems, disorientation, and impaired navigation. These effects have great consequence, as a bee can’t survive for more than 24 hours if she becomes disoriented and unable to find her way back to the hive.
Bayer Downplays EFSA’s “Death Knell” Report
Bayer, a leading manufacturer of the neonicotinoid pesticides at the heart of the debate, has gone on record stating EFSA’s report “did not alter existing risk assessments and warned against ‘over-interpretation of the precautionary principle.'”4 In other words, it sounds as though they’d rather farmers continue using their pesticides without question, even if there are major concerns that they’re decimating bee populations worldwide. Bayer also noted that they are ready to work with the European Commission to address any “perceived data gaps.”
In fact, Bayer plans to open the North American Bee Care Center by July 2013. The Center is intended to be a research hub as well as promote “the active promotion of bee-responsible use of Bayer products along with communication activities worldwide.”5
Of course, it’s highly unlikely that any forthcoming research from Bayer’s North American Bee Care Center will find pesticides at fault… already, a report funded by the chemical industry has come out stating that banning neonicotinoid pesticides would cost farmers more than $980 million in lost food production.6 Yet, if these chemicals truly are killing off bee colonies, we stand to lose much, much more than that…
Bees Pollinate 70 Percent of the World’s Food
There are about 100 crop species that provide 90 percent of food, globally. Of these, 71 are pollinated by bees.7 In the United States, a full one-third of the food supply depends on pollination from bees. Apple orchards, for instance, require one colony of bees per acre to be adequately pollinated. So if bee colonies continue to be devastated major food shortages could result. There is also concern that the pesticides could be impacting other pollinators as well, including bumblebees, hoverflies, butterflies, moths and others, which could further impact the environment.
If honeybees disappear, so, too, will all of these other innovations and any new developments that may be honeybee-inspired in the future, such as these contributions to human health, including:
Playing an important role in human medicine; raw honey, which has potent anti-inflammatory and anti-infective properties, is being used for wound healing and treating coughs, while “stun” chemicals from bee stings are being looked at as an effective anesthetic for humans.
Propolis, the “caulk” honey bees use to patch holes in their hives, may slow the growth of prostate cancer and has powerful immune-modulating effects, along with potent antioxidant and anti-microbial action, and healing, analgesic, anesthetic, and anti-inflammatory properties.
Bee pollen, which is often referred to as a superfood because it contains a broad range of nutrients required by your body. About half of its protein is in the form of free amino acids that are ready to be used directly by your body and can therefore contribute significantly to your protein needs.
Honeybees have helped make scientific discoveries in many fields, including the aeronautics industry, which used the design of the six-sided honeycomb to help design aircraft wings; honeybee communication systems have even been adopted by computer programmers to help run Internet servers more efficiently.8
Do You Want to Get Involved to Help Protect Honeybees?
The documentary film Vanishing of the Bees recommends four actions you can take to help preserve our honeybees:
Support organic farmers and shop at local farmer’s markets as often as possible. You can “vote with your fork” three times a day. (When you buy organic, you are making a statement by saying “no” to GMOs and toxic pesticides!)
Cut the use of toxic chemicals in your house and on your lawn, and use only organic, all-natural forms of pest control.
Better yet, get rid of your lawn altogether and plant a garden. Lawns offer very little benefit for the environment. Both flower and vegetable gardens provide excellent natural honeybee habitats.
Become an amateur beekeeper. Having a hive in your garden requires only about an hour of your time per week, benefits your local ecosystem, and you can enjoy your own honey!
If you are interested in more information about bee preservation, the following organizations are a good place to start.
Pesticide Action Network Bee Campaign9
The Foundation for the Preservation of Honey Bees10
American Beekeeping Federation11
Help the Honey Bees12<|endoftext|>
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Rainforests are defined by an excessive amount of rainfall. They are extremely vital organs of the global ecosystem and are responsible for nearly a third of the world’s oxygen provision. With over 30 million different species of plants and animals, they are believed to be the eldest and most complex land based ecosystems on the globe. Rainforests can be thought of as massive sponges, which suck up over half of the world’s rainfall and then release it back into the air in the forms of giant compounds of mist and water vapor. Throughout the 20th century tropical rainforests have become subject to great agricultural clearances and logging, which have shrunken the occupancy of rainforests dramatically. This can clearly be seen in West Africa where 90% of the original rainforest has been lost and on Madagascar where two thirds of the rainforest has been cut down. At the current rate scientists have estimated that it is only a matter of ten years before the Indonesian forest is completely gone and 13 to 16 years before the rainforest of New Guinea has followed its path to extinction.
Deforestation of the rainforest deal with lethal threats to the animals and the climate, Edward Osborne Wilson, biologist at Harvard University has said that 50,000 species could become extinct each year. He is also worried that due to the habitat destruction, that is deforestation of the rainforest, a quarter or more of the world’s animal species could become extinct within the coming 50 years.
The World Wildlife Fund (WWF) has defined deforestation a process in which natural forests are cut down through burning or logging, “either to use the timber or to replace the area for alternative uses”. According to WWF “Deforestation continues to be an urgent environmental issue that jeopardizes people’s livelihoods, threatens species, and intensifies global warming”. Deforestation is often the victim of palm, pulp, and soy plantations or other forms of infrastructure or logging, forest fires, climate change, and the harvesting of fuel-woods.
Landslides & Erosion
With the increase of logging and global deforestation of the rainforest, hazard related concerns have been expressed. Natural buffers provided by the forest have disappeared and the safety of local populations, questioned. Tree roots function as a natural method of keeping soil, rocks, and other unstable natural matter, in place, through binding with the bedrock. The removal of vegetation, specifically on steep hillsides with loose or shallow soil, raises the threat of landslides, which can be fatal to inhabitants.
Many of the nations occupied by the rainforest are categorized as being part of the developing world. An example of such a nation is Haiti, on the Caribbean island of Hispaniola. Haiti’s residents are amongst the poorest in the Americas, with a GNI of $760.00 as of 2012, this compares to the USA’s GNI of $50,120.00 in 2012. The rainforest is an essential part of the Haitian...<|endoftext|>
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Resurrectionofgavinstonemovie.com
Live truth instead of professing it
# How are sin values derived?
## How are sin values derived?
To find the value of sin 30 degree, we will use the following formula, Sinϴ = Perpendicular Hypotenuse. Thus, the value of Sin 30 degrees is equal to 12(half) or 0.5. Just like the way we derived the value of sin 30 degrees, we can derive the value of sin degrees like 0°, 30°, 45°, 60°, 90°,180°, 270° and 360°.
Who came up with sine?
Sine was introduced by Abu’l Wafa in 8th century, as a more convenient function, and gradually spread first in the Muslim world, and then to the West. (But apparently it was used in India centuries before him), as a more convenient function. However this new notation was adopted very slowly, it took centuries.
How are the values of sin cos and tan derived?
Sine θ = Opposite side/Hypotenuse = BC/AC. Cos θ = Adjacent side/Hypotenuse = AB/AC. Tan θ = Opposite side/Adjacent side = BC/AB.
### Why do we need sine rule?
The sine rule (or the law of sines) is a relationship between the size of an angle in a triangle and the opposing side. We can use the sine rule to work out a missing angle or side in a triangle when we have information about an angle and the side opposite it, and another angle and the side opposite it.
Why is sine opposite over hypotenuse?
The sine is always the measure of the opposite side divided by the measure of the hypotenuse. Because the hypotenuse is always the longest side, the number on the bottom of the ratio will always be larger than that on the top.
How to calculate law of sines?
– You only know the angle α and sides a and c; – Angle α is acute ( α < 90° ); – a is shorter than c ( a < c ); – a is longer than the altitude h from angle β, where h = c * sin (α) (or a > c * sin (α) ).
#### When to use law of sines?
– AAS: two angles are known, and a side which is not between them. – ASA: two angles are known, and the side between them. The third angle can be found using the rule for the sum of angles, and this angle must be opposite – SSA: two sides are known, and an angle which is not between them.
What does law of sines stand for?
Law of Sines. The Law of Sines is the relationship between the sides and angles of non-right (oblique) triangles . Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle. In Δ A B C is an oblique triangle with sides a, b and c
How do you use the law of sines?
The opposite side is the side opposite to the angle of interest,in this case side a.
• The hypotenuse is the side opposite the right angle,in this case side h. The hypotenuse is always the longest side of a right-angled triangle.
• The adjacent side is the remaining side,in this case side b.<|endoftext|>
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# Length of an Archimedean Spiral
By Murray Bourne, 21 Sep 2011
To find the total length of a flat spiral having outer end radius = 15.5 units, inner radius = 5 units & the increase in radius per turn = 0.81 unit, the total No. of turns in the spiral is 7.5.
This is an example of an Archimedean Spiral, otherwise known as an arithmetic spiral, where the arms get bigger by a constant amount for each turn.
We can see Archimedean Spirals in the spring mechanism of clocks and in vinyl records (used by the recording industry before CDs and MP3s); and in tightly coiled rope.
Watch mechanism [Image source: Hamilton Clock]
Coiled rope [Image source: Delderfield]
## However, her question can't be solved
This was another example of an impossible-to-solve reader question. (Some of them are not possible to solve either because there is not enough information given, the algebra is unreadable, or some key vocabulary is not used correctly.)
In this case, it's not possible because it has a fundamental flaw.
If the inner radius is 5 units and the increase in radius per turn is 0.81 units, then 7.5 turns will give us an outer radius of:
5 + 0.81 × 7.5 = 11.075
So the outer radius cannot be 15.5 units if the increase per turn is 0.81.
Here's what the spiral with 0.81 radius increase per turn looks like:
Archimedean spiral, 0.81 units between each arm;
On the other hand, if is it important to have the outer radius being 15.5 units, then the increase per turn would need to be:
(15.5 − 5)/7.5 = 10.5/7.5 = 1.4
Here's the spiral with inner radius 5, outer radius 15.5 after 7.5 turns:
distance between each arm is 1.4 units
The increase per turn is 1.4 units.
## Finding the Length of the Spiral
Before we can find the length of the spiral, we need to know its equation.
An Archimedean Spiral has general equation in polar coordinates:
r = a + bθ,
where
r is the distance from the origin,
a is the start point of the spiral and
b affects the distance between each arm.
(b is the distance between each arm.)
For both spirals given above, a= 5, since the curve starts at 5.
For the first spiral,
b = 0.81, giving us b = 0.12892
So the formula for the first spiral is
r = 5 + 0.12892θ
Using the same process,
b = 1.4, giving us b = 1.4/(2π) = 0.22282
So the formula for the second spiral is:
r = 5 + 0.22282θ
## Length of the first spiral
We'll use the formula for the Arc Length of a Curve in Polar Coordinates to find the length.
The starting value for θ is a = 0 and after 7.5 turns, the end point is b = 7.5 × 2π = 15π = 47.12389.
The derivative under the square root is straightforward:
Substituting everything into our formula gives us:
You can see where that final answer comes from in Wolfram|Alpha. It may be possible to find the actual integral on paper (it involves hyperbolic functions), but why waste our lives doing so? We are interested in the length, not pages of algebra!
## Length of the second spiral
I've solved the second one as an example in the section Arc Length of a Curve in Polar Coordinates.
### 60 Comments on “Length of an Archimedean Spiral”
1. Sami A. Melki says:
Excellent .
These plane geometry equations are simple and beautiful .
2. James says:
The explanation is very clear.
3. brent says:
Hi, thanks for this explanation.
I'm trying to work out the number of turns for a given length of spiral, (I'm an engineer for a garage door company, need to know turns to specify springs), I don't suppose you could give me some pointers getting N when I know L? I could use Solver in excel but I need to do it hundreds of times.
Brent.
4. Murray says:
@Brent: You can't solve it by any straightforward algebraic or calculus process - not that I can see, anyway.
Is your distance between each spiral always constant?
If so, I'm wondering if you can do it by getting a good approximation, then do a few actual calculations to zero in on the number of turns required.
For example, in the example in my post, the "average" radius of all the spirals is around 10 (1/2 way between 5 and 15). We know the length is 379, so the number of turns would be about 379 / (2 * pi * 10) = 6. This misses by quite a bit.
But a better "average" radius is 8 (obtained by working backwards from the known information), and this gives us the number of turns 379 / (2 * pi * 8 ) = 7.5.
Let's try 2 other lengths and see how we go, and I'm assuming our distance between spirals is constant at 1.4.
(1) Length 250 units. The average radius will be less - let's guess (for now) 7.5.
The number of turns will be 250 / (2 * pi * 7.5) = 5.3
Using this and the integration formula, gives 238, which is not bad for a reasonable guess. A small tweak or 2 would get you there.
(2) Going the other way, let's see what happens with length 500.
The average radius will be more, say 8.5.
Number of turns: 500 / (2 * pi * 8.5) = 9.4 turns
Applying the length formula, gives 520, which is once again pretty close. A few tweaks and we'll be very close.
Now, I don't know whether that would save you any work compared to what you are doing now. But maybe if you created a table with a bunch of "good" guesses and the resulting number of turns, and then used some kind of linear (or probably non-linear) interpolation, you would get close each time for a given length.
Does that help or am I describing what you're already doing?
5. brent says:
I'll describe what I ended up doing. I looked at the integral on the wolfram link in your blog post, saw it was a pretty regular shape - looks like a rectangle of width 2.pi.n and height average of top and bottom points. I put all this into an Excel table and used a data table to calculate length for n=1,2..10.
Then I graphed n-L and found that there's a quadratic trendline with and r^2=1 fit. So I used the linest function in excel to estimate the constants in the quadratic equations based on my table of n and L values, and used that to drive the rest of the excel table which calculates the required springs. (I used the quadratic equation formulae to find the roots of the quadratic. 5 years at uni learning maths, used it once now in 9 years of engineering...) I needed to have it in spreadsheet because I have to do this for every height and width combination of the door.
I couldn't have done any of that without your blog post. Thanks again. Brent.
6. Murray says:
Glad you got it to work, Brent.
And you're not the first engineer who has pointed out the lack of utility for much of the math learning at university. Of course, some engineers have got to know some of it, but maybe we need to re-think what is contained in the compulsory math components of engineering degrees.
7. brent says:
The thing about learning all the maths that you do in an engineering degree is not that you'll necessarily do all the things in your career, but that some of the subjects you'll do at uni require it. If you don't do the maths that lead up to it, then you're never going to be able to cope with thermodynamics or fluid mechanics. And you do need to be able to go through some of the maths in those subjects to truly say you understand the basic concepts.
The trouble is that there are simply so many fields that an engineer will go into, and you need to give an understanding of the fundamentals of just about everything: thermodynamics, fluid mechanics, structures, kinematics, control, electrical, materials etc etc etc. I might not have used the higher level maths much since leaving uni, but I've definitely used my understanding of all those subjects on a regular basis, and I gained my understanding by going through the maths.
What I would say to my 21 yr old self if I could: keep better track of your notes and text books, because it might be a decade before you need that information again but you will kick yourself if it's lost forever.
8. Neil says:
Brent,
I hope this helps. It is a couple days late. If you know your overall length (OL), ID of the spiral, and OD of the spiral then the equation for the number of turns (N) is:
N = OL/(PI*(ID+OD)).
Also, on the original posting with the spacing of .81 units, the OD of the spiral should have been 11.075 units not 10.67 units. I think you solved for 7 turns, not 7.5 turns.
9. Murray says:
Hmmm - seems a lot simpler that what Brent and I were talking about!
Thanks for spotting the error - I have amended the post.
10. brent says:
Neil,
Thanks a lot for the information.
As it turns out for a garage door I don't know the OD - it's going to change depending on N.
11. Jacob says:
This is a bit late, but new readers may still find it usefull.
Another way to make a quick approximation is to take the average 🙂 For the first spiral in the article, inner radius is 5, and outer radius is ~ 11. The average radius is then (5+11)/2 = 8. The circumference of a circle with radius 8 is 2*pi*8 = 50.27. 7.5 turns of that length is 8*50.27 = 377.25. This, IMHO at least, is reasonably close to the 'exact' 378.8.
No matter how you look at it, you need three of ID, OD, N and L to compute the fourth - you cannot 'just' deduce N from L if you dont know ID and OD. But knowing ID, OD and L, you can quickly make a good estimate based on circles with diameter (ID+OD)/2.
It is a little different if you have the increase per turn. Say you have a roll of tape and know ID, tape thickness t and tape length L. You can then use the following formula to compute the number of turns (or layers, if you will):
N = (t - ID + sqrt( (ID-t)^2+(4*t*L/pi) )/(2*t)
and from this you can also compute the outer diameter OD
OD = 2*N*t+ID
Note, that both of the above formulae are approximations based on concentric circles, rather than a true spiral. However, the end result is reasonably close. Also note that it is diameters, not radii, you need to plug in!
12. Murray says:
@Jacob - Thanks! We get closer and closer to a good approximation with each new post.
13. Rafael says:
Hi thanks a lot for your highly explanatory blog-entry! I still tried to solve the integral and after petty substitution I ended up with an integral of the form sqrt(x^2+1) ... which can be solved relatively easily once you know the trick (see http://www.matheboard.de/archive/30730/thread.html for example) ... however, in the blogpost from "wolfram" the integral is given by a much more complex formula (http://mathworld.wolfram.com/ArchimedesSpiral.html ) ... now I was wondering if you have any idea where my error lies with the relatively simple method? best regards and please keep up the good and informative posts 🙂
14. Ray Brown says:
Ok I'm trying to solve something for a fiction story and it's been years since I've done calculus. So maybe someone can solve this for me. If you wanted to travel to the galactic core from Earth you would want to travel down the spiral arm and not through the less dense regions if you needed to resupply. So Earth is about 7900Kpc from the galactic core. The start of the spiral arm at the end of the galactic bar is about 3000Kpc from the core. I do not know the distance between the arms. What I'm trying to figure out is how far you would travel down the spiral arm to get to the galactic bar. Any help would be greatly appreciated. Thanks.
15. Murray says:
Hi Ray
I think the dotted magenta distance is what you mean from your description. Before we go any further, is it correct?
[Image source: UniverseToday]
16. brent says:
I think for something with the precision like this you could use the pythagorean approximation just fine.
17. Chris Gavin says:
Hello,
I'm not a mathemetician, but I have an archimedian spiral question too.
I would like to create a plastic spiral for developing long lengths of cine film. If the film length is 15 metres (15000mm), and the inner radius of my spiral should be 20mm, and the outer radius maybe 150mm. Ideally I would like about 2mm spacing between the winds. How many turns would there be?
The spiral will always need to be 15metres long, but I might like to experiment with different radii, spacing and number of turns etc. is there a formula that helps me do the maths for this please?
With the answer, I should be able to draw the correct spiral in Adobe Illustrator, and from there, create the plans needed to have the spiral made out of plastic...
Many thanks if anyone can help.
18. Robert Livesey says:
If spiral equation is r = a + b*Theta, where a is start radius
Spiral length is L = (1/b)Integral(sqr(r^2 + b^2))dr
Solution:
L = .5*[r*sqr(r^2 + b^2) + b^2 * log(r + sqr(r^2 + b^2))]/b
In Excel, in the VB code pages add a module and enter following code:
Option Explicit
Const cPi As Double = 3.14159265358979
Public Function LSpiral(ByVal Theta As Double, Pitch As Double, Start As Double)
Dim b As Double
Dim r As Double
b = 0.5 * Pitch / cPi
r = Start + b * Theta
LSpiral = Integral(r, b) - Integral(Start, b)
End Function
Private Function Integral(x As Double, b As Double)
Dim Tmp As Double
Tmp = Sqr(x ^ 2 + b ^ 2)
Integral = 0.5 * (x * Tmp + b ^ 2 * Log(x + Tmp)) / b
End Function
Then in the spreadsheet have 3 cells for Theta, Pitch and Start - you could have a cell for Number of turns and get theta from Turns * 2Pi
You can play with values to get the length you want
19. Murray says:
@Robert - thanks for the Excel-based approach.
20. R.A,Maxwell says:
hi
I am a co op student and i am trying to make a roll of scotch tape in NX 7.5. I am attempting to use Law Curve to create a spiral line for an extrude to follow. how does the math (r = a + b?) equation or formula flow to create this? the roll needs to be roughly 15 mm thick and the total radius is 32.2 mm.
21. Murray says:
?Hello R.A.
This sounds like an interesting problem. Just to clarify a few things:
(1) What does "NX 7.5" mean?
(2) Does that mean the inner radius is 17.2 mm (since the roll itself is 15 mm thick and the total radius is 32.2?)
(3) How thick is the scotch tape? (I'll assume 0.07 mm, which I found here: http://www.tedpella.com/tape_html/tape.htm)
With these assumptions, we have
a = 17.2 (the inner radius)
b = (0.07 * number of turns)/ (2 pi)
So all together, the formula for your spiral for an angle theta would be:
r = 17.2 + ( (0.07 * number of turns)/ (2 pi)) theta
Does that help to get you started?
22. Singh says:
Hello Mr. Murray,
R.A means nx7.5 software. My query is similar. I want to draw a arithmatic spiral with the help of law curve in nx7.5. Pl. help. Thanks.
23. Murray says:
@Singh: Thanks for the heads up about NX7.5. It looks like great software.
You just need to use the same formula, r = a + b θ, as explained in the article. You can use t instead of θ.
24. Manufacturing Student says:
Hi Murray,
I have to turn a disc on lathe machine. The disc is rotating at 500 rpm. This disc has an Outer Diameter of 70 mm. The tool that will be used to turn this disc is advancing radially toward the disc with a feed rate of 0.1 mm /rev (this simply means the thickness of chip (or shaving) removed from disc would be 0.1 mm). I want to know that if after machining the dia of dic is reduced by 2 mm i.e. it has become 68 mm than what spiral length of material is removed from the disc. And secondly lets say that I want to remove the 2000 mm of spiral length of material from disc in my experiments and I know the starting diameter than is there any formula that I can calculate the dia of disc that would result after removing this much spiral length of material. Thanks
25. Murray says:
Hello
You just need to use the same formula given in the article.
To make sure it makes sense, let's first do 5 spirals:
For 5 rounds:
5 × outer circumference = 5 × (2πr) = 5 × (2π × 35) = 1099.5574
It makes sense, as our distance is slightly less than if it was 5 times around the outer rim.
For the case given, 68 mm final diameter means 10 revolutions, or 20π. The spiral length is:
I don't believe there is a formula for the other way around. But trial and error is quick and easy when using a computer algebra system.
So the amount of disk shaved is
9.215 × 0.1 = 0.9215
26. Manufacturing Student says:
Thanks Murray
It makes a lot of sense. I will use it in my experimental plan and will let you know.
Thanks a lot for your valuable time and effort
Regards
27. raul says:
hi Murray and people! maybe you can help me!
this is for an opensource hw project!
im trying to design a pcb heated bed for a 3d printer. my printer is a delta
This is basically a copper spiral path.
I want to define the overall Resistance and seize: OD (ID could be zero). Since the material is copper and given the standard pcb layer height of 35 um, this resistance just depends on the section width. Lets call it E.
This Width could be aproximated by taking the increase per turn, so
R = e*W*L = e*b*L
i'm trying to come up with some formula o procedure to be able to calculate
>spiral number of turns
given
> OD
> resistance
any ideas?
thanks so much!
R
maths with a purpose are such fun!
28. Hiep says:
using calculus, the answer is quite easy:
let s be length of an circular arc. Then
s = rθ; where θ is angle the arc spans, r=arc radius
In differential form we can write.
ds = rdθ then
s=integration(rdθ) (from 0 to arc angle)
Note that if r is constant, arc angle is 2*pi, then the integration will yield circumference of a circle,2*pi*r
Back to the spiral case:
r = r0 +n*θ where n is spiral constant, r0 is inner radius, r is spiral radius at any angle.
substitute r into ds equation and do integration, we have
spiral length = r0*θ + n*(θ^2)/2
Note: θ is in radian so for example if we have 10 turns spiral, then θ = 20*pi.
to find a spiral length from one angle to another use:
r0*(θ1-θ2) + n*(θ2^2-θ2^2)/2
29. Leah says:
is this a typo?
r0*(θ1-θ2) + n*(θ2^2-θ2^2)/2
(θ2^2-θ2^2) = 0
30. merry gil arroyo says:
How can i get the value of increase in radius?
31. Murray says:
@Merry: Near the top of the article, it says "the distance between each arm is 2πb divided by the number of turns".
Do you know your a and b values in this formula?
r = a + .
32. Clive Pottinger says:
Mr. Bourne
I was trying to solve a problem very similar to the 'galactic spiral' problem posed by Ray Brown. In my case I was thinking of an airplane travelling from a ground point A to ground point C climbing steadily until it reached it's highest altitude H at the midpoint B. The distance travelled from A to B would be an Archimedean spiral and this is the distance I wished to calculate.
However, it dawned on me that I had already calculated the great circle distances for A to B at ground level (gcg) and for A to B at the altitude H (gch). In other words I new how far a car would have to travel along the ground from A to B (gcg), and I knew how far a bird at altitude H would have to travel from A to B (gch). One can then think of the distance travelled by the airplane as the same as that of the car but with the horizontal scale smoothly changing from gcg to gch as it progresses. So the distance travelled would be the average: (gcg + gch) / 2.
So, then I thought of applying that to spirals.
It seems to me then that the length of a spiral segment covering θ degrees could be expressed as the length of the circular arc of θ degrees with a radius that is the average of the radii of the segment's end points.
Am I wrong? My math is not nearly good enough to be sure.
33. Murray says:
@Clive: Sorry about the delay in responding. This may be too late, but for what it's worth, here are some images that may help.
Let's assume our aircraft is partially going around the Earth, which I've drawn with radius = 1. The altitude of the aircraft is 0.5 (totally unrealistic, I know, but exaggerated so we can see what's going on).
One possible way to think about it is that the aircraft climbs and descends at a constant rate (in the x-y sense, relative to the horizon at that point), something like this:
The total distance traveled is:
2 × 0.8 + 1.3 = 2.9
However, aircraft take off horizontally, so maybe we should use tangents the the Earth, giving us this:
The distance travelled in this interpretation is:
2 × 1.12 + 0.32 = 2.56
Another possibility is your spiral suggestion (the second spiral is a bit off in my diagram).
This gives us a total flight of (very approx):
2 × 1.2 + 0.27 = 2.6
A third possibility is to consider the ascent and descent portions to be insignificant for the entire flight (we consider it just goes straight up), and just add the vertical climb plus ground distance. In this case we get:
2 × 0.5 + 1.9 = 2.9
Now to your final suggestion - taking the average of the 2 circular arc distances:
The distance now is 2.37.
I'm not sure we can conclude anything so far, but just wanted to indicate some possibilities.
BTW, I did the above diagrams using GeoGebra. It's an invaluable tool if you want to play with such questions, without getting bogged down in algebra.
34. Leo's Friend says:
Two scenario questions from someine who, obviously, is mathematically ignorant. The first concerns spirals made up of a set total length of "coil" and set distance between turns. The second may represent an infinite-length "coil" of percentage-proportionate distances.
Firstly, does your formula work for spirals going down to [tetminating at "a central-core point] zero (if such is accurately possible)? If such For example, if the spiral has 10 full turns, each turn is separated at an equal distance of 1. Would the formula work to determine the diffetent lengths of each full turn?
Secondarily, what if the distance between each turn differs by a static percentage relative to the length? Example, if the length of the first full turn is 10, the distance between its start point and the start point of the next inner turn is 1 (10% of the length). Ten, the distance between the 2nd and 3rd points would be 10% of the length of the 2nd full turn. Likewise, the distance between the 3rd and 4th would be 10% of the 3rd's full trun length. Etc, etc. Is there a formula yo determine the lengths of each arm in the series? I assume a spiral such as this would never realy achieve zero, but continue reducing in turn length and distance at each full turn.
35. Murray says:
Yes, you can use the formula to find the length of single spirals.
I'm not sure about your second case - I somehow doubt there will be a smooth join for each new arm of the spiral.
36. francesca says:
Hi there, I am trying to find the equation to a spiral I have but I havent been able to understand how to find parameter b.
My spiral has an outer radius of 14 and an inner radius of 2.05. The parameter a=1.7
I dont know if its necessary but the spiral has 10 turns.
Is there any way I can find parameter b with this info?
37. Murray says:
@Francesca: Since your inner radius is 2.05, it means a = 2.05.
In the article under the formula, it says (after I made a correction):
"b is the distance between each arm."
This means:
b = 11.95/10 = 1.195
and solving this gives:
b = 1.195/(2π) = 0.1902
So the equation for your spiral will be:
r = a +
r = 2.05 + 0.1902θ
In this image (made using GeoGebra) you can see it goes from 2.05 to 14 in 10 spirals.
Hope it helps.
38. Mark says:
With a clock hairspring the inner diameter and outer diameter of the coils remains constant at it’s attachments but so does the length of this spiral. So as the inner part of the coil is wound or unwound as in a watch then pitch cannot remain the same.
39. Murray says:
@Mark: Yes, in the case of a watch spring I imagine it will only form a spiral when it is completely unwound. In the wound state, the outer ring of the spiral becomes more straight, and the inner rings become more circular.
40. A7 says:
So.. I cannot solve In _0,^15pi in my Powercalc(my android device) D:
41. Murray says:
@A7: Do you mean ? Actually, you won't be able to solve that on any calculator because is not defined.
42. Alex says:
Can you solve for the amount of rotations when the only given information is the inner radius, the length between each arm, and the total length?
43. Murray says:
44. Mark says:
Hopefully, one will have an easier time by sticking with polar coordinates and not sneaking in rectangular / Cartesian ones. Starting in the same way with the equation of an Archimedean spiral, in this case one that does not start at the origin because :
To find the length of this curve from to , where is the number of turns of the spiral, leave the differential element of the curve as an infinitesimal circular arc instead of using a straight line segment, , then:
45. Murray says:
@MArk: Thanks for the alternate approach! I'm a big fan of simpler is better...
46. Abugheneda says:
Dear Engineers:
this post helped a lot and I am glad that one can find such a valuable information.
I have similar problem with different inputs.I have a disc rotating with certain r.p.m and a pin advancing with constant speed (mm/s) I know the staring Diameter and the stopping.
could I calculate the length of the track? and what effect the diameter of the pin have on the resulting length?
47. Murray says:
@Abugheneda: Yes, your situation can be calculated using the formulas given in the post. For the last part, I don't think the pin's diameter will make any difference to the length, unless the pin is very thick so that it affects how far apart each spiral is.
48. Abugheneda says:
@Murray thanks a lot I will give you feed back with the resulting machine working.....
49. Mark says:
@Murray: You are welcome. One thing i should point out, that got me stuck a bit, is if you are integrating say vectors at each point of this spiral curve (e.g, forces) then leaving them in purely polar form will not work. These vectors have to be projected into their components since integration is a simple addition type operation, while adding vectors uses a parallelogram law except for vector components that lie on the same axes. To bad because it makes for some very,very long equations when applying Castigliano's method to an Archimedean spiral spring.
50. Abugheneda says:
@Murray thanks for your help just one question
since my spiral goes from outside towards the center
will my equation be 'r=a-b*θ'?
51. Murray says:
I believe it would be correct. The distance will be the same whether you are thinking about it stating from the outside moving in, or inside moving out.
52. Chris Zuniga says:
@Murray Hello nice article. Mind lending me a help please?. I'm having some problems at trying to use your formula for calculating the length of a spiral in a CD. In my book the spacing between each arm is given as and the width of the reading sector spans between to .
To calculate the number of turns what I did was to divide the width of this sector by the spacing:
But for purposes of better precision I left it as indicated.
By going to your formula since:
In my case:
and:
, which is the distance from the origin.
In order to get the ending point of the integral you mentioned to
So with this upper boundary is established.
Appending the obtained constants my equations becomes into:
Therefore the required differential would be:
all that was left to do was to plug in the integral:
Now if I followed your steps correctly this solved by Wolfram Alpha renders:
Note: I couldn't make it to work with so instead I used as a variable. This part is odd, well I'm not very familiar with the use of Wolfram Alpha.
Here is the link from Wolfram Alpha
So the result is a value very close to what can be obtained from a formula given above.
which in my case would be:
Both answers seem to check with what my book says the answer.
However I don't understand in your example
Why the formula for the second spiral is given as:
Where the comes from?
If I were to calculate this for my problem what would be the value of the second spiral?
Why is the distance between each arm and not just .
When I followed your steps I felt mostly as reading a cookbook but I was left with the impression of keep investigating into this. At the end of your article you mentioned something about that the actual integral on paper would require hyperbolic functions. Does it mean that it cannot be solved using the ordinary integral methods found in most textbooks?. I'd like to know how could that integral be solved without much fuss as this left me intrigued.
I've attempted to reproduce the integral by replacing with letters A and B as constants but Wolfram Alpha rendered the results with log and none involving hyperbolic functions. Could it be that what I did was wrong?
I'll appreciate so much that you could help me clearing out these doubts.
53. Murray says:
@Chris: I'm impressed you entered all your math using LaTeX!
(1) Where the 0.22282 comes from?
I amended the post to more clearly indicate where that value came from.
(2) If I were to calculate this for my problem what would be the value of the second spiral?
I'm not sure what you mean here. I discussed 2 spirals in the post, but you only appear to be talking about one?
(3) Why is the distance between each arm and not just ?
I have seen it written both ways. I happen to be using the first.
(4) Does it mean that it cannot be solved using the ordinary integral methods found in most textbooks?
You can quite easily integrate hyperbolic functions - I chose not to as it was beyond the scope of this post.
(5) Wolfram Alpha rendered the results with log and none involving hyperbolic functions. Could it be that what I did was wrong?
Looks OK to me. But we just need a numerical answer, so no need to worry about the algebraic (indefinite) integral.
And finally...
(6) Why not ?
is actually a constant, related to the Golden Ratio. See https://www.wolframalpha.com/input/?i=phi
So Wolfram|Alpha thought you wanted that constant, and didn't realise you were using as a variable. Using x, t or theta are usually "safe" variables for computer algebra systems like Wolfram|Alpha.
Hope it helps!
54. Chris Zuniga says:
@Murray
Thanks for your kind words of compliment. In fact I have some experience by using LaTex due I'm constantly asking questions on math in different forums. But so far I found your post the most informative one.
Concerning to the value you referred to 0.22282 I totally overlooked the fact that the OP made a question which "cannot be solved" since apparently assigned an incorrect jump or spacing so that the No. of turns in the spiral to be 7.5 and that cannot be achieved using 0.81, so you made clear later if the distance is 1.4 units then the number of turns would be 7.5. I thought this would apply in my case. But the problem I posed was correctly made. The recent edit you made explained it better and this by extension answers the second question.
However the third question is where I'm still stuck at.
If I do understand correctly in the formula of the spiral what we have is
with and being real numbers. If I understood correctly. But if is the distance from the center to the point where the spiral begins and this is stated in my case
Why is not "transformed" or "equated" to whereas in problem is and not that's the thing which I'm not getting it.
Could it be that it can be related to this formula?
?
If so what is the justification? Is this used to transform from meters to radians? I'm just as explained confused about the meaning of parameter and how is it different from .
Again If that's the case (if it is transformed) I don't understand why one gets its value to stand as it is and the other requires the transformation?. Since they are both real numbers already why one has to change and the other not?. Mind you explaining this to me a bit better?.
You mentioned something about seen it written in both ways. This latter comment I don't understand. What did you meant by written in both ways? Where is the second one?
I am a bit rusty with integration methods but upon an inspection on what I recall I don't find what transformation or change of variable should I use to integrate the function obtained in the end.
When you mentioned that it is okay to you about the logarithmic functions in the resulting integral obtained by Wolfram Alpha did you meant that there was no discrepancy with the supposedly hyperbolic function which should appear in the answer?. How can this happen?.
As mentioned (and i can understand that it was beyond the scope of the question) I'm curious of how to obtain the indefinite integral. Thus I want to try out this excerise on my own using pen and paper.
Regarding the use of instead of it is a funny story. Back in high school I had an awful experience with trigonometry and as a result I developed an apprehension to and more specifically to . Years later when I studied chemistry in university I had this physics course in optics where the notation often used is and and since that memories bring me more joy (triggered also as I obtained good grades and developed a close friendship with the professor). I prefer to stick to those letters. Maybe it is just a subjective choice but that doesn't really change the meaning of what it was shown. I hope you understand me. However there is nothing wrong with using the "ordinary and usual" letters such as cosines law which is the only exception I do since it is easy to relate the arabic letters with their corresponding greek equivalents.
Out of curiosity one user @Mark suggested to use a much simpler approach
Now If you don't mind for purposes of brevity and to my mental health I'll rewrite the latter formula into:
(I'm letting , because it reminds me the value of inductive resistance from electronics which it happens to be my hobby as well)
Anyways... by plug in the values I obtained earlier this became into:
To which evaluated using Wolfram Alpha I obtain:
which amazingly it is very much exactly as what I obtained from your method.
The above method which I posted in my earlier comment was based on the fact that if you consider the area of the spiral to be a flat doughnut then this disk when unwind it results in a parallelogram and therefore the height is considered to be the separation between the arms or the tracks and the length is what it is to be calculated. As indicated this is an approximation and is very similar also to the result. However I'm concerned whether how accurate is this approximation since when you calculate the area by considering that it is a disk there is inevitable some sector which is added in both ends as the spiral doesn't end like a circle does.
I'm sorry If my questions seem redundant but I'd really hope you can help me to clear out these doubts. I'd like to write down this in my notes book to review it frequently.
55. Murray says:
Your questions are certainly not "redundant"! Good on you for pursuing it until you understand it.
"But if is the distance from the center to the point where the spiral begins and this is stated in my case
Why is not "transformed" or "equated" to whereas in problem is and not that's the thing which I'm not getting it."
Your value of 0.026 is the starting value for your spiral, and if it's divided by or anything else, then when (that is, when we substitute it), won't be your starting value any more.
Could it be that it can be related to this formula?
?
Hopefully this picture helps to explain it.
I've taken the most fundamental case possible, with and , so the curve is . As values of go from to , you can see we are at on the horizontal axis. Similarly if we go around again, we''ll be at and once more gets us to .
That's how polar coordinates work - for whatever value of , we need to move away from the origin by that much.
So since one turn is equivalent to radians, then we need to divide our in the equation by for it all to work.
If it's still not quite clear, I suggest substituting a bunch of values for and seeing what values of result.
There are often 2 or 3 ways to find an answer - I'm glad this post surfaced some of them!
56. Nagesh says:
Now for a process of oversimplification for the sake of market and the tools on our PC.
Inner dia is known D1, outer dia is known Dn and pitch is known as 'P'. well create a spreadsheet calculator and be done with it.
As many concentric circles as the pitch vs differential of outer dia and inner dia! (Dn-D1)/P etc., It shall be within .2% error, not good enough for space travel but enough for commerce.
I wonder if transformer winding fellows have got it figured?
I was trying to calculate as to how much area of 1.2mm thick leather hide can be packed in a Tube of 120 cm and 20cm dia!
Well, I was wondering if there was a formula for spiral length and here I am!
QED
57. Dogan says:
I have the following question:
This sequence formula 1/4 (12 n^2 - 6 n + (-1)^n (4 n - 1) + 1) should indicate the arc length each from zero point on an Archimedean spiral. A point should be created at each end of the arc length. The distances between the spiral tracks are sought. When trying it out experimentally, the points of the arc lengths are roughly arranged at an angle of 120 °. The probably distance is e + 1. how can i calculate it mathematically?
---------------------------------------------------------------
n | \frac{1}{4} (12 n^{2} + 4 (-1)^{n} n - 6 n + (-1)^{n+1} + 1)
1 | 1
2 | 11
3 | 20
4 | 46
5 | 63
6 | 105
7 | 130
8 | 188
9 | 221
10 | 295
11 | 336
12 | 426
13 | 475
14 | 581
15 | 638
58. Murray says:
@Dogan: I'm not sure where that sequence formula comes from, but it doesn't quite give us an Archimedean spiral.
I plotted the points given by the formula (with 120° spacing between each spiral arm) - the dark green segments - and then plotted an actual Archimedean spiral as close as I could to those points - in magenta.
In places it comes close, but mostly it doesn't work well.
The spiral gap is about 130 or so, but it would not be possible to calculate it for the sequence points as they are not evenly spaced.
59. Simon Geard says:
The integral for arc length is quite straightforward to obtain using standard techniques (substitution and either integration by parts of double-angle formula for hyperbolic functions). The result is:
where and are the inner and outer radius of the spiral and is the increase in radius per radian, is the increase in radius per turn:
Assuming >> :
For the case: a = 5 , b = 11.075 , p = 0.81 => L = 378.8095
This formula can be adapted for the number of turns by using the fact that
then using the approximation gives a quadratic in :
provided << since the log series is slow to converge.
For the values above the result is
n L(n) exact
1 33.9708 33.9703
2 73.0304 73.0287
3 117.1789 117.1755
4 166.4162 166.4109
5 220.7424 220.7351
6 280.1574 280.1482
7 344.6612 344.6502
8 414.2539 414.2411
9 488.9354 488.9212
10 568.7058 568.6903
11 653.5650 653.5486
12 743.5130 743.4961
13 838.5499 838.5327
14 938.6756 938.6585
15 1043.8902 1043.8736
16 1154.1936 1154.1779
17 1269.5859 1269.5714
18 1390.0670 1390.0542
19 1515.6369 1515.6263
20 1646.2957 1646.2877
If the spiral is modelled as a series of concentric circles the algebra is much more tedius. The result depends on the value of radius used for each ring. If the value for the i'th ring is
then
with corresponding to the average radius. Therefore the length of a spiral calculated using this value will always be an underestimate.
60. Bill Etienne says:
I have a problem that is similar to the original:
You have a perfect white sphere one meter in diameter and a remote control that will allow you to roll it in whatever direction you choose with mathematical precision. The sphere is at rest in a wide pool of black ink one centimeter deep. The floor of the pool is a perfectly level plane. What is the minimum distance the sphere must roll to be entirely covered in ink? What does its path look like? The ink does not bleed or run. Once a section of the sphere has touched ink, that section stays inked.
I believe I have the correct answer but I would like to run it by someone else to be sure.
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Permutations and Combinations
Dr. Math FAQ || Classic Problems || Formulas || Search Dr. Math || Dr. Math Home
For a basic review of concepts, see Introduction to Probability. And see Fast Food Combinations: How many possible combinations can be made from a special menu of eight items?
#### Permutations
Suppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.
Because order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:
CA CT AC AT TC TA
Now let's suppose we have 10 letters and want to make groupings of 4 letters. It's harder to list all those permutations. To find the number of four-letter permutations that we can make from 10 letters without repeated letters (10_P_4), we'd like to have a formula because there are 5040 such permutations and we don't want to write them all out!
For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040. This is part of a factorial (see note).
To arrive at 10 x 9 x 8 x 7, we need to divide 10 factorial (10 because there are ten objects) by (10-4) factorial (subtracting from the total number of objects from which we're choosing the number of objects in each permutation). You can see below that we can divide the numerator by 6 x 5 x 4 x 3 x 2 x 1:
``` 10! 10! 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
10_P_4 = ------- = ---- = --------------------------------------
(10 - 4)! 6! 6 x 5 x 4 x 3 x 2 x 1
= 10 x 9 x 8 x 7 = 5040
```
From this we can see that the more general formula for finding the number of permutations of size k taken from n objects is:
``` n!
n_P_k = --------
(n - k)!
```
For our CAT example, we have:
``` 3! 3 x 2 x 1
3_P_2 = ---- = ----------- = 6
1! 1
```
We can use any one of the three letters in CAT as the first member of a permutation. There are three choices for the first letter: C, A, or T. After we've chosen one of these, only two choices remain for the second letter. To find the number of permutations we multiply: 3 x 2 = 6.
Note: What's a factorial? A factorial is written using an exclamation point - for example, 10 factorial is written 10! - and means multiply 10 times 9 times 8 times 7... all the way down to 1.
#### Combinations
When we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:
CA CT AT
We say '3 choose 2' and write 3_C_2. But now let's imagine that we have 10 letters from which we wish to choose 4. To calculate 10_C_4, which is 210, we don't want to have to write all the combinations out!
Since we already know that 10_P_4 = 5040, we can use this information to find 10_C_4. Let's think about how we got that answer of 5040. We found all the possible combinations of 4 that can be taken from 10 (10_C_4). Then we found all the ways that four letters in those groups of size 4 can be arranged: 4 x 3 x 2 x 1 = 4! = 24. Thus the total number of permutations of size 4 taken from a set of size 10 is equal to 4! times the total number of combinations of size 4 taken from a set of size 10: 10_P_4 = 4! x 10_C_4.
When we divide both sides of this equation by 4! we see that the total number of combinations of size 4 taken from a set of size 10 is equal to the number of permutations of size 4 taken from a set of size 10 divided by 4!. This makes it possible to write a formula for finding 10_C_4:
``` 10_P_4 10! 10!
10_C_4 = -------- = ------- = ----------
4! 4! x 6! 4!(10-4)!
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= --------------------------------------
4 x 3 x 2 x 1 (6 x 5 x 4 x 3 x 2 x 1)
10 x 9 x 8 x 7 5040
= -------------- = ------ = 210
4 x 3 x 2 x 1 24
```
More generally, the formula for finding the number of combinations of k objects you can choose from a set of n objects is:
``` n!
n_C_k = ----------
k!(n - k)!
```
For our CAT example, we do the following:
``` 3! 3 x 2 x 1 6
3_C_2 = ------ = ----------- = --- = 3
2!(1!) 2 x 1 (1) 2
```
#### Pascal's Triangle
We can also use Pascal's Triangle to find combinations:
``` Row 0 1
Row 1 1 1
Row 2 1 2 1
Row 3 1 3 3 1
Row 4 1 4 6 4 1
Row 5 1 5 10 10 5 1
Row 6 1 6 15 20 15 6 1
```
Pascal's Triangle continues on forever - it can have an infinite number of rows. Each number is the sum of the two numbers just above it. For the 1 at the beginning of each row, we imagine that Pascal's triangle is surrounded by zeros: to get the first 1 in any row except row 0, add a zero from the upper left to the 1 above and to the right. To get the 3 in row 4, add the 1 left and above to the 2 right and above.
To find the number of combinations of two objects that can be taken from a set of three objects, all we need to do is look at the second entry in row 3 (remember that the 1 at the top of the triangle is always counted as row zero and that a 1 on the lefthand side of the triangle is always counted as entry zero for that row).
Looking at the triangle, we see that the second entry in row 3 is 3, which is the same answer we got when we wrote down all the two-letter combinations for the letters in the word CAT.
``` Row 0 1
Row 1 1 1
Row 2 1 2 1
Row 3 1 3 3 1
```
Suppose we want to find 10_C_4? To use Pascal's Triangle we would need to write out 10 rows of the triangle. This is a good time to use a formula.
More generally, to find n_C_k ("n choose k"), just choose entry k in row n of Pascal's Triangle.
One of the hardest parts about doing problems that use permutations and combinations is deciding which formula to use.<|endoftext|>
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Coral reefs feed and shelter thousands of species of fish, buffer coastal areas against waves and storms, and support local economies through tourism and fishing.
But they’re increasingly threatened by a dramatic and destructive stress response known as coral bleaching. Bleaching occurs when prolonged exposure to warmer ocean temperatures and other stressors causes corals to expel their symbiotic algae, leaving corals’ white skeletons visible. (These algae give corals their color and nourish them through photosynthesis.)
Use the slider to compare bleaching projections for the 2030s and 2050s under a business-as-usual emissions scenario. Data source: WRI Reefs at Risk, adapted from Donner, S.D., 2009 study in PLoS ONE.
The map above shows the projected frequency of what the National Oceanic and Atmospheric Organization (NOAA) calls Bleaching Alert Level 2 Events for the 2030s and 2050s. (Specifically, it shows the percentage of years in each decade that these events will likely occur.) A Bleaching Alert Level 2 Event indicates intense heat stress and a likelihood that corals will die.
The brighter the color, the more frequently these events are likely to occur. So, that wash of yellow and green across the map? It signifies a dramatic rise in both the range and frequency of the most severe levels of bleaching if carbon emissions continue unchecked.
An analysis in the World Resources Institute’s Reefs at Risk Revisited report found that under a business-as-usual scenario for greenhouse gas emissions, roughly half of the world’s reefs could experience enough thermal stress to induce severe bleaching in five out of 10 years during the 2030s. During the 2050s, this percentage is expected to grow to more than 95 percent.
NOAA announced in summer 2017 that the third global coral bleaching event—the longest and most widespread so far—appeared to be on the wane. But it’s a temporary reprieve. As the map above reveals, things are set to get much worse.
Further explore this data.<|endoftext|>
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We all know that greenhouse gas and carbon emission are the primary factors that catalyse global warming. That is why several natural environmental organisations from all over the world encourage everyone to reduce emissions.
Awareness is the first step we can do in order to reduce carbon footprint. We need to identify where and how these gasses are produced. We also need to know how we can change our human lifestyle so that we can reduce the production of such gasses. We can also slow down the rate of destruction of rainforests, so as to enhance the sequestration process. Under the sequestration process, on which scientists are working, carbon is locked up on a long-term basis.
What is carbon footprint? It is the amount of carbon dioxide put into the environment by households and businesses. A carbon footprint is composed of both a direct or primary footprint, and an indirect or secondary footprint. Our direct emissions arising from burning of fossil fuels when measured give us the primary footprint. Meanwhile emissions from the life-cycle of products we use compose the secondary footprint.
Typically, measuring carbon footprint production is done through extraction of raw materials, producing the goods, and transporting and distributing them for final disposal with the ultimate consumer. This measurement process is called a carbon audit. It will list various activities relating to the business and quantify them in terms of tonnes of carbon produced.
Once measuring the carbon footprint of a certain business, and found out that it is producing a lot, the next step is to educate them how to reduce that emission. Reduced use of gas, electricity, water, and oil apart from sourcing raw materials locally to reduce transportation will all help achieve this end.
Carbon emission has continued to increase and it can create hazard to the planet. According to certain energy agencies, global carbon emission hits record high last 2018. Read the full report here: https://www.abc.net.au/news/2019-03-26/global-carbon-emissions-hit-record-high-in-2018-according-to-iea/10941378?section=business<|endoftext|>
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The Danish Resistance movement during World War Two was in a curious position. In theory, Denmark was not officially at war with Nazi Germany (though clearly Denmark had been illegally occupied by the Germans in 1940) as the government had not declared war on Germany. The government and king, Christian X, had made a formal protest but agreed to a German decision that gave Denmark ‘independence’ despite having German troops stationed there against the wishes of the Danish government.
So any form of Danish resistance could not be ‘legalised’ by the Allies. Though the government in Copenhagen had accepted as a fait accompli that Denmark had been occupied, many Danes did not. Much of the Danish Navy had sailed to Allied ports and Danish ambassadors abroad had refused to accept their government’s decision.
A Danish Resistance movement did exist. Many of those in it had been in the Danish Army. Those in the resistance were willing to pass on intelligence to the Special Operations Executive (SOE) but refused to get involved in any sabotage operations called for by SOE. Any sabotage that did take place was sanctioned by resistance leaders within Denmark or based in Stockholm. There was an increase in acts of sabotage within Denmark from 1943 on.
Up to 1943, the Germans within Denmark had had a relatively easy time – for an occupying force. However, sabotage within Denmark led to a more marked hardening of attitude by the Germans. The arrest of resistance suspects usually led to strikes. This led to more arrests for civil disobedience, which caused more strikes.
By August 1943, the situation had become so bad, that the Germans sent the Danish government an ultimatum – they were to declare a state of emergency and they were to condemn to death all captured saboteurs. The government refused to do this and resigned. The Germans responded by formally seizing power and, legally, Denmark became an “occupied country”. It was only after this occurred that the Danish Resistance became legitimised as their actions were now against the Germans.
In September 1943, the ‘Danish Freedom Council’ was created. This attempted to unify the many different groups that made up the Danish resistance movement. The council was made up of seven resistance representatives and one member of SOE. The resistance movement grew to over 20,000 and in the lead-up to D-Day acts of sabotage markedly increased. Though the D-Day landings were to be in Normandy, SOE believed that the more German soldiers tied up elsewhere in Europe, the less that could be present in northern France. Therefore, the more acts of sabotage in Denmark, the more German troops would be tied down there.
The Danish Resistance used the country’s proximity to Sweden to great effect. Stockholm became an actual base for the Danish Resistance. Here they were far safer than in Denmark – but they could easily get back to their country. The sea route also allowed the Danish Resistance to get out of the country over 7,000 of Denmark’s 8,000 Jews. Because of this, Denmark had one of the lowest statistical casualty rates for Jews in the war.<|endoftext|>
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After doing the same digital citizenship lesson in 7 classes, grades 3-5, I’m not surprised to find out that not one student knew the definition of digital citizen. In all 7 classes, however, students were successful at giving a definition for citizen and digital. That discussion made it much easier to put the definitions together to correctly define digital citizen. It made for a great start to our lessons, and it allowed us to eventually start blogging in the classroom.
No matter what grade level you teach, digital citizenship is an important topic that needs to be covered in every K-12 class. Ideally, it’s a topic that you want to cover at the beginning of the school year, before having students interact with technology, but even though it is more than 6 months into the school year, it’s never too late to start.
Here are 5 reasons you should teach about digital citizenship:
- Digital citizenship is SO Common Core! – CCSS focuses on college and career readiness. Digital citizenship is a a 21st-century life skill that will definitely help students through their college years, career, and beyond.
- Free Lessons – Common Sense Media has a webpage dedicated to K-12 Digital Literacy and Classroom Curriculum.
- Protect Your Students’ Digital Footprint – Digital Citizenship gets students to really THINK before they post something online. Students don’t always remember that whatever they post online can follow them forever. Start teaching this to students early!
- Prevent Cyberbullying – Promote positive interactions between students and teach students how to deal with cyberbullying. You can give students the knowledge and power to be online upstanders.
- It’s Fun! – Digital Citizenship leads to fun, engaging classroom activities using social media in the classroom. Sites I love using are: Kidblog, Storybird, Padlet, and Today’s Meet. Eventually, I would love to start using Twitter, Instagram, and Google Hangouts with students.
It is now our job as educators to raise model digital citizens. Have fun with your 21st-century digital citizens! Once you do, the learning possibilities in your classroom are endless!<|endoftext|>
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Only icebergs that have a side measuring at least 19 kilometers (12 miles) long are named and tracked by the U.S. National Ice Center. That means nearly round or square icebergs—like the one pictured above—can be quite large and still not meet the criteria for naming and tracking.
That is the case with a piece of ice detected by satellite this week. On December 3, 2014, the Operational Land Imager (OLI) on Landsat 8 acquired an image (top) of an unnamed iceberg adrift in the South Atlantic Ocean. White clouds partially obscured the white surface of the ice. Two days later, the Moderate Resolution Imaging Spectroradiometer (MODIS) on NASA’s Aqua satellite acquired a cloud-free view of the berg (bottom).
On December 3, the iceberg was located in the South Atlantic about 240 kilometers (150 miles) west of South Georgia, and measured about 165 square kilometers (64 square miles)—approximately the same area as Galveston Island, Texas. That's where the similarities end. An Antarctic iceberg with the size and shape of Galveston would be long enough to be named and tracked.
Scientists are unsure about where on Antarctica the iceberg got its start. However, its current location east of the southern tip of South America is not unusual.
"Many bergs get trapped in, and then spin off of, the currents that go around Antarctica," said Kelly Brunt, a NASA glaciologist whose research includes the study of icebergs. "They often start heading north, especially where the currents get interrupted around South Georgia."
For example, an iceberg named C-16 floated more than halfway around Antarctica before moving north. The map below shows the trajectory of C-16 from March 2000 through February 2014. The berg moved counter-clockwise around the continent and stayed in the Antarctic Coastal Current until it turned north upon reaching the Weddell Sea. "My guess is that when it approached the Weddell Sea Gyre, following the coastal current became difficult," Brunt said.
NASA image (top) by Jeff Schmaltz, LANCE/EOSDIS Rapid Response, and by Jesse Allen, using Landsat data from the U.S. Geological Survey (middle). Map (bottom) by NASA/Kelly Brunt based on data provided by BYU/David Long. Caption by Kathryn Hansen.<|endoftext|>
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Dominic Ford, Editor
From the Earth feed
21 June will be the longest day of 2018 in the northern hemisphere, midsummer day.
This is the day of the year when the Sun's annual passage through the constellations of the zodiac carries it to its most northerly point in the sky, in the constellation of Cancer at a declination of 23.5°N.
On this day, the Sun is above the horizon for the longer than on any other day of the year in the northern hemisphere. This is counted by astronomers to be the first day of summer.
In the southern hemisphere, the Sun is above the horizon for less time than on any other day of the year and astronomers define this day to be the first day of winter.
Sunrise and sunset times for Cambridge|
Sunrise and sunset times
The table to the right lists the sunrise and sunset times in Cambridge around the solstice. In June the Sun's right ascension changes slightly faster than at other times of year, which means that each solar day lasts fractionally longer than 24 hours and the time of noon moves around a minute later each day.
This phenomenon is described by the equation of time.
The shift also affects sunrise and sunset times, and means that the latest sunset and earliest sunrise do not occur on the day of the solstice itself. Instead, the earliest sunrise occurs a few days beforehand, and the latest sunset is a few days later.
Solstices occur because the axis of the Earth's spin – its polar axis – is tilted at an angle of 23.5° to the plane of its orbit around the Sun.
The direction of the Earth's spin axis remains fixed in space as it circles around the Sun, while the Earth's sight line to the Sun moves through the constellations of the zodiac. As a result, sometimes the Earth's north pole is tilted towards the Sun (in June), and sometimes it is tilted away from it (in December). This gives rise to the Earth's seasons:
The date of the solstice
|Year||Time of solstice|
|2014||21 Jun 06:47 EDT|
|2015||21 Jun 12:31 EDT|
|2016||20 Jun 18:25 EDT|
|2017||21 Jun 00:12 EDT|
|2018||21 Jun 05:54 EDT|
|2019||21 Jun 11:40 EDT|
|2020||20 Jun 17:29 EDT|
|2021||20 Jun 23:18 EDT|
|2022||21 Jun 05:01 EDT|
The Earth orbits the Sun once every 365.242 days, and this is the time period over which the cycle of solstices and equinoxes, and consequently all the Earth's seasons, repeat from one year to the next.
In any year which is not a leap year, the solstices occur roughly 5 hours and 48 minutes – just under a quarter of a day – later from one year to the next.
This is why the seasons would drift later in the year if it was not for an additional day being inserted inserted into every fourth year on 29 February.
The chart below shows the day of the month when the June solstice falls in each year. The gradual drift of the four-year cycle earlier in the month is due to the equinoxes repeating 12 minutes less than a quarter of a day later each year.
In the Gregorian calendar, this is fixed by omitting leap years in three out of every four century years, e.g. 1700, 1800 and 1900, but not 2000.
Measuring the radius of the Earth
At the solstice, the Sun appears overhead at noon when observed from locations on the tropic of Cancer, at a latitude 23.5°N.
This fact was used by the ancient Greek astronomer Eratrosthenes in around 200 BC to work out the radius of the Earth for the first time. He knew that at midsummer, the Sun appeared exactly overhead in the Egyptian city of Swenet (now Aswan), because its light shone right down to the bottom of deep wells.
He travelled to Alexandria, on the Egyptian north coast, at a distance of 5,000 stades from Swenet. Here, he used a stick in the ground to determine that the Sun was seven degrees away from the zenith at midsummer, implying that a distance of 5,000 stades around the circumference of the Earth corresponded to a distance of seven degrees around the Earth's curved surface.
Thanks to this experiment, the ancient Greeks were well aware that the Earth was spherical and had a very good idea exactly how big it was, long before anyone had circumnavigated the globe.
The 2018 solstice
The exact position of the Sun when it reaches its most southerly declination in 2018 will be (J2000.0 coordinates):
|Object||Right Ascension||Declination||Constellation||Angular Size|
|The sky on 21 June 2018|
8 days old
All times shown in EDT.
The circumstances of this event were computed using the DE405 planetary ephemeris published by the Jet Propulsion Laboratory (JPL).
This event was automatically generated by searching the ephemeris for planetary alignments which are of interest to amateur astronomers, and the text above was generated based on an estimate of your location.<|endoftext|>
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You don't need to know advanced math to program, but you need to know how to use basic math functions to communicate with a computer. To create computer graphics, use functions and precise numbers to instruct the computer what to draw.
