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# Numbers 26 to 30
2 teachers like this lesson
Print Lesson
## Objective
Students will be able to count to 30 and recognize numbers up to 30.
#### Big Idea
Numbers don't stop at 20! We spend the last week or so of school looking at the numbers beyond 20.
## Problem of the Day
5 minutes
I start each math lesson with a Problem of the Day. I use the procedures outlined here on Problem of the Day Procedures.
Today's Problem of the Day:
What shape is this? How many are there?
For this problem, I set up some structure to help students. I added three ten frames for the students to use to organize the pictures before or as they count. I also have a line for the students to write the answer. On the Notebook file, the students can just drag the pictures into the ten frames. If you do not have a SMARTBoard, you can use the PDF file. You could also have the students solve the problem by using manipulatives.
I have one student come up and work on this problem. I remind student to check his or her work when they are finished and have the class tell if they agree or disagree by showing a thumbs up or thumbs down. I am also looking for students to explain how they can check their work. (Mathematical Practice 3: Construct viable arguments and critique the reasoning of others).
## Presentation of Lesson
25 minutes
I start this lesson by reviewing the numbers to 25. I show the students each number and say its name. I have the students repeat the number and practice writing it in the air. I introduce 26 to 30 in the same way.I show the students the Numbers 26-30 Worksheet and discuss how we can use three ten frames to show numbers in the in the 20s. By using ten frames, I'm continuing to focus students on the structure and patterns of our base 10 system.
I tell students that we will be practicing the numbers on a Numbers 26-30 Worksheet.
We are going to work on this paper together. When you get to your seat, you need to get out a pencil and put your name on your paper. When your name is on your paper hold your pencil in the air, that will let me know that you are ready to start.
I use the procedures outlined here on the Paper Procedures.
Count the objects and circle the correct number.
We work through the front of the paper with the students. For the back, I read the directions and allow the students to complete it on their own. I walk around and make sure that students are correctly counting circling the numbers. When students are finished with their paper, they can put it in the basket and get their center.
## Practice
20 minutes
This week's centers are:
I am not pulling groups this week because of end of the year assessments and activities.
Prior to clean up, I check in with each table to see how the centers are going. I have been using counting down from 20 slowly instead of a clean up song. Counting backwards is as critical as counting up. Students need to be able to know the number that comes before, as well as after, any given number (w/i 10, w/i 20, etc.). Counting back is a critical strategy for subtraction.
The students like to count backwards with me as they clean up and I can lengthen or reduce the clean up time based on how students are doing and how much time we have.
## Closing
10 minutes
To close, I put a student's paper on the document camera a project it on the SMART Board and have that student explain their work. I mention positive things noticed during centers as well as something that needs to be better next time.
I review what we did during our whole group lesson. "Today we learned about numbers to 30. Tomorrow we are going to look how to order number to 30."<|endoftext|>
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For most of the things, Slumber is a solution. And now there is a scientific theory to prove that. In a recent study published in the journal Nature Communications, researchers at Bar-Ilan University in Israel revealed the importance of sleep. With the use of 3D time-lapse imaging techniques in live zebrafish researchers were able to show how a single neuron needs sleep in order to maintain its nuclear balance. The subject used i.e. Zebra fish helped researchers to have a clear insight about the Brain and sleep patterns. Their transparent body structure and similar to human brain structure help while studying a single live cell in both the phases –awake and asleep. The findings that chromosomes were more active when the fish was asleep or in resting mode left scientists astonished.
DNA damages and misconduct that happen due to radiation, oxidative articulate, change neuronic activeness etc. received in wakefulness needs replenishment. And sleep comes to the rescue. What sleep does when a neuron sleeps is it increases chromosomal dynamics and normalizes the level of DNA damage. And this process of maintenance is not at its best when a neuron is active and awake.
So, sleep on because Sleep does consolidate and synchronize neuron repair within.<|endoftext|>
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Poultry housing helps to protect birds during unfavourable weather.
Chicken may look like rather happy-go-lucky animals, but there is a whole range of factors that may stress them out and render them more susceptible to diseases.
Food and Water
The feed may become a stressor if there is not enough, it is out of reach, there is too much competition for feed, feed particles are too fine or coarse, or the feed is high in salt. Any of these will have a negative impact on feed intake, resulting in birds having less energy to ward off diseases. The mouldy feed also contains poisonous substances that can suppress the immunity of birds.
The same applies when birds do not have access to enough water, or if the water is too hot or of a poor quality. According to the South African Poultry Association, dehydration will not only cause stress but also dry-out the respiratory tracts of birds, easing the penetration of disease-causing organisms. Dirty water is also a threat as it may contain a large variety of disease-causing organisms as well as bacteria that can weaken the immune system.
Birds need shade to protect them from the sun.
Lighting may become a stressor when birds do not have enough dark time to rest. According to the World Health Organisation, birds should also be given enough time to adapt to switches in day length manipulations during different stages of production as well as during daily switches from darkness to light and light to darkness. Lighting should be evenly dispersed, with no dark spots in between, to prevent overcrowding in brightly lit areas.
Climatic conditions and air quality may also be a challenge. It might be wise to choose breeds that perform better under warm climatic conditions, such as some of the indigenous breeds or naked necks, if birds will be produced outside in a region that is relatively warm. Extra care should then be taken to ensure the birds have access to enough shade and water and to prevent heat from having a negative impact on feed intake.
Cold stress has a negative impact on bird health because it results in birds needing more energy to stay warm. An easy way to identify whether birds are experiencing heat or cold stress is to look at their behaviour: Birds that are cold will crowd together to keep each other warm. They will also eat more than usual to make up for the energy they have lost in an attempt to keep warm.
Birds suffering from heat stress tend to pant and have laboured breathing. They will lift their wings and move away from other birds to keep cool and may become lethargic. Severe heat stress may lead to diarrhoea as well as seizures or convulsions.
When produced in poultry houses, climate control can be used to raise and lower temperatures and humidity levels. Special care should be taken to prevent over or under ventilation. The trick is to provide enough ventilation to allow moisture to move, but at the same time to avoid drafts.
Poor moisture removal from bedding material will lead to higher levels of ammonia and in turn, lead to a higher risk of respiratory diseases. If conditions in the poultry house are too dry, there is also a bigger chance that birds will become irritated by dust.
Exposure to Diseases and Disease Causing Vectors
Poor sanitation or pest management may cause stress by exposing birds to wild birds, rats, mice or flies. These animals will not only bother the birds but also present huge disease risks. Birds may also be stressed because they are ill, which is referred to as pathological stress. To reduce this stress and prevent serious complications, birds should be treated as soon as possible.
Personal Space and Handling
Space may become a problem when birds are overcrowded, which will lead to increased competition for feed and water as well as aggressive behaviour. Birds may also become stressed when they are caught, handled or treated. Producers should, therefore, plan the handling of birds carefully to ensure it is as stress-free as possible.
Signs of Stress
The following are signs that birds might be suffering from stress:
- aggression and fights
- feather plucking or shedding
- changes in appetite – birds eating more or less than usual
- lethargy or birds being sluggish
- weight loss
- drop in egg production for layers
By Glenneis Kriel<|endoftext|>
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This article is marked as 'retired'. The information here may be out of date and/or incomplete.
Astronomers have been keeping their eyes on a 1,200 foot wide chunk of space rock that has been hanging around the neighborhood lately. It’s an asteroid called Apophis, and it’s named for an ancient Egyptian spirit who was bent on plunging the world into eternal darkness. Scientists are concerned because this modern incarnation may succeed where its namesake failed.
Last year NASA gave Apophis a 1-in-37 chance of hitting the Earth in 2029, but after further observations of its path through space, NASA recently revised the odds to a 1-in-5,500 chance of colliding in 2036. Such an event, though unlikely, would cause an explosion about 100,000 times more powerful than the Hiroshima bomb and throw huge amounts of dust into the atmosphere, posing a grave risk to any humans who manage to survive the Avian Flu.
Scientists are now urging world governments to decide on a strategy for preventing the potential impact, because it will take years to develop and test any technology to deflect or destroy asteroids.
From the Guardian article:
At a recent meeting of experts in near-Earth objects (NEOs) in London, scientists said it could take decades to design, test and build the required technology to deflect the asteroid. Monica Grady, an expert in meteorites at the Open University, said: “It’s a question of when, not if, a near Earth object collides with Earth. Many of the smaller objects break up when they reach the Earth’s atmosphere and have no impact. However, a NEO larger than 1km [wide] will collide with Earth every few hundred thousand years and a NEO larger than 6km, which could cause mass extinction, will collide with Earth every hundred million years. We are overdue for a big one.”<|endoftext|>
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February 19, 1942 – Tuskegee Airmen Activated
On February 19,1942, the 100th Fighter Squardron was activated in the 332 Fighter Group which is better known as the Tuskegee Airmen.The 100th Fighter Squardron was the second African American combat unit of the Tuskegee Airmen and the Army Air Force. The first was the 99th which was activated on March 22, 1941. Tuskegee Airmen were the first African American soldiers to fly during World War II. These heroic soldiers fought valiantly in the war even when many initially doubted the flying capabilities of African Americans. The Tuskegee Airmen proved that not only could they fly, but they were as good as any of America’s pilots during the war. The Tuskegee Airmen whose chief role was the protection of large bombers from German fighter planes participated in over 15,000 sorties from May 1943 to June 1945. Approximately 1000 pilots were trained at Tuskegee Institute in Alabama.
Tribute To Black History Month
December 19, 1875 – April 3, 1950 – Carter Godwin Woodson
Dr. Carter G. Woodson began “Negro History Week” the forerunner to Black History Month. Dr. Woodson was a noted, historian, journalist, author and the founder of the Association for the Study of African American Life and History.<|endoftext|>
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# Counting Permutations
## Number of permutations of 3x3
To count number of permutations of 3x3, count permutations due to both the placement and orientation of each type of piece (corners and edges; centers are fixed):
• 3 possible rotations of 8 corners, 7 corners determine the 8th = ${\displaystyle 3^{7}}$
• 2 possible rotations of 12 edges = ${\displaystyle 2^{12}}$
• 8 different corner pieces to be distributed to 8 locations = ${\displaystyle 8!}$
• 12 different edge pieces to be distributed to 12 locations = ${\displaystyle 12!}$
• Half of the cube permutations (those requiring center faces to be swapped) cannot be reached without disassembling the cube = ${\displaystyle {\frac {1}{2}}}$
Total number of permutations of a 3x3 cube:
${\displaystyle N=3^{7}\times 8!\times 2^{12}\times 12!\times {\frac {1}{2}}=86504006548979712000\sim 8.65040\times 10^{19}}$
## Number of permutations of 4x4
Note: dedge = double edge
On a 4x4 cube, we use the same technique as the 3x3, counting permutations due to both placement and orientation of each type of piece. Now each face has four corner pieces, four double edges for a total of eight edge pieces, and four corner pieces. Counting:
Placing cubies:
• Placement: 8 corners being placed, for total of ${\displaystyle 8!}$
• Placement: 12 dedge left wing pieces that can go in 24 possible double-edge locations, for a total of ${\displaystyle {\dfrac {24!}{12!}}}$
• Placement: 12 dedge right-double-edge pieces that can go in 24 possible double-edge locations, for a total of ${\displaystyle {\dfrac {24!}{12!}}}$
• Placement: 24 center squares, which can be arranged in any configuration, for a total of ${\displaystyle 24!}$
Orienting cubies:
• Orientation: 3 orientations are possible on the 8 corners, placement of 7 determines the last, for total of ${\displaystyle 3^{7}}$
• Orientation: interestingly, the position of a dedge piece uniquely determines its orientation. If a dedge left wing piece is placed in a left dedge position, it will be oriented "correctly". If a dedge left wing piece is placed in a right dedge position, it will be oriented "backwards". And likewise for dedge right wing pieces being "correct" if placed in a right dedge spot and "backwards" if placed in a left dedge spot. Thus, there are 0 degrees of freedom due to orientation of double edge pieces. It's orientation is determined by the type of wing piece and the position being occupied.
• Orientation: Like the double-edge pieces, the center squares get 0 degrees of freedom due to orientation. Distributing 24 center squares among 24 center square positions uniquely determines the state of those center squares.
The grand total, then, is:
${\displaystyle T=3^{7}\times 8!\times \left({\dfrac {24!}{12!}}\right)^{2}\times 24!=91793597096746925044380089227138852984017051539472384000000000\sim 9.17935\times 10^{61}}$
## Number of permutations of 5x5
On a 5x5 cube, we apply the above techniques. We have fixed centers, so we don't consider those a type. The other cubie types are face cubies, corner cubies, and tredges, composed of tredge centers (the middle piece of the three pieces that make up a tredge) and tredge wings (the two end pieces of the three pieces that make up a tredge). There are 8 corner cubies, 9 face cubies on each face (and since 1 is fixed, actually 8 per face) for a total of 48 face cubies that can move, and 12 triple edges, for a total of 12 tredge centers, 12 tredge left wing pieces, and 12 tredge right wing pieces.
We run through enumeration of all the ways of placing each piece at a location on the cube, then multiply by the possible orientations, and reduce by the number of cube states that can't be reached.
Placing cubies:
• Placement: 8 corners being placed, for a total of ${\displaystyle 8!}$
• Placement: 48 face cubies being placed into 48 slots, any configuration, for a total of ${\displaystyle 48!}$
• Placement: 12 tredge left wings that can go into 24 possible double-edge locations for a total of ${\displaystyle {\dfrac {24!}{12!}}}$
• Placement: 12 tredge right wings that can go into 24 possible double-edge locations for a total of ${\displaystyle {\dfrac {24!}{12!}}}$
• Placement: 12 tredge centers that can go into 12 possible tredge center locations for a total of ${\displaystyle 12!}$
Orienting cubies:
• Orientation: 3 possible orientations of the 8 corners, but the placement of 7 of the corners determines the last, for a total of ${\displaystyle 3^{7}}$
• Orientation: the 48 face cubies have zero degrees of freedom because they have only one face and therefore no orientation.
• Orientation: the 24 triple edge wing cubies, like on the 4x4 cube, have no degrees of freedom due to orientation. This is because the position of the cubie and whether it is left or right wing piece determine its orientation.
• Orientation: the 12 triple edge (tredge) center cubies can be oriented in 2 ways each, ${\displaystyle 2^{12}}$ (note this is different from the case of 12 edges on the 3x3, where one single edge piece cannot be inverted by itself, so 11 edge cubies fix the state of the 12th cubie; on a 5x5, a single tredge center cubie can be inverted by itself.)
(Side note: how/why did we know to treat the tredge wings separately, instead of just tossing them all in a mix of 24 wing pieces that go into 24 locations? Because we know that in the solved state, solved tredges always consist of a left wing piece and a right wing piece.)
This makes the grand total:
${\displaystyle T=3^{7}\times 8!\times 48!\times \left({\dfrac {24!}{12!}}\right)^{2}\times 12!\times 2^{12}}$
This is equal to a number that is even larger than a googol (${\displaystyle 10^{100}}$):
${\displaystyle 3603399540205611914283185376146349701872354441978576086385485789093464996018186389353459508838400000000000000000}$
This is approximately
${\displaystyle \sim 3.6033995402\times 10^{111}}$
# Counting Cycles
## Length of Cycles on 3x3
One of the interesting features of a Rubik's Cube is that, starting from a solved cube, if any sequence of moves is repeated long enough on the cube, will eventually result in the cube returning to the solved state.
The simplest example is a single move (like U): after executing U four times, the cube is returned to the solved state.
It is a little less trivial with a sequence of moves like U'R'UR or LU, but these also eventually result in returning the cube to its solved state after 6 repetitions of the sequence U'R'UR or 105 repetitions of the LU sequence.<|endoftext|>
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1. ## Functions
Functions f and g are defined for x E R by
f:x --->e^x
g:x--->2x-3
(i) Solve the equation fg(x)=7
Function h is defined as gf
(ii)Express h in terms of x and state its range
(iii)Express h inverse in terms of x.
2. ## Re: Functions
(i)x=(ln7+3)/2...its easy. take ln both sides.
(ii)h(x)=2e^x-3
its minimum value will be when the term 2e^x is minimum since 3 is constant ..2e^x can attain a minimum zero(just attain) at x=-infinity.And its maximum value will be infinity.therefore range is (-3,infinity)
(iii)for finding h inverse take log again..
Hope it helps....
3. ## Re: Functions
Note that:
$f \circ g(x) = f(g(x))$
So, in this case, that means:
$f \circ g(x) = e^{g(x)} = e^{2x-3} = (e^{2x})(e^{-3}) = \frac{(e^x)^2}{e^3} = 7$
$e^x = \sqrt{7e^3}$, and taking logs, we get:
$x = \log(\sqrt{7e^3}) = \log((\sqrt{7})(e^{\frac{3}{2}}))$
$= \log(\sqrt{7}) + \log(e^{\frac{3}{2}}) = \log(\sqrt{7}) + \frac{3}{2} = \frac{\log(7) + 3}{2}$
Similarly,
$g \circ f(x) = 2(e^x) - 3$.
Since $e^x$ ranges from (0,∞), it's clear that $2(e^x)$ also ranges from (0,∞) and thus the range of $g \circ f$ is (-3,∞).
Consequently, $h^{-1} = f^{-1} \circ g^{-1}$ is only defined on (-3,∞). Can you figure out what it is, in terms of x?
4. ## Re: Functions
i solved it maybe it is : (ln(x+3)/2)=h-1(x)
here domain of h-1(x) is (-3,infinity) same as the range ..<|endoftext|>
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Cyberbullying refers to the use of technological communication tools and devices to intentionally inflict harm on an individual or group. This emerging public health problem is increasing around the world. In this entry, characteristics and consequences of cyberbullying are described. The prevalence of cyberbullying, and the various ways in which cyberbullying is perpetrated, are presented. Legal issues are noted, and suggestions for prevention and intervention are offered.
Research shows that victimization by cyberbullying is associated with decreased learning at school and with depression and anxiety. Individuals may be targeted based on their membership in diverse groups. A recent study found that intolerance was the motive for 16% of cases of cyberbullying. Students were targeted for their sexual orientation (most common), gender, disability, and religion. Other research found that about 70% of high school students in all racial groups had witnessed online racial discrimination; 29% of African American, 20% of White, and 42% of multiracial/other had been directly victimized based on race.
Several characteristics of cyberbullying increase the potential for harm. The anonymity in cyberspace means perpetrators can post hurtful content without fear of being identified. They feel free to say things that are unacceptable in usual interpersonal interactions. Anonymity also elevates fear and anxiety for the target. If a target does not know the source of the cyberbullying, it is easy to become suspicious of everyone, even friends.
The perpetrator of cyberbullying can act at any time and from any place, which denies the target a safe haven from the torment. Particularly for young people—whose social lives are increasingly conducted via technology—disconnecting from technology means ostracizing oneself, and is not generally considered a viable option. Another potent feature of cyberbullying is the enormous size of the potential audience. Rather than being seen by a few classmates, the cruelty can be witnessed by a seemingly limitless audience, intensifying the feelings of humiliation.
Nonverbal gestures, tone of voice, and emphasis give the listener in face-to-face communication critical information, but these clues are absent in online and text communications. Some messages intended to be humorous or joking are received as insulting and insensitive. Furthermore, the perpetrator does not witness the target's reaction, which deprives the perpetrator of feedback that could encourage empathy, remorse, or regret. Finally, the absence of any authority overseeing cyberspace diminishes perpetrators' concerns about consequences. Face-to-face bullying at school incurs the risk of being observed by a teacher, or by another student who could tell an authority; such risks are absent in cyberspace
It is difficult to estimate the prevalence of cyberbullying. Different researchers use different surveys, different definitions of cyberbullying, and different time frames for reporting (e.g., ever, in the last school term, in the last few months, in the last 2 weeks), leading to different prevalence rates. In addition, some studies have sampled children in school, whereas others have recruited participants from websites popular with young people. Thus, although one study shows much higher rates than another, the difference is likely to be caused by these measurement issues rather than a real difference in behavior. Although there are no absolutely reliable figures, one recent study, using a nationally representative sample of students in Grades 6 to 10, found 14% were involved in cyberbullying. Another study of youth ages 10 to 17 reported that 19% had been involved. These studies indicate that a significant number of students are involved in cyberbullying. Although younger students also are affected, no data from a nationally representative sample have been reported.
Research has also been inconclusive on gender differences in cyberbullying. One study found that boys were more likely to be cyberbullies and girls more likely to be victims, others have found girls were more involved than boys, and still others did not detect any gender differences. It may be that future studies will resolve the issue, or that researchers will discover that certain types of cyberbullying are more likely to be used by boys or girls.
Cyberbullies can target others in numerous ways. Hurtful messages can be conveyed via e-mail, text message, instant message (IM), in chat rooms, and even in online games. Some cyberbullies create websites in which unflattering photos (often doctored from benign originals), derogatory comments, and other content are posted to humiliate the target. Others gain access to the target's passwords, and then send nasty messages or post offensive content that appears to come from the target. This tactic can severely damage the target's relationships and reputation, and is very difficult to sort out even if the target eventually becomes aware of how his or her identity has been misappropriated.
The almost ubiquitous presence of photo and video cameras on cell phones, coupled with the small size of these devices, allows embarrassing photos or videos to be taken easily and surreptitiously and spread via the Internet or the phones themselves without the target's knowledge or permission. One study found that this form of cyberbullying was the most upsetting to targets. An offshoot of this form of bullying is sexting. This begins when someone voluntarily (usually in response to persuasion) sends a nude or revealing photo to a person he or she trusts to keep the photo private. Later (often after an argument or break-up), the recipient of the photo sends it to many people. In several cases, victims of sexting became so desperate that they committed suicide. About 20% of teenagers admit to engaging in this behavior. Some young men who were found to have images of underage females on their devices (even if they were sent by that female voluntarily) have been convicted of possession of child pornography, a sex offense.
Online games can also be used to cyberbully. Particularly in multiuser games, players communicate with each other, and that communication can easily be derogatory toward an individual. Websites for young children are designed to be fun and entertaining for children. However, there are features within the games that allow players to chat. Although there are site moderators and filters, children quickly figure out what words are acceptable, and find ways to get negative messages across. Social exclusion—not allowing some members to join activities—is easy and common on such sites.
Although many states have now passed legislation prohibiting cyberbullying, the effectiveness of such laws is uncertain. These laws typically require school districts to create and publicize policies related to cyberbullying. Only a few require educators to be trained to deal with this issue. The First Amendment of the U.S. Constitution protects free speech, which means even offensive and cruel speech is protected. For a school to take action, that speech must cause a “substantial disruption” of the educational environment. Existing laws against threatening someone's life, harassment, stalking, and impersonation can be used in some cases.
Cyberbullying emerged rather recently, and there has not been sufficient time to develop and test prevention programs. School districts interested in implementing programs are required to extrapolate from what is known about conventional bullying: whole school approaches are most effective, staff and students both need to be educated about cyberbullying, and the entire school community including parents and students need to be involved in shaping policies and planning programmatic changes. One hopes there will soon be rigorously evaluation studies of programs and curricula that will guide educators in their approach to this problem.
Digital Divide, Digital Learning, and Equal Access, Distance Education and Diversity, E-Inclusion and the Digital Divide, Free Speech and Diversity, Online Assessment and Diverse Learners
Summary—Cyberbullying and Other Online Safety Issues for Children, a Hearing Before the Subcommittee on Crime, Terrorism, and Homeland Security The
Antisocial behavior is intentionally acting in a way that can cause physical or emotional harm to other people or their property. Antisocial behavior
When most people think of bullying, they conjure up images of two or more young people on the school playground verbally and/or physically harassing<|endoftext|>
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The World of Tonality Part I: Scales
Now that we are getting into the Baroque period it is time to learn a lot more about how music works. In class we recently learned about scales.
The Piano Keyboard, Sharps and Flats
The piano keyboard is the most useful tool available for understanding the details of musical structure. This is because you can see every note, and they are all arranged into a distinctive pattern of white and black keys. The white-and-black pattern actually tells you a lot of information about how the musical universe works.
So, right off the bat we have to handle a little background information about how we name these notes.
If we just walk up and down the white keys on the piano, we hit all of the "plain" notes in the musical alphabet, which (in English) we call A, B, C and so forth. (If you studied music in another country you might be used to calling these notes do, re, mi etc.) This is also the result if we move a note up and down the lines and spaces of the musical staff.
If we want to refer to one of the black keys on the piano, we'll have to use a sharp or a flat. This modifies the basic note-names in a pretty simple way:
So we can name the notes around D like so:
Note that these black notes (and all notes, really) have more than one possible name. D-sharp can also be called E-flat, D-flat could also be C-sharp, and so on.
How do you know which notes are which
When we work with our scales, some students get a little anxious about knowing what all the keys are called. You don't really have to know this! But, nevertheless, here is a little guide you could use.
When you are getting started at the piano, really the most important note is C. You can kind of work everything out from there. Thus, you could get in the habit of looking for the group of two black keys - C is right next to that.
But, like I said, you don't really have to worry about what the keys are called.
Next, I usually do a little "stunt." In order to show how you can play almost any familiar tune on the white notes of the piano, I mime and sing a familiar melody as though I am playing the keys. The point is supposed to be that you should imagine (almost) all of the tunes in the world being played on the white keys.
If we take those notes from all of those tunes, sort them out, and put them in order, we get a scale. To start out we'll arrange them starting on C.
So this is our official definition of a scale: An ordered collection of notes that provides the basis for a piece of music.
The scale starts on C, then proceeds through D, E, F, and so on, until it loops around and arrives at the next C. This upper C is an octave higher than the first one. (This is the measure of distance between the two notes - it's called an octave because there are 8 white keys from C to C.) This is a somewhat magical property in music by which a higher note can sound somehow "the same" as a lower one.
In order to better understand what makes a scale a scale, we need to know about steps. A step is the distance from one note in the scale to the next. They come in two sizes - the "half step" is the smallest possible size, and a "whole step" is twice as big.
On the piano, a half step is the distance from one key to the immediate next one that touches it. This usually involves moving between the white and black keys, though there are two "cracks" in the pattern where a white key touches another white key - those are also a half step apart.
You can hear that a bunch of half steps sound "sneaky" or "slippery." It is somewhat unusual to hear so many half steps in a row.
A whole step is equal to two half steps. So when you make a whole step, you skip over a key on the way to the next one. For instance, if we go from the white key C to white key D, we'll "skip over" one black key in the middle. The majority of steps in music are actually whole steps, like this:
These sound a little more bold and substantial.
So, these are measures of musical distance. You could think of them like "inches" and "half-inches." Most of your notes are going to be an "inch" (= whole step) apart, but some of them are only a "half inch" (half step) apart.
Scales as a pattern of steps
The way to understand your major and minor scales is to think of them as a fixed pattern of whole steps and half steps. Major scales make one particular pattern, and minor scales make a different pattern.
By looking at the white keys on the piano, we can see the step pattern for our C major scale. White keys with an intervening black key make a whole step, and the "cracks" in the keyboard where there is no black key make a half step. The overall pattern could be memorized as "whole whole half, whole whole whole half."
If we want to make a different major scale, we just start on a different key and build the same WWHWWWH pattern. Here we'll go down to G and build a G major scale. We'll have to use one new black key (an F-sharp) in order to complete it.
And that's basically all you would need to know to figure out how to play your major scales! Different scales use different combinations of white and black keys, but they all make the same pattern of whole and half steps.
So far we've only made major scales. The other main flavor of scale is the minor scale. We can build a minor scale by going A to A on the white keys. It has a different pattern of steps (WHWWHWW) and a different sound.
We are not going to practice making minor scales in our class - we'll stick to the major.
Scales vs. Keys
For the sake of simplicity lets say that scales and keys are basically the same thing. If we say that a piece in "in the key of A major" that means it is based on the A major scale.
The most important property of a key is the home note, or tonic. This is also the note that our scales are based on, the C in C major. The tonic tends to sound "most important" and "most stable." (It is also, somewhat paradoxically, the least interesting-sounding tone - it is relatively colorless and "vanilla.") Any simple melody (like, say, a childrens' folk song) is probably going to end on the tonic note. (We usually test this in class - in each session I ask people to name simple songs, and then I show how you would play them in C major. Each one usually returns to C at the end.)
OK, that's it for now.
We will return to the world of music theory soon, to talk about chords.<|endoftext|>
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||The Mammals of Texas -
Cetacea : Family
Balaenopteridae : Balaenoptera musculus (Linnaeus)
Description. Largest of the whales;
upperparts slate gray with bluish cast, darker on head,
lips, and throat; paler on sides; underparts often
yellowish, sometimes spotted with white; dorsal fin
small, located posterior to vent; pectoral fin small,
about one-tenth of total length; commissure of mouth
nearly straight, except for downward curve near corner;
grooves on underparts numerous and extending posteriorly
past navel; whalebone black, broad basally, short (60-80
cm), and averaging 370 to each side; snout broad and
U-shaped. External measurements of one reported by
Scammon: length, 29 m; girth, 12 m; length of jawbone,
6.4 m. Estimated weight, up to 100 metric tons.
Distribution in Texas. Blue
whales occur in all oceans of the world. There are only
two records from the Gulf of Mexico; one stranded near
Sabine Pass, Louisiana, in 1924 and one stranded on the
Texas coast between Freeport and San Luis Pass in 1940.
Both of these identifications have been questioned,
however, and the occurrence of the blue whale in the Gulf
of Mexico is problematical. The current worldwide
population is only 11,000-12,000, with the current North
Atlantic population numbering 100-1,500 animals. Blue
whales are classified as "endangered."
Habits. Spring and summer finds
northern hemisphere blue whales migrating northward to
arctic feeding grounds. In fall and winter the whales
move back to temperate waters where mating and
parturition take place. This trend is also exhibited by
southern hemisphere blue whales although feeding grounds
are in the Antarctic. As the seasons are reversed between
the two hemispheres, northern and southern blue whales do
not interbreed in temperate and equatorial waters.
Feeding occurs mainly in higher
latitudes but is also common during migration. In the
North Pacific, for example, blue whales stop to feed off
California every fall of the year, on their way towards
northern waters. Small, shrimp-like crustaceans known as
"krill" predominate in the diet, tremendous
amounts of which are required to sustain a single whale.
An adult blue whale must consume about 3,000-5,000 kg of
krill daily to meet its energy requirements.
Female blue whales give birth to a
single calf in temperate or equatorial waters during the
winter months. Gestation is about 11 months and females
bear young every other year. The birth of a blue whale
has never been observed, but records from past whaling
activities indicate that a newborn blue whale is about 8
m in length and weighs 2-3 metric tons. The baby whale
nurses for about 8 months and during this time gains 90
kg per day, or 3.75 kg per hour! Sexual maturity
is reached at 5-6 years and the lifespan is unknown.
Illustration credit: Pieter<|endoftext|>
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...of the Dijon mustard
It is not quite as ancient as the world, but it has been part of the oldest Mediterranean cultures. For the Egyptians, Greeks and Romans mustard enhanced the flavors in food and enflamed the senses. The Greeks and the Romans called it Sinapsis. They used to grind and add the grains to food.
Originating in the Mediterranean basin, this herbaceous plant with bright green leaves and vibrant yellow petals, produces fruits containing several grains of mustard. The grains of wild mustard or black mustard, sinapis nigra, are ground and mixed with vinegar, grape must, wine or aromatic plants to make the dietary mustard. The word Mustard has its origins in the Latin "Mustum Ardens" which means burning grape must, because the Romans diluted the ground mustard grains in grape must. The grape must is the grape juice that has not been fermented yet, and which gives quite a spiced up flavor to mustard. In the Celt language it was known as Mwstardd, in England it became Mustard and in Norway Mustardhr whereas in Italy they kept the Latin word Sinapis.
Mustard is not quite as ancient as the world
In the first century AD it was found on every Roman table. Apicius even created mustard-based sauces to serve with certain dishes. Mustard was introduced in France when the Romans invaded Gaul. It was adopted easily as wine, vinegar and mustard got along very well.
Manuscripts exist from those days where the words Mustum, Mustardum and Mustarum appear, replacing the word Sinapis, as they substituted a new wine called Mustum to dilute the mustard, whereas the Romans used to use vinegar Acetum.
Mustard was also one of the earliest products to be used in medicine. Pythagoras praised it as an antidote to scorpion bites. Later, war wounds were treated with it. It was even believed that mustard protected against yellow fever epidemics.
In the 9th century Charlemagne recommended growing the spice in all of his estates. In the 17th century it was distributed to the poor of Dijon to protect them against chilblains (inflamed hands and feet as a result of the cold). Today mustard seeds are used in many preparations of preventive and natural medicine.
The 3 glorious dedicacies of Dijon...
Thriving in sunny conditions, the plant multiplies very quickly and is harvested hardly 2 months after dissemination. Ideal to restore the fertility of the ground, as its cultivation allows the soil to rest temporarily.
Of the forty different species of mustard, the most common are the Black mustard, the White, the Brown and the Dijon-Burgundy ones.
With Cream of Cassis and gingerbread, mustard is reputed to be one of the 3 glorious delicacies of Dijon.
Burgundy's chalky and densely wooded soil, rich in potassium and carbon has made possible the harvest of strong and piquant mustard seeds.
With the well earned reputation of fine gourmets, and the gastronomic knowledge and experience, the people of Burgundy were capable of making a condiment that soon became inseparably associated with the city of Dijon: the Dijon mustard.
The best Dijon mustard was the one that was ground with verjuice grape, which is a grape that is a little ripe and is more acidic. Already famous during the 13th century, Dijon mustard was sent to the provinces, dry or in flakes, and it was the user who use to dilute it with vinegar. This way it was able to conserve it for many years. It was in the 17th century that Dijon started to sell liquid mustard like the also renown Angers mustard.
Always present on the tables...
The name Moutarde de Dijon is reserved to mustards in paste made by sifted and sieved products. Always present on the tables of the Dukes of Burgundy, many barrels were regularly sent to the court of French kings, such as Louis XIV who never went anywhere without his mustard pot.