Using the Math block to perform simple math operations. It will return the result of the operation. For example, in this block, the value of
x will set to
time + 50:
3 . The 'time' block counts from 0 to 100. But this time you want the horizontal position of the red ball to start at 100 and go to 0. Can you figure out a simple mathematical formula that flips the direction?
4 . Create four circles on the four edges and make each circle move to its opposite side.
5 . It's straightforward to get the big circle to move correctly, but you'll need some math to move the small circles in the right location.
6 . These two lines have the same starting point, but you need to specify the motion of both their starting and ending points.
7 . When programming, you sometimes are given a specific function you need to implement. You'll need to nest many math blocks together to write the function. (Once you switch to text-based programming, these lines become easier to write!)
Try to solve the harder levels 8 and 9 as well, and make any animation you want for level 10!
More games at Blockly Games<|endoftext|>
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From 1998 to 2013, about one fifth of the Earth’s land surface covered by vegetation showed persistent and declining trends in productivity.
In some cases, advanced stages of land degradation there are leading to desertification in dryland areas, particularly in the grasslands and rangelands.
Land and soil degradation undermine the security and development of all countries. Reversing the effects of land degradation and desertification through sustainable land management is key to improving the lives and livelihoods of more than 1 billion people currently under threat.
Source - SDG Goal 15
KAIINOS, is using open satellite data and helping such organizations measure the impact their work quantitatively.
acres geographical area
land use practices
Satellite datasets have been used to understand temporal changes in land cover. Elevation datasets are used to derive the water stream networks. Based on these two datasets current land use is generated and compared to historical data.
This approach helps us to quantify the changes that have happened over a period of time and also enable us to plan the eco-restoration.<|endoftext|>
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A free online course for homeschooling families who wish to improve their children's math learning experience.
In parallel, this course is also offered in Spanish.
Note: This course is now self-paced. You can enroll, start and finish any time you wish.
If you enrolled already in this course at an earlier date, enter with that same account and continue.
(In case you forgot your password, on the login page click on "Forgot your Password?". You will then have to indicate your email address - the same you used for creating your account.)
In this course we will explore several creative ways of learning and teaching mathematics, corresponding to children's needs. Course participants (parents and teachers) will review and adjust their own concept about mathematics, and will practice mathematical thinking. At the same time they will realize some practical math projects with their children. So the course includes also the children's math classes. The methods applied here are mostly based on the concept of "Active School". The most important goals of the course are that participants:
For whom is this course?
This course is mostly designed for homeschooling families. This surrounding provides the best opportunities and the greatest freedom for practicing the projects suggested in this course. But other families, school teachers, students, and other interested participants may also benefit from this course, adapting the projects to their specific situation.
The projects are optimized for children from 6 to 12 years old; but you will also receive some hints of how to adapt them for smaller or older children.
Course Structure and Workload:
The course consists of five modules. Each module requires:
Additionally, it is much recommended to participate in the discussion forums related to the course.
The recommended total time required for completing each module is 15 hours distributed over two weeks. About half of this time will be invested into the practical project with the children.
Summary of Course Contents
Module 1: Let's start with a concrete activity
We will begin to acquire a new perspective about math; we will talk about the importance of concrete activities for learning; and we will realize our first practical project (Estimating and measuring quantities).
Module 2: Mathematics is understanding and applying principles
We will learn about principle-based math teaching, and we will realize a practical project about the principles which govern the decimal system, using an abacus and other related materials.
Module 3: Doing mathematics is creating, investigating and discovering.
How does the child's own investigative activity help to learn math? - The practical project of this module encourages children to discover mathematical properties by themselves.
Module 4: Continue thinking mathematically
We will go deeper into some elements of mathematical thinking, and we will realize another mathematical investigation.
Module 5: Your own project
We will talk about how to build a positive relationship with mathematics. - In this module, participants design a project based on their own ideas.
Some of the course videos are publicly accessible at https://youtube.com/user/educadorDiferente/.
Prerequisites for participants:
A disposition for changing your views about math, and for practicing new ways in learning and teaching math.
Opportunity for realizing the practical projects with your own children or with a comparable group.
No special mathematical knowledge is necessary.
Note: This is an introductory course. Other more advanced courses which build upon this one are planned for the future.
|প্রশিক্ষণ - "তথ্য অধিকার আইন ২০০৯”|<|endoftext|>
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# 2.4 – Zeros of Polynomial Functions
## Presentation on theme: "2.4 – Zeros of Polynomial Functions"— Presentation transcript:
2.4 – Zeros of Polynomial Functions
Rational Zero Theorem:
If f is a polynomial function of the form f(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 with degree n > 1, integer coefficients, and a0 ≠ 0, then every rational zero of f has the form p/q, where p is all possible factors of a0 and q is all possible factors of an.
Ex. 1 List all possible rational zeros of each function
Ex. 1 List all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 + 4x3 – 12x – 9
Ex. 1 List all possible rational zeros of each function
Ex. 1 List all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 + 4x3 – 12x – 9 p = ±9, ±3, ±1 = ±9, ±3, ±1 (six possible) q ±1
Ex. 1 List all possible rational zeros of each function
Ex. 1 List all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 + 4x3 – 12x – 9 p = ±9, ±3, ±1 = ±9, ±3, ±1 (six possible) q ±1 Test all possibilities using the Factor Theorem.
Ex. 1 List all possible rational zeros of each function
Ex. 1 List all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 + 4x3 – 12x – 9 p = ±9, ±3, ±1 = ±9, ±3, ±1 (six possible) q ±1 Test all possibilities using the Factor Theorem. g(9) = (9)4 + 4(9)3 – 12(9) – 9 g(-9) = (-9)4 + 4(-9)3 – 12(-9) – 9 g(3) = (3)4 + 4(3)3 – 12(3) – 9 g(-3) = (-3)4 + 4(-3)3 – 12(-3) – 9 g(1) = (1)4 + 4(1)3 – 12(1) – 9 g(-1) = (-1)4 + 4(-1)3 – 12(-1) – 9
Ex. 1 List all possible rational zeros of each function
Ex. 1 List all possible rational zeros of each function. Then determine which, if any, are zeros. g(x) = x4 + 4x3 – 12x – 9 p = ±9, ±3, ±1 = ±9, ±3, ±1 (six possible) q ±1 Test all possibilities using the Factor Theorem. g(9) = (9)4 + 4(9)3 – 12(9) – 9 = 9360 g(-9) = (-9)4 + 4(-9)3 – 12(-9) – 9 = 3744 g(3) = (3)4 + 4(3)3 – 12(3) – 9 = 144 g(-3) = (-3)4 + 4(-3)3 – 12(-3) – 9 = 0 g(1) = (1)4 + 4(1)3 – 12(1) – 9 = -16 g(-1) = (-1)4 + 4(-1)3 – 12(-1) – 9 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0 *Find real zeros using graphing calculator.
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0 *Find real zeros using graphing calculator. x = -2, x = -1, x = 3
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0 *Find real zeros using graphing calculator. x = -2, x = -1, x = 3 *Divide to find remaining zero.
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0 *Find real zeros using graphing calculator. x = -2, x = -1, x = 3 *Divide to find remaining zero. -2| |
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0
Ex. 2 Solve the equation. x4 + 2x3 – 7x2 – 20x – 12 = 0 *Find real zeros using graphing calculator. x = -2, x = -1, x = 3 *Divide to find remaining zero. -2| | 0 x3 – 7x – 6
-1| |
-1| | 0 x2 – x – 6
-1| 1 0 -7 -6 -1 1 6 1 -1 -6 | 0 x2 – x – 6 So (x + 2)(x + 1)(x2 – x – 6) = 0
-1| | 0 x2 – x – 6 So (x + 2)(x + 1)(x2 – x – 6) = 0 (x + 2)(x + 1)(x – 3)(x + 2) = 0
-1| | 0 x2 – x – 6 So (x + 2)(x + 1)(x2 – x – 6) = 0 (x + 2)(x + 1)(x – 3)(x + 2) = 0 (x + 1)(x – 3)(x + 2)2 = 0
-1| | 0 x2 – x – 6 So (x + 2)(x + 1)(x2 – x – 6) = 0 (x + 2)(x + 1)(x – 3)(x + 2) = 0 (x + 1)(x – 3)(x + 2)2 = 0 So x = -1, x = 3, and x = -2 (twice).
Ex. 3 Write each function as (a) the product of linear and irreducible factors and (b) the product of linear factors. Then (c) list all of its zeros. g(x) = x4 – 3x3 – 12x2 + 20x + 48
Ex. 3 Write each function as (a) the product of linear and irreducible factors and (b) the product of linear factors. Then (c) list all of its zeros. g(x) = x4 – 3x3 – 12x2 + 20x + 48 *Find real zeros on graphing calculator.
Ex. 3 Write each function as (a) the product of linear and irreducible factors and (b) the product of linear factors. Then (c) list all of its zeros. g(x) = x4 – 3x3 – 12x2 + 20x + 48 *Find real zeros on graphing calculator. x = -2, x = 3, x = 4
Ex. 3 Write each function as (a) the product of linear and irreducible factors and (b) the product of linear factors. Then (c) list all of its zeros. g(x) = x4 – 3x3 – 12x2 + 20x + 48 *Find real zeros on graphing calculator. x = -2, x = 3, x = 4 *Divide to find remaining zero.
Ex. 3 Write each function as (a) the product of linear and irreducible factors and (b) the product of linear factors. Then (c) list all of its zeros. g(x) = x4 – 3x3 – 12x2 + 20x + 48 *Find real zeros on graphing calculator. x = -2, x = 3, x = 4 *Divide to find remaining zero. -2| |
Ex. 3 Write each function as (a) the product of linear and irreducible factors and (b) the product of linear factors. Then (c) list all of its zeros. g(x) = x4 – 3x3 – 12x2 + 20x + 48 *Find real zeros on graphing calculator. x = -2, x = 3, x = 4 *Divide to find remaining zero. -2| | 0
3| |
3| | 0
3| | 0 x2 – 2x – 8
3| 1 -5 -2 24 3 -6 -24 1 -2 -8 | 0 x2 – 2x – 8 So g(x) = (x + 2)(x – 3)(x2 – 2x – 8)
3| | 0 x2 – 2x – 8 So g(x) = (x + 2)(x – 3)(x2 – 2x – 8) = (x + 2)(x – 3)(x + 2)(x – 4)
3| | 0 x2 – 2x – 8 So g(x) = (x + 2)(x – 3)(x2 – 2x – 8) = (x + 2)(x – 3)(x + 2)(x – 4) = (x + 2)2(x – 3)(x – 4)
3| | 0 x2 – 2x – 8 So g(x) = (x + 2)(x – 3)(x2 – 2x – 8) = (x + 2)(x – 3)(x + 2)(x – 4) (a) & (b) = (x + 2)2(x – 3)(x – 4) (c) x = -2 (twice), x = 3, x = 4
Ex. 4 Use the given zero to find all complex zeros of each function
Ex. 4 Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x ; -5i
Ex. 4 Use the given zero to find all complex zeros of each function
Ex. 4 Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x ; -5i -5i| |
Ex. 4 Use the given zero to find all complex zeros of each function
Ex. 4 Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x ; -5i -5i| -15i i i i i i i i | 0
Ex. 4 Use the given zero to find all complex zeros of each function
Ex. 4 Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x ; -5i -5i| -15i i i i i i i i | 0 5i| i i i i |
Ex. 4 Use the given zero to find all complex zeros of each function
Ex. 4 Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. h(x) = 3x5 – 5x4 – 13x3 – 65x2 – 2200x ; -5i -5i| -15i i i i i i i i | 0 5i| i i i i +15i -25i i i | 0
x3 – 5x2 – 88x + 60
3 -5 -88 60 3x3 – 5x2 – 88x + 60 *Find real zeros on graphing calculator.
3 -5 -88 60 3x3 – 5x2 – 88x + 60 *Find real zeros on graphing calculator. x = 5, x = 6, x = 2/3
x3 – 5x2 – 88x + 60 *Find real zeros on graphing calculator. x = 5, x = 6, x = 2/3 So x = -5i, x = 5i, x = 5, x = 6, x = 2/3
x3 – 5x2 – 88x + 60 *Find real zeros on graphing calculator. x = 5, x = 6, x = 2/3 So x = -5i, x = 5i, x = 5, x = 6, x = 2/3 And (x + 5i)(x – 5i)(x – 5)(x – 6)(x – 2/3)
x3 – 5x2 – 88x + 60 *Find real zeros on graphing calculator. x = 5, x = 6, x = 2/3 So x = -5i, x = 5i, x = 5, x = 6, x = 2/3 And (x + 5i)(x – 5i)(x – 5)(x – 6)(x – 2/3) or (x + 5i)(x – 5i)(x – 5)(x – 6)(3x – 2)
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|Nelson EducationSchoolMathematics 2|
Surf for More Math
Lesson 7 - Measuring Around Objects
Use these interactive activities to encourage students to have fun on the Web while learning about measuring around objects. Students can try these activities on their own or in pairs.
Measure and record distances around objects using both nonstandard and standard units.
Instructions for Use
Touch Pegs lets students investigate perimeter.
To use Touch Pegs, click the "Clear" button. Select a blue band under "Bands" then drag it to any location on the grid. Release the band to anchor it. A red node appears at the top of the band. Click on any part of the band and drag it in any direction. It will stretch like a rubber band. Release the band to secure it to a point on the grid. Secure the band to other points on the grid to form a shape.
To colour the area inside the shape, click on the shape's outline and then click one of the colour buttons to the left of the grid.
Calculate the perimeter of the shape you created. Click the "Measures" button to check your answer. Click on the "Clear" button to start the exercise over again.
Perimeter Explorer allows students to measure the perimeter of a random shape on a grid.
To use Perimeter Explorer, input a number (greater than 1 and less than 50) in the text box next to "This shape's area is:". Click the "Draw New Shape" button. Calculate the shape's perimeter and input the answer in the text box beside "What is this shape's perimeter?". Click the "Check Answer" button to check your answer.
Input a different area and click the "Draw New Shape" button to measure the perimeter of a different shape.<|endoftext|>
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WEEK 14- Growth Mindset - Neuroplasticity
Pen and Paper
Teachers to read and facilitate the following:
Not that long ago, it was believed that our brains ability was fixed and set after childhood. Recently however, neuroscientists have discovered that the brain never stops changing and adjusting. Our brain is actually malleable - it can be changed and shaped through training. This process is called neuroplasticity. Neuroplasticity is the ability of the nervous system to make large increases in the strengths of existing neural connections and also establish new connections (Pascual-Leone, et al., 2005). It is true that our thoughts and actions change our brain. This means that by directing our focus, attention and action towards certain things we can strengthen our neural connections. Stronger neural connections means that it is easier for us to do those things- like how we remember how to ride a bike. The plasticity of our brain allows us to improve skills through practice and repetition. For example a right handed person might write their name poorly with their left hand. However through practicing with the other hand, we can experience large improvements, as our neural connections strengthen. Neuroplasticity is one “super powers” of our brain- allowing us to learn many things as our brain adapts and changes.
Description of Positive Education Practice: Neuroplasticity- Opposite hand
Today we are going to practice writing our name with our less dominant hand. Write your full name 3 times on a piece of paper.
Tonight we are going to brush our teeth with our less dominant hand for 30 seconds. Then finish brushing your teeth with your preferred hand. Continue this practice each time you brush your teeth for one week and observe how much you improve. The improvement is due to strengthening neural connections. You can strengthen neural connections for just about any skill that you choose to practice.
“When it comes to learning a skill, repeated experiences are essential- as connections become stronger and more efficient through repeated use.” (Nagel, 2009)
UPP’s Positive Education Practices
A Positive Education Practice (PEP) is an evidenced-based positive psychology intervention, applied in school communities or other educational settings. At UPP, we have tried to make these PEP’s simple, concise and relevant for students and their teachers. The six pillars for the Positive Education Practices are: Positive emotion (P); Engagement (E); Relationships (R); Meaning (M); Accomplishment (A); and, Health (H).
We hope that these evidence-based tools of positive psychology will enhance help people to thrive and live their best life, both within and beyond the school gates.
For more activities like this (and much more), check out THRIVE Online Lesson Modules for Pastoral Care and Wellbeing.
Unleashing Personal Potential<|endoftext|>
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Problem Set 3 Solutions
# Problem Set 3 Solutions - Problem Set III Solutions 1 The...
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Problem Set III Solutions 1. The block is at rest which means that F x = F y = 0. From Figure 1, it is clear that F x = ma x = 0 = T 1 cos a = T 2 cos b (1) F y = ma y = 0 = T 1 sin a + T 2 sin b = mg. (2) Solving Equation 1 for T 2 and then plugging back into Equation 2 to solve for T 1 , T 2 = T 1 cos a cos b T 1 sin a + cos a cos b sin b = mg = T 1 = mg sin a + cos a tan b . Plugging in numbers: T 1 = (10 kg )(9 . 8 m/s 2 ) 1 2 + 3 2 3 = 49 N T 2 = T 1 3 2 2 = 49 3 N = 85 N 2. (i) For the case where there is no friction between the block and the table, the force on the two block combi- nation is F = ( M + m ) a = - kA = a = - kA M + m . Assuming the block of mass m does not slip, both blocks accelerate at the same rate. The force that is accelerating the smaller block is the force due to friction. Intuitively, if there were no friction between the blocks, the smaller mass would stay at the stretched position and not be pulled back by the spring when the mass M was released. (It would then, of course, fall to the ground.) Therefore, in this case, friction must act in the direction of the motion in order to keep the smaller block on top of the larger block. ma = F f μmg . Since we want to know the maximum distance the block can be pulled, we want the maximum value for the friction. Using the value for acceleration that we found above, ma = m kA m + M μmg A max = μg ( m + M ) k FIG. 1: Problem 1. FIG. 2: Problem 4
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2 (ii) When you add friction between the block and the table, the initial equation for the acceleration changes to F = ( M + m ) a = - kA + μ ( m + M ) g = a = - kA + μ ( m + M ) g M + m . Now the force you need to keep the little block from slipping is ma = m M + m ( kA - μ ( m + M ) g ) μmg A max = ( M + m ) g ( μ + μ ) k 3. The average force is the change in energy divided by the change in distance. Assuming the seat belt keeps the passenger safely in the seat, ¯ F = Δ( E ) Δ( x ) = 1 x f - x 0 1 2 mv 2 f - 1 2 mv 2 0 = 1 0 . 94 m - 1 2 (75 kg )(70 km/h ) 2 = 1 0 . 94 m - 1 2 (75 kg ) 70 km h 1000 m 1 km 1 h 3600 s 2 = - 15 kN 4. There are two forces acting on the mass, the force due to gravity and the force from the spring (see Figure 2). Writing down the forces on the mass, F = ma = kx - mg = x = m ( a + g ) k . The x I just solved for is the distance the spring stretches due to gravity and the acceleration of the elevator, but does not include it’s unstretched length. Therefore, the total length of the spring is given by X = 80 cm + m ( a + g ) k . (3) (a) Plugging into Equation 3 X = 80 cm + (7 . 2 kg )(0 . 95 + 9 . 8)( m/s 2 ) 150 N/m = 132 cm (b) In this case, a=0, so X = 80 cm + (7 . 2 kg )(9 . 8 m/s 2 ) 150 N/m = 127 cm (c) We need to calculate the acceleration and plug it into Equation 3. ¯ a = Δ( v ) Δ( t ) = 0 - 14 m/s 9 . 0 s X = 80 cm + (7 . 2 kg )( - 14 9 + 9 . 8)( m/s 2 ) 150 N/m = 120 cm (d) In this case, we are given X and need to solve for a . X = 3 . 2 m = 0 . 80 m + (7 . 2 kg )( a + 9 . 8 m/s 2 ) 150 N/m = a 40 . 2 m/s 2 To keep the mass from hitting the floor, a cannot be greater than 40 . 2 m/s 2 . (This is assuming that the spring continues to obey the same force law when stretched to this length.)
3 FIG. 3: Problem 5 FIG. 4: Problem 6 5. The forces acting on the mass are the force due to gravity and the tension in the rope, see Figure 3. The mass is accelerating in the x
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Swarm Steps — The Cheatsheet For Social Change
This was derived from ‘Swarmwise’ by Rick Falkvinge and edited and formatted by Rob Sutherland
What is a Swarm?
A swarm is a decentralized, collaborative social group. It looks like a traditional hierarchical organization from the outside and it’s built by a small core of people that make up a scaffolding of go-to people, enabling numbers of volunteers to cooperate on a common idea on a scale not possible before the net. A swarm will form if you present a compelling enough idea.
The idea needs to be tangible, credible, inclusive, and epic.
- Tangible: You need to post an outline of the goals you intend to meet, when, and how.
- Feasible: After having presented your daring goal, you need to present it as totally doable. Bonus points if nobody has done it before.
- Inclusive: There must be room for participation by every spectator who finds it interesting, and they need to realize this on hearing about the project.
- Positive: Finally, you must set out to change the entire world for the better — or at least make a major improvement for a lot of people.
Answer these questions before you swarm:
- How many people are affected by this idea?
- What is the critical threshold? Discover the threshold by identifying the group of people affected in a positive direction by your provocative idea, estimate the size of that group, and then make an educated guess how many of this group will engage in the swarm at the lowest level of activation.
- What event constitutes success, and what does it take to get there?
- What are the swarm’s values?
Gather your swarm around the idea!
A swarm excels at self-organization, but first you must assign the swarm this task.
Analyze your geography.
Look at the area your swarm covers and divide it into no more than thirty geographical areas since you will only be able to coordinate thirty groups at most. Create a discussion forum with that number of subgroups.
Understand “magic” numbers.
- Seven people in a work group
- Thirty people in an extended group.
- One hundred and fifty in a sub swarm
- When any group hits these limits, split it into two groups
These numbers aren’t magic but based on how people form social groups and communicate. By sticking to these limits you will avoid bottlenecks.
Recruit the scaffolding.
A swarm consists mainly of loosely organized activists but needs a scaffolding of organizers to provide support — janitors not administrators.
Each geographical area should have four function officers, one for PR/media, one for activism, one for swarm care and one for web, information, and infrastructure. In addition the area leader should have one or two deputies — making a group of at most seven in total. No area leader should have more than six people working directly with them. Each area leader is both at the bottom of one pyramid and the top of another. No subgroup should have more than thirty members — when it hits thirty it should divide into two subgroups.
The activism leader doesn’t lead activism as such, but rather supports it (like all of these roles). Whenever activists decide swarmwise that they want to stage a rally, hand out flyers, put up posters, or do some other form of visible activism, this is the person responsible for the practical details, such as PA equipment, permits, and other details on the ground to make things happen.
The person responsible for PR/media would be responsible for interactions with old media (newspapers, television, radio, etc.) in his or her particular geography. That includes sending press releases, making sure press kits with information are available, and other things related to serving old media with information about the swarm and its activities.
The person responsible for swarm care welcomes new activists into the swarm and continually measures the overall health of it. A typical task would be to call new activists just to make them feel welcome, and tell them when the next events — social as well as operational — take place. This is more than enough for one person to chew.
Finally, the information-and-web guy is the person who maintains the infrastructure of a blog or other web page that summarizes the relevant information of the swarm in this particular geography. (This person also communicates internally when events, such as rallies, happen. The swarm decides when and if they happen; it is the job of this person to communicate the consensus.)
One person should have one role in the scaffolding, Don’t have multi-role people this creates bottlenecks.
Direct everyone to go to the subgroup for their area and meet with the other members of that subgroup. Tell people to introduce themselves to one another, and to select a leader. Then set up a subforum where these subgroup leaders can discuss things between themselves and with you. As the swarm organizes into these subgroups by geography, it needs to be given a task that allows it to jell properly over the first four weeks or so of its existence.
Update the overall progress of the tasks at least daily.
The swarm consists only of relationships between people.
Getting people to know other people should be the main goal of your activities.
Make sure that newcomers feel welcome — for every new relationship that is created, the organization grows.
When the swarm hits 150 people, you must start breaking it up into smaller groups. Make sure that there are social sub-swarms everywhere that can attract and retain new people, and not just one centrally located chat channel. Sub-swarms will have the social maximum size of 150.
Communicate very clearly what you want to see happen and why. If people agree with you, they will make that happen, without you telling a single person anything else. Your role is to set goals and ambitions, ambitions that don’t stop short of changing the entire world for the better.
Don’t talk about abstract concepts, your prospective volunteers will just yawn. To grow the swarm to critical mass we’ll need a large recruitment surface with concepts that are easy to relate to people’s everyday lives.
Divide the people of the swarm into three groups by activity level: officers, activists, and passive supporters.
- Officers are the people in the scaffolding, people who have taken on the formal responsibility of upholding the swarm.
- Activists are the actual swarm, the people that make things happen on a huge scale.
- Passive supporters are people who agree with the goals as such, but haven’t taken any action
Control the Vision, but Never the Message.
As you build a swarm, it is imperative that everybody is empowered to act in the swarm on the basis of what they believe will further its goals — but no one is allowed to empower themselves to restrict others. People are allowed, encouraged, and expected to assume speaking and acting power for themselves in the swarm’s name, but never the kind of power that limits others’ right to do the same thing.
Swarms need a leader
Leaderless swarms are not capable of delivering a tangible change in the world at the end of the day. The scaffolding, the culture, and the goals of the swarm need to emanate from a founder.
You do the vision, the swarm does the talking
Communicate your vision to everybody, and let the thousands of activists translate your vision into words that fit their specific social context. Don’t make a one-size-fits-all message. activists not only are encouraged to translate your vision, but also to interpret and apply it to specific scenarios
Enforce the principle of “If you see something you don’t like, contribute with something you do like.”
When people in the swarm get criticized by the public and by influential people, that is a sign you’re on the right track. This is not something to fear, this is something to celebrate, and everybody in the swarm must know this.
After the Swarm forms
First, allow the swarm’s scaffolding to keep growing organically. Train your closest officers in swarm methodology and techniques and set subgoals that are spaced about eight weeks apart. This may seem like a contradiction to self-organization, but it’s not: you’re telling the swarm the things that need to happen to get from point A to point B. You’re not saying who should be doing what and when.
Each subgoal needs to be credible, relevant, achievable, and clearly contributing to the end success. leave 10 percent of the time of every subgoal unallocated for unforeseen events.
Using a visible mechanism with a competitive element and measuring things in internal competitions gets them done quickly and adds fun. Measuring the right thing is crucial — make sure you measure the right thing. Anything that you measure in public, people will strive and self-organize to improve. Also, routine activities that are the same from day to day require motivation that internal competitions provide.
Provide work areas where the activists can share work files with one another: posters, flyers, blog layouts, catchy slogans, campaign themes, anything related to spreading your ideas and vision. Also, they must have the ability to comment on and discuss these work files between them.
Have regular meetings over the phone or over a chat line These meetings should be limited to seven people if on the phone, or thirty people if in a chat channel. Timing and locations of physical meetings can serve to lock out activists from engaging in the swarm — often inadvertently.
Meetings are a necessary evil, because people who are eager to be part of the swarm can easily see meetings as the purpose of the swarm — they will tend to see meetings as work itself, rather than the short time frame where you report and synchronize the actual work
Focus in the swarm is always on what everybody can do, and never what people cannot or must do.
Communicate three values:
- We can do this.
- We are going to change the world for the better.
- This is going to be hard work for us, but totally worth it.
Shoot for the moon!
Once you’ve run the numbers and communicated to the swarm that your insane idea is actually achievable, blue sparks of energy will jolt across the swarm with loud, crackling noises. People will look high from the excitement of being a part of this. Feel high, too.
Managing the Swarm
No one has the right to limit what another can do. This would be the typical swarm-think, at least as far as plentiful resources are involved. (When it comes to money, in case the swarm has any, decisions need to be made.)
Ways of making decisions
- 51 percent of the swarm has the right to exercise power over 49 percent of the swarm — this is meeting-and-voting scenario. This is not only counter to swarm-think, but it also creates a culture of fear of losing rather than a culture of empowerment and action.
- Someone has the final decision. Ruling over others by decree is not only completely counter to swarm-think, but it doesn’t work in the first place, as people are volunteers and, quite frankly, do whatever they want.
- Everybody has the power of veto for decisions. This creates significant problems with who constitutes “everybody,” but it is one of the most inclusive ways to get volunteers on board once that problem has been solved. However, it only works well for smaller subgroups (30 or less people).
Ruling by decree is right out. Voting creates losers, losers are unhappy and fuck things up. Don’t assume that the collective makes better decisions than the individual activists — swarms rely on the exact opposite. The values we desire in a swarm are inclusion, diversity, and empowerment. But if we are voting on something, we are limiting the minority — not empowering them. A swarm is legitimate only because it lets every individual include themselves on their own terms in order to further the swarm’s goals. Use a consensus circle to make decisions and if your swarm can’t reach a consensus then:
Use “the law of two feet”: It is every activist’s right and responsibility to go where they feel they can contribute the most and, at the same time, get the most in return as an individual. If there is no such place within a particular swarm, an activist can leave the swarm and go elsewhere.
Activists can float in and out of organizations, networks, and swarms that best match the change they want to see in the world. One swarm fighting for a goal does not preclude more swarms doing the same, but perhaps with a slightly different set of parameters from a different founder. This is fundamentally good for the end cause.
The swarm’s rules are by and large that there are no rules. Some people will seek to impose them.You need to make absolutely clear that the swarm works by its own consensus, that decisions are made organically by individual activists flowing to and from initiatives of their own accord, and that this swarm is your initiative; if wannabe fixers don’t want to play by the swarm’s rules, they need to use the law of two feet and go somewhere else.
The swarm has a structure that can handle budgets and money, and that is the supporting scaffolding. It’s the duty of the officers of the swarm to distribute resources in the most effective way to support the end goals through the initiatives of the activists. In this particular aspect, the swarm will resemble a traditional top-down organization in terms of allocating its resources in a decentralized manner. You, in control of the swarm’s formal name and resources, allocate budgets to officers, who subdivide their budget in turn. Once the swarm has any money to speak of, most of it should be devoted to supporting individual activists’ initiatives where they can reclaim expenses. The swarm lives and dies with the creativity and initiatives of its activists.
Innovation and Activism
The advantages of the swarm are cost-efficiency and execution speed. In order to increase the advantage in execution speed you need to minimize the try-fail-learn-try again cycle — the time from a failure to the next attempt at succeeding. Make it possible to learn and try again, learn again and try again and say that this is not only allowed, but expected. Increase the number of experiments by using the Three Activist Rule.
The Three Activist Rule
If three activists agree that something is good for the swarm, they have a green light to act in the swarm’s name. It’s not that they don’t need to ask permission — it goes deeper than that. Rather, they should never ask permission if three activists agree that something is good. And beyond that: they are specifically banned from asking permission. Their own judgment is the best available in the organization for their own social context, and they have to use that judgment rather than hiding behind somebody else’s greenlighting.
Everything that we focus on, no matter how or why, will grow in the swarm. if there are behaviors we don’t want to see growing, we should ideally pretend they aren’t even there — block them out from our conscious radar, and spend time rewarding other kinds of behavior.
Trolls will create a group of followers determined to wreak havoc until they get their way. This can be very disruptive and goes counter to swarm-think, where the best ideas and the best arguments win, rather than the loudest mouths. Still, it is a significant disturbance.
The way to deal with this is not to agree to demands. If you do cave in to get rid of the disturbance, you will teach the entire organization that creating loud disturbances is a very effective way of getting influence in the swarm, and you will start going down a very bumpy road as other people start imitating that behavior. You will never be able to convince trolls that they have bad ideas (and especially so if all they want is attention for themselves, rather than recognition for ideas). You will never be able to win that person.
Rather, you need to identify the reward mechanisms within the subgroup that has formed around the troll. Odds are that they’re forming a group identity around not being recognized as individual activists. You can shatter this identity by recognizing good contributors in the troll support group; odds are that there are several good contributors in that group who are just temporarily wooed by the trolls charisma. If you pick away a couple of key people in this group and recognize them for good earlier work — unrelated to the troll’s yelling — you will isolate the trolls, and the disturbance will lose critical mass.
An organization is people, and attention is reward.
This works for most people but for those who troll as a way of life you’ll need to look at Troll Control in Swarms for further advice.
What behavior do we want to encourage?
- Initiatives. Even initiatives that fail.
- Supporting others. Actually, this one is quite important. Helping others excel is just as valuable as excelling on your own.
- Creativity and sharing ideas.
- Helping people get along.
At some point, you may want to adjust the goals of the swarm. For a political party, this is almost inevitable. For a single-issue swarm, it is more avoidable. Nevertheless, it creates very difficult problems in the face of the swarm’s disorganization and it may be easier to start an overlapping swarm.
Avoid Excluding People
Always work to include. It’s easy to inadvertently exclude people from participation and every exclusion is a failure. Just because you don’t see any people being formally excluded, that doesn’t mean people don’t feel excluded.
One way of getting around this, which the German Pirate Party has used very successfully, is to allow everybody with formal voting rights to select somebody to vote in his or her place. This voting right can be assigned differently for different issues, and also be assigned in turn, creating a chain of trust to make an informed vote. This taps into the heart of the swarm’s social mechanisms of trusting people and friends, rather than fearing to lose. “Trust over fear.” We like that. That’s swarm-think. The German Pirate Party calls this liquid democracy.
Dealing with surges
Getting 20,000 new colleagues and activists in a week isn’t a pipe dream. It happens. Quite rarely, but it does happen. You need to be prepared for it.
In a swarm organization, the organizational culture cannot be communicated from person to person as the organization grows — it must be actively communicated centrally, and repeatedly communicated as new people keep joining. You need to make a values document based on the swarm values you thought of at the beginning.
Here’s an example, used by the Swedish Pirate Party
Our organization is built on three different pillars: swarm work, traditional NGO structures, and a hierarchical top-down structure that distributes resources to support the swarm. These are roughly equally important, but fill completely different needs: the traditional NGO structure only resides at the General Assembly and the party board level, for the party’s legal foundation as an nonprofit organization; the hierarchic work distribute resources and associated mandates from the board into the organization, making decisions for effective opinion building and other operative work; and the spontaneous swarm work is the backbone of our activism.
We work under the following principles:
We make decisions.
We aren’t afraid to try out new things, new ways to shape opinion and drive the public debate. We make decisions without asking anybody’s permission, and we stand for them. Sometimes, things go wrong. It’s always okay to make a mistake in the Pirate Party, as long as one is capable of learning from that mistake. Here’s where the famous “three-pirate rule” comes into play: if three self-identified pirates are in agreement that some kind of activism is beneficial to the party, they have authority to act in the party’s name. They can even be reimbursed for expenses related to such activism, as long as it is reasonable (wood sticks, glue, and paint are reasonable; computer equipment and jumbotrons are not).
We are courageous.
If something goes horribly wrong, we deal with it then, and only then. We are never nervous in advance. Everything can go wrong, and everything can go right. We are allowed to do the wrong thing, because otherwise, we can never do the right thing either.
We advance one another. We depend on our cohesion. It is just as much an achievement to show solitary brilliance in results as it is to advance other activists or officers.
We trust one another. We know that each and every one of us wants the best for the Pirate Party.We take initiatives and respect those of others. The person who takes an initiative gets it most of the time. We avoid criticizing the initiatives of others, for they who take initiatives do something for the party. If we think the initiative is pulling the party in the wrong direction, we compensate by taking an initiative of our own more in line with our own ideals. If we see something we dislike, we respond by making and spreading something we like, instead of pointing out what we dislike. We need diversity in our activism and strive for it.
We respect knowledge.
In discussing a subject, any subject, hard measured data is preferable. Second preference goes to a person with experience in the subject. Knowing and having experience take precedence before thinking and feeling, and hard data takes precedence before knowing.
We respect the time of others and the focus of the organization.
If we dislike some activity or some decision, we discuss, we argue, we disagree, and/or we start an initiative of our own that we prefer. On the other hand, starting or supporting an emotional conflict with a negative focus, and seeking quantity for such a line of conflict, harms the organization as a whole and drains focus, energy, and enthusiasm from the external, opinion-shaping activities. Instead, we respect the time and focus of our co-activists, and the focus of the organization. When we see the embryo of an internal conflict, we dampen it by encouraging positive communication. When we see something we dislike, we produce and distribute something we like. We work actively to spread love and respect, and to dampen aggression and distrust.
We communicate positively.
If we see a decision we dislike, we make our point about why we dislike it without provoking feelings, or, better yet, we explain why an alternative would be better. We campaign outward and cohesively, not inward and divisively. Again, we communicate positively. We act with dignity. We’re always showing respect in our shaping of public opinion: respect toward each other, toward newcomers, and toward our adversaries. We act with courtesy, calm, and factuality, both on and off the record. In particular, we’re never disrespectful against our co-activists (one of the few things that officers in the Pirate Party will have zero tolerance with).
We’re in parliament.
We behave like the parliamentary party that we are. Related to the point above. We are long term. We depend on making the 2010 and 2014 elections, so our work is long term. As in “on a time span of several years.” The time span between elections, four years, is practically a geological era for many of us net activists.
We represent ourselves.
The Pirate Party depends on a diversity of voices. None of us represents the Pirate Party on blogs and social media: we’re a multitude of individuals that are self-identified pirates. The diversity gives us our base for activism, and multiple role models build a broader recruitment and inspiration base for activism. Internally, we’re also just ourselves, and never claim to speak for a larger group: if our ideas get traction, that’s enough; if they don’t get traction, the number of people agreeing with those ideas is irrelevant.
You should keep reminding the entire swarm about the organization values regularly, as part of your heartbeat messages both to reinforce the values to old activists and to introduce them to new activists. Describe one value in every or every other heartbeat message. Needless to say, you also need to practice what you preach.
However, having this document and continuously reminding people that it exists, in words and in action, is not enough. You also need leadership guidance and tons of empty positions in the organization that new activists can fill. As part of a surge like the ones described, you may discover that your organization has recruited an assistant local media manager in Backabeyond, Backwater, Ohio. If you don’t have an empty box for that position in advance, it can’t be filled. If the officers of the swarm’s scaffolding don’t know how to uphold and communicate the swarm values, it won’t happen.
So in addition to the values that go for the organization as a whole, you also need to communicate values for the leaders that take on formal responsibility in the scaffolding. Just like the overall values that apply to all activists, these need to be communicated over and over, and, of course, reinforced through action.
Here’s a sample set of leadership values for a working swarm.
Leading in the Pirate Party is a hard but rewarding challenge. It’s considerably harder than being a middle manager in a random corporation. On the other hand, it’s somewhat easier than sending letters by carrier mackerel across the Sahara. Above all, it is stimulating, exciting, and simply quite fun.
The challenges lie in the constant demands for transparency and influence from your area of responsibility, combined with the demands for results and accountability from those you report to.
Basically, this means that leadership in the Pirate Party is a social skill, rather than a management or technical skill. It is about making people feel secure in their roles.
Above all, we need to defend two things in all our actions:
- The organization’s focus. We’re going to make the parliamentary threshold. Everything we do must be aimed at that.
- The organization’s energy. It is incredibly easy to get drained of energy if you start feeling negative vibes. There is a need for a constantly reinforced we-can-do-this sentiment.
In order to sustain these two values, we who have taken on officers’ and leaders’ responsibility use the following means:
Monkey see, monkey do.
We are role models. We act just the way we want other people in the organization to act. One part of this is to always try to be positive.
We make decisions.
We have had decision-making authority delegated to us in some area of the organization, and we use it.
We lead by inspiring and suggesting, never by commanding.
We advance role models.
We reward our colleagues as often as we can, both in public and private, when they display a behavior we want to reinforce.
We reward with attention.
Every behavior that gets attention in an organization is reinforced. Therefore, we focus and give attention to good behavior, and, as far as possible, we completely ignore bad behavior. We praise the good and ignore the bad (with one exception below).
We assume good faith.
We assume that everybody wants the organization to succeed, even when they do things we don’t understand.
We react immediately against disrespect.
Even if we have great tolerance for mistakes and bad judgment, we do not show tolerance when somebody shows disrespect toward their colleagues, toward other activists.
We speak from our own position.
When we perceive somebody as being in the wrong, we never say “you’re stupid” or similar, but start from our own thoughts, feelings, and reactions.
Administration is a support and never a purpose.
We try to keep administrative weight and actions to a minimum, and instead prioritize activism.
We build social connections.
We meet, and we make others meet.
We develop our colleagues.
We help everybody develop and improve, both as activists and leaders.
Success in a swarm
Success doesn’t happen smoothly and fluidly. It happens in hard-to-predict enormous bursts. You have to grind along, sometimes for years. While grinding along without seeing any returns can feel disheartening at times, it’s important to understand that people are listening and do take notice to what you’re saying. Then one day, all of a sudden, the government announces new horrible legislation that confirms everything you’ve been saying for the past two years, and you find yourself with twenty thousand new followers and five thousand new activists overnight, as you’ve gone from a doomsday prophet to being a rallying point for well-needed change. That’s the way it works.
Don’t confuse persistent day-to-day grinding with a refusal to see roadblocks for the uptake of the swarm’s ideas. If people tell you that your website is confusing, that the officers of the swarm are inaccessible, or that new people who come to gatherings aren’t feeling welcome, those are real issues and should absolutely not be taken as a sign to just keep doing what you’re already doing. Everybody needs to listen for real blocks to adoption of the swarm’s ideas, all the time.
Above all maintain one value set, one value base — don’t subdivide and fight internally.
Keys to a successful swarm:
- Be better at understanding and using mass-scale social dynamics than your competitors.
- Use both online and offline social friendships. Offline friendships are much, much stronger than online friendships and connections and offline discussions much stronger in terms of emotional attachment and intensity between people.
- Understand that the swarm can only grow at its edges, where people who have joined the swarm know people who have not yet joined. There, and only there, are there social links that can be used to communicate the values, mission, and enthusiasm of the swarm to gain new recruits the people who are most active can’t recruit any new activists to the swarm themselves by talking to their friends. The people leading a swarm cannot influence a single individual directly to join the swarm.
- Communicate heartbeat messages to the entire swarm, typically once a week. Over Communicate the context of the news, the external news in particular
- Sample rhetoric. supply direct quotes that can initiate a conversation, or sample responses to typical questions.
- Confidence. enable them to use stickers or pins with the swarm’s symbols that in turn lead to conversations. If they’re not confident enough to initiate conversations, identifying with the swarm gets part of the way there.
- Sense of urgency. A swarm grows by people talking to one another, one conversation at a time. These conversations are the key to the long-term success of the swarm.
- Understand the activation ladder. The swarm can grow only on its edges. The activation ladder is equally important to understanding recruitment: the edges of the swarm are not sharp, but quite fuzzy, and it’s hard to define the moment when people decide to activate themselves in the swarm for the first time. Is it when they hear about the swarm? When they visit its web pages? When they first contact a human being in the swarm? I would argue that all three of these are different steps on the activation ladder.
- Identify as many steps as possible on the activation ladder, and make each of these steps as easy and accessible as possible. Asking activists to describe each step that led them to join and activate would be a good start to discovering the activation ladder for a particular swarm.
- Mobilize activists. Success for any swarm is its ability to mobilize activists; to activate its followers you shouldn’t do anything except contact the local leaders of the swarm and ask them to make something happen. The next thing to realize is that these local leaders must have the tools to make that something happen. When a message is sent to thousands of phones, hundreds of people show up. That is more than sufficient to look like a significant group of people, especially if you make sure that placards are available from a nearby stash so that the group looks like, well, a group.
- Don’t compete on resources, swarms don’t have enough — swarms are unbeatable on speed, reaction time, and cost efficiency.
- Setup a Call to Arms: Remember — Perception is Reality. Most people will match their actions and opinions to be at least compatible with their perception of the public opinion. Control the public perception of who’s the winning team, and you become the winning team. Therefore, you need some kind of call-to-arms mechanism to quickly relocate your swarm’s activity to where people are looking at that exact moment. Control perception of who’s the winning team, and you become the winning team. It’s not just that perception is reality. If you can shape perception, you can also shape reality. A swarm excels at this.
- Respect anonymity! The more information you require about your activists, the fewer activists you’ll have. If your opponents are rich in resources, they control a large enough part of society to be able to cause trouble in society for their opponents — their named opponents. You don’t need to know who your activists are. You just need them to talk about the swarm’s issues with their friends, show up at rallies, etc. Many will prefer to be anonymous, and honoring that will make the swarm immensely stronger.
Rewarding the long tail
The important thing is to get your swarm discussed and mentioned. Imagine you had one of these blogs, your traffic was in the low twenties of visitors a day, and all of a sudden you had a traffic spike of some five hundred visitors when you mentioned the Pirate Party in a blog post.
When you give up the illusory control of your brand — which you never had anyway — and reward people for discussing you, unconditional of the context, they will keep discussing you and your topics, services, or products. That is exactly what you want to happen. So reward the long tail with attention — that can tip an entire blogosphere toward discussing you, with the exception of the star bloggers, but they’re the few and the long tail are the many.
Using attention to build a community
When I led the Swedish Pirate Party, as soon as somebody mentioned the party by name on a blog, I would see if I could contribute anything to the discussion (did they ask a question out in the air or wonder aloud about anything?). When somebody mentioned on Twitter or their blog that they had joined the party, I would write a short “Welcome aboard!” signed by me personally.
This was easily accomplished with a folder of bookmarks containing search pages across blogs, Twitter, etc.: it was a one-click operation to see if anything had appeared that mentioned the party’s name. Still, this blew people’s minds.
Reward people for their interest in your swarm, and show them attention. It works wonders. Attention is reward. Unexpected attention is great reward. Engage with people who read what you write In many ways, success can be harder to handle than failure, because it sets expectations most people have never felt.
The danger lies in not realizing that people will regard everything you say as having much more weight than you place on it yourself at the time you say it. If your swarm is political, anything you do — or don’t do — will be interpreted as a political statement, be nice to all people, even to your adversaries. Doing so will not just benefit the culture of the swarm, where you lead by example and show people that being excellent to each other is the way to behave, but it will also catch your adversaries completely off guard. This is a good thing: “If you can’t convince them, confuse them.” You don’t have to agree with them — you just have to disagree nicely and politely.
The day after success
No time is as tough as the year after that year when you were the hottest thing in town.
This applies to every swarm as well. When we’ve been on a slowly upward trajectory for a couple of years, we tend to believe that any dings — any level-ups — are permanent ascensions to a new base level of popularity, acceptance, and visibility.
Keep the swarm on track, and do remind them of that saying in the entertainment business: no time is as tough as the year after the year you’re hot — and that year will come around, as certainly as the calendar tells you it will.
Having fun in the swarm is crucial to growing the activist base.
Don’t shoot for the moon. Shoot for Mars!
Links to other resources:
The Swarm 2.0 Project — https://github.com/gylany/swarm-2.0
‘Swarmwise’ by Rick Falkvinge — http://falkvinge.net/2013/02/14/swarmwise-the-tactical-manual-to-changing-the-world-chapter-one/
A slideshow on Swarms — https://www.youtube.com/watch?v=qp_gxtR6v4U<|endoftext|>
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The word ‘insult’ can be defined as an expression or any statement made by a person in order to putdown or offend another human being. The usage of such term can be mere accidental, i.e., without any actual intention of verbally hurting the person or; can be used with full knowledgde by a person that his words would affect the others in a negative way.
People have been referring to the insults throughout these past centuries. However, during the Elizabethan period, the usage of insults was different. In this era, insults were commonly seen or it can be said that they were used in the famous Shakespearian plays.
These insults unlike today were funny and very clever. Shakespeare used very simple language in his plays. We today find it difficult to understand the plays because of the different style in which English was spoken and written.
The Elizabethan English had only 24 vowels which made it difficult for writers like Shakespeare to express his ideas through his plays. During the course of his plays, Shakespeare came up with new words that were not only used the plays, but later on became a part of the spoken English and are still used till date.
The following are some of the sentences used in the Shakespearian plays:
- “Thou art a churlish, dismal-dreaming fustilarian.”
- “Thou cockered motley-minded hugger-mugger”
- “Thou roguish spur-galled baggage”<|endoftext|>
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Inquiry Strategies for the Journey North Teacher. How do you react to student answers in class? [[AAPT Session: Effects of variation of faculty practice on student perceptions, Chandra Turpen]] Many faculty and high school teachers use some form of peer instruction or student response system (like clickers) as promoted by Eric Mazur, but they’re used in a huge variety of ways in the classroom.
This has a sizeable impact on their effectiveness and how students respond to them, which has been the topic of study for a while. My program has created a document on Best Practices in Clicker Use that you can download. This current study focussed on how faculty responded to student answers to clicker questions. Did they focus on getting the right answer, or making sense of the answer? Why don't college students ask questions in class? Module 13: Behaving Equitably and Responding Affirmatively to Questions. By Robert J.
Marzano, Barbara B. Gaddy, Maria C. Foseid, Mark P. Questioning Strategies. Questioning, Listening and Responding - C. Roland Christensen Center for Teaching and Learning. It would be hard to name a more valuable pedagogical accomplishment than the mastery of questioning, listening, and response: three teaching skills as linked, though distinct, as the panels of a triptych.
(C. Roland Christensen, Education for Judgment, 1991) The three essential skills of questioning, listening and responding are the backbone of discussion-based teaching. While each is important in its own right, the skills are intricately interrelated: the potential effect of a good question is only fully realized if accompanied by active listening, which in turn is an essential prerequisite for the appropriate response, whether in the form of an acknowledgment or further questioning. Questioning Experienced case instructors employ different types of questions at various points in the class to shape the arc of the discussion toward student discovery and learning.
Listening. Responding. Education Week. Responding to Students' Comments. In Bertolt Brecht’s play Galileo, the scientist is asking a very young student to explain some complex scientific point.
The boy is incorrect and Galileo shouts, “Wrong! Stupid!” Responding to Student Comments and Using Praise Appropriately. Description: Simply informing a student that an answer is correct.
The best response to a correct answer is often a plain, unemotional statement that, yes, that answer is correct. Socrative. Culture of thinking. 5 Powerful Questions Teachers Can Ask Students. My first year teaching a literacy coach came to observe my classroom.
After the students left, she commented on how I asked the whole class a question, would wait just a few seconds, and then answer it myself. "It's cute," she added. Thought Questions - Asking the right questions is the answer. Why Do Teachers Ask the Questions They Ask? Although teacher questioning has received much attention in the past few years, studies on teacher questions in the ESL classroom have so far revolved around the ‘closed’/‘open’ or ‘display’/‘referential’ distinction.
Findings from classroom observations show excessive use of closed questions by teachers in the classroom. The argument that has been more or less accepted is that such questions seek to elicit short, restricted student responses and are therefore purposeless in the classroom setting. This paper attempts to conduct an analytical discussion of the argument. The questions of three non-native ESL teachers during reading comprehension in the upper secondary school in Brunei are analysed using a three-level question construct. Through this three-level question analysis, it is possible to challenge the argument concerning question types and purposes. Do We Really Have High Expectations for All Students? By Barbara Blackburn Do you have high expectations for your students?
I’ve never met a teacher who said, “I have low expectations for my students.” The challenge is that we sometimes have hidden low expectations of certain students. One year, early in my teaching career, several teachers “warned” me about Daniel, a new student in my room. 38 Question Starters based on Bloom’s Taxonomy - Curriculet.
The Question Game: A Playful Way To Teach Critical Thinking. The Question Game by Sophie Wrobel, geist.avesophos.de The Question Game: A Playful Way To Teach Critical Thinking.
Teacher Questions: An Alternative? Kant declared false the commonplace saying “That may be true in theory, but it won’t work in practice.”
He acknowledged that there might be difficulties in application, but he said that if a proposition is true in theory, it must work in practice. What about the proposition “If teachers don’t ask questions, students will ask more and better ones”? Who Wants to Know? Use Student Questions to Drive Learning. Martin Luther King, Jr. considered this to be life's most persistent and urgent question: "What are you doing for others? " As we approach the holiday that honors his legacy, here's another question worth pondering: How many of your students know how to ask persistent and urgent questions of their own? Knowing how to formulate a good question -- and having the courage to ask it -- is a skill with profound social justice implications.
Dan Rothstein and Luz Santana, founders of the Right Question Institute, first became interested in questioning techniques when they were working with parents in a low-income community. Parents told them they didn't participate in their children's education because they didn't know what to ask. 8 ways teachers can talk less and get kids talking more. If you do fewer teacher-directed activities, that means the kids will naturally do more talking, doesn’t it? Not necessarily. I have often found myself talking almost constantly during group work and student-directed projects because I’m trying to push kids’ thinking, provide feedback, and help them stay on task. Even when the learning has been turned over to the students, it’s still tempting to spend too much time giving directions, repeating important information, and telling students how they did instead of asking them to reflect on their work.
Here are 8 ways teachers can talk less and get students talking more: 1. A space: at the table. Throughout the year, as a class, we will be unpacking a few major skill sets. Understanding how to listen to and participate in a fruitful, engaging and critical discussion is one of those skills. Miss Guinto and I have been working together to create inviting environment in which you feel confident and comfortable to be yourself and share your thoughts, feels and idea. Miss G has done a great job on her blog Meta, but writing up a list of expectations guidelines on how to do this. For the most part, I have used her words below to explain the "Round Table" Expectations*. The majority of our discussions will be conducted this way and it is important that we are all on the same page regarding appropriate behavior at the table.
Slowing Down to Learn: Mindful Pauses That Can Help Student Engagement. The excerpt below is from the book “Mindfulness for Teachers: Simple Skills for Peace and Productivity in the Classroom,” by Patricia A. Jennings. 8 Strategies To Help Students Ask Great Questions. Questioning - Top Ten Strategies. “Learn from yesterday, live for today, hope for tomorrow.<|endoftext|>
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What is math readiness? Is it really possible to teach math readiness to very young children – toddlers and preschoolers? What does it look like? For several years Carolyn Galbraith ran a play and learning center, focusing on math readiness for preschool children. In this post, Carolyn describes some of her experiences and the outcomes she observed.
We are lying on our backs looking up at the ceiling, searching for shapes. We see squares, rectangles. The smoke detector is a circle. One of the kids points out that the joins between the ceiling panels make crosses, and excitedly: “Parallel lines!”
When my kids were small, I ran a play & learning center in a small town outside Sydney, Australia. It held mostly building blocks and was a place where parents and kids could come and build whatever they liked. Inspired by Papert’s constructionism, it was a space where people could learn by constructing items meaningful to them, and sharing them with others. That’s why I called it Kids Build Together.
On Monday mornings I ran a math readiness group. It had mostly 2 to 4 year-olds. There were three main areas of focus – enjoying patterns, enjoying problems, and enjoying the process. We read a lot of math-themed picture books, tussle with simple problems. I tried to always end with a math-themed poem. Here’s a sample of one of our sessions:
“Today we’ve sung “Round about the circle” and passed a bag around with different shapes. Now the kids are going to try to fit them together. There’s no right answer, and it’s fun to see the different ways that the kids turn and feel and explore the properties of their shapes.
Next we read a gorgeously illustrated book about spirals in nature called “Swirl by Swirl” by Joyce Sidman. Afterwards, the kids use different colored stones to make their own spirals. They’re beautiful.
We’ve been reading “Measuring Penny” by Loreen Leedy and using different ways to measure things. Today the kids use big blocks to build towers as tall as themselves. They’re predicting how many blocks they’ll need, and I write it down on the board. They’re surprised to discover it only needs 15 blocks, rather than 100, to reach their height! I give them a measuring tape as another way to check their height, and 100 cm is close – so their predictions weren’t far off, after all.
To finish off I read a poem about time “Twenty-Four Hours” by Charles Causley. The kids love listening to poetry, the rhythm captures them instantly.
When I think back to learning math when I was at school, music, stories and games were the things I wished we used (I remember trying to design my own curriculum back in high school!) Finding the incredible mathematical patterns in nature, from fractals to Fibonacci spirals, has been a wonderful discovery of adulthood, as well as learning the stories of the mathematicians themselves. I hoped to share the pleasures of discoveries like these with the kids and their parents, and let them in on a secret – that math is more than memorizing numbers, and that readiness for math can be readiness for enjoying life.
Now my two kids are at school, and my center is no more. I help run a math class for the little ones at the local school, and include some Math in Your Feet dance activities there. I also tutor some children with literacy and numeracy challenges. It’s been fun sharing some of the resources with the older ones, such as the book about Fibonacci and the ideas of fractals.
It’s now I realize why all the work on symmetry in the younger years is essential – so much of mathematics is symmetry and balance! There’s not much time to cover it in school, a lesson or so. The ones who’ve spent years building with blocks are immediately advantaged. I encourage the parents of the older ones to bring the Lego back out, and build together; to walk in nature and observe the wonderful math in flower petals and tree branches. The secret hasn’t changed with their ages; math is still more than memory, and is still an important part for enjoying life.
Enjoyed this story and feel inspired to try these math activities with your preschoolers? Let us know how it goes. Do you have a math story to share? We can’t wait to hear from you!
Make Valentine’s Day crafts to celebrate loving one another… in the context of mathematics!
Grow your math eyes on these little field trips inspired by topology, dynamical systems, and algebraic geometry for the young, the very young, and the young at heart.
Know another romantic math craft? Tell us in the comments!
Make math: Fold paper in two. Sketch a teardrop in such a way that the fold forms one side of its triangular top part. Cut it out, open, and decorate the heart-shaped greeting card. My teen and I decorated with a cardioid (see below).
Grow your math eyes: What interesting shapes can you fold or unfold into other interesting shapes? What if you fold more than once? Choose shapes to experiment, on paper or in your imagination.
Topic: Mirror Symmetry. Inspired by Big Math: #14 Algebraic Geometry, #53 Differential Geometry.
Make math: Cut out two strips of paper and glue them together to form the + sign. Attach the opposite ends, with a twist, as if making Mobius strips. Cut the strips through the middle, and Mobius Hearts will happen! Video instructions:
Grow your math eyes: Try twisting two or more times. Try cutting 2/3 of the way rather than through the middle. What other surfaces can you make by connecting, twisting, and cutting paper strips?
Topic: Surfaces. Inspired by Big Math: #54 General Topology, #57 Manifolds.
Roses are red. Violets are approximately blue.
A paracompact manifold with a Lorentzian metric,
can be a spacetime, if it has dimension greater than or equal to two.
By Sarah Kavassalis
Make math: Draw a heart. Mark two or more points around it, for example, up top. Draw smaller hearts there. Mark the same points on them, and draw even smaller hearts. Keep going. Decorate your heart fractal, put it on a wall, or use as a greeting card.
Grow your math eyes: How tiny can your hearts become? Will you run out of space on paper if you keep going and going and going? How many hearts do you add to your fractal at each step of the process? For more versions, choose how many hearts to add at each step, how much to shrink at each step, how to turn the hearts around – then observe what happens.
Topics: Fractals and exponential growth. Inspired by Big Math: #11 Number Theory, #37 Dynamical Systems.
Make math: Trace a plate on scratch paper and cut out the circle. Fold it in half a few times, as if for making a snowflake. Unfold; your circle is now split into equal parts.
Trace the same plate on cardboard. Then use your folded circle as a guide to mark your new circle. You may need to insert more dots by hand.
You can also print out a marked circle, or use a protractor to mark it. Then follow instructions from MathCraft for drawing, or Almost Unschooling for string art. The craft involves skipping the dots by one and two, again and again. The repetition feels meditative, and there is a beautiful surprise at the end. The heart-shaped curve emerges as if by magic! Decorate it and put it on the wall or use as a greeting card.
Grow your math eyes: What other curves can you make out of straight lines? Choose your own “recipe” for skipping dots around a circle, and see what happens. What if you used a triangle or a square instead?
Topic: Curves. Inspired by Big Math: #14 Algebraic Geometry, #52 Discrete Geometry.
Make math: Customize variables and make your own greeting cards out of pretty graphs. Go to Desmos Math-o-Gram site to play and share the love.
Grow your math eyes: Peek behind the scenes to experiment with the equations. Replace numbers or functions in them and see what happens. Young children can start experimenting without knowing much (or anything) about the functions.
Topic: Functions and graphs. Inspired by Big Math: #14 Algebraic Geometry, #33 Special Functions.
Doing math with your children’s friends or a small math circle? Chances are your students are of mixed ages and levels. How can you have good dynamics?<|endoftext|>
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# Triple integral solver
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In this blog post, we discuss how Triple integral solver can help students learn Algebra. In mathematics, a root of a polynomial equation is a value of the variable for which the equation satisfies. In other words, a root is a solution to the equation. Finding roots is a fundamental problem in mathematics, and there are a variety of ways to solve for them. One popular method is known as "factoring." Factoring is the process of breaking down an expression into its constituent factors. For example, if we have the expression x2+5x+6, we can factor it as (x+3)(x+2). Once we have factored an expression, we can set each factor equal to zero and solve for the roots. In our example, we would get two equations: x+3=0 and x+2=0. Solving these equations, we would find that the roots are -3 and -2. Another popular method for solving for roots is known as "graphical methods." These methods make use of the graphs of polynomials to find approximate values for the roots. While graphical methods can be useful, they are often less accurate than algebraic methods such as factoring. As a result, algebraic methods are typically preferred when finding roots.
This can be a useful tool for solving problems in physics or engineering, where you might need to find the total amount of energy in a system, for example. There are a variety of different methods that can be used to solve series, and the choice of method will depend on the particular problem you are trying to solve. However, some of the most popular methods include the Euler-Maclaurin formula and the Ricci identity. With a little practice, you should be able to use a series solver to solve a wide range of problems.
Solving trinomials can be a tricky business, but there are a few methods that can make the process a bit easier. One common method is to factor the trinomial into two binomials. This can be done by grouping the terms together in pairs, and then multiplying each pair to get the product. Another method is to use the quadratic formula. This involves plugging the values of the coefficients into a specific equation, and then solving for x. While these methods may seem daunting at first, with a little practice they can become second nature. With some patience and perseverance, solving trinomials can be a breeze.
There are many ways to solve problems involving interval notation. One popular method is to use a graphing calculator. Many graphing calculators have a built-in function that allows you to input an equation and then see the solution in interval notation. Another method is to use a table of values. This involves solving the equation for a few different values and then graphing the results. If the graph is a straight line, then the solution is simple to find. However, if the graph is not a straight line, then the solution may be more complicated. In either case, it is always important to check your work to make sure that the answer is correct.
How to solve partial fractions is actually not that difficult once you understand the concept. Partial fractions is the process of breaking up a fraction into simpler fractions. This is often done when dealing with rational expressions. To do this, you first need to find the greatest common factor of the numerator and denominator. Once you have found the greatest common factor, you can then divide it out of both the numerator and denominator. The next step is to take the remaining fraction and break it up into simpler fractions. This is often done by rewriting the fraction in terms of its simplest form. For example, if you have a fraction that is in the form of a/b, you can rewrite it as 1/b. In some cases, you may need to use more than one partial fraction to completely simplify a fraction. However, once you understand how to solve partial fractions, it should be a relatively straightforward process.
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# If the sum of the squares of three consecutive integers is 194, what are the numbers?
Jul 15, 2016
$7 , 8 , 9$ or $- 9 , - 8 , - 7$
#### Explanation:
${a}^{2} + {b}^{2} + {c}^{2} = 194$
There are two possible scenarios, either the three numbers go even-odd-even or they go odd-even-odd. Any even number can be written as $2 k , k \in \mathbb{Z}$ and any odd number can be written as $2 j + 1 , j \in Z$. For the first scenario, we have:
${\left(2 k\right)}^{2} + {\left(2 k + 1\right)}^{2} + {\left(2 \left(k + 1\right)\right)}^{2} = 194$
${\left(2 k\right)}^{2} + {\left(2 k + 1\right)}^{2} + {\left(2 k + 2\right)}^{2} = 194$
$4 {k}^{2} + 4 {k}^{2} + 4 k + 1 + 4 {k}^{2} + 8 k + 4 = 194$
$12 {k}^{2} + 12 k + 5 = 194$
We now have a quadratic we can solve for k.
$12 {k}^{2} + 12 k - 189 = 0$
$k = \frac{- 12 \pm \sqrt{144 - 4 \left(12\right) \left(- 189\right)}}{24} = \frac{7}{2} \mathmr{and} - \frac{9}{2}$
These are not integer values so this is not the case.
Trying the second case, we have:
${\left(2 k + 1\right)}^{2} + {\left(2 \left(k + 1\right)\right)}^{2} + {\left(2 \left(k + 1\right) + 1\right)}^{2} = 194$
${\left(2 k + 1\right)}^{2} + {\left(2 k + 2\right)}^{2} + {\left(2 k + 3\right)}^{2} = 194$
$4 {k}^{2} + 4 k + 1 + 4 {k}^{2} + 8 k + 4 + 4 {k}^{2} + 12 k + 9 = 194$
$12 {k}^{2} + 24 k + 14 = 194$
$12 {k}^{2} + 24 k - 180 = 0$
$12 \left({k}^{2} + 2 k - 15\right) = 0$
$12 \ne 0 \implies {k}^{2} + 2 k - 15 = 0$
$\left(k + 5\right) \left(k - 3\right) = 0$
so k is 3 or -5. This means our numbers are:
$7 , 8 , 9$ or $- 9 , - 8 , - 7$
Jul 15, 2016
Three consecutive integers are $\left\{- 9 , - 8 , - 7\right\}$ or $\left\{7 , 8 , 9\right\}$.
#### Explanation:
Let the three numbers be $x - 1$, $x$ and $x + 1$. As sum of their squares is $194$, we have
${\left(x - 1\right)}^{2} + {x}^{2} + {\left(x + 1\right)}^{2} = 194$ or
${x}^{2} - 2 x + 1 + {x}^{2} + {x}^{2} + 2 x + 1 = 194$ or
$3 {x}^{2} + 2 = 194$ or $3 {x}^{2} - 192 = 0$
Or ${x}^{2} - 64 = 0$ i.e. $\left(x + 8\right) \left(x - 8\right) = 0$
Hence $x = - 8$ or $x = 8$ and as this is middle number
Three consecutive integers are $\left\{- 9 , - 8 , - 7\right\}$ or $\left\{7 , 8 , 9\right\}$.<|endoftext|>
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Migration is an important, but threatened ecological process. Conserving migration requires the maintenance of functional connectivity across sufficiently large areas. Therefore, we need to know if, where and why species migrate. Elephants are highly mobile and can travel long distances but we do not know if they migrate. Here, we analysed the movement trajectories of 139 savanna elephants (Loxodonta africana) within eight clusters of protected areas across southern Africa to determine if elephants migrate, and if so, where, how and why they migrate. Only 25 of these elephants migrated. Elephants are a facultative partially migratory species, where only some individuals in a population migrate opportunistically, and not every year. Elephants migrated between distinct seasonal ranges corresponding to southern Africa’s dry and wet seasons. The timing of wet season migrations was associated with the onset of rainfall and the subsequent greening up of forage. Conversely, the duration, distance, and the timing of dry season migrations varied idiosyncratically. The drivers of elephant migration are likely a complex interaction between individual traits, density, and the distribution and availability of resources. Despite most migrations crossing administrative boundaries, conservation networks provided functional space for elephants to migrate.
Migration is an ecologically important process that can have consequences for individual fitness, population demography1,2,3, and the structure and dynamics of ecosystems4,5. However, migration is increasingly threatened by anthropogenic pressures, habitat fragmentation, and climate change1,6,7. Identifying migratory species and the ultimate and proximate drivers of their migratory tendencies is therefore of conservation importance. The savanna elephant (Loxodonta africana) is one of Africa’s most iconic and well-studied large mammal species. Yet, whether elephants migrate or not remains unsubstantiated in the scientific literature due to a lack of empirical evidence, small sample sizes, and inadequate analytical routines. Indeed, whether elephants do migrate, and if so, where, how and why they migrate needs to be investigated.
Migration, defined here as a repeated seasonal movement between two non-overlapping regions8, is an adaptive response to living in seasonal environments and to the spatiotemporal distribution of resources8. For large herbivorous mammals, migration is commonly linked to the seasonal distribution of resources related to forage availability and quality7,9,10,11. Additional reasons for migration may include predator avoidance12, intra-specific competition11 and parasite avoidance13,14. However, migration is multifaceted and there are a number of different types of migrations that take place at various scales8.
At the population scale, migration can be considered as complete, where all individuals migrate, or partial, where only some individuals in a population migrate15. Partial migration seems to be the norm amongst large mammals16. At the individual scale, migration can be obligate, where individuals migrate annually, or facultative, where individuals do not migrate annually but rather opportunistically in response to local environmental conditions8,10. Facultative migration might be more common than obligate migration in some taxa10, but supporting long-term studies are scarce.
Unravelling the drivers of migratory behaviour or tendencies can be complex as they often involve interactions between intrinsic, environmental, and density dependent factors15,17. For example, whether or not populations or individuals migrate may depend on the seasonality of the environment they live in. Large mammals that live in seasonal environments where resources vary spatiotemporally are more likely to migrate than those living in less seasonal environments10,18,19. If seasonal changes are predictable, then migration may evolve to be obligate, but in seasonal environments that are less predictable, facultative migration is expected20. Partial migration may also be density dependent15,21. To avoid intra-specific competition, some populations with high densities are more likely to have migratory individuals than others11,19. Density can also affect an individual’s tendency to switch between migratory and non-migratory movements17. Lastly, phenotypic characteristics, such as body size, sex, or age may influence an individual’s competitive ability or ability to avoid predation15, which in turn may alter the propensity of an individual to migrate15,17.
Savanna elephants are widely distributed and commonly occur where rainfall and primary productivity vary seasonally (Fig. 1). Across their distributional range, other co-occurring large mammal species such as Burchell’s zebra (Equus burchellii)7,11 and blue wildebeest (Connochaetes taurinus)7 migrate in response to seasonal environments. Therefore, like other co-occurring large mammals, elephants should exhibit a range of movement tactics which includes migration8,10,18. Early accounts of possible migratory behaviour in elephants stem from several count and telemetry studies documenting long distance movements in Namibia and East Africa22,23,24,25,26. Recent spatially explicit studies suggest spatial separation of seasonal ranges27,28,29,30,31,32. In these studies, migration seems to be induced by seasonal rainfall and the speculated changes in forage availability. However, these studies involved only a few individuals in a small part of the distributional range of savanna elephants. These studies also failed to give a clear definition of what type of migration they deduced. We therefore do not know if elephants are partially or fully migratory, if migration is facultative or obligatory, or what drives migration.
In the present study, we analysed the movements of 234 savanna elephants within eight clusters of protected areas distributed across southern Africa, to answer a) do elephants migrate? and if so, b) where and why do elephants migrate? The study area comprised of protected areas containing 74% of the continent’s estimated 352,271 elephants33 (Fig. 2). Habitats varied from predominantly arid shrublands in Namibia to mesic woodlands in Mozambique (Fig. 1). We defined migration as a movement between two non-overlapping seasonal ranges. Using two independent analytical routines, we classified movements within a year as migratory or not. We then investigated the patterns and drivers of elephant migration. Based on early accounts of possible elephant migration, we predicted that at the population scale elephants are partially migratory. Furthermore, we expected that only some populations would have migratory individuals, specifically those in more seasonal and arid environments. At the individual scale, in migratory populations, we expect that migratory elephants are facultative migrants due to the unpredictability of African savannas34. Additionally, we predict that more males would migrate than family herds, because, males are typically larger and their movements are uninhibited by young calves35.
Large herbivores typically migrate during periods of plant growth to access high quality forage and then return to avoid adverse weather conditions or limited resources when seasons change4. Elephants seek out greener than expected vegetation throughout the year36 and their home ranges are limited to permanent water sources in the dry season37,38. Therefore, we expect that the timing and duration of migrations would depend on the distribution of forage across a landscape and seasonal rainfall, as previously documented for large African ungulates11,39. In line with this expectation, we test the hypothesis that the migration of elephants is driven by the spatiotemporal distribution of resources, in particular food and water. We expect migrations to be associated with the onset of the wet season, when ephemeral water sources fill up and the distribution of surface water no longer limits the distribution of elephants. Furthermore, we expect elephant dry season ranges to be close to permanent water sources and the ranges that they migrate to, to be further away from permanent water sources where ephemeral water sources are likely available30,38. We also expect elephants to migrate to areas with higher primary productivity, particularly if water is no longer a limiting factor, for example during the wet season. Lastly, if migration is density dependent, we expect elephants to migrate from areas of high density to areas of low density during the wet season to avoid intra-specific competition.
The location data from the original sample size of 234 elephants was reduced to 139 elephants after discarding elephants that had insufficient location data (i.e. less than a full year of location data, please see Supplementary Information Fig. S1). Here and throughout this manuscript, we refer to a year as a unit of time to express duration i.e. 12 months. The 139 elephants yielded 234 years of location data (Table 1, Fig. 2). Of the 139 elephants, 72 had a single year worth of location data, while 67 had multiple years of data (41 individuals had two years, 24 had three years, and two had four years of location data). Of the 139 elephants, 97 were adult females and 42 were adult males.
The NSD (Net Squared Displacement) method (see Supplementary Information Fig. S2 for details) classified 120 of the 234 years as non-migratory (52%), and 114 as migratory (48%). The overlap method (see Supplementary Information Fig. S2 for details) classified 179 as non-migratory (76%), and 55 years as migratory (24%). A total of 31 years (20%) were classified as migratory in both the NSD and overlap method (Table 1). All results reported on from this point forward are based on the 31 migrations classified by both methods as migratory. The 31 years classified as migratory comprised of 25 individuals: 17 female and eight male elephants. There was no significant difference between the proportion of female and male elephants that were analysed (♂ 97; ♀ 42) in comparison to those that migrated (♂ 17; ♀ 8; Pearson’s Chi-square test, χ² = 1.6 × 10−30, P = 1.0).
Migrations occurred in six of the eight protected area clusters and no migrations took place in Kafue or Tembe (Fig. 3). The proportion of individuals that migrated differed between clusters, but remained low (Table 1). Most migrations took place within protected area networks with only seven migrations moving into unprotected areas (Fig. 3, Supplementary Information Table S1). However, 26 migrations out of the 31 extended beyond primary protected area (IUCN categories 1-IV) boundaries but not beyond those of secondary protected areas (IUCN categories V and VI) (Fig. 3). Ten migrations crossed international borders (Supplementary Information Table S1).
Most of the migratory elephants with more than one year of location data switched between being migratory and non-migratory. Of the 25 elephants that migrated, 15 had multiple years of location data (seven elephants had two, seven had three years, and one had four years of data). Nine of these 15 individuals switched between being non-migratory and migratory between years, four migrated every year, and two migrated twice but not during consecutive years (Table 1, Supplementary Information Table S1). For elephants that migrated more than once, the migratory route, distance and timing were similar. However, the migratory timing differed between years for an individual (migration ID 27 and 28) that migrated twice in the Luangwa cluster (Fig. 4). For elephants that switched between being migratory and non-migratory, non-migratory movements appeared idiosyncratic with no clear pattern between switching individuals.
Migratory characteristics varied between individuals (Figs 4, 5, Supplementary Information Table S1). One-way migration distances ranged from 20 to 249 km (Fig. 5) with no clear pattern between sex or cluster. The longest migration took place in Etosha (migration ID 22) while the shortest migration took place in Chobe (Fig. 3: migration ID 16). In total, 77% of all departures took place during November and January (Fig. 4). This pattern correlated with the onset of the wet season and the subsequent greening up of vegetation (Fig. 4). The duration of time elephants spent in their away migratory ranges also corresponded with the wet season for most migrations, except in six (migration ID 12, 20, 22, 23, 25, 28; Fig. 4). While departure dates were consistent, the dates elephants migrated back to their dry season ranges varied greatly but most returned before the end of the dry season (Fig. 4).
The selection of wet season ranges by migratory elephants could not be explained by differences in EVI, distance to water or elephant density (see Supplementary Information Table S3a). The model that best explained the selection of wet season ranges was the null model with an AICc weight of 0.32 (see Supplementary Information Table S3a). AICc differences across the candidate models were relatively low. Furthermore, none of the relationships were significant (primary productivity (P = 0.95); distance to water (P = 0.71); elephant density (P = 0.62), see Supplementary Information Table S3b).
We set out to investigate if elephants migrate, and if so where, how and why they migrate? Our assessment illustrates that only some savanna elephants do migrate, but that migrations take place in most regions where elephants are distributed and most migrations extend beyond the boundaries of primary protected areas. Elephants should be considered partial and facultative migrators that may migrate in response to seasonal rainfall. However, EVI, water availability, and population densities did not explain our recorded migratory patterns.
Migration is not a simple process and unravelling the migratory tendencies of a species requires investigating patterns at the individual and population scale. By analysing the yearly movement data of 139 savanna elephants from eight clusters of protected areas across southern Africa, we determined that like several other large mammals16, the elephant is a partially migratory species. In other words, only some individuals in a population migrate. Indeed, based on our strict analytical routines, where two prescribed methods were used to classify migrations, we found that overall very few elephants migrated (18%). This was also the case within each of the populations, albeit slight differences in the proportion of migratory elephants. For example, Etosha, a highly seasonal and arid environment, had the most migratory elephants (30%), as opposed to the less seasonal but more mesic Niassa (13%). At the population scale, migration is generally more common in seasonal and predictable environments18,19. Therefore, we expected this to be the case for elephants across their distributional range. These patterns may exist, but we could not verify this because so few individuals migrated. However, it is important to note that despite large inter-population differences in environmental conditions and elephant densities, most of the protected area clusters harboured migratory individuals.
Amongst partially migratory species, individuals can be obligate or facultative migrants8. Facultative migration, where individuals do not migrate every year, has been documented in several ungulate species10,17. Facultative migration is viewed as an adaptive response to living in variable seasonal environments and is as an opportunistic response driven by proximate local conditions10,20. Of the 67 elephants we monitored for more than one year, 11 switched between being migratory or non-migratory. This relatively low proportion is comparable to one of the only long-term studies to have addressed switching behaviour in another large mammal, the elk (Cervus elaphus), where only 16% of elk switched between migratory and resident17. Switching suggests that elephant migration is facultative and not a fixed obligate response. It also highlights a high degree of behavioural flexibility in elephants. African migratory mammals often rely on environmental cues such as rainfall to begin their wet season migration39. However, in highly variable environments, environmental cues at one’s present location may be unreliable predictors for distant locations. These cues may also vary between years. Therefore, movement behaviour needs to be flexible enough to cope with this uncertainty. To this end, elephants show a high degree of movement plasticity and likely only migrate opportunistically if conditions at a point in time are conducive for migration.
In our study, most of the migratory elephants (77%) migrated at the onset of the wet season, migrating towards a wet season range. For these elephants, their annual migration consisted of two distinct seasonal long-distance movements that were directional, occurring over a short period and taking place between two non-overlapping wet and dry season ranges, as have been documented by others27,30. Other migratory characteristics however were highly variable and idiosyncratic. This includes the time spent in their wet season ranges and the timing of return migrations towards their dry season ranges. Migrations also varied in distance and between clusters with inconsistencies between individuals and years. The elephant movement patterns that were classified as migratory and that took place within the dry and wet seasons support our conclusion that elephants do migrate seasonally. However, for the elephants that were classified as migratory but that did not migrate within the expected seasonal windows, nomadic or exploratory movements, where individuals undertake a round-trip or return journey, cannot be ruled out. To unravel these differences in movements, long term movement data over continuous years is required.
Theory suggests that benefits for migratory individuals include exploiting changes in forage abundance or quality, accessing spatiotemporally limited resources4, escaping competition19, avoiding predation12 or parasite pressure14. For elephants, predation and parasite pressure are likely not factors that influence migration and we could not address them in the present study. However, we could address the spatiotemporal distribution of resources, as well as the competition that may arise in competing for those resources. Elephants tend to congregate around permanent water sources during the dry season37,38,40, leading to a local increase in elephant density. We hypothesised that at the onset of rain, elephants migrate to areas further away from permanent water sources where they can avoid competition and access new growth forage higher in primary productivity than what they would have experienced if they had stayed within their dry season ranges. However, while our results do not support this hypothesis, for reasons listed below we cannot reject it.
Firstly, migratory herbivores generally track seasonal changes in food quality rather than food abundance4. This is apparent in several African ungulates that tend to select more nutritious open grasslands during the wet season11. Our measure of primary productivity (EVI) is more indicative of food abundance rather than food quality, and certainly is unable to detect the nutritional differences in vegetation types41. The consistent timing of wet season migrations at the onset of rainfall as well as the apparent areas that elephants are migrating to, suggests that elephants migrate to access new growth forage in wet season ranges where food quality is higher41. For example, during the wet season in northern Botswana, elephants migrated to the Makgadikgadi and Nxai Pan National Parks (see Fig. 3: migration ID’s 1, 2, 4, and 5), areas dominated by grasses, which typically have higher protein and mineral content during the wet season42. Secondly, migratory species often cannot remain in the areas they have migrated to due to limiting or constraining factors. For example, in temperate regions, snowfall often limits foraging ability and forage availability, forcing animals to migrate towards warmer climates or lower elevations10,20. In African savannas, water is considered to be the most limiting factor constraining seasonal habitat use in migratory species4. The availability of surface water within migratory wet season ranges is likely the limiting factor driving return migrations in elephants to dry season ranges where permanent water sources are available. In this regard, inter-annual variations in the amount of rainfall may explain the idiosyncratic patterns observed in the duration of time spent in wet season ranges and the highly inconsistent timing of return migrations. Our results show that elephants were not selecting areas further from permanent water sources during the wet season. However, we could not quantify the amount (volume or area) of permanent water available within their ranges. For instance, a small permanent water hole would affect the Euclidean distance to permanent water even though it may only be able to supply a few elephants with water. We also cannot rule out inter-population differences. For example, in the Luangwa, migrations took place in a north-south direction along the Luangwa River (Fig. 3: migration ID 27 and 28). Similarly, in Kruger National Park, the distribution of artificial water points means that elephants are almost always within close proximity to permanent water43. A more detailed analysis within each population could provide better insight.
Lastly, theory suggests that density dependence is an important driver of partial migration15,21. Migration is viewed as a tactic to avoid intra-specific competition and populations with high densities are often more likely to have migratory individuals10. This has been demonstrated in a number of large herbivores, including roe deer (Capreolus capreolus)10,19, elk (Cervus elaphus)17, zebra (Equus burchelli) and blue wildebeest (Connochaetes taurinus)11. Our results suggest that elephants do not migrate towards areas with fewer elephants. However, intra-specific competition is not only dependent on density but also on the availability of resources within a range. Density on its own is therefore not an accurate measure of possible intra-specific competition. We lacked sample size and density data at the right scales to thoroughly assess density dependent competition and its possible effect on migratory tendencies in elephants. Future studies should investigate density dependence at the population scale, with density measurements across the entire population and at the individual scale, with local density estimates.
While trying to unravel why elephants migrate, we also need to assess why only some elephants migrate and others do not. Sex, age, body size and even personality may influence an individual’s migratory tendencies10,15,17,44,45. In the present study, there were no differences in migration tendencies between sexes, and because all females collared in the present study were part of a breeding herd that typically consists of individuals of various ages, it is unlikely that age played a role. The reasons may come down to a combination of environmental variables and an individual’s characteristics or phenotypic traits. These traits may include body size that in turn may influence an individual’s energy demands, dominance, or competitive ability45,46. However, we also cannot rule out the possibility that elephant migration may be inherent, stemming from certain genetic traits.
Although our study focused on identifying and classifying migratory behaviour, one must recognise that elephants are very mobile and their movements can be highly variable. Elephants may employ a large continuum of movement behaviour that not only includes migration but may also include highly variable home range or resident behaviour40. Furthermore, environmental conditions conducive for migration, for example, broad landscape variability in resources, may also drive nomadic behaviour18. Particularly in less seasonal and unpredictable environments18. In several cases where migration was not identified, nomadic movements may have taken place. Investigating these movement types and other long-distance movements could be important for understanding the ecology and conservation of elephants. Our study represents the first attempt at identifying and investigating migration in savanna elephants across multiple populations. While our study is not complete, it is certainly a step in the right direction and forms a baseline for future studies. Primarily we wanted to identify with certainty whether elephants migrate. As such, we used two independent methods to classify migration. We acknowledge that this analytical routine was strict and that there was a large discrepancy between the two methods and less defined migrations may have been overlooked. As such, we were left with very few migratory elephants to analyse. Whether this was a true reflection of reality or a reflection of our method remains to be seen. Nonetheless, we then attempted to unravel fine scale individual patterns of migration from a sample of 25 animals, with data staggered over a 15-year period and across multiple populations. Moving forward, we suggest that the method should be refined and case studies are conducted within each population where individuals are tracked for multiple successive years.
The greatest challenge in conserving migration is maintaining functional connectivity across sufficiently large areas1,6,47. Migration is flexible in elephants; however, elephants can only migrate if protected areas have enough functional space for them to do so. Conservation networks consisting of a mosaic of primary and secondary protected areas are being developed to help link isolated populations and enlarge protected areas47,48,49. It is promising to note that while 26 migrations extended beyond primary protected area boundaries, only seven of the elephants entered unprotected areas. In many cases, secondary protected areas between national parks were utilised as corridors by elephants50 between their migratory ranges (e.g. between Chobe and Nxai Pan (migration ID 1); Makgadikgadi and Moremi (migration ID’s 2, 4, and 5); and North and South Luangwa (migration ID 27 and 28). The establishment of the Great Limpopo Transfrontier Park around Kruger National Park has also allowed elephants to utilise seasonal ranges beyond the national park boundaries into areas that were previously inaccessible due to fencing. Despite most migrations crossing administrative boundaries, conservation networks provided functional space for elephants to migrate, highlighting the success of conservation initiatives that are striving to maintain and increase connectivity between protected areas.
Materials and Methods
The study area comprised of protected areas containing 74% of the continent’s estimated 352,271 elephants33 (Fig. 2). Here we identified eight geographical clusters of protected areas within which elephant populations were known or suspected to be interconnected50,51,52 (Fig. 2). Protected areas were designated as either primary or secondary based on their designation in the World Database on Protected Areas53 (Fig. 2). Primary protected areas are national parks and game reserves. Secondary protected areas include game/wildlife management areas, communal conservancies, hunting reserves, and forest reserves.
Habitats varied from predominantly arid shrublands in Namibia to mesic woodlands in Mozambique (Fig. 1). The terrain was relatively flat across most of the study area, except near the Etendeka Mountains in western Namibia and the Muchinga Mountains in Zambia. Most elephants within this study roamed freely and were not confined by artificial boundaries51. However, at the time of the study Etosha National Park, Kruger National Park, and Kaudum were partially fenced and Tembe Elephant Reserve was fully fenced37.
Elephant data set
Elephants (n = 234) were captured and collared with African Wildlife Tracking GPS collars (model SM 2000E; African Wildlife Tracking, Pretoria, South Africa) between December 2002 and December 2014. All individual animals were fitted with GPS collars for at least one year. We, therefore, obtained a time series of GPS locations with a fixed interval between 1 and 24 hours. Subsequently, we resampled GPS data to a single location per day (median location).
All aspects of the study were subject to ethical review and were approved by the Animal Ethics Committee of the University of Pretoria (AUCC-040611-013) and were carried out in accordance with international and national guidelines.
The datasets analysed during the current study are available from the corresponding author upon reasonable request. The data are not publicly available due to the sensitivity of locational data of a highly-poached species.
Annual migration is a back and forth movement that takes place between two spatially distinct seasonal ranges in one year8. Therefore, the sampling unit was a full year of elephant location data and the objective was to classify every year of location data as migratory or non-migratory. Throughout this manuscript, we refer to a year as a unit of time to express duration i.e. 12 months. As suggested by Cagnacci54 we used more than one method to classify annual migration. The analytical procedure consisted of three steps: (1) filtering the location data to consist of elephants with full years of data; (2) classifying yearly location data as either migratory or non-migratory; and (3) quantifying and analysing the characteristics of each migration in terms of duration, timing, and distance. The methodological framework is outlined in a flow diagram in Supplementary Information Fig. S1.
Filtering the data into full years of location data
The starting date of a year may influence the ability of the methods to identify a migration54 e.g. if the starting date begun after the onset of migration, the start of a migration may be missed. Elephants are less mobile during the dry season and more faithful to dry season ranges across years30,37,40. They are therefore more likely to migrate during the wet season. Subsequently, for standardization and to maximise sample size, we assigned the starting date for each year as the earliest date in a wet season since the date of collaring. For all clusters the starting date was set as 01 November except for Etosha (01 December) and Zambezi (01 January). Based on the above starting dates, we filtered the elephant (n = 234) location data by discarding elephants that had less than one year of location data and less than 15 locations within a month.
Classifying migratory and non-migratory movement
To classify a year of location data as migratory or non-migratory we used two independent methods54: (1) overlap of seasonal ranges (overlap54), and (2) Net Squared Displacement (NSD55). A year of location data was classified as migratory if both the NSD method and the overlap method classified it as migratory.
We used the flexible approach of Cagnacci54 that delimits seasons by shifting time windows (resolution of one month) to obtain all possible combinations of three seasonal ranges within a year. For a year to be classified as migratory there must be: (1) a low degree of overlap between the first and second successive range i.e. spatial separation between seasonal ranges, and (2) a high degree of overlap between the first seasonal range and the third seasonal range i.e. the individual returned to the starting seasonal range (see Supplementary Information Fig. S2 for details). We computed the overlap between seasonal ranges using the kernaloverlaphr function of the R package adehabitat56 (href smoothing factor). The function calculates the Bhattacharyya’s affinity index (BA) index, which quantifies the degree of similarity among probability surface estimates on a scale from zero (no overlap) to one (complete overlap)57. We defined a successive season threshold value of BA = 0.1554. If the overlap between the first and second successive range was below 0.15 (i.e. no overlap), we further distinguished between migratory and non-migratory by assessing whether the individual returned to a range similar to that of its starting seasonal range. If there was high overlap between the possible return seasons (BA > 0.50), we defined the year as migratory.
Following Bunnefeld55, we used the NSD method to classify a year of location data as either migratory or non-migratory. NSD measures the cumulative squared displacement between locations from the starting date of a year over the period of a full year55. Using the nls function in R, we fitted nonlinear models (corresponding to migratory or non-migratory) to the plotted NSD of each year of location data for each elephant55. The best model for each year was selected using AICc and AICc weights17. If a migratory model was chosen as the best-fit model, the year was classified as migratory (see Supplementary Information Fig. S2 for details). Migratory years with migratory durations fewer than 30 days were discarded to avoid short exploratory movements being classified as migrations. This was consistent with the overlap method, which only considered ranges at a resolution of one month.
Quantifying and analysing migration patterns
For all years classified as migratory in both the NSD and overlap methods, we assessed the patterns of migration using the modelled parameters from the migratory NSD model. Patterns of the migrations that we extracted were duration, timing, and distance (see Supplementary Information Fig. S3 for details). The one-way migration distance was calculated as the asymptotic height of the modelled NSD and the leaving and return date were calculated as the day at which the migration reached half its asymptotic height54.
Analysing differences in seasonal ranges of migratory elephants
To determine if elephants that migrated selected ranges that differed from the ranges they migrated from we compared primary productivity, elephant density, and distance from water of the range they migrated from and the range they migrated to (Supplementary Information Table S2). The timing of each seasonal range was classified using the modelled timings from the migratory NSD model (see Supplementary Information Fig. S3). To quantify the seasonal ranges, we calculated the centroid point from the location data that fell between leave days and returns days of each migration. We then buffered the centroid point using a quarter of the migration distance to assess the core area of each range. The buffered areas were assumed to represent core seasonal ranges.
For each of the ranges elephant density, mean Enhanced Vegetation Index (EVI), and the mean distance from permanent water was extracted (see below). Elephant density (elephants/km2) was calculated as the total number of elephants estimated within a demarcated protected area divided by the area. Data on savanna elephant population estimates came from the African Elephant Database58 and our own databases (see59 for details). To minimize error within the dataset we only included estimates with a survey reliability of A or B60. Elephant counts were not conducted every year. Therefore, we used the count closest to the year the elephant migrated. If count data was unavailable for a certain area, then that data point was excluded from the analysis (Supplementary Information Table S2).
We used the Enhanced Vegetation Index (EVI) as an index of primary productivity61. We chose EVI over Normalised Difference Vegetation Index because it does not become saturated as easily in high-biomass areas61,62. We downloaded monthly EVI data from http://reverb.echo.nasa.gov/ and calculated mean EVI for each seasonal range. We excluded all water pixels from the analyses and set all EVI values < 0.05 (indicative of non-vegetated areas) to 0.0563 We used dry season Landsat 8 imagery and supervised classification to generate our own fine-scale (30 m) permeant water distribution estimates for each of the clusters (full details of the procedure can be found in64). Based on the water distribution we generated a Euclidean distance to water layer and calculated the mean distance to water for each of the seasonal ranges (see Supplementary Information Table S2 for all extracted values).
Generalised linear mixed models65 (GLMM) were used to model binary data represented by ‘0’ or ‘1’. Zero represented the range elephants migrated from and ‘1’ represented the range they migrated to. The explanatory variables were; mean EVI, mean distance to water (distance to water was log transformed to normalise the distribution), and elephant density. Only six animals migrated more than once. To account for these repeated observations of migration amongst individuals, one year was randomly selected from each of these individuals. To account for repeated observations of migration within each cluster, “cluster” was added as a random effect. We formulated each set of candidate models using an all subset approach and ranked each candidate mixed model using Akaike’s information criterion (AICc)66. AICc was used to account for small sample size. The strength of support for the best model and alternate best models was assessed using AICc differences between the approximate best model and alternate candidate models. The Akaike weight for each candidate model also was calculated66.
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Support for this study was provided by Billiton, Conservation Foundation Zambia, Conservation International’s southern Africa’s Wildlife Programme, the Conservation Lower Zambezi, the International Fund for Animal Welfare, the Mozal Community Development Trust, the National Research Foundation, the National Postcode Lottery of the Netherlands, Peace Parks Foundation, the US Fish and Wildlife Services, the University of Pretoria, the World Wildlife Fund (SARPO; Mozambique; SA), the Walt Disney Grant Foundation, and the Wildlifewins Lottery. Elephants Without Borders was funded by the Paul G. Allen Family Foundation, Jody Allen, Zoological Society of San Diego, Madeleine and Jerry Delman Cohen, Harry Ferguson, Botswana Government Conservation Trust Fund and Wilderness Trust. We acknowledge the kind logistical support of Bateleurs, South African National Parks, Wings for Wildlife, Cyril Taolo, Larry Patterson, Peter Perlstein, Mike Holding and Abu Camp. This research was sanctioned and supported by the Botswana Dept. of Wildlife & National Parks, Direcção Nacional de Areas de Conservação, the Namibian Ministry of Tourism & Environment, the Malawian Wildlife Dept., Ezemvelo KZN Wildlife of South Africa, and the Zambian Wildlife Authority. We also thank Dr Keith Leggett for his contribution of some telemetry data from the Etosha cluster.
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# GRE Math : How to find percentage from a fraction
## Example Questions
### Example Question #1 : How To Find Percentage From A Fraction
If one of the employees across both industries were to be selected at random, what is the probability that the employee will be a construction industry worker who stayed in the same role for 5 years or more?
17%
26%
65%
20%
10%
26%
Explanation:
The first step is to figure out the percentage of construction employees that have stayed in the same role for 5 years or more—this would include both the "5 to 9 year" and "10+ years" ranges. This would be 0.25 + 0.4 = 65% of all construction employees. To convert to the number of employees, we take the percentage of their total, 0.65 * 8,000,000 = 5,200,000 workers. However, since the probability we are attempting to find is of workers between both industries, we must add the 8 million to the 12 million = 20 million workers total. 5,200,000/20,000,000 = 0.26, or a 26% chance.
### Example Question #2 : How To Find Percentage From A Fraction
For every 1000 cookies baked, 34 are oatmeal raisin.
Quantity A: Percent of cookies baked that are oatmeal raisin
Quantity B: 3.4%
The two quantities are equal.
Quantity B is greater
Quantity A is greater
The relationship cannot be determined from the information given.
The two quantities are equal.
Explanation:
Simplify Quantity A by dividing the number of oatmeal raisin cookies by the total number of cookies to find the percentage of oatmeal raisin cookies. Since a percentage is defined as being out of 100, either multiply the resulting decimal by 100 or reduce the fraction until the denominator is 100. You will find that the two quantities are equal.
### Example Question #3 : How To Find Percentage From A Fraction
50 students took an exam. There were 4 A's, 9 B's, 15 C's, 8 D's, and the rest of the students failed. What percent of the students failed?
Explanation:
students failed.
which equals 28%.<|endoftext|>
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828 |
When the mosquito that carries the malaria parasite (Plasmodium falciparum) bites someone, the parasite must travel to the liver where it undergoes part of its lifecycle before infecting red blood cells and spreading to its next host. Until now, the first step of how the parasite gets to the liver hasn't been clear.
A group of researchers led by Dr. Justin Boddey, Dr. Sara Erickson, and Ms. Annie Yang, from the Division of Infection and Immunity at the Walter and Eliza Hall Institute in Victoria, Australia, just released new research in the journal Cell Reports detailing how the malaria parasite travels from the site of a mosquito bite into liver cells.
The researchers infected mice with human liver cells with malaria. They genetically engineered the parasite to lack certain components so they could figure out which ones were vital for entering into the liver cells.
They found that the parasites secrete two proteins, SPECT and PLP1, that stay on the parasite's surface and are needed for the parasite to be able to enter into a human cell. Without the ability to traverse into cells, the parasites couldn't get to the liver and establish an infection.
That critical step of cell transversal starts the cycle that results in a case of malaria, where the infected person feels the symptoms of the disease. And it may also be the target for stopping the disease.
"Our long-term goal is to eradicate malaria, so we have to look at ways of breaking the cycle of infection," Boddey said in a press release. "A vaccine or treatment that halts the liver-stage infection offers the best chance of eradication because it stops parasites before they take hold."
It all starts with a bite from an anopheles mosquito infected with the malaria parasite. Parasites in the sporozoite life stage are injected into humans from the insect bite.
Thanks to the new research from Boddey and colleagues, we now know how the sporozoites transverse cells and make their way to the liver, where they infect the liver hepatocyte cells.
Once the merozoite stage of the parasite bursts from the infected liver cells, they infect red blood cells. This is the stage where symptoms of the infection occur. Malaria symptoms appear seven to 15 days after being bitten by an infected mosquito. The first symptoms are fever, headache, chills, and vomiting.
Malarial infections are treated with quinine-type medications (chloroquine, mefloquine, quinine, or quinidine), atovaquone-proguanil, or artemether-lumefantrine.
Once inside red blood cells, the parasite can develop through the trophozoite stage to the schizont stage, where they can burst the red cells, releasing particles that can infect more red cells. Mosquitos that feed on an infected person's blood during this stage pick up more of the parasite it can inject into other people. And the cycle continues in another victim.
Most cases of malaria occur in tropical and subtropical areas throughout the world. Although there are five types of malarial parasites, P. falciparum is the type responsible for most of the deaths.
In 2015, the World Health Organization estimated that there were 212 million people infected with malaria and 429,000 deaths. Seventy percent of the deaths occur in children under age five.
Despite concentrated research and health agency efforts to develop a malaria vaccine, effective anti-malarials, and insecticides, malaria remains a global health threat. The Centers for Disease Control and Prevention cite several reasons for this: an efficient mosquito that transmits the infection, a high prevalence of the most deadly species of the parasite, favorable climate, weak infrastructure to address the disease, and high intervention costs that are difficult to bear in poor countries.
If a good vaccine or treatment target could be found, like the mechanism involved in cell transversal, that might open up an opportunity to get ahead of this global health threat.<|endoftext|>
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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses)
## Selina Textbook Solutions Chapter 7 - Ratio and Proportion (Including Properties and Uses)
Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 7 - Ratio and Proportion (Including Properties and Uses).
All Selina textbook questions of Chapter 7 - Ratio and Proportion (Including Properties and Uses) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.
Exercise/Page
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(C)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Given,
Applying componendo and dividendo,
Hence, a: b = c: d.
Solution 6
(i) x =
(ii)
Solution 7
Solution 8
Solution 9
Given,
Solution 10
Given, a, b and c are in continued proportion.
Solution 11
Solution 12
Since,
Applying componendo and dividendo, we get,
Squaring both sides,
Again applying componendo and dividendo,
3bx2 + 3b = 2ax
3bx2 - 2ax + 3b = 0.
Solution 13
Solution 14
Applying componendo and dividendo,
Solution 15
Applying componendo and dividendo,
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(D)
Solution 1
Given,
Solution 2
Solution 3
(3x - 4y): (2x - 3y) = (5x - 6y): (4x - 5y)
Solution 4
(i) Duplicate ratio of
(ii) Triplicate ratio of 2a: 3b = (2a)3: (3b)3 = 8a3 : 27b3
(iii) Sub-duplicate ratio of 9x2a4 : 25y6b2 =
(iv) Sub-triplicate ratio of 216: 343 =
(v) Reciprocal ratio of 3: 5 = 5: 3
(vi) Duplicate ratio of 5: 6 = 25: 36
Reciprocal ratio of 25: 42 = 42: 25
Sub-duplicate ratio of 36: 49 = 6: 7
Required compound ratio =
Solution 5
(i) (2x + 3): (5x - 38) is the duplicate ratio of
Duplicate ratio of
(ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25
Sub-duplicate ratio of 9: 25 = 3: 5
(iii) (3x - 7): (4x + 3) is the sub-triplicate ratio of 8: 27
Sub-triplicate ratio of 8: 27 = 2: 3
Solution 6
Let the required quantity which is to be added be p.
Then, we have:
Solution 7
Solution 8
15(2x2 - y2) = 7xy
Solution 9
(i) Let the fourth proportional to 2xy, x2 and y2 be n.
2xy: x2 = y2: n
2xy n = x2 y2
n =
(ii) Let the third proportional to a2 - b2 and a + b be n.
a2 - b2, a + b and n are in continued proportion.
a2 - b2 : a + b = a + b : n
n =
(iii) Let the mean proportional to (x - y) and (x3 - x2y) be n.
(x - y), n, (x3 - x2y) are in continued proportion
(x - y) : n = n : (x3 - x2y)
Solution 10
Let the required numbers be a and b.
Given, 14 is the mean proportional between a and b.
a: 14 = 14: b
ab = 196
Also, given, third proportional to a and b is 112.
a: b = b: 112
Using (1), we have:
Thus, the two numbers are 7 and 28.
Solution 11
Given,
Hence, z is mean proportional between x and y.
Solution 12
Since, q is the mean proportional between p and r,
q2 = pr
Solution 13
Given, a, b and c are in continued proportion.
a: b = b: c
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Ratio of number of boys to the number of girls = 3: 1
Let the number of boys be 3x and number of girls be x.
3x + x = 36
4x = 36
x = 9
Number of boys = 27
Number of girls = 9
Le n number of girls be added to the council.
From given information, we have:
Thus, 6 girls are added to the council.
Solution 19
7x - 15y = 4x + y
7x - 4x = y + 15y
3x = 16y
Solution 20
Solution 21
x, y, z are in continued proportion,
Therefore,
(By alternendo)
Hence Proved.
Solution 22
x =
By componendo and dividendo,
Squaring both sides,
By componendo and dividendo,
b2 =
Hence Proved.
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(A)
Solution 1
Solution 2
Solution 3
Solution 4
Hence, (5a + 4b + 15): (5a - 4b + 3) = 5: 1
Solution 5
Solution 6
Solution 7
x2 + 6y2 = 5xy
Dividing both sides by y2, we get,
Solution 8
Given,
Solution 9
Solution 10
Solution 11
Let x be subtracted from each term of the ratio 9: 17.
Thus, the required number which should be subtracted is 5.
Solution 12
Solution 13
Assuming that all the men do the same amount of work in one day and one day work of each man = 1 units, we have,
Amount of work done by (x - 2) men in (4x + 1) days
= Amount of work done by (x - 2)(4x + 1) men in one day
= (x - 2)(4x + 1) units of work
Similarly,
Amount of work done by (4x + 1) men in (2x - 3) days
= (4x + 1)(2x - 3) units of work
According to the given information,
Solution 14
According to the given information,
Increased (new) bus fare = original bus fare
(i) We have:
Increased (new) bus fare = Rs 245 = Rs 315
Increase in fare = Rs 315 - Rs 245 = Rs 70
(ii) We have:
Rs 207 = original bus fare
Original bus fare =
Increase in fare = Rs 207 - Rs 161 = Rs 46
Solution 15
Let the cost of the entry ticket initially and at present be 10 x and 13x respectively.
Let the number of visitors initially and at present be 6y and 5y respectively.
Initially, total collection = 10x 6y = 60 xy
At present, total collection = 13x 5y = 65 xy
Ratio of total collection = 60 xy: 65 xy = 12: 13
Thus, the total collection has increased in the ratio 12: 13.
Solution 16
Let the original number of oranges and apples be 7x and 13x.
According to the given information,
Thus, the original number of oranges and apples are 7 5 = 35 and 13 5 = 65 respectively.
Solution 17
Solution 18
(A)
(i)
(ii)
(B)
(i)
Solution 19(ii)
Solution 19(i)
3A = 4B = 6C
3A = 4B
4B = 6C
Hence, A: B: C = 4: 3: 2
Solution 20
(i) Required compound ratio = 2 9 14: 3 14 27
(ii) Required compound ratio = 2a mn x: 3b x2 n
(iii) Required compound ratio =
Solution 21
(i) Duplicate ratio of 3: 4 = 32: 42 = 9: 16
(ii) Duplicate ratio of
Solution 22
(i) Triplicate ratio of 1: 3 = 13: 33 = 1: 27
(ii) Triplicate ratio of
Solution 23
(i) Sub-duplicate ratio of 9: 16 =
(ii) Sub-duplicate ratio of(x - y)4: (x + y)6
=
Solution 24
(i) Sub-triplicate ratio of 64: 27 =
(ii) Sub-triplicate ratio of x3: 125y3 =
Solution 25
(i) Reciprocal ratio of 5: 8 =
(ii) Reciprocal ratio of
Solution 26
Solution 27
Solution 28
Solution 29
Reciprocal ratio of 15: 28 = 28: 15
Sub-duplicate ratio of 36: 49 =
Triplicate ratio of 5: 4 = 53: 43 = 125: 64
Required compounded ratio
=
Solution 30(b)
Solution 30(a)
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 7 - Ratio and Proportion (Including Properties and Uses) Page/Excercise 7(B)
Solution 1
(i) Let the fourth proportional to 1.5, 4.5 and 3.5 be x.
1.5 : 4.5 = 3.5 : x
1.5 x = 3.5 4.5
x = 10.5
(i) Let the fourth proportional to 3a, 6a2 and 2ab2 be x.
3a : 6a2 = 2ab2 : x
3a x = 2ab2 6a2
3a x = 12a3b2
x = 4a2b2
Solution 2
(i) Let the third proportional to 2 and 4 be x.
2, 4, x are in continued proportion.
2 : 4 = 4 : x
(ii) Let the third proportional to a - b and a2 - b2 be x.
a - b, a2 - b2, x are in continued proportion.
a - b : a2 - b2 = a2 - b2 : x
Solution 3
(i) Let the mean proportional between 6 + 3 and 8 - 4 be x.
6 + 3, x and 8 - 4 are in continued proportion.
6 + 3 : x = x : 8 - 4
x x = (6 + 3) (8 - 4)
x2 = 48 + 24- 24 - 36
x2 = 12
x= 2
(ii) Let the mean proportional between a - b and a3 - a2b be x.
a - b, x, a3 - a2b are in continued proportion.
a - b : x = x : a3 - a2b
x x = (a - b) (a3 - a2b)
x2 = (a - b) a2(a - b) = [a(a - b)]2
x = a(a - b)
Solution 4
Given, x + 5 is the mean proportional between x + 2 and x + 9.
(x + 2), (x + 5) and (x + 9) are in continued proportion.
(x + 2) : (x + 5) = (x + 5) : (x + 9)
(x + 5)2 = (x + 2)(x + 9)
x2 + 25 + 10x = x2 + 2x + 9x + 18
25 - 18 = 11x - 10x
x = 7
Solution 5
Solution 6
Let the number added be x.
(6 + x) : (15 + x) :: (20 + x) (43 + x)
Thus, the required number which should be added is 3.
Solution 7(iii)
Solution 7(ii)
Solution 7(i)
Solution 8
Let the number subtracted be x.
(7 - x) : (17 - x) :: (17 - x) (47 - x)
Thus, the required number which should be subtracted is 2.
Solution 9
Since y is the mean proportion between x and z
Therefore, y2 = xz
Now, we have to prove that xy+yz is the mean proportional between x2+y2 and y2+z2, i.e.,
LHS = RHS
Hence, proved.
Solution 10
Given, q is the mean proportional between p and r.
q2 = pr
Solution 11
Let x, y and z be the three quantities which are in continued proportion.
Then, x : y :: y : z y2 = xz ....(1)
Now, we have to prove that
x : z = x2 : y2
That is we need to prove that
xy2 = x2z
LHS = xy2 = x(xz) = x2z = RHS [Using (1)]
Hence, proved.
Solution 12
Given, y is the mean proportional between x and z.
y2 = xz
Solution 13
LHS = RHS
Hence proved.
Solution 14
Let a and b be the two numbers, whose mean proportional is 12.
Now, third proportional is 96
Therefore, the numbers are 6 and 24.
Solution 15
Let the required third proportional be p.
, p are in continued proportion.
Solution 16
Hence, mp + nq : q = mr + ns : s.
Solution 17
Hence, proved.
Solution 18
Then, a = bk and c = dk
Solution 19
a, b, c and d are in proportion
Then, a = bk and c = dk
Solution 20
Then, x = ak, y = bk and z = ck
## Browse Study Material
TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 7 - Ratio and Proportion (Including Properties and Uses) for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.
# Text Book Solutions
ICSE X - Mathematics
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# 10 Most Common 3rd Grade STAAR Math Questions
Preparing 3rd-grade students for the 3rd Grade STAAR Math test? Want a preview of the most common mathematics questions on the 3rd Grade STAAR Math test? If so, then you are in the right place.
The mathematics section of 3rd Grade STAAR can be a challenging area for some students, but with enough patience, it can be easy and even enjoyable!
Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions your students need to practice.
## 10 Sample 3rd Grade STAAR MathPractice Questions
1- Classroom A contains 8 rows of chairs with 4 chairs per row. If classroom B has three times as many chairs, which number sentence can be used to find the number of chairs in classroom B?
A. $$8 × 4 + 3$$
B. $$8 + 4 × 3$$
C. $$8 × 4 × 3$$
D. $$8 + 4 + 3$$
2- A cafeteria menu had spaghetti with meatballs for $8 and bean soup for$6. How much would it cost to buy three plates of spaghetti with meatballs and three bowls of bean soup? __________
3- There are 7 days in a week. There are 24 hours in day. How many hours are in a week?
A. 48
B. 96
C. 168
D. 200
4- Emily described a number using these clues:
Three-digit odd numbers that have a 6 in the hundreds place and a 3 in the tens place
A. 627
B. 637
C. 632
D. 636
5- This clock shows a time after 12:00 PM. What time was it 1 hour and 30 minutes ago?
A. 12:45 PM
B. 1:45 PM
C. 1:15 PM
D. 12:15 PM
6- Olivia has 84 pastilles. She wants to put them in boxes of 4 pastilles. How many boxes does she need?
A. 20
B. 21
C. 24
D. 28
7- A football team is buying new uniforms. Each uniform costs $20. The team wants to buy 11 uniforms. Which equation represents a way to find the total cost of the uniforms? A. $$(20 × 10) + (1 × 11) = 200 + 11$$ B. $$(20 × 10) + (10 × 1) = 200 + 10$$ C. $$(20 × 10) + (20 × 1) = 200 + 20$$ D. $$(11 × 10) + (10 × 20) = 110 + 200$$ 8- There are 92 students from Lexington Elementary School at the library on Tuesday. The other 54 students in the school are practicing in the classroom. Which number sentence shows the total number of students in Riddle Elementary School? A. $$92 + 54$$ B.$$92 -54$$ C. $$92 × 54$$ D. $$92 ÷54$$ 9- Use the picture below to answer the question. Which fraction shows the shaded part of this square? A. $$\frac{92}{100}$$ B. $$\frac{92}{10}$$ C. $$\frac{90}{100}$$ D. $$\frac{8}{100}$$ 10- Which number correctly completes the number sentence $$90 × 25 =$$ ? A. 225 B. 900 C. 1250 D. 2250 ## Best 3rd Grade STAAR Math Prep Resource for 2024 ## Answers: 1- C 3 times of 8 rows of chairs with 4 chairs per day is: $$3 × 8 × 4$$ 2- 42 3spaghetti with meatballs cost: $$3 × 8 = 24$$ 3bowls of bean soup cost: $$3 × 6 = 18$$ 3spaghetti with meatballs + 3bowls of bean soup cost: $$24 + 18 = 42$$ 3- C 1 day: 24 hours so 7 days $$= 7 × 24 = 168$$ hours 4- B Three-digit odd numbers with a 6 in the hundreds place and a 3 in the tens place is 637. 632 and 636 are even numbers. 5- A Subtract hours: $$14– 1 = 13$$ Subtract the minutes: $$15 – 30 = – 15$$ The minutes are less than 0, so: • Add 60 to minutes ( $$-15 +60 =45$$ minutes) • Subtract 1 from hours $$(13 – 1 =12)$$ 6- B Olivia wants to divide 84 pastilles into boxes of 4 pastilles, so $$84 ÷ 4 = 21$$ is the number of boxes. 7- C The football team should buy 11 uniforms and each uniform costs$20 so they should pay$$(11×20) 220$$.
Therefore, choice C is the correct answer:
$$(20 × 10) + (20 × 1) = 20(10+1) =20×11=220$$
8- A
Add 92 and 54 students to know the whole number of students.
9- A
the model for the fraction is divided into 100 equal parts. We shade 92 parts of these 100 parts that it’s equal to $$\frac{92}{100}$$
10- D
$$90 × 25 =2,250$$
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## Documents
TRANSCRIPT
CONFIDENTIAL 1
by Factoringby Factoring
CONFIDENTIAL 2
Warm UpWarm Up
Solve each equation by graphing the related function.
1) x2 - 49 = 0
2) x2 = x + 12
3) - x2 + 8x = 15
CONFIDENTIAL 3
You have solved quadratic equations bygraphing. Another method used to solvequadratic equations is to factor and use
the Zero Product Property.
Zero Product PropertyZero Product Property
Notice that when writing a quadratic function as its related quadratic equation,
you replace y with 0. So y = 0.
y = ax2 + bx + c
0 = ax2 + bx + c
ax2 + bx + c = 0
CONFIDENTIAL 4
One way to solve a quadratic equation in standard form is to graph the related function and find the x-values where y = 0. In other words, find the zeros of the related function. Recall that a quadratic function
may have two, one, or no zeros.
Using the Zero Product PropertyUsing the Zero Product Property
WORDS NUMBERS ALGEBRA
If the product of two quantities equals zero,
at least one of thequantities equals zero.
3 (0) = 0
0(4) = 0
If ab = 0,
then a = 0 or b = 0.
CONFIDENTIAL 5
Solving Quadratic Equations by GraphingSolving Quadratic Equations by GraphingUse the Zero Product Property to solve each
A) (x - 3)(x + 7) = 0
x - 3 = 0 or x + 7 = 0
x= 3 or x = -7
Use the Zero Product Property.
Solve each equation.
Check
(x - 3)(x + 7) = 0
(3 - 3)(3 + 7) 0
(0)(10) 0
0 0
(-7 - 3)(x + 7) = 0
(-7 - 3)(-7 + 7) 0
(10)(0) 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are 3 and -7.
CONFIDENTIAL 6
B) (x)(x - 5) = 0
x = 0 or x - 5 = 0
x= 5
Use the Zero Product Property.
Solve each equation.
Check
(x)(x - 5) = 0
(0)(0 - 5) 0
(0)(-5) 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are 0 and 5.
(x)(x - 5) = 0
(5)(5 - 5) 0
(5)(0) 0
0 0
CONFIDENTIAL 7
Now you try!
Use the Zero Product Property to solve each equation. Check your answer.
1a. (x)(x + 4) = 0
1b. (x + 4)(x - 3) = 0
CONFIDENTIAL 8
If a quadratic equation is written in standard form, a x 2 + bx + c = 0,
then to solve the equation, you may need to factor before using the
Zero Product Property.
CONFIDENTIAL 9
A) x2 + 7x + 10 = 0
(x + 5) (x + 2) = 0
x + 5 = 0 or x + 2 = 0 Use the Zero Product Property.
Solve each equation.
Check
x2 + 7x + 10 = 0
(-5)2 + 7(-5) + 10 0
25 - 35 + 10 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are -5 and -2.
x = -5 or x = -2
Factor the trinomial.
x2 + 7x + 10 = 0
(-2)2 + 7(-2) + 10 0
4 - 14 + 10 0
0 0
CONFIDENTIAL 10
B) x2 + 2x = 8
-8 -8
Use the Zero Product Property.
Solve each equation.
The solutions are -4 and 2.
x = -4 or x = 2
Factor the trinomial.
x2 + 2x = 8
x2 + 2x – 8 = 0
The equation must be written in standard form. So subtract 8 from both sides.
(x + 4) (x - 2) = 0
x + 4 = 0 or x - 2 = 0
CONFIDENTIAL 11
Check: Graph the related quadratic function. The zeros of the related function should be the
same as the solutions from factoring.
The graph of y = x2 + 2x - 8 shows two zeros appear to be -
4 and 2, the same as the solutions from factoring.
CONFIDENTIAL 12
C) x2 + 2x + 1 = 0
Use the Zero Product Property.
Solve each equation.
Both factors result in the same solution, so there is one solution, -1.
x = -1 or x = -1
Factor the trinomial.(x + 1) (x + 1) = 0
x + 1 = 0 or x + 1 = 0
CONFIDENTIAL 13
Check: Graph the related quadratic function. The zeros of the related function should be the
same as the solutions from factoring.
The graph of y = x2 + 2x + 1 shows that one zero appears
to be -1, the same as the solution from factoring.
CONFIDENTIAL 14
D) -2x2 = 18 - 12x
-2 (x - 3) (x - 3) = 0
-2 ≠ 0 or x - 3 = 0 Use the Zero Product Property.
Solve each equation.
Check
The only solution is 3.
x = 3
Factor the trinomial.
-2x2 = 18 - 12x
-2(3)2 18 - 12(3)
-18 18 - 36
0 0
Write the equation in standard form.-2x2 + 12x – 18 = 0
-2( x2 - 6x + 9) = 0 Factor out the GCF, -2.
Substitute 3 into the original equation.
CONFIDENTIAL 15
Now you try!
2a. x2 - 6x + 9 = 0
2b. x2 + 4x = 5
CONFIDENTIAL 16
Sports ApplicationSports ApplicationThe height of a diver above the water during a dive can be modeled by h = -16t2 + 8t + 48, where h is
height in feet and t is time in seconds. Find the time it takes for the diver to reach the water.
h = -16t2 + 8t + 48
0 = -16t2 + 8t + 48
Use the Zero Product Property.
Solve each equation.
Factor the trinomial.
The diver reaches the water when h = 0.
Factor out the GCF, -8.0 = -8(2t2 - t - 6)
0 = -8(2t + 3) (t -2)
-8 ≠ 0, 2t + 3 = 0 or t - 2 = 0
2t = -3 or t = 2
t = -3 2
Since time cannot be negative, (-3/2 ) does not make sense in this situation.
CONFIDENTIAL 17
It takes the diver 2 seconds to reach the water.
Check
0 = -16 t 2 + 8t + 48
0 -16(2)2 + 8(2) + 48
0 -64 + 16 + 48
0 0
Substitute 3 into the original equation.
CONFIDENTIAL 18
Now you try!
3.) The equation for the height above the water for another diver can be modeled by h = -16t2 + 8t + 24. Find the time it takes this diver
to reach the water.
CONFIDENTIAL 19
BREAK
CONFIDENTIAL 21
Assessment
1) (x + 2) (x - 8) = 0
Use the Zero Product Property to solve each equation. Check your answer.
2) (x - 6) (x - 5) = 0
3) (x + 7) (x + 9) = 0
4) (x) (x - 1) = 0
CONFIDENTIAL 22
6) 3x2 - 4x + 1 = 0
5) 30x = -9x2 - 25
8) x2 - 8x - 9 = 0
7) x2 + 4x - 12 = 0
CONFIDENTIAL 23
9) A group of friends tries to keep a beanbag from touching the ground without using their hands. Once the beanbag has been kicked, its height can be modeled by h = -16t2 + 14t + 2, where h is the height in feet above the ground and t is the time in seconds. Find the time it
takes the beanbag to reach the ground.
CONFIDENTIAL 24
You have solved quadratic equations bygraphing. Another method used to solvequadratic equations is to factor and use
the Zero Product Property.