In the 18th century the mustard range was much wider than it is now.
There were lists of types of mustard, such as mustard powder, red mustard, fine mustard with anchovies and câpres, garlic mustard, lemon mustard, à la grecque, à la marquise, à la reine, with truffles, champagne mustard, rose mustard, and the list goes on and on…
They also started using vinegar to dilute the seeds and that's why mustard making was adopted in Orléans too, as Orléans was a major producer of vinegar.
Many producers made both vinegar and mustard. Excellent mustards were also made in Bordeaux and many other towns of France.
Today, mustard is a condiment that is made and used all over the world, in many different ways and in many different recipes.
Authentic Dijon-style mustard however, must follow the original recipe established in Dijon.
To end it on a more spiritual note with mustard in general, here are a couple of sayings on the mustard seed:
« The kingdom of heaven is like a mustard seed that someone took and sowed in his field ; it is the smallest of all the seeds, but when it has grown it is the greatest of shrubs and becomes a tree, so that the birds of the air come and make nests in its branches.»
The Holly Bible.
« If you have faith the size of a mustard seed, you will say to this mountain, "Move from here to there," and it will move; and nothing will be impossible for you. »
The Holly Bible.
« He who has in his heart as much faith as a grain of mustard seed will not enter hell, and he who has in his heart as much pride as a grain of mustard seed will not enter paradise. »
(Sayings of Muhammad)<|endoftext|>
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The pediatric vital signs (respiratory rate, pulse rate, and blood pressure) give critical information about the condition of child having trouble breathing or having unusual drowsiness, irritability, confusion, or coma.
The following tables are from the 2006 edition of Pediatric Advanced Life Support.
Normal Respiratory Rates by Age
Age Breaths per Minute
Infant (less than one year) 30 to 60
Toddler (1 to 3 years) 24 to 40
Preschooler (4 to 5 years) 22 to 34
School Age (6 to 12 years) 18 to 30
Adolescent (13 to 18 years) 12 to 16
If a child of any age is breathing faster than 60 breaths per minute, this is abnormal and needs to be evaluated urgently. A breathing rate of below normal also needs to be evaluated urgently and may mean impending respiratory arrest. The treatment for impending respiratory arrest is to help the infant or child breath intially with bag-mask ventilation.
Normal Heart Rates (Pulses) per minute by age
Age Awake Rate Mean Sleeping rate
Newborn to 3 months 85 t0 205 140 80 to 160
3 months to 2 years 100 to 190 130 75 to 160
2 years to 10 years 60 to 140 80 60 to 90
greater than 10 years 60 to 100 80 50 to 90
To find the heart rate check the pulse, listen to the heart with a stethoscope, or view the heart rate on a monitor.
A faster than normal heart rate is called a tachycardia and a slower than normal heart rate is called a bradycardia. Fever, pain, and anxiety can cause a rapid heart rate (tachycardia) as can respiratory distress, respiratory failure, shock, and abnormal heart rythms. The most important cause of a slow heart rate is abnormally low oxygen in the blood. A slow heart rate in an infant or child struggling to breath requires immediate help to breath (usually with bag-mask ventilatory support).
Abnormally Low Blood Pressure (Hypotension) by age
Age Systolic Blood Pressure (mm Hg)
Term Neonate (0 to 28 days) Less than 60
Infants (1 to 12 moths) Less than 70
Children (1 to 10 years) Less than 70 plus 2 times the age in years
Children ( greater than 10 years) Less than 90
The blood pressure consists of two numbers: an upper or higher number called the systolic blood pressure and a lower number or bottom number called the diastolic blood pressure. In a seriously ill child we direct our therapy at the low systolic number. Low systolic blood pressure in a child means decompensated shock (inadequate blood flow to meet the needs the body). A child in decompensated shock needs vigorous intravenous fluid administration and airway and breathing treatment to prevent cardiac arrest.<|endoftext|>
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Relation and Function creates a lot of confusion to the students as they both are closely related to each other. One needs to have a clear knowledge about them so as to differentiate between them. So here we provide differences between both of them.
Relation- Two or more sets can be related to each other by any means. Consider for an example two sets A and B having m and n elements respectively, we can have a relation with any ordered pair which shows a relation between the two sets.
Functions- A functions can have a same Range mapped as that of in Relation, such that a set of inputs is related with exactly one output.
Consider for an example Set A & Set B are related in a manner that all the elements of Set A are related to exactly one element of Set B or many elements of set A are related to one element of Set B. Thus this type of relation is said to be a Function.
It is to be noted that a function cannot have One to Many Relation between the set A and B.<|endoftext|>
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# Show that the probability that he will never stops gambling is zero
• MHB
• Akea
In summary, we have shown that the probability of the gambler never stopping gambling is zero, and the probability of him having 500 dollars at the time of stopping is very small while the probability of having zero dollars is 0.5.
Akea
Can u help me with this question pls
Assume that a gambler plays a fair game where he can win or lose 1 dollar in each round . His initial stock is 200 dollar. He decides a priory to stop gambling at the moment when he either has 500 dollars or 0 dollars in his stock. Time is counted by the number of rounds played.
i) show that the probability that he will never stops gambling is zero
ii ) Compute the probability that at the time when he stops gambling he has 500 dollars and the probability that he has zero dollars
.
Sure, I can help you with this question. Let's break it down into two parts.
i) To show that the probability that the gambler will never stop gambling is zero, we can use the concept of infinite geometric series. In this case, the gambler's stock can either increase by 1 dollar or decrease by 1 dollar in each round, with equal probability. Therefore, the probability of him winning 1 dollar in a round is 1/2 and the probability of him losing 1 dollar is also 1/2.
Now, let's consider the probability that he will never stop gambling. This means that he will always have a stock of either 200 dollars or more. This can be represented as an infinite geometric series with a starting value of 200 dollars, a common ratio of 1/2 and an infinite number of terms. The sum of this series can be calculated as:
S = a / (1-r) = 200 / (1-1/2) = 400
Since this sum is finite, it means that the probability that the gambler will never stop gambling is zero.
ii) To compute the probability that the gambler has 500 dollars at the time when he stops gambling, we can use the concept of binomial distribution. The gambler can reach 500 dollars in two ways: either by winning 300 rounds (since he starts with 200 dollars) or by losing 200 rounds. The probability of winning 300 rounds out of 500 is given by:
P = (500 choose 300) * (1/2)^300 * (1/2)^200 = 0.000003
Similarly, the probability of losing 200 rounds out of 500 is also 0.000003. Therefore, the total probability of him having 500 dollars at the time of stopping gambling is 0.000006, which is a very small value.
On the other hand, the probability of him having zero dollars at the time of stopping gambling is 0.5, since he can either win or lose in each round with equal probability.
I hope this helps! Let me know if you have any further questions.
## 1. What does it mean to "never stop gambling"?
The phrase "never stop gambling" refers to a person continuously participating in gambling activities without ever taking a break or stopping.
## 2. How can someone prove that the probability of never stopping gambling is zero?
Statistically, it is impossible for a person to continuously gamble without ever stopping. This is because gambling involves random chance, and eventually, a person will either run out of money or reach a point where they no longer want to continue gambling. Therefore, the probability of never stopping gambling is zero.
## 3. Is there any scientific evidence to support this claim?
Yes, there have been numerous studies on gambling behavior and addiction that have shown that individuals who engage in continuous gambling often experience negative consequences and eventually stop gambling. Additionally, mathematical probability also supports the idea that the probability of never stopping gambling is zero.
## 4. Can someone become addicted to gambling even if they never stop?
Yes, someone can still develop a gambling addiction even if they never stop gambling. Addiction is often characterized by the inability to control one's behavior despite negative consequences, and this can happen even if a person never stops gambling.
## 5. Is there any way to prevent someone from never stopping gambling?
There are various measures that can be taken to help someone who may be at risk of continuously gambling. These include seeking professional help, setting limits and boundaries, and finding healthier alternatives to gambling. However, ultimately, it is up to the individual to make the decision to stop gambling.
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# MCQ Questions for Class 7 Maths Chapter 11 Perimeter and Area with Answers
If you’re looking for a way to enhance your Class 7 Maths, then look no further than the NCERT MCQ Questions for Class 7 Maths Chapter 11 Perimeter and Area with Answers. MCQ Questions for Class 7 Maths with Answers is perfect for those who are in Class 7 Maths and want to get ahead of everyone else by mastering their subject skills as soon as possible! You can practice and test your subject knowledge by solving these Class 7 Maths Chapter 11 Perimeter and Area Objective Questions.
## Perimeter and Area Class 7 MCQs Questions with Answers
Solving Class 7 Maths Chapter 11 MCQ with Answers can be of great help to students as they will be aware of all the concepts. These MCQ Questions on Perimeter Class 7 Maths with Answers pave for a quick revision, thereby helping you learn more about this subject.
Question 1.
Find the area of a circle having radius 14 cm.
(a) 196 cm2
(b) 308 cm2
(c) 616 cm2
(d) None of these
Area of a circle πr2 = $$\frac { 22 }{ 7 }$$ × 14 × 14 = 616 cm2.
Question 2.
Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m.
(a) 20 m
(b) 5 m
(c) 15 m
(d) 10 m
Question 3.
One of the sides and the corresponding height of a parallelogram are 4 cm and 3 cm respectively. Find the area of the parallelogram.
(a) 12 cm2
(b) 7 cm2
(c) 6 cm2
(d) None of these
Area of a parallelogram = side × height.
Question 4.
Find the area of a verandah 2.25 m wide constructed outside a room 5.5 m long and 4 m wide.
(a) 36sq.m
(b) 63sq.m
(c) 64sq.m
(d) 84sq.m
Question 5.
Find the cost of lamination in the above question at the rate of Rs. 2.00 per cm2.
(a) Rs. 1200
(b) Rs. 2200
(c) Rs. 2400
(d) None of these
For lamination, area of frame is considered and product of area and cost is found.
Question 6.
When the circumference and area of a circle are numerically equal, what is the diameter numerically equal to?
(a) Area
(b) Circumference
(c) 271
(d) 4
Question 7.
A gardener wants to fence a circular garden of diameter 14 m. Find the length of the rope he needs to purchase.
(a) 44 m
(b) 28 m
(c) 88 m
(d) None of these
Fence means circumference of the circle. First calculate radius and then find circumference of the given circular garden.
Question 8.
If the area of a circle is 2464 m2, find its diameter,
(a) 56m
(b) 154m
(c) 176m
(d) 206m
Question 9.
What is the circumference of a circle of radius 7 cm ?
(a) 44 cm
(b) 49 cm
(c) 14 cm
(d) None of these
Circumference of a circle = 2πr, 2 × $$\frac { 22 }{ 7 }$$ × r = 44 cm.
Question 10.
The height in the area of a triangle
(a) $$\frac { (2×area) }{ base }$$
(b) $$\frac { base }{ (2×area) }$$
(c) $$\frac { (2×area) }{ height }$$
(d) None of these
Answer: (a) $$\frac { (2×area) }{ base }$$
Question 11.
Ayush made his picture on a rectangular sheet of length 60 cm and breadth 20 cm wide. Area of picture is :
(a) 1200 cm2
(b) 1250 cm2
(c) 1100 cm2
(d) None of these
Area of rectangular picture is length × breadth.
Question 12.
A rectangular garden is 65 cm long and 50 cm wide. Two cross paths each 2 m wide are to be constructed parallel to the sides. If these paths pass through the centre of the garden, find the cost of constructing the paths at the rate Rs. 69 per m2.
(a) Rs. 15000
(b) Rs. 15594
(c) Rs. 15500
(d) None of these
Question 13.
If a side of a square is 4 cm then its perimeter is :
(a) 16 cm
(b) 8 cm
(c) 12 cm
(d) None of these
Perimetre of square is 4 × side
Question 14.
The area of triangle is
(a) ($$\frac { 1 }{ 2 }$$) × base × height
(b) ($$\frac { 1 }{ 2 }$$) × (base + height)
(c) base (d) height
(d) None of these
Answer: (a) ($$\frac { 1 }{ 2 }$$) × base × height
Question 15.
The length and the breadth of a rectangular piece of land are S00 m and 300 m. Find the cost of the land if 1 m2 of the land costs Rs. 10000.
(a) Rs. 15,00,00,00
(b) Rs. 1500,000
(c) Rs. 1500,00
(d) None of these
Area of a rectangular piece is multiplied by cost of land.
Question 16.
A wire bent in the form of a circle of radius 42 cm is again bent in the form of a square. What is the ratio of the regions enclosed by the circle and the square?
(a) 11:12
(b) 21:33
(c) 22:33
(d) 14:11
Question 17.
Find the cost of frame in the above question at the rate of Rs. 3.00 per cm.
(a) Rs. 480
(b) Rs. 3600
(c) Rs. 240
(d) None of these
For frame cost perimeter is multiplied by 3.
Question 18.
Find radius of a circle of diameter 9.8 m.
(a) 4.9 m
(b) 19.6 m
(c) 10 m
(d) None of these
Radius of a circle is half of its diametre.
Question 19.
What is the circumference of a circle of diameter 10 cm?
(a) 35 cm
(b) 30 cm
(c) 31.4 cm
(d) None of these
Question 20.
In the above questions, find the perimeter of the rectangular sheet.
(a) 80 cm
(b) 100 cm
(c) 90 cm
(d) None of these
Question 21.
The length and breadth of a rectangular hall in a model are 0.4 m and 30 cmrespectively. What is the distance between the opposite corners of the wall in the model?
(a) 34.16m
(b) 50m
(c) 34.16 cm
(d) 50cm
Question 22.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth.
(a) 35 cm
(b) $$\frac { 100 }{ 35 }$$
(c) 15 cm
(d) None of these
2(l + b) = 100, 2l + 2b = 100, 2 × 35 + 2b = 100, 2b = 30 ÷ 2 = 15 cm.
Question 23.
Which figure encloses more area: a square of side 2 cm ; a rectangle of side 3 cm and 2 cm ;An equilateral triangle of side 4 cm?
(a) equilateral triangle
(b) square
(c) triangle
(d) None of these
Question 24.
In the above question, what will be the perimeter of the picture frame ?
(a) 80 cm
(b) 160 cm
(c) 120 cm
(d) None of these
Perimeter of a rectangular picture is 2(length + breadth).
Question 25.
Find the area of the following triangle :
(a) 6 cm2
(b) 4 cm2
(c) 16 cm2
(d) None of these
Area of triangle ∑ = base × altitude.
Question 26.
If we cut a square along one of its diagonals, two triangles are obtained. Area of each triangle obtained = __________.
(a) ($$\frac { 1 }{ 2 }$$)×Area of the square
(b) ($$\frac { 1 }{ 4 }$$)×Area of the square
(c) Area of the square
(d) None of these
Answer: (a) ($$\frac { 1 }{ 2 }$$)×Area of the square
Question 27.
Find the perimeter of a triangle with sides 4 cm, 6 cm and 10 cm
(a) 20 cm
(b) 24 cm
(c) 9 cm
(d) 18 cm
Question 28.
The area of a rectangular sheet is 500 cm2. If the length of the sheet is 25 cm. What is its width ?
(a) 20 cm
(b) 25 cm
(c) 50 cm
(d) None of these
Breadth of a rectangle is area ÷ length.
Question 29.
The base in the area of parallelogram is
(a) $$\frac { area }{ height }$$
(b) area × base
(c) $$\frac { area }{ base }$$
(d) area × height
Answer: (a) $$\frac { area }{ height }$$
Question 30.
Find the area of a square park whose perimeter is 320 cm ?
(a) 6400 m2
(b) 6000 m2
(c) 1280 m2
(d) None of these
Side of square 320 ÷ 4 = 80, Area of square is 80 × 80 = 6400 m2.
Question 31.
The area of parallelogram is
(a) height × height
(b) base × height
(c) base + height
(d) base × base
Question 32.
If a side of a square is 5 cm then its area is :
(a) 20 cm2
(b) 25 cm2
(c) 10 cm2
(d) None of these
Area of square is side × side.
Question 33.
Find the area of a triangle with a base of 20 cm and a height of 30 cm
(a) 300
(b) 100
(c) 400
(d) 600
Question 34.
Find the height x, if the area of the parallelogram is 24 cm2 and the base is 4 cm.
(a) 4 cm
(b) 6 cm
(c) 5 cm
(d) None of these
Height of a parallelogram = area ÷ side.
Question 35.
The area of a square whose perimeter is 4 m
(a) 2 m2
(b) 1 m2
(c) 3 m2
(d) None of these
Question 36.
Perimeter of a square field is 400 cm, find its one side.
(a) 300 cm
(b) 200 cm
(c) 100 cm
(d) None of these
Side of square = perimeter ÷ four.
Question 37.
The length and the breadth of a rectangular piece of land are 400 m and 250 m respectively. What is the cost of the land at Rs. 1000 per square metre?
(a) Rs.10 lakhs
(b) Rs.1 crore
(c) Rs.10 crores
(d) Rs.10 thousands
Question 38.
Find x in the given figure :
(a) 4.5 cm
(b) 6 cm
(c) 2 cm
(d) None of these
x(6 × 3) ÷ 4.
Match the following :
1. Perimeter of square (a) 2(l + b) 2. Area of square (b) 4 × side 3. Perimeter of rectangle (c) l × b 4. Area of a rectangle (d) a2
1. Perimeter of square (b) 4 × side 2. Area of square (d) a2 3. Perimeter of rectangle (a) 2(l + b) 4. Area of a rectangle (c) l × b
Match the following :
1. Radius of circle (a) Base × Height 2. Area of triangle (b) πr2 3. Area of a ||gm (c) Half of its diametre 4. Area of circle (d) $$\frac { 1 }{ 2 }$$ × Base × Altitude
1. Radius of circle (c) Half of its diametre 2. Area of triangle (d) $$\frac { 1 }{ 2 }$$ × Base × Altitude 3. Area of a ||gm (a) Base × Height 4. Area of circle (b) πr2
Match the following:
1. 1 cm (a) 100 cm 2. 1 m (b) 100 mm2 3. 1 cm2 (c) 100,00 cm2 4. 1 m2 (d) 10 mm
1. 1 cm (d) 10 mm 2. 1 m (a) 100 cm 3. 1 cm2 (b) 100 mm2 4. 1 m2 (c) 100,00 cm2
State whether the given statements are True or False.
Question 1.
Question 2.
Perimeter of a square = a2.
Question 3.
Area of circle = πr2.
Question 4.
Perimeter of rectangle = 2(l + b)
Fill in the blanks.
1. The distance around a circular region is known as its …………. .
2. 1 hectare = …………..
3. Area of a circle = ……… × r2
4. Area of a rectangle = ……….. × Breadth
5. ………… is the part of plane occupied by the closed figure.
6. Perimetre of a rectangle = 2 (Length + ………)
7. Area of a ||gm = Base × ………..
8. ……….. is the distance around a closed figure.
9. Area of a square = Side × ………..
10. Perimetre of a square = 4 × ………..<|endoftext|>
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How should ELA teachers approach informational texts?
This question informs the lessons in this unit, which emphasize approaches to teaching informational texts in the context of literature study. Rather than replacing or superceding the study of literature, I see informational texts as ways to amplify literature so that students see it as relevant to their lives.
Thus, rather than a shift away from both the literary canon and contemporary imaginative texts, the CCSS offers teachers a unique opportunity to embrace innovative approaches to teaching both informational texts and the imaginative literature that led us to teach English.
Show the video montage of Billy Collins' "The Names." This is only a part of the poem, but I have included the entire poem in the resources.
After listening to and viewing "The Names," I introduced students to Thunder Dog, which tells the story of a blind man's survival of the World Trade Center attack.
Since I own a copy of the book, I showed it to students. Additionally, I have an on again-off again book blog called "Hanging by a Book," where I occasionally blog about books, particularly YA lit. I started this particular blog so that I would have exemplars to show students when I require them to review books. I wrote a post about Thunder Dog and had it projecting on the screen as I talked about the book.
I told students the book appeals to me because it tells the story of someone who not only survived 9-11 but whose life is about surviving against all odds. The author earned both an undergrad and grad degree, among many other successes.
I shared a personal story about my father, who lost his site when I was a child as a result of juvenile diabetes complications.
I ended by telling students that the book isn't great literature but that it is an inspiring story of an everyday hero who has much to say about overcoming life's obstacles.
After finishing the book talk, I told students we were going to look at a picture taken on 9-11 but not released until five years later. I explained that they would be "creating a narrative" about the picture. Then I projected the picture onto the screen so students could view it. 9-11 Photograph.
Next, I showed them the citation information, which includes the photographer and source of publication.
I then put the picture back on the screen and gave students ten minutes to write. Before they started, I told them they could get a closer look at the picture if they wanted.
The students wrote quietly. I reviewed the essays I would be sharing w/ them and didn't write because I already know the history of the photo.
After the students finished, I asked them to share their observations about the photo.
Three boys sitting close to the screen and who typically don't contribute to the dialogue eagerly shared their observations, which included these comments:
"They look like they are just chillin."
"They look like they don't have a care in the world."
"They look like they're in a relaxing place, like a park."
"That girl with the tramp stamp doesn't have a clue."
"They're having a relaxing time."
Perhaps their proximity to the image influenced their responses, but I found their responses compelling, and I suspect they might have silenced some in the class whose opinions differed with those of the boys.
One student asked, "Did you take that picture?" That was a surprising question.
When the students had finished offering ideas about the photo, I asked them to "burrito fold" a piece of paper I had given them when they entered the room. (Burrito fold is simply a trifold of the paper turned to the horizontal position).
Next, I asked the students to "hamburger fold" the paper, which is folding the paper in half.
I reminded them that we had talked about writing as having a beginning, middle, and end the previous day.
Next, I told the students that when the photo was published in 2006, Frank Rich wrote a New York Times opinion piece about the photo that I would be reading to them. I also explained that Rich's point of view evoked a response from a writer at SLATE and that I would also read that response. I gave the students listening directions:
1. Listen as I read.
2. Think about the beginning, middle, and end as you listen.
3. Put ideas from the beginning in the first column, from the middle in the second column, and from the end in the first column. Student Notes Example 1
4. Put Rich's ideas on the top of the page and the SLATE response on the bottom. Student Notes Example 2
I then read Rich's essay slowly. Because it is long, I paused in the middle to give the students a chance to rest their hands. Frank Rich Whatever Happened to the America.docx
I paused once again when I reached the end of Rich's piece and then proceeded to read the SLATE article, assuring students that it's much shorter. Frank Rich Is Wrong About That 9.docx
When I finished reading, I asked the students what they thought about the two pieces. Many students said they liked the SLATE piece better and noted that it gave those in the picture the "benefit of the doubt."
We talked about how both the Rich and SLATE pieces quoted the photographer, Thomas Hoepker.
The students wanted to know about the responses from those in the picture. I promised to put those on our LMS site for accessibility and explained that the SLATE article links to them.
I asked students if they know why I chose not to read the point of view of those in the picture. Eventually, we got to the idea that pictures express ideas, and that we consume those ideas and make judgments about them. I explained that I wanted them to think about the picture as observers on the outside rather than as participants in the action.
I asked students if they could see these ideas in their notes and as I read the contrasting opinions. Student Notes Example 3
Once the students grasped the idea that they could compare and contrast Rich's and SLATE's response to the Hoepker photo, I revisited the idea that they too, could compose essays that respond to multiple genres.
Finally, I told the students that we'll be going to the lab Friday so they can search for an essay w/ a topic and/or theme similar to their poem's and that I would help them find one if they want. I also announced that our new media specialist will be coming into the class tomorrow and sharing information about using the library resources to find an informational text on which to base their comparison/contrast essays.<|endoftext|>
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A staunch Zionist and one of the first Knesset members, Beba Trachtenberg Idelson was a champion of religious freedom and women’s rights in the new State of Israel. In 1916, Idelson joined a self-defense group in Ukraine run by the Zionist youth group Ze’iri Zion to protect herself from violent anti-Semitism, and in 1917 she married Israel Idelson, Ze’iri Zion’s leader. Although the couple was exiled to Siberia in 1923, the sentence was commuted to exile in Palestine. On their arrival in 1926, Beba worked odd jobs to pay the bills, then became politically active, serving as secretary of the Council of Working Women from 1931–1974, helping found the Organization of Working Women and becoming involved with the Council of Women’s Organizations. She represented the Mapai (Workers) Party as a delegate to the Zionist Congress four times between 1935 and 1946, became a member of the provisional government in 1948, and was elected to the Knesset for the Mapai Party in 1949. As a member of the Knesset’s committees for law and justice, foreign affairs, labor, and crafting a constitution, she created protections for women workers, fought the religious monopoly on marriage and divorce, and tried to define women’s equality in terms of universal human rights. She retired from the Knesset in 1965.
How to cite this page
Jewish Women's Archive. "Beba Idelson." (Viewed on May 25, 2019) <https://jwa.org/people/idelson-beba>.<|endoftext|>
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# What's the difference between 1000e^0.05t and 1000*1.05^t?
• B
• Karagoz
In summary, the difference between 1000e^0.05t and 1000*1.05^t is that the first expression represents exponential growth with a constant value of 1000 and a growth rate of 5%, while the second expression represents compound interest with a starting value of 1000 and a growth rate of 5% per time period. This means that the first expression will result in a higher value over time, as the growth rate is applied to the current value each time period, while the second expression will result in a slower growth rate as the growth is only applied to the initial value. Additionally, the first expression is continuous, while the second expression is discrete.
Karagoz
We have a population of y = 1000 at year 1980 (call it year 0).
Every year the population growth rate is 5% per year.
y' shows the growth rate of the y (population).
Since the population grows by 5% every year, the growth rate is:
y' = 0.05y.
This is a simple differential equation.
When y(0) = 1000
Then using a math software, the formula for the population is:
y(t) = 1000*e^(0.05t)
OR
We have a population of z = 1000 at year (1980) (call it year 0)
The population growth rate 5% per year.
Since the population grows by 5% per year, we can say:
z(t) = 1000*(1+0.05)^t = 1000*1.05^t
Derivation of z(t):
z’(t) = 1000(ln1.05)*e^(t*ln1.05)
Written as differential equation:
z’(t)=(ln1.05)*z(t)
The formula similar to z(t) is used when describing the growth of a money (in a bank at a interest rate of 5%).
Both the formula y(t) and formula z(t) describes growth rate by 5% per year.
But it’s obvious that z(t) ≠ y(t)
What is the difference between y(t) = 1000*e^(0.05t) and z(t) = 1000*1.05^t when both describes a growth rate of 5% per year?
What does z(t) describe and what does y(t) describe, and what’s the difference between what each formula describe?
Karagoz said:
What is the difference between y(t) = 1000*e^(0.05t) and z(t) = 1000*1.05^t when both describes a growth rate of 5% per year?
The difference is that y(t) is the exact description and z(t) is an approximation. If you write ##y(t)=y_0e^{\lambda~t}## and do a series expansion for the exponential, you get ##y(t)=y_0 (1 +\lambda t+\lambda t^2/2+\lambda t^3/6 +...)=y_0 \sum_{k=0} ^{\infty} (\lambda t)^k/k!##
Here ##y_0=1000## and ##\lambda = 0.05##. Your expression for z(t) tosses out all terms higher than first order, therefore it is approximately correct and not equal to y(t).
To clarify my question:
When calculating what the value of the money C will be after x years, when having an interest rate at n, we use the formula:
f(x) = C*(1+n)^x
E.g. z(t) = 1000*(1+0.05)^t = 1000*1.05^t
But why don't we use the other formula: g(x) = C*e^(0.05n)?
What would it mean if we used g(x) instead of f(x) when calculating how the value C in a bank account will after x years with an interest rate n?
Karagoz said:
What would it mean if we used g(x) instead of f(x) when calculating how the value C in a bank account will after x years with an interest rate n?
The plot below shows the difference 1000*e0.05*t - 1000*1.05t for 0 < t < 50 years. At the end of 50 years, you will be able to withdraw $12182.49 according to the exponential calculation and$11467.40 according to the approximation, a shortfall of \$715.09.
Karagoz said:
But why don't we use the other formula: g(x) = C*e^(0.05n)?
It's not that "we" don't use the exponential formula, it's that the banks don't use it. Any guesses why? If you choose a bank that compounds interest more often than yearly, your interest will be reinvested sooner so the bank keeps less of your money.
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• InterestDifference.png
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• InterestDifference.png
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So the formula f(x) means yearly interest of n is compounded once a year.
But if banks did use g(x), that would mean the yearly interest of n is compounded far more frequently?
Karagoz said:
But if banks did use g(x), that would mean the yearlt interest of n was compounded every second?
Or millisecond, or microsecond, or ...
Edit: Some banks compound interest quarterly, some monthly and some daily.
## 1. What do the variables in the equations represent?
The variable "t" represents time in years, while the number "1000" represents an initial value or amount.
## 2. What is the significance of the "^" symbol in the equations?
The "^" symbol represents an exponent, meaning that the number that follows it is raised to a certain power.
## 3. How do the two equations differ?
The first equation, 1000e^0.05t, uses the mathematical constant "e" which is approximately equal to 2.71828. The second equation, 1000*1.05^t, uses a constant multiplier of 1.05. This results in a difference in the rate of growth or decay over time.
## 4. Which equation should be used in different situations?
The choice of which equation to use depends on the specific situation and the type of growth or decay being measured. For example, if the growth or decay is continuous, such as in a biological population, the first equation may be more appropriate. If the growth or decay is discrete, such as in a financial investment, the second equation may be more applicable.
## 5. Can these equations be used to predict future values?
Yes, these equations can be used to predict future values by plugging in a specific value for "t" and solving for the resulting value. However, the accuracy of the prediction may depend on the accuracy of the initial values and the assumptions made about the growth or decay process.
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Today’s class will be dedicated to open discussion on McGregor’s The Party. Questions below will guide our discussion and may be amended or supplemented based on your interests:
How are the leading party figures profiled by McGregor similar to – and different from – leading political figures in the United States or other political context with which you are familiar?
Why is there confusion on the part of visitors as to whether there are Communists in China today? In what sense is the Chinese Communist Party (CCP) still Marxist? In what sense has it transformed its approach?
The article notes that Nixon sought to “expunge the use of the term ‘communist’ when talking about the Chinese.” Meanwhile, Chinese leaders have made subtle but noticeable effort to downplay their socialist credentials, for example by introducing Xi Jinping as China’s “president” or by swapping the English name of the People’s University of China for Renmin University of China. Why might this be the case for either – or both – side?
What meaning(s) does “democracy” have in China today? What does the CCP mean when it declares itself “democratic”?
The political scientist Archie Brown notes that within different political cultures, citizens have a different assumed relationship to power. In some places, they feel they have a right – or even an obligation – to advocate for group or community interests to help shape public policy. In other places, they feel that they should defer to leaders to make good choices on their behalf. Which approach better describes China’s political culture? What evidence can you draw upon to support your view? How does such a distinction relate to the discussion of “democracy,” above?
Why has the party retained elements of the secrecy of the pre-1949 period? What might be the advantages and disadvantages of this approach?
How would you characterize the party’s relationship to the law, the Constitution of the People’s Republic, and the idea of “rule by law”?
McGregor asks, “If the Party, locked in its ossified Leninist ways, is secretive, corrupt, hostile to the rule of law and vindictive in pursuit of its enemies, it begs the question: how on earth did it manage to preside over one of the greatest spurts of economic growth and wealth creation in recorded history?” How does he go about answering this question? Do you find his response convincing? Why or why not?
What are the most important challenges that lie ahead for the party? Do you think the party is likely to adapt to meet these challenges? Why or why not?<|endoftext|>
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The friction present in a sleeve bearing depends upon several factors. For example, the constant value of the coefficient of friction depends upon what materials comprise the sleeve and bearing. Other important factors include the size of the shafts, the rotational speed and the lubricant viscosity. In a rolling-element bearing, its static friction (and the torque needed to overcome that force) usually exceeds its running friction. Consider all these factors to calculate friction in a given sleeve bearing.
Determine the materials of which the inner bearing and outer sleeve are composed. Refer to tables of standard coefficients of friction to determine a rough value for the coefficient of friction between those two materials in particular. Note this dimensionless constant using the greek letter "mu" (µ).
Determine the sizes of the bearing and of the sleeve. Note the radius of the shaft using the letter "R."
Sciencing Video Vault
Subtract the area of the bearing shaft from the area of the sleeve to calculate the radial clearance between them. Note this clearance, using the same units as for R, but use the letter "c."
Determine the viscosity of the lubricant in the bearing. Note the force per area multiplied by time with the letter "P."
Determine the speed at which the bearing revolves in the shaft. Note the revolutions per second with the letter "n."
Multiply 2 by pi squared (π^2) by µ (the coefficient of friction) by n (the speed of revolution) by R (the radius of the shaft).
Multiply P (the viscosity of the lubricant) by c (the radial clearance between shaft and sleeve).
Finally, divide the value calculated in Step 6 by the value calculated in Step 7 to complete Petroff's equation. The result is the force of the friction present in the sleeve bearing.
Friction = (2 * π^2) * (µ * c * R) / (P * c)
π (pi) has no dimensions or units. µ (coefficient of friction) has no dimensions or units, and varies depending on the materials in question. Often, assuming a range of values here proves more useful than pinpointing one. c (radial clearance) uses units of area, such as meters squared. n (speed) uses units of movement / time, such as revolutions per second. R (shaft radius) uses units of space, such as meters. P (viscosity) uses units of force / area * time, such as kilogram-meters / meters squared * seconds.<|endoftext|>
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# Addition of Fractions with Unlike Denominators
Examples, solutions, and videos to help Grade 4 students learn how use visual models to add two fractions with related units using the denominators 2, 3, 4, 5, 6, 8, 10, and 12.
Common Core Standards: 4.NF.3a, 4.NF.3d, 4.NF.1, 4.MD.2
### NYS Common Core Grade 4 Module 5, Lesson 20
Lesson 20 Concept Development
Problem 1: Add unit fractions with related denominators using tape diagrams.