Zero Product PropertyZero Product Property
Notice that when writing a quadratic function as its related quadratic equation,
you replace y with 0. So y = 0.
y = ax2 + bx + c
0 = ax2 + bx + c
ax2 + bx + c = 0
Let’s review
CONFIDENTIAL 25
One way to solve a quadratic equation in standard form is to graph the related function and find the x-values where y = 0. In other words, find the zeros of the related function. Recall that a quadratic function
may have two, one, or no zeros.
Using the Zero Product PropertyUsing the Zero Product Property
WORDS NUMBERS ALGEBRA
If the product of two quantities equals zero,
at least one of thequantities equals zero.
3 (0) = 0
0(4) = 0
If ab = 0,
then a = 0 or b = 0.
CONFIDENTIAL 26
Solving Quadratic Equations by GraphingSolving Quadratic Equations by GraphingUse the Zero Product Property to solve each
A) (x - 3)(x + 7) = 0
x - 3 = 0 or x + 7 = 0
x= 3 or x = -7
Use the Zero Product Property.
Solve each equation.
Check
(x - 3)(x + 7) = 0
(3 - 3)(3 + 7) 0
(0)(10) 0
0 0
(-7 - 3)(x + 7) = 0
(-7 - 3)(-7 + 7) 0
(10)(0) 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are 3 and -7.
CONFIDENTIAL 27
A) x2 + 7x + 10 = 0
(x + 5) (x + 2) = 0
x + 5 = 0 or x + 2 = 0 Use the Zero Product Property.
Solve each equation.
Check
x2 + 7x + 10 = 0
(-5)2 + 7(-5) + 10 0
25 - 35 + 10 0
0 0
Substitute each
solution for xinto the original
equation.
The solutions are -5 and -2.
x = -5 or x = -2
Factor the trinomial.
x2 + 7x + 10 = 0
(-2)2 + 7(-2) + 10 0
4 - 14 + 10 0
0 0
CONFIDENTIAL 28
B) x2 + 2x = 8
-8 -8
Use the Zero Product Property.
Solve each equation.
The solutions are -4 and 2.
x = -4 or x = 2
Factor the trinomial.
x2 + 2x = 8
x2 + 2x – 8 = 0
The equation must be written in standard form. So subtract 8 from both sides.
(x + 4) (x - 2) = 0
x + 4 = 0 or x - 2 = 0
CONFIDENTIAL 29
Check: Graph the related quadratic function. The zeros of the related function should be the
same as the solutions from factoring.
The graph of y = x2 + 2x - 8 shows two zeros appear to be -
4 and 2, the same as the solutions from factoring.
CONFIDENTIAL 30
Sports ApplicationSports ApplicationThe height of a diver above the water during a dive can be modeled by h = -16t2 + 8t + 48, where h is
height in feet and t is time in seconds. Find the time it takes for the diver to reach the water.
h = -16t2 + 8t + 48
0 = -16t2 + 8t + 48
Use the Zero Product Property.
Solve each equation.
Factor the trinomial.
The diver reaches the water when h = 0.
Factor out the GCF, -8.0 = -8(2t2 - t - 6)
0 = -8(2t + 3) (t -2)
-8 ≠ 0, 2t + 3 = 0 or t - 2 = 0
2t = -3 or t = 2
t = -3 2
Since time cannot be negative, (-3/2 ) does not make sense in this situation.
CONFIDENTIAL 31
It takes the diver 2 seconds to reach the water.
Check
0 = -16 t 2 + 8t + 48
0 -16(2)2 + 8(2) + 48
0 -64 + 16 + 48
0 0
Substitute 3 into the original equation.
CONFIDENTIAL 32
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# Read Standard Form Numbers with Negative Indices
In this worksheet, students read and write standard form numbers with negative indices as normal numbers.
Key stage: KS 3
Curriculum topic: Number
Curriculum subtopic: Interpret/Compare Standard Form Numbers
Difficulty level:
### QUESTION 1 of 10
Standard (index) form is a shorthand way of writing a very large or a very small number.
It is written as a number between 1 and 10 (but not equal to 10) multiplied by a power of 10, for example 2.94 × 102 or 4.8 × 10-3.
In this worksheet, we will just look at very small numbers.
Example
Change 5.83 × 10-6 into a normal number.
We divide 5.83 by 10 six times.
Each digit therefore moves 6 places to the right.
This gives the number 0.00000583.
Other Examples
1.122 × 10-7 = 0.0000001122
6 x 10-8 = 0.00000006
Write the following standard form number as a normal number:
3.24 x 10-4
Write the following standard form number as a normal number:
3.24 x 10-6
Write the following standard form number as a normal number:
3.4 x 10-6
Write the following standard form number as a normal number:
1.253 x 10-3
Write the following standard form number as a normal number:
1 x 10-2
Write the following standard form number as a normal number:
1.00773 x 10-3
Write the following standard form number as a normal number:
8.06 x 10-9
Write the following standard form number as a normal number:
5 x 10-10
Write the following standard form number as a normal number:
9.9 x 10-5
Write the following standard form number as a normal number:
4.444 x 10-4
• Question 1
Write the following standard form number as a normal number:
3.24 x 10-4
0.000324
EDDIE SAYS
The power in this question is 10-4 This means each digit in the number moves four places to the right.
• Question 2
Write the following standard form number as a normal number:
3.24 x 10-6
0.00000324
EDDIE SAYS
The power in this question is 10-6 This means each digit in the number moves six places to the right.
• Question 3
Write the following standard form number as a normal number:
3.4 x 10-6
0.0000034
EDDIE SAYS
The power in this question is 10-6 This means each digit in the number moves six places to the right.
• Question 4
Write the following standard form number as a normal number:
1.253 x 10-3
0.001253
EDDIE SAYS
The power in this question is 10-3 This means each digit in the number moves three places to the right.
• Question 5
Write the following standard form number as a normal number:
1 x 10-2
0.01
EDDIE SAYS
The power in this question is 10-2 This means each digit in the number moves two places to the right.
• Question 6
Write the following standard form number as a normal number:
1.00773 x 10-3
0.00100773
EDDIE SAYS
The power in this question is 10-3 This means each digit in the number moves three places to the right.
• Question 7
Write the following standard form number as a normal number:
8.06 x 10-9
0.00000000806
EDDIE SAYS
The power in this question is 10-9 This means each digit in the number moves nine places to the right.
• Question 8
Write the following standard form number as a normal number:
5 x 10-10
0.0000000005
EDDIE SAYS
The power in this question is 10-10 This means each digit in the number moves ten places to the right.
• Question 9
Write the following standard form number as a normal number:
9.9 x 10-5
0.000099
EDDIE SAYS
The power in this question is 10-5 This means each digit in the number moves five places to the right.
• Question 10
Write the following standard form number as a normal number:
4.444 x 10-4
0.0004444
EDDIE SAYS
The power in this question is 10-4 This means each digit in the number moves four places to the right.
---- OR ----
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of 11/11
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 -Great Minds. eureka math.org This file derived from G3-M1-TE-1.3.0-06.2015 Lesson 17: Model the relationship between multiplication and division. Lesson 17 NYS COMMON CORE MATHEMATICS CURRICULUM 3 1 217 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 17 Objective: Model the relationship between multiplication and division. Suggested Lesson Structure Fluency Practice (9 minutes) Application Problem (5 minutes) Concept Development (36 minutes) Student Debrief (10 minutes) Total Time (60 minutes) Fluency Practice (9 minutes) Sprint: Multiply or Divide by 4 3.OA.7 (9 minutes) Sprint: Multiply or Divide by 4 (9 minutes) Materials: (S) Multiply or Divide by 4 Sprint Note: Framing division through missing factors in multiplication sentences builds a strong foundation for understanding the relationships between multiplication and division. See Lesson 2 for directions for administering a Sprint. Between Sprints, include the following group counts in place of movement exercises. Count by twos to 20 forward and backward. Count by threes to 30, hum/talk forward and backward. (Hum as you think 1, 2, say 3, hum 4, 5, say 6, etc.) Count by fives to 50 forward and backward. Application Problem (5 minutes) Mrs. Peacock bought 4 packs of yogurt. She had exactly enough to give each of her 24 students 1 yogurt cup. How many yogurt cups are there in 1 pack? Note: This problem is designed to lead into the Concept Development. In Problem 1, students will analyze how a number bond represents the division expression 24 ÷ 4.
# Lesson 17 - EngageNY
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### Text of Lesson 17 - EngageNY
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 -Great Minds. eureka math.org This file derived from G3-M1-TE-1.3.0-06.2015
Lesson 17: Model the relationship between multiplication and division.
Lesson 17 NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
217
Lesson 17
Suggested Lesson Structure
Sprint: Multiply or Divide by 4 (9 minutes)
Materials: (S) Multiply or Divide by 4 Sprint
Note: Framing division through missing factors in multiplication sentences builds a strong foundation for understanding the relationships between multiplication and division. See Lesson 2 for directions for administering a Sprint.
Between Sprints, include the following group counts in place of movement exercises.
Count by twos to 20 forward and backward.
Count by threes to 30, hum/talk forward and backward. (Hum as you think 1, 2, say 3, hum 4, 5, say 6, etc.)
Count by fives to 50 forward and backward.
Application Problem (5 minutes)
Mrs. Peacock bought 4 packs of yogurt. She had exactly enough to give each of her 24 students 1 yogurt cup. How many yogurt cups are there in 1 pack?
Note: This problem is designed to lead into the Concept Development. In Problem 1, students will analyze how a number bond represents the division expression 24 ÷ 4.
Lesson 17: Model the relationship between multiplication and division.
Lesson 17 NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
218
NOTES ON
MULTIPLE MEANS
Problem 3 of the Problem Set. Because
of the duplication, the suggested
process for completing the Problem Set
is to save Problem 3 until the end.
However, for some classes, it may
prove useful to preview the example
here and have students complete it as
one of the first problems they do
independently on the Problem Set.
This will build confidence by giving
students an immediate sense of
success.
Materials: (S) Personal white board
Problem 1: Use the number bond to relate multiplication and division.
T: (Draw or project the number bond shown to the right.) The number bond represents the division equation you wrote to solve the Application Problem. Turn and tell your partner how it shows 24 ÷ 4.
S: (Discuss.)
T: Look back at the Application Problem. Is the unknown in the number bond the same as the unknown in the division problem? What does it represent?
S: They’re the same. The unknown represents the size of the groups.
T: (Project a second number bond where the total and one part are drawn. Write __ × 4 = 24.) Skip-count by fours to find the unknown factor. Each time you say a four, I will make a new part of my number bond. (Draw the parts as students count.)
S: 4, 8, 12, 16, 20, 24.
T: How many fours make 24?
S: 6 fours!
S: 6.
T: The division equations are the same. How do the quotients in the two number bonds represent different things?
S: The 6 in the first number bond represents the size of the groups. The 6 in the second number bond represents the number of groups.
Repeat the process with 32 ÷ 4. (Model how the quotient can represent the number of groups or the size of the groups.)
T: How do the multiplication and division equations relate in each example?
S: I thought of the division equation like a multiplication equation with an unknown factor and skip-counted by fours until I reached the total.
Lesson 17: Model the relationship between multiplication and division.
Lesson 17 NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
219
Problem 2: Solve word problems to illustrate the relationship between multiplication and division.
Write or project the following problem: A classroom has tables that seat a total of 20 students. Four students are seated at each table. How many tables are in the classroom?
T: Draw and label a tape diagram to represent the problem.
S: (Draw diagram shown to the right.)
T: Without solving, write a division equation and a multiplication equation with an unknown factor to represent your drawing.
S: (Write 20 ÷ 4 = __ and __ × 4 = 20.)
T: What does the unknown in both problems represent?
S: The number of groups.
S: To solve the division, I will add units of 4 to the tape diagram until I get to 20. That is just skip-counting by fours. Skip-counting is a way to solve the multiplication, too. The strategies are the same for both equations because you can use one to solve the other.
T: Solve both equations now.
Repeat the process with 16 ÷ 4. (Problem 2 models division where the quotient represents the number of groups.)
Problem Set (10 minutes)
Students should do their personal best to complete the Problem Set within the allotted 10 minutes. For some classes, it may be appropriate to modify the assignment by specifying which problems they work on first. Some problems do not specify a method for solving. Students should solve these problems using the RDW approach used for Application Problems.
Student Debrief (10 minutes)
Lesson Objective: Model the relationship between multiplication and division.
The Student Debrief is intended to invite reflection and active processing of the total lesson experience.
Invite students to review their solutions for the Problem Set. They should check work by comparing answers with a partner before going over answers as a class. Look for misconceptions or misunderstandings that can be addressed in the Debrief. Guide students in a conversation to debrief the Problem Set and process the lesson.
Lesson 17: Model the relationship between multiplication and division.
Lesson 17 NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
220
Any combination of the questions below may be used to lead the discussion.
In the first problem on the Problem Set, what patterns did you notice in the array?
How did the patterns you noticed help you solve the multiplication and division sentences?
Share student work from Problems 3 and 4. Students may have solved using number bonds or tape diagrams, multiplication, or division. Compare approaches.
How can a number bond show both multiplication and division?
Discuss: Division is an unknown factor problem.
In Problems 3 and 4, the unknown is the size of each group. What is different about Problem 4? (It is a two-step problem.)
Exit Ticket (3 minutes)
After the Student Debrief, instruct students to complete the Exit Ticket. A review of their work will help with assessing students’ understanding of the concepts that were presented in today’s lesson and planning more effectively for future lessons. The questions may be read aloud to the students.
Lesson 17: Model the relationship between multiplication and division. 221
Lesson 17 Sprint NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
Multiply or Divide by 4
1. 2 × 4 = 23. __ × 4 = 40
2. 3 × 4 = 24. __ × 4 = 8
3. 4 × 4 = 25. __ × 4 = 12
4. 5 × 4 = 26. 40 ÷ 4 =
5. 1 × 4 = 27. 20 ÷ 4 =
6. 8 ÷ 4 = 28. 4 ÷ 1 =
7. 12 ÷ 4 = 29. 8 ÷ 4 =
8. 20 ÷ 4 = 30. 12 ÷ 4 =
9. 4 ÷ 1 = 31. __ × 4 = 16
10. 16 ÷ 4 = 32. __ × 4 = 28
11. 6 × 4 = 33. __ × 4 = 36
12. 7 × 4 = 34. __ × 4 = 32
13. 8 × 4 = 35. 28 ÷ 4 =
14. 9 × 4 = 36. 36 ÷ 4 =
15. 10 × 4 = 37. 24 ÷ 4 =
16. 32 ÷ 4 = 38. 32 ÷ 4 =
17. 28 ÷ 4 = 39. 11 × 4 =
18. 36 ÷ 4 = 40. 44 ÷ 4 =
19. 24 ÷ 4 = 41. 12 ÷ 4 =
20. 40 ÷ 4 = 42. 48 ÷ 4 =
21. __ × 4 = 20 43. 14 × 4 =
22. __ × 4 = 24 44. 56 ÷ 4 =
A Number Correct: _______
Lesson 17: Model the relationship between multiplication and division. 222
Lesson 17 Sprint NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
Multiply or Divide by 4
1. 1 × 4 = 23. __ × 4 = 8
2. 2 × 4 = 24. __ × 4 = 40
3. 3 × 4 = 25. __ × 4 = 12
4. 4 × 4 = 26. 8 ÷ 4 =
5. 5 × 4 = 27. 4 ÷ 1 =
6. 12 ÷ 4 = 28. 40 ÷ 4 =
7. 8 ÷ 4 = 29. 20 ÷ 4 =
8. 16 ÷ 4 = 30. 12 ÷ 4 =
9. 4 ÷ 1 = 31. __ × 4 = 12
10. 20 ÷ 4 = 32. __ × 4 = 24
11. 10 × 4 = 33. __ × 4 = 36
12. 6 × 4 = 34. __ × 4 = 28
13. 7 × 4 = 35. 32 ÷ 4 =
14. 8 × 4 = 36. 36 ÷ 4 =
15. 9 × 4 = 37. 24 ÷ 4 =
16. 28 ÷ 4 = 38. 28 ÷ 4 =
17. 24 ÷ 4 = 39. 11 × 4 =
18. 32 ÷ 4 = 40. 44 ÷ 4 =
19. 40 ÷ 4 = 41. 12 × 4 =
20. 36 ÷ 4 = 42. 48 ÷ 4 =
21. __ × 4 = 16 43. 13 × 4 =
22. __ × 4 = 20 44. 52 ÷ 4 =
B
[KEY]
Lesson 17: Model the relationship between multiplication and division. 223
Lesson 17 Problem Set NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
Name Date
× = 24 24 ÷ =
× 4 = ÷ 4 =
× 4 = ÷ 4 =
2 × 4 = ÷ 4 = 2
× 4 = 12 12 ÷ 4 =
Lesson 17: Model the relationship between multiplication and division. 224
Lesson 17 Problem Set NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
2. The baker packs 36 bran muffins in boxes of 4. Draw and label a tape diagram to find the number of boxes he packs.
3. The waitress arranges 32 glasses into 4 equal rows. How many glasses are in each row?
4. Janet paid \$28 for 4 notebooks. Each notebook costs the same amount. What is the cost of 2 notebooks?
Lesson 17: Model the relationship between multiplication and division. 225
Lesson 17 Exit Ticket NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
Name Date
1. Mr. Thomas organizes 16 binders into stacks of 4. How many stacks does he make? Draw and label a number bond to solve.
2. The chef uses 28 avocados to make 4 batches of guacamole. How many avocados are in 2 batches of guacamole? Draw and label a tape diagram to solve.
Lesson 17: Model the relationship between multiplication and division. 226
Lesson 17 Homework NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
Name Date
× 4 = 16 16 ÷ 4 =
× 4 = 12 12 ÷ 4 =
Lesson 17: Model the relationship between multiplication and division. 227
Lesson 17 Homework NYS COMMON CORE MATHEMATICS CURRICULUM 3 1
2. The teacher puts 32 students into groups of 4. How many groups does she make? Draw and label a tape diagram to solve.
3. The store clerk arranges 24 toothbrushes into 4 equal rows. How many toothbrushes are in each row?
4. An art teacher has 40 paintbrushes. She divides them equally among her 4 students. She finds 8 more brushes and divides these equally among the students, as well. How many brushes does each student receive?
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France suffered a notoriously quick defeat at the hands of its German neighbour in World War II.
The country capitulated just six weeks after Hitler’s land invasion began on May 10, 1940.
France had a system of defence, built in the more than ten years leading up to 1939, but it failed miserably.
The problem was that Maginot Line, a great line of fortifications that spanned France’s borders with several neighbours, was essentially a glorified trench.
And like any trench, it belonged to the age of the First World War, not the mechanised warfare known as blitzkrieg that Hitler brought to the Second.
The Wehrmacht simply went around the line, borrowing the low plains of Belgium to France’s north.
The line’s fortifications were built to various degrees. The dotted lines across from Belgium and Switzerland in this map designate “rural fortifications.”
Invasion was even easier for Germany’s forces because its main route, Belgium, falls within the open land of the European Plain (in the Cold War that followed, it was always imagined that the plains of Europe would host the battles of a third World War).
The line’s failure had huge implications. In a war that would last another five years, France’s role was early on reduced to one of resistance. It would not significantly contribute Germany and its allies as it had down in World War I.
Because of its place in history, the Maginot line has come to mean “a defensive barrier or strategy that inspires a false sense of security,” according to Merriam-Webster.
But recent historians have suggested a revision on what the line did or did not accomplish.
The line was a failure “in the eyes of the average French person. Yet the most modern fortification system of its day actually fulfilled its mission,” according to Michaël Seramour, a French author who wrote a book about the line. “It obliged the German Wehrmacht to attack through the Belgian plains again, as in 1914, and immobilized part of its forces.”
In this reading, concentrating the German attack in Belgium was actually part of the plan. Moreover, though France’s strategy was one of defensive over offensive development, it actually had roughly the same amount of armoured tanks on the western front as its German rival (both had nearly 2,500).
But according to analysis by West Point graduate Robert A. Doughty, “the Germans recognised the potential of massed armoured forces in conducting rapid, mobile operations, [while] French armoured units were committed to battle in a piecemeal fashion.”
Whether the Maginot line had the intended effect or not, it did not have the hoped for result.
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# Go Math Grade 6 Answer Key Chapter 2 Fractions and Decimals
Are you looking for the best material to score top in the exams? Then, you are in the right place. HMH Go Math Grade 6 Answer Key Chapter 2 Fractions and Decimals is the best material for the 6th standard students. Here you can find the explanations for each and every question in different methods. Refer to Go Math Grade 6 Chapter 2 Fractions and Decimals Solution Key to learn the concepts of the chapter. So, Download HMH Go Math Grade 6 Answer Key Chapter 2 Fractions and Decimals for free.
The Go Math Grade 6 Chapter 2 Fractions and Decimals Solution Key consists of various topics like compare and order fractions and decimals, multiply fractions, Divide Fractions, Model Mixed Number Division, etc. We have provided detailed explanations in simple methods here. All the solutions are prepared according to the topics in the Fractions and Decimals Chapter. So, access the links and start your preparation for the exams.
Lesson 1: Fractions and Decimals
Lesson 2: Compare and Order Fractions and Decimals
Lesson 3: Multiply Fractions
Lesson 4: Simplify Factors
Mid-Chapter Checkpoint
Lesson 5: Investigate • Model Fraction Division
Lesson 6: Estimate Quotients
Lesson 7: Divide Fractions
Lesson 8: Investigate • Model Mixed Number Division
Lesson 9: Divide Mixed Numbers
Lesson 10: Problem Solving • Fraction Operations
Chapter 2 Review/Test
### Share and Show – Page No. 71
Write as a fraction or as a mixed number in simplest form.
Question 1.
95.5
_____ $$\frac{□}{□}$$
$$\frac{1}{2}$$
Explanation:
95.5
95.5 is 95 ones and 5 tenths.
5 tenths = $$\frac{5}{10}$$
Simplify using the GCF.
The GCF of 5 and 10 is 10.
Divide the numerator and the denominator by 10
$$\frac{5 ÷ 10}{10 ÷ 10}$$ = $$\frac{1}{2}$$
Write as a decimal.
Question 4.
$$\frac{7}{8}$$
_____
0.875
Explanation:
Use division to rename the fraction part as a decimal.
7/8 = 0.875
The quotient has 3 decimal places.
Add the whole number to the decimal.
0 + 0.875 = 0.875.
So, $$\frac{7}{8}$$ = 0.875
Question 5.
$$\frac{13}{20}$$
_____
0.65
Explanation:
Use division to rename the fraction part as a decimal.
$$\frac{13}{20}$$ = 0.65
The quotient has 2 decimal places.
Add the whole number to the decimal.
0 + 0.65 = 0.65.
So, $$\frac{13}{20}$$ = 0.65
Write as a fraction or as a mixed number in simplest form.
Question 7.
0.27
$$\frac{□}{□}$$
$$\frac{27}{100}$$
Explanation:
0.27 is 0 ones and 27 hundredths.
27 hundredths = $$\frac{27}{100}$$
Simplify using the GCF.
The GCF of 27 and 100 is 1.
Divide the numerator and the denominator by 1
$$\frac{27 ÷ 1}{100 ÷ 1}$$ = $$\frac{27}{100}$$
Question 8.
0.055
$$\frac{□}{□}$$
$$\frac{11}{200}$$
Explanation:
0.055 is 0 ones and 55 thousandths.
55 thousandths = $$\frac{55}{1000}$$
Simplify using the GCF.
The GCF of 55 and 1000 is 5.
Divide the numerator and the denominator by 5
$$\frac{55 ÷ 5}{1000 ÷ 5}$$ = $$\frac{11}{200}$$
Question 9.
2.45
_____ $$\frac{□}{□}$$
$$\frac{9}{20}$$
Explanation:
2.45 is 2 ones and 45 hundredths.
45 hundredths = $$\frac{45}{100}$$
Simplify using the GCF.
The GCF of 45 and 100 is 5.
Divide the numerator and the denominator by 1
$$\frac{45 ÷ 5}{100 ÷ 5}$$ = $$\frac{9}{20}$$
Write as a decimal.
Question 10.
$$\frac{3}{8}$$
_____
0.375
Explanation:
Use division to rename the fraction part as a decimal.
$$\frac{3}{8}$$ = 0.375
The quotient has 3 decimal places.
Add the whole number to the decimal.
0 + 0.375 = 0.375.
So, $$\frac{3}{8}$$ = 0.375
Question 11.
3 $$\frac{1}{5}$$
_____
3.2
Explanation:
Use division to rename the fraction part as a decimal.
$$\frac{1}{5}$$ = 0.2
The quotient has 1 decimal place.
Add the whole number to the decimal.
3 + 0.2 = 3.2.
So, 3 $$\frac{1}{5}$$ = 3.2
Identify a decimal and a fraction in simplest form for the point.
Question 13.
Point A
Type below:
__________
0.2
Question 14.
Point B
Type below:
__________
0.9
Explanation:
Point B is between 0.8 and 1.0. Every point is separated by 0.1. So, Point B is at 0.9
Question 15.
Point C
Type below:
__________
0.5
Explanation:
Point C is between 0.4 and 0.6. Every point is separated by 0.1. So, Point C is at 0.5
Question 16.
Point D
Type below:
__________
0.1
Explanation:
Point D is between 0 and 0.2. Every point is separated by 0.1. So, Point D is at 0.1
### Problem Solving + Applications – Page No. 72
Use the table for 17 and 18.
Question 17.
Members of the Ozark Trail Hiking Club hiked a steep section of the trail in June and July. The table shows the distances club members hiked in miles. Write Maria’s July distance as a decimal.
_____ miles
2.625 miles
Explanation:
Maria’s July distance = 2 $$\frac{5}{8}$$
Use division to rename the fraction part as a decimal.
$$\frac{5}{8}$$ = 0.625
The quotient has 3 decimal places.
Add the whole number to the decimal.
2 + 0.625 = 2.625.
2 $$\frac{5}{8}$$ = 2.625
Question 19.
What’s the Error? Tabitha’s hiking distance in July was 2 $$\frac{1}{5}$$ miles. She wrote the distance as 2.02 miles. What error did she make?
Type below:
__________
Tabitha’s hiking distance in July was 2 $$\frac{1}{5}$$ miles.
2 $$\frac{1}{5}$$
Use division to rename the fraction part as a decimal.
$$\frac{1}{5}$$ = 0.2
The quotient has 1 decimal place.
Add the whole number to the decimal.
2 + 0.2 = 2.2.
2 $$\frac{1}{5}$$ = 2.2
She wrote the distance as 2.02 miles in mistake.
Question 21.
Identify a decimal and a fraction in simplest form for the point.
Type below:
__________
Point A: 0.5
Point B: 0.7
Point C: 0.3
Point D: 0.8
Explanation:
Every point is differentiated by 0.1 distance.
The A is between 0.4 and 0.6 which is 0.5
The B is between 0.6 and 0.8 which is 0.7
The C is between 0.1 and 0.6 which is 0.53
### Fractions and Decimals – Page No. 73
Write as a fraction or as a mixed number in simplest form.
Question 1.
0.52
$$\frac{□}{□}$$
$$\frac{13}{25}$$
Explanation:
0.52
0.52 is 52 hundredths.
52 hundredths = $$\frac{52}{100}$$
Simplify using the GCF.
The GCF of 52 and 100 is 4.
Divide the numerator and the denominator by 4
$$\frac{52 ÷ 4}{100 ÷ 4}$$ = $$\frac{13}{25}$$
Question 2.
0.02
$$\frac{□}{□}$$
$$\frac{1}{50}$$
Explanation:
0.02
0.02 is 2 hundredths.
2 hundredths = $$\frac{2}{100}$$
Simplify using the GCF.
The GCF of 2 and 100 is 2.
Divide the numerator and the denominator by 2
$$\frac{2 ÷ 2}{100 ÷ 2}$$ = $$\frac{1}{50}$$
Question 3.
4.8
______ $$\frac{□}{□}$$
$$\frac{4}{5}$$
Explanation:
4.8
4.8 is 4 ones and 8 tenths.
8 tenths = $$\frac{8}{10}$$
Simplify using the GCF.
The GCF of 8 and 10 is 2.
Divide the numerator and the denominator by 2
$$\frac{8 ÷ 2}{10 ÷ 2}$$ = $$\frac{4}{5}$$
Write as a decimal.
Question 5.
$$\frac{17}{25}$$
______
0.68
Explanation:
Use division to rename the fraction part as a decimal.
17/25 = 0.68
The quotient has 2 decimal places.
Add the whole number to the decimal.
0 + 0.68 = 0.68.
So, $$\frac{17}{25}$$ = 0.68
Question 6.
$$\frac{11}{20}$$
______
0.55
Explanation:
Use division to rename the fraction part as a decimal.
11/20 = 0.55
The quotient has 2 decimal places.
Add the whole number to the decimal.
0 + 0.55 = 0.55.
So, $$\frac{11}{20}$$ = 0.55
Question 8.
7 $$\frac{3}{8}$$
______
7.375
Explanation:
Use division to rename the fraction part as a decimal.
$$\frac{3}{8}$$ = 0.375
The quotient has 3 decimal places.
Add the whole number to the decimal.
7 + 0.375 = 7.375.
So, 7 $$\frac{3}{8}$$ = 7.375
Identify a decimal and a fraction or mixed number in simplest form for each point.
Question 9.
Point A
Type below:
__________
0.4
Explanation:
Point A is between 0 and 0.5. Every point is separated by 0.1. So, Point A is at 0.4
Question 10.
Point D
Type below:
__________
1.9
Explanation:
Point D is between 1.5 and 2. Every point is separated by 0.1. So, Point D is at 1.9
Question 11.
Point C
Type below:
__________
1.2
Explanation:
Point C is between 1 and 1.5. Every point is separated by 0.1. So, Point C is at 1.2
Question 12.
Point B
Type below:
__________
0.6
Explanation:
Point C is between 0.5 and 1. Every point is separated by 0.1. So, Point C is at 0.6
Problem Solving
Question 13.
Grace sold $$\frac{5}{8}$$ of her stamp collection. What is this amount as a decimal?
______
0.625
Explanation:
Grace sold $$\frac{5}{8}$$ of her stamp collection.
Use division to rename the fraction part as a decimal.
$$\frac{5}{8}$$ = 0.625
The quotient has 3 decimal places.
Add the whole number to the decimal.
0 + 0.625 = 0.625.
So, $$\frac{5}{8}$$ = 0.625
Question 14.
What if you scored an 0.80 on a test? What fraction of the test, in simplest form, did you answer correctly?
$$\frac{□}{□}$$
$$\frac{4}{5}$$
Explanation:
0.80 is 0 ones and 8 tenths.
8 tenths = $$\frac{8}{10}$$
Simplify using the GCF.
The GCF of 8 and 10 is 2.
Divide the numerator and the denominator by 2
$$\frac{8 ÷ 2}{10 ÷ 2}$$ = $$\frac{4}{5}$$
### Lesson Check – Page No. 74
Question 1.
After a storm, Michael measured 6 $$\frac{7}{8}$$ inches of snow. What is this amount as a decimal?
______ inches
6.875 inches
Explanation:
Michael measured 6 $$\frac{7}{8}$$ inches of snow.
Use division to rename the fraction part as a decimal.
$$\frac{7}{8}$$ = 0.875
The quotient has 3 decimal places.
Add the whole number to the decimal.
6 + 0.875 = 6.875.
So, 6 $$\frac{7}{8}$$ = 6.875.
Spiral Review
Question 3.
Gina bought 2.3 pounds of red apples and 2.42 pounds of green apples. They were on sale for $0.75 a pound. How much did the apples cost altogether?$ ______
$3.54 Explanation: Gina bought 2.3 pounds of red apples and 2.42 pounds of green apples. They were on sale for$0.75 a pound.
$0.75 x 2.3 = 1.725$0.75 x 2.42 = 1.815
1.725 + 1.815 = 3.54
So the apples cost $3.54 Question 4. Ken has 4.66 pounds of walnuts, 2.1 pounds of cashews, and 8 pounds of peanuts. He mixes them together and divides them equally among 18 bags. How many pounds of nuts are in each bag? ______ pounds Answer: 0.82 pounds Explanation: Ken has 4.66 pounds of walnuts, 2.1 pounds of cashews, and 8 pounds of peanuts. 4.66 + 2.1 + 8 = 14.76 He mixes them together and divides them equally among 18 bags. 14.76/18 = 0.82 Question 5. Mia needs to separate 270 blue pens and 180 red pens into packs. Each pack will have the same number of blue pens and the same number of red pens. What is the greatest number of packs she can make? How many red pens and how many blue pens will be in each pack? Type below: __________ Answer: There are 2 red pens and 3 blue pens in each pack. Explanation: Mia needs to separate 270 blue pens and 180 red pens into packs. The GCF of 270 and 180 is 90 The greatest number of packs she can make is 90. Divide the total number of red pens by the total number of packs. 180/90 = 2 Divide the total number of blue pens by the total number of packs. 270/90 = 3 There are 2 red pens and 3 blue pens in each pack. Question 6. Evan buys 19 tubes of watercolor paint for$50.35. What is the cost of each tube of paint?
$______ Answer:$2.65
Explanation:
Evan buys 19 tubes of watercolor paint for $50.35.$50.35/19 = $2.65 ### Share and Show – Page No. 77 Order from least to greatest. Write <, >, or =. Question 2. 0.8 _____ $$\frac{4}{12}$$ Answer: 0.8 < latex]\frac{4}{12}[/latex] Explanation: Write the decimal form of $$\frac{4}{12}$$ = 0.3333 0.8 > 0.333 So, 0.8 < latex]\frac{4}{12}[/latex] Question 3. 0.22 _____ $$\frac{1}{4}$$ Answer: 0.22 < $$\frac{1}{4}$$ Explanation: Write the decimal form of $$\frac{1}{4}$$ = 0.25 0.22 < 0.25 So, 0.22 < $$\frac{1}{4}$$ Question 4. $$\frac{1}{20}$$ _____ 0.06 Answer: $$\frac{1}{20}$$ < 0.06 Explanation: Write the decimal form of $$\frac{1}{20}$$ = 0.05 0.05 < 0.06 So, $$\frac{1}{20}$$ < 0.06 Use a number line to order from least to greatest. On Your Own Order from least to greatest. Question 6. 0.6, $$\frac{4}{5}$$, 0.75 Type below: __________ Answer: 0.6, 0.75, $$\frac{4}{5}$$ Explanation: Write the decimal form of $$\frac{4}{5}$$ = 0.8 0.6, 0.8, 0.75 Compare decimals. All ones are equal. Compare tenths: 6 < 7 < 8 So, from least to greatest, the order is 0.6, 0.75, 0.8 So, 0.6, 0.75, $$\frac{4}{5}$$ Question 7. $$\frac{1}{2}$$, $$\frac{2}{5}$$, $$\frac{7}{15}$$ Type below: __________ Answer: $$\frac{2}{5}$$, $$\frac{7}{15}$$, $$\frac{1}{2}$$ Explanation: Write the decimal form of $$\frac{1}{2}$$ = 0.5 Write the decimal form of $$\frac{2}{5}$$ = 0.4 Write the decimal form of $$\frac{7}{15}$$ = 0.466 0.5, 0.4, 0.466 Compare decimals. All ones are equal. Compare tenths: 4 < 5 Compare hundredths of 0.4 and 0.466; 0 < 6 So, from least to greatest, the order is 0.4 < 0.466 < 0.5 So, $$\frac{2}{5}$$, $$\frac{7}{15}$$, $$\frac{1}{2}$$ Question 8. 5 $$\frac{1}{2}$$, 5.05, 5 $$\frac{5}{9}$$ Type below: __________ Answer: 5.05, 5 $$\frac{1}{2}$$, 5 $$\frac{5}{9}$$ Explanation: Write the decimal form of 5 $$\frac{1}{2}$$ = 5.5 Write the decimal form of 5 $$\frac{5}{9}$$ = 5.555 5.5, 5.05, 5.5555 Compare decimals. All ones are equal. Compare tenths: 0 < 5 Compare hundredths of 5.5 and 5.55; 0 < 5 So, from least to greatest, the order is 5.05 < 5.5 < 5.55 So, 5.05, 5 $$\frac{1}{2}$$, 5 $$\frac{5}{9}$$ Question 9. $$\frac{5}{7}$$, $$\frac{5}{6}$$, $$\frac{5}{12}$$ Type below: __________ Answer: $$\frac{5}{12}$$, $$\frac{5}{7}$$, $$\frac{5}{6}$$ Explanation: $$\frac{5}{7}$$, $$\frac{5}{6}$$, $$\frac{5}{12}$$ To compare fractions with the same numerators, compare the denominators. So, from least to greatest, the order is $$\frac{5}{12}$$, $$\frac{5}{7}$$, $$\frac{5}{6}$$ Question 10. $$\frac{7}{15}$$ _____ $$\frac{7}{10}$$ Answer: $$\frac{7}{15}$$ < $$\frac{7}{10}$$ Explanation: $$\frac{7}{15}$$ and $$\frac{7}{10}$$ To compare fractions with the same numerators, compare the denominators. So, $$\frac{7}{15}$$ < $$\frac{7}{10}$$ Question 11. $$\frac{1}{8}$$ _____ 0.125 Answer: $$\frac{1}{8}$$ = 0.125 Explanation: Write the decimal form of $$\frac{1}{8}$$ = 0.125 0.125 = 0.125 Question 12. 7 $$\frac{1}{3}$$ _____ 6 $$\frac{2}{3}$$ Answer: 7 $$\frac{1}{3}$$ > 6 $$\frac{2}{3}$$ Explanation: Compare the whole numbers first. 7 > 6. So, 7 $$\frac{1}{3}$$ > 6 $$\frac{2}{3}$$ Question 13. 1 $$\frac{2}{5}$$ _____ 1 $$\frac{7}{15}$$ Answer: 1 $$\frac{2}{5}$$ < 1 $$\frac{7}{15}$$ Explanation: 1 $$\frac{2}{5}$$ _____ 1 $$\frac{7}{15}$$ If the whole numbers are the same, compare the fractions. Compare $$\frac{2}{5}$$ and $$\frac{7}{15}$$ 5 and 15 are multiples of 15. So, $$\frac{2 x 3}{5 x 3}$$ = $$\frac{6}{15}$$ $$\frac{6}{15}$$ < $$\frac{7}{15}$$ Use common denominators to write equivalent fractions. 1 $$\frac{2}{5}$$ < 1 $$\frac{7}{15}$$ Question 14. Darrell spent 3 $$\frac{2}{5}$$ hours on a project for school. Jan spent 3 $$\frac{1}{4}$$ hours and Maeve spent 3.7 hours on the project. Who spent the least amount of time? Show how you found your answer. Then describe another possible method. Type below: __________ Answer: Jan spent the least amount of time. Explanation: Darrell spent 3 $$\frac{2}{5}$$ hours on a project for school. Jan spent 3 $$\frac{1}{4}$$ hours and Maeve spent 3.7 hours on the project. Write the decimal form of 3 $$\frac{2}{5}$$ = 3.4 Write the decimal form of 3 $$\frac{1}{4}$$ = 3.25 3.4, 3.25, 3.7 3.25 is the least one. So, Jan spent the least amount of time. ### Problem Solving + Applications – Page No. 78 Use the table for 15–18. Question 15. In one week, Altoona, PA, and Bethlehem, PA, received snowfall every day, Monday through Friday. On which days did Altoona receive over 0.1 inch more snow than Bethlehem? Type below: __________ Answer: Altoona received over 1 inch more snow than Bethlehem on Friday Explanation: Altoona (converted to decimal form): 2.25, 3.25, 2.625, 4.6, 4.75 Bethlehem: 2.6, 3.2, 2.5, 4.8, 2.7 Altoona received over 1 inch more snow than Bethlehem on Friday Question 16. What if Altoona received an additional 0.3 inches of snow on Thursday? How would the total amount of snow in Altoona compare to the amount received in Bethlehem that day? Type below: __________ Answer: Altoona received 0.1 inches more snow than Bethlehem on Thursday Explanation: Altoona received an additional 0.3 inch of snow on Thursday = 4.6 + 0.3 = 4.9 Bethlehem received on Thursday = 4.8 Altoona received 0.1 inches more snow than Bethlehem on Thursday Question 17. Explain two ways you could compare the snowfall amounts in Altoona and Bethlehem on Monday. Type below: __________ Answer: Explanation: Altoona received on Monday = 2.25 Bethlehem received on Monday = 2.6 Bethlehem received 0.35 inches more snow than Altoona on Monday. As the whole numbers are equal compare 1/4 and 0.6. 0.25 < 0.6 So, Altoona received less snow compared to Bethlehem on Monday. Question 18. Explain how you could compare the snowfall amounts in Altoona on Thursday and Friday. Type below: __________ Answer: Altoona received on Thursday = 4.6 Altoona received on Friday = 4.75 4.6 < 4.75 Altoona received less snow on Thursday compared to Friday. Question 19. Write the values in order from least to greatest. Type below: __________ Answer: 1/3, 0.39, 2/5, 0.45 Explanation: 1/3 = 0.333 0.45 0.39 2/5 = 0.4 Compare tenths: 3 < 4 Compare hundredths: 0.33 < 0.39 0.4 < 0.45 So, 1/3, 0.39, 2/5, 0.45 ### Compare and Order Fractions and Decimals – Page No. 79 Write <, >, =. Question 1. 0.64 _____ $$\frac{7}{10}$$ Answer: 0.64 < $$\frac{7}{10}$$ Explanation: Write the decimal form of $$\frac{7}{10}$$ = 0.7 Compare tenths: 6 < 7 So, 0.64 < 0.7 0.64 < $$\frac{7}{10}$$ Question 4. 7 $$\frac{1}{8}$$ _____ 7.025 Answer: 7 $$\frac{1}{8}$$ > 7.025 Explanation: Write the decimal form of 7 $$\frac{1}{8}$$ = 7.125 Compare tenths: 1 > 0 7 $$\frac{1}{8}$$ > 7.025 Order from least to greatest. Question 5. $$\frac{7}{15}$$, 0.75, $$\frac{5}{6}$$ Type below: __________ Answer: $$\frac{7}{15}$$, 0.75, $$\frac{5}{6}$$ Explanation: Write the decimal form of $$\frac{7}{15}$$ = 0.466 0.75 Write the decimal form of $$\frac{5}{6}$$ = 0.833 Order from least to greatest: $$\frac{7}{15}$$, 0.75, $$\frac{5}{6}$$ Question 7. 3.25, 3 $$\frac{2}{5}$$, 3 $$\frac{3}{8}$$ Type below: __________ Answer: 3.25, 3 $$\frac{2}{5}$$, 3 $$\frac{3}{8}$$ Explanation: Write the decimal form of 3 $$\frac{2}{5}$$ = 3.4 Write the decimal form of 3 $$\frac{3}{8}$$ = 3.375 Compare tenths: Order from least to greatest: 3.25, 3 $$\frac{2}{5}$$, 3 $$\frac{3}{8}$$ Question 8. 0.9, $$\frac{8}{9}$$, 0.86 Type below: __________ Answer: 0.86, $$\frac{8}{9}$$, 0.9 Explanation: Write the decimal form of $$\frac{8}{9}$$ = 0.88 Compare tenths: 0.86, 0.88, 0.9 Order from least to greatest: 0.86, $$\frac{8}{9}$$, 0.9 Order from greatest to least. Question 9. 0.7, $$\frac{7}{9}$$, $$\frac{7}{8}$$ Type below: __________ Answer: $$\frac{7}{8}$$, $$\frac{7}{9}$$, 0.7 Explanation: 0.7 = 7/10 To compare fractions with the same numerators, compare the denominators. 7/10, 7/9, 7/8 Order from greatest to least: 7/8, 7/9, 7/10 Question 10. 0.2, 0.19, $$\frac{3}{5}$$ Type below: __________ Answer: $$\frac{3}{5}$$, 0.2, 0.19 Explanation: Write the decimal form of $$\frac{3}{5}$$ = 0.6 Compare tenths: 0.6, 0.2, 0.19 Order from greatest to least: $$\frac{3}{5}$$, 0.2, 0.19 Question 11. 6$$\frac{1}{20}$$, 6.1, 6.07 Type below: __________ Answer: Explanation: Write the decimal form of 6$$\frac{1}{20}$$ = 121/20 = 6.05 Compare tenths: 6.1, 6.07, 6.05 Order from greatest to least: 6.1, 6.07, 6$$\frac{1}{20}$$ Question 12. 2 $$\frac{1}{2}$$, 2.4, 2.35, 2 $$\frac{1}{8}$$ Type below: __________ Answer: 2 $$\frac{1}{2}$$, 2.4, 2.35, 2 $$\frac{1}{8}$$ Explanation: Write the decimal form of 2 $$\frac{1}{2}$$ = 2.5 Write the decimal form of 2 $$\frac{1}{8}$$ = 2.125 Compare tenths: 2.5, 2.4, 2.35, 2.125 Order from greatest to least: 2 $$\frac{1}{2}$$, 2.4, 2.35, 2 $$\frac{1}{8}$$ Question 14. Malia and John each bought 2 pounds of sunflower seeds. Each ate some seeds. Malia has 1 $$\frac{1}{3}$$ pounds left, and John has 1 $$\frac{2}{5}$$ pounds left. Who ate more sunflower seeds? __________ Answer: Malia Explanation: Malia and John each bought 2 pounds of sunflower seeds. Each ate some seeds. Malia has 1 $$\frac{1}{3}$$ pounds left, and John has 1 $$\frac{2}{5}$$ pounds left. 2 – 1 $$\frac{1}{3}$$ = 0.667 2 – 1 $$\frac{2}{5}$$ = 0.6 0.667 > 0.6 So, Malia ate more sunflower seeds Question 15. Explain how you would compare the numbers 0.4 and $$\frac{3}{8}$$. Type below: __________ Answer: Write the decimal form of $$\frac{3}{8}$$ = 0.375 Compare tenths: 0.4 > 0.375 ### Lesson Check – Page No. 80 Question 1. Andrea has 3 $$\frac{7}{8}$$ yards of purple ribbon, 3.7 yards of pink ribbon, and 3 $$\frac{4}{5}$$ yards of blue ribbon. List the numbers in order from least to greatest. Type below: __________ Answer: Andrea has 3 $$\frac{7}{8}$$ yards of purple ribbon, 3.7 yards of pink ribbon, and 3 $$\frac{4}{5}$$ yards of blue ribbon. Write the decimal form of 3 $$\frac{7}{8}$$ = 3.875 3.7 Write the decimal form of 3 $$\frac{4}{5}$$ = 3.8 Least to greatest : 3.7, 3 $$\frac{4}{5}$$, 3 $$\frac{7}{8}$$ Spiral Review Question 3. Tyler bought 3 $$\frac{2}{5}$$ pounds of oranges. Graph 3 $$\frac{2}{5}$$ on a number line and write this amount using a decimal. Type below: __________ Answer: Tyler bought 3 $$\frac{2}{5}$$ pounds of oranges. Decimal Form: 17/5 = 3.4 Question 4. At the factory, a baseball card is placed in every 9th package of cereal. A football card is placed in every 25th package of the cereal. What is the first package that gets both a baseball card and a football card? Type below: __________ Answer: 225th package Explanation: Look for the first number where both 25 and 9 are a factor of. 25 x 1 = 25 which isn’t a factor of 9, so it won’t be 25. 25 x 2 = 50, which isn’t a factor of 9. 75 is not a factor of 9. (you know because you don’t get a whole number when you divide 75 into 9.) 100 is not a factor of 9, nor is 125, 150, 175, or 200. However, 225 is a factor of both 25 and 9. This makes sense because 25 x 9 is 225. This means that the first package with both will be the 225th package. Question 6. Carrie buys 4.16 pounds of apples for$5.20. How much does 1 pound cost?
$_____ Answer:$1.25
Explanation:
Carrie buys 4.16 pounds of apples for $5.20.$5.20/4.16 = $1.25 1 pound cost =$1.25
### Share and Show – Page No. 83
Find the product. Write it in simplest form.
Question 1.
6 × $$\frac{3}{8}$$
$$\frac{□}{□}$$
$$\frac{9}{4}$$
Explanation:
$$\frac{6 × 3}{1 × 8}$$
$$\frac{18}{8}$$
Simplify using the GCF.
The GCF of 18 and 8 is 2.
Divide the numerator and the denominator by 2.
$$\frac{18 ÷ 2}{8 ÷ 2}$$ = $$\frac{9}{4}$$
Question 2.
$$\frac{3}{8}$$ × $$\frac{8}{9}$$
$$\frac{□}{□}$$
$$\frac{1}{3}$$
Explanation:
Multiply the numerators and Multiply the denominators.
$$\frac{3 × 8}{8 × 9}$$ = $$\frac{24}{72}$$
Simplify using the GCF.
The GCF of 24 and 72 is 24.
Divide the numerator and the denominator by 24.
$$\frac{24 ÷ 24}{72 ÷ 24}$$ = $$\frac{1}{3}$$
Attend to Precision Algebra Evaluate using the order of operations.
Write the answer in the simplest form.
Question 4.
$$\left(\frac{3}{4}-\frac{1}{2}\right) \times \frac{3}{5}$$
$$\frac{□}{□}$$
$$\frac{3}{20}$$
Explanation:
$$\left(\frac{3}{4}-\frac{1}{2}\right) \times \frac{3}{5}$$
Perform operations in parentheses.
$$\frac{3}{4}$$ – $$\frac{1}{2}$$ = $$\frac{1}{4}$$
$$\frac{1}{4}$$ x $$\frac{3}{5}$$ = $$\frac{1 x 3}{4 x 5}$$ = $$\frac{3}{20}$$
Question 5.
$$\frac{1}{3}+\frac{4}{9} \times 12$$
$$\frac{□}{□}$$
$$\frac{28}{3}$$
Explanation:
$$\frac{1}{3}$$ + $$\frac{4}{9}$$ = $$\frac{7}{9}$$
$$\frac{7 x 12}{9 x 1}$$ = $$\frac{84}{9}$$
Simplify using the GCF.
The GCF of 84 and 9 is 3.
Divide the numerator and the denominator by 3.
$$\frac{84 ÷ 3}{9 ÷ 3}$$ = $$\frac{28}{3}$$
Question 6.
$$\frac{5}{8} \times \frac{7}{10}-\frac{1}{4}$$
$$\frac{□}{□}$$
$$\frac{11}{16}$$
Explanation:
$$\frac{5 x 7}{8 x 10}$$ = $$\frac{35}{80}$$
$$\frac{35}{80}$$ – $$\frac{1}{4}$$ = $$\frac{11}{16}$$
Practice: Copy and Solve Find the product. Write it in simplest form.
Question 8.
$$1 \frac{2}{3} \times 2 \frac{5}{8}$$
$$\frac{□}{□}$$
$$\frac{35}{8}$$
Explanation:
1 $$\frac{2}{3}$$ = $$\frac{5}{3}$$
2 $$\frac{5}{8}$$ = $$\frac{21}{8}$$
$$\frac{5 × 21}{3 × 8}$$ = $$\frac{105}{24}$$
Simplify using the GCF
The GCF of 105 and 24 is 3.
Divide the numerator and the denominator by 3.
$$\frac{105 ÷ 3}{24 ÷ 3}$$ = $$\frac{35}{8}$$
Question 9.
$$\frac{4}{9} \times \frac{4}{5}$$
$$\frac{□}{□}$$
$$\frac{16}{45}$$
Explanation:
$$\frac{4 × 4}{9 × 5}$$ = $$\frac{16}{45}$$
Question 10.
$$\frac{1}{6} \times \frac{2}{3}$$
$$\frac{□}{□}$$
$$\frac{1}{9}$$
Explanation:
$$\frac{1 × 2}{6 × 3}$$ = $$\frac{2}{18}$$
Simplify using the GCF
The GCF of 2 and 18 is 2.
Divide the numerator and the denominator by 2.
$$\frac{2 ÷ 2}{18 ÷ 2}$$ = $$\frac{1}{9}$$
Question 12.
$$\frac{5}{6}$$ of the 90 pets in the pet show are cats. $$\frac{4}{5}$$ of the cats are calico cats. What fraction of the pets are calico cats? How many of the pets are calico cats?
Type below:
__________
60 calico cats
Explanation:
5/6 x 90 = 450/6 = 150/2
150/2 x 4/5 = 60
Question 13.
Five cats each ate $$\frac{1}{4}$$ cup of cat food. Four other cats each ate $$\frac{1}{3}$$ cup of cat food. How much food did the nine cats eat?
Type below:
__________
$$\frac{31}{12}$$
Explanation:
5 x 1/4 = 5/4
4 x 1/3 = 4/3
5/4 + 4/3 = 31/12
Attend to Precision Algebra Evaluate using the order of operations.
Write the answer in the simplest form.
Question 14.
$$\frac{1}{4} \times\left(\frac{3}{9}+5\right)$$
$$\frac{□}{□}$$
$$\frac{4}{3}$$
Explanation:
3/9 + 5 = 16/3
1/4 x 16/3
1 x 16 = 16
4 x 3 = 12
16/12
Simplify using the GCF
The GCF of 16 and 12 is 4.
Divide the numerator and the denominator by 4.
$$\frac{16 ÷ 4}{12÷ 4}$$ = $$\frac{4}{3}$$
Question 16.
$$\frac{4}{5}+\left(\frac{1}{2}-\frac{3}{7}\right) \times 2$$
$$\frac{□}{□}$$
$$\frac{33}{35}$$
Explanation:
1/2 – 3/7 = 1/14
1/14 x 2 = 1/7
4/5 + 1/7 = 33/35
Question 17.
$$15 \times \frac{3}{10}+\frac{7}{8}$$
$$\frac{□}{□}$$
$$\frac{141}{8}$$
Explanation:
3/10 + 7/8 = 47/40
15 x 47/40 = 141/8
$$\frac{141}{8}$$
### Page No. 84
Question 18.
Write and solve a word problem for the expression $$\frac{1}{4} \times \frac{2}{3}$$. Show your work.
Type below:
__________
$$\frac{1}{6}$$
Explanation:
$$\frac{1}{4} \times \frac{2}{3}$$ = $$\frac{1 X 2}{4 X 3}$$ = $$\frac{2}{12}$$
Simplify using the GCF
The GCF of 2 and 12 is 2.
Divide the numerator and the denominator by 2.
$$\frac{2 ÷ 2}{12 ÷ 2}$$ = $$\frac{1}{6}$$
Question 19.
Michelle has a recipe that asks for 2 $$\frac{1}{2}$$ cups of vegetable oil. She wants to use $$\frac{2}{3}$$ that amount of oil and use applesauce to replace the rest. How much applesauce will she use?
Type below:
__________
$$\frac{10}{6}$$
Explanation:
2 1/2 * 2/3 = 5/2 * 2/3 = 10/6 She will use 10/6 or 1 2/3 cups of vegetable oil
Question 20.
Cara’s muffin recipe asks for 1 $$\frac{1}{2}$$ cups of flour for the muffins and $$\frac{1}{4}$$ cup of flour for the topping. If she makes $$\frac{1}{2}$$ of the original recipe, how much flour will she use for the muffins and topping?
Type below:
__________
Cara will use 1$$\frac{1}{8}$$ cups of flour.
Explanation:
For first we will find how many cups of flours need to makes the original recipe. Cara uses 1 1/2 cups of flour for the muffins and 1/4 cup off flour for the topping.
So, 1 1/2 + 1/4 cups of flour to make the original recipe.
1 1/2 = 3/2
3/2 + 1/4 = 7/4
To make the original recipe Cara needs 7/4 cups of flour.
If she makes $$\frac{1}{2}$$ of the original recipe, then
7/4 x 1/2 = 7/8 = 1 1/8
Cara will use 1 1/8 cups of flour.
### Multiply Fractions – Page No. 85
Find the product. Write it in simplest form.
Question 1.
$$\frac{4}{5} \times \frac{7}{8}$$
$$\frac{□}{□}$$
$$\frac{7}{10}$$
Explanation:
Multiply the numerators and Multiply the denominators.
$$\frac{4 × 7}{5 × 8}$$ = $$\frac{28}{40}$$
Simplify using the GCF.
The GCF of 28 and 40 is 4.
Divide the numerator and the denominator by 4.
$$\frac{28 ÷ 4}{40 ÷ 4}$$ = $$\frac{7}{10}$$
Question 2.
$$\frac{1}{8} \times 20$$
$$\frac{□}{□}$$
$$\frac{5}{2}$$
Explanation:
$$\frac{1 × 20}{1 × 8}$$
$$\frac{20}{8}$$
Simplify using the GCF.
The GCF of 20 and 8 is 4.
Divide the numerator and the denominator by 4.
$$\frac{20 ÷ 4}{8 ÷ 4}$$ = $$\frac{5}{2}$$
Question 3.
$$\frac{4}{5} \times \frac{3}{8}$$
$$\frac{□}{□}$$
$$\frac{3}{10}$$
Explanation:
Multiply the numerators and Multiply the denominators.
$$\frac{4 × 3}{5 × 8}$$ = $$\frac{12}{40}$$
Simplify using the GCF.
The GCF of 12 and 40 is 4.
Divide the numerator and the denominator by 4.
$$\frac{12 ÷ 4}{40 ÷ 4}$$ = $$\frac{3}{10}$$
Question 6.
Karen raked $$\frac{3}{5}$$ of the yard. Minni raked $$\frac{1}{3}$$ of the amount Karen raked. How much of the yard did Minni rake?
$$\frac{□}{□}$$
$$\frac{1}{3}$$
Explanation:
Minni raked 1/5 of the yard.
So, minni raked 3/5 of 1/3 means 3/5 x 1/3
Multiply the numerators and Multiply the denominators.
$$\frac{3 × 1}{5 × 3}$$ = $$\frac{3}{15}$$
Simplify using the GCF.
The GCF of 3 and 15 is 3.
Divide the numerator and the denominator by 3.
$$\frac{3 ÷ 3}{15 ÷ 3}$$ = $$\frac{1}{3}$$
Question 7.
$$\frac{3}{8}$$ of the pets in the pet show are dogs. $$\frac{2}{3}$$ of the dogs have long hair. What fraction of the pets are dogs with long hair?
$$\frac{□}{□}$$
$$\frac{1}{4}$$ are dogs with long hair
Explanation:
$$\frac{3}{8}$$ of the pets in the pet show are dogs. $$\frac{2}{3}$$ of the dogs have long hair.
$$\frac{3}{8}$$ of $$\frac{2}{3}$$ = $$\frac{3 × 2}{8 × 3}$$ = $$\frac{6}{24}$$
The GCF of 6 and 24 is 6.
Divide the numerator and the denominator by 6.
$$\frac{6 ÷ 6}{24 ÷ 6}$$ = $$\frac{1}{4}$$
$$\frac{1}{4}$$ are dogs with long hair
Evaluate using the order of operations.
Question 8.
$$\left(\frac{1}{2}+\frac{3}{8}\right) \times 8$$
______
7
Explanation:
1/2 + 3/8 = 7/8
7/8 × 8 = 7
Question 9.
$$\frac{3}{4} \times\left(1-\frac{1}{9}\right)$$
$$\frac{□}{□}$$
$$\frac{2}{3}$$
Explanation:
1 – 1/9 = 8/9
3/4 × 8/9 = 24/36
The GCF of 24 and 36 is 12.
Divide the numerator and the denominator by 12.
$$\frac{24 ÷ 12}{36 ÷ 12}$$ = $$\frac{2}{3}$$
Question 10.
$$4 \times \frac{1}{8} \times \frac{3}{10}$$
$$\frac{□}{□}$$
$$\frac{3}{20}$$
Explanation:
Multiply the numerators and Multiply the denominators.
$$\frac{4 × 1 × 3}{1 × 8 × 10}$$ = $$\frac{12}{80}$$
Simplify using the GCF.
The GCF of 12 and 80 is 4.
Divide the numerator and the denominator by 4.
$$\frac{12 ÷ 4}{80 ÷ 4}$$ = $$\frac{3}{20}$$
Question 11.
$$6 \times\left(\frac{4}{5}+\frac{2}{10}\right) \times \frac{2}{3}$$
______
4
Explanation:
4/5 + 2/10 = 1
6 × 1 × 2/3 = 12/3
The GCF of 12 and 3 is 4.
Divide the numerator and the denominator by 3.
$$\frac{12 ÷ 3}{3 ÷ 3}$$ = $$\frac{4}{1}$$ = 4
Problem Solving
Question 12.
Jason ran $$\frac{5}{7}$$ of the distance around the school track. Sara ran $$\frac{4}{5}$$ of Jason’s distance. What fraction of the total distance around the track did Sara run?
$$\frac{□}{□}$$
$$\frac{4}{7}$$
Explanation:
Jason ran $$\frac{5}{7}$$ of the distance around the school track. Sara ran $$\frac{4}{5}$$ of Jason’s distance.
$$\frac{5}{7}$$ × $$\frac{4}{5}$$ = 20/35
The GCF of 20 and 35 is 5.
Divide the numerator and the denominator by 5.
$$\frac{20 ÷ 5}{35 ÷ 5}$$ = $$\frac{4}{7}$$
Question 13.
A group of students attends a math club. Half of the students are boys and $$\frac{4}{9}$$ of the boys have brown eyes. What fraction of the group are boys with brown eyes?
$$\frac{□}{□}$$
$$\frac{2}{9}$$ group are boys with brown eyes
Explanation:
A group of students attends a math club. Half of the students are boys and $$\frac{4}{9}$$ of the boys have brown eyes.
$$\frac{4}{9}$$ × $$\frac{1}{2}$$ = 4/18 = 2/9
2/9 group are boys with brown eyes
Question 14.
Write and solve a word problem that involves multiplying by a fraction.
Type below:
__________
A group of students attends a math club. Half of the students are boys and $$\frac{6}{9}$$ of the boys have brown eyes. What fraction of the group are boys with brown eyes?
$$\frac{□}{□}$$
A group of students attends a math club. Half of the students are boys and $$\frac{6}{9}$$ of the boys have brown eyes.
$$\frac{6}{9}$$ × $$\frac{1}{2}$$ = 6/18 = 1/3
1/3 group are boys with brown eyes.
### Lesson Check – Page No. 86
Question 1.
Veronica’s mom left $$\frac{3}{4}$$ of a cake on the table. Her brothers ate $$\frac{1}{2}$$ of it. What fraction of the cake did they eat?
$$\frac{□}{□}$$
$$\frac{2}{4}$$
Explanation:
Veronica’s mom left $$\frac{3}{4}$$ of a cake on the table. Her brothers ate $$\frac{1}{2}$$ of it.
Since the fraction of the eaten cake is 1/2, you can multiply the numerator and denominator by and get an equivalent fraction, which is 2/4.
Question 2.
One lap around the school track is $$\frac{5}{8}$$ mile. Carin ran 3 $$\frac{1}{2}$$ laps. How far did she run?
_____ $$\frac{□}{□}$$
2$$\frac{3}{16}$$
Explanation:
One lap around the school track is $$\frac{5}{8}$$ mile. Carin ran 3 $$\frac{1}{2}$$ laps.
3 $$\frac{1}{2}$$ = $$\frac{7}{2}$$
Therefore, the total distance covered = 7/2 × 5/8 = 35/16 = 2 3/16
Spiral Review
Question 5.
Naomi went on a 6.5-mile hike. In the morning, she hiked 1.75 miles, rested, and then hiked 2.4 more miles. She completed the hike in the afternoon. How much farther did she hike in the morning than in the afternoon?
_____ miles
Naomi went on a 6.5-mile hike. In the morning, she hiked 1.75 miles, rested, and then hiked 2.4 more miles. She completed the hike in the afternoon.
To find how many miles she walked in the afternoon you just subtract the morning miles 4.15 from the total miles 6.5.
6.5 – 4.15 = 2.35
To find how many more miles she walked in the morning you just subtract the morning from the afternoon 4.15 – 2.35=1.8 miles.
She hiked 1.8 more miles in the morning
Question 6.
A bookstore owner has 48 science fiction books and 30 mysteries he wants to sell quickly. He will make discount packages with one type of book in each. He wants the most books possible in each package, but all packages must contain the same number of books. How many packages can he make? How many packages of each type of book does he have?
Type below:
__________
18 packages
Explanation:
The bookstore owner can make 18 possible packages
48 – 30 = 18 packages
### Share and Show – Page No. 89
Find the product. Simplify before multiplying.
Question 1.
$$\frac{5}{6} \times \frac{3}{10}$$
$$\frac{□}{□}$$
$$\frac{1}{4}$$
Explanation:
$$\frac{5}{6} \times \frac{3}{10}$$
Multiply the numerators and Multiply the denominators.
$$\frac{5 × 3}{6 × 10}$$ = $$\frac{15}{60}$$
Simplify using the GCF.
The GCF of 15 and 60 is 15.
Divide the numerator and the denominator by 15.
$$\frac{15 ÷ 15}{60 ÷ 15}$$ = $$\frac{1}{4}$$
Question 3.
$$\frac{2}{3} \times \frac{9}{10}$$
$$\frac{□}{□}$$
$$\frac{3}{5}$$
Explanation:
$$\frac{2}{3} \times \frac{9}{10}$$
Multiply the numerators and Multiply the denominators.
$$\frac{2 × 9}{3 × 10}$$ = $$\frac{18}{30}$$
Simplify using the GCF.
The GCF of 18 and 30 is 6.
Divide the numerator and the denominator by 6.
$$\frac{18 ÷ 6}{30 ÷ 6}$$ = $$\frac{3}{5}$$
Question 4.
After a picnic, $$\frac{5}{12}$$ of the cornbread is left over. Val eats $$\frac{3}{5}$$ of the leftover cornbread. What fraction of the cornbread does Val eat?
$$\frac{□}{□}$$
$$\frac{1}{4}$$
Explanation:
After a picnic, $$\frac{5}{12}$$ of the cornbread is left over. Val eats $$\frac{3}{5}$$ of the leftover cornbread.
$$\frac{5}{12} \times \frac{3}{5}$$
Multiply the numerators and Multiply the denominators.
$$\frac{5 × 3}{12 × 5}$$ = $$\frac{15}{60}$$
Simplify using the GCF.
The GCF of 15 and 60 is 15.
Divide the numerator and the denominator by 15.
$$\frac{15 ÷ 15}{60 ÷ 15}$$ = $$\frac{1}{4}$$
Question 5.
The reptile house at the zoo has an iguana that is $$\frac{5}{6}$$ yd long. It has a Gila monster that is $$\frac{4}{5}$$ of the length of the iguana. How long is the Gila monster?
$$\frac{□}{□}$$
$$\frac{2}{3}$$
Explanation:
The reptile house at the zoo has an iguana that is $$\frac{5}{6}$$ yd long. It has a Gila monster that is $$\frac{4}{5}$$ of the length of the iguana.
$$\frac{5}{6} \times \frac{4}{5}$$
Multiply the numerators and Multiply the denominators.
$$\frac{5 × 4}{6× 5}$$ = $$\frac{20}{30}$$
Simplify using the GCF.
The GCF of 20 and 30 is 10.
Divide the numerator and the denominator by 10.
$$\frac{20 ÷ 10}{30 ÷ 10}$$ = $$\frac{2}{3}$$
Find the product. Simplify before multiplying.
Question 6.
$$\frac{3}{4} \times \frac{1}{6}$$
$$\frac{□}{□}$$
Explanation:
$$\frac{3}{4} \times \frac{1}{6}$$
Multiply the numerators and Multiply the denominators.
$$\frac{3 × 1}{4 × 6}$$ = $$\frac{3}{24}$$
Simplify using the GCF.
The GCF of 3 and 24 is 3.
Divide the numerator and the denominator by 3.
$$\frac{3 ÷ 3}{24 ÷ 3}$$ = $$\frac{1}{8}$$
Question 7.
$$\frac{7}{10} \times \frac{2}{3}$$
$$\frac{□}{□}$$
$$\frac{7}{15}$$
Explanation:
$$\frac{7}{10} \times \frac{2}{3}$$
Multiply the numerators and Multiply the denominators.
$$\frac{7 × 2}{10 × 3}$$ = $$\frac{14}{30}$$
Simplify using the GCF.
The GCF of 14 and 30 is 2.
Divide the numerator and the denominator by 2.
$$\frac{14 ÷ 2}{30 ÷ 2}$$ = $$\frac{7}{15}$$
Question 8.
$$\frac{5}{8} \times \frac{2}{5}$$
$$\frac{□}{□}$$
$$\frac{1}{4}$$
Explanation:
$$\frac{5}{8} \times \frac{2}{5}$$
Multiply the numerators and Multiply the denominators.
$$\frac{5 × 2}{8 × 5}$$ = $$\frac{10}{40}$$
Simplify using the GCF.
The GCF of 10 and 40 is 10.
Divide the numerator and the denominator by 10.
$$\frac{10 ÷ 10}{40 ÷ 10}$$ = $$\frac{1}{4}$$
Question 9.
$$\frac{9}{10} \times \frac{5}{6}$$
$$\frac{□}{□}$$
$$\frac{3}{4}$$
Explanation:
$$\frac{9}{10} \times \frac{5}{6}$$
Multiply the numerators and Multiply the denominators.
$$\frac{9 × 5}{10 × 6}$$ = $$\frac{45}{60}$$
Simplify using the GCF.
The GCF of 45 and 60 is 15.
Divide the numerator and the denominator by 15.
$$\frac{45 ÷ 15}{60 ÷ 15}$$ = $$\frac{3}{4}$$
Question 10.
$$\frac{11}{12} \times \frac{3}{7}$$
$$\frac{□}{□}$$
$$\frac{11}{28}$$
Explanation:
$$\frac{11}{12} \times \frac{3}{7}$$
Multiply the numerators and Multiply the denominators.
$$\frac{11 × 3}{12 × 7}$$ = $$\frac{33}{84}$$
Simplify using the GCF.
The GCF of 33 and 84 is 3.
Divide the numerator and the denominator by 3.
$$\frac{33 ÷ 3}{84 ÷ 3}$$ = $$\frac{11}{28}$$
Question 12.
Mr. Ortiz has $$\frac{3}{4}$$ pound of oatmeal. He uses $$\frac{2}{3}$$ of the oatmeal to bake muffins. How much oatmeal does Mr. Ortiz have left?
$$\frac{□}{□}$$
$$\frac{1}{2}$$
Explanation:
Mr. Ortiz has $$\frac{3}{4}$$ pound of oatmeal. He uses $$\frac{2}{3}$$ of the oatmeal to bake muffins.
$$\frac{3}{4} \times \frac{2}{3}$$
Multiply the numerators and Multiply the denominators.
$$\frac{3 × 2}{4 × 3}$$ = $$\frac{6}{12}$$
Simplify using the GCF.
The GCF of 6 and 12 is 6.
Divide the numerator and the denominator by 6.
$$\frac{6 ÷ 6}{12 ÷ 6}$$ = $$\frac{1}{2}$$
Question 13.
Compare Strategies To find $$\frac{16}{27}$$ × $$\frac{3}{4}$$, you can multiply the fractions and then simplify the product or you can simplify the fractions and then multiply. Which method do you prefer? Explain.
Type below:
__________
$$\frac{16}{27}$$ × $$\frac{3}{4}$$
$$\frac{16 × 3}{27 × 4}$$ = $$\frac{16 × 3}{4 × 27}$$
$$\frac{48}{96}$$
Simplify using the GCF.
The GCF of 48 and 96 is 48.
Divide the numerator and the denominator by 48.
$$\frac{48 ÷ 48}{96 ÷ 48}$$ = $$\frac{1}{2}$$
### Problem Solving + Applications – Page No. 90
Question 14.
Three students each popped $$\frac{3}{4}$$ cup of popcorn kernels. The table shows the fraction of each student’s kernels that did not pop. Which student had $$\frac{1}{16}$$ cup unpopped kernels?
__________
Mirza
Explanation:
Three students each popped $$\frac{3}{4}$$ cup of popcorn kernels. The table shows the fraction of each student’s kernels that did not pop.
Katie = 3/4 x 1/10 = 3/40
Mirza = 3/4 x 1/12 = 1/16
Question 15.
The jogging track at Francine’s school is $$\frac{3}{4}$$ mile long. Yesterday Francine completed two laps on the track. If she ran $$\frac{1}{3}$$ of the distance and walked the remainder of the way, how far did she walk?
____ mile
1 mile
Explanation:
Length of jogging track at Francine’s school = 3/4 mile
Let the distance covered by running be = x
Let the distance covered by walking be = y
Total number of laps completed by Francine = 2
Total distance covered by Francine = number of laps X distance covered in one lap
2 x 3/4 = 3/25 mile
Now,
distance covered by running = 1/3 of the total distance
x = 1/3 x 3/2
distance covered by walking y = total distance – distance covered by running
3/2 – x = 3/2 – 1/2 = 1 mile
Hence, Francine walked for 1 mile.
Question 16.
At a snack store, $$\frac{7}{12}$$ of the customers bought pretzels and $$\frac{3}{10}$$ of those customers bought low-salt pretzels. Bill states that $$\frac{7}{30}$$ of the customers bought low-salt pretzels. Does Bill’s statement make sense? Explain.
Type below:
__________
Bill’s statement does not make sense because it is incorrect:
7/12 customers bought pretzels.
3/10 Of those customers bought low-salt pretzels (x)
3/10 of 7/12 = x
21/120 = x
Simplify: 7/40
To be correct, Bill would have to say that 7/40 of the customers bought low-salt pretzels, but instead, he had said 7/30.
Question 17.
The table shows Tonya’s homework assignment. Tonya’s teacher instructed the class to simplify each expression by dividing the numerator and denominator by the GCF. Complete the table by simplifying each expression and then finding the value.
Type below:
__________
### Simplify Factors – Page No. 91
Find the product. Simplify before multiplying.
Question 1.
$$\frac{8}{9} \times \frac{5}{12}$$
$$\frac{□}{□}$$
$$\frac{10}{27}$$
Explanation:
$$\frac{8}{9} \times \frac{5}{12}$$
Multiply the numerators and Multiply the denominators.
$$\frac{8 × 5}{9 × 12}$$ = $$\frac{40}{108}$$
Simplify using the GCF.
The GCF of 40 and 108 is 4.
Divide the numerator and the denominator by 4.
$$\frac{40 ÷ 4}{108 ÷ 4}$$ = $$\frac{10}{27}$$
Question 3.
$$\frac{15}{20} \times \frac{2}{5}$$
$$\frac{□}{□}$$
$$\frac{3}{10}$$
Explanation:
$$\frac{15}{20} \times \frac{2}{5}$$
Multiply the numerators and Multiply the denominators.
$$\frac{15 × 2}{20 × 5}$$ = $$\frac{30}{100}$$
Simplify using the GCF.
The GCF of 30 and 100 is 10.
Divide the numerator and the denominator by 10.
$$\frac{30 ÷ 10}{100 ÷ 10}$$ = $$\frac{3}{10}$$
Question 4.
$$\frac{9}{18} \times \frac{2}{3}$$
$$\frac{□}{□}$$
$$\frac{1}{3}$$
Explanation:
$$\frac{9}{18} \times \frac{2}{3}$$
Multiply the numerators and Multiply the denominators.
$$\frac{9 × 2}{18 × 3}$$ = $$\frac{18}{54}$$
Simplify using the GCF.
The GCF of 18 and 54 is 18.
Divide the numerator and the denominator by 18.
$$\frac{18 ÷ 18}{54 ÷ 18}$$ = $$\frac{1}{3}$$
Question 5.
$$\frac{3}{4} \times \frac{7}{30}$$
$$\frac{□}{□}$$
$$\frac{7}{40}$$
Explanation:
$$\frac{3}{4} \times \frac{7}{30}$$
Multiply the numerators and Multiply the denominators.
$$\frac{3 × 7}{4 × 30}$$ = $$\frac{21}{120}$$
Simplify using the GCF.
The GCF of 21 and 120 is 3.
Divide the numerator and the denominator by 3.
$$\frac{21 ÷ 3}{120 ÷ 3}$$ = $$\frac{7}{40}$$
Question 6.
$$\frac{8}{15} \times \frac{15}{32}$$
$$\frac{□}{□}$$
$$\frac{1}{4}$$
Explanation:
$$\frac{8}{15} \times \frac{15}{32}$$
Multiply the numerators and Multiply the denominators.
$$\frac{8 × 15}{15 × 32}$$ = $$\frac{120}{480}$$
Simplify using the GCF.
The GCF of 120 and 480 is 120.
Divide the numerator and the denominator by 120.
$$\frac{120 ÷ 120}{480 ÷ 120}$$ = $$\frac{1}{4}$$
Question 7.
$$\frac{12}{21} \times \frac{7}{9}$$
$$\frac{□}{□}$$
$$\frac{4}{9}$$
Explanation:
$$\frac{12}{21} \times \frac{7}{9}$$
Multiply the numerators and Multiply the denominators.
$$\frac{12 × 7}{21 × 9}$$ = $$\frac{84}{189}$$
Simplify using the GCF.
The GCF of 84 and 189 is 21.
Divide the numerator and the denominator by 21.
$$\frac{84 ÷ 21}{189 ÷ 21}$$ = $$\frac{4}{9}$$
Question 8.
$$\frac{18}{22} \times \frac{8}{9}$$
$$\frac{□}{□}$$
$$\frac{8}{11}$$
Explanation:
$$\frac{18}{22} \times \frac{8}{9}$$
Multiply the numerators and Multiply the denominators.
$$\frac{18 × 8}{22 × 9}$$ = $$\frac{144}{198}$$
Simplify using the GCF.
The GCF of 144 and 198 is 18.
Divide the numerator and the denominator by 18.
$$\frac{144 ÷ 18}{198 ÷ 18}$$ = $$\frac{8}{11}$$
Problem Solving
Question 9.
Amber has a $$\frac{4}{5}$$-pound bag of colored sand. She uses $$\frac{1}{2}$$ of the bag for an art project. How much sand does she use for the project?
$$\frac{□}{□}$$ pounds
$$\frac{2}{5}$$ pounds
Explanation:
Amber has a $$\frac{4}{5}$$-pound bag of colored sand. She uses $$\frac{1}{2}$$ of the bag for an art project.
4/5 X 1/2 = 2/5
### Lesson Check – Page No. 92
Find each product. Simplify before multiplying.
Question 1.
At Susie’s school, $$\frac{5}{8}$$ of all students play sports. Of the students who play sports, $$\frac{2}{5}$$ play soccer. What fraction of the students in Susie’s school play soccer?
$$\frac{□}{□}$$
$$\frac{1}{4}$$
Explanation:
At Susie’s school, $$\frac{5}{8}$$ of all students play sports. Of the students who play sports, $$\frac{2}{5}$$ play soccer.
Multiply 5/8 X 2/5, and the answer is 0.25, which converts to 25/100 or 1/4
Question 2.
A box of popcorn weighs $$\frac{15}{16}$$ pounds. The box contains $$\frac{1}{3}$$ buttered popcorn and $$\frac{2}{3}$$ cheesy popcorn. How much does the cheesy popcorn weigh?
$$\frac{□}{□}$$
$$\frac{5}{8}$$
Explanation:
Total weight of a box of popcorn =15/16 pounds.
We are given two types of popcorns are there, butter popcorns and cheesy popcorns.
Butter popcorn is the one-third of the total weight = 1/3 of the Total weight
Plugging the value of the total weight, we get
= 1/3 * 15/16 = 5/16 pounds.
Cheesy popcorn = 2/3 of Total weight
Plugging the value of total weight, we get
= 2/3 * 15/16 = 10/16 or 5/8 pounds.
Therefore, cheesy popcorn weighs is 5/8 pounds.
Spiral Review
Question 4.
A 1.8-ounce jar of cinnamon costs $4.05. What is the cost per ounce?$ ______
$2.25 per ounce Explanation: If a 1.8-ounce jar costs$4.05, do $4.05 divided by 1.8.$4.05 / 1.8 = $2.25 per ounce. Question 5. Rose bought $$\frac{7}{20}$$ kilogram of ginger candy and 0.4 kilograms of cinnamon candy. Which did she buy more of? Explain how you know. Type below: __________ Answer: Rose bought ginger candy = 7/20 kilogram = 0.35 Kilogram She bought cinnamon candy = 0.4 kilogram 0.4 > 0.35 Therefore, She bought cinnamon candy more. Question 6. Don walked 3 $$\frac{3}{5}$$ miles on Friday, 3.7 miles on Saturday, and 3 $$\frac{5}{8}$$ miles on Sunday. List the distances from least to greatest. Type below: __________ Answer: 3 $$\frac{3}{5}$$, 3 $$\frac{5}{8}$$, 3.7 Explanation: 3 $$\frac{3}{5}$$ = 18/5 = 3.6 3 $$\frac{5}{8}$$ = 29/8 = 3.625 3.6 < 3.625 < 3.7 3 $$\frac{3}{5}$$, 3 $$\frac{5}{8}$$, 3.7 ### Mid-Chapter Checkpoint – Vocabulary – Page No. 93 Choose the best term from the box to complete the sentence. Question 1. The fractions $$\frac{1}{2}$$ and $$\frac{5}{10}$$ are _____. Type below: __________ Answer: Equivalent fractions Question 2. A _____ is a denominator that is the same in two or more fractions. Type below: __________ Answer: Common Denominator Concepts and Skills Write as a decimal. Tell whether you used division, a number line, or some other method. Question 3. $$\frac{7}{20}$$ _____ Answer: 0.35 Explanation: By using Division, $$\frac{7}{20}$$ = 0.35 Question 4. 8 $$\frac{39}{40}$$ _____ Answer: 8.975 Explanation: By using Division, 8 $$\frac{39}{40}$$ = 359/40 = 8.975 Question 5. 1 $$\frac{5}{8}$$ _____ Answer: 1.625 Explanation: By using Division, 1 $$\frac{5}{8}$$ = 13/8 = 1.625 Question 6. $$\frac{19}{25}$$ _____ Answer: 0.76 Explanation: By using Division, $$\frac{19}{25}$$ = 0.76 Order from least to greatest. Question 7. $$\frac{4}{5}, \frac{3}{4}, 0.88$$ Type below: __________ Answer: $$\frac{3}{4}$$, $$\frac{4}{5}$$,0.88 Explanation: Write the decimal form of 4/5 = 0.8 Write the decimal form of 3/4 = 0.75 0.88 0.75 < 0.8 < 0.88 Question 8. 0.65, 0.59, $$\frac{3}{5}$$ Type below: __________ Answer: 0.59, $$\frac{3}{5}$$, 0.65 Explanation: Write the decimal form of 3/5 = 0.6 0.59 < 0.6 < 0.65 Question 10. 0.9, $$\frac{7}{8}$$, 0.86 Type below: __________ Answer: 0.86, $$\frac{7}{8}$$, 0.9 Explanation: Write the decimal form of $$\frac{7}{8}$$ = 0.875 0.86 < 0.875 < 0.9 Find the product. Write it in simplest form. Question 11. $$\frac{2}{3} \times \frac{1}{8}$$ $$\frac{□}{□}$$ Answer: $$\frac{1}{12}$$ Explanation: $$\frac{2}{3} \times \frac{1}{8}$$ Multiply the numerators and Multiply the denominators. $$\frac{2 × 1}{3 × 8}$$ = $$\frac{2}{24}$$ Simplify using the GCF. The GCF of 2 and 24 is 2. Divide the numerator and the denominator by 2. $$\frac{2 ÷ 2}{24 ÷ 2}$$ = $$\frac{1}{12}$$ Question 12. $$\frac{4}{5} \times \frac{2}{5}$$ $$\frac{□}{□}$$ Answer: $$\frac{8}{25}$$ Explanation: $$\frac{4}{5} \times \frac{2}{5}$$ Multiply the numerators and Multiply the denominators. $$\frac{4 × 2}{5 × 5}$$ = $$\frac{8}{25}$$ Question 13. 12 × $$\frac{3}{4}$$ _____ Answer: 9 Explanation: 12 × $$\frac{3}{4}$$ Multiply the numerators and Multiply the denominators. $$\frac{12 × 3}{1 × 4}$$ = $$\frac{36}{4}$$ = 9 Question 14. Mia climbs $$\frac{5}{8}$$ of the height of the rock wall. Lee climbs $$\frac{4}{5}$$ of Mia’s distance. What fraction of the wall does Lee climb? $$\frac{□}{□}$$ Answer: $$\frac{7}{40}$$ Explanation: find the LCM (least common denominator) for 5/8 and 4/5. 5/8= 25/40 and 4/5= 32/40. Subtract and you get 7/40. ### Page No. 94 Question 15. In Zoe’s class, $$\frac{4}{5}$$ of the students have pets. Of the students who have pets, $$\frac{1}{8}$$ have rodents. What fraction of the students in Zoe’s class have pets that are rodents? What fraction of the students in Zoe’s class have pets that are not rodents? Type below: __________ Answer: $$\frac{1}{10}$$ of the students in Zoe’s class have pets that are rodents $$\frac{7}{10}$$ of the students in Zoe’s class have pets that are not rodents Explanation: In Zoe’s class, $$\frac{4}{5}$$ of the students have pets. Of the students who have pets, $$\frac{1}{8}$$ have rodents. 4/5 X 1/8 = 1/10 4/5 – 1/10 = 7/10 Question 16. A recipe calls for 2 $$\frac{2}{3}$$ cups of flour. Terell wants to make $$\frac{3}{4}$$ of the recipe. How much flour should he use? _____ cups Answer: 2 cups Explanation: 2 $$\frac{2}{3}$$ = 8/3 8/3 * 3/4 = 2 Question 17. Following the Baltimore Running Festival in 2009, volunteers collected and recycled 3.75 tons of trash. Graph 3.75 on a number line and write the weight as a mixed number. Type below: __________ Answer: Volunteers collected and recycled 3.75 tons of trash. We need to convert 3.75 as a mixed number. The mixed number consists of a whole number and a proper fraction. In the given number 3.75, 3 as the whole number and convert 0.75 to a fraction. 3.75 = 3 + 0.75 = 3 + 75/100 We can reduce the fraction 75/ 100 = 3+ 3/4 = 3 3/4 Question 18. Four students took an exam. The fraction of the total possible points that each received is given. Which student had the highest score? If students receive a whole number of points on every exam item, can the exam be worth a total of 80 points? Explain. Type below: __________ Answer: 22/25 = 0.88 17/20 = 0.85 4/5 = 0.8 3/4 = 0.75 Monica had the highest score Let x be the total number of points: (22/25 + 17/20 + 4/5 + 3/4)x = 80 x = 24.39 That is not a whole number of points. ### Share and Show – Page No. 97 Use the model to find the quotient. Question 1. $$\frac{1}{2}$$ ÷ 3 $$\frac{□}{□}$$ Answer: $$\frac{1}{6}$$ Explanation: 1/2 groups of 3 $$\frac{1}{2}$$ ÷ 3 1/2 × 1/3 = 1/6 Question 2. $$\frac{3}{4} \div \frac{3}{8}$$ ______ Answer: 2 Explanation: 3/4 groups of 3/8 3/4 × 8/3 = 2 Use fraction strips to find the quotient. Then draw the model. Question 3. $$\frac{1}{3}$$ ÷ 4 $$\frac{□}{□}$$ Answer: $$\frac{1}{12}$$ Explanation: $$\frac{1}{3}$$ ÷ 4 $$\frac{1}{3}$$ × $$\frac{1}{4}$$ $$\frac{1}{12}$$ Question 4. $$\frac{3}{5} \div \frac{3}{10}$$ ______ Answer: 2 Explanation: $$\frac{3}{5} \div \frac{3}{10}$$ $$\frac{3}{5}$$ × $$\frac{10}{3}$$ 2 Draw a model to solve. Then write an equation for the model. Interpret the result. Question 5. How many $$\frac{1}{4}$$ cup servings of raisins are in $$\frac{3}{8}$$ cup of raisins? Type below: __________ Answer: 1.5 Explanation: 3/8 × 1/4 = 1.5 Question 6. How many $$\frac{1}{3}$$ lb bags of trail mix can Josh make from $$\frac{5}{6}$$ lb of trail mix? Type below: __________ Answer: 2 Explanation: Multiply 1/3 with 2 1/3 × 2 = 2/6. 2/6 can go into 5/6 twice so the answer is two bags. Question 7. Pose a Problem Write and solve a problem for $$\frac{3}{4}$$ ÷ 3 that represents how much in each of 3 groups. Type below: __________ Answer: $$\frac{1}{4}$$ Explanation: $$\frac{3}{4}$$ ÷ 3 $$\frac{3}{4}$$ × $$\frac{1}{3}$$ = 1/4 ### Problem Solving + Applications – Page No. 98 The table shows the amount of each material that students in a sewing class need for one purse. Use the table for 8–10. Use models to solve. Question 8. Mrs. Brown has $$\frac{1}{3}$$ yd of blue denim and $$\frac{1}{2}$$ yd of black denim. How many purses can be made using denim as the main fabric? _____ purses Answer: 5 purses Explanation: Mrs. Brown has $$\frac{1}{3}$$ yd of blue denim and $$\frac{1}{2}$$ yd of black denim. 3 + 2 = 5 Question 9. One student brings $$\frac{1}{2}$$ yd of ribbon. If 3 students receive an equal length of ribbon, how much ribbon will each student receive? Will each of them have enough ribbon for a purse? Explain. Type below: __________ Answer: One student brings $$\frac{1}{2}$$ yd of ribbon. If 3 students receive an equal length of ribbon, $$\frac{1}{2}$$ ÷ 3 1/2 × 1/3 = 1/6 They don’t have enough ribbon for a purse Question 10. Make Arguments There was $$\frac{1}{2}$$ yd of purple and pink striped fabric. Jessie said she could only make $$\frac{1}{24}$$ of a purse using that fabric as the trim. Is she correct? Use what you know about the meanings of multiplication and division to defend your answer. Type below: __________ Answer: There was $$\frac{1}{2}$$ yd of purple and pink striped fabric. Jessie said she could only make $$\frac{1}{24}$$ of a purse using that fabric as the trim. 1/2 × 12 = 1/24 So, 12 is the answer Question 11. Draw a model to find the quotient. $$\frac{1}{2}$$ ÷ 4 = Type below: __________ Answer: Explanation: 1/2 × 1/4 = 1/8 ### Model Fraction Division – Page No. 99 Use the model to find the quotient Question 1. $$\frac{1}{4}$$ ÷ 3 = $$\frac{□}{□}$$ Answer: $$\frac{1}{12}$$ Explanation: $$\frac{1}{4}$$ ÷ 3 $$\frac{1}{4}$$ × $$\frac{1}{3}$$ = $$\frac{1}{12}$$ Question 2. $$\frac{1}{2} \div \frac{2}{12}=$$ ______ Answer: 3 Explanation: $$\frac{1}{2} \div \frac{2}{12}=$$ $$\frac{1}{2}$$ × $$\frac{12}{2}$$ = $$\frac{12}{4}$$ = 3 Use fraction strips to find the quotient. Question 3. $$\frac{5}{6} \div \frac{1}{2}=$$ ______ $$\frac{□}{□}$$ Answer: $$\frac{5}{3}$$ Explanation: $$\frac{5}{6} \div \frac{1}{2}=$$ $$\frac{5}{6}$$ × $$\frac{2}{1}$$ = $$\frac{5}{3}$$ Question 4. $$\frac{2}{3}$$ ÷ 4 = $$\frac{□}{□}$$ Answer: $$\frac{1}{6}$$ Explanation: $$\frac{2}{3}$$ ÷ 4 $$\frac{2}{3}$$ × $$\frac{1}{4}$$ = $$\frac{2}{12}$$ = 1/6 Question 5. $$\frac{1}{2}$$ ÷ 6 = $$\frac{□}{□}$$ Answer: $$\frac{1}{12}$$ Explanation: $$\frac{1}{2}$$ ÷ 6 $$\frac{1}{2}$$ × $$\frac{1}{6}$$ = $$\frac{1}{12}$$ Question 6. $$\frac{1}{3} \div \frac{1}{12}$$ ______ Answer: 4 Explanation: $$\frac{1}{3} \div \frac{1}{12}$$ $$\frac{1}{3}$$ × $$\frac{12}{1}$$ = $$\frac{12}{3}$$ = 4 Draw a model to solve. Then write an equation for the model. Interpret the result. Question 7. If Jerry runs $$\frac{1}{10}$$ mile each day, how many days will it take for him to run $$\frac{4}{5}$$ mile? ______ days Answer: 8 days Explanation: If Jerry runs $$\frac{1}{10}$$ mile each day, $$\frac{4}{5}$$ ÷ $$\frac{1}{10}$$ $$\frac{4}{5}$$ × $$\frac{10}{1}$$ = $$\frac{40}{5}$$ = 8 Problem Solving Question 8. Mrs. Jennings has $$\frac{3}{4}$$ gallon of paint for an art project. She plans to divide the paint equally into jars. If she puts $$\frac{1}{8}$$ gallon of paint into each jar, how many jars will she use? ______ jars Answer: 6 jars Explanation: Mrs. Jennings has 3/4 Gallons of paint for an art project. In 1 jar she puts 1/8 gallon of paint. The number of jars in which she plans to to divide the paint equally is given by, n= 3/4 ÷ 1/8 n = $$\frac{3}{4}$$ × $$\frac{8}{1}$$ = $$\frac{24}{4}$$ = 6 Question 9. If one jar of glue weighs $$\frac{1}{12}$$ pound, how many jars can Rickie get from $$\frac{2}{3}$$ pound of glue? ______ jars Answer: 8 jars Explanation: The weight of glue in one jar = 1/12 pound To get 2/3 pound of glue Rickie can get the number of jars 2/3 ÷ 1/12 2/3 × 12/1 = 24/3 = 8 Question 10. Explain how to use a model to show $$\frac{2}{6} \div \frac{1}{12}$$ and $$\frac{2}{6}$$ ÷ 4. Type below: __________ Answer: Explanation: $$\frac{2}{6} \div \frac{1}{12}$$ 2/6 = 1/3 1/3 x 12/1 = 4 $$\frac{2}{6}$$ ÷ 4 1/3 x 1/4 = 1/12 ### Lesson Check – Page No. 100 Question 1. Darcy needs $$\frac{1}{4}$$ yard of fabric to make a banner. She has 2 yards of fabric. How many banners can she make? ______ banners Answer: 8 banners Explanation: Darcy needs $$\frac{1}{4}$$ yard of fabric to make a banner. She has 2 yards of fabric. 2 ÷ $$\frac{1}{4}$$ = 2 x 4 = 8 Spiral Review Question 3. Letisha wants to read 22 pages a night. At that rate, how long will it take her to read a book with 300 pages? ______ nights Answer: 14 nights Explanation: Letisha wants to read 22 pages a night. It takes her to read a book with 300 pages 300/22 = 13.6 13.6 is near to 14 So, it is for 2 weeks. Question 5. Each block in Ton’s neighborhood is $$\frac{2}{3}$$ mile long. If he walks 4 $$\frac{1}{2}$$ blocks, how far does he walk? ______ miles Answer: 3 miles Explanation: If each block is 2/3 miles long, and he walks 4 1/2 blocks, we can simply multiply by two. It looks like this: (2/3)(4 1/2) to multiply, make 4 1/2 into an improper fraction and multiply normally (2/3)(9/4) Ton walks 3 miles in total. Question 6. In Cathy’s garden, $$\frac{5}{6}$$ of the area is planted with flowers. Of the flowers, $$\frac{3}{10}$$ of them are red. What fraction of Cathy’s garden is planted with red flowers? $$\frac{□}{□}$$ Answer: $$\frac{1}{4}$$ Explanation: In Cathy’s garden, $$\frac{5}{6}$$ of the area is planted with flowers. Of the flowers, $$\frac{3}{10}$$ of them are red. 5/6 x 3/10 = 1/4 ### Share and Show – Page No. 103 Estimate using compatible numbers. Question 1. $$22 \frac{4}{5} \div 6 \frac{1}{4}$$ _______ Answer: 4 Explanation: 22 $$\frac{4}{5}$$ = 114/5 = 22.8 6 $$\frac{1}{4}$$ = 25/4 = 6.25 22.8 is closer to 24 6.25 is closer to 6 24/6 = 4 Question 2. $$12 \div 3 \frac{3}{4}$$ _______ Answer: 3 Explanation: 3 $$\frac{3}{4}$$ = 15/4 = 3.75 3.75 is closer to 4 12/4 = 3 Question 3. $$33 \frac{7}{8} \div 5 \frac{1}{3}$$ _______ Answer: 7 Explanation: 33 $$\frac{7}{8}$$ = 271/8 = 33.875 5 $$\frac{1}{3}$$ = 16/3 = 5.333 33.875 is closer to 35 5.333 is closer to 5 35/5 = 7 Question 4. $$3 \frac{7}{8} \div \frac{5}{9}$$ _______ Answer: 4 Explanation: 3 $$\frac{7}{8}$$ = 31/8 = 3.875 $$\frac{5}{9}$$ = 0.555 3.875 is closer to 4 0.555 is closer to 1 4/1 = 4 Question 5. $$34 \frac{7}{12} \div 7 \frac{3}{8}$$ _______ Answer: 5 Explanation: 34 $$\frac{7}{12}$$ = 415/12 = 34.583 7 $$\frac{3}{8}$$ = 59/8 = 7.375 34.583 is closer to 35 7.375 is closer to 7 35/7 = 5 Question 6. $$1 \frac{2}{9} \div \frac{1}{6}$$ _______ Answer: 5 Explanation: 1 $$\frac{2}{9}$$ = 11/9 = 1.222 $$\frac{1}{6}$$ = 0.1666 1.222 is closer to 1 0.1666 is closer to 0.2 1/0.2 = 5 On Your Own Estimate using compatible numbers. Question 7. $$44 \frac{1}{4} \div 11 \frac{7}{9}$$ _______ Answer: 4 Explanation: 44 $$\frac{1}{4}$$ = 177/4 = 44.25 11 $$\frac{7}{9}$$ = 106/9 = 11.77 44.25 is closer to 44 11.77 is closer to 11 44/11 = 4 Question 8. $$71 \frac{11}{12} \div 8 \frac{3}{4}$$ _______ Answer: 8 Explanation: 71 $$\frac{11}{12}$$ = 863/12 = 71.916 8 $$\frac{3}{4}$$ = 35/4 = 8.75 71.916 is closer to 72 8.75 is closer to 9 72/9 = 8 Question 9. $$1 \frac{1}{6} \div \frac{1}{8}$$ _______ Answer: 12 Explanation: 1 $$\frac{1}{6}$$ = 7/6 = 1.166 $$\frac{1}{8}$$ = 0.125 1.166 is closer to 1.2 0.125 is closer to 0.1 1.2/0.1 = 12 Estimate to compare. Write <, >, or =. Question 10. $$21 \frac{3}{10} \div 2 \frac{5}{6}$$ _______ $$35 \frac{7}{9} \div 3 \frac{2}{3}$$ Answer: $$21 \frac{3}{10} \div 2 \frac{5}{6}$$ < $$35 \frac{7}{9} \div 3 \frac{2}{3}$$ Explanation: 21 $$\frac{3}{10}$$ = 213/10 = 21.3 2 $$\frac{5}{6}$$ = 17/6 = 2.833 21.3 is closer to 21 2.833 is closer to 3 21/3 = 7 35 $$\frac{7}{9}$$ = 322/9 = 35.777 3 $$\frac{2}{3}$$ = 11/3 = 3.666 35.777 is closer to 36 3.666 is closer to 4 36/4 = 9 7 < 9 So, $$21 \frac{3}{10} \div 2 \frac{5}{6}$$ < $$35 \frac{7}{9} \div 3 \frac{2}{3}$$ Question 13. Marion is making school flags. Each flag uses 2 $$\frac{3}{4}$$ yards of felt. Marion has 24 $$\frac{1}{8}$$ yards of felt. About how many flags can he make? About _______ flags Answer: About 8 flags Explanation: Marion is making school flags. Each flag uses 2 $$\frac{3}{4}$$ yards of felt. Marion has 24 $$\frac{1}{8}$$ yards of felt. 2 $$\frac{3}{4}$$ = 11/4 24 $$\frac{1}{8}$$ = 193/8 193/8 ÷ 11/4 193/8 x 4/11 = 8.77 About 8 flags Question 14. A garden snail travels about 2 $$\frac{3}{5}$$ feet in 1 minute. At that speed, how many hours would it take the snail to travel 350 feet? About _______ hours Answer: About 2 hours Explanation: 2 $$\frac{3}{5}$$ = 2.6 That’s how long he travels in one minute. There are 60 minutes in an hour so multiply it by 60 and see if that gets you close to 350. 60 x 2.6 = 156 Now let’s add one more hour. 156 + 156 = 312 14 x 2.6 = 36.4 312 + 36.4 = 348.4 348.4 + 2.6 = 351 So two hours and fourteen minutes ### Problem Solving + Applications – Page No. 104 What’s the Error? Question 15. Megan is making pennants from a piece of butcher paper that is 10 $$\frac{3}{8}$$ yards long. Each pennant requires $$\frac{3}{8}$$ yard of paper. To estimate the number of pennants she could make, Megan estimated the quotient 10 $$\frac{3}{8}$$ ÷ $$\frac{3}{8}$$. Look at how Megan solved the problem. Find her error Estimate: 10 $$\frac{3}{8}$$ ÷ $$\frac{3}{8}$$ 10 ÷ $$\frac{1}{2}$$ = 5 Correct the error. Estimate the quotient. So, Megan can make about _____ pennants. Describe the error that Megan made Explain Tell which compatible numbers you used to estimate 10 $$\frac{3}{8}$$ ÷ $$\frac{3}{8}$$. Explain why you chose those numbers. Type below: __________ Answer: 10 $$\frac{3}{8}$$ ÷ $$\frac{3}{8}$$ 10 $$\frac{3}{8}$$ = 83/8 = 10.375 $$\frac{3}{8}$$ = 0.375 She had written 10 ÷ $$\frac{1}{2}$$ = 5 10.375 is closer to 10 0.375 is closer to 0.5 10/0.5 = 20 But she has written 5 instead of 20. Megan can make about 20 pennants. For numbers 16a–16c, estimate to compare. Choose <, >, or =. Question 16. 16a. 18 $$\frac{3}{10} \div 2 \frac{5}{6}$$ ? $$30 \frac{7}{9} \div 3 \frac{1}{3}$$ _____ Answer: 16a. 18 $$\frac{3}{10} \div 2 \frac{5}{6}$$ < $$30 \frac{7}{9} \div 3 \frac{1}{3}$$ Explanation: 18 $$\frac{3}{10}$$ = 183/10 = 18.3 2 $$\frac{5}{6}$$ = 17/6 = 2.833 18.3 is closer to 18 2.833 is closer to 3 18/3 = 6 30 $$\frac{7}{9}$$ = 277/9 = 30.777 3 $$\frac{1}{3}$$ = 10/3 = 3.333 30.777 is closer to 30 3.333 is closer to 3 30/3 = 10 6 < 10 Question 16. 16b. 17 $$\frac{4}{5} \div 6 \frac{1}{6}$$ ? $$19 \frac{8}{9} \div 4 \frac{5}{8}$$ _____ Answer: 17 $$\frac{4}{5} \div 6 \frac{1}{6}$$ < $$19 \frac{8}{9} \div 4 \frac{5}{8}$$ Explanation: 17 $$\frac{4}{5}$$ = 89/5 = 17.8 6 $$\frac{1}{6}$$ = 37/6 = 6.1666 17.8 is closer to 18 6.1666 is closer to 6 18/6 = 3 19 $$\frac{8}{9}$$ = 179/9 = 19.888 4 $$\frac{5}{8}$$ = 37/8 = 4.625 19.888 is closer to 20 4.625 is closer to 5 20/5 = 4 3 < 4 17 $$\frac{4}{5} \div 6 \frac{1}{6}$$ < $$19 \frac{8}{9} \div 4 \frac{5}{8}$$ Question 16. 16c. 17 $$\frac{5}{6} \div 6 \frac{1}{4}$$ ? $$11 \frac{5}{7} \div 2 \frac{3}{4}$$ _____ Answer: 17 $$\frac{5}{6} \div 6 \frac{1}{4}$$ < $$11 \frac{5}{7} \div 2 \frac{3}{4}$$ Explanation: 17 $$\frac{5}{6}$$ = 107/6 = 17.833 6 $$\frac{1}{4}$$ = 25/4 = 6.25 17.833 is closer to 18 6.25 is closer to 6 18/6 = 3 11 $$\frac{5}{7}$$ = 82/7 = 11.714 2 $$\frac{3}{4}$$ = 11/4 = 2.75 11.714 is closer to 12 2.75 is closer to 3 12/3 = 4 3 < 4 17 $$\frac{5}{6} \div 6 \frac{1}{4}$$ < $$11 \frac{5}{7} \div 2 \frac{3}{4}$$ ### Estimate Quotients – Page No. 105 Estimate using compatible numbers. Question 1. $$12 \frac{3}{16} \div 3 \frac{9}{10}$$ ______ Answer: 3 Explanation: 12 $$\frac{3}{16}$$ = 195/16 = 12.1875 3 $$\frac{9}{10}$$ = 39/10 = 3.9 12.1875 is closer to 12 3.9 is closer to 4 12/4 = 3 Question 2. $$15 \frac{3}{8} \div \frac{1}{2}$$ ______ Answer: 30 Explanation: 15 $$\frac{3}{8}$$ = 123/8 = 15.375 $$\frac{1}{2}$$ = 0.5 15.375 is closer to 15 0.5 is closer to 0.5 15/0.5 = 30 Question 3. $$22 \frac{1}{5} \div 1 \frac{5}{6}$$ ______ Answer: 11 Explanation: 22 $$\frac{1}{5}$$ = 111/5 = 22.2 1 $$\frac{5}{6}$$ = 11/6 = 1.8333 22.2 is closer to 22 1.8333 is closer to 2 22/2 = 11 Question 4. $$7 \frac{7}{9} \div \frac{4}{7}$$ ______ Answer: 16 Explanation: 7 $$\frac{7}{9}$$ = 70/9 = 7.777 $$\frac{4}{7}$$ = 0.571 7.777 is closer to 8 0.571 is closer to 0.5 8/0.5 = 16 Question 5. $$18 \frac{1}{4} \div 2 \frac{4}{5}$$ ______ Answer: 6 Explanation: 18 $$\frac{1}{4}$$ = 73/4 = 18.25 2 $$\frac{4}{5}$$ = 14/5 = 2.8 18.25 is closer to 18 2.8 is closer to 3 18/3 = 6 Question 6. $$\frac{15}{16} \div \frac{1}{7}$$ ______ Answer: 10 Explanation: $$\frac{15}{16}$$ = 0.9375 $$\frac{1}{7}$$ = 0.1428 0.9375 is closer to 1 0.1428 is closer to 0.1 1/0.1 = 10 Question 7. $$14 \frac{7}{8} \div \frac{5}{11}$$ ______ Answer: 30 Explanation: 14 $$\frac{7}{8}$$ = 119/8 = 14.875 $$\frac{5}{11}$$ = 0.4545 14.875 is closer to 15 0.4545 is closer to 0.5 15/0.5 = 30 Problem Solving Question 10. Estimate the number of pieces Sharon will have if she divides 15 $$\frac{1}{3}$$ yards of fabric into 4 $$\frac{4}{5}$$ yard lengths. About ______ pieces Answer: About 3 pieces Explanation: Sharon will have if she divides 15 $$\frac{1}{3}$$ yards of fabric into 4 $$\frac{4}{5}$$ yard lengths. 3 7/36 is the answer. So, about 3 pieces Question 11. Estimate the number of $$\frac{1}{2}$$ quart containers Ethan can fill from a container with 8 $$\frac{7}{8}$$ quarts of water. About ______ containers Answer: About 18 containers Question 12. How is estimating quotients different from estimating products? Type below: __________ Answer: To estimate products and quotients, you need to first round the numbers. To round to the nearest whole number, look at the digit in the tenth place. If it is less than 5, round down. If it is 5 or greater, round up. Remember that an estimate is an answer that is not exact, but is approximate and reasonable. Let’s look at an example of estimating a product. Estimate the product: 11.256×6.81 First, round the first number. Since there is a 2 in the tenth place, 11.256 rounds down to 11. Next, round the second number. Since there is an 8 in the tenth place, 6.81 rounds up to 7. Then, multiply the rounded numbers. 11×7=77 The answer is 77. Let’s look at an example of estimating a quotient. Estimate the quotient: 91.93÷4.39 First, round the first number. Since there is a 9 in the tenth place, 91.93 rounds up to 92. Next, round the second number. Since there is a 3 in the tenth place, 4.39 rounds down to 4. Then, divide the rounded numbers. 92÷4=23 The answer is 23. ### Lesson Check – Page No. 106 Question 1. Each loaf of pumpkin bread calls for 1 $$\frac{3}{4}$$ cups of raisins. About how many loaves can be made from 10 cups of raisins? About ______ loaves Answer: About 5 loaves Explanation: Divide 10 by 1 3/4. The answer is 5.714285 So you can make about 5 loaves of bread with 10 cups of raisins if each loaf needs 1 3/4 cups of raisins. Spiral Review Question 3. A recipe calls for $$\frac{3}{4}$$ teaspoon of red pepper. Uri wants to use $$\frac{1}{3}$$ of that amount. How much red pepper should he use? $$\frac{□}{□}$$ teaspoon Answer: $$\frac{1}{4}$$ teaspoon Explanation: A recipe calls for $$\frac{3}{4}$$ teaspoon of red pepper. Uri wants to use $$\frac{1}{3}$$ of that amount. $$\frac{1}{3}$$ of $$\frac{3}{4}$$ = $$\frac{1}{4}$$ Question 4. A recipe calls for 2 $$\frac{2}{3}$$ cups of apple slices. Zoe wants to use 1 $$\frac{1}{2}$$ times this amount. How many cups of apples should Zoe use? ______ cups Answer: 4 cups Explanation: A recipe calls for 2 2/3 cups of apple slices. Zoe wants to use 1 1/2 times this amount. We will multiply the number of apple slices to 1 1/2 2 2/3 X 1 1/2 8/3 X3/2 = 24/6 = 4 cups Zoe will use 4 cups of apple slices. Question 5. Edgar has 2.8 meters of rope. If he cuts it into 7 equal parts, how long will each piece be? ______ meters Answer: 0.4 meters Explanation: 2.8/7 = 0.4 meters Question 6. Kami has 7 liters of water to fill water bottles that each hold 2.8 liters. How many bottles can she fill? ______ bottles Answer: 2 bottles Explanation: 7/2.8 = 2.5 she can only fill 2 because anything over that would 8.4 liters of water ### Share and Show – Page No. 109 Estimate. Then find the quotient. Question 1. $$\frac{5}{6}$$ ÷ 3 $$\frac{□}{□}$$ Answer: $$\frac{3}{10}$$ Explanation: 5/6 = 0.8333 is closer to 0.9 0.9/3 = 0.3 = 3/10 Use a number line to find the quotient. Question 2. $$\frac{3}{4} \div \frac{1}{8}$$ _______ Answer: Explanation: 3/4 x 8 = 3 x 2 = 6 Question 3. $$\frac{3}{5} \div \frac{3}{10}$$ _______ Answer: Explanation: 3/5 x 10/3 = 2 Estimate. Then write the quotient in simplest form. Question 4. $$\frac{3}{4} \div \frac{5}{6}$$ $$\frac{□}{□}$$ Answer: $$\frac{1}{1}$$ Explanation: 3/4 = 0.75 is closer to 0.8 5/6 = 0.8333 is closer to 0.8 0.8/0.8 = 1 Question 5. $$3 \div \frac{3}{4}$$ _______ Answer: 4 Explanation: 3/4 = 0.75 3/0.75 = 4 Question 6. $$\frac{1}{2} \div \frac{3}{4}$$ $$\frac{□}{□}$$ Answer: $$\frac{625}{1000}$$ Explanation: 1/2 = 0.5 3/4 = 0.75 is closer to 0.8 0.5/0.8 = 0.625 = 625/1000 Question 7. $$\frac{5}{12} \div 3$$ $$\frac{□}{□}$$ Answer: $$\frac{2}{10}$$ Explanation: 5/12 = 0.4166 is closer to 0.6 0.6/3 = 0.2 = 2/10 On Your Own Practice: Copy and Solve Estimate. Then write the quotient in simplest form Question 8. $$2 \div \frac{1}{8}$$ _______ Answer: 20 Explanation: 1/8 = 0.125 is closer to 0.1 2/0.1 = 20 Question 9. $$\frac{3}{4} \div \frac{3}{5}$$ $$\frac{□}{□}$$ Answer: $$\frac{1}{1}$$ Explanation: 3/4 = 0.75 is closer to 0.8 3/5 = is 0.6 closer to 0.8 0.8/0.8 = 1 Question 10. $$\frac{2}{5} \div 5$$ $$\frac{□}{□}$$ Answer: $$\frac{1}{10}$$ Explanation: 2/5 = 0.4 is closer to 0.5 0.5/5 = 0.1 = 1/10 Question 11. $$4 \div \frac{1}{7}$$ _______ Answer: 40 Explanation: 1/7 = 0.1428 is closer to 0.1 4/0.1 = 40 Practice: Copy and Solve Evaluate using the order of operations. Write the answer in simplest form. Question 12. $$\left(\frac{3}{5}+\frac{1}{10}\right) \div 2$$ $$\frac{□}{□}$$ Answer: $$\frac{7}{20}$$ Explanation: 3/5 + 1/10 = 7/10 = 0.7 0.7/2 = 7/20 Question 13. $$\frac{3}{5}+\frac{1}{10} \div 2$$ $$\frac{□}{□}$$ Answer: $$\frac{13}{20}$$ Explanation: $$\frac{3}{5}+\frac{1}{10} \div 2$$ (1/10)/2 = 1/20 3/5 + 1/20 = 0.65 = 13/20 Question 14. $$\frac{3}{5}+2 \div \frac{1}{10}$$ _______ $$\frac{□}{□}$$ Answer: Explanation: 2/(1/10) = 1/5 3/5 + 1/5 = 4/5 Question 15. Generalize Suppose the divisor and the dividend of a division problem are both fractions between 0 and 1, and the divisor is greater than the dividend. Is the quotient less than, equal to, or greater than 1? Type below: __________ Answer: Divisor and Dividend are fractions lying between 0 and 1 Also, Divisor > Dividend A smaller number is being divided by a larger number Whenever a smaller number is divided by a larger number, the quotient is less than 1 Example: 0,5/0,6 Here, they are both numbers between 0 and 1, and the divisor is greater than the dividend. The result is 0,8333, LESS THAN 1 Hence, the answer is that the quotient will be less than 1 ### Problem Solving + Applications – Page No. 110 Use the table for 16–19. Question 16. Kristen wants to cut ladder rungs from a 6 ft board. How many ladder rungs can she cut? _______ ladder rungs Answer: 8 ladder rungs Explanation: Kristen wants to cut ladder rungs from a 6 ft board. ladder rungs = 3/4 ft 6/(3/4) = 8 rungs Question 17. Pose a Problem Look back at Problem 16. Write and solve a new problem by changing the length of the board Kristen is cutting for ladder rungs. Type below: __________ Answer: Kristen wants to cut ladder rungs from a 9 ft board. How many ladder rungs can she cut? Kristen wants to cut ladder rungs from a 9 ft board. ladder rungs = 3/4 ft 9/(3/4) = 12 rungs Question 18. Dan paints a design that has 8 equal parts along the entire length of the windowsill. How long is each part of the design? $$\frac{□}{□}$$ yards Answer: $$\frac{1}{16}$$ yards Explanation: Dan paints a design that has 8 equal parts along the entire length of the windowsill. (1/2)/8 = 1/2 x 1/8 = 1/16 yards Question 19. Dan has a board that is $$\frac{15}{16}$$ yd. How many “Keep Out” signs can he make if the length of the sign is changed to half of the original length? _______ signs Answer: 3 signs Explanation: Dan has a board that is $$\frac{15}{16}$$ yd. If the length of the sign is changed to half of the original length, (5/8)/2 = 5/16 (15/16) ÷ 5/16 = 15/16 x 16/5 = 3 Question 20. Lauren has $$\frac{3}{4}$$ cup of dried fruit. She puts the dried fruit into bags, each holding $$\frac{1}{8}$$ cup. How many bags will Lauren use? Explain your answer using words and numbers. Type below: __________ Answer: 6 Explanation: Lauren has $$\frac{3}{4}$$ cup of dried fruit. She puts the dried fruit into bags, each holding $$\frac{1}{8}$$ cup. 3/4 ÷ 1/8 = 3/4 x 8 = 6 Lauren has 3/4 and in 1/4 there are 2 1/8s. That 3 fourths times two = 6 so 6 one eights ### Divide Fractions – Page No. 111 Estimate. Then write the quotient in simplest form. Question 1. $$5 \div \frac{1}{6}$$ _____ Answer: 25 Explanation: 1/6 = 0.166 is closer to 0.2 5/0.2 = 25 Question 2. $$\frac{1}{2} \div \frac{1}{4}$$ _____ Answer: 5 Explanation: 1/2 = 0.5 is closer to 1 1/4 = 0.25 is closer to 0.2 1/0.2 = 5 Question 3. $$\frac{4}{5} \div \frac{2}{3}$$ _____ $$\frac{□}{□}$$ Answer: 1 $$\frac{1}{5}$$ Explanation: 4/5 = 0.8 is closer to 0.8 2/3 = 0.66 is closer to 0.6 0.8/0.6 = 1 1/5 Question 4. $$\frac{14}{15} \div 7$$ $$\frac{□}{□}$$ Answer: $$\frac{2}{15}$$ Explanation: 14/15 = 0.9333 0.9/7 = 2/15 Question 5. $$8 \div \frac{1}{3}$$ _____ Answer: 20 Explanation: 1/3 = 0.33 is closer to 0.4 8/0.4 = 20 Question 6. $$\frac{12}{21} \div \frac{2}{3}$$ $$\frac{□}{□}$$ Answer: $$\frac{1}{1}$$ Explanation: 12/21 = 0.571 is closer to 0.6 2/3 = 0.666 is closer to 0.6 0.6/0.6 = 1 Question 7. $$\frac{5}{6} \div \frac{5}{12}$$ _____ Answer: 2 Explanation: 5/6 = 0.833 is closer to 0.8 5/12 = 0.416 is closer to 0.4 0.8/0.4 = 2 Question 8. $$\frac{5}{8} \div \frac{1}{2}$$ _____ $$\frac{□}{□}$$ Answer: 1 $$\frac{2}{10}$$ Explanation: 5/8 = 0.625 is closer to 0.6 1/2 = 0.5 is closer to 0.5 0.6/0.5 = 1.2 = 1 2/10 Question 10. Hideko has $$\frac{3}{5}$$ yard of ribbon to tie on balloons for the festival. Each balloon will need $$\frac{3}{10}$$ yard of ribbon. How many balloons can Hideko tie with ribbon? _____ balloons Answer: 2 balloons Explanation: 3/10 yard of ribbon required to tie = 1 balloon 3/5 yard of ribber can tie = (3/5) ÷ (3/10) = 2 ballons With 3/5 yard, Hideko can tie 2 balloons Problem Solving Question 11. Rick knows that 1 cup of glue weighs $$\frac{1}{18}$$ pound. He has $$\frac{2}{3}$$ pound of glue. How many cups of glue does he have? _____ cups Answer: 12 cups Explanation: For 1/18lb, 1 cup For 2/3lb, x cups. 1/8x = 1 x 2/3 1/8x = 2/3 x = 2/3 x 18 x = 2 x 6 = 12 cups Question 12. Mrs. Jennings had $$\frac{5}{7}$$ gallon of paint. She gave $$\frac{1}{7}$$ gallon each to some students. How many students received paint if Mrs. Jennings gave away all the paint? _____ students Answer: 4 students Explanation: Mrs. Jennings had $$\frac{5}{7}$$ gallon of paint. She gave $$\frac{1}{7}$$ gallon each to some students. $$\frac{5}{7}$$ ÷ $$\frac{1}{7}$$ = 25/7 = 3.571 is closer to 4 Question 13. Write a word problem that involves two fractions. Include the solution. Type below: __________ Answer: Mrs. Jennings had $$\frac{5}{7}$$ gallon of paint. She gave $$\frac{1}{7}$$ gallon each to some students. How many students received paint if Mrs. Jennings gave away all the paint? Answer: Mrs. Jennings had $$\frac{5}{7}$$ gallon of paint. She gave $$\frac{1}{7}$$ gallon each to some students. $$\frac{5}{7}$$ ÷ $$\frac{1}{7}$$ = 25/7 = 3.571 is closer to 4 ### Lesson Check – Page No. 112 Question 1. There was $$\frac{2}{3}$$ of a pizza for 6 friends to share equally. What fraction of the pizza did each person get? $$\frac{□}{□}$$ Answer: $$\frac{1}{9}$$ Explanation: There was $$\frac{2}{3}$$ of a pizza for 6 friends to share equally. $$\frac{2}{3}$$ ÷ 6 = 2/3 x 1/6 = 2/18 = 1/9 Question 2. Rashad needs $$\frac{2}{3}$$ pound of wax to make a candle. How many candles can he make with 6 pounds of wax? _____ candles Answer: 9 candles Explanation: Rashad needs 2/3 pound a wax to make candles. 1 Candle = 2/3 pounds. So, for 2 pounds, 3 x 2/3 = 3 candles 2 pounds = 3 candles 1 pound = 3/2 candles So, for 6 pounds, 6 x 3/2 = 9 candles Spiral Review Question 4. Ebony walked at a rate of 3 $$\frac{1}{2}$$ miles per hour for 1 $$\frac{1}{3}$$ hours. How far did she walk? _____ $$\frac{□}{□}$$ Answer: 4 $$\frac{2}{3}$$ Explanation: Ebony walked at a rate of 3 $$\frac{1}{2}$$ miles per hour for 1 $$\frac{1}{3}$$ hours. 3 1/2 miles = 7/2 miles … 1 hour x miles = ? … 1 1/3 hours = 4/3 hours 7/2 x 4/3 = 1 x x x = 7/2 x 4/3 x = 14/3 = 4 2/3 miles The correct result would be 4 2/3 miles. Question 5. Penny uses $$\frac{3}{4}$$ yard of fabric for each pillow she makes. How many pillows can she make using 6 yards of fabric? _____ pillows Answer: 8 pillows Explanation: Penny uses $$\frac{3}{4}$$ yard of fabric for each pillow she makes. Using 6 yards of fabric 6/(3/4) = 24/3 = 8 Question 6. During track practice, Chris ran 2.5 laps in 81 seconds. What was his average time per lap? _____ seconds Answer: 32.4 seconds Explanation: During track practice, Chris ran 2.5 laps in 81 seconds. 81/2.5 = 32.4 seconds ### Share and Show – Page No. 115 Use the model to find the quotient. Question 1. $$3 \frac{1}{3} \div \frac{1}{3}$$ _____ Answer: 21 Explanation: Model 3 with 3 hexagonal blocks. Model 1/2 with 1 trapezoid block. For 1/6, 6 triangle blocks are equal to 1 hexagon. So, a triangle block shows 1/6. Count the triangles. There are 21 triangle blocks. So, 3 1/2 ÷ 1/6 = 21. Question 2. $$2 \frac{1}{2} \div \frac{1}{6}$$ _____ Answer: 15 Explanation: Model 2 with 2 hexagonal blocks. Model 1/2 with 1 trapezoid block. For 1/6, 6 triangle blocks are equal to 1 hexagon. So, a triangle block shows 1/6. Count the triangles. There are 15 triangle blocks. So, $$2 \frac{1}{2} \div \frac{1}{6}$$ = 15. Use pattern blocks to find the quotient. Then draw the model. Question 3. $$2 \frac{2}{3} \div \frac{1}{6}$$ _____ Answer: Explanation: 2 2/3 = 8/3 8/3 ÷ 1/6 = 16 Question 4. $$3 \frac{1}{2} \div \frac{1}{2}$$ _____ Answer: Explanation: 3 1/2 = 7/2 7/2 ÷ 1/2 = 7 Draw a model to find the quotient. Question 5. $$3 \frac{1}{2} \div 3$$ _____ $$\frac{□}{□}$$ Answer: Explanation: 3 1/2 = 7/2 7/2 ÷ 3 = 21/2 Question 6. $$1 \frac{1}{4} \div 2$$ $$\frac{□}{□}$$ Answer: Explanation: 1/4 ÷ 2 = 1/2 Question 7. Use Appropriate Tools Explain how models can be used to divide mixed numbers by fractions or whole numbers Type below: __________ Answer: Multiply the whole number part by the fraction’s denominator. Add that to the numerator. Then write the result on top of the denominator. ### Problem Solving + Applications – Page No. 116 Use a model to solve. Then write an equation for the model. Question 8. Use Models Eliza opens a box of bead kits. The box weighs 2 $$\frac{2}{3}$$ lb. Each bead kit weighs $$\frac{1}{6}$$ lb. How many kits are in the box? What does the answer mean? Type below: __________ Answer: 16 kits are in the box Explanation: Eliza opens a box of bead kits. The box weighs 2 $$\frac{2}{3}$$ lb. Each bead kit weighs $$\frac{1}{6}$$ lb, 2 $$\frac{2}{3}$$ ÷ $$\frac{1}{6}$$ = 8/3 ÷ 1/6 = 16. 16 kits are in the box Question 10. Sense or Nonsense? Steve made this model to show $$2 \frac{1}{3} \div \frac{1}{6}$$. He says that the quotient is 7. Is his answer sense or nonsense? Explain your reasoning Type below: __________ Answer: $$2 \frac{1}{3} \div \frac{1}{6}$$ = 7/3 ÷ 1/6 = 14. He said the quotient is 7. His answer is Nonsense. Question 11. Eva is making muffins to sell at a fundraiser. She has 2 $$\frac{1}{4}$$ cups of flour, and the recipe calls for $$\frac{3}{4}$$ cup of flour for each batch of muffins. Explain how to use a model to find the number of batches of muffins Eva can make. Type below: __________ Answer: 3 Explanation: Eva is making muffins to sell at a fundraiser. She has 2 $$\frac{1}{4}$$ cups of flour, and the recipe calls for $$\frac{3}{4}$$ cup of flour for each batch of muffins. 