1/3 + 1/6 =
Lesson 20 Problem Set
1. Use a tape diagram to represent each addend. Decompose one of the tape diagrams to make like units. Then write the complete number sentence. Part (a) is partially completed.
a. 1/4 + 1/8
b. 1/4 + 1/12
c. 2/6 + 1/3
d. 1/2 + 3/8
e. 3/10 + 3/5
f. 2/3 + 2/9
2. Estimate to determine if the sum is between 0 and 1 or 1 and 2. Draw a number line to model the addition. Then write a complete number sentence. Part (a) has been completed for you.
a. 1/2 + 1/4
b. 1/2 + 4/10
c. 6/10 + 1/2
d. 2/3 + 3/6
e. 3/4 + 6/8
f. 4/10 + 6/5
3. Solve the following addition problem without drawing a model. Show your work.
2/3 + 4/6
Lesson 20 Homework
1. Use a tape diagram to represent each addend. Decompose one of the tape diagrams to make like units. Then write the complete number sentence. Part (a) is partially completed.
a. 1/3 + 1/6
b. 1/2 + 1/4
c. 3/4 + 1/8
d. 3/4 + 1/8
e. 1/4 + 5/12
f. 3/5 + 3/10
2. Estimate to determine if the sum is between 0 and 1 or 1 and 2. Draw a number line to model the addition. Then write a complete number sentence. Part (a) has been completed for you.
a. 1/3 + 1/6
b. 3/5 + 7/10
c. 5/12 + 1/4
d. 3/4 + 5/8
e. 7/8 + 3/4
f. 1/6 + 5/3
3. Solve the following addition problem without drawing a model. Show your work.
5/6 + 1/3
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.<|endoftext|>
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# Principle of Data Reduction
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## Transcription
1 Chapter 6 Principle of Data Reduction 6.1 Introduction An experimenter uses the information in a sample X 1,..., X n to make inferences about an unknown parameter θ. If the sample size n is large, then the observed sample x 1,..., x n is a long list of numbers that may be hard to interpret. Any statistic, T (X), defines a form of data reduction or data summary. For example, the sample mean, the sample variance, the largest observation, and the smallest observation are four statistics that might be used to summarize some key features of the sample. The statistic summarizes the data in that, rather than reporting the entire sample x, it reports only that T (x) = t. For example, two samples x and y will be treated as equal, if T (x) = T (y) is satisfied. The advantages and consequences of this type of data reduction are the topics of this chapter. We study three principles of data reduction. 1. The Sufficiency Principle promotes a method of data reduction that does not discard information about θ while achieving some summarization of the data. 2. The Likelihood Principle describes a function of the parameter determined by the observed sample, that contains all the information about θ that is available from the sample. 3. The Equivariance Principle prescribes yet another method of data reduction that still preserves some important features of the model. 11
2 12 CHAPTER 6. PRINCIPLE OF DATA REDUCTION 6.2 The Sufficiency Principle Sufficiency Principle: If T (X) is a sufficient statistic for θ, then any inference about θ should depend on the sample X only through the value T (X). That is, if x and y are two sample points such that T (x) = T (y), then the inference about θ should be the same whether X = x or Y = y is observed Sufficient Statistics Definition A statistic T (X) is a sufficient statistic for θ if the conditional distribution of the sample X given the value of T (X) does not depend on θ. To use this definition to verify that a statistic T (X) is a sufficient statistic for θ, we must verify that P (X = x T (X) = T (x)) does not depend on θ. Since {X = x} is a subset of {T (X) = T (x)}, P θ (X = x T (X) = T (x)) = P θ(x = x and T (X) = T (x)) P θ (T (X) = T (x)) P θ (X = x) = P θ (T (X) = T (x)) = p(x θ) q(t (x)) θ), where p(x θ) is the joint pmf/pdf of the sample X and q(t θ) is the pmf/pdf of T (X). Thus, T (X) is a sufficient statistic for θ if and only if, for every x, the above ratio is constant as a function of θ. Theorem If p(x θ) is the joint pdf or pmf of X and q(t θ) is the pdf or pmf of T (X), then T (X) is a sufficient statistic for θ if, for every x in the sample space, the ratio p(x θ)/q(t (x) θ) is constant as a function of θ. Example (Binomial sufficient statistic) Let X 1,..., X n be iid Bernoulli random variables with parameter θ, 0 < θ < 1. Then T (X) = X X n is a sufficient statistic for θ. Note that T (X) counts the number of X i s that equal 1, so T (X) Bi(n, θ). The ratio of pmfs is thus p(x θ) θ x q(t (x) θ) = i (1 θ) 1 x i ( n t) θ t (1 θ) n t (define t = n x i) = θp x i (1 θ) P (1 x i ) ( n ) t θ t (1 θ) n t = ( 1 n ) = ( 1 n t P xi ) Since this ratio does not depend on θ, by Theorem 6.2.1, T (x) is a sufficient statistic for θ.
3 6.2. THE SUFFICIENCY PRINCIPLE 13 Example (Normal sufficient statistic) Let X 1,..., X n be iid N(µ, σ 2 ), where σ 2 is known. We wish to show that the sample mean, T (X) = X, is a sufficient statistic for µ. n f(x µ) = (2πσ 2 ) 1/2 exp( (x i µ) 2 /(2σ 2 )) = (2πσ 2 ) n/2 exp{ [ (x i x) 2 + n( x µ) 2 ]/(2σ 2 )} Recall that the sample mean X N(µ, σ 2 /n). Thus, the ratio of pdf is p(x θ) q(t (x)) θ) = (2πσ2 ) n/2 exp{ [ n (x i x) 2 + n( x µ) 2 ]/(2σ 2 )} (2πσ 2 /n) 1/2 exp{ n( x µ) 2 /(2σ 2 )} = n 1/2 (2πσ 2 ) (n 1)/2 exp{ (x i x) 2 /(2σ 2 )}, which does not depend on µ. By Theorem 6.2.1, the sample mean is a sufficient statistic for µ. Example (Sufficient order statistics) Let X 1,..., X n be iid from a pdf f, where we are unable to specify any more information about the pdf. It then follows that n! n f X(1),...,X (n) (x) = f X(x i ), if x 1 <... < x n 0 otherwise where (x (1),..., x (n) is the order statistic. statistic. By Theorem 6.2.1, the order statistic is a sufficient Of course, this is not much of a reduction, but we should not expect more with so little information about the density f. However, even if we specify more about the density, we still may not be able to get much of a sufficient reduction. For example, suppose that f is the Cauchy pdf f(x θ) = 1 π(x θ) 2 or the logistic pdf f(x θ) = e (x θ) (1+e (x θ) ) 2. It turns out that outside of the exponential family of distributions, it is rare to have a sufficient statistic of smaller dimension than the size of the sample, so in many cases it will turn out that the order statistics are the best that we can do. Theorem (Factorization Theorem) Let f(x θ) denote the joint pdf or pmf of a sample X. A statistic T (X) is a sufficient statistic for θ if and only if there exist functions g(t θ) and h(x) such that, for all sample points x and all parameter points θ, f(x θ) = g(t (x) θ)h(x). (6.1)
4 14 CHAPTER 6. PRINCIPLE OF DATA REDUCTION Proof: We give the proof only for discrete distributions. Suppose T (X) is a sufficient statistic. Choose g(t θ) = P θ (T (X) = t) and h(x) = P (X = x T (X) = T (x)). Because T (X) is sufficient, the conditional probability h(x) does not depend on θ. Thus, f(x θ) = P θ (X = x) = P θ (X = x and T (X) = T (x)) = P θ (T (X) = T (x))p (X = x T (X) = T (x)) = g(t (x) θ)h(x) So factorization (6.1) has been exhibited. Also, the last two lines above imply that g(t (x) θ) is the pmf of T (X). Now assume the factorization (6.1) exists. Let q(t θ) be the pmf of T (X). Define A T (x) = {y : T (y) = T (x)}. Then f(x θ) g(t (x) θ)h(x) = q(t (x) θ) q(t (x) θ) g(t (x) θ)h(x) = A g(t (y) θ)h(y) (density transformation) T (x) = g(t (x) θ)h(x) g(t (x) θ) A T (x) h(y) h(x) = A h(y) T (x) (since T is a constant on A T (x)) Since the ratio does not depend on θ, by Theorem 6.2.1, T (X) is a sufficient statistic for θ. To use the Factorization Theorem to find a sufficient statistic, we factor the joint pdf of the sample into two parts. One part does not depend on θ and it constitutes the h(x) function. The other part depends on θ, and usually it depends on the sample x only through some function T (x) and this function is a sufficient statistic for θ. Example (Normal sufficient statistic) Let X 1,..., X n be iid N(µ, σ 2 ), where σ 2 is known. The pdf can be factored as f(x µ) = (2πσ 2 ) n/2 exp{ We can define (x i x) 2 /(2σ 2 )} exp{ n( x µ) 2 /(2σ 2 )}. g(t θ) = exp{ n(t µ) 2 /(2σ 2 )}
5 6.2. THE SUFFICIENCY PRINCIPLE 15 by defining T (x) = x, and h(x) = (2πσ 2 ) n/2 exp{ (x i x) 2 /(2σ 2 )}. Thus, by the Factorization Theorem, T (X) = X is a sufficient statistic for µ. Example (Uniform sufficient statistic) Let X 1,..., X n be iid observations from the discrete uniform distribution on 1,..., θ. That is, the unknown parameter, θ, is a positive integer and the pmf of X i is 1 θ x = 1, 2,..., θ f(x θ) = 0 otherwise. Thus, the joint pmf of X 1,..., X n is θ n x i {1,..., θ} for i = 1,..., n f(x θ) = 0 otherwise. Let I A (x) be the indicator function of the set A; that is, it is equal to 1 if x A and equal to 0 otherwise. Let N = {1, 2,...} be the set of positive integers and let N θ = {1, 2,..., θ}. Then the joint pmf of X 1,..., X n is n n f(x θ) = θ 1 I Nθ (x i ) = θ n I Nθ. Defining T (x) = max i x i, we see that n n I Nθ = ( I N (x i ))I Nθ (T (x)). Thus we have the factorization n f(x θ) = θ n I Nθ (T (x))( I N (x i )). By the factorization theorem, T (X) = max i X i is a sufficient statistic for θ. Example (Normal sufficient statistic, both parameters unknown) Assume that X 1,..., X n are iid N(µ, σ 2 ), and that both µ and σ 2 are unknown so the parameter vector is θ = (µ, σ 2 ). Let T 1 (x) = x and T 2 (x) = s 2 = n (x i x) 2 /(n 1). f(x θ) =(2πσ 2 ) n/2 exp{ [ (x i x) 2 + n( x µ) 2 ]/(2σ 2 )} = (2πσ 2 ) n/2 exp{ (n(t 1 µ) 2 + (n 1)t 2 )/(2σ 2 )}.
6 16 CHAPTER 6. PRINCIPLE OF DATA REDUCTION Let h(x) = 1. By the factorization theorem, T (X) = (T 1 (X), T 2 (X)) = ( X, S 2 ) is a sufficient statistic for (µ, σ 2 ). The results can be generalized to the exponential family of distributions. Theorem Let X 1,..., X n be iid observations from a pdf or pmf f(x θ) that belongs to an exponential family given by k f(x θ) = h(x)c(θ) exp{ w i (θ)t i (x)} where θ = (θ 1,..., θ d ), d k. Then T (X) = ( t 1 (X j ),..., t k (X j )) is a sufficient statistic for θ. j=1 j= Minimal Sufficient Statistics In any problem, there are many sufficient statistics. For example, 1. It is always true that the complete sample, X, is a sufficient statistic. 2. Any one-to-one function of a sufficient statistic is also a sufficient statistic. Recall that the purpose of a sufficient statistic is to achieve data reduction without loss of information about the parameter θ; thus, a statistic that achieves the most data reduction while still retaining all the information about θ might be considered preferable. Definition A sufficient statistic T (X) is called a minimal sufficient statistic if, for any other sufficient statistic T (X), T (x) is a function of T (x). To say that T (x) is a function of T (x) simply means that if T (x) = T (y), then T (x) = T (y). Let T = {t : t = T (x) for some x X } be the image of X under T (x). Then T x partitions the sample space into sets A t, t T, defined by A t = {x : T (x) = t}. If {B t : t T } are the partition sets for T (x) and {A t : t T } are the partition sets for T (x), then Definition states that every B t is a subset of some A t. Thus, the partition associates with a minimal sufficient statistic, is the coarsest possible partition for a sufficient statistic, and a minimal sufficient statistic achieves the greatest possible data reduction for a sufficient statistic.
7 6.2. THE SUFFICIENCY PRINCIPLE 17 Example (Two Normal sufficient statistics) Suppose that X 1,..., X n are observations from N(µ, σ 2 ) with σ 2 known. We know that T (X) = X is a sufficient statistic for µ. The factorization theorem shows that T (X) = ( X, S 2 ) is also a sufficient statistic for µ. T (X) can be written as a function of T (X) by defining the function r(a, b) = a. Then T (x) = x = r( x, S 2 ) = r(t (x)). Theorem let f(x θ) be the pmf or pdf of a sample X. Suppose there exists a function T (x) such that, for every two sample points x and y, the ratio f(x θ)/f(y θ) is a constant as a function of θ if and only if T (x) = T (y). Then T (x) is a minimal sufficient statistic for θ. Proof: To simply the proof, we assume f(x θ) > 0 for all x X and θ. First we show that T (X) is a sufficient statistic. Let T = {t : t = T (x) for some x X } be the image of X under T (x). Define the partition sets induced by T (x) as A t = {x : T (x) = t}. For each A t, choose and fix one element x t A t. For any x T, x T (x) is the fixed element that is in the same set, A t, as x. Since x and x T (x) are in the same set A t, T (x) = T (x T (x)) and, hence, f(x θ)/f(x T x θ) is constant as a function of θ. Thus, we can define a function on X by h(x) = f(x θ)/f(x T x θ) and h does not depend on θ. Define a function on T by g(t θ) = f(x t θ). Then it can be seen that f(x θ) = f(x T x θ)f(x θ) f(x T x θ) = g(t (x) θ)h(x) and, by the factorization theorem, T (X) is a sufficient statistic for θ. Now to show that T X is minimal, let T (X) be any other sufficient statistic. By the factorization theorem, there exist function g and h such that f(x θ) = g (T (x) θ)h (x). Let x and y be any two sample points with T (x) = T (y). Then f(x θ) f(y θ) = g (T (x) θ)h (x) g (T (y) θ)h (y) = h (x) h (y). Since this ratio does not depend on θ, the assumptions of the theorem imply that T (x) = T (y). Thus, T (x) is a function of T (x) and T (x) is minimal. Example (Normal minimal sufficient statistic) Let X 1,..., X n be iid N(µ, σ 2 ), both µ and σ 2 unknown. Let x and y denote two sample points, and let ( x, S 2 x) and (ȳ, S 2 y) be the sample means and variances corresponding to the x and y samples, respectively. Then the ratio of densities is f(x µ, σ 2 ) f(y µ, σ 2 ) = exp{[ n( x2 ȳ) 2 + 2nµ( x ȳ) (n 1)(s 2 x s 2 y)]/(2σ 2 )}.
8 18 CHAPTER 6. PRINCIPLE OF DATA REDUCTION This ratio will be constant as a function of µ and σ 2 if and only x = ȳ and s 2 x = s 2 y. Thus, by Theorem 6.2.4, ( X, S 2 ) is a minimal sufficient statistic for (µ, σ 2 ). Example (Uniform minimal sufficient statistic) Suppose X 1,..., X n are iid uniform observations on the interval (θ, θ + 1), < θ <. Then the joint pdf of X is 1 θ < x i < θ + 1, i = 1,..., n, f(x θ) = 0 otherwise, which can be written as 1 max i x i 1 < θ < min i x i f(x θ) = 0 otherwise Thus, for two sample points x and y, the numerator and denominator of the ratio f(x θ)/f(y θ) will be positive for the same values of θ if and only if min i x i = min i y i and max i x i = max i y i. Thus, we have that T (X) = (X (1), X (n) ) is a minimal sufficient statistic. This example shows the dimension of a minimal sufficient statistic may not match the dimension of the parameter. A minimal sufficient statistic is not unique. Any one-to-one function of a minimal sufficient statistic is also a minimal sufficient statistic. For example, T (X) = (X (n) X (1), (X (n) + X (1) )/2) is also a minimal sufficient statistic in Example 6.2.9, and T (X) = ( n X i, n X2 i ) is also a minimal sufficient statistic in Example Ancillary Statistics Definition A statistic S(X) whose distribution does not depend on the parameter θ is called an ancillary statistic. Alone, an ancillary statistic contains no information about θ. An ancillary statistic is an observation on a random variable whose distribution is fixed and known, unrelated to θ. Paradoxically, an ancillary statistic, when used in conjunction with other statistics, sometimes does contain valuable information for inferences about θ. Example (Uniform ancillary statistic) Let X 1,..., X n be iid uniform observations on the interval (θ, θ + 1), < θ <. The range statistic R = X (n) X (1) is an ancillary statistic.
9 6.2. THE SUFFICIENCY PRINCIPLE 19 Recall that the cdf of each X i is 0 x θ F (x θ) = x θ θ < x < θ θ + 1 x. Thus, the joint pdf of X (1) and X (n) is n(n 1)(x (n) x (1) ) n 2 θ < x (1) < x (n) < θ + 1 g(x (1), x (n) θ) = 0 otherwise Making the transformation R = X (n) X (1) and M = (X (n) + X (1) )/2, we see the joint pdf of R and M is n(n 1)r n 2 0 < r < 1, θ + r/2 < m < θ + 1 r/2 h(r, m θ) = 0 otherwise Thus, the pdf for R is h(r θ) = θ+1 r/2 θ+r/2 n(n 1)r n 2 dm = n(n 1)r n 2 (1 r), 0 < r < 1. This is a beta pdf with α = n 1 and β = 2. Thus, the distribution of R does not depend on θ, and R is ancillary. Example (Location and scale family ancillary statistic) Let X 1,..., X n be iid observations from a location family with cdf F (x θ), < θ <. Show that the range statistic R = X (n) X (1) is an ancillary statistic. Let X 1,..., X n be iid observations from a scale family with cdf F (x/σ), σ > 0. Show that any statistic that depends on the sample only through the n 1 values X 1 /X n,..., X n 1 /X n is an ancillary statistic Sufficient, Ancillary, and Complete Statistics A minimal sufficient statistic is a statistic that has achieved the maximal amount of data reduction possible while still retaining all the information about the parameter θ. Intuitively, a minimal sufficient statistic eliminates all the extraneous information in the sample, retaining only that piece with information about θ. Since the distribution of an ancillary statistic does not depend on θ, it
10 20 CHAPTER 6. PRINCIPLE OF DATA REDUCTION might be suspected that a minimal sufficient statistic is unrelated to an ancillary statistic. However, this is not necessarily the case. Recall Example in which X 1,..., X n were iid obs from a uniform(θ, θ + 1) distribution. We have pointed out that the statistic (X (n) X (1), (X (n) + X (1) )/2) is a minimal sufficient statistic, and in Example we showed that X (n) X (1) is an ancillary statistic. Thus, in this case, the ancillary statistic is an important component of the minimal sufficient statistic. Certainly, the ancillary statistic and the minimal sufficient statistic are not independent. The following example shows that an ancillary statistic can sometimes give important information for inference about θ. Example (Ancillary precision) Let X 1 and X 2 be iid obs from the discrete distribution that satisfies P (X = θ) = P (X = θ + 1) = P (θ + 2) = 1/3, where θ, the unknown parameter, is any integer. It can be shown with an argument similar to that in Example that (R, M), where R = X (n) X (1) and M = (X (n) + X (1) )/2, is a minimal sufficient statistic. To see how R might give information about θ, consider a sample point (r, m), where m is an integer. First we consider only m, θ must be one of three values, either θ = m or θ = m 1 or θ = m 2. With only the information that M = m, all three θ values are possible values. But now suppose we get the information that R = 2. Then it must be the case that X (1) = m 1 and X (2) = m + 1, and the only possible value for θ is θ = m 1. Thus, the knowledge of the value of the ancillary statistic R has increased our knowledge about θ. Of course, the knowledge of R alone would give us no information about θ. For many important situations, however, our intuition that a minimal sufficient statistic is independent of any ancillary statistic is correct. A description of situations in which this occurs relies on the next definition. Definition Let f(t θ) be a family of pdfs or pmfs for a statistic T (X). The family of probability distributions is called complete if E θ g(t ) = 0 for all θ implies P θ (g(t ) = 0) = 1 for all θ. Equivalently, T (X) is called a complete statistic.
11 6.2. THE SUFFICIENCY PRINCIPLE 21 Example (Binomial complete sufficient statistic) Suppose that T has a binomial (n, p) distribution, 0 < p < 1. Let g be a function such that E p g(t ) = 0. Then 0 = E p g(t ) = ( ) n g(t ) p t (1 p) n t = (1 p) n t t=0 ( ) n ( p g(t) t 1 p for all p, 0 < p < 1. Let r = p/(1 p), 0 < r <. The last expression is a polynomial of degree n in r, where the coefficient of r t is g(t) ( n t). For the polynomial to be 0 for all r, each coefficient must be 0. Thus, g(t) = 0 for t = 0, 1,..., n. Since T takes on the values 0, 1..., n with probability 1, this yields that P p (g(t ) = 0) = 1 for all p. Hence, T is a complete statistic. t=0 ) t Example Let X 1,..., X n be iid uniform (0, θ) observations, 0 < θ <. Show that T (X) = max i X i is a complete statistic. Using an argument similar to that used in one of previous examples, we can see that T (X) = max i X i is a sufficient statistic and the pdf of T (X) is nt n 1 θ n 0 < t < θ f(t θ) = 0 otherwise. Suppose g(t) is a function satisfying E θ g(t ) = 0 for all θ. Since E θ g(t ) is a constant of θ, its derivative with respect to θ is 0. Thus we have that 0 = d dθ E θg(t ) = d dθ θ 0 = (θ n ) d dθ g(t)nt n 1 θ n dt θ = θ n ng(θ)θ n = θ 1 ng(θ) 0 ng(t)t n 1 dt + ( d θ dθ θ n ) ng(t)t n 1 dt 0 Since θ 1 n 0, it must be g(θ) = 0. This is true for every θ > 0, hence, T is a complete statistic. Theorem (Basu s Theorem) If T (X) is a complete and minimal sufficient statistic, then T (X) is independent of every ancillary statistic. Proof: We give the proof only for discrete distributions. Let S(X) be any ancillary statistic. Then P (S(X) = s) does not depend on θ since S(X) is ancillary. Also the conditional probability, P (S(X) = s T (X) = t) = P (X {x : S(x) = s} T (X) = t),
12 22 CHAPTER 6. PRINCIPLE OF DATA REDUCTION does not depend on θ because T (X) is a sufficient statistic (recall the definition!). Thus, to show that S(X) and T (X) are independent, it suffices to show that P (S(X) = s T (X) = t) = P (S(X) = s) (6.2) for all possible values t T. Now, P (S(X) = s) = t T P (S(X) = s T (X) = t)p θ (T (X) = t). Furthermore, since t T P θ(t (X) = t) = 1, we can write P (S(X) = s) = t T P (S(X) = s)p θ (T (X) = t) Therefore, if we define the statistic g(t) = P (S(X) = s T (X) = t) P (S(X) = s), the above two equations show that E θ g(t ) = t T g(t)p θ (T (X) = t) = 0 for all θ. Since T (X) is a complete statistic, this implies that g(t) = 0 for all possible values t T. Hence, (6.2) is verified. It should be noted that the minimality of the sufficient statistic was not used in the proof of Basu s theorem. Indeed, the theorem is true with this word omitted, because a fundamental property of a complete statistic is that it is minimal. Theorem (Complete statistics in the exponential family) Let X 1,..., X n be iid observations from an exponential family with pdf or pmf of the form f(x θ) = h(x)c(θ) exp ( k w j (θ)t j (x) ), where θ = (θ 1, θ 2,..., θ k ). Then the statistic T (X) = ( t 1 (X i ), t 2 (X i ),..., t k (X i )) j=1 is complete if {w 1 (θ),..., w k (θ)) : θ Θ} contains an open set in R k.
13 6.3. THE LIKELIHOOD PRINCIPLE 23 Example (Using Basu s theorem) Let X 1,..., X n be iid exponential observations with parameter θ. Consider computing the expected value of g(x) = X n X X n. We first note that the exponential distributions form a scale parameter family and thus g(x) is an ancillary statistic. The exponential distributions also form an exponential family with t(x) = x and so, by Theorem 6.2.6, T (X) = is a complete statistic and by, Theorem 6.2.3, T (X) is a sufficient statistic. theorem, T (X) and g(x) are independent. Thus we have X i Hence, by Basu s θ = E θ X n = E θ T (X)g(X) = (E θ T (X))(E θ g(x)) = nθe θ g(x). Hence, for any θ, E θ g(x) = 1/n. 6.3 The likelihood Principle The likelihood function Definition Let f(x θ) denote the joint pdf or pmf of the sample X = (X 1,..., X n ). Then, given that X = x is observed, the function of θ defined by L(θ x) = f(x θ) is called the likelihood function. This definition almost seems to be defining the likelihood function to be the same as the pdf or pmf. The only distinction between these two functions is which variable is considered fixed and which is varying. When we consider the pdf or pmf f(x θ), we are considering θ as fixed and x as the variable; when we consider the likelihood function L(θ x), we are considering x to be the observed sample point and θ to be varying over all possible parameter values. If we compare the likelihood function at two parameter points and find that L(θ 1 x) > L(θ 2 x),
14 24 CHAPTER 6. PRINCIPLE OF DATA REDUCTION then the sample we actually observed is more likely to have occurred if θ = θ 1 than if θ = θ 2, which can be interpreted as saying that θ 1 is a more plausible value for the true value of θ than is θ 2. We carefully use the word plausible rather than probable because we often think of θ as a fixed value. Furthermore, although f(x θ), as a function of x, is a pdf, there is no guarantee that L(θ x), as a function of θ, is a pdf. LIKELIHOOD PRINCIPLE: If x and y are two sample points such that L(θ x) is proportional to L(θ y), that is, there exists a constant C(x, y) such that L(θ x) = C(x, y)l(θ y) for all θ, then the conclusions drawn from x and y should be identical. Note that C(x, y) may be different for different (x, y) pairs but C(x, y) does not depend on θ. 6.4 The Equivariance Principle The first type of equivariance might be called measurement equivariance. It prescribes that the inference made should not depend on the measurement scale that is used. The second type of equivariance, actually an invariance, might be called formal invariance. It states that if two inference problems have the same formal structure in terms of the mathematical model used, then the same inference procedure should be used in both problems. The elements of the model that must be the same are: Θ, the parameter space; {f(x θ) : θ Θ}, the set of pdfs or pmfs for the sample; and the set of allowable inferences and consequences of wrong inferences. For example, Θ may be Θ = {θ : θ > 0} in two problems. But in one problem θ may be the average price of a dozen eggs and in another problem θ may refer to the average height of giraffes. Yet, formal invariance equates these two parameter spaces since they both refer to the same set of real numbers. Equivariance Principle: If Y = g(x) is a change of measurement scale such that the model for Y has the same formal structure as the model for X, then an inference procedure should be both measurement equivariant and formally equivariant. Example (Binomial equivariance) Let X Bionmial(n, p) with known n and p. Let T (x) be the estimate of p that is used when X = x is observed. Rather than using the number of successes, X, to make an inference about p, we could use the number of failures, Y = n X. We
15 6.4. THE EQUIVARIANCE PRINCIPLE 25 can see that Y Binomial(n, q = 1 p). Let T (y) be the estimate of q that is used when Y = y is observed, so that 1 T (y) is the estimate of p when Y = y is observed. If x successes are observed, then the estimate of p is T (x). But if there are x successes, then there are n x failures and 1 T (n x) is also an estimate of p. Measurement equivariance requires that these two estimates be equal, that is, T (x) = 1 T (n x), since the change from X to Y is just a change in measurement scale. Furthermore, the formal structures of the inference problems based on X and Y are the same. X and Y both have binomial(n, θ) distributions, 0 θ 1. So formal invariance requires that T (z) = T (z) for all z = 0,..., n. Thus, measurement and formal invariance together require that T (x) = 1 T (n x) = 1 T (n x). If we consider only estimators satisfying the above equality, then we have greatly reduced and simplified the set of estimators we are wiling to consider. Whereas the specification of an arbitray estimator requires the specification of T (0),..., T (n), the specification of an estimator satisfying the above equality requires the specification only of T (0),..., T ([n/2]), where [n/2] is the greatest integer not larger than n/2. This is the type of data reduction that is always achieved by the equivariance principle. The inference to be made for some sample points determines the inference to be made for other sample points.
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On November 26, MSL, carrying its Curiosity rover, blasted into space. At last, two years late, it was on its way to the Red Planet. Thirteen thousand onlookers watched it soar from Cape Canaveral. It would take eight months for the spacecraft to journey the 352 million miles to Mars. Its goal was to search for evidence that microscopic life might once have lived on Mars—or be capable of living there now. It also contained sensors that would detect radiation affecting the ability of astronauts to land there some day.64 Administrator Bolden declared, “We are very excited about sending the world’s most advanced scientific laboratory to Mars. MSL will tell us critical things we need to know about Mars, and while it advances science, we’ll be working on the capabilities for a human mission to the red planet and to other destinations where we’ve never been.”65
Everyone connected with the mission was elated, but they also knew the risk of failure. A Russian probe to Phobos, a Mars moon, had launched on November 9 and failed to escape Earth’s orbit. The U. S. spacecraft was now on a trajectory to Mars. But the landing, several months hence, would be daunting. And what would the rover find? Time would tell. The future of the Mars program—and maybe NASA—depended greatly on the answer.<|endoftext|>
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Gabriel Maybank || October 25, 2016
What is Compression?
A compressor is a fine-tune processor that controls the dynamic range of an instrument/voice. Dynamics are the variations in volume between softer and louder parts of a song, recording, instrument, etc. The goal of a compressor is to maintain musical dynamics. Have you ever listened to a live singer and noticed that when he or she sings quiet or talks it’s almost not audible, but when he or she sings they are near blowing the speakers? A compressor is used to control that range between soft and loud by bringing up the quiet levels and compressing or lowering the louder levels. The compressor acts as an automatic fader. The compressor reduces or boosts the signal at a specified ratio above or below a set threshold.
Input Gain: The amount of the signal that is sent to the compressor’s input stage.
Threshold: Level at which the compressor proportionately begins to reduce or boost the signal. If the threshold is set for -15dB, all signals above this level will be attenuated and anything below will be unaffected.
Output Gain: The amount of signal sent to the device’s output or speakers after compression.
Slope Ratio: The input-to-output gain ratio determines how much signal is needed to cause the signal to be reduced by a certain –dB. For example, a 2:1 ratio will produce a 1-dB increase in the output for every 2-dB increase in the input.
Attack: Determines how fast or slowly the compressor will reduce the signals beyond the threshold.
Release: Determines how slowly the compressor will restore the signal back to its dynamic level once it has fallen below the threshold.
Knee: “Soft-Knee” is the gradual and smooth transitioning between no treatment and full treatment of the compressor making the compression more transparent. “Hard-Knee” is the strict limit and sharp transition between no treatment and full treatment of the compression and is a more obvious transition to achieve a desired effect.
Gain Reduction Meter: Displays how much gain reduction is applied, when the reduction starts and stops, and a visual indication of the attack and release activity.
Input Meter: Displays input gain
Output Meter: Displays what we hear and ensures output does not exceed a specific limit.
Types of Compressors
Downward Compressor: Reduces the level of signals above the threshold, making loud sounds quieter.
Upward Compressor: Boost the level of signal below the threshold, making quiet sounds louder.
Limiter: Ensures no signal exceeds the threshold by reducing any signals above the threshold down to the threshold level
Expander: Reduces any signal below the threshold, making quiet sounds quieter.
Upward Expander: Boost any signal above the threshold, making loud sounds louder
Gate: Attenuates all signals below the threshold by a fixed amount.
Ducker: Attenuates all signals above the threshold by a fixed amount.
Setting Up a Compressor
Setting the compressor is not a visual action; an audio engineer has to train their ears to hear the differences being made while adjusting settings.
DO NOT MIX WITH YOUR EYES.
- Threshold all the way up so there is no compression
- Ratio all the way up or 1:1
- Low or short attack and release
Extent: Set the threshold below the highest signal
Degree: Set the ratio down from maximum
Attack: Set the attack longer
Release: Set the release longer
- Compression can be used to bring instruments forward or backward in the mix by bringing up or lowering the attack.
- Compression can be used to lengthen the sustain of instruments for a creative effect. Result is unnatural reverb decay. This achieved by having a low threshold, high ratio, fast attack and fast release so the gain reduction recovers as quickly as possible
- Compression can be used as a De-esser, or sibilance attenuator.
- Compression can be used to add punch to an instrument like a bass guitar by starting from a medium attack moving towards a slower attack and medium threshold and slower release time.
- Use a fast attack of 1 to 5ms and a ratio between 6:1 to 15:1 and moderate to heavy gain reduction to control brass instruments.
- A faster attack time and a slower release can add to an electric guitar’s “bite” in the mix, but because of the guitar amp and pedals they do not need much compression.
- For the kick, a 4:1 ratio with an attack setting of 10ms or slower can emphasize the initial attack and add depth and presence.
- A snare’s attack settings should be faster to catch the initial transient (the hit). The threshold should be set for minimal reduction during a quiet part and a larger amount of gain reduction during the louder sections.
- Compression on vocals should begin with a threshold setting of 0dB, a ratio of 4:1, and the attack and release at their midpoints. Gain reductions that are somewhere in between 3 and 6 dB will often sit well in the mix.
- A gate can be used to reduce or remove a hum coming from a faulty direct box or over bleed from another louder instrument
- A ducker can be used to combat masking. It is an effective way to clear space for more the important instruments.<|endoftext|>
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The working of a hot air balloon is actually much simpler then you could imagine. The basic workings of a hot air balloon are based on the fact that heated air rises in cooler air because it is lighter. Because hot air rises it means that it can’t escape from the bottom of the envelope.
There are 3 main parts to a balloon:
The envelope – This is effectively the balloon, which is the fabric where all the hot air is contained in order to lift the balloon, vital for hot air balloon rides! All the heated air is stored in this area to lift the balloon, as the air in here cools, the burner is used to heat it up.
The Burner – This is where the air in the envelope is heated. The pilot controls the amount of heat which propels the balloon, in order to ascent the pilot increases the flame which is shot up into the envelope.
The basket – this is where the pilot and passengers stand during a balloon ride, surrounding them with breath-taking views. This is usually made from wicker, which works well because it is sturdy, flexible, and relatively lightweight. This flexibility absorbs some of the impacts on landing so the passengers don’t feel the impact force.
The burner uses propane gas to heat up the air which essentially is what keep the balloon in flight. It must be fired regularly throughout hot air balloon flights in order to keep the balloon stable. You might be wondering how the balloon moves from side to side? Well, it’s very simple. The wind blows in different directions, depending on the altitude the balloon is at, so in order to move around, the pilot simple ascends of descends to the right height and will be blown along by the wind.
The final step of any hot air balloon ride is the descent; this is equally as simple as the rest of the flight! There is a valve at the top of the balloon which is covered by a piece of fabric. When the pilot decides it is time to descent he simple pulls on a chord which opens the value. This lets cool air into the envelope which lowers the temperature inside the balloon, slowly bringing it back down to earth.
Of course all this depends on the weather, balloons cannot fly in rainy or windy conditions as it will damage the balloon and make for unsafe flying. In order to check this pilots send up a pilball to double check the safety of the wind, and to check which direction the balloon is going to take off.<|endoftext|>
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Parametric Equations
Butterfly Curve
The Butterfly Curve is one of many beautiful images generated using parametric equations.
Basic Description
We often graph functions by letting one coordinate be dependent on another. For example, graphing the function $f(x) = y = x^2$ has y values that depend upon x values. However, some complex functions are best described by having the coordinates be described using an equation of a separate independent variable, known as a parameter. Changing the value of the parameter then changes the value of each coordinate variable in the equation. We choose a range of values for the parameter, and the values that our function takes on as the parameter varies traces out a curve, known as a parametrized curve. Parametrization is the process of finding a parametrized version of a function.
Parametrized Circle
One curve that can be easily parametrized is a circle of radius one:
We use the variable t as our parameter, and x and y as our normal Cartesian coordinates.
We now let $x = cos(t)$ and $y = sin(t)$, and let t take on all values from $0$ to $2\pi$.
When $t=0$, the coordinate $(1,0)$ is hit. As t increases, a circle is traced out as x initially decreases, since it is equal to the cosine of t, and y initially increases, since it is equal to the sine of t. The circle continues to be traced until t reaches $2\pi$, which gives the coordinate $(1,0)$ once again.
It is also useful to write parametrized curves in vector notation, using a coordinate vector: $\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} cos(t) \\ sin(t)\\ \end{bmatrix}$
The butterfly curve in this page's main image uses more complicated parametric equations as shown below.
A More Mathematical Explanation
Note: understanding of this explanation requires: *Linear Algebra
Sometimes curves which would be very difficult or even impossible to graph in terms of elementary fun [...]
Sometimes curves which would be very difficult or even impossible to graph in terms of elementary functions of x and y can be graphed using a parameter. One example is the butterfly curve, as shown in this page's main image.
This curve uses the following parametrization:
$\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} \sin(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right) \\ \cos(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right)\\ \end{bmatrix}$
Parametric construction of the butterfly curve
Parametrized Surfaces and Manifolds
In the above cases only one independent variable was used, creating a parametrized curve. We can use more than one independent variable to create other graphs, including graphs of surfaces. For example, using parameters s and t, the surface of a sphere can be parametrized as follows: $\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} sin(t)cos(s) \\ sin(t)sin(s) \\cos(t) \end{bmatrix}$
While two parameters are sufficient to parametrize a surface, objects of more than two dimensions will require more than two parameters. These objects, generally called manifolds, may live in higher than three dimensions and can have more than two parameters, so cannot always be visualized. Nevertheless they can be analyzed using the methods of vector calculus and differential geometry.<|endoftext|>
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Conjunctive use of water relates to the combined use of ground and surface water. Due to the augmented water source, higher water reliability can be achieved. Conjunctive use therefore functions as a buffer for periods of water scarcity. The idea of this management approach is to use surface water when the water table is high and change to groundwater when the water table is low. This technique might be especially important as a buffer function for mitigating impacts of climate change, such as increased heat and drought (FOSTER et al. 2010).
Even though there is no consistent definition of conjunctive use of water, it is most commonly referred to as the combined use of ground and surface water (FOSTER et al. 2010). The natural hydrologic connection between groundwater and surface water is used in order to improve water reliability (see also the water cycle). This practice can become particularly important for mitigating problems related to climate change, such as increased drought or overexploitation and decreases in groundwater availability (see also water resources assessment and water pollution) (DUDLEY & FULTON 2005).
Two main usage phases of conjunctive use are differentiated:
- Recharge: when the water table is high, the use of surface water is to be maximised. The recharge of groundwater can be enhanced artificially by surface water recharge and subsurface water recharge.
- Recovery: During dry season, water is drawn from groundwater resources (for possible measures to draw groundwater, see groundwater sources).
It is important to keep recharge and recovery in balance (see also water balance estimation). However, as there are usually several users (e.g. farmers) relying on the same water source, individual use may lead to overexploitation of groundwater. This is a widespread problem, because conjunctive use often occurs unplanned and by default. To enhance the benefits of this practice, better coordination between the users is needed, which is part of conjunctive water management discussed below (WORLD BANK 2006).
Conjunctive water management
Adapted from DUDLEY & FULTON (2005)
Conjunctive water management is an improved form of conjunctive use. Additional to the combined use of surface water and groundwater, components of groundwater management tools, such as water balance estimation and water resources assessment, are added. In comparison to conjunctive use, conjunctive water management helps to better manage water use between different parties, leading to well-prepared recharge and recovery phases and therefore to a lower probability of harming groundwater sources.
Today, conjunctive water management is less frequently implemented than conjunctive use, because of lacking institutions or coordination mechanisms (see also building an institutional framework) (WORLD BANK 2006).
If usage is done unplanned, groundwater can be exploited or salinity of groundwater may increase, leading to diminished agricultural productivity. Increased salinity may also be a threat to the security of drinking water supplies (FOSTER et al. 2010).
Conjunctive use of water is often done by default and therefore without special knowledge. To optimise the effectiveness of conjunctive use, system managers should be defined and well trained (see also developing human resources). Management of conjunctive use is best done at the local level where the special needs and conditions are known and the interested actors can be involved. Also, the potential for conjunctive use varies with differing hydrogeological settings, including such factors as average rainfall and geomorphological position. For further information on variations of the dynamics and constraints of conjunctive use (see FOSTER et al. 2010).
Achieving well-managed conjunctive use can be impeded by institutional constraints, economic considerations or the political climate (see also creating policies and a legal framework and building an institutional framework). It is therefore necessary to consider these factors during the implementation process. Participatory monitoring and evaluation can help to enhance the needed learning process and coordination within the community. On-going monitoring is recommended in order to constantly improve conjunctive use.
Shared use of groundwater and surface water resources.
Conjunctive use is often part of an overall water resources management based on water balance estimations. It may be used for both urban and rural areas. Its effectiveness differs due to different hydrogeological settings.
High performance, which may have long-term impacts on water availability and quality.
Special knowledge is needed. Otherwise, groundwater sources may be exploited.
On-going monitoring and adaptations are needed.
Only reliable if monitoring and coordination is done well.
Improves sustainability of water sources.
The potential for conjunctive use varies with differing hydrogeological settings, including such factors as average rainfall and geomorphological position. The implementation choice therefore needs to be adapted to the prevalent conditions (FOSTER et al. 2010). It is best implemented on the local level, as it is easier to identify the special needs and circumstances of the community and monitoring on local level is more easily applied.
Conjunctive Water Management. Economic Tools for Evaluating Alternative Policy and Management Options
A paper describing the role and key components of hydro-economic models for effective conjunctive water management. Three such models, the steady state lumped parameter model, the dynamic lumped parameter model and the distributed parameter model are described more precisely.HAFI, A. (2006): Conjunctive Water Management. Economic Tools for Evaluating Alternative Policy and Management Options. Canberra: Australian Bureau of Agricultural and Resource Economic (ABARE) URL [Accessed: 19.09.2012] PDF
This paper discusses six different water diversion possibilities applied in Colorado, USA. Insights are given on how the concept of conjunctive use can add to the reliability of a more standardised water rights plan and may also increase the yield.JEHN, L. J. (n.y): Conjunctive Use Issues. Optimizing Resources and Minimizing Adverse Effects. Denver, CO: Jehn Water Consultants, Inc. URL [Accessed: 19.09.2012] PDF
This paper explores the reasons underpinning the apparent poor approach to full integration in the management and use of both water sources, and the absence of more coordinated planning. This paper is intended to provide insight into these barriers to adoption and hence provide a new focus on an old paradigm; a focus intended to make progress with the objective of improved water management and water use efficiency and so support longer term outcomes in the form of improved food security in critical parts of the world.FAO GEF IAH IHP Worldbank (2012): Conjunctive use and management of groundwater and surface water. Australia: Sinclair Knight Merz URL [Accessed: 28.03.2013] PDF
A case study of poor water related projects done in the River Rokel (Seli) basin in Sierra Leone, West Africa. This paper describes the problems that were faced and also places recommendations for future planning.AKIWUMI, F. A. (1997): Conjunctive Water Use in an African River Basin. A Case Study in poor Planning. In: Sustainability of Water Resources under Increasing Uncertainty: Volume 240 URL [Accessed: 19.09.2012] PDF
This study identifies trends in conjunctive use of canal and groundwater in the Punjab area of Pakistan and its implications for sustainability of agricultural performance. A main problem of the use of surface (canal) water arises due impact of farmer of the head end on the water supply for farmers on the tail end of the canal.MURRAY-RUST, D. H. ; VELDE, E. J. van der (1994): Conjunctive Use of Canal and Groundwater in Punjab, Pakistan. Management and Policy Options. In: Irrigation and Drainage Systems : Volume 8 , 201-231. URL [Accessed: 19.09.2012] PDF
Sustaining Surface and Groundwater Resources. Proceedings of the International Workshop on Conjunctive Water Management for Sustainable Irrigated Agriculture in South Asia, April 2002
Groundwater is the largest source of irrigation in South and Southeast Asia and China, its use for cities growing rapidly, leading to higher salinity of groundwater resources. Conjunctive water management is seen as a potential solution to ensure environmentally sustainable and financial viable irrigation agriculture. This paper delivers an overview on this thematic.QURESHI, A. S. ; BHATTI, A. ; JEHANGIR, W. A. (2002): Sustaining Surface and Groundwater Resources. Proceedings of the International Workshop on Conjunctive Water Management for Sustainable Irrigated Agriculture in South Asia, April 2002. Lahore, Pakistan: International Water Management Institute (IWMI) URL [Accessed: 19.09.2012] PDF
Conjunctive Use of Surface and Ground Water for Improving Water Productivity in Coastal Area of Orissa
This study on conjunctive use was carried out in an Eastern part of India, which has plenty of water but poor utilisation compared to other parts of India.SINGANDHUPE, R. B. SETHI R. R. CHAKRAVARTY, H. MOHANTY, R. K. KUMAR, A. (2012): Conjunctive Use of Surface and Ground Water for Improving Water Productivity in Coastal Area of Orissa. (= Proceedings of: India Water Week 2012 - Water, Energy and Food Security: Call for Solutions, April 2012 ). New Delhi: India Water Week URL [Accessed: 19.09.2012] PDF<|endoftext|>
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The most significant known geomagnetic storm to blast Earth occurred in 660 B.C., researchers say, based on traces of the storm's particles preserved in both ice cores and tree rings. Though it had no impact on the pre-industrial and pre-technological world, such an event today would trigger widespread power outages and collapse communication and navigation systems. In short, modern society would come to a screeching halt.
The study titled Multiradionuclide evidence for an extreme solar proton event around 2,610 B.P. (∼660 BC) was published online March 11 in Proceedings of the National Academy of Sciences.
Lead author Raimund Muscheler, a professor of Quaternary Sciences at Lund University in Sweden, said a burst of solar energy from the sun could overwhelm power grids, air travel, and disrupt satellite communications if this were to happen.
"Today, we have a lot of infrastructures that could be badly damaged, and we travel in air and space where we are much more exposed to high-energy radiation," Muscheler told LiveScience.
Researchers analyzed ice cores from Greenland to uncover the mystery of Earth's most powerful solar storms. One ice core, in particular, dated back more than 100,000 years and contained radioactive isotopes that showed a massive solar storm struck the planet around 2,700 years ago.
"If that solar storm had occurred today, it could have had severe effects on our high-tech society," said Muscheler, a geologist at Lund University in Sweden.
One example of a documented severe geomagnetic storm was known as ‘the Carrington Event,’ occurred in 1859 and knocked out telegraph circuits around the world, starting fires and causing massive auroras as far south as Hawaii.
Another example was a solar storm that knocked out power stations across Quebec, Canada, in 1989 and Malmö, Sweden, in 2003.
The Daily Telegraph said more recently, a blast of plasma from the sun, narrowly missed Earth during London’s 2012 Olympic Games, which at the time, the London region was Earth-facing, could have crippled communication networks and caused widespread panic.
Earth’s magnetic field protects the plant from the sun's harmful blasts of energy, but sometimes the sun overpowers the planet’s defenses.
In April 2017, we reported that San Fransisco, New York, and Los Angeles, simultaneously experienced a power grid failure.
It was convenient at the time for many to blame the Russians; however, our report showed readers how the broad power outage was likely caused by a geomagnetic storm.
The report ends by warning that the next severe solar storm could be immient and "threaten modern society."<|endoftext|>
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Topic 1: Statistical Analysis
Terms in this set (...)
how many data points do we need to develop an accurate representation of the overall data
a measurement of how data spreads around the mean
Standard deviation represents how much of the data within +/- 1 standard deviation of the mean?
Standard deviation represents how much of the data within +/- 2 standard deviation of the mean?
each thing must have an equal chance of being selected
has two sets of data; more degrees of freedom (which strengthens the argument); and sees if chance played a role in the data results
graphical representation of the variability of data; used to show the range or standard deviation on a graph
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Language is an important skill that allows a person to communicate. A child begins to develop language even before she can use words, as seen by a baby who cries to get her needs met. A delay in language skills can cause frustration for a child as well as miscommunication about what she may be trying to convey. Language development is important to a child in order to adequately exchange information with others in a meaningful way.
Importance of Language Skills
According to the American Speech, Language and Hearing Association, language is not the same as speech. Language consists of a set of social standards that shows comprehension of the meanings behind words, putting words together in a sentence in order to communicate and understanding commands, directions and information given by others. Children must develop language skills to relate with their parents and peers, as well as to grow into a person who can socially interact with others through life.
Hitting Specific Speech Milestones
Each child follows his own milestones for development of language, but overall he should be exhibiting certain behaviors by a specific age. By 12 months, a baby recognizes and reacts to the sound of his name and uses inflection with sounds. By 2 years, a child has a vocabulary of almost 300 words and can respond to simple commands. By 3 years, he has almost 1,000 words, 90 percent of which is understandable and can give his name and age. By 5 years, he can understand the concept of opposites and can speak in sentences of up to nine words.
A parent can encourage the development of language skills with a child by interacting regularly, singing songs and reading simple stories. As a child grows, she can talk about her day or explain her likes and dislikes. A parent can play games with a child, such as naming objects or practicing concepts of direction; for example: in, out, over and through. Visiting a library encourages reading stories, and a parent can teach a child simple stories and rhymes to encourage repetition and language.
Things to Consider
While most children follow their own timetable for developing language, there are some signs that can indicate a concern. A child between 18 and 24 months who is unable to follow very basic commands or who has trouble making vocalizations may be showing signs of a language development difficulty. For a child older than 2 years, professional help may be needed if his speech is unintelligible most of the time or if he does not try to copy sounds or gestures.
When You Need Professional Help
Parents that suspect a problem with language development in their child should contact their pediatrician. Some children have undiagnosed hearing problems that can impact language, and medical testing may be necessary. A physician can also direct parents to appropriate resources. Speech and language pathologists can assess where delays lie in expressive and receptive language and work with a child to teach him skills to communicate better.<|endoftext|>
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How did a single species of rodent invade South America and then quickly branch off into 350 new species?
The answer is simple -- the rodents were able to move quickly across the continent unencumbered by geographic boundaries that can't be easily crossed such as an ocean.
FSU Professor of Biological Science Scott Steppan and his former postdoctoral researcher John Schenk, now at Georgia Southern University, developed a new model that shows how geography can play a major role in how families of animals evolve and result in many species. The research was part of a $500,000 grant from the National Science Foundation to understand why the superfamily Muroidea (which includes the subfamily Sigmodontinae) is the most diverse branch of the mammal family tree.
The research is published in the journal American Naturalist.
"Biologists have long been fascinated by the process of adaptive radiations, which occurs when a lineage undergoes rapid speciation that is correlated with variation in habitats that species evolve into, resulting in numerous species that differ in the environments in which they live," Schenk said. "Our study was the first to directly measure how movement across the landscape facilitated the speciation process."
Scientists in the past had believed that when species evolved in a short period of time they diverged because of different ecological niches, not necessarily because of the ability to cross geographic boundaries.
Schenk and Steppan found that the species Sigmodontinae moved into South America and quickly diversified across all the regions within the continent. In about 8 million years, they diversified into about 400 new species that covered South America.
"That's really fast," Steppan said.
As time passed, the movement between regions and diversification slowed because the regions and their associated ecological niches were essentially full.
"This [geography] largely tracks with what speciation is doing," Steppan said. "After that initial burst, it slows down."
Steppan studies evolutionary biology and has always been interested in how species diversify. Most studies of species diversification involve islands where different groups of terrestrial species are walled off from competing interests by ocean water allowing for animals to diversify based simply on the ecology of the island.
But little is known about how species diversify across the continuous geography of a continent.
Steppan said that he and Schenk are looking at several possibilities for follow-up studies that focus on DNA work and measuring other aspects of the rodents, such as the structure of their limbs and skulls, to see how these structures adapted to new environments over time.<|endoftext|>
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During the 18th century, criticism of the French monarchy also came from people who worked for the Old Regime. Some of the king’s own ministers criticized past practices and proposed reforms, but a more influential source of dissent was the parlements, 13 regional royal courts led by the Parlement of Paris. The parlements were empowered to register royal decrees, and all decrees had to be registered by the parlements before becoming law. In this capacity, the parlements frequently protested royal initiatives that they believed to threaten the traditional rights and liberties of the people. In widely distributed publications, they held up the image of a historically free France and denounced the absolute rule of the crown that in their view threatened traditional liberties by imposing religious orthodoxy and new taxes.
These protests blended with those of others, most notably an influential group of professional intellectuals called the philosophes. Like those who supported the parlements, the philosophes did not advocate violent revolution. Yet, they claimed to speak on behalf of the public, arguing that people had certain natural rights and that governments existed to guarantee these rights. In a stream of pamphlets and treatises—many of them printed and circulated illegally—they ridiculed the Old Regime’s inefficiencies and its abuses of power. During this time, the parlementaires and the philosophes together crafted a vocabulary that would be used later to define and debate political issues during the Revolution. They redefined such terms as despotism, or the oppression of a people by an arbitrary ruler; liberty and rights; and the nation.
The discontent of the French people might not have brought about a political revolution if there had not been a fiscal crisis in the late 1780s. Like so much else in the Old Regime, the monarchy’s financial system was inefficient and antiquated. France had neither a national bank nor a centralized national treasury. The nobility and clergy—many of them very wealthy—paid substantially less in taxes than other groups, notably the much poorer peasantry. Similarly, the amount of tax charged varied widely from one region to another.
Furthermore, the monarchy almost always spent more each year than it collected in taxes; consequently, it was forced to borrow, which it did increasingly during the 18th century. Debt grew in part because France participated in a series of costly wars—the War of the Austrian Succession (1740-1748), the Seven Years’ War (1756-1763), and the American Revolution (1775-1783). Large existing debts and a history of renouncing earlier ones meant that the country was forced to borrow at higher interest rates than some other countries, further adding to the already massive debt. By 1789 the state was forced to spend nearly half its yearly revenues paying the interest it owed.
Financial reform was attempted before 1789. Upon his accession to the throne in 1774, Louis XVI appointed the reform-minded Anne Robert Jacques Turgot as chief finance minister. Between 1774 and 1776 Turgot sought to cut government expenses and to increase revenues.
He removed government restrictions on the sale and distribution of grain in order to increase grain sales and, in turn, government revenue. Jacques Necker, director of government finance between 1777 and 1781, reformed the treasury system and published an analysis of the state of government finance in 1781 as a means to restore confidence in its soundness. But most of these reforms were soon undone as the result of pressure from a variety of financial groups, and the government continued to borrow at high rates of interest through the 1780s. Charles Alexandre de Calonne was appointed minister of finance in 1783, and three years later he proposed a new general plan resembling Turgot’s. He wanted to float new loans to cover immediate expenses, revoke some tax exemptions, replace older taxes with a new universal land tax and a stamp tax, convene regional assemblies to oversee the new taxes, and remove more restrictions from the grain trade. "France" © Emmanuel BUCHOT, Encarta, Wikipedia
Photos of European countries to visit
Photos of Asian countries to visit
Photos of America<|endoftext|>
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A padded linen cap was the most basic form of medieval head protection. It was constructed in the same way as the aketon, with a layer of padding, such as flax or horsehair, stuffed in between two layers of material. This formed a tough, dense cushion to help protect the skull of the wearer.
Some arming caps may have had hidden interlinings of mail if the wearer could afford it.
The arming cap could not provide very much protection on its own, but it was certainly better than nothing.
But often the arming cap, like the aketon, formed the base-layer over which other pieces of armour were worn. A mail coif was often worn over it, but the most vital protection was some kind of rigid helmet made of hardened leather or, preferably, iron or steel.
Layering was the secret to the best head protection, building out from the foundation provided by the arming cap.<|endoftext|>
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The word “genius” is universally associated with the name of Leonardo da Vinci – a true Renaissance man, he embodied scientific spirit, artistic talent and humanist sensibilities. Exactly 500 years have passed since Leonardo died in his home at Château du Clos Lucé, outside Tours, France. Yet far from fading to insignificance, his thinking has carried down the centuries – and still surprises today.
The Renaissance marked the transition from the Middle Ages to modernity, after the spread of the plague caused a global crisis resulting in some 200m deaths across Europe and Asia. Today, the world is on the cusp of a climate crisis, which is predicted to cause widespread displacement, extinctions and death, if left unaddressed. Then, as now, radical solutions were called for to revolutionise the way people live and safeguard humanity against catastrophe.
Around 1486 – after a pestilence that killed half the population in Milan – Leonardo turned his thoughts to urban planning problems. Following a typical Renaissance trend, he began to work out an “ideal city” project, which – due to its excessive costs – would remain unfulfilled. Yet given that unsustainable urban models are a key cause of global climate change today, it’s only natural to wonder how Leonardo might have changed the shape of modern cities.
The birth of urbanism
Although the Renaissance is renown as an era of incredible progress in art and architecture, it’s rarely noted that the 15th century also marked the birth of urbanism as true discipline.
The rigour and method behind the conscious conception of a city had been largely missing in Western thought until the moment when prominent Renaissance men pushed forward large-scale urban projects, such as the reconfiguration of Pienza, the expansion of Ferrara and the construction of the fort town Palmanova.
These works surely inspired Leonardo’s decision to rethink the design of medieval cities, with their winding and overcrowded streets and with houses piled against one another.
Discovering Leonardo’s city
It is not easy to identify a coordinated vision of Leonardo’s ideal city because of his disordered way of working with notes and sketches. But from sources including the Paris manuscript B and the Codex Atlanticus – the largest collection of Leonardo’s papers ever assembled – a series of innovative thoughts can be reconstructed, regarding the foundation of a new city along the Ticino River, designed for the easy transport of goods and clean urban spaces.
Leonardo wanted a comfortable and spacious city, with well-ordered streets and architecture. He recommended “high, strong walls”, with “towers and battlements of all necessary and pleasant beauty”, and felt the place needed “the sublimity and magnificence of a holy temple” and “the convenient composition of private homes”.
His plans for a “modern” and “rational” city were consistent with Renaissance ideals. But, in keeping with his unconventional personality, Leonardo included several innovations in his urban design. Leonardo wanted the city to be built on several levels, linked with vertical staircases. This design can be seen in today’s high-rise buildings, but was absolutely unconventional at the time.
Indeed, his idea of taking full advantage of the interior spaces by positioning flights of stairs on the outside of the buildings wasn’t implemented until the 1920s and 1830s, with the birth of the Modernist movement. While in the upper layers of the city, people could walk undisturbed between elegant palaces and streets, the lower layer was the place for services, trade, transport and industry.
But the true originality of Leonardo’s vision was its fusion of architecture and engineering. Leonardo made designs for extensive hydraulic plants to create artificial canals throughout the city. The canals, regulated by locks and basins, were supposed to make it easier for boats to navigate inland and transport goods.
Leonardo also thought that the width of the streets ought to match the average height of the adjacent houses: a rule still followed in many contemporary cities across Italy, to allow access to sun and reduce the risk of damage from earthquakes.
Fiction and the future
Although some of these features existed in Roman cities, before Leonardo’s drawings there had never been a multi-level, compact modern city which was thoroughly technically conceived. Indeed, it wasn’t until the 19th century that some of his ideas were applied. For example, the subdivision of the city by function – with services and infrastructures located in the lower levels and wide and well-ventilated boulevards and walkways above for residents – is an idea that can be found in Haussmann’s renovation of Paris under Emperor Napoleon III.
It is necessary to wait even until the 20th century to see the same ideas represented in the vertical city of Futurist architects, or in the modern city of Hilbeseimer or Le Corbusier – as well as in dystopian tales such as Fritz Lang’s Metropolis and Philip Dick’s Blade Runner. Certainly, creating a city with different levels opens up the possibility of greater inequality between city-dwellers.
Today, Leonardo’s ideas are not simply valid – they actually suggest a way forward for urban planning. Many scholars think that the compact city – built upwards instead of outwards, integrated with nature (especially water systems) with efficient transport infrastructure – could help modern cities become more efficient and sustainable. This is yet another reason why Leonardo was aligned so closely with modern urban planning – centuries ahead of his time.<|endoftext|>
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Moving on the Moon
Europe is preparing to go forward to the Moon, but how will astronauts move once they get there? Despite the Apollo missions, little is known about what lunar gravity may mean for our bodies. ESA’s space medicine team is working to find out through a series of studies.
The level of gravity on the Moon is about one sixth of Earth’s so while Apollo astronauts did not float as astronauts do on the International Space Station, they tended to hop rather than walk.
Education coordinator at ESA’s astronaut centre in Cologne, Germany, David Green is leading this research alongside science operations engineer Tobias Weber. He says, though much research has been carried out into the impacts of microgravity as experienced on the International Space Station, the physiological impact of working in lunar gravity remains unknown.
Studying the effects of lunar gravity will help identify potential risks and create measures to keep astronauts fit and healthy.
“How microgravity influences our bodies is also investigated through bedrest studies that recreate some of the changes we associate with living in space by putting people in bed with their head below horizontal,” David explains.
“These studies show the way in which the body adapts to life in weightlessness, resulting in bone weakness and muscle loss, and it is why astronauts are prescribed daily exercise when in orbit.”
Do we need a gym on the Moon?
To help answer some of their questions such as how lunar gravity might impact the biomechanics of walking, running and hopping, the team – in collaboration with German Aerospace Center DLR and academic partners – conducted the first “Movement in low gravity study” in 2017. This study used a vertical treadmill to simulate various levels of reduced gravity.
During the study, researchers looked at movement patterns, muscle activity, ground reaction forces and aspects of Achilles tendon function.
While the initial study showed that jumping may be the best way to prevent muscle and bone loss, the vertical treadmill did not allow subjects to jump as high as they would be able to on the Moon.
“We believe jumping and hopping on the Moon may provide forces similar to walking and running on Earth. This would allow astronauts to maintain their bone and muscle condition through everyday movement,” Tobias says. “This may reduce the need for training equipment such as on the Space Station and it is something we hope to explore further.”
The second phase of the study will be conducted using NASA’s Active Response Gravity Offload System (ARGOS) at the Johnson Space Center in Houston, Texas, which allows a greater range of vertical movement.
This will enable researchers to determine maximum jump heights, alongside what forces and strains will be placed on an astronaut’s muscles and bones.
Simulating Moon, Mars and asteroids
A similar system to ARGOS will form part of a new facility at ESA’s astronaut centre known as Luna 2. Research into movement in low gravity will be used to build expertise and prepare for surface operations on the Moon, Mars, asteroids and beyond.<|endoftext|>
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Physics Solution Manual for 1100 and 2101
7 chapter 26 the refraction of light lenses and
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Unformatted text preview: tance. 1296 THE REFLECTION OF LIGHT: MIRRORS SOLUTION According to the mirror equation, we have 1 1 1 + = d o1 d i1 f 14 244 4 3 and First position of object 1 1 1 + = d o2 d i2 f 14 244 4 3 Second position of object Since the focal length is the same in both cases, it follows that 1 1 1 1 + = + d o1 d i1 d o2 d i2 1 1 1 1 1 1 1 = + − = + − = −0.071 cm −1 d i2 d o1 d i1 d o2 25 cm −17 cm 19 cm b g d i2 = −14 cm The negative value for di2 indicates that the image is located 14 cm behind the mirror. 21. REASONING a. We are dealing with a concave mirror whose radius of curvature is 56.0 cm. Thus, the focal length of the mirror is f = 1 R = 28.0 cm (Equation 25.1) The object distance is 2 do = 31.0 cm. With known values for f and do, we can use the mirror equation to (Equation 25.3) find the image distance. b. To determine the image height hi, recall that it is related to the object height ho by the magnification m; hi = mho. The magnification is related to the image and object distances by the magnification equation, m = −di / d o (Equation 25.4). Since we know values for ho, di, and do, we can find the image height hi. SOLUTION a. The image distance is given by the mirror equation as follows: 1 1 1 1 1 = − = − di f do 28.0 cm 31.0 cm so di = 290 cm (25.3) b. Using the magnification equation, we find that the image height is 290 cm d hi = mho = − i ho = − ( 0.95 cm ) = 31.0 cm do −8.9 cm (25.4) Chapter 25 Problems 1297 c. Since hi is negative, the image is inverted relative to the object. Thus, to make the picture on the wall appear normal, the slide must be oriented upside down in the projector. 22. REASONING For an image that is in front of a mirror, the image distance is positive. Since the image is inverted, the image height is negative. Given the image distance, the mirror equation can be used to determine the focal length, but to do so a value for the object distance is also needed. The object and image heights, together with the knowledge that the image is inverted, allows us to calculate the magnification m. The magnification m is given by m = –di/do (Equation 25.4), where di and do are the image and object distances, respectively. SOLUTION According to Equation 25.4, the magnification is m= hi d =− i ho do or do = − d iho hi Substituting this result into the mirror equation, we obtain FI GJ HK − F 1 IL b1.5 cm g 1P 0.11 cm =G − = JMb cm g+ O H cm K 3.5 13 N Q h h 1 1 1 1 1 = + =− i + = − i +1 f do di d i ho d i d i ho −1 or f = 9 .1 cm 23. REASONING Since the image is behind the mirror, the image is virtual, and the image distance is negative, so that di = −34.0 cm . The object distance is given as do = 7.50 cm . The mirror equation relates these distances to the focal length f of the mirror. If the focal length is positive, the mirror is concave. If the focal length is negative, the mirror is convex. SOLUTION According to the mirror equation (Equation 25.3), we have 1 11 = + f d o di or f= 1 1 = = 9.62 cm 11 1 1 + + do di 7.50 cm ( −34.0 cm ) Since the focal length is positive, the mirror is concave . 1298 THE REFLECTION OF LIGHT: MIRRORS 24. SSM REASONING The object distance do is the distance between the object and the 111 mirror. It is found from += (Equation 25.3), where di is the distance between the d o di f mirror and the image, and f is the focal length of the mirror. We are told that the image appears in front of the mirror, so, according to the sign conventions for spherical mirrors, the image distance must be positive: di = +97 cm. SOLUTION Solving 111 + = (Equation 25.3) for do yields d o di f 1 11 =− d o f di do = or 1 11 − f di = 1 1 1 − ( +42 cm ) ( +97 cm ) = 74 cm 25. REASONING Since the focal length f and the object distance do are known, we will use the mirror equation to determine the image distance di. Then, knowing the image distance as well as the object distance, we will use the magnification equation to find the magnification m. SOLUTION a. According to the mirror equation (Equation 25.3), we have 111 =− di f do or di = 1 1 = = −4.3 m 11 1 1 − − f d o ( −7.0 m ) 11 m The image distance is negative because the image is a virtual image behind the mirror. b. According to the magnification equation (Equation 25.4), the magnification is m=− di do =− ( −4.3 m ) = 11 m 0.39 26. REASONING The magnification m is given by m = –di/do (Equation 25.4), where di and do are the image and object distances, respectively. The object distance is known, and we can obtain the image distance from the mirror equation: 1 11 += do di f (25.3) Chapter 25 Problems 1299 SOLUTION Solving the mirror equation (Equation 25.3) for the image distance di gives 1 1 1 + = do di f 1 1 1 do − f =− = di f do fd o or di = or fd o do − f Substituting this result into the magnification equation (Equation 25.4) gives m=− di do =− c fd o / d o − f do h = f f − do Using this result with the given values for the focal length and object distances, we find Smaller object distance m= f −27 .0 cm = = 0.750 f − do −27 .0 cm − 9 .0 cm Greater object distance m= f −27 .0 cm = = 0...