2 $$\frac{1}{4}$$ ÷ $$\frac{3}{4}$$ = 9/4 ÷ 3/4 = 3 ### Model Mixed Number Division – Page No. 117 Use the model to find the quotient. Question 1. $$4 \frac{1}{2} \div \frac{1}{2}$$ _____ Answer: 9 Explanation: Count the number of trapezoids to find the answer. Question 2. $$3 \frac{1}{3} \div \frac{1}{6}$$ _____ Answer: 20 Use pattern blocks or another model to find the quotient. Then draw the model. Question 3. $$2 \frac{1}{2} \div \frac{1}{6}$$ _____ Answer: Explanation: Model 2 with 2 hexagonal blocks. Model 1/2 with 1 trapezoid block. For 1/6, 6 triangle blocks are equal to 1 hexagon. So, a triangle block shows 1/6. Count the triangles. There are 15 triangle blocks. So, 212÷16 = 15. Question 4. $$2 \frac{3}{4} \div 2$$ _____ Answer: Explanation: 2 3/4 ÷ 2 = 11/2 Problem Solving Question 5. Marty has 2 $$\frac{4}{5}$$ quarts of juice. He pours the same amount of juice into 2 bottles. How much does he pour into each bottle? _____ $$\frac{□}{□}$$ quarts Answer: 1$$\frac{2}{5}$$ quarts Explanation: Marty has 2 $$\frac{4}{5}$$ quarts of juice. He pours the same amount of juice into 2 bottles. 2 $$\frac{4}{5}$$ = 14/5 = 2.8 2.8/2 = 1.4 = 1 2/5 Question 6. How many $$\frac{1}{3}$$ pound servings are in 4 $$\frac{2}{3}$$ pounds of cheese? _____ pounds Answer: 14 pounds Explanation: 4 2/3 = 14/3 (14/3)/(1/3) = 14 Question 7. Write a word problem that involves dividing a mixed number by a whole number. Solve the problem and describe how you found the answer. Type below: __________ Answer: How many $$\frac{1}{3}$$ pound servings are in 4 $$\frac{2}{3}$$ pounds of cheese? Explanation: 4 2/3 = 14/3 (14/3)/(1/3) = 14 ### Lesson Check – Page No. 118 Sketch a model to find the quotient. Question 1. Emma has 4 $$\frac{1}{2}$$ pounds of birdseed. She wants to divide it evenly among 3 bird feeders. How much birdseed should she put in each? _____ $$\frac{□}{□}$$ pounds Answer: 1$$\frac{1}{2}$$ pounds Explanation: Emma has 4 1/2 pounds of birdseed. Convert this to an improper fraction. 4 1/2 = 9/2 Emma wants to divide it evenly among 3 bird feeders. So, she should put (9/2)/3 = 3/2 = 1 1/2 Spiral Review Question 3. The Ecology Club has volunteered to clean up 4.8 kilometers of highway. The members are organized into 16 teams. Each team will clean the same amount of highway. How much highway will each team clean? _____ kilometers Answer: 0.3 kilometers Explanation: The Ecology Club has volunteered to clean up 4.8 kilometers of highway. The members are organized into 16 teams. The total length of the highway is given to clean = 4.8 kilometers If the members are organized into 16 teams. 4.8/16 = 0.3 Hence, each team will clean 0.3 kilometers of the highway. Question 4. Tyrone has$8.06. How many bagels can he buy if each bagel costs $0.65? _____ bagels Answer: 12 bagels Explanation:$8.06/$0.65 = 12.4 12 bagels Question 5. A nail is 0.1875 inch thick. What is its thickness as a fraction? Is 0.1875 inch closer to $$\frac{1}{8}$$ inch or $$\frac{1}{4}$$ inch on a number line? Type below: __________ Answer: 0.1875 = 3/16 which is at the same distance to 1/4 and 1/8 It is the same distance apart. Question 6. Maria wants to find the product of 5 $$\frac{3}{20}$$ × 3 $$\frac{4}{25}$$ using decimals instead of fractions. How can she rewrite the problem using decimals? Type below: __________ Answer: 16.274 Explanation: The decimal for 5 3/20 is 5.15 The decimal for 3 4/25 is 3.16 5.15 × 3.16 = 16.274 ### Share and Show – Page No. 121 Estimate. Then write the quotient in simplest form. Question 1. $$4 \frac{1}{3} \div \frac{3}{4}$$ ______ $$\frac{□}{□}$$ Answer: 5$$\frac{375}{1000}$$ Explanation: 4 1/3 = 13/3 = 4.333 is closer to 4.3 3/4 = 0.75 is closer to 0.8 4.3/0.8 = 5.375 = 5 375/1000 Question 2. Six hikers shared 4 $$\frac{1}{2}$$ lb of trail mix. How much trail mix did each hiker receive? $$\frac{□}{□}$$ Answer: $$\frac{75}{100}$$ Explanation: 6 hikers = 4.5 lbs of trail mix 4.5/6= .75 lbs each hiker. Question 3. $$5 \frac{2}{3} \div 3$$ ______ $$\frac{□}{□}$$ Answer: 2$$\frac{947}{1000}$$ Explanation: 5 2/3 = 17/3 = 5.666 is closer to 5.6 5.6/3 = 1.866 is closer to 1.9 5.6/1.9 = 2.947 = 2 947/1000 Question 4. $$7 \frac{1}{2} \div 2 \frac{1}{2}$$ ______ Answer: 3 Explanation: 7 1/2 = 15/2 = 7.5 2 1/2 = 5/2 = 2.5 7.5/2.5 = 3 On Your Own Estimate. Then write the quotient in simplest form. Question 5. $$5 \frac{3}{4} \div 4 \frac{1}{2}$$ ______ $$\frac{□}{□}$$ Answer: 1$$\frac{27}{100}$$ Explanation: 5 3/4 = 23/4 = 5.75 4 1/2 = 9/2 = 4.5 5.75/4.5 = 1.27 = 1 27/100 Question 6. $$5 \div 1 \frac{1}{3}$$ ______ $$\frac{□}{□}$$ Answer: 3$$\frac{84}{100}$$ Explanation: 1 1/3 = 4/3 = 1.33 is closer to 1.3 5/1.3 = 3.84 = 3 84/100 Question 7. $$6 \frac{3}{4} \div 2$$ ______ $$\frac{□}{□}$$ Answer: 3$$\frac{2}{5}$$ Explanation: 6 3/4 = 27/4 = 6.75 is closer to 6.8 6.8/2 = 3.4 = 3 2/5 Question 8. $$2 \frac{2}{9} \div 1 \frac{3}{7}$$ ______ $$\frac{□}{□}$$ Answer: 1$$\frac{571}{1000}$$ Explanation: 2 2/9 = 20/9 = 2.22 is closer to 2.2 1 3/7 = 10/7 = 1.428 is closer to 1.4 2.2/1.4 = 1.571 = 1 571/1000 Question 9. How many 3 $$\frac{1}{3}$$ yd pieces can Amanda get from a 3 $$\frac{1}{3}$$ yd ribbon? ______ Answer: 1 Explanation: (3 1/3) ÷ (3 1/3) = 1 Evaluate Algebra Evaluate using the order of operations. Write the answer in simplest form. Question 11. $$1 \frac{1}{2} \times 2 \div 1 \frac{1}{3}$$ _____ $$\frac{□}{□}$$ Answer: 2$$\frac{1}{4}$$ Explanation: (1 1/2) × 2 = 3/2 × 2 = 3 1 1/3 = 4/3 3/(4/3) = 9/4 = 2.25 = 2 1/4 Question 12. $$1 \frac{2}{5} \div 1 \frac{13}{15}+\frac{5}{8}$$ _____ $$\frac{□}{□}$$ Answer: 1$$\frac{3}{8}$$ Explanation: (1 2/5)/(1 13/15) = (7/5)/(28/15) = 3/4 = 0.75 0.75 + 0.625 = 1.375 = 1 3/8 Question 13. $$3 \frac{1}{2}-1 \frac{5}{6} \div 1 \frac{2}{9}$$ _____ Answer: 2 Explanation: (1 5/6)/(1 2/9) = (11/6)/11/9 = 3/2 = 1 1/2 = 1.5 3 1/2 = 7/2 = 3.5 3.5 – 1.5 = 2 Question 14. Look for a Pattern Find these quotients: $$20 \div 4 \frac{4}{5}$$, $$10 \div 4 \frac{4}{5}$$, $$5 \div 4 \frac{4}{5}$$. Describe a pattern you see. Type below: __________ Answer: 20 ÷ 4 4/5 = 20 ÷ 24/5 = 20/4.8 = 4.1666 10 ÷ 4 4/5 = 10 ÷ 24/5 = 10/4.8 = 2.08333 5 ÷ 4 4/5 = 5 ÷ 24/5 = 5/4.8 = 1.04166 The pattern is multiplied by 2 every time. ### Page No. 122 Question 15. Dina hikes $$\frac{1}{2}$$ of the easy trail and stops for a break every 3 $$\frac{1}{4}$$ mile. How many breaks will she take? a. What problem are you asked to solve? Type below: __________ Answer: How many breaks Dina will take when hikes $$\frac{1}{2}$$ of the easy trail and stops for a break every 3 $$\frac{1}{4}$$ mile. Question 15. b. How will you use the information in the table to solve the problem? Type below: __________ Answer: Dina easy trail length, break time Question 15. c. How can you find the distance Dina hikes? How far does she hike? ______ $$\frac{□}{□}$$ miles Answer: 9$$\frac{3}{4}$$ miles Explanation: 19 1/2 × 1/2 = 39/2 × 1/2 = 39/4 = 9 3/4 Question 15. d. What operation will you use to find how many breaks Dina takes? Type below: __________ Answer: Division Question 15. e. How many breaks will Dina take? ______ breaks Answer: 3 breaks Explanation: 39/4 ÷ 13/4 = 3 Question 17. Rex’s goal is to run 13 $$\frac{3}{4}$$ miles over 5 days. He wants to run the same distance each day. Jordan said that Rex would have to run 3 $$\frac{3}{4}$$ miles each day to reach his goal. Do you agree with Jordan? Explain your answer using words and numbers. Type below: __________ Answer: Rex’s goal is to run 13 $$\frac{3}{4}$$ miles over 5 days. He wants to run the same distance each day. 13 $$\frac{3}{4}$$ ÷ 5 = 55/4 ÷ 5 = 11/4 or 2 3/4. Jordan answer is wrong ### Divide Mixed Numbers – Page No. 123 Estimate. Then write the quotient in simplest form. Question 1. $$2 \frac{1}{2} \div 2 \frac{1}{3}$$ ______ $$\frac{□}{□}$$ Answer: 1$$\frac{1}{2}$$ Explanation: 2 1/2 = 5/2 = 2.5 is closer to 3 2 1/3 = 7/3 = 2.333 is closer to 2 3/2 = 1.5 = 1 1/2 Question 2. $$2 \frac{2}{3} \div 1 \frac{1}{3}$$ ______ Answer: 2 Explanation: 2 2/3 = 8/3 = 2.666 is closer to 2.6 1 1/3 = 4/3 = 1.333 is closer to 1.3 2.6/1.3 = 2 Question 3. $$2 \div 3 \frac{5}{8}$$ $$\frac{□}{□}$$ Answer: $$\frac{1}{2}$$ Explanation: 3 5/8 = 29/8 = 3.625 is closer to 3.6 2/3.6 = 0.5 = 1/2 Question 4. $$1 \frac{13}{15} \div 1 \frac{2}{5}$$ $$\frac{□}{□}$$ Answer: $$\frac{126}{100}$$ Explanation: 1 13/15 = 28/15 = 1.8666 is closer to 1.9 1 2/5 = 7/5 = 1.4 is closer to 1.5 1.9/1.5 = 1.266 126/100 Question 5. $$10 \div 6 \frac{2}{3}$$ ______ $$\frac{□}{□}$$ Answer: 1$$\frac{1}{2}$$ Explanation: 6 2/3 = 20/3 = 6.666 is closer to 6.7 10/6.7 = 3/2 = 1 1/2 Question 6. $$2 \frac{3}{5} \div 1 \frac{1}{25}$$ ______ $$\frac{□}{□}$$ Answer: 2$$\frac{3}{5}$$ Explanation: 2 3/5 = 13/5 = 2.6 1 1/25 = 26/25 = 1.04 is closer to 1 2.6/1 = 13/5 or 2 3/5 Question 7. $$2 \frac{1}{5} \div 2$$ ______ $$\frac{□}{□}$$ Answer: 1$$\frac{1}{10}$$ Explanation: 2 1/5 = 11/5 = 2.2 is closer to 2.2 2.2/2 = 1.1 = 11/10 = 1 1/10 Question 8. Sid and Jill hiked 4 $$\frac{1}{8}$$ miles in the morning and 1 $$\frac{7}{8}$$ miles in the afternoon. How many times as far did they hike in the morning as in the afternoon? ______ $$\frac{□}{□}$$ times Answer: 2$$\frac{1}{5}$$ times Explanation: Sid and Jill hiked 4 $$\frac{1}{8}$$ miles in the morning and 1 $$\frac{7}{8}$$ miles in the afternoon. 4 $$\frac{1}{8}$$ = 33/8 1 $$\frac{7}{8}$$ = 15/8 (33/8) ÷ (15/8) = 33/15 = 11/5 or 2 1/5 Problem Solving Question 9. It takes Nim 2 $$\frac{2}{3}$$ hours to weave a basket. He worked Monday through Friday, 8 hours a day. How many baskets did he make? ______ baskets Answer: 15 baskets Explanation: he worked (Mon – Fri) 5 days at 8 hrs per day = 5 × 8= 40 hrs 40/ (2 2/3) = 40 / (8/3) = 40 × 3/8 = 120/8 = 15 baskets Question 10. A tree grows 1 $$\frac{3}{4}$$ feet per year. How long will it take the tree to grow from a height of 21 $$\frac{1}{4}$$ feet to a height of 37 feet? ______ years Answer: 9 years Explanation: A tree grows 1 3/4 = 7/4 feet per year. If you would like to know how long will it take the tree to grow from a height of 21 1/4 = 85/4 feet to a height of 37 feet, 37 – 21 1/4 = 37 – 85/4 = 148/4 – 85/4 = 63/4 = 15 3/4 15 3/4 / 1 3/4 = 63/4 / 7/4 = 63/4 × 4/7 = 9 years Question 11. Explain how you would find how many 1 $$\frac{1}{2}$$ cup servings there are in a pot that contains 22 $$\frac{1}{2}$$ cups of soup. Type below: __________ Answer: Given that, Total number of cups = 22 1/2 The number of cups required for each serving = 1 1/2 The number of servings = 22 1/2 ÷ 1 1/2 = 45/2 ÷ 3/2 = 45/3 = 15 ### Lesson Check – Page No. 124 Question 1. Tom has a can of paint that covers 37 $$\frac{1}{2}$$ square meters. Each board on the fence has an area of $$\frac{3}{16}$$ square meters. How many boards can he paint? ______ boards Answer: 200 boards Explanation: Tom has a can of paint that covers 37 $$\frac{1}{2}$$ square meters. Each board on the fence has an area of $$\frac{3}{16}$$ square meters. 37 $$\frac{1}{2}$$ ÷ $$\frac{3}{16}$$ = 200 square meters Spiral Review Question 3. The three sides of a triangle measure 9.97 meters, 10.1 meters, and 0.53 meter. What is the distance around the triangle? ______ meters Answer: 20.6 meters Explanation: The distance around the triangle is call perimeter, to get it we must add the 3 sides. So, 9.97 + 10.1 + 0.53 = 20.6 meters Question 4. Selena bought 3.75 pounds of meat for$4.64 per pound. What was the total cost of the meat?
$______ Answer:$17.40
Explanation:
Selena bought 3.75 pounds of meat.
The cost of meat of one pound = $4.64 The total cost of the meat = 4.64 × 3.75 =$17.40
The total cost of 3.75 lb of meat was $17.40. Question 5. Melanie prepared 7 $$\frac{1}{2}$$ tablespoons of a spice mixture. She uses $$\frac{1}{4}$$ tablespoon to make a batch of barbecue sauce. Estimate the number of batches of barbecue sauce she can make using the spice mixture. Type below: __________ Answer: 30 batches of sauce Explanation: Melanie prepared 7 $$\frac{1}{2}$$ tablespoons of a spice mixture. She uses $$\frac{1}{4}$$ tablespoon to make a batch of barbecue sauce. 4 X 1/4 tbsp = 1 tbsp. 4 X 7 1/2 = 30. she can make 30 batches of sauce Question 6. Arturo mixed together 1.24 pounds of pretzels, 0.78 pounds of nuts, 0.3 pounds of candy, and 2 pounds of popcorn. He then packaged it in bags that each contained 0.27 pounds. How many bags could he fill? ______ bags Answer: 16 bags Explanation: Arturo mixed together 1.24 pounds of pretzels, 0.78 pounds of nuts, 0.3 pounds of candy, and 2 pounds of popcorn. 1.24 + 0.78 + 0.3 + 2 = 4.32 4.32/0.27 = 16 ### Page No. 127 Question 1. There is $$\frac{4}{5}$$ lb of sand in the class science supplies. If one scoop of sand weighs $$\frac{1}{20}$$ lb, how many scoops of sand can Maria get from the class supplies and still leave $$\frac{1}{2}$$ lb in the supplies? Type below: __________ Answer: 16 scoops Explanation: There is $$\frac{4}{5}$$ lb of sand in the class science supplies. If one scoop of sand weighs $$\frac{1}{20}$$ lb, $$\frac{4}{5}$$ ÷ $$\frac{1}{20}$$ = 4/5 × 1/20 = 16 scoops Question 2. What if Maria leaves $$\frac{2}{5}$$ lb of sand in the supplies? How many scoops of sand can she get? ______ scoops Answer: 8 scoops Explanation: There is $$\frac{2}{5}$$ lb of sand in the class science supplies. If one scoop of sand weighs $$\frac{1}{20}$$ lb, $$\frac{2}{5}$$ ÷ $$\frac{1}{20}$$ = 2/5 × 20 = 8 Question 3. There are 6 gallons of distilled water in the science supplies. If 10 students each use an equal amount of the distilled water and there is 1 gal left in the supplies, how much will each student get? $$\frac{□}{□}$$ gallon Answer: $$\frac{1}{2}$$ gallon Explanation: There are 6 gallons of distilled water in the science supplies. There is 1 gal left in the supplies, 6 – 1 = 5 10 students each use an equal amount of the distilled water = 5/10 = 1/2 .5 gal for each student ### On Your Own – Page No. 128 Question 4. The total weight of the fish in a tank of tropical fish at Fish ‘n’ Fur was $$\frac{7}{8}$$ lb. Each fish weighed $$\frac{1}{64}$$ lb. After Eric bought some fish, the total weight of the fish remaining in the tank was $$\frac{1}{2}$$ lb. How many fish did Eric buy? ______ fish Answer: 386 fish Explanation: The total weight of the fish in a tank of tropical fish at Fish ‘n’ Fur was $$\frac{7}{8}$$ lb. Each fish weighed $$\frac{1}{64}$$ lb. After Eric bought some fish, the total weight of the fish remaining in the tank was $$\frac{1}{2}$$ lb. 386 is the answer Question 5. Fish ‘n’ Fur had a bin containing 2 $$\frac{1}{2}$$ lb of gerbil food. After selling bags of gerbil food that each held $$\frac{3}{4}$$ lb, $$\frac{1}{4}$$ lb of food was left in the bin. If each bag of gerbil food sold for$3.25, how much did the store earn?
$______ Answer:$9.75
Explanation:
The store would earn 9.75$because 3 bags of gerbil food is sold. Then you would multiply 3 by 3.25. Question 6. Describe Niko bought 2 lb of dog treats. He gave his dog $$\frac{3}{5}$$ lb of treats one week and $$\frac{7}{10}$$ lb of treats the next week. Describe how Niko can find how much is left. Type below: __________ Answer: Niko bought 2 lb of dog treats. He gave his dog $$\frac{3}{5}$$ lb of treats one week and $$\frac{7}{10}$$ lb of treats the next week. Let us find the amount of dog food eaten by dogs in two months. 3/5 + 7/10 = 13/10 Now we will subtract the amount of food eaten by the dog from the amount of food initially to find the remaining amount of dog food. 2 – 13/10 = 7/10 Therefore, 7/10 pounds of food was remaining in the bag at the end of the two months. Question 7. There were 14 $$\frac{1}{4}$$ cups of apple juice in a container. Each day, Elise drank 1 $$\frac{1}{2}$$ cups of apple juice. Today, there is $$\frac{3}{4}$$ cup of apple juice left. Derek said that Elise drank apple juice on nine days. Do you agree with Derek? Use words and numbers to explain your answer. Type below: __________ Answer: Derek is correct. Explanation: An apple juice container had 14 1/2 =14.25 She drank per day 1 1/2= 1.5 The left part in the container 3/4= .75 14.25 cups – .75 cup = 13.5 cups 13.5 cups ÷ 1.5 cups per day= 9 days ### Problem Solving Fraction Operations – Page No. 129 Read each problem and solve. Question 1. $$\frac{2}{3}$$ of a pizza was left over. A group of friends divided the leftover pizza into pieces each equal to $$\frac{1}{18}$$ of the original pizza. After each friend took one piece, $$\frac{1}{6}$$ of the original pizza remained. How many friends were in the group? ______ friends Answer: 9 friends Explanation: Let us say that there are x friends. Each one gets 1/18 of the original pizza: but this in turn leaves 1/6 of the 2/3 leftover. 1x/18 = 2/3 – 1/6 x = 12 – 3 = 9 Question 2. Sarah’s craft project uses pieces of yarn that are $$\frac{1}{8}$$ yard long. She has a piece of yarn that is 3 yards long. How many $$\frac{1}{8}$$ -yard pieces can she cut and still have 1 $$\frac{1}{4}$$ yards left? ______ pieces Answer: 14 pieces Explanation: Sarah’s craft project uses pieces of yarn that are $$\frac{1}{8}$$ yard long. She has a piece of yarn that is 3 yards long. If she left 1 $$\frac{1}{4}$$ yards left, 3 – 1 $$\frac{1}{4}$$ = 7/4 7/4 ÷ $$\frac{1}{8}$$ = 14 Question 3. Alex opens a 1-pint container of orange butter. He spreads $$\frac{1}{16}$$ of the butter on his bread. Then he divides the rest of the butter into $$\frac{3}{4}$$ -pint containers. How many $$\frac{3}{4}$$ -pint containers is he able to fill? ______ $$\frac{□}{□}$$ containers Answer: 1$$\frac{1}{4}$$ containers Explanation: Alex opens a 1-pint container of orange butter. He spreads $$\frac{1}{16}$$ of the butter on his bread. 1 – 1/16 = 15/16 Then he divides the rest of the butter into $$\frac{3}{4}$$ -pint containers. (15/16) ÷ (3/4) = 5/4 = 1 1/4 Question 4. Kaitlin buys $$\frac{9}{10}$$ a pound of orange slices. She eats $$\frac{1}{3}$$ of them and divides the rest equally into 3 bags. How much is in each bag? ______ lb Answer: 17/90 lb Explanation: Kaitlin buys $$\frac{9}{10}$$ a pound of orange slices. She eats $$\frac{1}{3}$$ of them and divides the rest equally into 3 bags. If she starts with 9/10 pounds and has eaten 1/3 of them, 9/10 – 1/3 = 17/30 This is the amount she has left. Let’s divide this value by 3 to see how many pounds are in one bag. (17/30)/3 = 17/90 There are 17/90 pounds in one bag. Question 5. Explain how to draw a model that represents $$\left(1 \frac{1}{4}-\frac{1}{2}\right) \div \frac{1}{8}$$. Type below: __________ Answer: Divide 2 bars into 8 quarters. Below that draw 1 1/4 or 5 quarters. Remove 1/2 or 2 quarters Divide each of the 3 quarters left into 2 eighths Explanation: $$\left(1 \frac{1}{4}-\frac{1}{2}\right) \div \frac{1}{8}$$ 1 1/4 -1/2 = 5/4 – 1/2 = 3/4 3/4 ÷ 1/8 = 6 ### Lesson Check – Page No. 130 Question 2. John has a roll containing 24 $$\frac{2}{3}$$ feet of wrapping paper. He wants to divide it into 11 pieces. First, though, he must cut off $$\frac{5}{6}$$ foot because it was torn. How long will each piece be? ______ $$\frac{□}{□}$$ feet Answer: 2$$\frac{4}{25}$$ feet Explanation: John had a roll containing wrapping paper = 24 2/3 = 74/3 First, he must cut off 5/6 feet because it was torn. He wants to divide it into 11 pieces. 74/3 – 5/6 Taking the L.C.M of 3 and 6 is 6 (148-5)/6 = 143/6 = 23.83 feet He wants to divide it into 11 pieces. length of the each piece = 23.83/11 = 2.16 feet Spiral Review Question 3. Alexis has 32 $$\frac{2}{5}$$ ounces of beads. How many necklaces can she make if each uses 2 $$\frac{7}{10}$$ ounces of beads? ______ necklaces Answer: 12 necklaces Explanation: Alexis has 32 $$\frac{2}{5}$$ ounces of beads. If each uses 2 $$\frac{7}{10}$$ ounces of beads, 32 $$\frac{2}{5}$$ × 2 $$\frac{7}{10}$$ 32 $$\frac{2}{5}$$ = 162/5 2 $$\frac{7}{10}$$ = 27/10 162/5 × 27/10 = 12 necklaces Question 4. Joseph has$32.40. He wants to buy several comic books that each cost $2.70. How many comic books can he buy? ______ comic books Answer: 12 comic books Explanation: Joseph has$32.40. He wants to buy several comic books that each cost $2.70.$32.40/\$2.70 = 12 comic books
Question 5.
A rectangle is 2 $$\frac{4}{5}$$ meters wide and 3 $$\frac{1}{2}$$ meters long. What is its area?
______ $$\frac{□}{□}$$ m2
9$$\frac{4}{5}$$ m2
Explanation:
2 $$\frac{4}{5}$$ = 14/5
3 $$\frac{1}{2}$$ = 7/2
14/5 × 7/2 = 9 4/5
Question 6.
A rectangle is 2.8 meters wide and 3.5 meters long. What is its area?
______ m2
9.8 m2
Explanation:
A rectangle is 2.8 meters wide and 3.5 meters long.
2.8 × 3.5 = 9.8
### Chapter 2 Review/Test – Page No. 131
Question 1.
Write the values in order from least to greatest.
Type below:
__________
0.45, 0.5, 5/8, 3/4
Explanation:
3/4 = 0.75
5/8 = 0.625
0.45, 0.5
0.45 < 0.5 < 0.625 < 0.75
Question 2.
For numbers 2a–2d, compare. Choose <, >, or =.
2a. 0.75 _____ $$\frac{3}{4}$$
2b. $$\frac{4}{5}$$ _____ 0.325
2c. 1 $$\frac{3}{5}$$ _____ 1.9
2d. 7.4 _____ 7 $$\frac{2}{5}$$
2a. 0.75 = $$\frac{3}{4}$$
2b. $$\frac{4}{5}$$ > 0.325
2c. 1 $$\frac{3}{5}$$ < 1.9
2d. 7.4 = 7 $$\frac{2}{5}$$
Explanation:
2a. 3/4 = 0.75
0.75 = 0.75
2b. $$\frac{4}{5}$$ = 0.8
0.8 > 0.325
2c. 1 $$\frac{3}{5}$$ = 8/5 = 1.6
1.6 < 1.9
2d. 7 $$\frac{2}{5}$$ = 37/5 = 7.4
7.4 = 7.4
Question 3.
The table lists the heights of 4 trees.
For numbers 3a–3d, select True or False for each statement.
3a. The oak tree is the shortest. True False
3b. The birch tree is the tallest. True False
3c. Two of the trees are the same height. True False
3d. The sycamore tree is taller than the maple tree. True False
Type below:
__________
3a. The oak tree is the shortest. True
3b. The birch tree is the tallest. False
3c. Two of the trees are the same height. False
3d. The sycamore tree is taller than the maple tree. False
Explanation:
Sycamore = 15 2/3 = 47/3 = 15.666
Oak = 14 3/4 = 59/4 = 14.75
Maple = 15 3/4 = 63/4 = 15.75
Birch = 15.72
### Page No. 132
Question 4.
For numbers 4a–4d, choose Yes or No to indicate whether the statement is correct.
4a. Point A represents 1.0. Yes No
4b. Point B represents $$\frac{3}{10}$$. Yes No
4c. Point C represents 6.5. Yes No
4d. Point D represents $$\frac{4}{5}$$. Yes No
Type below:
__________
4a. Point A represents 1.0. Yes
4b. Point B represents $$\frac{3}{10}$$. Yes
4c. Point C represents 6.5. No
4d. Point D represents $$\frac{4}{5}$$. Yes
Question 5.
Select the values that are equivalent to one twenty-fifth. Mark all that apply.
Options:
a. 125
b. 25
c. 0.04
d. 0.025
c. 0.04
Explanation:
one twenty-fifth = 1/25 = 0.04
Question 6.
The table shows Lily’s homework assignment. Lily’s teacher instructed the class to simplify each expression by dividing the numerator and denominator by the GCF. Complete the table by simplifying each expression and then finding the product.
Type below:
___________
a. Simplified Expression: 1/10
Product: 0.1
b. Simplified Expression: 1/2
Product: 0.5
c. Simplified Expression: 15/56
Product: 0.267
d. Simplified Expression: 1/12
Product: 0.083
Explanation:
a. 2/5 × 1/4 = 2/20
Simplify using the GCF.
The GCF of 2 and 20 is 2.
Divide the numerator and the denominator by 2.
Product: 0.1
b. 4/5 × 5/8 = 1/2
Product: 0.5
c. 3/7 × 5/8 = 15/ 56
Product: 0.267
d. 4/9 × 3/16 = 1/12
Product: 0.083
### Page No. 133
Question 7.
Two-fifths of the fish in Gary’s fish tank are guppies. One-fourth of the guppies are red. What fraction of the fish in Gary’s tank are red guppies? What fraction of the fish in Gary’s tank are not red guppies? Show your work.
Type below:
___________
1/10 of the fish are red guppies.
and 9/10 of the fish are not red guppies.
Explanation:
two-fifths of the fish in Gary’s fish tank are guppies.
One-fourth of the guppies are red.
Let the total number of fish in Gary’s fish tank be x.
It is given that two-fifths of the fish in Gary’s fish tank are guppies.
So, the number of guppies in Gary’s fish tank is 2/5 × x
Given that One-fourth of the guppies are red.
number of red guppies = 1/4 × 2x/5 = x/10
So, 1/10 of the fish are red guppies.
1 – 1/10 = 9/10 of the fish are not red guppies.
Question 8.
One-third of the students at Finley High School play sports. Two-fifths of the students who play sports are girls. What fraction of all students are girls who play sports? Use numbers and words to explain your answer.
Type below:
___________
One-third of the students at Finley High School play sports. Two-fifths of the students who play sports are girls.
1/3 × 2/5 = 2/15 of the girls in the school play sports.
Question 9.
Draw a model to find the quotient.
$$\frac{3}{4}$$ ÷ 2 =
$$\frac{3}{4}$$ ÷ $$\frac{3}{8}$$ =
How are your models alike? How are they different?
Type below:
___________
Explanation:
$$\frac{3}{4}$$ ÷ 2 = 3/4 × 1/2 = 3/8
$$\frac{3}{4}$$ ÷ $$\frac{3}{8}$$ = 3/4 × 8/3 = 2
Both models are multiplying with the 3/4.
The number line model shows how many groups of 3/8 are in 3/4.
Question 10.
Explain how to use a model to find the quotient.
2 $$\frac{1}{2}$$ ÷ 2 =
Type below:
___________
5/4
Explanation:
2 1/2 = 5/2
5/2 groups of 2
5/2 ÷ 2 = 5/2 × 1/2 = 5/4
### Page No. 134
Question 11.
$$\frac{7}{8}$$ ÷ $$\frac{3}{5}$$ =
_______ $$\frac{□}{□}$$
1 $$\frac{11}{24}$$
Explanation:
$$\frac{7}{8}$$ ÷ $$\frac{3}{5}$$
$$\frac{7}{8}$$ × $$\frac{5}{3}$$ = 35/24 = 1 $$\frac{11}{24}$$
Question 12.
$$2 \frac{1}{10} \div 1 \frac{1}{5}=$$ =
_______ $$\frac{□}{□}$$
1 $$\frac{3}{4}$$
Explanation:
2 $$\frac{1}{10}$$ = 21/10
1 $$\frac{1}{5}$$ = 6/5
(21/10) ÷ (6/5) = 7/4 or 1 3/4
Question 13.
Sophie has $$\frac{3}{4}$$ quart of lemonade. If she divides the lemonade into glasses that hold $$\frac{1}{16}$$ quart, how many glasses can Sophie fill? Show your work
_______ glasses
12 glasses
Explanation:
Let x be the number of glasses
1/16x = 3/4
x = 3/4 × 16 = 3 × 4 = 12 glasses
Question 14.
Ink cartridges weigh $$\frac{1}{8}$$ pound. The total weight of the cartridges in a box is 4 $$\frac{1}{2}$$ pounds. How many cartridges does the box contain? Show your work and explain why you chose the operation you did.
_______ cartridges
36 cartridges
Explanation:
Weight of ink cartridges = 1/8 pounds
Total weight of the cartridges in a box = 4 1/2 = 9/2 pounds
so, the number of cartridges that box contain is given by
9/2 ÷ 1/8 = 36
Hence, there are 36 cartridges that the box contains.
### Page No. 135
Question 16.
Complete the table by finding the products. Then answer the questions in Part A and Part B.
Part A
Explain how each pair of division and multiplication problems are the same, and how they are different.
Type below:
___________
1/5 ÷ 3/4 = 4/15; 1/5 × 4/3 = 4/15
2/13 ÷ 1/5 = 10/13; 2/13 × 5/1 = 10/13
4/5 ÷ 3/5 = 4/3; 4/5 × 5/3 = 4/3
the product of each pair of division and multiplication problems is the same.
They are different from the operation performed.
Question 16.
Part B
Explain how to use the pattern in the table to rewrite a division problem involving fractions as a multiplication problem.
Type below:
___________
First, since it’s the division you have to change the second fraction which is called the reciprocal. That means the second fraction has to be flipped before you can multiple the fractions.
### Page No. 136
Question 17.
Margie hiked a 17 $$\frac{7}{8}$$ mile trail. She stopped every 3 $$\frac{2}{5}$$ miles to take a picture. Martin and Tina estimated how many times Margie stopped.
Type below:
___________
Margie hiked a 17 7/8 mile trail.
Distance hiked by Margie = 17 7/8 = 143/8 mile.
She stopped every 3 2/5 miles to take a picture = 17/5 mile
Number of pictures = (143/8) ÷ (17/5) = 715/136 = 5.28
So she can take a maximum of 6 pictures and a minimum of 5 pictures.
Question 18.
Brad and Wes are building a tree house. They cut a 12 $$\frac{1}{2}$$ foot piece of wood into 5 of the same length pieces. How long is each piece of wood? Show your work.
_______ $$\frac{□}{□}$$ foot
2 $$\frac{1}{2}$$ foot<|endoftext|>
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2,790 |
Courses
# 03 - Let's Recap - Circles - Class 10 - Maths Class 10 Notes | EduRev
## Class 10 : 03 - Let's Recap - Circles - Class 10 - Maths Class 10 Notes | EduRev
``` Page 1
CIRCLES
Circle: A circle is a collection of all points in a plane which are at a constant distance
from a fixed point.
Some parts of a circle:
Chord: A line segment joining any two end points on the circle is called a chord.
Diameter is the longest chord.
Secant: A line which intersects the circle in two distinct points is called secant of the
circle.
Tangent: A line which touches the circle at only one point is called tangent to the
circle.
Three different situations that can arise when a circle and a line are given in a
plane, consider a circle and a line PQ. There can be three possibilities as given:
(i) The line PQ and the circle have no common point. Here,
PQ is called a non-intersecting line with respect to the
circle.
(ii) There are two common points A and B that the line PQ and
the circle have. Here, we call the line PQ a secant of the
circle.
(iii) There is only one point A which is common to the line PQ
and the circle. Here, the line PQ is called a tangent to the
circle.
Page 2
CIRCLES
Circle: A circle is a collection of all points in a plane which are at a constant distance
from a fixed point.
Some parts of a circle:
Chord: A line segment joining any two end points on the circle is called a chord.
Diameter is the longest chord.
Secant: A line which intersects the circle in two distinct points is called secant of the
circle.
Tangent: A line which touches the circle at only one point is called tangent to the
circle.
Three different situations that can arise when a circle and a line are given in a
plane, consider a circle and a line PQ. There can be three possibilities as given:
(i) The line PQ and the circle have no common point. Here,
PQ is called a non-intersecting line with respect to the
circle.
(ii) There are two common points A and B that the line PQ and
the circle have. Here, we call the line PQ a secant of the
circle.
(iii) There is only one point A which is common to the line PQ
and the circle. Here, the line PQ is called a tangent to the
circle.
Tangent to the circle:
A tangent to a circle
The word ‘tangent’
touch and was introduced
1583.
In the above figure,
A'
B' is a tangent to the circle
The tangent to a circle
two end points of its
Important Theorems:
Theorem 1: The tangent at any point of a circle is perpendicular to the
through the point of contact.
circle is a line that intersects the circle at
‘tangent’ comes from the Latin word ‘tangere’,
introduced by the Danish Mathematician Thomas
, there is only one tangent at a point of the circle.
is a tangent to the circle.
circle is a special case of the secant, when
its corresponding chord coincide.
Important Theorems:
The tangent at any point of a circle is perpendicular to the
through the point of contact.
at only one point.
‘tangere’, which means to
Thomas Fineke in
only one tangent at a point of the circle.
when the
The tangent at any point of a circle is perpendicular to the radius
Page 3
CIRCLES
Circle: A circle is a collection of all points in a plane which are at a constant distance
from a fixed point.
Some parts of a circle:
Chord: A line segment joining any two end points on the circle is called a chord.
Diameter is the longest chord.
Secant: A line which intersects the circle in two distinct points is called secant of the
circle.
Tangent: A line which touches the circle at only one point is called tangent to the
circle.
Three different situations that can arise when a circle and a line are given in a
plane, consider a circle and a line PQ. There can be three possibilities as given:
(i) The line PQ and the circle have no common point. Here,
PQ is called a non-intersecting line with respect to the
circle.
(ii) There are two common points A and B that the line PQ and
the circle have. Here, we call the line PQ a secant of the
circle.
(iii) There is only one point A which is common to the line PQ
and the circle. Here, the line PQ is called a tangent to the
circle.
Tangent to the circle:
A tangent to a circle
The word ‘tangent’
touch and was introduced
1583.
In the above figure,
A'
B' is a tangent to the circle
The tangent to a circle
two end points of its
Important Theorems:
Theorem 1: The tangent at any point of a circle is perpendicular to the
through the point of contact.
circle is a line that intersects the circle at
‘tangent’ comes from the Latin word ‘tangere’,
introduced by the Danish Mathematician Thomas
, there is only one tangent at a point of the circle.
is a tangent to the circle.
circle is a special case of the secant, when
its corresponding chord coincide.
Important Theorems:
The tangent at any point of a circle is perpendicular to the
through the point of contact.
at only one point.
‘tangere’, which means to
Thomas Fineke in
only one tangent at a point of the circle.
when the
The tangent at any point of a circle is perpendicular to the radius
Remarks:
1. By theorem mentioned
there can be one and only one tangent.
2. The line containing the radius through the point of contact is also sometimes
called the ‘normal’ to the circle at the point.
Number of Tangents from a Point on a Circle
Case 1: There is no tangent to a circle passing
circle.
Case 2: There is one and only one tangent to a
lying on the circle
Case 3: There are exactly two
outside the circle
mentioned above, we can also conclude that at any point on a circle
there can be one and only one tangent.
The line containing the radius through the point of contact is also sometimes
’ to the circle at the point.
Number of Tangents from a Point on a Circle:
There is no tangent to a circle passing through a point lying inside the
There is one and only one tangent to a circle passing through a point
lying on the circle
There are exactly two tangents to a circle through a point lying
above, we can also conclude that at any point on a circle
The line containing the radius through the point of contact is also sometimes
through a point lying inside the
circle passing through a point
circle through a point lying
Page 4
CIRCLES
Circle: A circle is a collection of all points in a plane which are at a constant distance
from a fixed point.
Some parts of a circle:
Chord: A line segment joining any two end points on the circle is called a chord.
Diameter is the longest chord.
Secant: A line which intersects the circle in two distinct points is called secant of the
circle.
Tangent: A line which touches the circle at only one point is called tangent to the
circle.
Three different situations that can arise when a circle and a line are given in a
plane, consider a circle and a line PQ. There can be three possibilities as given:
(i) The line PQ and the circle have no common point. Here,
PQ is called a non-intersecting line with respect to the
circle.
(ii) There are two common points A and B that the line PQ and
the circle have. Here, we call the line PQ a secant of the
circle.
(iii) There is only one point A which is common to the line PQ
and the circle. Here, the line PQ is called a tangent to the
circle.
Tangent to the circle:
A tangent to a circle
The word ‘tangent’
touch and was introduced
1583.
In the above figure,
A'
B' is a tangent to the circle
The tangent to a circle
two end points of its
Important Theorems:
Theorem 1: The tangent at any point of a circle is perpendicular to the
through the point of contact.
circle is a line that intersects the circle at
‘tangent’ comes from the Latin word ‘tangere’,
introduced by the Danish Mathematician Thomas
, there is only one tangent at a point of the circle.
is a tangent to the circle.
circle is a special case of the secant, when
its corresponding chord coincide.
Important Theorems:
The tangent at any point of a circle is perpendicular to the
through the point of contact.
at only one point.
‘tangere’, which means to
Thomas Fineke in
only one tangent at a point of the circle.
when the
The tangent at any point of a circle is perpendicular to the radius
Remarks:
1. By theorem mentioned
there can be one and only one tangent.
2. The line containing the radius through the point of contact is also sometimes
called the ‘normal’ to the circle at the point.
Number of Tangents from a Point on a Circle
Case 1: There is no tangent to a circle passing
circle.
Case 2: There is one and only one tangent to a
lying on the circle
Case 3: There are exactly two
outside the circle
mentioned above, we can also conclude that at any point on a circle
there can be one and only one tangent.
The line containing the radius through the point of contact is also sometimes
’ to the circle at the point.
Number of Tangents from a Point on a Circle:
There is no tangent to a circle passing through a point lying inside the
There is one and only one tangent to a circle passing through a point
lying on the circle
There are exactly two tangents to a circle through a point lying
above, we can also conclude that at any point on a circle
The line containing the radius through the point of contact is also sometimes
through a point lying inside the
circle passing through a point
circle through a point lying
Note: The length of the segment of the tangent from the external point P and the
point of contact with the circle is called the
to the circle.
Theorem 2: The lengths of tangents drawn from an external point to a circle are
equal.
From the above diagram,
OQ = OR (Radius of the same circle with centre O)
OP = OP (Common side for both the triangles)
Also, ? OQP = ? ORP
? By RHS congruence condition,
?OQP
Which gives us PQ = PR
Remarks:
1. The theorem can also be proved by using the Pythagoras Theorem as follows:
PQ
2
= OP
2
– OQ
2
= OP
2. Note that ? OPQ =
? QPR, i.e. the centre
The length of the segment of the tangent from the external point P and the
point of contact with the circle is called the length of the tangent
The lengths of tangents drawn from an external point to a circle are
From the above diagram, i.e. in right triangles OQP and ORP,
(Radius of the same circle with centre O)
(Common side for both the triangles)
By RHS congruence condition,
OQP ? ?ORP
Which gives us PQ = PR (Corresponding side of congruent triangles)
The theorem can also be proved by using the Pythagoras Theorem as follows:
= OP
2
– OR
2
= PR
2
(As OQ = OR) which gives PQ = PR.
OPQ = ? OPR. Therefore, OP is the angle bisector of
centre lies on the bisector of the angle between the two tangents.
The length of the segment of the tangent from the external point P and the
length of the tangent from the point P
The lengths of tangents drawn from an external point to a circle are
n right triangles OQP and ORP,
(Corresponding side of congruent triangles)
The theorem can also be proved by using the Pythagoras Theorem as follows:
(As OQ = OR) which gives PQ = PR.
OPR. Therefore, OP is the angle bisector of
lies on the bisector of the angle between the two tangents.
```
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## Mathematics (Maths) Class 10
62 videos|363 docs|103 tests
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;<|endoftext|>
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# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 163: 34
(a) $F = 6750~N$ (b) $F = 1.35\times 10^6~N$ (c) seat belt and air bag: The force is 11.5 times the person's weight. no seat belt and no air bag: The force is 2300 times the person's weight.
#### Work Step by Step
(a) We can find the rate of deceleration as: $a = \frac{0-v_0^2}{2x}$ $a = \frac{-(15~m/s)^2}{(2)(1~m)}$ $a = -112.5~m/s^2$ We can use the magnitude of deceleration to find the net force on the person. $F = ma$ $F = (60~kg)(112.5~m/s^2)$ $F = 6750~N$ (b) We can find the rate of deceleration as: $a = \frac{0-v_0^2}{2x}$ $a = \frac{-(15~m/s)^2}{(2)(0.005~m)}$ $a = -22500~m/s^2$ We can use the magnitude of deceleration to find the net force on the person. $F = ma$ $F = (60~kg)(22500~m/s^2)$ $F = 1.35\times 10^6~N$ (c) We can find the person's weight; $weight = mg = (60~kg)(9.80~m/s^2)$ $weight = 588~N$ When the seat belt and air bag stops the person: $F = \frac{6750~N}{588~N} = 11.5$ The force is 11.5 times the person's weight. When the person is not restrained: $F = \frac{1.35\times 10^6~N}{588~N} = 2300$ The force is 2300 times the person's weight.
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.<|endoftext|>
| 4.5625 |
1,078 |
LCM the 4 and also 6 is the smallest number amongst all common multiples the 4 and also 6. The first few multiples of 4 and also 6 are (4, 8, 12, 16, 20, 24, . . . ) and (6, 12, 18, 24, . . . ) respectively. There room 3 commonly used approaches to discover LCM the 4 and 6 - through listing multiples, by element factorization, and also by department method.
You are watching: Find the least common multiple of 4 and 6.
1 LCM the 4 and also 6 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM that 4 and also 6 is 12.
Explanation:
The LCM of 2 non-zero integers, x(4) and y(6), is the smallest positive integer m(12) that is divisible by both x(4) and also y(6) without any type of remainder.
The methods to discover the LCM that 4 and 6 are described below.
By Listing MultiplesBy element Factorization MethodBy division Method
### LCM the 4 and 6 by Listing Multiples
To calculation the LCM that 4 and 6 by listing the end the usual multiples, we can follow the given listed below steps:
Step 1: list a couple of multiples the 4 (4, 8, 12, 16, 20, 24, . . . ) and 6 (6, 12, 18, 24, . . . . )Step 2: The usual multiples native the multiples that 4 and 6 are 12, 24, . . .Step 3: The smallest common multiple of 4 and 6 is 12.
∴ The least usual multiple that 4 and 6 = 12.
### LCM of 4 and also 6 by element Factorization
Prime administrate of 4 and also 6 is (2 × 2) = 22 and (2 × 3) = 21 × 31 respectively. LCM that 4 and also 6 deserve to be acquired by multiplying prime factors raised to their respective highest possible power, i.e. 22 × 31 = 12.Hence, the LCM the 4 and also 6 by prime factorization is 12.
### LCM that 4 and 6 by department Method
To calculation the LCM of 4 and 6 through the division method, we will divide the numbers(4, 6) by their prime factors (preferably common). The product of these divisors gives the LCM that 4 and also 6.
Step 3: continue the procedures until just 1s are left in the critical row.
The LCM that 4 and 6 is the product of every prime numbers on the left, i.e. LCM(4, 6) by department method = 2 × 2 × 3 = 12.
☛ likewise Check:
## FAQs ~ above LCM the 4 and also 6
### What is the LCM of 4 and 6?
The LCM of 4 and also 6 is 12. To uncover the least usual multiple (LCM) that 4 and 6, we need to uncover the multiples of 4 and 6 (multiples the 4 = 4, 8, 12, 16; multiples the 6 = 6, 12, 18, 24) and also choose the smallest multiple the is precisely divisible through 4 and also 6, i.e., 12.
### If the LCM the 6 and 4 is 12, find its GCF.
LCM(6, 4) × GCF(6, 4) = 6 × 4Since the LCM of 6 and also 4 = 12⇒ 12 × GCF(6, 4) = 24Therefore, the GCF = 24/12 = 2.
### What is the Relation between GCF and LCM that 4, 6?
The complying with equation have the right to be offered to refer the relation in between GCF and also LCM of 4 and 6, i.e. GCF × LCM = 4 × 6.
See more: Where To Find Strength In Pokemon Diamond ? How Do I Obtain The Hm Strength In Diamond
### What is the the very least Perfect Square Divisible through 4 and 6?
The the very least number divisible by 4 and 6 = LCM(4, 6)LCM that 4 and also 6 = 2 × 2 × 3 ⇒ least perfect square divisible by every 4 and also 6 = LCM(4, 6) × 3 = 36 Therefore, 36 is the required number.
### What room the techniques to discover LCM that 4 and 6?
The frequently used approaches to find the LCM that 4 and 6 are:<|endoftext|>
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Chemistry describes the connections (a.k.a. bonds) among atoms. At this level, it is the sharing or exchange of only the outermost electrons which facilitates an increase in complexity from smaller atoms to molecules, crystals or metals. In this sharing or exchange of electrons, properties are completely changed. As we will soon see, community at every level is transformative!
BONDING BY SHARING BETWEEN ADJACENT ATOMS (COVALENT BONDING)
To begin, we start with two hydrogen atoms in isolation from one another. Their energy in this state is defined as zero as shown above on the right side of the picture. As the two atoms approach one another they are attracted to one another and so their combined energy is gradually reduced to a lower value until the lowest energy is achieved at the bottom of the curve.
This distance where the two atoms have the most stable energy is called the bonding distance or bond length and signals the formation of the resulting molecule. If the two atoms continue past this point to come closer together, they repel one another increasing the molecule’s overall energy once again. In response to this repulsion, the atoms return to the “sweet spot” where the energy is the lowest.
A single bond distance defines the hydrogen molecule and a specific lowering or release of energy is associated with the formation of that bond, the bonding energy, for two hydrogen atoms. To break the bond and again separate the atoms requires an input equal to that energy. Hence, the molecule is more “stable” than two atoms existing separately. Coming together is about sacrifice yes, but also synergistic benefit.
A molecule is not the mere juxtaposition of more than one atom with others. It is about a new entity where the outer electrons are shared by the two nuclei making up the molecule. This sharing of electrons is so important, chemists use different terms to name the locations of the electrons. Before the two hydrogen atoms begin sharing the electrons, we say they exist in “atomic” orbitals, emphasizing their isolation from one another.
After the association of the two hydrogen atoms we say they now exist in “molecular” orbitals emphasizing instead their new identity as a hydrogen molecule instead of two hydrogen atoms. It is in this sharing of electrons that the two atoms are transformed into a single molecule. We call the sharing of the electrons, “covalent”, a co-operative sharing of the outer shell, or valence electrons from each of the atoms.
The combination of two atoms to form a hydrogen molecule is transformative for the atoms. The new hydrogen molecules have different physical and chemical properties than the originally separate hydrogen atoms. New possibilities for reactions now exist for the molecule that were not available before and other possibilities have been given up. The molecule is a trade-off with greater complexity.
Covalent bonds not only exist between atoms of the same type, but also between atoms of different types. These almost always share their bonding electrons somewhat unevenly, because the original atoms each have a different attraction for the electrons. Examples include water, H-O-H, better known as H2O, and carbon dioxide, CO2, or O=C=O. In both these bonds, the oxygen atom is known to attract the electrons more strongly than either the hydrogen or carbon atoms, resulting in uneven or polar bonds.
What is perhaps more amazing is that because water is a bent molecule it will interact with electric and magnetic fields, but carbon dioxide will not because it is linear. The details are not important here, except to say even the three-dimensional shape produced by the coming together of atoms makes the difference in the properties of the molecule.
Sharing of these outer shell electrons between atoms and the formation of a
3-dimensional molecule from atoms leads to an abundance of possibilities not open to just atoms alone. In fact, in the world of practical chemistry, it is the rearrangement of these very bonds that leads to the transformation of one compound to another, what we call chemical reactions.
Covalent compounds exist as molecules, literally “little lumps” of matter, very small communities where electrons are shared as needed among the member atoms. The elements called nonmetals such as oxygen (O), hydrogen (H), and carbon (C) (see above), participate in this type of sharing. They are known to hold onto these outer electrons quite tightly (have high ionization energies) and so although they are not likely to give them away, sharing is possible.
Metal atoms by contrast hold onto their electrons more loosely than nonmetals (have low ionization energies), so we will see a completely different type of bonding as atoms of metals and non-metals combine as discussed in the second section and as atoms of metals come together themselves (the third section)
In the periodic table notice the existence of “metalloids” which are intermediate in their properties between metals and nonmetals. These elements sometimes behave like metals and at other times like nonmetals, depending on what other elements are around them. (Sounds like some people whose behavior is dependent on their environment, eh?)
BONDING BY COMPLETE EXCHANGE (IONIC BONDING)
Another possibility for the combination of atoms involves the transfer of electrons rather than their equal or unequal sharing. A beautiful example of this is common table salt: NaCl, sodium chloride. Sodium (Na) is a highly reactive metal; indeed, it will violently react with water and form a solution of lye [drain cleaner], NaOH, in the process. Chlorine (Cl2), a yellow-green gas is poisonous, and was used in World War I as a chemical weapon before its ban by the Geneva Convention.
Yet, when sodium and chlorine are existent together as sodium chloride (NaCl), we obtain again an entirely new set of properties. Some of its most common uses include, but are not limited to: flavoring our food, melting snow and ice on our roads and sidewalks, softening our well water, and making paper and rubber.
How is it that this transfer of electrons comes about? Sodium metal, Na, begins with 11 electrons, in what we call “shells” of 2, 8, and 1. Each non-metal chlorine atom, Cl, begins with 17 electrons, with 2, 8, and 7 electrons in its shells. Shells listed first are nearest the nucleus and not likely to be exchanged, but those listed last, the valence, or outer-shell electrons, are furthest from the nucleus and are more loosely held.
As shown in the diagram, by the simple transfer of the one valence electron in the last shell from sodium to the last shell of 7 electrons in chlorine we are left with the sodium ion, Na+, with 10 electrons in shells of 2 and 8 and the chloride ion, Cl–, with 18 electrons in shells of 2, 8, and 8.
For reasons we won’t explore here these octets, or sets of 8 electrons, in the last shell are much more stable than the electron arrangement in the neutral atoms, not containing these octets in the last shells. The resulting Na+ and Cl– ions, being opposite in charge, are strongly attracted to one another.