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Anger management is a range of steps and processes that can help someone to manage and reduce their anger. This can involve altering the way they think about certain things that make them angry and changing the ways they react to anger to be constructive, rather than counterproductive.…
What is bullying?
Bullying is the use of power by a person or group of people to intentionally cause physical or psychological harm to another person or group of people. It includes humiliating, demeaning or threatening behavior and can occur anywhere - at school, home, at work or online .
Signs your child might be being bullied include:
- Unexplained bruises, scratches or other injuries;
- Missing or damaged belongings and clothes;
- Changes in sleeping or eating patterns;
- Frequent fluctuations in mood, or anger problems;
- Feeling ill in the morning or not wanting to go to school;
- Changes in the route they take to school;
- Appearing insecure or frightened;
- Their grades at school fall;
- They are often 'losing' money or stealing, and;
- They refuse to talk about what is wrong and become withdrawn.
Although anyone can be a victim of bullying, some people are at greater risk of bullying than others, including people who are:
- Lesbian, gay, bisexual, transgender and intersex;
- Perceived as being weak and unable to defend themselves;
- Less popular than others;
- From a poor socio-economic background;
- Low in confidence and self-esteem;
- Experiencing depression or anxiety;
- Disabled, or;
- Perceived as annoying, antagonistic or attention-seeking.
When people think of bullying, a lot of people think of the physical element of it, but there is also an important psychological element to bullying. You don't have to punch someone to hurt them. Verbal abuse can also be very hurtful.
Cyberbullying can occur on the Internet in social sites such as social media sites, chat rooms, video chat and blogs. Cyberbullying can often be difficult to identify, but it is just as important to address. It can include things such as:
- Impersonating someone else online;
- Spreading rumors or lies about someone;
- Tricking others into revealing private information;
- Posting pictures of others without their consent, and;
- Sending people mean messages.
Examples of workplace bullying include:
- Refusal by colleague(s) or a boss to acknowledge your achievements and contributions;
- Isolation or separation from colleagues;
- Having unrealistic goals set for you, which then change as you near accomplishing them;
- Being singled out or treated differently to others;
- Having your work or credit stolen or plagiarized;
- Being given more responsibility, but less authority;
- Being given too much work causing an overload;
- Having significant work replaced with menial tasks, or not getting enough work;
- Not being trained to perform duties you are expected to perform, and;
- Being humiliated, threatened, patronized or demeaned, either in private or in front of others.
Dos and don'ts
If your child has experienced bullying, do not:
- Tell them to ignore it;
- Blame a young person for being bullied, or assume they have provoked it;
- Criticize how they have dealt with being bullied;
- Contact the bully or their parents, or;
- Encourage your child to retaliate or 'fight back'.
- Let them know you are glad they told you about it;
- Listen to what they have to say;
- Reassure them that you will take helpful action;
- Collect evidence such as screenshots, if the bullying is online;
- Contact a teacher and/or principal to discuss finding a solution, and;
- Contact school authorities if bullying continues after speaking with the principal.
If your child is bullying other children, do:
- Stay calm and talk to your child, focusing on this specific aspect of their behavior rather than on them generally;
- Explain why their behavior is inappropriate and make sure they understand;
- Give them clear boundaries;
- Encourage them to think from the other child's perspective ('would you like it if someone did that to you?');
- Teach them there are more positive ways to interact with others than asserting their dominance or control over them;
- Teach them social and conflict resolution skills, and;
- Take the opportunity to reflect on the behavior of others within your family, as children tend to copy their family members.
If you are being bullied at school, do:
- Look the bully in the eye and tell them to leave you alone;
- Walk away from the bully, and;
- Tell an adult or teacher.
If you are being bullied in the workplace politely ask for the behavior to stop. If this does not work, have a neutral person mediate a discussion and try to arrive at a resolution. If this, too, does not work, or if the bullying is serious, file a report as specified in your organization's policies and ask for a formal investigation.
Schools and workplaces can work to prevent or stop bullying by:
- Creating policies and rules, including a mission statement, code of conduct and reporting system to assess how often bullying happens and how it is responded to;
- Providing training to staff members or teachers about the school or workplace policies and rules, as well as training in the conflict resolution skills needed to intervene to enforce them when necessary, and;
- Establishing a culture of acceptance, respect and inclusiveness, and reinforcing it at staff meetings and school assemblies.
- Australian Psychological Society?: Parent guide to helping children manage conflict aggression and bullying. Accessed 21 August 2014 from http://www.psychology.org.au/publications/tip_sheets/bullying/
- Best Practices for Preventing or Reducing Bullying in Schools. Accessed 21 August 2014 from http://cs.oxfordjournals.org/content/27/3/167.short
- BMC Public Health | Full text | Early Risk Factors for Being a Bully Victim or Bully/Victim in Late Elementary and Early Secondary Education. The longitudinal TRAILS study. Accessed 22 August 2014 from http://www.biomedcentral.com/1471-2458/11/440
- Cyber-Bullying - cyber-bullying-no-crops.ashx. Accessed 21 August 2014 from http://www.afp.gov.au/policing/cybercrime/~/media/afp/pdf/c/cyber-bullying-no-crops.ashx
- Lesbian Gay Bisexual Trans and Intersex Equality | Bullying. Accessed 22 August 2014 from https://bullying.humanrights.gov.au/lesbian-gay-bisexual-trans-and-intersex-equality-1
- National Centre Against Bullying. Accessed 21 August 2014 from http://www.ncab.org.au/
- Preventing and Responding to Bullying at Work - bullying_at_work_2054.pdf. Accessed 22 August 2014 from http://www.workcover.nsw.gov.au/formspublications/publications/Documents/bullying_at_work_2054.pdf
- RACGP - Bullying - Effects prevalence and strategies for detection. Accessed 21 August 2014 from http://www.racgp.org.au/afp/2011/march/bullying/
- Risk Factors | StopBullying.gov. Accessed 22 August 2014 from http://www.stopbullying.gov/at-risk/factors/
- Workplace bullying - Workplace_bullying.pdf. Accessed 21 August 2014 from https://www.comcare.gov.au/__data/assets/pdf_file/0020/75215/Workplace_bullying.pdf
FAQ Frequently asked questions
What is bullying?
Bullying is any action that intimidates, threatens or harms another person or group of people, either physically or emotionally.
How can I raise awareness of bullying prevention in our school?
Some approaches taken to raise awareness include social media campaigns, morning or assembly announcements, posters and student surveys.
Are there different types of bullying?
Bullying can be verbal, physical, or virtual (online). It can be racially, sexually or socially motivated, or in some other form.
Is bullying really that harmful?
People who are bullied are negatively affected in a number of ways. They often lose confidence in themselves and become insecure. If subjected to chronic and severe bullying, some people even take their own lives.
Who is more at risk of being bullied?
People who are more at risk of being bullied include gay, lesbian, bisexual, transgender and intersex people, as well as people with disabilities or who are viewed as easy targets because they are seen as being weak and unable or unwilling to defend …
What can I do if my child is being bullied?
If your child is being bullied you can listen to what they are saying and take action to help them by contacting their teacher and principal, not the parents of the child who is bullying them. If the bullying has occurred online, you can take …
How can I help prevent bullying at my child's school?
Schools can work to prevent bullying by creating policies supported by a mission statement, a code of conduct and a reporting system that allows people to assess how often bullying happens and the responses to it. Schools have to work …
How do I respond to a bully in the workplace?
If you are being bullied in the workplace, you can politely ask them to stop their bullying behavior, or if it continues, or is serious, lodge a formal complaint and start an investigation into their behavior through your organization's regular …
What can I do if my child is bullying others?
Stay calm and focus on their behavior and not them. Let them know their behavior is not acceptable and explain why. Ask them to think what would happen if things were the other way around and they were the one being bullied, to help put things …<|endoftext|>
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# Algebra 1 Study Guide
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Algebra 1 study guide is used to solve the problems related to algebra 1 topics. It involves various such as exponents, linear inequalities, quadratic equations, radical expression, rational expression, etc. This tool provides the detailed description about the steps involved. It also tells about the relevant formulas used. This study guide basically helps in understanding the typical questions as well as the methods to solve these types of typical questions.
This can be more illustrated in scope clarity by taking suitable examples. The examples are as follows:-
Question 1:- Solve for x if 16 = x2.
Solution 1:- We can rewrite this in this way,
x2 = 16 ------equation 1
We know that,
16 = 4 * 4 = 42 ------equation 2
Now by comparing equation no. 1 and equation no. 2, so we will get the following:-
x2 = 16 = 42
That is x2 = 42
So by comparing, we find that x =4
So x = 4 is the answer to the above problem
Question 2:- Find x if 2x+ 3 = 10
Solution 2:-
To find: - the value of unknown variable x
From given equation, 2x+3 = 10 ------ equation 1
In equation 1, to find the value of x, subtract 3 both sides, we will get
2x + 3 - 3 = 10-3
2x = 7
Now the last step to check the value of x is, we need to divide both sides by 2, we will get
2x/2 = 7/2
x = 7/2 = 3.5
Therefore, x = 3.5 is the solution.<|endoftext|>
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# Search by Topic
#### Resources tagged with Patterned numbers similar to Volume of a Pyramid and a Cone:
Filter by: Content type:
Stage:
Challenge level:
### There are 45 results
Broad Topics > Numbers and the Number System > Patterned numbers
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Pattern Power
##### Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Sum Equals Product
##### Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Oranges and Lemons, Say the Bells of St Clement's
##### Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
### Pyramids
##### Stage: 3 Challenge Level:
What are the missing numbers in the pyramids?
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### Icosagram
##### Stage: 3 Challenge Level:
Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . .
### Colour Building
##### Stage: 3 Challenge Level:
Using only the red and white rods, how many different ways are there to make up the other colours of rod?
### Tower of Hanoi
##### Stage: 3 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### Pinned Squares
##### Stage: 3 Challenge Level:
The diagram shows a 5 by 5 geoboard with 25 pins set out in a square array. Squares are made by stretching rubber bands round specific pins. What is the total number of squares that can be made on a. . . .
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Hidden Rectangles
##### Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
### Odd Differences
##### Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = nĀ² Use the diagram to show that any odd number is the difference of two squares.
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Back to Basics
##### Stage: 4 Challenge Level:
Find b where 3723(base 10) = 123(base b).
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Paving Paths
##### Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### A One in Seven Chance
##### Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### Investigating Pascal's Triangle
##### Stage: 2 and 3 Challenge Level:
In this investigation, we look at Pascal's Triangle in a slightly different way - rotated and with the top line of ones taken off.
### Four Coloured Lights
##### Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
### Rolling Coins
##### Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### Chameleons
##### Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
### Harmonic Triangle
##### Stage: 4 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Lastly - Well
##### Stage: 3 Challenge Level:
What are the last two digits of 2^(2^2003)?
### Power Crazy
##### Stage: 3 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
### Magic Squares
##### Stage: 4 and 5
An account of some magic squares and their properties and and how to construct them for yourself.
### Score
##### Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
### Sept 03
##### Stage: 3 Challenge Level:
What is the last digit of the number 1 / 5^903 ?
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
##### Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
### Counting Binary Ops
##### Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
### Generating Number Patterns: an Email Conversation
##### Stage: 2, 3 and 4
This article for teachers describes the exchanges on an email talk list about ideas for an investigation which has the sum of the squares as its solution.
### One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.<|endoftext|>
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New research shows that the heliosphere is dominated by two giant jets of material shooting backwards over the north and south poles of the sun, which are confined by the interaction of the sun’s magnetic field with the interstellar magnetic field.
As the sun skims through the galaxy, it emits charged particles in a stream of plasma called the solar wind. The solar wind, in turn, creates a bubble known called the heliosphere that extends far beyond the planets of the solar system. For decades, scientists have visualized the heliosphere as shaped like a comet, with a very long tail extending some 464 billion miles, which is thousands of times as far as the distance from Earth to the sun.
New NASA-funded research now suggests that the heliosphere is actually dominated by two giant jets of material shooting backwards over the north and south poles of the sun, which are confined by the interaction of the sun’s magnetic field with the interstellar magnetic field. These curve around in two—relatively short – tails toward the back. The end result is a heliosphere without that long tail; a heliosphere that looks a lot more like a crescent moon than a comet. What’s more, the two jets are similar to other astrophysical jets seen in space, so studying them locally could open doors to understanding such jets throughout the universe. The research is described in a paper in Astrophysical Journal Letters, which appeared online on February 19, 2015.
“Everyone’s assumption has been that the shape of the heliosphere was molded by the flow of interstellar material passing around it,” said Merav Opher, an astronomer at Boston University, who is lead author on the paper. “Scientists thought the solar wind flowing down the tail could easily pull the magnetic fields in the heliosphere along as it flowed by, creating this long tail. But it turns out the magnetic fields are strong enough to resist that pull – so instead they squeeze the solar wind and create these two jets.”
Opher and her colleagues found the jets and determined the new shape when they adjusted simulations of the heliosphere based on observations collected from NASA’s Voyager 1 spacecraft, which recently moved outside of the heliosphere into interstellar space. As the first man-made object outside of our solar system, Voyager provided our only glimpse so far of the interstellar medium and it provided one giant surprise: The magnetic fields out there were aligned pretty much the same as the ones in here, though it had long been expected they would be oriented in a different direction.
Opher — along with space scientist Jim Drake at the University of Maryland in College Park who is a co-author on the paper – had previously created models based on computer code developed by space physicists at the University of Michigan of the heliosphere. Their previous work focused on the nose of the heliosphere, trying to understand the physics there as we hurtle through space. To see if they could replicate the unexpected Voyager results, the team created a higher resolution simulation.
The new simulation described a heliosphere unlike one considered before.
“The Voyagers had a flashlight in the kitchen, and nobody was looking in the attic,” she remarks. “We noticed, while studying the draping of the galaxy’s magnetic field around the nose, that the heliosphere was much shorter than we anticipated.”
Instead of being dominated solely by the flow of the interstellar material to create a long tail, the shape of the heliosphere is also affected by the solar wind jets emanating from the sun, said Drake.
“If there were no interstellar flow, then the magnetic fields around the sun would shape the solar wind into two jets pointing straight north and south,” said Drake. “The magnetic fields contract around these jets, shooting the solar wind out like squishing a tube of toothpaste.”
In the presence of the interstellar flow, these jets are bowed backwards, creating a crescent shape, as seen from the side of the sun. The jets erode in the presence of the strong interstellar flow, leading to two attenuated, short tails. This leads to a much shorter heliosphere of only about 250 times the distance between Earth and the sun, or about 23 billion miles.
“Not only is the heliosphere’s shape different from what people thought,” said Drake. “But the mechanism for these jets is the same as in many astrophysical systems. Astrophysical jets elsewhere produce energetic particles, but they are remote and hard to diagnose. Our jets are nearby so we might be able to figure out how they produce the energetic particles measured in the heliosphere.”
To support their heliosphere model, the scientists turned to additional observations of the tail. Both NASA’s Cassini and Interstellar Boundary Explorer, or IBEX, have gathered information about the tail end of the heliosphere by looking at what’s called energetic neutral atoms, or ENAs. ENAs are created by energetic particle collisions in space and conveniently travel in straight lines, unlike many other particles in space. Observing ENAs traveling in from a certain area, therefore, can be used to map that region.
“Cassini data showed a similar amount of ENAs from the tail and the nose,” said Opher. “Suggesting that the size of both sides was similar, which means a short tail.”
An IBEX paper from 2013 also described as a two-lobed shape down the tail. Opher and Drake suggest that the lobes observed might actually have been the two jets with interstellar, non-heliospheric material in between. The paper on the IBEX results, however, interpreted the heliosphere as having a long tail.
With such previous results, Opher expects the new model to be controversial. “This is going to be heavily, heavily debated,” she said, pointing out that many scientists work from the traditional comet-shaped model of the heliosphere. But, said Opher, the out-of-the-box results coming from spacecraft observations demand a similarly unconventional explanation.
In the meantime, these newly postulated jets look like baby versions of the super-powered jets that exist around exotic objects like black holes and pulsars. They are also seen around proto-stars just being born. Being able to study these jets in our own backyard provides a homegrown laboratory in which to study a structure that is seen everywhere in the universe.
“If we’re right about all of this, it gives us a local test bed for exploring some very important physics,” said Drake.
Publication: M. Opher, et al., “Magnetized Jets Driven By the Sun: the Structure of the Heliosphere Revisited,” 2015, ApJ, 800, L28; doi:10.1088/2041-8205/800/2/L28
PDF Copy of the Study: Magnetized Jets Driven By the Sun: the Structure of the Heliosphere Revisited
Images: M. Opher/Reproduced by permission of the AAS; NASA/Casalegno/GALEX; M. Opher<|endoftext|>
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Contact The Learning Centre
### Arc Length
The arc length can be found using the formula
$\mbox{Arc Length} = \theta r$ where $$\theta$$ is measured in radians and $$r$$ is the radius of the circle.
Example: for the diagram below, calculate the radius of the sector, given that the length of the arc is $$12$$ cm and the angle subtended by the arc is $$35^{\circ}$$.
To find the radius, we firstly need to convert the angle to be in radians:
$35^{\circ} = \frac{\pi}{180} \times 35 = \frac{7\pi}{36}$
Secondly, we need to rearrange the arc length formula to make the radius the subject:
\begin{eqnarray*}
\mbox{Arc Length} &=& \theta r \\
r &=& \frac{\mbox{Arc Length}}{\theta}
\end{eqnarray*}
Finally, substitute in the given information:
\begin{eqnarray*}
r &=& \frac{\mbox{Arc Length}}{\theta} \\
&=& \frac{12}{\frac{7\pi}{36}} \\
&\approx& 19.6\mbox{ cm}
\end{eqnarray*}
Therefore, the radius is approximately $$19.6\mbox{ cm.}$$<|endoftext|>
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# Maharashtra Board 11th Maths Solutions Chapter 9 Probability Ex 9.5
Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 9 Probability Ex 9.5 Questions and Answers.
## Maharashtra State Board 11th Maths Solutions Chapter 9 Probability Ex 9.5
Question 1.
If odds in favour of X solving a problem are 4 : 3 and odds against Y solving the same problem are 2 : 3. Find the probability of:
(i) X solving the problem
(ii) Y solving the problem
Solution:
(i) Odds in favour of X solving a problem are 4 : 3.
∴ The probability of X solving the problem is
P(X) = $$\frac{4}{4+3}=\frac{4}{7}$$
(ii) Odds against Y solving the problem are 2 : 3.
∴ The probability of Y solving the problem is
P(Y) = 1 – P(Y’)
= 1 – $$\frac{2}{2+3}$$
= 1 – $$\frac{2}{5}$$
= $$\frac{3}{5}$$
Question 2.
The odds against John solving a problem are 4 to 3 and the odds in favour of Rafi solving the same problem are 7 to 5. What is the chance that the problem is solved when both of them try it?
Solution:
The odds against John solving a problem are 4 to 3.
Let event P(A’) = P (John does not solve the problem)
= $$\frac{4}{4+3}$$
= $$\frac{4}{7}$$
So, the probability that John solves the problem
P(A) = 1 – P(A’) = 1 – $$\frac{4}{7}$$ = $$\frac{3}{7}$$
Similarly, Let P(B) = P(Rafi solves the problem)
Since the odds in favour of Rafi solving the problem are 7 to 5,
P(B) = $$\frac{7}{7+5}$$ = $$\frac{7}{12}$$
Required probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Since A, B are independent events,
P(A ∩ B) = P(A) . P(B)
∴ Required probability = P(A) + P(B) – P(A) . P(B)
Question 3.
The odds against student X solving a statistics problem are 8 : 6 and odds in favour of student Y solving the same problem are 14 : 16. Find the chance that
(i) the problem will be solved if they try it independently.
(ii) neither of them solves the problem.
Solution:
The odds against X solving a problem are 8 : 6.
Let P(X’) = P(X does not solve the problem) = $$\frac{8}{8+6}$$ = $$\frac{8}{14}$$
So, the probability that X solves the problem
P(X) = 1 – P(X’) = 1 – $$\frac{8}{14}$$ = $$\frac{6}{14}$$
Similarly, let P(Y) = P(Y solves the problem)
Since odds in favour of Y solving the problem are 14 : 16,
P(Y) = $$\frac{14}{14+16}=\frac{14}{30}$$
So, the probability that Y does not solve the problem
P(Y’) = 1 – P(Y)
= 1 – $$\frac{14}{30}$$
= $$\frac{16}{30}$$
(i) Required probability
P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y)
Since X and Y are independent events,
P(X ∩ Y) = P(X) . P(Y)
∴ Required probability = P(X) + P(Y) – P(X) . P(Y)
= $$\frac{6}{14}+\frac{14}{30}-\frac{6}{14} \times \frac{14}{30}$$
= $$\frac{73}{105}$$
(ii) Required probability = P(X’ ∩ Y’)
Since X and Y are independent events, X’ and Y’ are also independent events.
∴ Required probability = P(X’) . P(Y’)
= $$\frac{8}{14} \times \frac{16}{30}$$
= $$\frac{32}{105}$$
Question 4.
The odds against a husband who is 60 years old, living till he is 85 are 7 : 5. The odds against his wife who is now 56, living till she is 81 are 5 : 3. Find the probability that
(i) at least one of them will be alive 25 years hence.
(ii) exactly one of them will be alive 25 years hence.
Solution:
The odds against her husband living till he is 85 are 7 : 5.
Let P(H’) = P(husband dies before he is 85) = $$\frac{7}{7+5}=\frac{7}{12}$$
So, the probability that the husband would be alive till age 85
P(H) = 1 – P(H’) = 1 – $$\frac{7}{12}$$ = $$\frac{5}{12}$$
Similarly, P(W’) = P(Wife dies before she is 81)
Since the odds against wife will be alive till she is 81 are 5 : 3.
∴ P(W’) = $$\frac{5}{5+3}=\frac{5}{8}$$
So, the probability that the wife would be alive till age 81
P(W) = 1 – P(W’) = 1 – $$\frac{5}{8}$$ = $$\frac{3}{8}$$
(i) Required probability
P(H ∪ W) = P(H) + P(W) – P(H ∩ W)
Since H and W are independent events,
P(H ∩ W) = P(H) . P(W)
∴ Required probability = P(H) + P(W) – P(H) . P(W)
= $$\frac{5}{12}+\frac{3}{8}-\frac{5}{12} \times \frac{3}{8}$$
= $$\frac{40+36-15}{96}$$
= $$\frac{61}{96}$$
(ii) Required probability = P(H ∩ W’) + P(H’ ∩ W)
Since H and W are independent events, H’ and W’ are also independent events.
∴ Required probability = P(H) . P(W’) + P(H’) . P(W)
= $$\frac{5}{12} \times \frac{5}{8}+\frac{7}{12} \times \frac{3}{8}$$
= $$\frac{25+21}{96}$$
= $$\frac{46}{96}$$
= $$\frac{23}{48}$$
Question 5.
There are three events A, B, and C, one of which must, and only one can happen. The odds against event A are 7 : 4 and odds against event B are 5 : 3. Find the odds against event C.
Solution:
Since odds against A are 7 : 4,
P(A) = $$\frac{4}{7+4}=\frac{4}{11}$$
Since odds against B are 5 : 3,
P(B) = $$\frac{3}{5+3}=\frac{3}{8}$$
Since only one of the events A, B and C can happen,
P(A) + P(B) + P(C) = 1
$$\frac{4}{11}$$ + $$\frac{3}{8}$$ + P(C) = 1
∴ P(C) = 1 – ($$\frac{4}{11}$$ + $$\frac{3}{8}$$)
= 1 – $$\left(\frac{32+33}{88}\right)$$
= $$\frac{23}{88}$$
∴ P(C’) = 1 – P(C)
= 1 – $$\frac{23}{88}$$
= $$\frac{65}{88}$$
∴ Odds against the event C are P(C’) : P(C)
= $$\frac{65}{88}$$ : $$\frac{23}{88}$$
= 65 : 23
Question 6.
In a single toss of a fair die, what are the odds against the event that number 3 or 4 turns up?
Solution:
When a fair die is tossed, the sample space is
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Let event A: 3 or 4 turns up.
∴ A = {3, 4}
∴ n(A) = 2
∴ P(A) = $$\frac{n(A)}{n(S)}$$ = $$\frac{2}{6}=\frac{1}{3}$$
P(A’) = 1 – P(A) = 1 – $$\frac{1}{3}$$ = $$\frac{2}{3}$$
∴ Odds against the event A are P(A’) : P(A)
= $$\frac{2}{3}: \frac{1}{3}$$
= 2 : 1
Question 7.
The odds in favour of A winning a game of chess against B are 3 : 2. If three games are to be played, what are the odds in favour of A’s winning at least two games out of the three?
Solution:
Let event A: A wins the game and event B: B wins the game.
Since the odds in favour of A winning a game against B are 3 : 2,
the probability of occurrence of event A and B is given by
P(A) = $$\frac{3}{3+2}=\frac{3}{5}$$ and P(B) = $$\frac{2}{3+2}=\frac{2}{5}$$
Let event E: A wins at least two games out of three games.
∴ P(E) = P(A) . P(A) . P(B) + P(A) . P(B) . P(A) + P(B) . P(A) . P(A) + P(A) . P(A) . P(A)
∴ Odds in favour of A’s winning at least two games out of three are P(E) : P(E’)
= $$\frac{81}{125}: \frac{44}{125}$$
= 81 : 44<|endoftext|>
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Aeration, Circulation, and Fountains
Circulating and aerating water in stormwater ponds provides two critical benefits
1) oxygenation of the water and
2) mixing of water to prevent stratification.
All living organisms consume oxygen to live, including aquatic species. As aquatic organisms grow and multiply, their demand for oxygen from the water increases. Stormwater ponds are often extremely productive systems because they receive large loads of nutrients from lawn fertilizers, people feeding fish, turtles, and ducks, and pet waste in the community. These nutrients encourage microbes, invertebrates, and fish to grow and multiply, increasing the demand for oxygen. Under normal conditions plenty of oxygen is supplied by diffusion from the atmosphere and by photosynthesis of phytoplankton to cover the demand; however, when nutrients become excessive or when ponds stratify, the supply of oxygen may not compensate for the demand. When the supply cannot keep up with demand, then oxygen becomes depleted, leading to stagnancy and fish kills.
The primary source of oxygen in ponds is the atmosphere, the air above the pond. Oxygen diffuses into the water from the air. As long as the water can be exposed to the air, then an ample supply of oxygen can enter the pond. One of the leading obstacles for the diffusion of oxygen into the water is stratification of the water column. Stratification occurs when the sun heats the surface water while deeper water remains cool. The warm and cool layers do not mix. This disrupts the circulation of oxygenated water from the surface of the pond to the bottom. While the warm surface water receives oxygen from the atmosphere as well as from photosynthesis of phytoplankton, the cooler bottom layer becomes depleted of oxygen. As the summer progresses, oxygen levels drop severely in the bottom layer. Stratification can be a significant threat to fish health in small ponds because it increases the potential for a "turnover."
What causes a turnover?
Turnovers occur naturally in the fall. The incremental drop in air temperature slowly decreases the surface water temperature, and the layers in the pond mix gradually. This gradual mixing rarely causes a fish kill. On the other hand, rapid turnover can cause a fish kill by quickly diluting the oxygen in the pond. Rapid turnovers can occur anytime during the warmer months of the year when ponds are stratified, and they most often coincide with storm events or windy days. A cold rain or the blowing winds of a storm front can cause rapid mixing and result in a fish kill.
How do I prevent a turnover?
- Proper pond construction is important. The ratio of surface water (epilimnion) to deep water (hypolimnion) is a major determining factor for the frequency and severity of turnovers. Deep ponds with small surface areas often are more prone to severe turnovers because there is a larger volume of deep, low-oxygen water relative to the amount of oxygenated surface water. Broad, shallow ponds have fewer turnovers because less of the pond's volume is tied up in low oxygen deep water. Most stormwater ponds are shallow (6ft deep or less), but some are deeper. It is important to know the topography of the pond's bottom (bathymetry) as well as the average depth of the pond. Over time, the depth of the pond will change as it fills with sediment, so the bottom contours may not be the same as when the pond was constructed originally. (See section on Pond Construction and Sedimentation)
- Prevent aquatic weeds from covering more than 20% of the pond surface. Floating and submersed weeds impede circulation and the diffusion of oxygen into the pond. They also remove oxygen from the water as they decay. Controlling aquatic plants will improve circulation and reduce stagnancy.
- Reduce nutrients and algae growth. Loading ponds with nutrients from yard fertilizers, exposed sediment, pet waste, and feeding of fish/ducks/turtles will increase algal growth which can further deplete oxygen.
- Mechanical circulation can prevent turnovers. The primary purpose of mechanical circulation is not aeration/oxygenation (injecting oxygen into the water). Its purpose is to prevent stratification and bring water to the surface so that it can be exposed to oxygen in the atmosphere. By mixing the water, circulation systems never allow the pond to form layers, thus they do not develop a low oxygen layer on the bottom. In this way a circulation system becomes an insurance investment to prevent a fish kill.
What is the best way to circulate the water?
The primary goal of a circulation system should be to bring deep water to the surface and expose it to the atmosphere. By far, diffusion systems circulate water most efficiently. Diffusion systems are a lot like airstones in an aquarium. They use an air compressor to pump air into the bottom of the pond. The diffuser breaks the air into tiny bubbles which expand as they rise. The rising of the bubbles pushes water and forms a current that lifts bottom water up to the surface. The resulting current mixes the entire water column and prevents stratification in the pond.
Are there other ways to circulate the water?
Yes. Water pumps such as centrifugal irrigation pumps and sump pumps can be used to push water but they must be designed correctly to work. Remember, the goal of a circulation system should be to prevent stratification. Irrigation pumps can be mounted on land as long as they draw water through pipes from near the bottom and shoot it over the surface to create a current. Unlike diffuser systems, which are self-cleaning, water pumps are prone to becoming clogged by vegetation and debris and require frequent maintenance, especially when submerged plant growth is not controlled. Water pumps also consume more energy than air compressors.
What about fountains... Do they work?
Surface fountains are the least efficient ways to circulate ponds, and they do very little to prevent stratification. They move water, but typically fountains are floating devices that draw surface water into a shallow pump and spray it above the surface. Functionally speaking, they are circulating surface water that is already oxygenated and are not preventing stratification. As a result, fountains are often deployed more for aesthetic reasons than functional circulation and aeration. That being said, surface fountains do provide for limited circulation that is beneficial to fish and other aquatic species. The amount of circulation they create is dependent on the size of the motor and the shape of the sprayer.
Will circulation systems control aquatic weeds?
No, not necessarily. Circulations systems will not control existing aquatic weed problems, except for some blue-green algae (cyanobacteria) which thrive in still water. Some cyanobacteria require stagnant waters to grow large blooms. Circulation systems can disrupt these harmful algae blooms from growing out of control. Also, in ponds that are overloaded with nutrients (eutrophic) and consistently low in oxygen, circulation system may improve water chemistry and make dissolved nutrients less available to algae and weeds. By reducing the availability of nutrients, circulation systems potentially can slow but not prevent the growth of aquatic weeds. If the pond is consistently loaded with new nutrients, then a circulation system will have little or no effect on reducing weed growth.
Are there any other benefits to having a circulation system?
Yes. Oxygenating the sediments and organic matter that collect on the pond bottom helps reduce the production of Hydrogen Sulfide, the gas that makes the rotten egg odor that most folks associate with stagnant swamps and sewers. This gas is produced by anaerobic bacteria that thrive in sediments with little oxygen. Also, oxygenating organic sediments helps accelerate the microbial breakdown of the "muck" that collects in ponds as leaves and debris decompose. Last, oxygenating bottom sediments helps slow the release of nutrients from organic sediments that contribute to algae blooms.
Are there any problems caused by circulation systems?
Yes. Circulation systems can keep sediment suspended in the water and reduce water clarity. They can increase shoreline erosion especially in small ponds with no shoreline protection using native plants or hardscapes (See section on Shoreline Protection) or where the system is placed too close to the bank. Last, circulation systems require electricity, so they are an ongoing utility expense and may require additional hardware to run electricity to the areas where it is needed.
Do I really need a circulation system?
As mentioned previously, most stormwater ponds are shallow basins that do not develop a large volume of low oxygen water in the bottom. Also, they are designed to exchange water with each passing storm. On the other hand, some stormwater ponds seem to have chronic problems with fish kills and would benefit from a circulation system. Stormwater ponds are prone to excessive weed and algae growth. If large masses of these plants are killed using aquatic herbicides, the resulting decay will remove oxygen from the water and may cause a fish kill. Aeration can help prevent such a fish kill. Finally, if your community has stocked the pond with grass carp or tilapia to control aquatic vegetation or other fish for recreational benefits, having a circulation system is good protection for that investment. Circulation systems may not be necessary unless your pond develops chronic problems with fish kills or noxious odors. It is important for the community to document fish kills so that they can be tracked for frequency and timing. This will help the community when deciding whether or not to invest in a circulation system.
When is the best time to install a circulation system?
It is best to install and start circulation systems before ponds become stratified, which usually means late fall through early spring (April). If your community is considering installing a system in the warm summer months, you should be careful not to cause a man-made turnover when the system is turned on. If the system is installed in the summer it must be started in short increments (10 minutes per day) over the first two weeks in order to gradually mix the water. Consult with your system manufacturer for start-up recommendations. You may need to test oxygen levels during start-up.
How large does the system need to be?
A general rule is that one diffuser unit can circulate a one to two acre pond, but sizing diffuser systems has a lot to do with surface area, depth, and the shape of the pond. Diffusers are less efficient in shallow ponds because the bubbles travel a shorter distance and create less current, so more diffusers may be needed. Also, ponds with odd shapes, bottlenecks, and narrow canals may need several diffusers to adequately circulate water in each of the sections of the pond. The same goes for fountains and water pumps. For water pumps, it is suggested that 1 horsepower of pump be used for each surface acre of pond. You should always consult with the circulation system manufacturer or professional installer for guidance on sizing and designing your system.
Do I have to run the system all the time?
If circulation and preventing stratification is your goal, then, yes, you should run your system all day and night. Running the system intermittently may allow the pond to temporarily stratify. The longer the pond remains stratified (layered), the greater the chance that turning the circulation system on will actually cause a turnover and fish kill. You do not need to run it all year. Ponds naturally destratify (mix) in the fall when air temperatures drop. As the temperature of the pond surface cools and matches the temperature of the deeper water, the pond will mix gradually. Once this occurs, the pond will remain mixed until the following spring when air temperatures rise. Circulation systems can remain off from November through March across most of South Carolina.
Is there anything else I can do to prevent low oxygen conditions?
Yes. Reducing the amount of nutrients entering the pond is essential to reducing the demand for oxygen in the system. Here are the best practices for reducing nutrients in runoff...
1) test your soil before fertilizing (you might be applying too much)
2) do not fertilize grass on bank slope around pond
3) after fertilizing lawns, sweep or blow fertilizer off of roads and driveways into the grass (if it is on the road, it will wash to the pond in the next storm)
4) do not feed fish, turtles, ducks, or geese (food in their mouths is fertilizer in the pond)
5) wash vehicles with phosphate free soaps or take vehicles to car washes
6) do not throw yard waste in ponds
7) plant shoreline vegetation to absorb nutrients
8) prevent sediment from washing off of construction sites
Reducing the nutrients carried in runoff will also help reduce aquatic weed problems, turbidity of the water, noxious odors, and other unfavorable conditions in residential stormwater ponds.<|endoftext|>
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# Lesson video
In progress...
Hello and welcome to today's lesson, my name's Miss Thomas, I'll be going through the lesson with you today.
We've got some really exciting learning coming up.
We're looking at how we can use known multiplication facts to divide.
Which we can do a lot that mentally.
So we'll get some really good helpful strategies today.
Before we start though I want to show you something really cool, so this is a pop up card.
That somebody I know made and it's like a lily flower inside.
Well, not so much like a.
Well kind of like a waterbed lily.
How exciting is that? Okay, well let's get started with our maths for today.
In today's lesson first we'll be partitioning numbers to divide mentally, then we'll go to talk task, where you can have a practise.
After that we'll be using known multiplication facts to divide mentally and finally you'll finish with your end of lesson quiz.
The equipment you will need for this lesson is a pencil, paper and a ruler.
Pause video now if you need to gather your equipment.
Let's get started.
Question says how many groups of 6 are in 78? And the equation for that is, 78 divided by 6.
We're trying to find out how many groups of 6 are in 78.
Well I could use my counters to do that, so I've got 78 blue counters here and I can group them all into sixes, to see how many groups of 6 I've got.
If I take out 6 counters and I put them in a group and I'll take out another 6 counters and I'll put them in a group.
You might of said this will take me a long time and if I had a greater number than 78, it would take me even longer.
I would not be able to solve this quickly so it's not really an efficient method.
What multiples of 6 are in 78, that we know? Well, I know that 60 and 18 are multiples of 6 and 78, let's see how we could use our multiples to help us.
So we know that 60 and 18 are multiples of 6 and 78, okay let's have a look at how we can use that.
Well I know that 6 multiplied by 10 is equal to 60, so I know that there are 10 groups of 6 in 60.
I know that 6 multiplied by 3 is 18, so therefore I know there are 3 groups of 6 in 18.
I can then add how many groups of 6 are in 60, so I can add 10 plus the three groups of 6 that are in 18, so 10 plus 3 which is equal to 13.
So I know that there are 13 groups of 6 in 78, 78 divided by 6 is equal to 13.
Let's have a look at another one, let's do this one together.
The question says first, what multiples of 4 do we know? Can you call out some multiples of 4 to your screen please? Great, I'm sure lots of you said multiples of 4 that are in the times table we know up to 12.
What multiples of 4 could we partition 432 into? This time you might need to derive some facts as well as using your known facts.
Pause the video and have a go at finding multiples of 4 that we could use to partition 432 into.
Great, I'm sure you've found a way, I'm going to show you the way that I did it and if you did it differently that's absolutely fine.
So I partitioned 432 into 400 and 32.
I'm going to derive some facts about 400, so I know that 4 times 10 is equal to 40, so I know that 4 times 100 is 10 times greater.
So my products 40 will be 10 times greater, it will be 400.
So I know that there are a 100 groups of 4 in 400.
Let's take a look at 32.
4 multiplied by 8 is equal to 32, so I know that there are 8 groups of 4 in 32.
100 groups of 4 in 400 and 8 groups of 4 in 32.
So the last thing to do is add the number of groups of 4 there are, so 100 plus 8 is equal to 108.
So I know that there are 108 groups of 4 in 432, 432 divided by 4 is equal to 108.
Let's take a look at the next one, you need to decide what multiples of 6 do we know and what multiples of 6 could we partition 372 into.
Pause the video and decide.
Welcome back, I'm sure you found a way, so the way that I did it is I partition 372 into 300 and 72.
So with my 300 I'm going to use my derived facts, I know that 6 times 5 is equal to 30, so I know that 6 multiplied by 50 is equal to 300.
It's 10 times greater, 50 is 10 times greater than 5, so my answer will be 10 times greater than 30 will be 300.
So I know that there are 50 groups of 6 in 300.
Let's take a look at the 72, I know that 6 multiplied by 12 is 72.
So I know that there are 12 groups of 6 in 72, there are 50 groups of 6 in 300 and there are 12 groups of 6 in 72.
So now I need to add the number groups of 6, so 50 plus 12 which is equal to 62, so I know that there are 62 groups of 6 in 372.
372 divided by 6 is equal to 62.
Using your known multiplication facts solve the equations, use jottings to record the steps of your divisions.
What are the fewest number of steps, you can use? So often we think of the fewest number of steps, the most efficient way, because its quicker.
So you've got the equation 545 divided by 5 and 927 divided by 3.
Use the sentence stems to explain out loud your thinking, once you've solved the equations.
Welcome back, we've got our answers here, so we know that 545 divided by 5 is equal to 109 and we know that 927 divided by 3 is equal to 309.
There are many ways you could have solved these equations, hopefully you tried to use the sentence stems to explain out loud your method.
If you didn't do this yet pause the video and have a go, hopefully you found methods that didn't require lots of steps to solve it.
These are the most efficient methods for the equation.
Okay next were going to be dividing on a number line, to find how many groups of 6 are in 324.
So we want, 324 I'm going to think back my known facts.
well first of all I know that 6 multiplied by 50 is 300 because I know that 6 times 5 is 30, 6 times 50 is going to be 10 times greater.
So I'm going to take my 300 on my number line to get to 24, so I know that there are 50 groups of 6 in 300.
Now I need to go from 24 down to 0, so I know that 6 multiplied by 4 is equal to 24, so I can take away 24.
So I know that there are 4 groups of 6 in 24 and that will take me to 0.
The final step is to add how many groups of 6 are in 24 and how many groups of 6 are in 300.
So 50 plus 4 is equal to 54, I know that 324 divided by 6 is equal to 54, I've used my known multiplication facts to divide 324 by 6.
Let's take a look at the next one, we'll do this one together.
824 divided by 4, so we're going to have our number line, with 824 at the end and we want to get down to 0, using our known multiplication facts.
Can you shout out a fact that you think we could use to help us? Call out your answer.
Great, there are many you could have said, the one that I'm going to choose is that we know that 4 multiples by 200 is 800 because we know that 4 times 2 is equal to 8, so we can use our derived facts 200 is a one hundred times greater, so our answer 8 will be one hundred times greater it will be 800.
And that will take us all the way down, we'll have to take away 800 and it will take us all the way down to 24.
Next we need to get from 24 to 4, could you divide 24 divided by 4, what would that give you? Call out your answer.
You might have used your known fact, of 4 multiplied by 6 is 24, so 24 divided by 4 is equal to 6.
So you can take away 24 that will take us to 0, so we know that there are 6 groups of 4 in 24 and that there are 200 groups of 4 in 800.
So the final thing we add how many groups of 4 are in 800 and how many groups of 4 are in 24.
So 200 plus 6 is 206, 824 divided by 4 is equal to 206.
Here we've got Tia and Joe working it out, you've got to decide what's the same and what's different.
Pause the video and explain your thinking out loud to your screen.
Welcome back, you might have found that what's the same is that Tia and Joe are both solving the equation to 824 divided by 4 and both children got the same answer of 206.
However you might have found that they've used different know and derived facts to get their answers and Tia she's calculated hers in two steps and Joe he's calculated his in four different steps.
We can use different know facts to derive and to get the same answer, so both children have got it correct.
Now you've reached your independent task, for each question you need to do number 1, number 2 and number 3, so number 1, what's the fewest number of jumps you can complete it in? Number 2, can you complete it in an even number of jumps? Number 3, can you complete it in and odd number of jumps? So you might want to partition your number, you could have a go on the number line, have a go at solving the equations.<|endoftext|>
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5.1.2 The human genome
All life is ‘encoded’ chemically in genes. What this means is that the structure of an organism, the organs it possesses, its colouring, and so on are all determined by different genes. A very simple organism may have just a few genes, and a complex one tens of thousands. The ‘map’ of an organism's genes is referred to as its genome. It shows, in essence, which genes give rise to which characteristics or traits of the organism. The word ‘template’ would describe the genome better than ‘map’.
Figure 13 shows the 23 pairs of human chromosomes that constitute the structure of the human genome. These chromosomes contain between 30,000 and 40,000 genes in total. For each human characteristic, such as eye or hair colour, the human genome shows where the genes are that control that characteristic.
Until recently, the idea of mapping the genome of even a simple organism was just that, an idea. The work involved in extracting genetic material, examining it and mapping it to known traits, would be analogous to sitting a dozen people down at typewriters and asking them to write a multi-volume encyclopedia. It could have been done, but it would have been time consuming (and therefore costly).
Why do it? DNA acts like a computer program. Just as programs instruct a computer to produce certain outputs, DNA instructs the body to develop proteins that make up tissues, cells, antibodies, and so on in a certain way. If there is a defect in a person's genetic makeup then problems can occur; for example, that person might be more susceptible than average to certain diseases. Mapping the human genome offered some enticing possibilities:
better understanding of diseases, particularly complex and threatening diseases like cancers;
an understanding of the relationship between different human groups. For example, are we descended from one pair of proto-humans, or did different groups have many different origins?
International effort to map the human genome began in 1995, when it was estimated that the project would require US$3 billion and take eight years. But, due to the development of computer-controlled robotic laboratory techniques and improvements in information technology (IT) systems, the Sanger Centre announced in 2000 the first draft of the human genome.<|endoftext|>
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# practice_ch1 - EXERCISE 1.1 4 Second order nonlinear...
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EXERCISE 1.1 4. Second order; nonlinear because of cos( r + u ) 8. Second order; nonlinear because of x ˙ 2 10. Writing the differential equation in the form u ( dv/du ) + (1 + u ) v = ue u we see that it is linear in v . However, writing it in the form ( v + uv − ue u )( du/dv ) + u = 0, we see that it is nonlinear in u . 11. From y = e− x/ 2 we obtain y’ = 1/2 e −x/ 2 . Then 2 y’ + y = −e −x/ 2 + e −x/ 2 = 0. 14. From y = cos x ln(sec x + tan x ) we obtain y’ = 1 + sin x ln(sec x + tan x ) and y” = tan x + cos x ln(sec x + tan x ). Then y” + y = tan x . 18. The function is y = 1 /√ 1 sin x , whose domain is obtained from 1 sin x 0 or sin x 1. Thus, the domain is {x | x π/ 2 + 2 nπ} . From y’ = 1/2 (1 sin x ) 3 / 2 ( cos x ) we have 2 y’ = (1 sin x ) 3 / 2 cos x = [(1 sin x ) 1 / 2 ] 3 cos x = y 3 cos x. An interval of definition for the solution of the differential equation is ( π/ 2 , 5 π/ 2). Another one is (5 π/ 2 , 9 π/ 2), and so on. 21. Differentiating P = c 1 e t / (1 + c 1 e t ) we obtain 27. (a) From y = e mx we obtain y’ = me mx . Then y’ + 2 y
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practice_ch1 - EXERCISE 1.1 4 Second order nonlinear...
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Effect of Climate Change on Fruits and Vegetables
Climate change can have serious consequences on fruit and vegetables. As temperatures rise and extreme weather conditions such as damaging rainfalls become more frequent, there are more “false springs.” In the past 5 years, we have had 2 record highs and 2 false springs with damaging late freezes. We have also had record high temperatures in February 2017.
When temperatures get hotter, fruit and vegetable yields can be reduced due to lower photosynthesis and reduced pollination. Heat also causes more internal defects and increased tissue damage to the fruit and vegetables.
Growers are challenged to find ways to adapt to the increasing temperatures. Managing mulch to reduce soil temperatures is one method to reduce the stress on many crops. The use of reflective materials or radiation blocks is also a way to reduce sunburn on fruits and vegetables. The use of wax reflective materials is also a way to help fruits such as apples keep their color.
The use of shade netting is also an effective technique to keep plants cooler during the hottest periods of the day. It has proven to be very effective for both tomatoes and peppers.
Another strategy is to use water-based cooling which can reduce the effect of heat on crops. Low water volume sprinklers have been proven to be successful during the hottest times of the day for plant cooling.
Growers are now altering the planting dates. Planting earlier in the spring or later in the summer allows growers to avoid the hottest times of the season and improve crop production.
The strategy that is most promising is to breed new, stronger, more heat resistant fruits and vegetables. This is a long term strategy and will require years of testing and experimentation.
Climate change can have a negative effect on both the quality and quantity of the fruit and vegetable crops. Lower production also leads to higher costs to the consumer. This is why it is so important to prepare now with longterm strategies to ensure our farmers can continue to have healthy and plentiful harvests in the future no matter what the temperature is.
Fruits and vegetables are essential to everyone’s diet. Eating a diet high in fruits and vegetables is an effective tool to protect against some types of cancer, type 2 diabetes, and heart disease. Fruit at work is an awesome way to keep employees healthier, more motivated and it helps them maintain a healthier weight. Eat fruit and vegetables every day!<|endoftext|>
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## Sunday, 10 June 2012
### Challenge Question 1: Solutions
This is the worked solution to Challenge Question 1.
To begin thinking about the problem it is a must to visualise or draw out what is actually going on and begin labelling what we know.
As all that matters in this question is the ratio from side to side we can say that the largest square has a side length of 1, the next is 1/2, the next 1/4, etc. This means that the area of the first square is 1, then next is 1/4, the next 1/16, etc.
You can see that there is a ratio of 1/4 between the area of a square and the square before it. So to find the total area of all the squares you can use the formula for the sum of an infinite geometric series a/(1 - r) (explained here), multiply it by 4 (for each branch) and then add 1 (for the central square). So we get [4 × 0.25/(1 - 0.25)] + 1, which gives a total area for the squares as 7/3.
Now for the area of the shaded squares. The first shaded square has a side length of 1/4, so it has an area of 1/16, the next shaded square has a side length of 1/8 and an area of 1/64. You can see that the ratio from the area of one square to the next is 1/16. The total area of the shaded squares uses the sum of an infinite geometric series again and multiplies it by 4 (for each branch). So we get 4 × 0.125/(1 - 1/16), which equals 8/15.
So the area of all the overlapping squares is 8/15 and the total area of all the squares is 7/3, so the proportion of the shaded squares is 8/15 ÷ 7/3 = 0.22857142857... So the final answer is that the shaded squares occupy 22.86% of the total area of the squares.
#### 1 comment:
1. These rules are general — they work at the property level. (Intuitively, I have a chemical analogy that “evenness” is a molecule some numbers have, and cannot be removed by multiplication.) help me with maths<|endoftext|>
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Though better known for his treatments of philosophy, history, and politics, the Scottish philosopher David Hume also made several essential contributions to economic thought. His empirical argument against British mercantilism formed a building block for classical economics. His essays on money and international trade published in Political Discourses strongly influenced his friend and fellow countryman adam smith.
British mercantilists believed that economic prosperity could be realized by limiting imports and encouraging exports in order to maximize the amount of gold in the home country. The American colonies facilitated this policy by providing raw materials that Britain manufactured into finished goods and reexported back to the colonial consumers in America. Needless to say, the arrangement was short-lived.
But even before the American Revolution intervened in mercantilistic pursuits, David Hume showed why net exporting in exchange for gold currency, hoarded by Britain, could not enhance wealth. Hume’s argument was essentially the monetarist quantity theory of money: prices in a country change directly with changes in the money supply. Hume explained that as net exports increased and more gold flowed into a country to pay for them, the prices of goods in that country would rise. Thus, an increased flow of gold into England would not necessarily increase England’s wealth substantially.
Hume showed that the increase in domestic prices due to the gold inflow would discourage exports and encourage imports, thus automatically limiting the amount by which exports would exceed imports. This adjustment mechanism is called the price-specie-flow mechanism. Surprisingly, even though Hume’s idea would have bolstered Adam Smith’s attack on mercantilism and argument for free trade, Smith ignored Hume’s argument. Although few economists accept Hume’s view literally, it is still the basis of much thinking on balance-of-payments issues.
Considering Hume’s solid grasp of monetary dynamics, his misconceptions about money behavior are all the more noteworthy. Hume erroneously advanced the notion of “creeping inflation”—the idea that a gradual increase in the money supply would lead to economic growth.
Hume made two other major lasting contributions to economics. One is his idea, later elaborated by friedrich hayek in The Road to Serfdom, that economic freedom is a necessary condition for political freedom. The second is his assertion that “you cannot deduce ought from is”—that is, value judgments cannot be made purely on the basis of facts. Economists now make the same point by distinguishing between normative (what should be) and positive (what is).
Hume died the year The Wealth of Nations was published, and in the presence of its author, Adam Smith.<|endoftext|>
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# What is the imaginary unit i ? What is a complex number? Identify the imaginary and real parts of a complex number. Practice - see textbook p.103 #7 –
## Presentation on theme: "What is the imaginary unit i ? What is a complex number? Identify the imaginary and real parts of a complex number. Practice - see textbook p.103 #7 –"— Presentation transcript:
What is the imaginary unit i ? What is a complex number? Identify the imaginary and real parts of a complex number. Practice - see textbook p.103 #7 – 42 1.3 Complex Numbers http://www.youtube.com/watch?v=NeTRNpBI17I
Add and Subtract Complex Numbers Practice - see textbook p.103 #43 - 50 Add the real parts and the imaginary parts, just like combining like terms. Be sure to distribute the negative sign when subtracting! http://www.youtube.com/watch?v=eyy_SR1JfyQ&feature=youtu.be
http://www.youtube.com/watch?v=O9xQaIi0NX0 Multiplying easy complex numbers * Remember i 2 = -1
Multiplying complex numbers Practice - see textbook p.103 #51- 68 http://www.youtube.com/watch?v=Fmr3o2zkwLM&feature=youtu.be
Complex numbers and their conjugate The conjugate of a complex number is simply the same complex number with the sign changed on the imaginary part. When you multiply a complex number by the conjugate, the product simplifies to a real number with no imaginary part. Complex conjugates are needed for division of complex numbers. http://www.youtube.com/watch?v=VbOzEkvMb_A&feature=youtu.be
Dividing complex numbers To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator and simplify. Practice - see textbook p.103 #69 – 80 http://www.youtube.com/watch?v=XBJjbJAwM1c
Practice - see textbook p.103 #85 - 96 Simplifying powers of i http://www.youtube.com/watch?v=sfP6SmEYHRw
What you should know: Add and Subtract complex numbers Imaginary part Real part Complex number Imaginary unit Divide complex numbers Multiply complex numbers Simplify radicals with negative radicands Standard form Quotient Complex conjugate Numerator Denominator Product Powers of i 1.3 Complex Numbers
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# Module 4: MATH AND DATA ANALYSIS
## Learning Objectives
After reading this lesson, you should be able to:
• solve linear equations in MATLAB,
• solve nonlinear equations in MATLAB,
• manipulate polynomials and polynomial coefficients,
• plot data with the syms data type.
## How do I solve for roots of a linear equation?
In this section, we will go over using MATLAB to solve for the roots of an equation using solve(). For example, $$x-2=0$$ has a root at $$x=2$$.
Example 1 shows the procedure for solving for the root of a linear equation in MATLAB. Note that when the equation is defined in Example 1, we write y = 9*x +7 == 10. This means that the equation 9*x + 7 == 10 is stored in the variable y, which we call subsequently in the program like we would any other variable. The right-hand side of the equation that you enter does not have to be equal to zero. For example, you could also write 9*x == 3 and still get the same answer from MATLAB.
You might be asking at this point, “Why do we need double equals when entering the equation?” The simple answer to this is that MATLAB needs to recognize that you are entering an equation rather than storing something in a variable. Remember that y = 9*x + 7 == 10 has two parts. Entering the equation 9*x + 7 == 10 is stored in the variable y. MATLAB needs a way to differentiate between these two operations. All you need to remember is to put the double equals (==) when entering an equation. MATLAB will return an error if you fail to do so.
### Example 1
Solve for the root of the $$9x+7=10$$.Display the equation and the root of the equation using a single fprintf() function.
Solution
## What is a nonlinear equation?
A nonlinear equation is an equation that has nonlinear terms of the unknowns. One of the simplest examples of a nonlinear equation is the quadratic equation that has the form
$$ax^{2} + bx + c = 0$$ (1)
where,
$$a \neq 0$$.
The values of a, b, and c are constant coefficients and x is the independent variable. If one were to solve a nonlinear equation using traditional methods they would need to be provided at least two items: the nonlinear equation and the unknown variable to solve for.
Take the case of the equation given in the above Equation (1). The nonlinear equation is a quadratic equation and the variable to solve for is x. This example can be solved using the widely known solution of a quadratic equation as
$$x = \displaystyle\frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$$ (2)
However, other more complex nonlinear equations can take longer to solve exactly, while most nonlinear equations are impossible to solve analytically. Another example of a nonlinear equation is $$\text{sin}(2x) = \displaystyle\frac{3x}{7}$$. In these cases, MATLAB can be used to solve the nonlinear equations or even to verify the results obtained from an analytical method.
## How can I use MATLAB to solve nonlinear equations?
As we know, MATLAB is a tool. Just like a mechanic might use a wrench to solve a problem, an engineer, scientist or mathematician might use MATLAB to solve a problem. In the case of solving a nonlinear equation, MATLAB can be used to solve nearly any equation (limited more so by the operating computer than the program itself).
We used solve() to find the roots of a linear equation as a demonstration of symbolic variables. However, solve() is also used to find the roots of nonlinear equations. In this case, we may get multiple roots. Unique, none and infinite number of solutions are other possibilities for nonlinear equations. Multiple roots will be stored in vector form, which you can see in Example 2. The variable soln is a vector that holds the three roots we found from the third-order polynomial equation.
You can see in Example 2 that the solution to the nonlinear equation has three possible values. We know this is correct because the equation is a cubic equation. The outputs to the solve command are stored in a vector and can each be pulled out of the vector by using matrix references, which we covered in Lesson 2.6.
### Example 2
Solve the nonlinear equation $$x^3-15x^2+47x=33$$ for x using MATLAB (that is, find its roots).
Solution
MATLAB also provides a method to solve an equation numerically using vpasolve() (note that solve() uses analytical methods). Although polynomial equations have a finite number of solutions (which is equal to the order of the polynomial), other nonlinear equations can possibly have an infinite number of solutions (e.g., $$\text{tan}\left(x \right) = x$$). Therefore, MATLAB does not report all solutions in the same way. Check the vpasolve() documentation for specifics.
In Example 3, we show an application that shows a real-world problem. The only difference here is the prerequisite mathematical knowledge to set up the problem (find the appropriate equation). The programming portion is more or less the same, which is the beauty of MATLAB and many other similar programming languages. Once written, a program like the one in Example 2 can be quickly edited for a completely different equation like the one in Example 3. We chose to use the aforementioned vpasolve() in Example 3.
### Example 3
Find the depth x to which a ball is floating in water (see Figure 1) based on the following cubic equation
$4R^{3}S - 3x^{2}R = - x^{3}$
where,
R = radius of the ball,
S = specific gravity of the ball.
Use values of R and S to be 0.055 and 0.6, respectively.
Figure 1: Diagram of a partially submerged sphere in water for use with Example 3.
Solution
Note the three possible solutions to the third order polynomial equation for Example 2. Since the limits of x are given by $$0 \leq x \leq 2R$$ and R = 0.055, the limits of x are $$0 \leq x \leq 0.11$$. Hence, the only physically acceptable solution is $$x = 0.0624$$.
## Is there a faster way to work with polynomial equations in MATLAB?
We can create a symbolic polynomial from a vector of coefficients using the function poly2sym(). (You can do the reverse operation with sym2poly().) This can be especially helpful when using functions that return a vector of coefficients directly such as polyfit(), which we will cover later in Lessons 4.7 and 4.8. You can see in Example 4 that the coefficient output (coefs) is not very intuitive. However, we can directly convert the coefficients to polynomial form, which is more intuitive, and find the roots of the corresponding polynomial with the roots() function.
### Example 4
Given any size vector of polynomial coefficients, generate the full polynomial in symbolic form and find its roots.
Test the program using the coefficient vector $$\left\lbrack \begin{matrix} 0.003 & -0.0748 & -4.2660 & 1.8370 \end{matrix} \right\rbrack$$.
Solution
## Can I plot with symbolic variables?
Plotting with syms has all the same considerations as any other function you wish to plot with the additional step of converting to a data type that is accepted by the function you are using to plot (e.g., plot(), [bar()](https://www.mathworks.com/help/matlab/ref/bar.html?s_tid=srchtitle), etc.). That is, you will likely need to convert from sym to a numeric data type to plot the data.
To get a vector of data points to plot, we can use the subs() function. Although it was used in the previous examples to replace a variable with just a single number or character, it also accepts vector inputs as the replacement value(s). When the input is a vector, the output is a corresponding vector as well.
### Example 5
Plot the function $$\displaystyle{y(x)=4x^3-20x^2+9x}$$ and its zeros on the same plot. Choose the domain to plot the function based upon the solutions found for the zeros of the function. Include title, axis labels, legend, and plot grid.
Solution
Figure 6: The MATLAB figure output for Example 5.
## Lesson Summary of New Syntax and Programming Tools
Find the roots of an equation solve() solve(eq,x)
Numerically find the roots of an equation vpasolve() vpasolve(eq,x)
Form a symbolic polynomial function from coefficients poly2sym() poly2sym(coefs,x)
Extract the coefficients from a polynomial sym2poly() sym2poly(polynomial)
Get the roots of a polynomial using only is coefficients as an input roots() roots(coefs)
## Multiple Choice Quiz
(1). A MATLAB function for solving an equation analytically is
(a) fsolved()
(b) nonlinear()
(c) vpasolve()
(d) solve()
(2). The MATLAB function to find the solution to a polynomial equation given only its coefficients is
(a) solve()
(b) coefs()
(c) roots()
(d) polysolve()
(3). To solve $$x^2+2x=0$$, which line of code should be added to the program
(a) solve(x^2+2x==0,x)
(b) solve(x^2-2x,x)
(c) solve(x^2+2*x=0,x)
(d) solve(x^2+2*x==0,x)
(4). The MATLAB substitution function subs() has __________ input variables
(a) 1
(b) 2
(c) 3
(d) 4
(5). The output solutions for a single nonlinear equation given by the solve() function
(a) gives only one solution to the equation.
(b) gives only the physically acceptable solution(s).
(c) provides the solutions as separate outputs.
(d) stores solutions in a single vector.
## Problem Set
(1). Use the solve() function to find the solution of the equation $$3x^{2} + 2x = 5$$. Display the answers in the Command Window.
(2). Given that the value of the two variables a and b can be changed, set up the equation $$ax^{2} + 2x = b$$ so that it can be solved for any real values of a and b.
(3). Solve the nonlinear equation $$2x+e^{-x}=5$$. Output both the original equation and at least one of its roots using a single fprintf() function.
(4). You are working for ‘DOWN THE TOILET COMPANY Inc.’ that makes floats for H3H3 toilets. The floating ball has a specific gravity of 0.6 and has a radius of 5.5 cm. You are asked to find the depth to which the ball is submerged when floating in water (Figure A).
The equation that gives the depth x in meters to which the ball is submerged underwater is given by
$x^{3} - 0.165x^{2} + 3.993 \times 10^{- 4} = 0$
Find the depth, $$x$$, to which the ball is submerged underwater.
Figure A: Floating ball in water for Exercise 4.
(5). You have a spherical storage tank containing oil (Figure B). The tank has a diameter of 6 ft. You are asked to calculate the height, h, to which a dipstick 8 ft long would be wet with oil when immersed in the tank when it contains 4 $$\text{f}\text{t}^{3}$$ of oil. The equation that gives the height, $$h$$, of the liquid in the spherical tank for the above-given volume and radius is given by
$f(h) = h^{3} - 9h^{2} + 3.8197 = 0$
Find the height, $$h$$, to which the dipstick is wet with oil.
Figure B: Dipstick inside spherical tank which contains some oil. Used in Exercise 5.
(6). You are making a cool bookshelf to carry your interesting books that range from $$8\frac12$$” to 11” in height. The bookshelf material is wood, which has a Young’s Modulus of $$3.667 \text{ Msi}$$, length (L) of 29”, a thickness of 3/8”, and width of 12”. You want to find the maximum vertical deflection of the bookshelf. The vertical deflection of the shelf is given by $v(x) = 0.42493\ \times \text{ 1}\text{0}^{- 4}x^{3} - 0.13533 \times \text{ 1}\text{0}^{- 8}x^{5} - 0.66722\ \times \text{ 1}\text{0}^{- 6}x^{4} - 0.018507x$
where $$x$$ is the position along the length of the beam (Figure C). To find the maximum deflection we need to find where
$f(x) = \frac{dv}{dx} = 0$
and conduct the second derivative test. The equation that gives the position,$$x$$, where the deflection is extreme (minimum or maximum) is given by $- 0.67665\ \times \text{ 1}\text{0}^{- 8}x^{4} - 0.26689\ \times \text{ 1}\text{0}^{- 5}x^{3} + 0.12748\ \times \text{ 1}\text{0}^{- 3}x^{2} - 0.018507 = 0$
(a) Plot the deflection of the beam, $$v(x)$$, for values of x from 0 to L.
(b) Find the position $$x$$ where the deflection is maximum.
(c) Find the value of the maximum deflection.
Figure C: Schematic of bookshelf used in Exercise 6.
(d) A trunnion has to be cooled from a temperature of $$80{^\circ}F$$before it is shrink-fitted into a steel hub (See Figure D on the next page). The equation that gives the temperature, $$T_{f}$$, to which the trunnion has to be cooled to obtain the desired contraction is given by
${f(T_{f}) = - 0.50598\ \times \text{ 1}\text{0}^{- \text{10}}T_{f}^{3} + 0.38292\ \times \text{ 1}\text{0}^{- 7}T_{f}^{2} + }{0.74363\ \times \text{ 1}\text{0}^{- 4}T_{f} + 0.88318\ \times \text{ 1}\text{0}^{- 2} = 0}$
Find the roots of the equation to find the temperature, $$T_{f}$$, to which the trunnion has to be cooled. Choose the physically acceptable root of the equation.
Figure D: Trunnion and hub shown prior to shrink fitting for Exercise 7.<|endoftext|>
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Rocks are a lot like people… they have a past, they are made up of smaller bits and pieces, they respond to stress in different ways — some deform, some break, some change in structure and make-up. Some rocks are straight forward, like a homogeneous sandstone, and other rocks are more complicated like a really old gneiss (pronounced “nice”) or a metaconglomerate.
Rocks are roughly categorized by how they came into being: did molten rock make its way into the earth or on to the earth? Igneous! Did layers of sediments get covered over and compressed (lithified)? Sedimentary! Were the pressure and temperature just too much and the rock’s mineral crystals changed? Metamorphic!
The rock I work with the most is a sedimentary rock called shale. According to some that is not a real name for rock. Shale is a catchall term to describe a lithified chunk of extremely tiny sediment. Depending on whether you are speaking with a geologist, a sedimentologist (a geologist who specializes in sedimentology), a petroleum engineer, a civil engineer, or a soil mechanic for that matter, you will get a different definition of precisely what a shale is and what it is not. Or if it even exists.
The term “shale” was originally used to describe a very fine-grained, laminated, clayey rock. It can also describe a formation, such as the Bakken Shale, Pierre Shale or Marcellus Shale. It can also be and has been used to describe any very fine-grained rock regardless of lamination or clay content.
There are any number of rock names used to describe what is typically referred to as shale: mudstone, claystone, siltstone, argillite, marl, and lutite to name a few. These names bear specific reference to measurable attributes: grain size, clay type and amount, lamination or bedding planes in the rock, and the amount and type of organic content in the rock.
Why does defining a rock by grain size, clay type, lamination, or even organic content matter? Scientists like precision, which requires the definition of terms. Engineers like repeatability, which means predictability, which means that a process can be created and then refined for the best possible outcome. Researchers like definitions and predictability because we use these to generate a hypothesis. If the outcome diverges wildly from the hypothesis, we get very excited and want to figure out why this is.
In the picture above you can see the Marcellus Shale in outcrop behind me. The visible horizontal layers means it is laminated. You cannot determine how clay-rich the shale is just by looking at it, a mineralogy study needs to be done for that. As for grain size, very-fine-grained rock grains are less than 0.0625 mm. Too tiny for the human eye to distinguish. The blacker a shale is the higher the total organic content (TOC). The higher the TOC, the greater the chance for hydrocarbons.
And this, my friends, is when it gets interesting.<|endoftext|>
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Basic Physics
Springs in series and parallel – problems and solutions
1. A 160-gram object attaches at one end of a spring and the change in length of the spring is 4 cm. What is the change in length of three springs connected in series and parallel, as shown in the figure below?