In fact, we call sodium chloride, NaCl, for simplicity’s sake. This one-to-one ratio is simply the lowest ratio (1:1) of atoms in sodium chloride. The actual number of pairs is overwhelmingly great at room temperature. Atoms come together in such great numbers that we can see a single crystal of sodium chloride at room temperature. Think of that the next time you hold a crystal of salt in your hand!
Ionic compounds are almost always composed of a metallic element with a nonmetallic clement. The loosely held electrons are donated from the metal to the nonmetals, which hold on tightly to them. Examples include
- calcium chloride, CaCl2 (used to melt snow from our sidewalks in winter)
- sodium nitrate, NaNO3 (used to preserve some lunch meats and bacon), and
- aluminum sulfate, Al2(SO4)3 (used in the purification of drinking water).
So, what difference does a transfer of electrons cause compared to sharing? Check out the contrasting properties of covalent and ionic compounds:
|Properties||Covalent Compounds||Ionic Compounds|
|representative unit||isolated molecule||ions in crystals|
|types of elements||nonmetals with nonmetals||metals with nonmetals|
|melting/boiling point||usually low||usually high|
|physical state (250C)||usually gases, also liquid & solid||usually solids|
|solubility in water||usually lower, except for acids||usually higher|
|electrical conducitivity of solid||does not conduct electricity||does not conduct electricity|
|electrical conductivity of solution with water||does not conduct electricity, except for acids||does conduct electricity|
BONDING BY OPEN AND WIDESPREAD SHARING (METALLIC BONDING)
Metals have a third way of coming together. Individual atoms of metals have loosely held electrons in their outermost shells as we know from their easy donation of electrons to nonmetals in ionic bonding. Yet when metals come together with other metals the sharing is remarkably different: they now share the electrons over a vast community. The outermost electrons are free to move in a matrix of nuclei with only their inner shells of electrons. We call these freely wandering valence electrons, the “sea” of electrons.
This property is what makes metals such wonderful conductors of electricity. The electrons are free to move throughout the metal from one place to another, when moved by an electric field. This sea of electrons is also the reason that metals are malleable and can be pounded in sheets, like aluminum or gold foil. Metals are also ductile and can be drawn in wires for the same reason. We use copper, for instance, to wire our homes and businesses.
|representative unit||widespread connection among metal atoms|
|types of elements||metal with metal|
|melting/boiling point||usually very high|
|physical state (250C)||solid, except for mercury (Hg)|
|solubility in water||does not dissolve in water|
|electrical conductivity of solid||conducts electricity easily|
|electrical conductivity of solution with water||no electrical conductivity due to dissolution|
The way electrons are either shared or exchanged among atoms and how that leads to the creation of molecules, metals, or crystals, can remind us of the benefit for us as human beings to freely exchange or share our resources. For us too, the possibility of greater complexity and new possibilities awaits.
I don’t know about you, but I’m good at donating that to which I’m not particularly attached to someone in need. And, I have become better at receiving what seems to be “extra” from other people, especially if I really need it. I do the transfer of resources to and from myself (an ionic process) fairly well.
And, I suppose I’m even OK at sharing what little I have with others if they put in a similar share and none of us is particularly attached to what we’re sharing. My instinct to follow the wisdom of the metallic atoms seems intact too.
But, when I look at what I’m attached to, especially when I think a resource is scarce, I have grown accustomed to keeping “extra” for myself. My first inclination is the opposite of sharing, hoarding. So it makes me wonder: what new possibilities I’ve been missing because I haven’t participated in as much sharing (a covalent process) as the atoms do naturally? It’s a good question!
There’s a lot to learn from atoms, besides academic chemistry!
Bless the Lord, all you molecules that share electrons,
All you atoms that transfer electrons, bless the Lord.
All you metals with your seas of electrons, bless the Lord,
Praise and exult God forever!<|endoftext|>
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Emerald ash borer (Agrilus plannipennis Fairmaire) is a federally regulated pest and poses a tremendous threat towards natural forested areas and municipalities across North America and Canada. Emerald ash borer (EAB) is a highly destructive, non-native invasive beetle responsible for killing more than 100 million ash trees throughout the United States. Originally from Asia, the metallic wood-boring beetle was first discovered in Michigan in 2002, where it was presumably thought to be present for roughly 10 years or longer before being detected. Since then, EAB has now become established in 35 states across North America, including Colorado. In September 2013, EAB was positively identified in the City of Boulder, making it the first identification within the state of Colorado. Since then, it has now been positively detected within the City of Longmont, area of Gunbarrel, and the City of Lafayette, Town of Lyons, and the City of Superior.
It is suggested that EAB is the most destructive insect to hit North America. At first, when
populations are low, neglect damage from the beetle is noticed; however, EAB’s “fitness” (ability
to survive and reproduce) is substantially high allowing for populations to quickly increase in
number. Therefore, once established and populations become highly dense, healthy trees can
die within several years if untreated.
Though EAB can infest new areas naturally, the main method of EAB introduction is unwilling by humans. The transportation of infected firewood to other areas for use or sale, along with other ash material, is the number one factor contributing to new infestations. Once EAB is detected, federal agencies establish quarantine areas regulating the movement of hardwood materials within the infested area. Considering Boulder County is the own County with confirmed identifications, a quarantine has been placed in Boulder County, along with additional portions of Weld and Jefferson Counties.
Out of the +4,000 street and public area trees in Berthoud, roughly 25% of trees are ash (Fraxinus spp.). Ash trees, though not native to Colorado, have been heavily planted throughout the state as street and shade trees due to their fast growth rate and high adaptability. The two most common species of ash planted in the Front Range are green ash (Fraxinus pennsylvanica) and white ash (Fraxinus americana). However, all species of ash in the genus Fraxinus are at significant risk of elimination from this devastating insect.
Since its discovery in 2002, research on chemical treatments has resulted in success. The most widely used treatment are micro injections to protect the tree, these injections can protect the tree for up to two years costing several hundred dollars per tree depending on its size. Chemical soil drenches around the base of the tree have also been somewhat effective as have bark sprays. Bees are thought not to be affected since they do not visit the pollen on ash trees, their pollen is wind distributed.
So where do we go from here? The Town of Berthoud and the Tree Advisory Committee have been proactively working since the discovery of EAB within the City of Boulder limits. Town Staff and Tree Advisory Committee Members attend state conferences and workshops regarding EAB management and control, and are remaining up-to-date with the latest recommendations and literature publications for chemical treatments. Town Staff has developed an Emerald Ash Borer Management and Response Plan intended to provide best management practices for Town of Berthoud Staff as well as residents within the Town to reference as needed. Town Staff and Tree Advisory Committee Members also plan to hold community workshops and make available printed material to educate the public regarding recommended treatment protocols, how to identify an ash tree, and most importantly, what EAB infestation symptoms to look for.<|endoftext|>
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Introduction
PART I: FIRST ORDER EQUATIONS Section 7 Finding the direction for shock curves James V. Herod*
Web page maintained by Evans M. Harrell, II, [email protected].
We need to handle the problem which occurred in Example 4.2 and again in Example 5.4. In both these examples, the characteristic curves crossed. Here's an understanding for why this crossing of the characteristics causes a problem.
Suppose the characteristic curves cross at {a,b}. We had found the value of u(a,b) by tracing back along the characteristic to the line where the initial value of u was specified and determined how u evolved by following above that characteristic curve. The problem is that we find there might be two characteristic curves going through {a,b}. Tracing back to the initial values may put us at different initial values and u may evolve differently as it moves along the different characteristic curves. Thus, when we come to {a,b} along the two curves, we may have different values as candidates for the values of u(a,b).
This induces a shock curve. It is developing one understanding of the shock curve that we consider in this section. Two examples will be used to illustrate the problem again. Both use the same partial differential equation, with different initial values. We find that one has crossing characteristics and the other does not.
Example 7.1 (Burger's Equation)
The partial differential equation is
u_t + u u_x = 0. (7.1)
We take the initial value to by
u(0,x) = f(x).
The reader who has worked through these notes to this point can derive the solution for the characteristic curves:
x = f([[xi]]) t + [[xi]].
At first, one might think this is a good situation; the characteristic curves are straight lines. Alas, they don't have the same slope. The nature of the initial value f is the important key. For this equation, if the values of f never decrease, the characteristics will not cross. In problem (7.2), you are asked to examine the characteristics for two different initial distributions f:
f1(x) = BLC{(A(1 if x < 0, 1-x if 0 < x < 1, 0 if 1 < x))
and
f2(x) = BLC{(A(0 if x < 0, x if 0 < x < 1, 1 if 1 < x)).
For one of these the characteristic curves cross, for the other, they do not.
Warning: This property of the initial distribution being non-decreasing implying the characteristics do not cross is not necessarily true for equations other than the Burger's equation. Recall, for example, equation (5.1), example 5.2, and figure 5.3.
Having this illustration of how the characteristic curves might cross for a partial differential equation with one set of initial conditions and not cross with another, we address how to determine the value of u in this region where it might be multivalued. This can be resolved with some tools from the calculus.
There is information from the integral calculus which we should recall. Some of the important ideas in that topic involve comparisons of integrals over areas with integrals around contours.
The following is a fundamental idea in that comparison.
STOKES THEOREM IN 2D. Suppose that D is a bounded region in the plane with a piecewise smooth boundary denoted [[partialdiff]]D. Suppose also that {P(x,y), Q(x,y)} has continuous second partial derivatives and is a function from R^2 to R^2. Then
integral [[partialdiff]]D Pdx + Qdy = integral D ([[partialdiff]]Q/[[partialdiff]]x - [[partialdiff]]P/[[partialdiff]]y) dx dy.
This theorem will be used to determine the direction the shock curve travels if characteristic lines cross and it appears that u should be multivalued.
(Click here if you prefer the divergence theorem of Gauss to the Stokes theorem - comments by Evans Harrell.)
In the previous sections we found it was reasonable to have solutions of a partial differential equation that were not differentiable at all points in their domain. (Re-look at Figure 5.4.) A wider notion of what a "solution" is seems appropriate.
To formulate this larger notion of solution, we want to develop the notion of the adjoint of the linear mapping L defined by
L(u) = au_x +bu_y + cu.
First, note the calculus:
v L(u) = v (au_x + bu_y + cu)
= (vau)x + (vbu)y - u(av)x - u(bv)y + vcu.
This implies
integral D v(au_x + bu_y + cu) dx dy + integral D [(va)x + (vb)y - vc]u dx dy
= integral D [(avu)x + (bvu)y] dx dy = integral [[partialdiff]]D [-bvu dx + avu dy].
Use Stokes theorem in 2D and the fact that v = 0 on the boundary of D to get that this last integral is zero. It follows that
integral D v(au_x + bu_y + cu) dx dy = integral D [-(va)x - (vb)y + vc]u dx dy.
We define
L*(v) to be -(va)x - (vb)y + vc
and say that L* is the adjoint of L.
We see that if u is differentiable and satisfies L(u) = 0, then also
integral D L*(v) u dx dy = 0
for all v in the domain of L*.
Thus, an alternate definition of a solution for L(u) = 0 is to say that
u is a "weak" solution of L(u) = 0 provided
integral D L*(v) u dx dy = 0.
for all v that are differentiable in D and 0 on the boundary
of D.
The above makes the notion of a function u being a solution for a partial differential equation when u is not even differentiable a little more comfortable.
The question of what is the direction that the "shock" travels is next. The sets D, D^+, and D^- are suggested in Figure 7.3. We will establish each of the following steps:
1. G(u) and H(u) are defined so that
F([[partialdiff]]G(u),[[partialdiff]]x) = a u_x and F([[partialdiff]]H(u),[[partialdiff]]y) = b u_y .
We assume that
0 = a u_x + b u_y = F([[partialdiff]]G(u),[[partialdiff]]x) + F([[partialdiff]]H(u),[[partialdiff]]y) .
2. If v = 0 on [[partialdiff]]D then
0 = integral D B(G(u) vx + H(u) vy) dx dy.
3. integral D+ B(G(u) vx + H(u) vy) dx dy =
integral L v B(G(u^+) dy - H(u^+)dx)
4. integral D- B(G(u) vx + H(u) vy) dx dy =
integral L v B(-G(u^-) dy + H(u^-)dx)
5. Add 3 and 4 above to get 2 so that
0 = integral L v B((G(u^+)-G(u^-)) dy - (H(u^+) - H(u^-))dx).
Conclude: F(dy,dx) = F(H(u^+) - H(u^-),G(u^+) - G(u^-)) .
Proof of 2:
F([[partialdiff]][G(u) v],[[partialdiff]]x) = F([[partialdiff]]G(u),[[partialdiff]]x) v + G(u) F([[partialdiff]]v,[[partialdiff]]x) and F([[partialdiff]][H(u) v],[[partialdiff]]y) = F([[partialdiff]]H(u),[[partialdiff]]y) v + H(u) F([[partialdiff]]v,[[partialdiff]]y) .
Hence if v = 0 on [[partialdiff]]D then
0 = integral [[partialdiff]]D [-H(u) v dx + G(u) v dy ]
= integral D [ F([[partialdiff]][G(u) v],[[partialdiff]]x) + F([[partialdiff]][H(u) v],[[partialdiff]]y) ]dxdy
= integral D [G(u) F([[partialdiff]]v,[[partialdiff]]x) + H(u) F([[partialdiff]]v,[[partialdiff]]y) ]dx dy + integral D [ F([[partialdiff]]G(u),[[partialdiff]]x) v + F([[partialdiff]]H(u),[[partialdiff]]y) v ] dx dy.
The last of these two integrals is 0 because of (1) above.
Proof of 3:
Figure 7.3
The curve is given by y = s(x) and we suppose that s^-1(y) = x. Suppose that v vanishes on the boundary of the rectangle. Then
integral D+ G(u) F([[partialdiff]]v,[[partialdiff]]x) dx dy = I(y1,y2, ) I(x1, s^-1(y), ) (G(u) v)x - G(u)x v)dx )dy
= I(y1,y2, ) (I(x1, s^-1(y), ) (G(u)v)x dx) dy - integral D+ G(u)x v dx dy
= I(y1,y2, ) G(u)+ v dy - integral D+ G(u)x v dx dy.
Also,
integral D+ H(u) F([[partialdiff]]v,[[partialdiff]]y) dx dy
= I(x1,x2, ) I(s(x), y2, ) (H(u) v)y dy dx -integral D+ H(u)y v dx dy
= - I(x1,x2, ) H(u)+ v dy - integral D+ H(u)y v dx dy
Thus,
integral D+ G(u) F([[partialdiff]]v,[[partialdiff]]x) dx dy + integral D+ H(u) F([[partialdiff]]v,[[partialdiff]]y) dx dy = integral s v ( G(u^+)dy - H(u^+) dx)
Proof of 4:
In a similar manner, 4 can be established.
^
^Example: Here is a graphical representation of a solution for the equation
u_x + 2 u u_y = 0
with u(0,y) given by the continuous function
u(0,y) = 1 if y < 0, 1- y if 0 < y < 1, and 0 if 1 < y.
The extension past the shock point used the Rankine-Hagonoit jump condition. The shock associated with the solution can be seen in the graph.
* u:=proc(x,y)
if x <=1/2 then
if y < 2*x then 1
elif 2*x<= y and y < 1 then (1-y)/(1-2*x)
elif 1 <= y then 0
fi
elif 1/2 < x then
if y < x+1/2 then 1
else 0
fi fi
end:
* plot3d(u,0..2,-2..2,orientation=[-160,50],numpoints=900);
Solution with shock
Exercise
7.1 Application of Stokes Theorem in the plane: Integrate < curl F, [[eta]] > over D.
a. F(x,y) = {x,y}, D = D1(0) (= the unit disk).
b. F(x,y) = {-y,x}, D = D1(0),
c. F(x,y) = {3y, 5x}, D = D1(0), ans:2[[pi]]
d. F(x,y) = {0,x^2}, D is the rectangle with vertices at {0,0}, {a,0}, {a,b}, and {0,b}. ans: a^2b
e. F(x,y) = {3xy + y^2, 2xy + 5x^2}, D= D1({1,2}) (= the disk with radius 1 and center {1,2} ). ans:7[[pi]].
7.2 Examine the characteristics for Burger's equation (7.1) with two initial distributions f:
f1(x) = BLC{(A(1 if x < 0, 1-x if 0 < x < 1, 0 if 1 < x))
and
f2(x) = BLC{(A(0 if x < 0, x if 0 < x < 1, 1 if 1 < x)).
7.3. Solve Burger's equation (2.1) with both initial condition given in Exericse 7.2 above.
7.4. Solve the equation
u_t + 2 u u_x = 0
with both initial condition given in Exericse 7.2 above.
7.5. Draw the solution for Burger's equation with
u(0,x) = BLC{(A( 1-x^2 if |x| <= 1, 0 if |x| > 1)).<|endoftext|>
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Strategies and Resources to Create a Trauma-Sensitive School
Resources Schools Can Use to Incorporate Trauma-Sensitive Practices
- Calmer Classrooms: A Guide to Working with Traumatized Children
Child Safety Commission, Melbourne, Victoria, Australia
Calmer Classrooms: A Guide to Working with Traumatized Children assists teachers and other educators in understanding and working with children and youth whose lives have been affected by trauma, particularly related to child maltreatment. It explains the effect of abuse on learning and attachment, providing teachers with strategies involving relationship-based practices for challenges and conflicts in the classroom, as well as self-care.
- Child Trauma Toolkit for Educators
National Child Traumatic Stress Network
The Child Trauma Toolkit for Educators provides information for educators, parents and caretakers, including facts, suggestions, psychological and behavioral impact of trauma by grade level, and self-care.
- Creating Sanctuary in the School
Creating Sanctuary in Schools describes the basis for the process of providing a safe and healing environment for children in schools who need to recover from the effects of trauma, as well as for less traumatized children. Basic assumptions, values, goals and the process that must be shared by all members of the system are described.
- Helping Traumatized Children Learn
Trauma Learning and Policy Initiative
Volume 1 – A Report and Policy Agenda summarizes the research from psychology and neurobiology that documents the impact trauma from exposure to violence can have on children’s learning, behavior and relationships in school. The report also introduces the Flexible Framework, a tool organized according to six core operational functions of schools that can help any school create a trauma-sensitive learning environment for all children.
Volume 2 – Creating and Advocating for Trauma-Sensitive Schools offers a guide to a process for creating trauma-sensitive schools and a policy agenda to provide the support schools need to achieve this goal. Grounded in theory and practice in schools and with families, the guide is intended to be a living document that will grow and change as more schools become trauma-sensitive and add their ideas. The policy agenda calls for changes in laws, policies, and funding streams to support schools in this work. Together, the online learning community and the book are designed to complement each other, helping to build a growing and increasingly visible trauma-sensitive learning community.
- Making SPACE for Learning: Trauma-Informed Practice in Schools
Australian Childhood Foundation
Making SPACE for Learning is a resource guide to assist schools to unlock the potential of traumatized children and young people to grow and develop at school. This publication 1) explains how trauma can impact child development and functioning, including learning; 2) promotes five principles for trauma-informed schools using the acronym SPACE (Staged, Predictable, Adaptive, Connected, and Enabled); and 3) lists many concrete, specific trauma-sensitive strategies schools can implement.
- Supporting and Educating Traumatized Students: A Guide for School-Based Professionals
Eric Rossen and Robert Hull, Editors
Oxford University Press
Supporting and Educating Traumatized Students: A Guide for School-Based Professionals provides practical, effective, and implementable strategies and resources for adapting and differentiating instruction, modifying the classroom and school environments, and building competency for students affected by trauma. The book offers techniques and strategies designed for different educational environments within the context of multiple potential sources of trauma.
- The Heart of Learning and Teaching Compassion, Resiliency, and Academic Success
Office of the Superintendent of Public Instruction, State of Washington
The Heart of Learning: Compassion, Resiliency, and Academic Success is a handbook written and compiled by the State of Washington Office of the Superintendent of Public Instruction and Western Washington University staff. It contains valuable information for educators to help them on a daily basis as they work with students whose learning has been adversely impacted by trauma in their lives.
Websites With Information about Trauma and Trauma-Sensitive Practices For Schools
- Australian Child and Adolescent Trauma, Loss and Grief Network
The Australian Child and Adolescent Trauma, Loss and Grief Network website provides information, videos, and podcasts about traumatized children. The website contains links to documents and other websites that are useful to educators and other professionals working in schools.
- The Language of Trauma and Loss
The Language of Trauma and Loss provides teachers with information about the effect of trauma and loss on children, and the teacher’s role in identifying and referring appropriate students. The program also helps teachers establish a safe classroom and improve language arts skills using trauma and loss as a vehicle. Web-streamed videos offer professional development information for teachers, and age-specific lessons for elementary, middle and high school students, and can be used as a vehicle to help students express their concerns.
- National Child Traumatic Stress Network
Established by Congress in 2000, the National Child Traumatic Stress Network (NCTSN) is a unique collaboration of academic and community-based service centers whose mission is to raise the standard of care and increase access to services for traumatized children and their families across the United States. Combining knowledge of child development, expertise in the full range of child traumatic experiences, and attention to cultural perspectives, the NCTSN serves as a national resource for developing and disseminating evidence-based interventions, trauma-informed services, and public and professional education.
- Trauma and Learning Policy Initiative
The Trauma and Learning Policy Initiative’s mission is to ensure that children traumatized by exposure to family violence and other adverse childhood experiences succeed in school.
- Treatment and Services Adaptation Center
The Treatment and Services Adaptation Center for Resiliency, Hope and Wellness in Schools promotes trauma-informed school systems that provide prevention and early intervention strategies to create supportive and nurturing school environments.
- Professional Quality of Work Life
Professional quality of life is the quality one feels in relation to their work as a helper. Both the positive and negative aspects of doing one’s job influence professional quality of life. This site covers self-care for secondary stress and trauma and vicarious traumatization that comes from doing your job. The PROQOL Test provides educators with a score meaned to their level of Compassion Satisfaction, Compassion Fatigue, and Burnout.
- Tips for Helping Students Recovering from Traumatic Events
US Department of Education
This brochure, which is based on discussions with some three dozen experts who work with students, provides practical information for parents and students who are coping with the aftermath of a natural disaster, as well as teachers, coaches, school administrators and others who are helping those affected.
- Cognitive Behavioral Intervention for Trauma in Schools (CBITS)
The Cognitive Behavioral Intervention for Trauma in Schools (CBITS) program is a school-based, group and individual intervention. It is designed to reduce symptoms of post-traumatic stress disorder (PTSD), depression, and behavioral problems, and to improve functioning, grades and attendance, peer and parent support, and coping skills. CBITS has been used with students from 5th grade through 12th grade who have witnessed or experienced traumatic life events such as community and school violence, accidents and injuries, physical abuse and domestic violence, and natural and man-made disasters. CBITS uses cognitive-behavioral techniques (e.g., psychoeducation, relaxation, social problem solving, cognitive restructuring, and exposure).
- Trauma-Informed Care
Wisconsin Department of Health Services
This website provides information about efforts to bring trauma-informed care to Wisconsin organizations.
For questions about this information, contact [email protected] (608) 266-8960<|endoftext|>
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1. Math 12 logarithms
Hi all,
I just a few questions on Math 12, not sure how to do these questions (I need to know the how-to more than the answer itself):
1) "The population of a type of bacteria triples every 20 hours. In how many hours will a population of 30 become a population of 1000?"
2) "Solve for x: log(subcript a) a ^ 2x = log (subscript b^2) b^(3x-3)
3) Simplify: [log (subscript: 1/x) 1/y ] - [log (subscript 1/x) y ] - [log (subscript X) 1/y]
(This is from the Logarithms unit).
2. Originally Posted by funnytim
Hi all,
I just a few questions on Math 12, not sure how to do these questions (I need to know the how-to more than the answer itself):
1) "The population of a type of bacteria triples every 20 hours. In how many hours will a population of 30 become a population of 1000?"
2) "Solve for x: log(subcript a) a ^ 2x = log (subscript b^2) b^(3x-3)
3) Simplify: [log (subscript: 1/x) 1/y ] - [log (subscript 1/x) y ] - [log (subscript X) 1/y]
(This is from the Logarithms unit).
2) $\log_a(a^{2x})=log_b(b^{3x-3})$.
Do you remember the rule $\log_a(a^m)=m$?
So we get $2x=3x-3$. Can you solve for x?
3) $\log_{\frac{1}{x}}(\frac{1}{y})-\log_{\frac{1}{x}}(y)-\log_x(\frac{1}{y})$
Do you remember the rule $\log_a(m)-\log_a(n) = \log_a(\frac{m}{n})$? Using this we get
$\log_{\frac{1}{x}}(\frac{\frac{1}{y}}{y})-\log_x(\frac{1}{y})$
$=\log_{\frac{1}{x}}(\frac{1}{y^2})-\log_x(\frac{1}{y})$
We can't do any more as the bases of the logarithms are not the same.
3. Hold on there, Pardner.
$log_{1/x}(a) = \frac{log(a)}{log(1/x)} = \frac{log(a)}{-log(x)} = -log_{x}(a)$<|endoftext|>
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Functions and Algebra Competition Questions
Chapter
Chapter 1
Section
Functions and Algebra Competition Questions
Solutions 10 Videos
The graph of y = -x^2-5x + 36 intersects the x-axis at two points, A and B. The length of line segment AB is
A. 9
B. 13
C. 5
D. 36
Q1
If f(x)=x^2+5x+3k and f(k)=-16, f(2) equals
A. -4
B. -16
C. -2
D. 2
Q2
The minimum distance between the parabolas y = -5x^2 - 8 and y = 7x^2 + 6 is
A. 14
B. 12
C. 8
D. 6
Q3
A quadratic function of the form f(x) = ax^2 + bx + c has roots x = \frac{-3 \pm \sqrt{31}}{2}. The graph of the function passes through the point (1, -3). What is the equation of the function?
A. \displaystyle f(x) = 2x^2 -6x + 11
B. \displaystyle f(x) = 2x^2 +6x - 11
C. \displaystyle f(x) = 3x^2 - 2\sqrt{5}x - 4
D. \displaystyle f(x) = 2x^2 + 2\sqrt{5}x -3
Q4
The sum, S, and product, P, of the roots of the function f(x) = -3x^2 + 24x + 477 are
A. S = 24, P = 477
B. S = 24, P = 1437
C. S = 8, P = -159
D. S = -8, P = 159
Q5
If x^y = 4, then the value of x^{3y}-x^{2y} is
A. 48
B. 64
C. 4
D. 12
Q6
If M = 5^x + 5^{-x} and N = 5^x =5^{-x}, then the value fo M^2 -N^2 is
A. 2(5^{2x})
B. 2(5^{-2x})
C. 4
D. 0
Q7
In the given diagram, XY = X2 = 15 and YZ = 18. The value of \sin Z is
A. \displaystyle \frac{3}{5}
B. \displaystyle \frac{4}{5}
C. \displaystyle \frac{4}{9}
D. \displaystyle \frac{4}{3}
Explain why the minimum value of (x - a)^2 + (x - b)^2 occurs when x = \frac{a + b}{2}. What is the minimum value.<|endoftext|>
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|Terms & Symbols|
Stars are so unimaginably far away that the light we receive from them arrives in rays that are perfectly parallel. Your eye is designed to focus these parallel rays to a point, allowing you to identify where the light is coming from.
A telescope, in its original configuration (refractor), consists of two lenses. The first one, the objective lens, collects light and focuses it to a point. (Note that the objective mirror in a reflecting telescope does exactly the same thing.) The second lens, the eyepiece, catches the light as it diverges away from the focal point and bends it back to parallel rays, so your eye can re-focus it to a point.
Notice how the telescope has taken all the light passing through the objective lens and compressed it down to a column of light that will pass through the pupil of the eye. This is one of the three major tasks of the telescope, the full list being:
The equations on this page permit you to find just exactly how well the telescope will perform these tasks, and along the way I also show how the tasks are accomplished, by explaining both the theory and the practice. We will be talking specifically about visual observing through the telescope – how the telescope and your eye work together. Understanding photography starts with understanding these ideas, and here we are going to stick to visual observation.
CAUTION - telescope manufacturers will often advertise the magnification of the scope, and give really big, impressive numbers. The problem is that the number is essentially meaningless.
The magnification of a telescope is a combined function of the scope and the eyepiece that is used, so the user can set the magnification to almost any arbitrary value by selecting a suitable eyepiece. Whether the resulting image is clear, or barely visible, depends on other properties of the telescope. Therefore the magnification is not the most important measure of a telescope.
What actually is the most important measure is the diameter of the objective, or more simply the scope diameter, because that determines both the resolving power (the smallest detail you can see) and the light-gathering power (the faintest objects you can see). How the scope diameter determines the performance of your telescope is explained through the equations below.
Of course you can use the equations below however you like. Just in case it seems like a lot of equations to you, or maybe looks a bit overwhelming, here's a procedure I can suggest for you. It's in two parts: first finding the intrinsic telescope properties, then determining its operating points.
These are properties of the scope that depend only on the diameter of the objective, so are intrinsic and fundamental to the scope.
The goal is to bracket the range of eyepieces - largest and smallest focal lengths - to use with the scope and determine the performance of the scope at each end of the range.
Click on the chalk board to see the process in action with various random interesting examples.
|Dep||Diameter of the exit pupil. The exit pupil is where the light leaving the eyepiece converges to its smallest circle -- you find the exit pupil when you bring your eye up to the eyepiece until you can see the whole image.|
|DO||Diameter of the objective. The "objective" can be either the large lens at the front of the telescope (in a refractor) or the large mirror at the back of the telescope (in a reflector).|
|fe||Focal length of the eyepiece. The distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.|
|fO||Focal length of the objective. The distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.|
|fR||f-Ratio. Simply the ratio of the focal length to the diameter
of the objective, or fO/DO. That means it's the
number of lens diameters from the lens to its focal point, as shown
below. This is written "f/" and then the value. Often given along
with the diameter of the objective to describe the scope.
|FOVe||Field of view of the eyepiece. A measure of the area you can see when looking through the eyepiece alone. This is expressed as the angle from one side of the area to the other (with you at the vertex). The two parameters fe and FOVe are the two primary specifications for the eyepiece.|
|FOVscope||Field of view of the scope. Tells you how much of the sky you see in the image in the telescope. This is the distance from one side of the eyepiece image to the other, expressed in degrees or minutes of arc across the sky.|
|Lmag||Star magnitude limit. The faintest star you can see in the scope, expressed in terms of star magnitude. The majority of visible stars have magnitudes in the range of 1-6, where the brightest stars have the lowest magnitude number, near 1, and the faintest you can see by eye are magnitude 6. The scope can show stars as faint as magnitude 16 or even higher (i.e. fainter), depending on its diameter.|
|M||Magnification. The apparent increase in size of an object when looking through the telescope, compared with viewing it directly.|
|PR||Resolving Power. The smallest separation between two stars that can possibly be distinguished with the scope. This is an indication of the finest detail the scope is capable of seeing -- regardless of the magnifying power.|
|SB||Surface Brightness. The brightness, or more correctly the brightness density, of bodies that cover an area, such as planets, nebulae and galaxies. These are distinct from bodies that appear as points, such as stars or asteroids. When an object has area, that area gets larger as magnification is increased, so the light is spread thinner. As a result surface brightness decreases with increasing magnification.|
So here they are, collected all together. Click on the "Explain This" button next to an equation to see the theory and practice to back it up -- how it came to be, what it means, and how to use it.
Note: I've made these equations as simple and easy to use as possible, so you can do most, if not all, of the calculations in the field in your head.
An important pointer: always work in millimeters. The fact that scope values -- especially the diameter -- are sometimes given in inches and sometimes in centimeters can make things more complicated than they need to be. Millimeters work out really well, so always convert if necessary. Multiply inches by 25 and centimeters by 10 to convert to millimeters.
|Term Computed||Equation||Theory & Practice|
|Resolving Power (in arcseconds)||Explain This|
|Scope Field of View||Explain This|
What the Eye Resolves at Magnification M
|Explain This||Diameter of the Exit Pupil||or||Explain This|
|Focal Length of the Eyepiece||Explain This|
|Star Magnitude Limit
Faintest Star Magnitude the Scope Can See
As a Percent of Maximum Brightness
|Term||Exit Pupil||M||fe||SB||Theory & Practice|
Highest Brightness the Scope Can Deliver
|7 mm||DO/7||7×fR||100%||Explain This|
Balance of Performance for Deep Sky Observing
|5 mm||DO/5||5×fR||50%||Explain This|
Best Match of Scope & Eye
|2 mm||DO/2||2×fR||8%||Explain This|
Highest Detail the Scope Can Deliver
|1 mm||DO||fR||2%||Explain This|
Your questions and comments regarding this page are welcome.
You can e-mail Randy Culp for inquiries,
suggestions, new ideas or just to chat.
Updated 11 May 2019<|endoftext|>
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Figure 1. Covalent bonding in oxygen molecule, O2.
Figure 2. Covalently bonded molecules have individual molecular orbitals holding different atoms together, unlike the balanced geometric cluster of an ionic compound. Because of the forces of repulsion between the electron clouds surrounding adjacent nuclei, even small molecules have distinct shapes, which may differ quite markedly from one another as in the half-dozen common molecules here. (A) nitrogen dioxide, NO2. (B) Carbon dioxide, CO2. (C) Phosgene, COCl2. (D) Sulfur dioxide, SO2. (E) Water, H2. (F) Ammonia, NH3.
A covalent bond is a chemical bond formed when two atoms share two valence electrons, one contributed by each of the atoms (see Figure 1). Unlike ionic bonds, covalent bonds connect specific atoms and so can give rise to true, discrete molecules. Substances with covalent bonds – mostly organic compounds – tend to have low melting and boiling points and to be soluble in nonpolar solvents.
The electrons in a covalent bond are shared equally only when the atoms are identical. In most covalent bonds the electrons are held to a greater extent by one atom than by the other. See also electronegativity.
The covalent bond is the classical electron pair or homopolar bond of chemistry, particularly of organic chemistry. It is a strong bond: the bond between two carbon atoms in diamond, for example, has a cohesive energy of 7.3 eV with respect to separated neutral atoms. This is comparable with the bond strength in ionic crystals, in spite of the fact that the covalent bond acts between neutral atoms. The covalent bond has strong directional properties. Thus carbon, silicon, and germanium have the diamond structure, with atoms joined to four nearest neighbors at tetrahedral angles, even though this arrangement gives a low filling of space.
The covalent bond is usually formed from two electrons, one from each atom participating in the bond. The electrons forming the bond tend to be partly localized in the region between the two atoms joined by the bond. The spins of the two electrons in the bond are antiparallel.
Shapes of covalent compounds
The shape of a covalently bonded molecule (see Figure 2) depends on the shapes of the orbitals occupied by the electrons, both those involved in chemical bonding and any others in the outermost shells of the individual atoms. A water molecule, for example, in which two hydrogen atoms are each singly bonded to a central carbon atom, might be visualized as the three atoms joined in a straight line. Electrons, all being negatively charged, repel each other, so this in-line arrangement might seem best. It would mean that the volumes of electron probability around the two hydrogen atoms are as far apart from each other as they possibly could be, thus reducing repulsion to a minimum. But, in addition to the two electrons from the oxygen's outer shell, which are involved in bonding, there are four other electrons in this shell, situated as two 'lone pairs' in filled orbitals. Their effect on neighboring electron clouds has to be taken into account. When this is done it is found that a shape almost like that of one segment of a diamond is adopted. It is a tetrahedral structure in which the hydrogen atoms and the two pairs of oxygen electrons (not involved in bonding) all lie as far away from each other as possible.
Methane, in which one carbon atom is surrounded by four carbon atoms and all the outermost carbon electrons are involved in bonding, adopts a regular tetrahedral configuration. Intermediate in structure between methane and water is ammonia (NH3(, in which there are three nitrogen-hydrogen bonds and one filled orbital containing a lone pair of nitrogen electrons.<|endoftext|>
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# How to find the minimum value of $\frac{x}{2x+3y}+\frac{y}{y+z}+\frac{z}{z+x}$?
Let $x,y,z\in [1,4]$ such that $x \geq y$ and $x \geq z$.
Find the minimum value of this expression: $$P=\frac{x}{2x+3y}+\frac{y}{y+z}+\frac{z}{z+x}$$
• Since $x \not= 0$, denote $y= u x$ and $z=v x$. You are, then, looking for minimum in $\frac{1}{4} \le u \le 1$ and $\frac{1}{4} \le v \le 1$. Aug 26, 2011 at 6:06
• @Sasha: following to your suggestion, we get $P=\frac{1}{2+3u}+\frac{u}{u+v}+\frac{v}{v+1}$ What then shall we do? Aug 26, 2011 at 9:48
• or you can transform into polar coordinates, reduce the variables to two(same as DKhanh does), then apply second partial derivative test Aug 26, 2011 at 10:54
As has been mentioned in comments, we want to find the minimum value of $$P=\frac{1}{2+3u}+\frac{u}{u+v}+\frac{v}{v+1}$$ for $u,v\in[\frac{1}{4},1]$. Take partials of $P$ with respect to $u$ and $v$: \begin{align} \frac{\partial P}{\partial u}&=-\frac{3}{(2+3u)^2}+\frac{v}{(u+v)^2}\\ \frac{\partial P}{\partial v}&=-\frac{u}{(u+v)^2}+\frac{1}{(v+1)^2} \end{align} To find an interior extremum, we need $\frac{\partial P}{\partial u}=\frac{\partial P}{\partial v}=0$. In that case, we need $$0=u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}=-\frac{3u}{(2+3u)^2}+\frac{v}{(v+1)^2}$$ However, for $u,v\in[\frac{1}{4},1]$, we have $$\begin{array}{c} \frac{12}{121}\le\frac{3u}{(2+3u)^2}\le\frac{3}{25}\\ \frac{4}{25}\le\frac{v}{(v+1)^2}\le\frac{1}{4} \end{array}$$ Thus, $u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}\ge\frac{1}{25}$, so there can be no interior extremum.
Because $(u,v)\cdot\nabla P=u\frac{\partial P}{\partial u}+v\frac{\partial P}{\partial v}\ge\frac{1}{25}$ everywhere in $[\frac{1}{4},1]\times[\frac{1}{4},1]$, the minimum must be taken on the left or bottom edge of that square; i.e. $u=\frac{1}{4}$ or $v=\frac{1}{4}$.
$u=\frac{1}{4}$: $\frac{\partial P}{\partial v}=-\frac{4}{(1+4v)^2}+\frac{1}{(v+1)^2}=\frac{12v^2-3}{(1+4v)^2(v+1)^2}$ which vanishes at $v=\frac{1}{2}$.
Because $P(\frac{1}{4},\frac{1}{4})=\frac{117}{110}$ and $P(\frac{1}{4},\frac{1}{2})=\frac{34}{33}$ and $P(\frac{1}{4},1)=\frac{117}{110}$, $P(\frac{1}{4},\frac{1}{2})$ is a local minimum.
$v=\frac{1}{4}$: $\frac{\partial P}{\partial u}=-\frac{3}{(2+3u)^2}+\frac{4}{(4u+1)^2}=\frac{-12u^2+24u+13}{(2+3u)^2(4u+1)^2}$ which does not vanish on $[\frac{1}{4},1]$.
Because $P(\frac{1}{4},\frac{1}{4})=\frac{117}{110}$ and $P(1,\frac{1}{4})=\frac{33}{20}$, $P(\frac{1}{4},\frac{1}{4})$ is a local minimum.
Thus, $P(\frac{1}{4},\frac{1}{2})=\frac{34}{33}$ is the minimum of P for $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$.
• I just noticed the algebra-precalculus tag. Using partials might not be good.
– robjohn
Aug 26, 2011 at 15:02
• why don't you apply the condition $x>y , x>z$? Aug 27, 2011 at 2:29
• @DKhanh: Following Sasha's comment, $u = y/x$, $v = z/x$. Since $y\le x$ and $x,y\in[1,4]$, we have that $u\in[\frac{1}{4},1]$. Since $z\le x$ and $x,z\in[1,4]$, we have that $v\in[\frac{1}{4},1]$. I used that $(u,v)\in[\frac{1}{4},1]\times[\frac{1}{4},1]$.
– robjohn
Aug 27, 2011 at 3:58
The answer is $\frac{34}{33}$.
To see this, start with the fact that the derivative $\partial_zP(x,y,z)$ of $P(x,y,z)$ with respect to $z$ has the sign of $(x-y)(z^2-xy)$.
1. Assume that $x>y$.
Then $\partial_zP<0$ at $z=1$ hence $P$ is not minimal at $(x,y,1)$ and $\partial_zP>0$ at $z=x$ hence $P$ is not minimal at $(x,y,x)$. Thus $z^2=xy$.
Define $Q$ by $Q(w)=\dfrac1{2+3w^2}+\dfrac{2w}{1+w}$, then $P(x,y,\sqrt{xy})=Q(u)$ with $u=\sqrt{y/x}$ hence $u\in[\frac12,1]$. Now, $Q''>0$ on the interval $[\frac12,1]$ and $Q'(\frac12)>0$ hence $Q'>0$ on $[\frac12,1]$. Thus, $Q(u)\ge Q(\frac12)$ for every $u$ in $[\frac12,1]$.
2. Assume that $x=y$.
Then $P(x,x,z)=1+\frac15$ for every $z$.
Finally $Q(\frac12)=1+\frac1{33}<1+\frac15$ hence $P$ is minimum at $(4,1,2)$ where its value is $P(4,1,2)=Q(\frac12)=1+\frac1{33}$.<|endoftext|>
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# Problems on Time and Work
Q. 1. A completes a work in 12 days and B complete the same work in 24 days. If both of them work together,
then the number of days required to complete the work will be
(a) 8 (b) 6 (c) 7 (d) 5 (e) None of these
Ans: (a) 8
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Therefore, here, the required number of days = 12 24/ 36 = 8 days. Ans.
Q. 2. If 4 men can colour 48 m long cloth in 2 days, then 6 men can colour 36 m long cloth in
(a) 1 day (b) 1 days (c) day (d) day (e) None of these
Ans: (a) 1 day
Explanation:
The length of cloth painted by one man in one day = 48 / 4 2 = 6 m
No. of days required to paint 36 m cloth by 6 men = 36/ 6 6 = 1 day. Ans.
Q. 3. If 3 persons can do 3 times of a particular work in 3 days, then, 7 persons can do 7 times of that work in
(a) 7 days (b) 6 days (c) 4 days (d) 3 days (e) None of these
Ans: (d) 3 days.
Explanation:
That is, 1 person can do one time of the work in 3 days.
Therefore, 7 persons can do 7 times work in the same 3 days itself. Ans.
Q. 4. Mangala completes a piece of work in 10 days, Raju completes the same work in 40 days. If both of them
work together, then the number of days required to complete the work is
(a) 15 (b) 10 (c) 9 (d) 8 (e) None of these
Ans: ( d) 8
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
That is, the required No. of days = 10 40/50 = 8 days. Ans.
Q. 5. 12 men work 8 hours per day to complete the work in 10 days. To complete the same work in 8 days,
working 15 hours a day, the number of men required
(a) 4 (b) 5 (c) 6 (d) 8 (e) None of these
Ans: (d) 8
Explanation:
That is, 1 work done = 12 8 10
Then, 12 8 10 = ? 15 8
? (i.e. No. of men required) = 12 8 10/15 10 = 8 days. Ans.
Q. 6. If 5 people undertook a piece of construction work and finished half the job in 15 days. If two people drop
out, then the job will be completed in
(a) 25 days (b) 20 days (c) 15 days (d) 10 days (e) None of these
Ans: (a) 25 days
Explanation:
That is, half the work done = 5 15
Then, 5 15 = 3 ? 1/2
i.e. 5 15 = 3 ?
therefore, ? (No. days required) = 5 15/3 = 25 days. Ans.
Q. 7. 30 labourers working 7 hours a day can finish a piece of work in 18 days. If the labourers work 6 hours a
day, then the number of labourers required to finish the same piece of work in 30 days will be
(a) 15 (b) 21 (c) 25 (d) 22 (e) None of these
Ans: (b) 21
Explanation:
That is, 1 work done = 30 7 18 = ? 6 30
? (No. of labourers) = 30 7 18/6 30
= 21. Ans.
Q. 8. If 5 girls can embroider a dress in 9 days, then the number of days taken by 3 girls will be
(a) 20 days (b) 10 days (c) 14 days (d) 15 days (e) None of days
Ans: (d) 15 days
Explanation:
That is, 5 9 = 3 ?
? = 5 9/3 = 15 days. Ans.
Q. 9. A and B together can plough a field in 10 hours but by himself A requires 15 hours. How long would B
take to plough the same field?
(a) 10 hours (b) 20 hours (c) 30 hours (d) 40 hours (e) None of these
Ans: (c) 30 hours
Explanation:
If A and B together can do a piece of work in x days and A alone can do the same work in y days, then B alone
can do the same work in x y/ y x days.
Therefore, the No. of hours required by B = 10 15/ 15 10 = 150/5 = 30 hours. Ans.
Q. 10. 16 men or 20 women can finish a work in 25 days. How many days 28 men and 15 women will take to
finish this job?
(a) 19 13/43 days (b) 11 27/43 days (c) 8 days (d) 10 days (e) None of these
Ans: (d) 10 days
Explanation:
16 men = 20 women
Therefore, 1 women = 16/20 men = 4/5 men
15 women = 4/5 15 men = 12 men
i.e. 28 men + 15 women = 28 men + 12 men = 40men
1 work done by men = 16 25
16 25 = 40 ?
? ( no. of days) = 16 25/40 = 10 days. Ans.
Q. 11. A can do a piece of work in 5 days and B can do the same work in 10 days. How many days will both
take to complete the work?
(a) 5 days (b) 3 1/3 days (c) 3 days (d) 6 days (e) None of these
Ans: (b) 3 1/3 days
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
That is, the required No. of days = 5 10/15 = 3 1/3 days. Ans.
Q. 12. If 12 men can do a piece of work in 24 days, then in how many days can 18 men do the same work?
(a) 36 (b) 20 (c) 18 (d) 16 (e) None of these
Ans: (d) 16
Explanation:
1 work done = 12 24
Then, 12 24 = 18 ? days
## ? days = 12 24/18 = 16. Ans.
Q. 13. A group of workers accepted to do a piece of work in 25 days. If 6 of them did not turn for the work and
the remaining workers did the work in 40 days, then the original number of workers was
(a) 22 (b) 20 (c) 18 (d) 16 (e) None of these
Ans: (d) 16
Explanation:
Let the original number of workers be x
Then, x 25 = (x 6) 40
25x = 40x -240
240 = 40x -25x = 15x
x = 240/15 = 16. Ans.
Q. 14. If 8 men or 12 women can do a piece of work in 10 days, then the number of days required by 4 men
and 4 women to finish the work is
(a) 8 (b) 10 (c) 12 (d) 4 (e) None of these
Ans: (c) 12
Explanation:
8 men = 12 women
1 woman = 8/12 men = 2/3 men
4 women = 2/3 4 men = 8/3 men
4 men + 4 women = 4 + 8/3 men = 20/3 men
1 work done = 8 10
8 10 = 20/3 ?days
?days = 8 10 3/20 = 12 days. Ans.
Q. 15. If 8 men can dig a well in 18 days, then the number of days, 12 men will take to dig the same well will be
(a) 12 days (b) 10 days (c) 8 days (d) 16 days (e) None of these
Ans: (a) 12 days
Explanation:
Work done = 8 18
Then, 8 18 = 12 ? days
? days = 8 18/12 = 12 days. Ans.
Q. 16. 39 men can repair a road in 12 days working 5 hours a day. In how many days will 30 men working 6
hours peer day complete the work?
(a) 10 (b) 13 (c) 14 (d) 15 (e) None of these
Ans: (b) 13
Explanation:
1 work done = 39 12 5
39 12 5 = 30 6 ?days
? days = 39 12 5/ 30 6 = 13 days. Ans.
Q. 17. A certain number of men can do a work in 40days. If there were 8 men more, it could be finished in 10
days less. How many men were there initially?
(a) 30 (b) 24 (c) 16 (d) 20 (e) None of these
Ans: (b) 24
Explanation:
Let x be the initial number of men
Then, 1 work done = x 40
Then, x 40 = (x + 8 ) (40 10)
40x = 30x + 240
10x = 240
Therefore, x = 240/10 = 24 men. Ans.
Q. 18. If 4 men or 6 boys can finish a work in 20 days. How long will 6 men and 11 boys take to finish the same
work?
(a) 10 days (b) 6 days (c) 4 days (d) 3 days (e) None of these
Ans: (b) 6 days
Explanation:
4 men = 6 boys
Then, 1 boy = 4/6 men = 2/3 men
11 boys = 2/3 11 men = 22/3 men
Then, 6 men + 11 boys = 6 + 22/3 men = 40/3 men
1 work done = 4 men 20days
That is, 4 20 = 40/3 ? days
? days = 4 20 3/40 = 6 days. Ans.
Q. 19. A works twice as fast as B. if B can complete a work in 12 days independently. The number of days in
which A and B can together finish the work?
(a) 18 days (b) 8 days (c) 6 days (d) 4 days (e) None of these
Ans: (d) 4 days
Explanation:
If B takes 12 days to finish the work, then A takes 6 days to finish the same work.
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Therefore, the Number of days taken by A and B together to finish the same work = 6 12/18
= 4 days. Ans.
Q. 20. A and B can do a piece of work in 4 days. If A can do it alone in 12 days, B will finish the work in
(a) 4 days (b) 6 days (c) 8 days (d) 10 days (e) None of these
Ans: (b) 6 days
Explanation:
If A and B together can do a piece of work in x days and A alone can do the same work in y days, then B alone
can do the same work in x y/ y x days.
Therefore, the number of days B will take to finish the work = 4 12/ 12 4
= 48/8 = 6 days. Ans.
Q. 21. 5 men can do a piece of work in 6 days while 10 women can do it in 5 days. In how many days can 5
women and 3 men do it?
(a) 4 (b) 5 (c) 6 (d) 8 (e) None of these
Ans: (b) 5
Explanation:
5men 6 = 10 women 5
30 men = 50 women
1 woman = 30/50 men = 3/5 men
5 women = 3/5 5 men = 3 men
5 women + 3 men = 3 + 3 = 6 men
That is, 1 work done = 5 6
5 6 = 6 ?days
? days = 5 6/6 = 5 days. Ans.
Q. 22. A work could be completed in 100 days. However, due to the absence of 10 workers, it was completed in
110 days then, the original number of workers was
## (a) 100 (b) 110 (c) 55 (d) 50 (e) None of these
Ans: (b) 110
Explanation:
Let x be the original number of workers
Then, x 100 = ( x 10) 110
100 x = 110x 1100
1100 = 10 x
x = 110. Ans.
Q. 23. A particular job can be completed by a team of 10 men in 12 days. The same job can be completed by a
team of 10 women in 6 days. How many days are needed to completed the job if the two teams work together?
(a) 4 (b) 6 (c) 9 (d) 18 (e) None of these
Ans: (a) 4
Explanation:
Consider mens work days as one group and womens working days as other group
Then, the required No. of days = 12 6/18 = 4 days. Ans.
Q. 24. A can do a piece of work in 12 days, while B can do it in 8 days with the help of C, they finish the work in
4 days. Then C alone can do the work in
(a) 24 days (b) 20 days (c) 16 days (d) 4 days (e) None of these
Ans: (a) 24 days
Explanation:
A and B together can do the work in 12 8/20 = 96/20 days
A +B and C together do the work in 4 days.
Then C alone can do the work in 96/20 4/96/20 4 days
= 96/5/16/20 = 96/5 20/16 = 24 days. Ans.
Q. 25. A job can be completed by 12 men in 12 days. How many extra days will be needed to complete the job
if 6 men leave after working for 6 days?
(a) 3 (b) 12 (c) 6 (d) 24 (e) None of these
Ans: (c) 6
Explanation:
Half the work is done by 12 men in 6 days.
Remaining half should be done by 6 men in ? days
Then, 12 6 = 6 ?days
36 = 3 ?days
? days = 36/3 =12 days
Extra days required = 12 -6 = 6 days. Ans.
Q. 26. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long should it
take, worker A and worker B, working together but independently, do the same job?
(a) 4 1/9 hours (b) 4 2/9 hours (c) 4 4/9 hours (d) 4 5/9 hours (e) None of these
Ans: (c) 4 4/9 hours
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Then 8 10/18 = 4 4/9 hours. Ans.
Q. 27. Three workers working all days can do a work in 10 days. But one of them having other employment can
work only half time. In how many days the work can be finished?
(a) 15 days (b) 16 days (c) 12 days (d) 12.5 days (e) None of these
Ans: (c) 12 days.
Explanation:
Sum of each persons one days work = 1/3 10 + 1/3 10 + 1/3 20 = 1/30 + 1/30 + 1/60
= 5/60 = 1/12
Therefore, the work can be finished in 12 days. Ans.
Q. 28. Sam, Bob and Kim can do a job alone in 15 days, 10 days and 30 days respectively. Sam is helped by
Bob and Kim every third day. In how many days will the job be completed?
(a) 9 days (b) 8 1/3 days (c) 8 days (d) 6 1/3 days (e) None of these
Ans: (a) 9 days
Explanation:
The work done by the three persons in 3 days ( because Sam works only on every third day)
= 3/15 + 1/10 + 1/30 = 6 + 3 + 1/30 = 10/30
= 1/3 of the work
Therefore, the job will be completed in 3 3 = 9 days. Ans.
Q. 29. Prakash alone can complete a job in 12 hours, Jayant alone can complete the same job in 6 hours. How
many hours will they take together to complete the job?
(a) 4 (b) 6 (c) 2 (d) 8 (e) None of these
Ans: (a) 4
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Here, the No. of hours required to complete the work , working together = 12 6/18
= 4 hours. Ans.
Q. 30. 56 men completes a work in 12 days. How many men will be required to complete the work in 16 days?
(a) 38 (b) 24 (c) 42 (d) 48 (e) None of these
Ans: (c) 42
Explanation:
1 work done = 56 men 12 days
Then, 56 12 = ? men 16
? men = 56 12/16 = 42 men. Ans.
Q. 31. Six men or tern boys can do a piece of work in fifteen days. How long would it take for 12 men and 5
boys to do the same piece of work?
(a) 6 days (b) 28 days (c) 25 days (d) 7 days (e) None of these
Ans: (a) 6 days
Explanation:
6 men = 10 boys
Then, 1 boy = 6/10 men = 3/5 men
Then, 5 boys = 3/5 5 = 3 men
12 men + 5 boys = 15 men
1 work done = 6 men 15 days
Therefore, 6 15 = 15 ? days
? days = 6 15/15 = 6 days. Ans.
Q. 32. If 4 workers can make 42 toys in 6 days, how many toys can 12 workers make in 3 days?
(a) 63 (b) 28 (c) 252 (d) 7 (e) None of these
Ans: (a) 63
Explanation:
4 workers in 3 days make 21 toys,
Then, 1 worker can make 21/4 toys in 3 days
## Therefore, No. of toys made by 12 workers in 3 days = 21/4 12 = 63 toys. Ans.
Q. 33. 10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days.
If all the 10 men and 15 women work together, in how many days will work get completed?
(a) 6 (b) 7 2/3 (c) 6 2/3 (d) 6 1/3 (e) None of these
Ans: (c) 6 2/3
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Here, mens working days can be taken as A and that of women as B
Then, 15 12/27 = 6 2/3 days.Ans.
Q. 34. Three men, four women and six children can complete a work in 7 days . A woman does double the work
a man does and a child does half the work a man does. How many women can complete this work in 7 days?
(a) 8 (b) 7 (c) 12 (d) 10 (e) None of these
Ans: ( b) 7
Explanation:
3 men = 3/2 women
1 child = 1/2/2 = woman
Then, 6 children = 6 = 6/4 = 3/2 women
Then, 3/2 women + 4 women + 3/2 women = 7 women
7 women does the work in 7 days.
Therefore, the No. of women required to finish the work in 7days = 7. Ans
Q. 35. A sum of money is sufficient to pay Ps wages for 25 days or Qs wages for 30 days. The money is
sufficient to pay the wages of both for
(a) 13 5/11 days (b) 13 days (c) 13 6/11 days (d) 13 7/11 days (e) None of these
Ans: (d) 13 7/11
Explanation:
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Here, wages can be taken as days,
Then, 25 30/ 55 = 13 7/11 days. Ans.
Q. 36. A certain number of men complete a piece of work in 50 days. If there were 8 men more the work could
be finished in 10 days less. How many men were originally there?
(a) 40 (b) 50 (c) 32 (d) 25 (e) None of these
Ans: (c) 32
Explanation:
Let the original No. of men = x
Then, I work done = 50x
Then, 50x = (x + 8) 40
50x = 40x + 320
10 x = 320
x = 320/10 = 32 men. Ans.
Q. 37. A and B together finish a work in 30 days they worked for it for 20 days and then B left. The remaining
work was done by A alone in 20 more days. B alone can finish the work in
(a) 48 days (b) 50 days (c) 54 days (d) 60 days (e) none of these
Ans: ( d) 60 days
Explanation:
(A + B)s 1 days work = 1/30
## A+B s 20 days work = 1/30 20 = 2/3
Remaining work = 1 -2/3 = 1/3
1/3 of the work is done by A in 20 days
Therefore, A can complete the full work in 3 20 = 60 days
So, As 1 days work = 1/60
Then, Bs 1 day work = 1/30 1/60 = 1/60
Therefore, B can complete the work in 60 days. Ans.
Q. 38. A can complete a job in 9 days, B in 10 days and C in 15 days. B and C start the work together and are
forced to leave after 2 days. The time taken to complete the remaining work is
(a) 13 days (b) 10 days (c) 9 days (d) 6 days (e) None of these
Ans: (d) 6 days
Explanation:
As 1 day work = 1/9
Bs 1 day work = 1/10
Cs 1 day work = 1/15
B + C s 1 day work = 1/10 + 1/15 = 1/6
B + Cs 2 days work = 1/6 2 = 1/3
The work remaining = 1 1/3 = 2/3
A can do the full work in 9 days.
So, the remaining 2/3 of the work is done by A in 2/3 9 days = 6 days. Ans.
Q. 39. If 40 men working 9 hours a day can finish a work in 21 days, in how many days will 27 men working 10
hours a day do the same work?
(a) 20 days (b) 28 days (c) 29 days (d) 25 days (e) None of these
Ans: (b) 28 days
Explanation:
That is, 1 work done = 40 9 21
Then, 40 9 21 = 27 10 ? days
Therefore, ? days = 40 9 21 /27 10 = 28 days. Ans.
Q. 40. 8 children and 12 men complete a certain piece of work in 9 days. Each child takes twice the time taken
by a man to finish the work. In how many days will 12 men finish the same work?
(a) 8 (b) 15 (c) 9 (d) 12 (e) None o f these
Ans: (d) 12
Explanation:
8 children = 4men
Then, 4 + 12 = 16 men completes the job in 9 days
Then, 16 9 = 12 ? days
? days = 16 9/12 = 12 days. Ans.
Q. 41. Rajesh and Ajay can complete a job in 16 days. Rajesh alone can do it in 24 days. How long will Ajay
alone take to finish the whole work?
(a) 20 days (b) 48 days (c) 30 days (d) 36 days (e) 28 days
Ans: (b) 48 days
Explanation:
If A and B together can do a piece of work in x days and A alone can do the same work in y days, then B alone
can do the same work in x y/ y x days.
Then, Ajay alone can finish the whole work in, 16 24/24 16 = 16 24/8 = 48 days.Ans.
Q. 42. Ram, Shyam and Mohan can do a piece of work in 12, 15 and 20 days respectively. How long will they
take to finish it together?
(a) 10 days (b) 12 days (c) 14 days (d) 8 days (e) 5 days
Ans: (e) 5 days
Explanation:
When A, B and C can do a work in x, y and z days respectively. Then, the three of them together can finish the
work in xyz/ x y + y z + x z days
Therefore, the No. of days taken by them together to finish the whole work
= 12 15 20/12 15 + 15 20 + 12 20
= 12 15 20/720 = 5 days. Ans.
Q. 43. Tapas works twice as much as Mihir. If both of them finish the work in 12 days, Tapas alone can do it in:
(a) 20 days (b) 24 days (c) 18 days (d) 20 days (e) None of these
Ans: (c)18 days
Explanation:
Mihir = 2 Tapas (Consider, Tapas as T and Mihir as M)
So, T 2 T/ T + 2T = 12
2 T T/3 T = 12
2 T = 3 12 = 36
T = 36/2 = 18 days i.e. No. of days taken by Tapas alone = 18 days. Ans.