Known :
The change in length of a spring (Δx) = 4 cm = 0.04 m
Mass (m) = 160 gram = 0.16 kg
Acceleration due to gravity (g) = 10 m/s2
Weight (w) = m g = (0.16)(10) = 1.6 Newton
Wanted : The change in length of three spring (Δx)
Solution :
The equation of Hooke’s law :
k = w / Δx = 1.6 / 0.04 = 40 N/m
The three springs have the same constant, k = 40 N/m.
Determine the equivalent constant :
Spring 2 (k2) and spring 3 (k3) tare connected in parallel. The equivalent constant :
k23 = k2 + k3 = 40 + 40 = 80 N/m
Spring 1 (k1) and spring 23 (k23) are connected in series. The equivalent constant :
1/k = 1/k1 + 1/k23 = 1/40 + 1/80 = 2/80 + 1/80 = 3/80
k = 80/3
Determine the change in length of three springs :
Δx = w / k = 1.6 : 80/3 = (1.6)(3/80) = 4.8 / 80 = 0.06 m = 6 cm
Read : Hooke's law and elasticity – problems and solutions
2. Three springs with the same constant connected in series and parallel, and a 2-kg object attached at one end of a spring, as shown in figure below. Spring constant is k1 = k2 = k3 = 300 N/m. What is the change in length of the three springs. Acceleration due to gravity is g = 10 m.s-2.
Known :
Spring constant k1 = k2 = k3 = 300 N.m-1
Acceleration due to gravity (g) = 10 m.s-2
Object’s mass (m) = 2 kg
Object’s weight (w) = m g = (2)(10) = 20 Newton
Wanted : The change in length of the three springs (Δx)
Solution :
Determine the equivalent constant :
Spring 1 (k1) and spring 2 (k2) are connected in parallel. The equivalent constant :
k12 = k1 + k2 = 300 + 300 = 600 N/m
Spring 3 (k3) and spring 12 (k12) are connected in series. The equivalent constant :
1/k = 1/k3 + 1/k12 = 1/300 + 1/600 = 2/600 + 1/600 = 3/600
k = 600/3 = 200 N/m
Determine the change in length of the three springs :
Δx = w / k = 20/200 = 2/20 = 1/10 = 0.1 m
Read : The magnitude of net torque – problems and solutions
3. Three springs are connected in series and parallel, as shown in figure below. If spring constant k = 50 Nm-1 and a mass of 400 gram attached at one end of a spring. What is the change in length of the three springs.
Known :
Spring constant 1 (k1) = k = 50 Nm-1
Spring constant 2 (k2) = k = 50 Nm-1
Spring constant 3 (k3) = 2k = 2 (50 Nm-1) = 100 Nm-1
Object’s mass (m) = 400 gram = 0.4 kg
Acceleration due to gravity (g) = 10 m/s2
Object’s weight (w) = m g = (0.4)(10) = 4 Newton
Wanted : The change in length (Δx)
Solution :
Determine the equivalent constant :
Spring 1 (k1) and spring 2 (k2) are connected in parallel. The equivalent constant :
k12 = k1 + k2 = 50 + 50 = 100 N/m
Spring 3 (k3) and spring 12 (k12) are connected in series. The equivalent constant :
1/k = 1/k3 + 1/k12 = 1/100 + 1/100 = 2/100
k = 100/2 = 50 N/m
Determine the change in length of the three springs :
Δx = w / k = 4 / 50 = = 0.08 m = 8 cm
Read : Calculation of the electrical energy usage by electrical tool – problems and solutions
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• Jul 17th 2009, 05:09 AM
flyinhigh123
The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangments of 2n different objects taken n at a time as 14:5. Find the value of n.
If 2nPn = 8.(2n-1)P(n-1), find the value of n
the P refers to the formula in permutations and combinations that is nPr = n!/(n-r)!
i dont understand how to solve these questions. some help is greatly appreciated. thankyou =D
• Jul 17th 2009, 07:49 AM
Soroban
Hello, flyinhigh123!
You're expected to know the formula for Permutations
. . and be able to handle factorials.
Quote:
(1) The ratio of the number of arrangements of $2n+2$ different objects taken $n$ at a time
to the number of arrangments of $2n$ different objects taken $n$ at a time is $14:5$.
Find the value of $n$.
$2n+2$ objects taken $n$ at a time: . $_{2n+2}P_n \:=\:\frac{(2n+2)!}{(n+2)!}$
$2n$ objects taken $n$ at a time: . $_{2n}P_n \:=\:\frac{(2n)!}{n!}$
The ratio is: . $R \;=\;\frac{\dfrac{(2n+2)!}{(n+2)!}} {\dfrac{(2n)!}{n!}} \;=\;\frac{(2n+2)!}{(n+2)!}\cdot\frac{n!}{(2n)!} \;=\;\frac{(2n+2)!}{(2n)!}\cdot\frac{n!}{(n+2)!}$
And we have: . $\frac{(2n+2)(2n+1)}{(n+2)(n+1)} \;=\;\frac{14}{5} \quad\Rightarrow\quad 5(2n+2)(2n+1) \;=\;14(n+2)(n+1)$
. . which simplifies to: . $n^2 - 2n - 3 \:=\:0 \quad\Rightarrow\quad (n+1)(n-3) \:=\:0 \quad\Rightarrow\quad n \:=\:-1,3$
Therefore: . $\boxed{n \:=\:3}$
• Jul 17th 2009, 07:35 PM
flyinhigh123
thankyou for your help soroban =D i realli appreciate your clear explaining and solution<|endoftext|>
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Lower Dauphin School District
Subject: English/Language Arts Grade Level(s): Fourth Grade
Synopsis of Curriculum
The English/Language Arts curriculum is divided into five general areas: Foundational Skills, Speaking and Listening, Reading Informational Text, Reading Literature, and Writing. Fourth grade students will learn to identify the main idea of a text, summarize the text, refer to details in the text to make inferences, and explain events, procedures, ideas or concepts and why they happened based on specific information from the text. The students will be able to compare and contrast two different texts on the same topic. The students will also learn to integrate information from two different texts to demonstrate understanding of the topic. Fourth grade students will learn to describe the theme of a text, provide relevant details to support what the text says explicitly, and, describe in depth a character, setting or event using specific details from the text. The students will learn to compare and contrast different themes, topics and pattern of events in literature including texts from different cultures. Students will write informative, explanatory, opinion and narrative pieces conveying ideas or topics clearly. The students will group related information in paragraphs and sections linking ideas within categories of information using words and phrases. When writing opinion pieces, the students will learn to use an organizational structure that groups related ideas and supports the writer’s purpose. The students will learn to write a narrative that uses dialogue and descriptions to develop experiences and events or show the characters’ responses to situations.
Units of Study Titles<|endoftext|>
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Pigeon post is an obsolete method of sending messages by using homing pigeons. The method was used from antiquity until the early 20th century. The use of' homing pigeons to carry messages is as old as the ancient Persians from whom the art of training the birds probably came. The Greeks conveyed the names of the victors at the Olympic Games to their various cities by this means. Before the telegraph this method of communication had a considerable vogue amongst stockbrokers and financiers. The Dutch government established a civil and military system in Java and Sumatra early in the 19th century, the birds being obtained from Baghdad.
The pigeon post which was in operation while Paris was besieged during the Franco-Prussian War of 1870–1871 is probably the most famous. Barely six weeks after the outbreak of hostilities, the Emperor Napoleon III and the French Army of Chalons surrendered at Sedan on September 2, 1870. The normal channels of communication into and out of Paris were interrupted during the four-and-a-half months of the siege. With the encirclement of the city on 18th September, the last overhead telegraph wires were cut the next day, and the secret telegraph cable in the bed of the Seine was located and cut on 27th September. For an assured communication into Paris, the only successful method was by the time-honoured carrier-pigeon, and thousands of messages, official and private, were thus taken into the besieged city. Pigeons were regularly taken out of Paris by balloon. Soon a regular service was in operation, based first at Tours and later at Poitiers. The first despatch was dated 27th September and reached Paris on 1st October, but it was only from 16th October, when an official control was introduced, that a complete record was kept.
Major-General Donald Roderick Cameron, Commandant of the Royal Military College of Canada in Kingston, Ontario from 1888–1896, recommended an international pigeon service for marine search and rescue and military service. A pigeon post between look-out stations at lighthouses on islands and the mainland at the citadel in Halifax, Nova Scotia provided a messenger service from 1891 until it was discontinued in 1895.
A magnifying glass is a convex lens which is used to produce a magnified image of an object. The lens is usually mounted in a frame with a handle though other designs are produced. A magnifying glass works by creating a magnified virtual image of an object behind the lens. Stamp collectors frequently use magnifying glasses to inspect their stamps. This photograph shows the magnified image of the Deutsche Post 1 Reichsmark stamp issued on May 12 1946.
Henry Bishop (1611–1691) was Postmaster General of the United Kingdom and inventor of the first postmark used on mail. In 1660, at The Restoration, Henry Bishop paid £21,500 per year to farm the Post Office for a term of seven years. Bishop was the first officially appointed Postmaster General to Charles II, but within a year of taking office he was accused of abuses. Bishop gave up the remainder of his lease to Daniel O'Neill.
The "Bishop Mark", which takes his name, introduced in 1661, was designed to show the date on which a letter was received by the post and to ensure that the dispatch of letters would not be delayed.
Things you can do
Did you know...
... that the first Penny Post was established in London in 1680 by William Dockwra nearly 200 years before the better known Uniform Penny Post that was part of the postal reforms of 1839 and 1840 in Great Britain.
... that Czesław Słania (1921-2005) is the most prolific stamp engraver, with more than 1,000 post stamps for 28 postal administrations?
... that a forerunner is a postage stamp used during the time period before a region or territory issues stamps of its own?
... that the Royal Philatelic Society is the oldest philatelic society in the world, founded in London in 1869?
... that Marcophily is the specialised study and collection of postmarks, cancellations and postal markings applied by hand or machine on mail?
... that throughout U.S. history, different types of mail bags have been called mail pouch, mail sack, mail satchel, catcher pouch, mochila saddle mailbag, and portmanteau depending on form, function, place and time?
... that Non-denominated postage are postage stamps that do not show a monetary value on the face?
... that the Daguin machine was a cancelling machine first used in post offices in Paris in 1884?
... that the first airmail of the United States was a personal letter from George Washington carried on an aerial balloon flight from Philadelphia by Jean Pierre Blanchard?
Stamp of the month
Epaulettes are the colloquial name of the first series of postage stamp issued by Belgium that depicted King Leopold I and his prominent epaulettes from which the name derives. They were first usable on 1 July 1849 when two values, with the same design were issued: a brown 10 centimes and a blue 20 centimes. Their introduction proved extremely successful and led to a large expansion of the postage system. The Epaulettes were soon superseded by new types.
The brown 10 centimes denomination, could be used to send a letter up to a distance of 30 kilometres (19 mi); the blue 20 centimes could be used on all other ordinary national mail. The stamps depicted Leopold I wearing military uniform, with highly visible epaulettes, and were printed using the Intaglio method. They were inscribed "POSTES" ("post") at the top, along with the stamp's value in numbers. At the bottom was the stamps face value in French language text. No Dutch language version was produced.
Purge server cache<|endoftext|>
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# How do you find the cartesian graph of r = 5sin(θ)?
Nov 27, 2015
You basically get a circle.
#### Explanation:
Consider the following diagram:
We can see that the relationships between rectangular and polar coordinates are:
$r = \sqrt{{x}^{2} + {y}^{2}}$
$\theta = \arctan \left(\frac{y}{x}\right)$
and:
$x = r \cos \left(\theta\right)$
$y = r \sin \left(\theta\right)$
Given our expression:
$r = 5 \sin \left(\theta\right)$
multiply by $r$ both sides:
${r}^{2} = 5 r \sin \left(\theta\right)$
so that you get, using our relationships of conversion:
$\textcolor{red}{{x}^{2} + {y}^{2} = 5 y}$ (1)
which is the equation of a circle centered at $\left({x}_{c} = 0 , {y}_{c} = \frac{5}{2}\right)$ and with radius $r = \frac{5}{2}$, whose equation is found from the general form of a circle:
${\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$
(try to substitute the values of the center and radius and you'll find (1)).
You can now plot it directly.
Next, to have some fun, I used Excel to evaluate, using our relationships of conversion, the coordinates $x$ and $y$ (in red) and plot them:<|endoftext|>
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## SOLVING QUADRATIC EQUATION BY GRAPHING
Solving quadratic equation by graphing :
Here we are going to see how to how to solve quadratic equation by graphing.
A quadratic function has standard form f(x) = ax2 + bx + c. In a is not equal to zero, the value of the related quadratic function is 0.
## How to Solve quadratic equation by graphing
• To solve quadratic equation by graphing first we need to find the vertex of the given quadratic equation.
• By using the table and giving some random values of x, we can get the values of y.
• Two points of the graph which intersects the x-axis at two distinct points is known as solution.
Note :
Quadratic equations always have two roots. However, these roots are not always two distinct numbers. Sometimes the two roots are the same number.
Let us see some example problems of solving quadratic equation by graphing
## Solving quadratic equation by graphing - Examples
Example 1 :
Solve x2 + 6x + 7 = 0 by graphing
Solution :
Step 1 :
Graph the related function f(x) = x2 + 6x + 7
Let the quadratic function as y = x2 + 6x - 7
x-coordinate of the vertex = -b/2a
here a = 1, b = 6 and c = -7
x = -6/2(1) ==> -6/2 ==> -3
Apply x = -3 in the given equation, we get
y = (-3)2 + 6(-3) - 7
y = 9 - 18 - 7
y = -16
Hence the vertex is (-3, -16)
Make a table of values to find other points to sketch the graph.
x-8-6-4-3-202 y9-7-15-16-15-79 Set of ordered pairs :(-8, 9)(-6, -7)(-4, -15)(-3, -16)(-2, -15)(0, -7)(2, 9)
Hence the solution of the quadratic equation are (-7, 1).
Checking :
x2 + 6x - 7 = 0
x2 + 7x - 1x - 7 = 0
x(x + 7) - 1(x + 7) = 0
(x - 1) (x + 7) = 0
x - 1 = 0 x + 7 = 0
x = 1 and x = -7
Let us see next example on"Solving quadratic equation by graphing".
Example 2 :
Solve x2 + x + 4 = 0 by graphing.
Solution :
Step 1 :
Graph the related function f(x) = x2 + x + 4
Let the quadratic function as y = x2 + x + 4
x-coordinate of the vertex = -b/2a
here a = 1, b = 1 and c = 4
x = -1/2(1) ==> -1/2
Apply x = -1/2 in the given equation, we get
y = (-1/2)2 + (-1/2) + 4
y = 1/4 - 1/2 + 4
y = (1 - 2 + 16)/4
y = 15/4
Hence the vertex is (-1/2, 15/4)
Make a table of values to find other points to sketch the graph.
x-1012 y6446 Set of ordered pairs :(-1, 6)(0, 4)(1, 4)(2, 6)
The graph has no x-intercept. Thus, there are no real number solutions for this equation.
Checking :
x2 + x + 4 = 0
We cannot factor the quadratic equation . Hence it has no real roots.
Let us see next example on"Solving quadratic equation by graphing".
Example 3 :
Solve x2 - 7x + 6 = 0 by graphing.
Solution :
Step 1 :
Graph the related function f(x) = x2 - 7x + 6
Let the quadratic function as y = x2 - 7x + 6
x-coordinate of the vertex = -b/2a
here a = 1, b = -7 and c = 6
x = -(-7)/2(1) ==> 7/2 ==> 3.5
Apply x = 7/2 in the given equation, we get
y = (7/2)2 - 7(7/2) + 6
y = 49/4 - 49/2 + 6
y = (49 - 98 + 24)/4
y = -25/4
Hence the vertex is (7/2, -25/4)
Make a table of values to find other points to sketch the graph.
x-2-1012 y-41460-4 Set of ordered pairs :(-2, -4)(-1, 14)(0, 6)(1, 0)(2, -4)
Hence the solution of the quadratic equation are (1, 6).
Checking :
x2 - 7x + 6 = 0
x2 - 6x - 1x + 6 = 0
x(x - 6) - 1(x - 6) = 0
(x - 1) (x - 6) = 0
x - 1 = 0 x - 6 = 0
x = 1 and x = 6
After having gone through the stuff given above, we hope that the students would have understood "Solving quadratic equation by graphing".
Apart from the stuff given above, if you want to know more about "Solving quadratic equation by graphing", please click here
Apart from the stuff "How to find nth term of arithmetic sequence" given in this section, if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Algebra word problems
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Word problems on direct variation and inverse variation
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Word problems on comparing rates
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Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
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Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Multiplying Binomials Mentally
## Multiply using FOIL and graphical representations
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Practice Multiplying Binomials Mentally
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Multiplying Binomials
Have you ever been to a community garden? Take a look at this dilemma.
The students loved walking through the community garden which was adjacent to the town hall. All of the flowers were in full bloom. One of the community workers was there and he presented the students with the following problem. The students suspected that this was the work of their teacher Mr. Travis, but they began to work on solving the problem anyway. Here it is.
A farmer has two rectangular fields. One measures 3x+7\begin{align*}3x + 7\end{align*} by 2x4\begin{align*}2x - 4\end{align*}. The other measures x2+1\begin{align*}x^2 + 1\end{align*} by 6x+5\begin{align*}6x + 5\end{align*}. Find the combined area of the two fields.
To figure this out, you will need to understand multiplying binomials. Because we find the area of a rectangle through multiplication and both of these fields have measurements written in binomials, understanding binomials will be important in solving this problem.
### Guidance
We defined binomials as two-term polynomials.
When we added and subtracted polynomials, we were careful to combine like terms. When we multiply polynomials we will carefully apply the rules of exponents, as well.
When we multiply binomials, we can use a table to help us to organize and keep track of the information. This table will help you to organize and keep track of your work. Let’s take a look.
Multiply the binomials (x+5)(x+3)\begin{align*}(x+5)(x+3)\end{align*}.
We can use a table like a rectangle, as if each of the binomials were a dimension of the rectangle.
We will insert the two binomials along the sides of the table like a rectangle.
Now, we will find the area of the four separate rectangles.
The dimensions of the first rectangle is xx\begin{align*}x \cdot x\end{align*}, the second is 5x\begin{align*}5 \cdot x\end{align*}, the third is 3x\begin{align*}3 \cdot x\end{align*}, and the fourth is 35\begin{align*}3 \cdot 5\end{align*}.
In order to find the total, we will add the four areas: x2+5x+3x+15\begin{align*}x^2 + 5x +3x +15\end{align*}
Now, combine like terms carefully: x2+8x+15\begin{align*}x^2 + 8x + 15\end{align*}.
Here is one that is a little different.
Multiply (5x8)2\begin{align*}(5x - 8)^2\end{align*}.
Remember that the exponent applies to the entire binomial such that (5x8)2=(5x8)(5x8)\begin{align*}(5x -8)^2 = (5x - 8)(5x - 8)\end{align*}.
5x\begin{align*}5x\end{align*} 8\begin{align*}-8\end{align*}
5x\begin{align*}5x\end{align*} 25x2\begin{align*}25x^2\end{align*} 40x\begin{align*}-40x\end{align*}
8\begin{align*}-8\end{align*} 40x\begin{align*}-40x\end{align*} 64\begin{align*}64\end{align*}
25x240x40x+64=25x280x+64
A second method for multiplying binomials is similar to the algorithm that we commonly use for multiplying two-digit numbers.
thousandshundredstensones73 70 +3 ×81 81 +173 70 +3 +4580 4000+ 500+ 80 +04653 4000+ 500+150 +3
When you expand the multiplication like is done on the right, you can see that in our multiplication algorithm for two-digit numbers, we line up numbers by similar places.
We will use this same idea to multiply binomials but instead of using decimal places, we will line up the products by like terms.
Take a look at this one.
Multiply \begin{align*}(3x + 2)(5x + 4)\end{align*}
Just as in multiplication with numerals, we first multiply 4 by 2 and get 8. We then multiply 4 by \begin{align*}3x\end{align*} and get \begin{align*}12x\end{align*}. When we multiply \begin{align*}5x\end{align*} by 2, we get \begin{align*}10x\end{align*}. This is not a like term of 8 but of \begin{align*}12x\end{align*} so the \begin{align*}10x\end{align*} was aligned beneath the \begin{align*}12x\end{align*}. Finally, the product of \begin{align*}5x\end{align*} and \begin{align*}3x\end{align*} is \begin{align*}15x^2\end{align*}. This is to the \begin{align*}2^{nd}\end{align*} power so how no like terms. Our sum on the bottom-right shows the final product.
A third way is to use “FOIL”. “FOIL” is an acronym which tells us which terms to multiply in order to get our product—
F—First terms in the binomials
O—Outside terms in the binomials
I—Inside terms in the binomials
L—Last terms in the binomials
Let’s multiply \begin{align*}(2x+8)(5x-13)\end{align*} using the FOIL method.
F First terms are \begin{align*}2x\end{align*} and \begin{align*}5x\end{align*} \begin{align*}(\mathbf{\underline{2x}}+8)(\mathbf{\underline{5x}}-13)\end{align*} \begin{align*}10x^2\end{align*}
O Outside terms are \begin{align*}2x\end{align*} and -13 \begin{align*}(\mathbf{\underline{2x}}+8)(5x\underline{\mathbf{-13}})\end{align*} \begin{align*}-26x\end{align*}
I Inside terms are 8 and \begin{align*}5x\end{align*} \begin{align*}(2x\underline{\mathbf{+8}})(\mathbf{\underline{5x}}-13)\end{align*} \begin{align*}40x\end{align*}
L Last terms are 8 and -13 \begin{align*}(2x\underline{\mathbf{+8}})(5x\underline{\mathbf{-13}})\end{align*} -104
The table above helps to illustrate which terms must be multiplied. However, we don’t need to make a table like that for each multiplication. We can show it like this:
Of the three methods we’ve seen for multiplication, you might agree that this is the quickest method. Of course, all three methods should give us the same product.
Write the acronym FOIL and what each letter stands for in your notebook.
Take a look at one more.
Multiply \begin{align*}(5x^3+2x)(7x^2+8)\end{align*}
Multiply the following binomials.
#### Example A
\begin{align*}(x+2)(x+4)\end{align*}
Solution: \begin{align*}x^2+6x+8\end{align*}
#### Example B
\begin{align*}(x-6)(x+5)\end{align*}
Solution: \begin{align*}x^2-x-30\end{align*}
#### Example C
\begin{align*}(x+3)(x-3)\end{align*}
Solution: \begin{align*}x^2-9\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
Remember, we need to work on figuring out the area of both rectangles and then the sum of those two areas will give us the total area. Here is how we can solve this problem.
### Vocabulary
Binomials
polynomials with two terms in them.
Perfect Square Trinomial
a trinomial where the first term and the last term are perfect squares because the binomial was to the second power or squared.
FOIL
firsts, outers, inners, lasts – binomial multiplication
### Guided Practice
Here is one for you to try on your own.
Multiply by using a table.
Multiply \begin{align*}(x-4)(x+6)\end{align*}
Solution
\begin{align*}x\end{align*} \begin{align*}-4\end{align*}
\begin{align*}x\end{align*} \begin{align*}x^2\end{align*} \begin{align*}-4x\end{align*}
\begin{align*}+6\end{align*} \begin{align*}6x\end{align*} \begin{align*}-24\end{align*}
Notice the careful work with the negative and positive signs.
### Practice
Directions: Use a table to multiply the following binomials.
1. \begin{align*}(x+3)(x+5)\end{align*}
2. \begin{align*}(x-3)(x-5)\end{align*}
3. \begin{align*}(x+3)(x-3)\end{align*}
4. \begin{align*}(x+2)(x-8)\end{align*}
5. \begin{align*}(3x^2+3x)(6x-2)\end{align*}
6. \begin{align*}(2x-7y)(5x+4y)\end{align*}
7. \begin{align*}(2x-9)^2\end{align*}
Directions: Multiply the following binomials vertically.
1. \begin{align*}(d+2)(4d-1)\end{align*}
2. \begin{align*}(5x+7)(5x-7)\end{align*}
3. \begin{align*}(4b^2+3c)(2b-5c^2)\end{align*}
Directions: Multiply the following binomials using the FOIL method:
1. \begin{align*}(p+6)(5p+2)\end{align*}
2. \begin{align*}(-7y^2-4y)(6y+2)\end{align*}
3. \begin{align*}(x^3+3x)^2\end{align*}
4. \begin{align*}(2x+1)(x-4)\end{align*}
5. \begin{align*}(3x-3)(5x+9\end{align*})
6. \begin{align*}(x+5)^2\end{align*}
### Vocabulary Language: English
Binomial
Binomial
A binomial is an expression with two terms. The prefix 'bi' means 'two'.
FOIL
FOIL
FOIL is an acronym used to remember a technique for multiplying two binomials. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms.
Perfect Square Trinomial
Perfect Square Trinomial
A perfect square trinomial is a quadratic expression of the form $a^2+2ab+b^2$ (which can be rewritten as $(a+b)^2$) or $a^2-2ab+b^2$ (which can be rewritten as $(a-b)^2$).<|endoftext|>
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Hearing is an important function in human communication. Although there are volumes written on the anatomical and physiological processes of the auditory system, I’d like to give you just a brief description of how people hear.
Sound enters the auditory system through the external ear. The Pinna is made of cartilage and it collects, funnels and even amplifies sound waves into the ear canal. The pinna is more than just a cute place to hang earrings. Its shape aids in localization of sound and boosts high-frequency energy.
Sound waves then travel down the ear canal and hit the eardrum. The ear drum is a tight flap of skin (like the face of a drum) at the end of the ear canal. As the eardrum vibrates back and forth, it transduces the acoustical signal into a mechanical one. The vibrations of the ear drum are then sent through three little bones in the middle ear known as the Maleus, Incus and Stapes. The last bone (Stapes) is like a little piston fitted into a window. The movement of the stapes back and forth displaces a fluid on the other side of the window in the inner ear.
Within the inner ear, the fluid movement causes the deflection of hair cells. You can think of these hair cells as nerve-endings. As the cilica on top of the hair cells deflect, an electrical charge is created inside the cell, transmitting an electrical signal up the auditory nerve to the brainstem, and on to the brain. And, Viola, we perceive the sound.<|endoftext|>
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# If 15 cans hold 135 litres of milk , how many such cans will be needed to store 180 litres of milk
If 15 cans hold 135 litres of milk , how many such cans will be needed to store 180 litres of milk
### 2 thoughts on “If 15 cans hold 135 litres of milk , how many such cans will be needed to store 180 litres of milk”
20 cans
Step-by-step explanation:
since, 135 litres hold by 15 cans
therefore 1 litre hold by 15/135
therefore 180 litre hold by 1/9*180
therefore 20 cans needed
2. ⇒ Given:
15 cans can hold 135 litres of milk.
⇒ To Find:
The no. of cans that are needed to store 180 litres of milk.
⇒ Solution:
In order to solve this question, we will be using the concept of unitary method where all the measures are found out in their singular form and then multiplied to the required measures as per the question.
Here, it is given that 15 cans can hold 135 litres of milk. To move forward, we will have to to find the amount of milk a single can can hold.$$\sf{\longrightarrow\:Capacity\:of\:1\:can=\dfrac{Total\:milk\:held\:by\:total\:cans}{Total\:cans}}$$
$$\sf{\longrightarrow\:Capacity\:of\:one\:can=\dfrac{135}{15}}$$
$$\sf{\longrightarrow\:Capacity\:of\:one\:can=\dfrac{27}{3}}$$
$$\bf{\longrightarrow\:Capacity\:of\:one\:can=9\:litres}$$
Now, we know that a single can has a capacity of 9 litres. Here, it is asked to find the no. of cans which can be found as follows:
$$\sf{\longrightarrow\:No.\:of\:cans=\dfrac{Total\:litres}{Capacity\:of\:1\:can}}$$
$$\sf{\longrightarrow\:No.\:of\:cans=\dfrac{180}{9}}$$
$$\bf{\longrightarrow\:No.\:of\:cans=20}$$
Therefore 20 cans are required to hold 180 litres of milk.
### Unitary Method:
By applying the unitary method, we find the measure of one single unit and apply that measure to find the total value of the units as per the question provided. Unitary method is used in a wide variety of mathematical branches including ratios. There are two types of results of unitary method:
✳ Direct variation
As the unit becomes singular (decreased), the value also decreases.
✳ Indirect variation
When either of the two terms is in an increase, the other term is on a decrease.<|endoftext|>
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Home / The science / Mathematics / How to calculate the volume of the barrel
# How to calculate the volume of a barrel
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948 Views
</a>
Many building materials, fuel, foodstuffs come to wholesale bases in containers of cylindrical or close to them forms.
It can be drums, cisterns or cans.
In all such cases, to determine the amount of a substance, you must first calculate the volume.
The shape of such containers may differ slightly from the classical cylinder, and therefore the calculation methods will differ.
You will need
• - tools for measuring length-
• - calculator.
Instructions
1
The barrel can have a regular cylindrical shape. In this case, its base is a circle. If possible, find the center of the base, draw a diameter through it and measure it. Divide it by 2. This will give you a radius. This can be done with a flank lying on its side or inverted. When this is not possible, measure the circumference of the base and find the radius by the formula R = P / 2 ?, where R is the radius of the base and P is the circumference.
2
Measure the height of the barrel. This is most easily done with a tape measure. For accuracy, draw a line on the side that will be strictly perpendicular to the base. Calculate the volume of the barrel by the same formula as the volume of any other cylinder. V =? * R2 * H, where V - volume, R - radius of the base of the drum, H - height.
3
The bottom of the barrel does not necessarily represent itselfThe circumference, and the lateral surface may be curved. In this case, you need to make a few more measurements. If its sides are curved along a parabola or along an arc of a circle, calculate or measure the diameter of the middle section. Mark the diameter of the base as D ', and the area of the middle section as D ".
4
The volume of the barrel, the lateral surface of which is curvedOn the arc of a circle, calculate by multiplying the number? To 1/12 of the height and the sum of the doubled square of the diameter of the average section and the square of the base. This can be expressed by the formula V =? * H / 12 * (2D2 + D'2), where H is the height, D 'and D "are the diameters of the base and the mean cross-section, respectively, and V is the volume.
5
By a similar principle, the barrel volume is calculated,The lateral surface of which is curved along a parabola. Is it equal to the product of a number? 1/15 of the height and the sum of the doubled square of the diameter of the base, the product of both diameters and 0, 75 of the square of the diameter of the base). This action can be designated by the formula V =? * H / 15 * (2D? 2 + D '* D "+ 0,75D'2).
How to calculate the volume of a barrel Was last modified: June 21st, 2017 By
It is main inner container footer text<|endoftext|>
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One of the most useful measures in statistics is that of variance. It is a measure that is used to quantify the amount of variation or dispersion of a set of data values. I’ll try to explain what this means with an example, but there are plenty of other online sources if you want to explore the math behind it.
Let’s assume the following population of numbers. These numbers are the exam grades for a student.
The average of this grades is 6.5. And the variance is 2.3. Variance and the mean average are not expressed in the same unit of measurement. To have a comparable measure in the same unit we have to calculate the square root of the variance. This provides us with the standard deviation, which we discuss in another of our lessons.
When we have only a sample of the data from a population, then the variance we calculate is called sample variance and when the data from the whole population is available then it is called population variance.
Excel has 6 functions to calculate variance. 3 of them calculate the sample variance and 3 the population variance.
Although VAR and VARP functions can still be used in office 2016 for compatibility, they have been replaced by VAR.S and VAR.P respectively, and should be avoided, since they might not be supported in future versions of Excel.
VAR.S calculates the sample variance and VAR.P the population variance.
VARA and VARPA also calculate the sample and population variance respectively. Their difference to the previous functions is to the way they treat logical values and text when they are part of our population.
VAR.S and VAR.P ignore text and logical values, while VARA and VARPA count the text as zero and the logical values as 1 and zero for TRUE and False respectively.
Let’s see an example.
This example has mostly the same data with some minor differences. We have already calculated the mean average as 6.4.
Since this table consists of all of the grades for a certain student we assume this table is our whole population.
So in cell B10 we will calculate the Variance using the VAR.P and in cell B11 its alternative VARPA.
In B10 we type the name of the function and select the data set. The values of SICK and FALSE are ignored during the calculation.
In B11 as before we type the function and the cell range. The value FALSE is calculated now as 0 and the same goes with value SICK. This is the reason for the different result. Otherwise both the functions should produce the same result.
Let’s assume, for the sake of this example, that we only had the exam results for the months from August to December. That means we would have only a sample of the whole population.
So in cells B12 and B13 we will calculate the sample variance using the data for these months.
In B12 we type the name of the function same as before and the cell range, I3 to M6.
We do the same in B13 but using the VARA function instead.
Notice the difference between the 2 sets of functions. The ones that calculate the sample and the ones that calculate the population variance.
If this was a real life statistical problem, we should choose to use the functions that consider the absences of the student as zero-graded instead of ignoring them. So we should go with VARA and VARPA functions<|endoftext|>
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How do you calculate the sum of interior angles?
To find the sum of interior angles of a polygon, multiply the number of triangles in the polygon by 180°. The formula for calculating the sum of interior angles is ( n − 2 ) × 180 ∘ where is the number of sides. All the interior angles in a regular polygon are equal.
How do you find the sum of the interior angles of a polygon?
(n-2)x 180 degrees : The formula for finding the sum of all angles in a polygon (REGULAR).
360°
What is the sum of the interior angles in a shape with 27 sides?
The sum of any icositetragon’s interior angles is 3960 degrees.
1260°
180°
900°
What is the sum of the interior angles of a 15 sided polygon?
Each triangle has an angle sum of 180 degrees , so the sum of the interior angles of the 15-gon must be 13 × 180 = 2340 degrees . Since the 15-gon is regular, this total is shared equally among the 15 interior angles. Each interior angle must have a measure of 2340 ÷ 15 = 156 degrees . an octagon we can draw 5 diagonals.