Q. 44. Xavier can do a job in 40 days. He worked on it for 5 days and then Paes finished it in 21 days. In how
many days Xavier and Paes can finish the work?
(a) 10 (b) 15 (c) 20 (d) 840/61 (e) 12
Ans: (b) 15
Explanation:
Xaviers one days work = 1/40
His 5 days work = 1/40 5 = 1/8
The work remaining = 1 1/8 = 7/8
7/8 of the work is done by Paes in 21 days
Then, his 1 days work = 7/8/21 = 7/21 8 = 1/24
Therefore, the time taken by Paes to finish the whole work = 24 days.
Then. The No. of days two of them together take to finish the work = 40 24/40 + 24
= 40 24/64 = 15 days. Ans.
Q. 45. A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working
together, what part of the same work they can finish in a day?
(a) 1/6 (b) 1/9 (c) 2/5 (d) 2/7 (e) None of these
Ans: (a) 1/6
Explanation:
B alone can do the same work in 9 days
So, the No. of days both of them together take to do the work = 18 9/ 18 + 9
=18 9/27 = 6 days.
The part of the work, they together finish in one day = 1/6. Ans.
Q. 46. A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will
complete the same work in:
(a) 1/24 day (b) 7/24 day (c) 3 3/7 days (d) 4 days (e) None of these
Ans: (c) 3 3/7 days
Explanation:
When A, B and C can do a work in x, y and z days respectively. Then, the three of them together can finish the
work in xyz/ x y + y z + x z days
Therefore, the No. of days taken by them together = 24 6 12/24 6 + 6 12 + 24 12
## = 24 6 12/ 504 = 3 3/7 days. Ans.
Q. 47. A man can do a piece of work in 5 days, but with the help of his son, he can do it in 3 days. In what time
can the son do it alone?
(a) 6 days (b) 7 days (c) 7 days (d) 8 days (e) None of these
Ans: (c) 7 days
Explanation:
If A and B together can do a piece of work in x days and A alone can do the same work in y days, then B alone
can do the same work in x y/ y x days.
Therefore, his son alone can do the work in 3 5/ 5 3 = 15/2 = 7 days. Ans.
Q. 48. A works twice as fast as B. if B can complete a work in 12 days independently, the number of days in
which A and B can together finish the work is:
(a) 4 days (b) 6 days (c) 8 days (d) 18 days (e) None of these
Ans: (a) 4 days
Explanation:
That is, if b alone can finish the work in 12 days,
A alone can finish the work in 6 days.
If A can complete a work in x days and B can complete the same work in y days, then, both of them together
can complete the work in x y/ x+ y days.
Therefore,
A and B can together finish the work in 6 12/18 days = 4 days. Ans.
Q. 49. A is twice as good a workman as B and together they finish a piece of work in 14 days. The number of
days taken by A alone to finish the work is:
(a) 11 (b) 21 (c) 28 (d) 42 (e) None of these
Ans: (b) 21
Explanation:
As work is twice as Bs work
So, we can consider 2A = B
Then, A 2A/ A+ 2A = 14
2A A/ 3A = 14
2 A = 14 3
A = 21 days. i.e. A alone can finish the work in 21 days. Ans.
Q. 50. A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. Working
together, they can do it in:
(a) 20 days (b) 22 days (c) 25 days (d) 30 days (e) None of these
Ans: (b) 22 days
Explanation:
That is, 3A = B, replacing 3 A in place of B, we get
3A A = 60 days
A = 30 days, i.e. A alone can finish the work in 30 days
Then, B alone can finish the work in 3 30 = 90 days
Therefore, working together they can finish it in 30 90/ 120 days. = 22 days. Ans.
Q. 51. A and B can do a job together in 7 days. A is 1 times as efficient as B. the same job can be done by A
alone in:
(a) 9 1/3 days (b) 11 days (c) 12 days (d) 16 1/3 days (e) None of these
Ans: (b) 11 days
Explanation:
1 = 7/4
## 7/4 A = B, replacing B by 7/4 A, we get
A 7/4 A/ A + 7/4 A = 7days
7 A A/4 / 11A/4 = 7 A A/ 11 A = 7A /11 = 7
A = 77/7 = 11, So, A alone can finish the work in 11 days. Ans.
Q. 52. Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days
taken by Tanya to do the same piece of work is:
(a) 15 (b) 16 (c) 18 (d) 25 (e) None of these
Ans: (b) 16
Explanation:
The ratio of their efficiency, i.e. Sakshi : Tannya = 100 : 125 = 4 : 5
Therefore, ratio of time taken by them = 5 : 4
That is, No. of days taken by Sakshi is 5 units of ratio = 20 days
So, 1 unit of ratio = 20/5 = 4 days
Therefore, No. of days taken by Tannya = 4 4 = 16 days. Ans.
Q. 53. A is 30% more efficient than B. How much time will they, working together, take to complete a job which
A alone could have done in 23 days?
(a) 11 days (b) 13 days (c) 20 3/17 days (d) 14 days (e) None of these
Ans: (b) 13 days
Explanation:
The ratio of their efficiency = A : B = 130 : 100 = 13 : 10
Therefore, the ratio of the time taken by them = 10 : 13
A takes 23 days = 10 units of ratio
So, 1 unit of ratio = 23/10
Time taken by B alone to finish the work = 23/10 13 = 29.9 = 30 days
Therefore, the No. of days taken to finish the work working together = 23 30/ 53
= 13 days. Ans.
Q. 54. A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the
work, how much time shall B take to do it ?
(a) 30 days (b) 35 days (c) 40 days (d) 36 days (e) None of these
Ans: (a) 30 days
Explanation:
That is , A = 3B/2/1/2 = 3B/2
Replacing A by 3B/2, we get
3B/2 B/3B/2 + B = 18 days
3 B B/5B =18
3B = 90
B = 30 days, B alone can finish the work in 30 days. Ans.
Q. 55. A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of
the work that is left is:
(a) (b) 1/10 (c) 7/15 (d) 8/15 (e) None of these
Ans: (d) 8/15
Explanation:
Working together, the No. of days taken by them to finish the work = 15 20/35
= 60/7 days.
Then, their 1 days work = 7/60
Their, 4 days work = 7 4/60 = 7/15 of the total work
Therefore, the fraction of the work, that is left = 1- 7/15 = 8/15. Ans.
Q. 56. A can finish a work in 18 days and B can do the work in 15 days. B worked for 10 days and left the job.
In how many days, A alone can finish the remaining work
(a) 5 (b) 5 (c) 6 (d) 8 (e) None of these
Ans: (c) 6 days
Explanation:
Bs one days work = 1/15
Bs 10 days work = 1/15 10 = 2/3
Remaining work = 1/3
As 1 day work = 1/18
No. of days A take to finish 1/3 of the work = 1/3/1/18 = 1/3 18 = 6 days. Ans.
Q. 57. A and B can complete a work in 15 days and 10 days respectively. They started doing the work together
but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in:
(a) 8 days (b) 10 days (c) 12 days (d) 15 days (e) None of these
Ans: (c) 12 days
Explanation:
No. of days taken by them working together = 15 10/25 = 6 days
Their 1 days work = 1/6
Their 2 days work = 1/6 2 = 1/3
The work remaining = 2/3
As 1 day work = 1/15
Time taken by A to finish 1/3 of the work = 2/3/1/15 = 2/3 15 = 10 days
Therefore, the total number of days taken to finish the whole work = 10 +2 = 12 days. Ans.
Q. 58. A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to
leave after 3 days. The remaining work was done by A in:
(a) 5 days (b) 6 days (c) 10 days (d) 10 days (e) None of these
Ans: (c) 10 days
Explanation:
No. of days taken by B and C together to finish the work = 9 12/21 = 36/7 days
Their 1 day work = 7/36
So, their 3 days work = 7/36 3 = 7/12
The remaining work = 5/12
As 1 days work = 1/24
To finish the remaining 5/12 of the work, the No. of days taken by A = 5/12/1/24
= 5/12 24= 10 days. Ans.
Q. 59. A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10
hours while machine R can print them in 12 hours. All the machines are started at 9 a.m. while machine P is
closed at 11 a.m. and the remaining two machines complete the work. Approximately at what time will the work
be finished?
(a) 11: 30 a.m. (b) 12 noon (c) 12:30 p.m. (d) 1 p.m. (e) None of these
Ans: (d) 1 p.m.
Explanation:
The three machines 1 hours work = 1/8 + 1/10 + 1/12 = 37/120
Their, work from 9 a.m. to 11 a.m. = 2 hours work = 37/120 2= 27/60
The remaining work = 23/60
Machines Q and Rs 1 hours work = 1/10 + 1/12 = 11/60
Therefore, the time taken by Q and R to finish the remaining 23/60 work = 23/60/11/60
= 23/60 60/11 = 23/11 = 2 hours approx.
Therefore, the approximate time at which the work is finished = 11 + 2 = 1 p.m. Ans.
Q. 60. A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A
in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish
work?
(a) 18 days (b) 24 days (c) 30 days (d) 36 days (e) None of these
Ans: (a) 18 days
Explanation:
i.e. 2 ( A + B + C) s 1 days work = 1/30 + 1/24 + 1/20 = 1/8
their 1 days work = 1/8/2 = 1/16
their 10 days work = 1/16 10 = 5/8
the remaining work = 3/8
As 1 days work = 1/16 1/24 = 1/48
Therefore, the No. of days A alone will take to complete 3/8 of the work = 3/8/ 1/48
= 3/8 48 = 3 6 = 18 days. Ans.
Q. 61. X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then
after 4 days Y joined him till the completion of the work. How long did the work last?
(a) 6 days (b) 10 days (c) 15 days (d) 20 days (e) None of these
Ans: (b) 10 days
Explanation:
Xs 1 days work = 1/20
Xs 4 days work = 1/20 4 = 1/5
The remaining work = 4/5
X and Ys 1 day work = 1/20 + 1/12 = 4/30 = 2/15
Therefore, Both together finish the remaining work in 4/5/2/15 days
= 4/5 15/2 = 6 days
Therefore, the total number of days taken to finish the work = 4 + 6 = 10 days. Ans.
Q. 62. A and B can together finish a work in 30 days. They worked together for 20 days and then B left. After
another 20 days, A finished he remaining work. In how many days A alone can finish the job?
(a) 40 (b) 50 (c) 54 (d) 60 (e) None of these
Ans: (d) 60
Explanation:
A + B s 1 days work = 1/30
Their 20 days work = 1/30 20 = 2/3
Remaining work = 1/3
1/3 of the work is done by A in 20 days.
Then, whole work can be done by A in 3 20 = 60 days. Ans.
Q. 63. X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long
will they together take to complete the work?
(a) 13 1/3 days (b) 15 days (c) 20 days (d) 56 days (e) None of these
Ans: (a) 13 1/3 days
Explanation:
As 8 days work = 1/40 8 = 1/5
Remaining work = 4/5
4/5 of the work is finished by Y in 16 days
So, Y can finish the whole work alone in 16 5/4 days = 20days
They both together can finish it in 40 20/60 =13 1/3 days. Ans.
Q. 64. A does 4/5 of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days.
## How long B alone would take to do the whole work?
(a) 23 days (b) 37 days (c) 37 days (d) 40 days (e) None of these
Ans: (c) 37 days
Explanation:
A alone can finish the work in 20 5/4 days = 25 days
As 1 days work = 1/25
A and B together can finish the whole work in 5 3 days = 15 days
Their 1 days work = 1/15
Therefore, Bs 1 days work = 1/15 1/25 = 1/37
Therefore, B alone can finish the whole work in 37 days.Ans.
Q. 65. A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the
remaining work alone in 44 days. In how many days shall B finish the whole work alone?
(a) 30 days (b) 40 days (c) 60 days (d) 70 days (e) None of these
Ans: (c) 60 days
Explanation:
A + Bs 1 days work = 1/30
i.e. A + B = 1/30 --------- (i)
16 A + 44 B = 1 ------ (ii) ( i.e. 1 = the whole work done)
Multiplying (i) by 16 and subtracting it from (ii), we get
i.e. 16 A + 44 B = 1
16 A + 16B = 8/15
28 B =7/15
B = 1/60 i.e. Bs 1 days work = 1/60
Therefore, B alone can finish the work in 60 days. Ans.
Q. 66. A and B together can do a piece of work in 12 days, which B and C together can do in 16 days. After A
has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days C alone will do
the work?
(a) 16 (b) 24 (c) 36 (d) 48 (e) None of these
Ans: (b) 24
Explanation:
According to the question,
A + Bs 1 days work = 1/12
B+ Cs 1 days work = 1/16
A worked for 5 days, B for 7 days and C for 13 days. So, we can assume that,
A+ B has been working for 5 days and B+ C has been working for 2 days and C alone for 11 days.
i.e. As 5 days work + Bs 7 days work + Cs 13 days work = 1
(A+ B)s 5 days work + (B + C)s 2 days work + Cs 11 days work = 1
5/12 + 2/16 + Cs 11 days work = 1
So, Cs 11 days work = 1 (5/12 + 2/16) = 11/24
Cs 1 days work = 11/24/11 = 1/24
Therefore, C alone can finish the work in 24 days. Ans.
Q. 67. A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together
but A leaves after some days and then B completed the remaining work in 23 days. The number of days after
which A left the work was:
(a) 6 (b) 8 (c) 9 (d) 12 (e) None of these
Ans: (c) 9
Explanation:
## A and B together can finish the work in 45 40/85 = 360/17 days
A and Bs 1 days work = 17/360
As 1 days work = 1/45
Bs 1 days work = 1/40
Bs 23 days work 1/40 23 = 23/40
Remaining work = 1 23/40 = 17/40
17/40 of the work is done by A and B together
17/40 of the work is done by A and B together in = 17/40/17/360 = 17/40 360/17 days
= 9 days.
Therefore, A left after 9 days. Ans.
Q. 68. A can do a piece of work in 20 days which B can do in 12 days. B worked at it for 9 days. A can finish the
remaining work in:
(a) 3 days (b) 5 days (c) 7 days (d) 11 days (e) None of these
Ans: (b) 5 days
Explanation:
A s 1 days work = 1/20
Bs 1 days work = 1/12
Bs 9 days work = 1/12 9 =
Remaining work =
A can finish the remaining work in = 1/4/1/20 = 20 = 5 days. Ans.
Q. 69. A and B together can complete a work in 3 days. They start together. But, after 2 days, B left the work. If
the work is completed after 2 more days, B alone could do the work in
(a) 5 days (b) 6 days (c) 9 days (d) 10 days (e) None of these
Ans: (b) 6 days
Explanation:
A and Bs 1 days work = 1/3
Their 2 days work = 1/3 2 = 2/3
Remaining work = 1/3
1/3 of the work is finished by A in 2 days
Then, the whole work can be finished by A alone in 3 2 = 6 days
So, As 1 days work = 1/6
Therefore, Bs 1 days work = 1/3 1/6 = 1/6
Therefore, the whole work can be done B alone in 6 days. Ans.
Q. 70. A man, a woman and a boy together complete a piece of work in 3 days. If a man alone can do it in 6
days and a boy alone in 18 days, how long will a woman take to complete the work?
(a) 9 days (b) 21 days (c) 24 days (d) 27 days (e) None of these
Ans: (a) 9 days
Explanation:
(Man + Boy + Woman)s 1 days work = 1/3
Mans 1days work = 1/6
Boys 1 days work = 1/18
Then, Woman 1 days work = 1/3 ( 1/6 + 1/18) = 1/3 4/18 =2/18 = 1/9
Therefore, the Woman alone can finish the work in 9 days. Ans.
Q. 71. A can do 1/3 of a work in 5 days and B can do 2/5 of the work in 10 days. In how many days both A and
B together can do the work?
(a) 7 (b) 8 4/5 (c) 9 3/8 (d) 10 (e) None of these
Ans: (c) 9 3/8
Explanation:
A can do finish the whole work in 3 5 days = 15 days
B can finish the whole work in 5 /2 10 days = 25 days
A and B together can finish the work in 15 25/ 40 days = 9 3/8 days. Ans.
Q. 72. A and B can do a piece of work in 12 days; B and C can do it in 15 days and C and A can do it in 20
days. A alone can do the work in:
(a) 15 2/3 days (b) 24 days (c) 30 days (d) 40 days (e) None of these
Ans: (c) 30 days
Explanation:
2 (A + B + c)s 1 days work = 1/12 + 1/15 + 1/20 = 12/60 = 1/5
A + B + Cs 1 days work = 1/5/2 = 1/10
As 1 days work = 1/10 (B + C)s 1 days work = 1/10 1/15 = 1/30
Therefore, A alone can finish the work in 30 days. Ans.
Q. 73. A and B together can complete a piece of work in 8 days while B and C together can do it in 12 days. All
the three together can complete the work in 6 days. In how much time will A and C together complete the work?
(a) 8 days (b) 10 days (c) 12 days (d) 20 days (e) None of these
Ans: (a) 8 days
Explanation:
As 1 days work = A + B + Cs 1 days work B + Cs 1 days work = 1/6 1/12
= 1/12
Bs 1 days work = A + Bs 1 days work As 1 days work = 1/8 1/12= 1/24
Cs 1 days work = B+ Cs 1 days work Bs 1 days work = 1/12 1/24 = 1/24
A +B 1 days work = 1/12 + 1/24 = 3/24
Therefore, A and C together will finish the whole work in 24/3 = 8 days. Ans.
Q. 74. A can do a work in 4 days, B can do it in 5 days and C can do it in 10 days. A, B and C together can do
the work in
(a) 1 3/5 days (b) 1 9/11 days (c) 2 5/6 days (d) 3 days (e) None of these
Ans: (b) 1 9/11 days
Explanation:
When A, B and C can do a work in x, y and z days respectively. Then, the three of them together can finish the
work in xyz/ x y + y z + x z days
That is, A, B and C together ca do the work in 4 5 10/ 20 + 50 + 40
= 4 5 10/110 = 20/11 = 1 9/11 days. Ans.
Q. 75. 3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to finish the work?
(a) 4 days (b) 10 days (c) 15 days (d) 20 days (e) None of these
Ans: (a) 4 days
Explanation:
3 men = 5 women
1 woman = 3/5 men
So, 5 women = 3/5 5 = 3 men
6 men and 5 women = 6 + 3 = 9 men
1 work done = 3 men 12 days
3 12 = 9 ? days
? days = 3 12/9 = 4 days. Ans.
Q. 76. Kamal can do a work in 15 days. Bimal is 50% more efficient than Kamal. The number of days, Bimal will
take to do the same piece of work, is
(a) 10 (b) 10 (c) 12 (d) 14 (e) None of these
Ans: (a) 10
Explanation:
Kamals 15 days is 150% of No. days taken by Bimal to finish the job.
Therefore, the No. of days taken by Bimal = 15/150 100 = 10 days. Ans.
Q. 77. A does 20% less work than B. if A can complete a piece of work in 7 hours, then B can do it in
(a) 5 hours (b) 5 hours (c) 6 hours (d) 6 hours (e) None of these
Ans: (c) 6 hours
Explanation:
B can finish the work in 80% of 7 hours of time
7 = 15/2
Therefore, time taken by B = .8 15/2 = 30/5 = 6 hours. Ans.
Q. 78. Ram and Shyam together can finish a job in 8 days. Ram can do the same job on his own in 12 days.
How long will Shyam take to do the job by himself?
(a) 16 days (b) 20 days (c) 24 days (d) 30 days (e) None of these
Ans: (c) 24 days
Explanation:
If A and B together can do a piece of work in x days and A alone can do the same work in y days, then B alone
can do the same work in x y/ y x days.
Therefore, the No. of days Shyam take to finish the job alone = 8 12/12 8
= 8 12/4 = 24 days. Ans.
Q. 79. The rates of working of A and B are in the ratio 3 : 4. The number of days taken by them to finish the
work are in the ratio:
(a) 3 : 4 (b) 9 : 16 (c) 4 : 3 (d) 3 : 2 (e ) None of these
Ans: (c) 4 : 3
Explanation:
The No. of days taken by them = inverse ratio = 4 : 3. Ans.
Q. 80. A certain number of men complete a piece of work in 60 days. If there were 8 men more, the work could
be finished in 10 days less. How many men were originally there?
(a) 30 (b) 40 (c) 32 (d) 36 (e) None of these
Ans: (b) 40
Explanation:
Let x be the original number of men
Then 1 work done = x 60
Then, 60x = (x + 8) 50
60x = 50x + 400
10 x = 400
x = 40. Ans.
Q. 81. A does half as much work as B in three- fourth of the time. If together they take 18 days to complete the
work, how much time shall B take to do it?
(a) 30 days (b) 35 days (c) 40 days (d) 42 days (e) None of these
Ans: (a) 30 days
Explanation:
Let x be the number of days taken by B alone to finish the whole work
Then, A alone will finish the whole work in 3x/4 2 days = 3x/2 days.
Then, working together,
3x/2 x/3x/2 + x = 18 days
3x x/5x = 18
3x/5 = 18
Therefore, the number of days taken by B = x = 18 5/3 = 30 days. Ans.
Q. 82. A and B can finish a piece of work in 12 days and 18 days respectively. A begins to do the work and they
work alternately one at a time for one day each. The whole work will be completed in
(a) 14 1/3 days (b) 15 2/3 days (c) 16 1/3 days (d) 18 2/3 days (e) None of these
Ans: (a) 14 1/3 days
Explanation:
Both of them together can finish the work (daily working) in 12 18/30 days = 36/5
= 7 1/5 days.
By working in alternate days, their 2 days work = 5/36
Their, 14 days work (because they take more than double of 7 days) = 5/36 7
= 35/36
Remaining work = 1 35/36 = 1/36
Because A started the work, the remaining work is finished by A
As 1 days work = 1/12
Therefore, the No. of days taken by A to finish 1/36 work = 1/36/1/12 = 1/3 days.
Therefore, the total work is completed in 14 + 1/3 = 14 1/3 days. Ans.
Q. 83. A can do a piece of work in 14 days which B alone can do in 21 days. They begin together but 3 days
before the completion of the work, A leaves off. The total number of days for the work to be completed is
(a) 6 3/5 days (b) 8 days (c) 10 1/5 days (d) 13 days (e) None of these
Ans: (c) 10 1/5 days
Explanation:
Bs 1 days work = 1/21
Bs 3 days work = 1/21 3 = 1/7
The remaining work finished by both of them = 1 1/7 = 6/7
Both together can finish it in 14 21/35 days = 42/5 days
Their, 1 days work = 5/42
Therefore, the No. of days taken by them together to finish 6/7 work
= 6/7/5/42 = 6/7 42/5 = 36/5 = 7 1/5 days
Therefore, the total No. of days for the work to be completed = 3 + 7 1/5 = 10 1/5 days. Ans.
Q. 84. A and B can finish a piece of work in 30 days. They worked at it for 20 days and then B left. The
remaining work was done by A alone in 20 more days. A alone can finish the work in
(a) 48 days (b) 50 days (c) 54 days (d) 60 days (e) None of these
Ans: (d) 60 days
Explanation:
(A + B)s 1 days work = 1/30
Their 20days work = 1/30 20 = 2/3
Remaining 1/3 work is done by A in 20 days
Therefore, A alone can finish the work in 3 20 = 60 days. Ans.
Q. 85. 10 women can complete a work in 8 days and 10 children take 12 days to complete the work. How many
days will 6 women and 3 children together take to complete the work?
(a) 7 (b) 8 (c) 9 (d) 12 (e) None of these
Ans: (e) None of these
Explanation:
1 womens 1 days work = 1/8/10 = 1/80
1 childs 1 days work = 1/12/10 = 1/120
6 womens 1 days work = 1/80 6 = 3/40
## 3 childrens 1 days work = 1/120 3 = 1/40
6 womens + 3 childrens 1 days work = 3 /40 + 1/40 = 1/10
Therefore, they will finish the whole work in 10 days. Ans.
Q. 86. 9 children can complete a piece of work in 360 days; 18 men can complete the same piece of work in 72
days and 12 women can complete it in 162 days. In how many days can 4 men, 12 women and 10 children
together complete the piece of work?
(a) 68 days (b) 81 days (c) 96 days (d) 124 days (e) None of these
Ans: (b) 81 days
Explanation:
1 childs 1 days work = 1/360 9 = 1/3240
10 childrens 1 days work = 1/324
1 mans 1 days work = 1/72 18 = 1/1296
4 mens 1 days work = 1 4/1296 = 1/324
12 womens 1 days work = 1/162 given
Then, (4 men + 12 women + 10 children)s 1 days work = 1/324 + 1/162 + 1/324
= 1/324 + 2/324 + 1/324 = 4/324 = 1/81
Therefore, the required No. of days = 81 days. Ans.
Q. 87. 9 men working 7 hours a day can complete a piece of work in 15 days. How many days can 6 men
working for 9 hours a day, complete the same piece of work?
(a) 15 days (b) 16 days (c) 16 days (d) 17 days (e) None of these
Ans: (d) 17 days
Explanation:
1 work done = 9 7 15
9 7 15 = 6 9 ? days
? days = 9 7 15/6 9
= 35/2 = 17 days. Ans.
Q. 88. A man, a woman and a boy can together complete a piece of work in 3 days. If a man alone can do it in
6 days and a boy alone in 18 days, how long will a woman take to complete the work?
(a) 9 days (b0 21 days (c) 24 days (d) 27 days (e) None of these
Ans: (a) 9 days
Explanation:
(1 Man + 1 woman + 1 boy)s 1days work = 1/3
1 mans 1 day work = 1/6
1boys 1 days work = 1/18
(1 Man + 1 boy) s 1 days work = 1/6 + 1/18 = 2/9
Therefore, 1 womans 1 days work = 1/3 2/9 = 3-2/9 = 1/9
Therefore, the woman alone can finish the work in 9 days. Ans.
Q. 89. 8 men can do a piece of work in 12 days. 4 women can do it in 48 days and 10 children can do it in 24
days. In how many days can 10 men, 4 women and 10 children together complete the piece of work?
(a) 5 days (b) 15 days (c) 28 days (d) 6 days (e) None of these
Ans: (d) 6 days
Explanation:
1 mans 1 days work = 1/8 12 = 1/96
10 mens 1 days work = 1 10/96 = 5/48
1 womans 1 days work = 1/192
4 womens 1 days work = 1/192 4 = 1/48
1 childs 1 days work = 1/240
## 10 childrens 1 days work = 1/24
Therefore, (10 men + 4 women + 10 children)s 1 days work = 5/48 + 1/48 + 1/24
= 8/48 =1/6
The required No. of days = 6 days. Ans.
Q. 90. 8 men can dig a pit in 20 days. If a man works half as much again a s a boy, then 4 men and 9 boys can
dig a similar pit in:
(a) 10 days (b) 12 days (c) 15 days (d) 16 days (e) None of these
Ans: (d) 16 days
Explanation:
1 work done = 8 20
1 man = 3/2 Boys
1 boy = 2/3 men
Then, 9 boys = 9 2/3 men = 6 men
Then, 4 men + 9 boys = 10 men
Then, 8 20 = 10 ?days
? days = 8 20/10 = 16 days. Ans.
Q. 91. A man and a boy complete a work together in 24 days. If for the last 6 days man alone does the work
then it is completed in 26 days. How long the boy will take to complete the work alone?
(a) 72 days (b) 20 days (c) 24 days (d) 36 days (e) None of these
Ans: (a) 72 days
Explanation:
(man + boy)s 1 days work = 1/24
Their 20 days work = 1/24 20 = 5/6
The remaining 1/6 work is done by the man in 6days
Therefore, the man alone will finish the work in 6 6 days = 36 days
Mans 1 days work = 1/36
Therefore, boys 1 days work = 1/24 1/36 = 3 2 /72 = 1/72
Therefore, the Boy alone will finish the work in 72 days. Ans.
Q. 92. A, B and C completed a piece of work costing Rs.1800. A worked for 6 days, B for 4 days and C for 9
days. If their daily wages are in the ratio 5 : 6 : 4, how much amount will be received by A?
(a) Rs.800 (b) Rs. 600 (c) Rs.900 (d) Rs.750 (e) None of these
Ans: (b) Rs. 600
Explanation:
The new ratio = wages ratio No. of days
A = 5 6 = 30
B = 6 4 = 24
C = 4 9 = 36
Therefore, the ratio of the amount received = 30 : 24 : 36 = 5 : 4 :6
Total ratio = 15
1 unit of ratio = 1800/15 = Rs. 120
Therefore, amount received by A = 5 units = 5 120 = Rs.600. Ans.
Q. 93. 12 men take 36 days to do a work while 12 women complete 3/4th of the same work in 36 days. In how
many days 10 men and 8 women together will complete the same work?
(a) 6 days (b) 12 days (c) 27 days (d) Data inadequate (e) None of these
Ans: (c) 27 days
Explanation:
1 mans 1 days work = 1/ 36 12 = 1/432
## 10 mens 1 days work = 10/432
12 women can finish the whole work in 4 36/3 = 48 days
1 womans 1 days work = 1/ 48 12 = 1/576
8 womens 1 days work = 8/572 = 1/72
(10 men + 8 women)s 1 days work = 10/432 + 1/72 = 10 + 6/432 = 16/432
= 1/27
Therefore, the required No. of days = 27 days. Ans.
Q. 94. A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. working
together, they can do it in
(a) 20 days (b) 22 days (c) 25 days (d) 30 days (e) None of these
Ans: (b) 22 days
Explanation:
B = 3A
3A A =60 days
A = 30days
Then, B = 90 days
(A + B) = 30 90/ 120 = 45/2 = 22 days . Ans.
Q. 95. A works twice as fast as B. if both of them can together finish a piece of work in 12 days, then B alone
can do it in
(a) 24 days (b) 27 days (c) 36 days (d) 48 days (e) None of these
Ans: (c) 36 days
Explanation:
B = 2A
2A A/3A = 12
2A/3 = 12
A = 18 days
B = 2 18 = 36 days
Therefore, B alone can finish the work in 36 days. Ans.
Q. 96. A man, a woman and a boy can complete a job in 3 days, 4 days and 12 days respectively. How many
boys must assist 1 man and 1 woman to complete the job in 1/4th of a day?
(a) 10 (b) 14 (c) 19 (d) 41 (e) None of these
Ans: (d) 41
Explanation:
1 mans 1 days work = 1/3, days work = 1/3 = 1/12
1 womans 1 days work = , days work = = 1/16
1 boys 1 days work = 1/12, days work = 1/12 = 1/48
Let x be the No. of boys required.
Then, (1 man + 1 woman + x boy)s days work 1/12 + 1/16 + x/48 = 1
=4+3+x=1
48
i.e. 7 + x= 48 and x = 41. Ans.
Q. 97. A does 4/5th of a work in 20 days. He then calls in B and they together finish the remaining work in 3
days. How long B alone would take to do the whole work?
(a) 23 days (b) 37 days (c) 37 days (d) 40 days (e) None of these
Ans: (c) 37 days
Explanation:
A can finish the whole work in 20 5/4 days = 25 days
## A and B together finish the whole work in 5 3 days = 15 days
Therefore, B can finish the whole work in 25 B/ 25 + B = 15
25 B = 15 ( 25 + B)= 375 + 15B
10B = 375 and B = 375/10 = 37 days. Ans.
Q. 98. A can do 1/4th part of the work in 10 days, B can do 40% of the work in 40 days and C can do 1/3rd of
the work in 13 days. Who will complete the work first?
(a) A (b) B (c) C (d) A and C both (e) None of these
Ans: (c) C
Explanation:
The whole work is done by A in 4 10 = 40 days
B can do the whole work in 40/40 100 days = 100 days
C can do the whole work in 3 13 = 39 days.
Therefore, the 1st person to complete the work is C. Ans.
Q. 99. A alone can finish a work in 10 days which B alone can finish in 15 days. If they work together and finish
it, then out of a total wages of Rs.3000, A will get:
(a) Rs.1200 (b) Rs.1500 (c) Rs. 1800 (d) Rs.2000 (e) None of these
Ans: (c) Rs. 1800
Explanation:
Ratio of working days of A : B = 10 : 15
Therefore, their wages ratio = reverse ratio = 15 : 10
Therefore, A will get 15 units of ratio
Total ratio = 25
1 unit of ratio =3000/25 = 120
So, As amount = 120 15 = Rs.1800. Ans.
Q. 100. Directions: (Questions 1 to 5):
Each of the following questions consists of a question followed by three statements I, II and III. You have to
study the question and the statements and decide which of the statement(s) is /are necessary to answer the
question.
1. In how many days can A and B working together complete a job?
I. A alone can complete the job in 30 days.
II. B alone can complete the job in 40 days
III. B takes 10 days more than A to complete the job.
(a) I and II only (b) II and III only (c) I and III only (d) Any two of the three (e) All I , II and III
2. In how many days can the work be completed by A and B together ?
I. A alone can complete the work in 8 days.
II. If A alone works for 5 days and B alone works for 6 days, the work gets completed.
III. B alone can complete the work in 16 days.
(a) I and II only (b) II and III only (c) Any two of the three (d) II and either I or III (e) None of these
3. How many workers are required to complete the construction work in 10 days?
I. 20% of the work can be completed by 8 workers in 8 days.
II. 20 workers can complete the work in 16 days.
III. One-eighth of the work can be completed by 8 workers in 5 days.
(a) I only (b) II and III only (c) III only (d) I and III only (e ) Any one of the three.
4. In how many days can the work be done by 9 men and 15 women?
I. 6 men and 5 women can complete the work in 6 days.
II. 3 men and 4 women can complete the work in 10 days.
III. 18 men and 15 women can complete the work in 2 days.
(a) III only (b) All I, II and III (c) Any two of the three (d) Any one of the three (e) None of these
5. In how many days can 10 women finish the work?
I. 10 men can complete the work in 6days
II. 10 men and 10 women together can complete the work in 3 3/7 days.
III. If 10 men work for 3 days and thereafter 10 women replace them, the remaining work is completed in 4
days.
(a) Any two of the three (b) I and II only (c) II and III only (d) I and III only (e) None of these
1. Ans: (d) Any two of the three
Explanation:
I. As 1days work = 1/30
II. Bs 1 days work = 1/40
III. Bs 1 days work = 1/40
I and II gives the answer,
I and III gives the answer,
II and III gives the answer,
Therefore, the correct answer is (d). Ans.
2. Ans: (c) Any two of the three
Explanation:
I. As 1 days work = 1/8
II. 1 work done = As 5 days work + Bs 6 days work
III. Bs 1 days work = 1/16
So, I and II gives the answer,
I and III gives the answer,
II and III gives the answer.
Therefore, the correct answer is (c) .Ans.
3. Ans: (e) Any one of the three
Explanation:
I. 1 workers 1 days work = 1/320, required No. of workers = 320/10 =32.
II. 1 workers 1 days work = 1/320
III. 1 workers 1 days work = 1/320
Therefore, I alone gives the answer
Therefore, the correct answer is (e). Ans.
4. Ans: (c) Any two of the three
Explanation:
I. Gives, the linear equation 6x + 5y = 6
II. Gives, the linear equation 3x + 4 y = 10
III. Gives, the linear equation 18x + 15 y = 2
i.e. Any two of the equations gives the correct answer.
Therefore, the correct answer is (c) Ans.
5. Ans: (b) I and II only
Explanation:
I. Gives, 1 mans 1 days work = 1/60
II. I and II gives 1 womans 1 days work = 1/80
III. (10 men + 10 women)s 1 days work = 1/7
## Therefore, I and II gives the answer
So, the correct answer is (b). Ans.
Solution to Mitul Sharma's Question on Time and Work
Question: A certain work can be done in a certain time by 25 men. With 5 men less it could have been done in
3 days more. In what time can it be done by 40 men.
Solution:
let x be the time taken by 25 men to complete the work in days.
Then, 25 * x = 20 * (x + 3)
25x = 20x + 60
x = 12
Then, 40 men will compete the work in = 25 * 12/40 = 15/2 = 7 1/2 days.<|endoftext|>
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Understanding the periodic table through the lens of the volatile Group I metals
The news broke that a railroad car, loaded with pure sodium, had just derailed and was spilling its contents. A television reporter called me for an explanation of why firefighters were not allowed to use water on the flames bursting from the mangled car. While on the air I added some sodium to a bit of water in a petri dish and we observed the vicious reaction. For further dramatic effect, I also placed some potassium into water and astonished everyone with the explosive bluish flames.
Because Group I metals, also known as alkali metals, are very reactive, like the sodium from the rail car or the potassium, they are not found in nature in pure form but only as salts. Not only are they very reactive, they are soft and shiny, can easily be cut even with a dull knife and are the most metallic of all known elements.
I am a chemist who spent his career building new molecules, sometimes using Group I elements. By studying the behavior and trends of Group I elements, we can get a glimpse of how the periodic table is arranged and how to interpret it.
The arrangement of the periodic table and the properties of each element in it is based of the atomic number and the arrangement of the electrons orbiting the nucleus. The atomic number describes the number of protons in the nucleus of the element. Hydrogen's atomic number is 1, helium's is 2, lithium's is 3 and so on.
Each of the 18 columns in the table is called a group or a family. Elements in the same group share similar properties. And the properties can be assumed based on the location within the group. Going from the top of Group I to the bottom, for example, the atomic radii – the distance from the nucleus to the outer electrons – increases. But the amount of energy needed to rip off an outer electron decreases going from the top to the bottom because the electrons are farther from the nucleus and not held as tightly.
This is important because how elements interact and react with each other depends on their ability to lose and gain electrons to make new compounds.
The horizontal rows of the table are called periods. Moving from the left side of the period to the right, the atomic radius becomes smaller because each element has one additional proton and one additional electron. More protons means that electrons are pulled in more tightly toward the nucleus. For the same reason electronegativity – the degree to which an element tends to gain electrons – increases from left to right.
The force required to remove the outermost electron, known as the ionization potential, also increases from the left-hand side of the table, which has elements with a metallic character, to the right side, which are nonmetals.
Electronegativity decreases from the top of the column to the bottom. The melting point of the elements within a group also decreases from the top to the bottom of a group.
Applying the basics to Group I elements
As its name implies, Group I elements occupy the first column in the periodic table. Each element starts a new period. Lithium is at the top of the group and is followed by sodium, Na; potassium, K; rubidium, Rb; cesium, Cs and ends with the radioactive francium, Fr. Because it is highly radioactive, virtually no chemistry is performed with this element.
Because each element in this column has a single outer electron in a new shell, the volumes of these elements are large and increase dramatically when moving from the top to the bottom of the group.
Of all the Group I elements, cesium has the largest volumes because the outermost single electron is loosely held.
In spite of these trends, the properties of the elements of Group I are more similar to each other than those of any other group.
Alkali metals through history
Using chemical properties as his guide, Russian chemist Dimitri Mendeleev correctly ordered the first Group I elements into his 1869 periodic table. It is called periodic because every eighth element repeats the properties of the one above it in the table. After arranging all of the then known elements, Mendeleev took the bold step of leaving blanks where his extrapolation of chemical properties showed that an element should exist. Subsequent discovery of these new elements proved his prediction correct.
Some alkali metals have been known and put to good use long before Mendeleev created the periodic table. For instance, the Old Testament mentions salt – a combination of the alkali metal sodium with chlorine – 31 times. The New Testament refers to it 10 times and calls sodium carbonate "neter" and potassium nitrate "saltpeter."
People have known since antiquity that wood ashes produce a potassium salt which, when combined with animal fat, will yield soap. Samuel Hopkins obtained the first U.S. patent on July 31, 1790, for soap under the new patent statute just signed into law by President George Washington a few months earlier.
The pyrotechnic industry loves these Group I elements for their vibrant colors and explosive nature. Burning lithium produces a vivid crimson red color; sodium a yellow one; potassium lilac; rubidium red; and cesium violet. These colors are produced as electrons jump from their home environment orbiting the nucleus and returning back again.
The cesium atomic clock, the most accurate timepiece ever developed, functions by measuring the frequency of cesium electrons jumping back and forth between energy states. Clocks based on electrons jumping provide an extremely precise way to count seconds.
Other applications include sodium vapor lamps and lithium batteries.
In my own research I have used Group I metals as tools to perform other chemistry. Once I was in need of absolutely dry alcohol, and the driest I could buy still contained minute traces of water. The only way to get rid of the last remnant of water was by treating the water-containing alcohol with sodium – a rather dramatic way to remove water.
The alkali elements not only occupy the first column in the periodic table, but they also show the most reactivity of all groups in the entire table and have the most dramatic trends in volume and ionization potential, while maintaining great similarity among themselves.<|endoftext|>
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# Lesson 1
Share Sandwiches
### Lesson Purpose
The purpose of this lesson is for students to relate equal shares of objects to division and to fractions.
### Lesson Narrative
In previous grades, students learned to interpret products of whole numbers, such as $$3 \times 5$$, as the total number of objects in 3 groups each containing 5 objects. They interpreted division, such as $$15 \div 3$$, to be either the number of groups when 15 things are put in groups of 3 or as the number of things in each group when 15 things are put in 3 equal groups. They also solve word problems posed with whole numbers and having whole-number answers, including problems in which remainders must be interpreted. The goal of the next several lessons is to extend this understanding of division to quotients like $$15 \div 6$$ where the result is not a whole number. Students learned to interpret fractions such as $$\frac{15}{6}$$ in a previous grade and this unit will establish that $$\frac{15}{6}$$ is the value of the quotient $$15 \div 6$$
In this lesson, students use what they know about division to make sense of situations where people equally share sandwiches. This lesson is meant to be an invitation to explore the relationships between division and fractions. The problems were written so students can revisit the meaning of division and be curious about how division applies to situations when the quotient represents a fractional quantity without having to name the quantity. Although students discuss how the situations in the lesson can be represented with division expressions, they do not need to write them or formally explain them, as that will be the focus of upcoming lessons. Throughout this unit, it is assumed that the sharing is always equal sharing, whether explicitly stated or not.
• Representation
• MLR1
### Learning Goals
Teacher Facing
• Interpret and represent contexts relating division and fractions in a way that makes sense to them.
### Student Facing
• Let’s share sandwiches.
Building On
Building Towards
### Lesson Timeline
Warm-up 10 min Activity 1 20 min Activity 2 15 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
What evidence did you see that each student felt they belonged in math class today?
### Suggested Centers
• Rolling for Fractions (3–5), Stage 2: Multiply a Fraction by a Whole Number (Supporting)
• Compare (1–5), Stage 4: Divide within 100 (Supporting)
### Print Formatted Materials
For access, consult one of our IM Certified Partners.<|endoftext|>
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Circle or Semi Circle MCQ Quiz - Objective Question with Answer for Circle or Semi Circle - Download Free PDF
Last updated on Dec 6, 2022
The practice test has Circle or Semicircle Objective Questions provided with the detailed solutions along with shortcuts and tricks. Candidates will be able to solve Circle or Semicircle Question Answers with accuracy by the end of the quizz. These questions will clear all concepts related to circle and semicircle which will help you ace the interviews, entrance exams and competitive exams. Regular practise of the Circle or Semicircle MCQ Quiz will get you a good score.
Latest Circle or Semi Circle MCQ Objective Questions
Circle or Semi Circle Question 1:
The area of a square field is 484 m2 and a circular park, whose radius is 14/22th of the side of the square field. What is the area (in m2) of the circular park?
1. 648m2
2. 616 m2
3. 267m2
4. 512m2
Option 2 : 616 m2
Circle or Semi Circle Question 1 Detailed Solution
Given:
Area of a square field = 484 m2
Radius = 14/22th of the side.
Circumference of circle = Perimeter of a square
a = side of the square
r = radius of the circle
Formula:
Area of square = a2
Perimeter of Square = 4a
Circumference = 2πr
Calculation
a2 = 484
a = √484
a = 22m
r = 14/22 × 22
r = 14m
Area of circle = πr2
⇒ πr2 = 22/7 × 14 × 14
⇒ 22 × 2 × 14
⇒ 22 × 28
⇒ 616m2.
Circle or Semi Circle Question 2:
If the arc of a circle of radius 56 cm has a length of 44 cm, then the angle (in degrees) subtended at the centre of the circle is:
1. 40°
2. 55°
3. 43°
4. 45°
Option 4 : 45°
Circle or Semi Circle Question 2 Detailed Solution
Given:
Radius of the circle = 56 cm
Length of the arc = 44 cm
Concept used:
Length of an arc that subtended an angle θ at the centre of the circle = $$\frac {2\pi R} {360} \times θ$$ (where R is the radius of the circle)
Calculation:
Let the subtended angle at the centre by θ.
According to the concept,
$$\frac {2\pi \times 56} {360} \times θ$$ = 44
⇒ $$\frac {2\times 22 \times 56} {7 \times 360} \times θ$$ = 44
⇒ θ = $$\frac {360 \times 7} {56}$$
⇒ θ = 360/8
⇒ θ = 45°
∴ The angle subtended at the centre of the circle is 45°.
Circle or Semi Circle Question 3:
In the given figure, the length of arc AB is equal to twice the length of radius r of the circle. Find the area of sector OAB in terms of the radius r.
1. 3r
2. 2r
3. πr2
4. r2
Option 4 : r2
Circle or Semi Circle Question 3 Detailed Solution
GIVEN:
Length of the arc AB = 2 r
Where r = radius of the circle
FORMULA USED:
l = r × $$\theta$$
Area of an arc = $$\theta$$/360 × $$\pi$$ r2
Where l = length of an arc
r = radius of the circle
CALCULATION:
Here, we have length of the arc = 2r
⇒ l = r × $$\theta$$
⇒ 2r = r × $$\theta$$
⇒ $$\theta$$ = 2
Area of the sector = $$\theta$$/360 × $$\pi$$ r2
⇒ 2/360 × 180 × r2 = r2 [As $$\pi$$ = 180 ]
Hence, area of sector in terms of an area is r2 .
Circle or Semi Circle Question 4:
The area of a circle inscribed in an equilateral triangle is 154 cm2. What is the perimeter of the triangle?
1. 21 cm
2. 42√3 cm
3. 21√3 cm
4. 42 cm
Option 2 : 42√3 cm
Circle or Semi Circle Question 4 Detailed Solution
Given:
The area of a circle inscribed in an equilateral triangle = 154 cm2
Concept:
In an equilateral triangle:
• The perpendicular bisectors of an equilateral triangle are also angle bisectors, altitudes, and medians.
• The circumcenter of an equilateral triangle is also the incenter, orthocenter & centroid of the triangle.
• Centroids divide each median into two pieces such that one piece is twice the length of the other piece.
Formula used:
Area of circle = πr2
Height of an equilateral triangle (h) = $$\frac{√3}{{2}}$$a
The perimeter of a triangle = 3a
Where, a is a side
Calculation:
According to the question
Area of circle = 154 cm2
⇒ πr2 = 154 cm2
⇒ ($$\frac{22}{{7}}$$ × r2) = 154
⇒ r2 = 49
⇒ r = 7 cm
Now, by above concept used, AO : OD = 2 : 1
⇒ OD = r = $$\frac{h}{{3}}$$
⇒ h = 3r = (3 × 7) = 21 cm
Now, Height of an equilateral triangle
⇒ 21 = $$\frac{√3}{{2}}$$a
⇒ a = $$\frac{42}{{√3}}$$
Now, by rationalizing √3 by numerator and denominator
⇒ a = ($$\frac{42}{{√3}}$$ × $$\frac{√3}{{√3}}$$) = 14√3
Now, perimeter of triangle = 3a = (3 × 14√3) = 42√3 cm.
∴ The require perimeter of triangle is 42√3 cm.
Circle or Semi Circle Question 5:
The circumferences of two circles are in the ratio 2 ∶ 3. What is the ratio of their areas?
1. 2 ∶ 3
2. 4 ∶ 9
3. 1 ∶ 3
4. 8 ∶ 27
Option 2 : 4 ∶ 9
Circle or Semi Circle Question 5 Detailed Solution
Given:
The ratio of circumferences of two circles = 2 : 3
Formula used:
Circumference of circle = 2πr
Area of circle = πr2
Calculation:
Let the radius of the circle be r1 and r2 respectively.
⇒ ($$\frac{2\pi r_1}{{2\pi r_2}}$$) = $$\frac{2}{{3}}$$
⇒ $$\frac{ r_1}{{ r_2}}$$ = $$\frac{2}{{3}}$$
Now, the ratio of the area is
⇒ ($$\frac{\pi r^2_1}{{\pi r^2_2}}$$) = ($$\frac{2}{{3}}$$)2
⇒ ($$\frac{4x^2}{{9x^2}}$$) = $$\frac{4}{{9}}$$
∴ The required ratio is 4 : 9.
Top Circle or Semi Circle MCQ Objective Questions
Circle or Semi Circle Question 6
Six chords of equal lengths are drawn inside a semicircle of diameter 14√2 cm. Find the area of the shaded region?
1. 7
2. 5
3. 9
4. 8
Option 1 : 7
Circle or Semi Circle Question 6 Detailed Solution
Given:
Diameter of semicircle = 14√2 cm
Radius = 14√2/2 = 7√2 cm
Total no. of chords = 6
Concept:
Since the chords are equal in length, they will subtend equal angles at the centre. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.
Formula used:
Area of sector = (θ/360°) × πr2
Area of triangle = 1/2 × a × b × Sin θ
Calculation:
The angle subtended by each chord = 180°/no. of chord
⇒ 180°/6
⇒ 30°
Area of sector AOB = (30°/360°) × (22/7) × 7√2 × 7√2
⇒ (1/12) × 22 × 7 × 2
⇒ (77/3) cm2
Area of triangle AOB = 1/2 × a × b × Sin θ
⇒ 1/2 × 7√2 × 7√2 × Sin 30°
⇒ 1/2 × 7√2 × 7√2 × 1/2
⇒ 49/2 cm2
∴ Area of shaded region = 6 × (Area of sector AOB - Area of triangle AOB)
⇒ 6 × [(77/3) – (49/2)]
⇒ 6 × [(154 – 147)/6]
⇒ 7 cm2
Area of shaded region is 7 cm2
Circle or Semi Circle Question 7
AB is a diameter of a circle with center O. A tangent is drawn at point A. C is a point on the circle such that BC produced meets the tangent at P. If ∠APC = 62º, then find the measure of the minor arc AC(i.e.∠ ABC).
1. 31º
2. 62º
3. 28º
4. 66º
Option 3 : 28º
Circle or Semi Circle Question 7 Detailed Solution
Given:
AB is a diameter of a circle with a center O
∠APC = 62º
Concept used:
The radius/diameter of a circle is always perpendicular to the tangent line.
Sum of all three angles of a triangle = 180°
Calculation:
Minor arc AC will create angle CBA
∠APC = 62º = ∠APB
∠BAP = 90° (diameter perpendicular to tangent)
In Δ APB,
∠APB + ∠BAP + PBA = 180°
⇒ PBA = 180° - (90° + 62°)
⇒ PBA = 28°
∴ The measure of minor arc AC is 28°
Mistake PointsMeasure of the minor arc AC is asked,
∠ABC marks arc AC,
∴ ∠ABC is the correct angle to show a measure of arc AC
This is a previous year's question, and according to the commission, this is the correct answer.
Circle or Semi Circle Question 8
Quadrilateral ABCD circumscribed a circle. If AB = 8 cm, BC = 7 cm and CD = 6 cm, then the length of AD is:
1. 7 cm
2. 6.8 cm
3. 7.5 cm
4. 6 cm
Option 1 : 7 cm
Circle or Semi Circle Question 8 Detailed Solution
AB = 8 cm, BC = 7 cm and CD = 6 cm.
As we know,
AD + BC = CD + AB
⇒ AD + 7 = 6 + 8
⇒ AD = 14 – 7
Circle or Semi Circle Question 9
If the area of a semi circular field is 30800 sq. m. then the perimeter of the field is (use π = 22/7)
1. 240 m
2. 720 m
3. 460 m
4. 840 m
Option 2 : 720 m
Circle or Semi Circle Question 9 Detailed Solution
Given:
The area of semi-circular field = 30800 m2
Formula used:
Area of semi-circle = (πr2)/2
The perimeter of the semi circle = πr + 2r
Where, r = Radius of the circle
Calculation:
According to the question,
The area of a semi-circular field = 30800 m2
⇒ (πr2)/2 = 30800
⇒ r2 = (30800 × 2 × 7)/22
⇒ r2 = 1400 × 2 × 7
⇒ r2 = 19600
⇒ r = 140 m
Perimeter of the semi-circular field = πr + 2r
⇒ Perimeter of the semi-circular field = (22/7) × 140 + 2 × 140
⇒ Perimeter of the semi-circular field = 440 + 280
⇒ Perimeter of the semi-circular field = 720 m
∴ The perimeter of the semi circular field is 720 m.
Circle or Semi Circle Question 10
Length of curved part of a semi-circular field is 24 m more than its linear part. Find the area of the semi-circular field?
1. 660 m2
2. 672 m2
3. 638 m2
4. 693 m2
Option 4 : 693 m2
Circle or Semi Circle Question 10 Detailed Solution
GIVEN:
Length of curved part of a semi-circular field is 24 m more than its linear part
CONCEPT:
Mensuration
FORMULA USED:
Length of curved part of semicircle = πr
Length of linear part of semicircle = 2r
CALCULATION:
Let the radius of semi-circular field = ‘r’ m
So, length of curved part of the field = πr
And length of linear part of the field = 2r
So,
πr – 2r = 24
⇒ (22r/7) – 2r = 24
⇒ r = 21 m
So, the area of semi-circular field = (1/2) × (22/7) × (21)2 = 693 m2
Circle or Semi Circle Question 11
A circular park has a path of a certain width around it. If the ratio between the perimeter of outer circle and inner circle is 3 : 1 and radius of outer circle is 6 cm, then find the width of the path.
1. 4 cm
2. 5 cm
3. 7 cm
4. 2 cm
Option 1 : 4 cm
Circle or Semi Circle Question 11 Detailed Solution
Given:
A circular park has a path of a certain width around it.
Ratio between the perimeter of outer circle and inner circle is 3 : 1
Radius of outer circle = 6 cm
Formula used:
Perimeter of a circle = 2πr unit
where r → radius of circle
Calculation:
Let r cm and R cm be the radius of inner circle and outer circle respectively.
It is given that, R = 6 cm
Ratio between the perimeter of outer circle and inner circle = 3 : 1
⇒ 2πR/2πr = 3/1
⇒ R/r = 3/1
⇒ 3r = 6
⇒ r = 2
So width of the path = (R - r) cm
⇒ (6 - 2) cm
⇒ 4 cm
∴ The width of the path is 4 cm.
Circle or Semi Circle Question 12
The area of a circular hall is 154 hectares. Find the cost of fencing the hall at the rate of 50 paise per meter.
1. Rs. 2200
2. Rs. 2000
3. Rs. 1500
4. Rs. 1800
Option 1 : Rs. 2200
Circle or Semi Circle Question 12 Detailed Solution
Given:
The area of a circular hall is 154 hectares.
The rate of fencing per meter is 50 paise.
Formula:
Let r be the radius of a circle then,
Area of a circle = πr2
Circumference of a circle = 2πr.
Concept Used:
1 hectare = 10000 m2.
Calculation:
Let A and r be the area and radius of a circular hall respectively.
A = πr2 = 154 hectares
πr2 = 1540000 m2
r = 700 m.
Circumference of the circle
2πr = 2 × (22/7) × 700 m
⇒ 2 × (22/7) × 700 m = 4400 m
Cost of fencing = 4400 × 0.5 = 2200
The cost of fencing the hall is Rs. 2200.
Circle or Semi Circle Question 13
If the side of a square increases by 20%, then what will be a percent increase in its perimeter?
1. 44%
2. 20%
3. 80%
4. 40%
Option 2 : 20%
Circle or Semi Circle Question 13 Detailed Solution
Given:
Let the side of the square be x
And Perimeter of square = 4x
Formula Used:
Perimeter of square = 4 × (side)
Area of Square = (side)2
Area of Circle = π × (radius)
Perimeter of Circle = 2 × π × (radius)
Concept :
If sides increase by 20%. So Perimeter also increases by 20%
Sides become 1.2 times original
So, Perimeter becomes 4 × 1.2 = 1.2 times original
Alternative solution:
Let the side of the square be 100 cm
Side of the square is increased by 20% becomes = 100 + 100 × 20/100
⇒ 120 cm
Perimeter before increase = 4 × 100 = 400 cm
After 20% increased = 4 × 120 = 480 cm
Increase in percentage = [(480 - 400)/400]× 100
⇒ 20%
Mistake PointsPlease attention there asks the Perimeter, not the area, mostly students mark the 44% suppose to the area, but ask the perimeter.
Circle or Semi Circle Question 14
A circle is inscribed in ΔABC, touching AB, BC and AC at the points P, Q and R, respectively: If AB - BC = 4 cm, AB - AC = 2 cm and the perimeter of ΔABC = 32 cm, then $$BC \over 2$$ (in cm) = ?
1. $$20 \over 3$$
2. $$13 \over 3$$
3. $$11 \over 3$$
4. $$10 \over 3$$
Option 2 : $$13 \over 3$$
Circle or Semi Circle Question 14 Detailed Solution
Given data:
AB - BC = 4 cm
AB - AC = 2 cm
Perimeter = 32 cm
Calculation:
According to the question-
BC = AB - 4 cm
AC = AB - 2 cm
AB + BC + AC = 32 cm
Substituting the value of BC and AC-
⇒ AB + (AB - 4) + (AB - 2) = 32 cm
⇒ 3AB - 6 = 32
⇒ AB = $$38\over 3$$ cm.
Substituting the value of AB in equation of BC-
BC = $$38\over 3$$ - 4 = $$38-12\over 3$$ = $$26\over 3$$ cm
∴ $$BC\over 2$$ = $$13\over 3$$ cm.
The answer is $$13\over 3$$ cm.
Circle or Semi Circle Question 15
In a circle with centre O, chords PR and QS meet at the point T, when produced, and PQ is a diameter. If $$\angle$$ROS = 42º, then the measure of $$\angle$$PTQ is
1. 58º
2. 59º
3. 69º
4. 48º
Option 3 : 69º
Circle or Semi Circle Question 15 Detailed Solution
Given:
ROS = 42º
Concept used:
The sum of the angles of a triangle = 180°
Exterior angle = Sum of opposite interior angles
Angle made by an arc at the centre = 2 × Angle made by the same arc at any point on the circumference of the circle
Calculation:
Join RQ and RS
According to the concept,
∠RQS = ∠ROS/2
⇒ ∠RQS = 42°/2 = 21° .....(1)
Here, PQ is a diameter.
So, ∠PRQ = 90° [∵ Angle in the semicircle = 90°]
In ΔRQT, ∠PRQ is an exterior angle
So, ∠PRQ = ∠RTQ + ∠TQR
⇒ 90° = ∠RTQ + 21° [∵ ∠TQR = ∠RQS = 21°]
⇒ ∠RTQ = 90° - 21° = 69°
⇒ ∠PTQ = 69°
∴ The measure of ∠PTQ is 69°<|endoftext|>
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Table of Contents
Definition of Anthracosis
Anthracosis – Black Lung – a lung condition resulting from long-term exposure to coal dust, also called coal worker’s pneumonoconiosis (cwp) and black lung disease.
There are two types: simple and complicated. In simple anthracosis the coal dust coats the lungs in wide distribution, and the immune response encapsulates the dust particles without causing scarring (fibrosis).
This is the most common type of anthracosis and may generate no symptoms or mild symptoms such as dyspnea (shortness of breath) with exertion and chronic cough. The doctor may detect simple anthracosis during routine medical examination or screening for lung disease.
Diagnosis generally considers X-RAY findings in combination with history of exposure to coal dust. The doctor may also conduct bronchoscopy to examine the inner bronchial structures, which have a characteristic black appearance.
Complicated anthracosis becomes aggressively fibrotic, though doctors do not know what causes it to do so. It continues to progress even after exposure ends, and may result in disabling obstructive lung disease. Symptoms include worsening cough and dyspnea. About 15 percent of coal workers who have simple anthracosis develop complicated anthracosis.
Improvements in mining techniques and conditions, including environmental filtration systems, have greatly reduced the amount of dust coal mining operations produce. Those who work as cutters, loaders, and continuous mining operators face the highest risk for exposure. In the United States, the Occupational Safety and Health Administration (OSHA) regulates permissible levels of dust and worker exposure. Diagnosis of new cases of anthracosis is steadily declining as a result. Federal health programs provide medical care and other benefits for coal workers who have anthracosis.
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