1440°
720°
Does the length of a side of a regular polygon affect the sum of the interior angles?
Does the length of a side of a regular polygon affect the sum of the interior angle measures ? Therefore, the angle sum is dependent only on the number of sides n, not the length of the sides . For example, for a pentagon the angle sum is 3(180) = 540 degrees.
You might be interested: How long does it take to paint a house interior
1800°
What is the sum of the interior angles in a 10 sided shape?
In geometry, a decagon (from the Greek δέκα déka and γωνία gonía, “ten angles “) is a ten- sided polygon or 10 -gon. The total sum of the interior angles of a simple decagon is 1440°. A self-intersecting regular decagon is known as a decagram.
What is the sum of interior angles of a 19 Gon?
Regular Polygons
SidesNameInterior Angles
18Octdecagon160.00°<|endoftext|>
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# Into Math Grade 4 Module 5 Lesson 7 Answer Key Use Equations to Solve Multistep Problems
We included HMH Into Math Grade 4 Answer Key PDF Module 5 Lesson 7 Use Equations to Solve Multistep Problems to make students experts in learning maths.
## HMH Into Math Grade 4 Module 5 Lesson 7 Answer Key Use Equations to Solve Multistep Problems
I Can Use equations to solve multistep problems.
The apple trees grew so many apples that Lia filled 7 baskets with 13 apples each to take to the school picnic. Once there, Lia gave apples to everyone who passed by. At the end of the day, she had 46 apples left. How many apples did she pass out?
Explanation:
Lia passed out 45 apples.
Turn and Talk How were your steps for solving the problem alike or different?
Build Understanding
1. Malik collects stamps in an album that has 48 pages. He puts 8 stamps on each of the first 6 pages. He puts 15 stamps on each of the next 9 pages. How many stamps are in Malik’s album?
A. How can you show what is happening in the problem?
______________________________
______________________________
We can show what is happening in the problem by solving it step by step.
B. Is there any extra information in the problem that you do not need?
______________________________
The total number of pages is not needed to solve the problem.
C. What will the answer tell us?
______________________________
The answer tells us the number of stamps in Malik’s album.
D. What do you need to do first to solve the problem?
______________________________
______________________________
First i need to multiply number of pages with number of stamps put on them.
E. About how many stamps do you think there are? Why?
______________________________
______________________________
I think there are about 200 stamps.
F. Show how you can solve the problem.
Multiply 8 stamps with 6 pages = 8 x 6 = 48 stamps
Multiply 15 stamps with 9 pages = 15 x 9 = 135 stamps
135 + 48 = 183
G. There are ___ stamps in Malik’s album.
There are 183 stamps in Malik’s album.
Step It Out
2. On Monday and Tuesday, Anne bikes to the beach and back home. On Thursday and Friday, Anne bikes to the park and back home. How many miles does Anne ride her bike?
A. Underline the information you need to solve the problem.
B. Write an equation to model the number of miles Anne bikes to and from each location.
C. How can you write an equation to find the number of miles Anne bikes? Use the order of operations to solve the problem. Let t = the total number of miles Anne bikes.
D. Anne rides her bike ___.
The number of miles Anne bikes on Monday and Tuesday to and from home to beach is
Beach = 2 x 2 x 12 = 48 miles
The number of miles Anne bikes on Thursday and Friday to and from home to park is
Park = 2 x 2 x 8 = 32 miles
Total number of miles Anne bikes = t = 48 + 32 = 80.
Anne rides her bike 32 miles.
Connect to Vocabulary
The order of operations is a set of rules that gives the order in which calculations are done From left to right first multiply and divide, and then add and subtract.
Check Understanding Math Board
Find the product. Show your work.
Question 1.
Corey goes to a store with $25 to spend. He buys 3 packages of baseball cards. Each package costs$6. How much money does Corey have now? ___
Let c = the money Corey has left.
$7 is the money Corey has left Explanation: Corey goes to a store with$25 to spend.
He buys 3 packages of baseball cards.
Each package costs $6 3 packages costs 3 x$6 = $18 Let c = the money Corey has left c =$25 – $18 =$7.
Question 2.
Model with Mathematics Evan has 4 pieces of rope that are the same length. He needs 75 feet of rope to build a bridge. How much more rope does Evan need?
• What do you need to do first to solve the problem?
__________________________________
• Write an equation to find the answer to the first step. Then solve. Let f = the total length of rope Evan already has.
__________________________________
• About how much more rope does Evan need?
__________________________________
• Write an equation to find exactly how much more rope Evan needs. Then solve. Let n = the length of rope Evan needs.
__________________________________
Evan needs ____ more feet of rope.
__________________________________
I need to find the length of the rope Evan need.
The equation that tells the length of the rope Evan already has is f = 13 x 4
Evan need 75 feet of rope
The equation that tells the length of the rope Evan need is n = 75 – 13 x 4
75 – 52 = 23
Evan needs 23 more feet of rope.
Use the order of operations and find the value of n.
Question 3.
40 × 8 – 7 × 3 = n
___ = n
299
Explanation:
(40 x 8) – (7 x 3)
= 320 – 21
299 = n.
Question 4.
32 + 2 × 35 – 8 = n
___ = n
n = 94
Explanation:
32 + (2 x 35) – 8
= (32 + 70) – 8
= 102 – 8
= 92 = n.
I’m in a Learning Mindset!
What do I do when I do not know what to do?
______________________________
Scroll to Top
Scroll to Top<|endoftext|>
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The Industrial Revolution that took place in Great Britain radically changed the structure of its society and launched the British Empire into near global hegemony. The well-being of a large percentage of the nation’s children were casualties of this revolution as they increasingly found themselves working in crowded and unhealthy factories, mines, textile mills, farms, and as chimney sweeps and matchmakers. During the time period, there were many that despised the use of child labor. But even reformers, while calling for some regulation, did not advocate for its abolition as they didn’t believe the industry nor the children’s families could survive without it. Child labor had more or less always been present in England as children had worked on family farms throughout history. This new outrage at the use of child labor resulted in part because it was now “more readily observable in the factory than in the obscurity of the cottage” (Hewitt, 406).
Before the Industrial Revolution, children worked hard and long hours, but usually with their family members in domestic industries and agriculture. Some scholars argue that there may not have even been an increase in child employment during the Industrial Revolution. However, for many of the children, their work became “more regular and more intensive” (Cunningham, 117). As the Industrial Revolution progressed, the percentage of a family’s income provided by the father could have been as low as 25%. This follows the steady drop in real wages that took place during the time period. Now, whether this drop in real wages was due to an increase in cheap child labor or child labor came about as a result of the effect a drop in real wages had on the family unit, the two are nevertheless correlated.
Additionally, the Industrial Revolution originally centered around water powered mills from the cotton industry. These were often far from population centers and lacked enough traditional labor sources. Therefore, children were brought on as labors, similar to apprentices, and housed on site.
Written by Taylor Egbert
Cunningham, Hugh. “The employment and unemployment of children in England c. 1680-1851.” Past & Present 126 (1990): 115-150.
Hewitt, Margaret, and Ivy Pinchbeck. “Children in English Society, 2 vols.” (1973): 546-581.
Horrell, Sara, and Jane Humphries. “”The Exploitation of Little Children”: Child Labor and the Family Economy in the Industrial Revolution.” Explorations in Economic History 32.4 (1995): 485-516.
Nardinelli, Clark. “Child labor and the factory acts.” The Journal of Economic History 40.04 (1980): 739-755.<|endoftext|>
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# Difference between revisions of "1997 AIME Problems/Problem 8"
## Problem
How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
## Solution
### Solution 1
For more detailed explanations, see related problem (AIME I 2007, 10).
The problem is asking us for all configurations of $4\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:
• The first two columns share no two numbers in the same row. There are ${4\choose2} = 6$ ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are ${4\choose 2}$ ways. This gives $6^2 = 36$.
• The first two columns share one number in the same row. There are ${4\choose 1} = 4$ ways to pick the position of the shared 1, then ${3\choose 2} = 3$ ways to pick the locations for the next two 1s, and then $2$ ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in $2$ ways. This gives $4 \cdot 3 \cdot 2 \cdot 2 = 48$.
• The first two columns share two numbers in the same row. There are ${4\choose 2} = 6$ ways to pick the position of the shared 1s. Everything is then fixed.
Adding these cases up, we get $36 + 48 + 6 = \boxed{090}$.
### Solution 2
Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ = $\boxed{090}$.
-pi_is_3.141
## See also
1997 AIME (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Over the centuries, trade centers grew to prominence, flourished and then declined and were replaced by a new center. By the 9th Century, the Mayan trade center in Tikal was starting to decline. The Ancestral Puebloan trade center at Chaco Canyon and the Mississippian trade center at Cahokia were just becoming power centers and the Trade center for the Toltecs in Mexico was in its prime.
|Panoramic view of the ruins of Teotihuacan|
The city of Tula rose to prominence after the fall of Teotihuacan around 900 and reached its height
|Tula, ancient Toltec capital|
At that point, the Toltec city of Tula picked up the slack and became the dominant center in Mexico. Unlike Teotihuacan, it was well fortified and supported a powerful military. So, in the 9th century, it would have been the regional trade center and in my hypothetical trade route, it would have been one of the main stops.
It never grew to the size of Teotihuaca. At its height, the population was probably only 60,000 with another 20,000 in the surrounding area. The factors lending to Tula becoming the dominant center in its region were its fertile farmland, obsidian mines, and location (along the trade route).
The population would have been made up of the ruling and elite class, craftsmen, merchants and a large number of farmers. About half the population was involved in mining and crafting of obsidian and the working of travertine and ceramics. They became so well-known for their craftsmanship that later the Aztec words for craftsmanship were synonymous with the Toltec and Tula.
But, agriculture also played a prominent roll in the economics of Tula. They traded chili peppers, amaranth, squash and maguey along with corn and beans. They also harvested a number of wild plants like mesquite beans and cactus fruit. They even domesticated dogs and turkeys. The skilled farmers used irrigation to produce bounteous crops of corn.
Tula didn’t last as long as many of the Mexican empires and likely suffered the same fate as Teotihuacan. The leeched soil and drying climate led to the decline of their farming culture. Like Teotihuacan, the ceremonial center was burned and looted around 1179 A.D. At that point, the trade center probably moved to Tenochtitlan.
I think it is clear that commerce and trade played a major role in the prominence and decline of the major empires of Native America just as it does in the world today and did in the Old World. The centers of power in the world seem to be constantly changing, but the principles of prosperity seem to be unchanged throughout history.<|endoftext|>
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St. Ann’s, Nova Scotia
1819 – 1851
Associated name: the Normanite settlement
Location: on the northern coast of Cape Breton Island, approximately 30 kilometres west of the City of Sydney. 46.2 N, -60.6 W.
St. Ann’s was founded by Norman McLeod in order to create a community committed to a rigorous, somewhat “fire and brimstone” approach to religious practice. Besides being its religious leader, McLeod also became the community’s civilian leader, magistrate, and teacher. Born in 1780 in the Highlands of northern Scotland, Norman McLeod was educated at the University of Aberdeen and later studied for the ministry at the University of Edinburgh. However, he soon left, having become highly critical of the Church of Scotland for what he saw as a lack of discipline among the clergy and a softening of church doctrine. After teaching and preaching unofficially for a short period, McLeod decided to emigrate to North America in order to have greater religious freedom.
In 1817, he traveled to Pictou, Nova Scotia aboard the Frances Ann. In Pictou, his energetic sermons conducted in the old Highland style soon attracted a number of followers who were dubbed the “Normanites” by locals. By 1819, finding Pictou already quite developed and too similar to Scotland for his liking, McLeod decided to seek out new land where he could establish a more isolated community for his followers. McLeod had been in correspondence with Presbyterians in Ohio and it was there that he hoped to create his new community. In 1819, McLeod and a small group departed Pictou for Ohio on an exploratory journey aboard the schooner, The Ark. Their voyage was to have taken them via the Gulf of Mexico and the Mississippi River. However, during a severe storm shortly after departure, the group was forced to take shelter in St. Ann’s Harbour on Cape Breton Island.
In the morning the group discovered plentiful fish stocks and uninhabited land and quickly decided to proceed no further. After some preliminary work on the land, the group returned to Pictou for their families and possessions. McLeod petitioned the government for a 300- to 400-acre land grant for the new community and by 1820, some 700 followers had arrived in St. Ann’s.
The community of St. Ann’s was created on the slopes along the harbour. Settlers erected their own log cabins. A homestead for McLeod was also constructed by the settlers on a plot of land along the peninsula overlooking Black Cove. Within a few years, his original log cabin was replaced with a three-storey house. A school house was constructed east of the McLeod homestead and in 1822, the first Calvinist church on Cape Breton Island was built from logs. A larger church with seating for 1,000 was constructed years later.
For 30 years, as magistrate, minister and teacher, McLeod influenced nearly every aspect of community life. He was perhaps best known for his religious zeal and fiery sermons which drew followers from St. Ann’s and beyond. It was McLeod’s belief that he had been chosen by God and his first concern was to ensure his followers’ moral and spiritual welfare. In exchange for his ministering, women were expected to provide domestic labour, while men were expected to help clear land and construct structures on his property.
Conditions at St. Ann’s were primitive. Winters were long and harsh, limiting the agricultural potential and predictability of the community. For example, in 1848, the community was affected by a particularly difficult winter resulting in famine because the potato crop failed.
At about this time, McLeod received a letter from his eldest son, Donald, who had been living in Australia for some years. He wrote about the good soil and climate in Australia which seeded in Norman McLeod the idea of resettling in more hospitable conditions. In 1851, McLeod sailed with some of his followers first to Australia, and then to New Zealand. In all, about 800 settlers left St. Ann’s for New Zealand between 1851 and 1859, settling mainly in the Waipu River area north of Auckland.
Elements of the physical settlement and life within it are found in Choyce (2007); Lotz (1974); Lamb (1975); Robinson (1974). Norman McLeod, himself, is also described in those books, and a helpful summary biography of him by R. MacLean is found in the online Dictionary of Canadian Biography.
by Jennifer L. Wardle
- Choyce, L. 2007. Nova Scotia: Shaped by the sea: A living history. East Lawrencetown, NS: Pottersfield Press.
- Lotz, Jim and Pat. 1974. Cape Breton Island. Vancouver, BC: Douglas, David and Charles.
- Lamb, James B. 1975. The hidden heritage: Buried romance at St. Ann’s, N.S.. Windsor, NS: Lancelot Press.
- MacLean, R. 2003. “McLEOD, NORMAN,” in Dictionary of Canadian Biography, vol. 9. University of Toronto/Université Laval. Accessed April 7, 2016. http://www.biographi.ca/en/bio/mcleod_norman_9E.html
- Robinson, Neil. (1974). Lion of Scotland, being an account of Norman McLeod’s forty years’ search for a land where he and his followers could live as they wished. Auckland, NZ: Hoddler & Stoughton.<|endoftext|>
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B cells (aka B lymphocytes) are one of many types of cells in the immune system. There are many millions of them spread throughout the body. They are the only cells in the body that produce factors called antibodies (aka immunoglobulins) that do many important things. People with too few B cells or the antibodies they produce can have trouble defending against common microbes (viruses, bacteria, etc) and may get very serious and life-threatening infections.
Antibodies can bind to proteins and other biomolecules that are floating around in the body or that are attached to other cells. When each B cell develops, it makes a series of changes to its genetic code that end up producing an antibody with a random binding specificity. This makes every new B cell likely to recognize something different than every other B cell in the body.
That's important because it leads to the total population of B cells being able to recognize almost any kind of protein or biomolecule in the world, including those that are on dangerous microbes that the person has never been exposed to before. When a B cell finds something that it recognizes through its surface antibody (aka B cell receptor) it can start to become activated.
If B cells get activated, they can divide to form daughter cells with the same specificity leading to increased number of them. They can also make other genetic changes that allow their antibodies to leave the cell and spread throughout the rest of the body. This is the basis of how most vaccines work. By activating B cells that are specific toward the proteins on the particular microbe or factor (flu virus, measles virus, tetanus toxin, etc) that is being targeted by the vaccine, the immune system gets more prepared to recognize and eliminate those microbes and factors.<|endoftext|>
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In Our Universe (Belknap), cosmologist Jo Dunkley clearly explains many of the big things we know about the universe, and how scientists came to discover them, from black holes, to distant galaxies, expanding space, and more. Complemented by simple, effective illustrations of complex astrophysical concepts and techniques, Our Universe is an engaging introduction to the nature of our cosmic home.The author: Jo Dunkley is professor of physics and astrophysical sciences at Princeton. She has won awards from the Royal Astronomical Society, the Institute of Physics, and the Royal Society for her work on the origins and evolution of the Universe.
Opening Lines: On a clear night the sky above us is strikingly beautiful, filled with stars and lit by the bright and changing Moon. The darker our vantage point, the more stars come into view, numbering from the tens or hundreds into the many thousands. We can pick out the familiar patterns of the constellations and watch them slowly move through the sky as the Earth spins around. The brightest lights we can see in the night sky are planets, changing their positions night by night against the backdrop of the stars. Most of the lights look white, but with our naked eyes we can notice the reddish tint of Mars, and the red glow of stars like Betelgeuse in the Orion constellation. On the clearest nights we can see the swathe of light of the Milky Way and, from the southern hemisphere, two shimmery smudges of the Magellanic Clouds.
Beyond its aesthetic appeal, the night sky has long been a source of wonder and mystery for humans around the world, inspiring questions about what and where the planets and stars are, and how we on Earth fit into the larger picture revealed by the sky above us. Finding out the answers to those questions is the science of astronomy, one of the very oldest sciences, which has been at the heart of philosophical inquiry since ancient Greece. Meaning ‘law of the stars’, astronomy is the study of everything that lies outsides our Earth’s atmosphere, and the quest to understand why those things behave the way they do.
Humans have been practicing astronomy in some form for millennia, tracking patterns and changes in the night sky and attempting to make some sense of them. For most of human history astronomy has been limited to those objects visible to the naked eye: the Moon, the brightest planets of our Solar System, the nearby stars, and some transient objects like comets. In just the last 400 years humans have been able to use telescopes to look deeper into space, opening up our horizons to studying moons around other planets, stars far dimmer than the naked eye can see, and clouds of gas where stars are born. In the last century our horizon has moved outside our Milky Way galaxy, allowing the discovery and study of a multitude of galaxies, and find entirely new planets around other stars. In doing so, modern astronomy continues to seek solutions to the age-old questions of how we came to be here on Earth, how we fit into our larger home, what will be the fate of Earth far in the future, and whether there are other planets that could be home to other forms of life.
Reviews: “This luminous guide to the cosmos encapsulates myriad discoveries. Astrophysicist Jo Dunkley swoops from Earth to the observable limits, then explores stellar life cycles, dark matter, cosmic evolution and the soup-to-nuts history of the Universe. No less a thrill are her accounts of tenth-century Persian astronomer Abd al-Rahman al-Sufi, twentieth- and twenty-first-century researchers Subrahmanyan Chandrasekhar, Jocelyn Bell Burnell and Vera Rubin, and many more.”—Nature<|endoftext|>
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# 3rd Class Mathematics Multiplication Problem Based on Multiplications
## Problem Based on Multiplications
Category : 3rd Class
### Commutative or Order Property of Multiplication
The numbers can be multiplied in either order or on changing the order of the numbers then the resulting product of the numbers does not change. The product of 56 and 67, 67 x 56 = 3752. The product of the numbers 56 and 67 on changing their order = 56 x 67 = 3752 is same as the product in the previous order. Therefore, multiplication is commutative.
The product of two numbers M and N is same as the product of the numbers A and B. Which one of the following options is correct about the above statement?
(a) M x N = A x B
(b) M x A = N x B
(c) Both A and B
(d) M x B = N x A
(e) None of these
Explanation
Explications Product of M and N is equal to the product of A and B. Hence,
M x M = A x B.
Associative or Grouping Property of Multiplication
The product of three numbers remains same even if the group of the numbers is changed. The product of the numbers 21, 4 and 25 when written in the order of (21 x 4) x 25 = 84 x 25 = 2100 and the product of the same numbers 21, 4 and 25 is same when written in the order of 21 x (4 x 25) = 21 x 100 = 2100. If three numbers are denoted by a, b and c then their product (a x b) x c = a x (b x c).
The product of group of the numbers A and B with a number C is same as the another group of A and C with B. Which one of the following options is correct about the above statement?
(a)
(b)
(c)
(d) All of these
(e) None of these
Explanation According to the associative property of multiplication (A x B) x C = (A x C) x B is a true statement.
Multiplicative Property of Zero (Or Null property)
The product of a number and zero is always zero. Therefore, 34 x 0 = 0, 0 x 45 = 0.
For a number N, the product of N by zero is zero or N x 0 = 0, 0 x N = 0.
What should be multiplied with the number 58765 to get zero?
(a) 2
(b) 1
(c) 0
(d) All of these
(e) None of these
Explanations: The given number should be multiplied with zero to get zero as their product.
Multiplicative Property of 1 or Multiplicative Identity Property
The product of a number with 1 is the number itself. Therefore, 34 x 1 = 31. For a number N, the product of N and 1 is the number itself or N x l = N, 1 x N = N.
What should be multiplied with the number 10876 to get the number itself?
(a) 1
(b) 2
(c) 0
(d) All of these
(e) None of these
Explication To get the multiplicand itself the multiplicand should be multiplied with 1.
Multiplicative Series
When a group of numbers is arranged in such a way that there are some common relation between the numbers are obtained, the numbers in the group are called in series. In the group of numbers: 2, 4, 6, 8, 10, 12, the group of the numbers are starting from 2 and every next number is 2 more than the previous number. Therefore, the next number after 12 is 12 + 2 = 14.
Find odd one out from the numbers: 16, 48, 8, 26, 64
(a) 16
(b) 48
(c) 26
(d) 64
(e) None of these
Explanation 26 is odd because except 26 all the numbers are the multiples of 8.
Multiplication of Decimal Numbers
The following steps are used for the multiplication of the decimal numbers by decimal numbers (345.67 and 234.54):
Step 1: Convert the decimal numbers into whole numbers, 5.67 into 567 and 4.54 into 454.
Step 2: Multiply the converted numbers by simple multiplication method, 567 x 454 = 257418
Step 3: Put the decimal point in the resulting product by counting the number of digits form left which is equal to the sum of number of decimal digits of multiplicand and multiplier = 2 + 2 = 4. Therefore, the decimal point will be placed in the product of the both decimal numbers by counting 4 from ones or right = 25.7418
What is the product of the decimal numbers 638.23 and 0.76?
(a) 485.0548
(b) 485.4805
(c) 4.8548
(d) All of these
(e) None of these
8 calendars were distributed in 10 factories. How many calendars were distributed?
(a) 80
(b) 81
(c) 82
(d) 83
(e) None of these
Explanation The total number of distributed calendars to 10 factory = 10 x 8 = 80.
A bucket contains 5 litres of water. How much water do 1324 such buckets can contain?
(a) 5545 litres
(b) 6620 litres
(c) 6045 litres
(d) All of these
(e) None of these
Explanation Quantity of water in 1324 such buckets = 1324 x 5 = 6620 litres of water.
In a function, there are 30 chairs in each row. If the number of rows is 40 then how many number of chairs are there in all?
(a) 1250
(b) 1300
(c) 1200
(d) 1400
(e) None of these
Explanations Total number of chairs = Number of rows x Number chairs in one row = 40 x 30 = 1200.
A bike travels a distance of 65 km in 1 hour. How many kilometers will it cover in 10 days if bike runs 4 hours a day?
(a) 2600km
(b) 2649km
(c) 2660km
(d) All of these
(e) None of these
Explanation The distance travelled by the bike in a day = 65 x 4 = 260 km.
Hence, distance travelled by bike in 10 days = 260 x 10 = 2600 km.
The speed of an aeroplane is 249 km in 1 hour. If the aeroplane takes 3.00 hours to cover the distance between two towns then the distance between the towns is:
(a) 815km
(b) 747km
(c) 810km
(d) 801km
(e) None of these
Explanation Distance covered by aeroplane in 3 hours = 249 x 3 = 747 km
There are 87 books each with 220 pages and 50 other books each with 350 pages. What is the total number of pages?
(a) 36640
(b) 36635
(c) 36639
(d) 36630
(e) None of these
Explanation Total number of pages = (220 x 87) + (350 x 50) = 19140 + 17500 = 36640. Therefore, option (a) is correct and rest of the options is incorrect.
There are 987 students in Junior sections in a school and 453 students are in senior sections. Each student contributes Rs 2 for the Teacher's day. What is the total collection?
(a) Rs 2890
(b) Rs 2880
(c) Rs 2810
(d) Rs 2800
(e) None of these
Explication Total number of students = 987 + 453 = 1440. Total collection = Rs 2 x 1440 = Rs 2880.
What is the total salary of 10 workers, if each gets Rs 2756.50 excluding the amount of Rs 34.75 for ESI?
(a) Rs 28902.50
(b) Rs 28803.50
(c) Rs 27912.50
(d) Rs 28003.50
(e) None of these
Explanation: Salary of each worker = 2756.50 + 34.75 = Rs 2791.25.
Therefore, total salary of 10 workers = 2791.25x10 = Rs 27912.50.
• The product of the numbers does not change on reversing their order.
• If a number is multiplied by 0 then the result is zero.
• In a multiplication the number which is multiplied is known as multiplicand.
• The number by which it is multiplied is known as multiplier and the answer or the result of multiplication is known as product.
• In a series of numbers, one number is related to another
• The product of two decimal numbers is the product of their whole and decimal parts.
• The product of a group of numbers does not change on changing the order of the numbers.
• The product of a whole number by 10, 100 and 1000 is obtained by placing the same number of zero after the number.
• The product of a number by 1 is the number itself.
• Multiplication sign is also denoted by simply a dot between two numbers.
• On the key board of computers symbol for multiplication is denoted by
• The word product comes from the Latin word produce.
• Babylonians method for multiplication was very similar to the method that is used today.
• The modern method of multiplications is based on Hindu Arabic numerals which was first discovered by Indian mathematicians.
#### Other Topics
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by Irma Patterson
Kwanzaa was created by Dr. Karenga, professor and chair of black studies at California State University at Long Beach. He says, "Kwanzaa is a synthesis of both the cultural values and practice of Africans on the Continent and in the United States with strict attention to cultural authenticity and values for a meaningful, principled, and productive life." The origins of Kwanzaa on the African continent are in the agricultural celebrations called the "first-fruit" celebrations or (to a lesser degree) the full or general harvest celebrations. At least five sets of values and practices central to African first-fruit celebrations have influenced the development of Kwanzaa.
- Ingathering of the people. The ingathering of the crops become another occasion to pull the people together and to reaffirm and reinforce the bonds between them.
- Reverence for the Creator and creation. A profound reverence for Creator and creation is a central focus for the ingathering of the people in African first-fruit celebrations.
- Commemoration of the past, especially the ancestors. To honor the ancestors is to honor heritage, roots, and lineage.
- Recommitment to cultural ideals. Recommitment involves a reaffirmation and rededication in thought and practice to cooperation, peace, truth, justice, righteousness, sisterhood, brotherhood, harmony, reciprocity, creativity, collective work, responsibility, sharing, mutual care, and confidence. Self-determination and cultural integrity serve as grounding and social glue for the community.
- Celebration of the good. The celebration is of the Creator and the creation of life, of the people, of their history and culture, of a good harvest, and of the promise of the next year. In a word, it is a celebration of all that is good in the widest sense of the word, i.e., divine, natural, cosmic, and social.
African American Origins
Kwanzaa was created to serve several functions in the African American community. First, Kwanzaa was created to reaffirm and restore the African heritage and culture.
Second, Kwanzaa was created to introduce and reinforce the Nouzo Saba* (the seven principles). These are:umoja (unity), kujichagulia (self-determination), ujima (collective work and responsibility), yhanai (creativity), and imani (faith). These principles are projected as the minimum set of African moral values that African Americans need in order to rebuild and strengthen family, community, and culture, as well as to become a self-conscious social force in the struggle to control their destiny and daily life.
Third, Kwanzaa was created to address the absence of non-heroic holidays in the national African American community.
Fourth, Kwanzaa was created to serve as a regular communal celebration that reaffirms and reinforces the bonds between African Americans as a people.
Finally, Kwanzaa was created as an act of cultural self-determination, as a self-conscious statement of our cultural truth as an African people.
Symbols of Kwanzaa
Kwanzaa symbols serve as instructive and inspirational objects that represent and reinforce desirable principles, concepts, and practices. Kwanzaa has seven basic symbols and two supplementary symbols. These seven basic symbols are:
- Mazao (crops). As the central symbols of Kwanzaa, the mazao represent the historical roots of the holiday itself and the rewards of collective productive labor.
- Mkeka (mat). The mkeka is the symbol of tradition and, by extension, history. Tradition and history are foundations for correct knowledge and understanding of self, society, and the world. In recognition of this fact, all other Kwanzaa symbols are placed on the mkeka, and it too becomes a foundation.
- Kinara (candleholder). The kinara is symbolic of our parent people, the continental Africans.
- Muhindi (corn). The muhindi represents children and all the hopes and challenges attached to them. Each house or family places in the Kwanzaa set as many ears of corn as it has children. But each house always places at least one ear of corn, whether there are children present or not. In African society, parenthood is both specific and general, biological and social; and whether one is specifically responsible for a child or not, he or she has general and social responsibility for the children of the community.
- Zawadi (gifts). The zawadi are symbolic of the seeds sown by the children (i.e., commitments made and kept) and of the fruits of the labor of the parents.
- Kikombe Cha Umoja (unity cup). The kikombe serves two basic functions: to pour libation for ancestors and to drink from, serving as a ritual to reinforce unity in the family and community. At Kwanzaa, pouring libation for the ancestors is a very important ritual. It is poured for us a) to remember and honor those who walked and worked before us and thus paved the path down which we now walk, b) to reaffirm our link to them and our life through them, c) to raise models before the community that instruct and inspire, d) to express recommitment to the legacy they left by preserving and expanding it, and e) to cultivate and sustain a cultural practice that sets a model of how our children will act toward us.
- Mishumaa Saba (the seven candles). The mishumaa saba represent the seven principles that are the heart and spirit of Kwanzaa. The candles are placed securely in the kinara, the symbol of ancestry, to symbolize the rootedness of the principles in the way of the ancestors. The lighting of the candles is a daily ritual during Kwanzaa, which symbolizes the light and life principles themselves and the general ancient African concept of raising up light to lessen darkness in both the spiritual and intellectual sense. To light the candles is to give light and life to the seven principles. This means to bring illumination to them, to make them clear, to explain them, to elucidate them, and to animate them.
Kwanzaa is celebrated from December 26 to January 1. It is not a reaction to or substitute for anything. It is the outgrowth of a normal need of people to shape their world in their own cultural image and interests.
Kwanzaa is above all a cultural choice, distinct from a religious one. Africans of all faiths can and do celebrate Kwanzaa. What Kwanzaa offers is not an alternative to their religion or faith but a common ground of African culture which they all share and cherish. It is this common ground of culture on which they all meet, in which they find ancient and enduring meaning, and by which they are thus reaffirmed and reinforced.
Let me share with you how we celebrate Kwanzaa at First Reformed Church in Jamaica, New York:
On December 26 we spend the day decorating the fellowship hall of the church with banners, signs, and other decorations, all in black, red, and green. Members of the congregation arrive around 7:30 p.m. Some are dressed in African clothes. We begin with the lighting of the candle for Umoja (unity) and a brief explanation of Kwanzaa. A drop or two of water is sprinkled around. The principle Umoja is then discussed. This is first done in small groups, and then the ideas are brought to the larger group. A prayer is then said and the candle is snuffed out, indicating the end of the celebration for that day. A potluck supper of African or Caribbean dishes ends the evening.
This celebration is continued for the next four evenings with visits to homes of members of the congregation. We take homemade gifts or books for the children.
On December 31 a lavish feast, the Kwanzaa Karamu is celebrated, once again by members of the congregation at the fellowship hall of the church. We all dress in African clothes and bring dishes from the cuisines of the Caribbean, Africa, or South America. The program for the evening consists of the following:
- Welcoming: the person in charge welcomes everyone who came and makes introductory remarks about Kwanzaa.
- Remembering: someone may read poems by Paul Lawrence Dunbar or Langston Hughes or may read about Martin Luther King Jr. or any other black hero. People may dance, sing, form unity circles, or tell traditional stories.
- Reassessment or Recommitment: a guest lecturer is introduced. One year we had the Rev. Dr. William Tolbert III, West African regional director for Religions for Peace, from Liberia as our speaker.
- Rejoicing: this part of the program may include the passing of the unity cup and the pouring of a libation, forming a circle and recalling names of family ancestors and black heroes, a performance by drummers and dancers, a feast, and group dancing.
- Farewell statement: someone may challenge the group about commitment for the following year.
* The Kwanzaa terms are Swahili.
Dr. Maulana Karenga's website: www.officialkwanzaawebsite.org<|endoftext|>
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The first exoplanet was discovered in 1995, orbiting close to the sun-like star 51 Pegasus. Since then more than 3,4000 other planets have been found by astronomers (as of Feb 2017) and most appear a similar size to Jupiter. There are numerous ways to detect if a star has planets orbiting around it, and the different methods all contribute to the number we know about today.
Around 20% of the planets found so far have very short orbital periods, and take less than 10 days to go around their parent stars. The laws of gravity tell us that planets with shorter orbital periods are closer to their central star, whereas those with longer periods are further out.
In the Solar System, Mercury, the closest planet, has an 88 day orbital period, whereas Neptune, the furthest, takes an incredible 165 years to go around the Sun. As surface temperatures on Mercury are around 400°C, these much closer planets are expected to be well above 1000°C; far too hot for life to exist.
With the high temperatures, it would also be difficult for gas-giant planets to have formed so close, as all the gases needed to create it would have been vaporised and blown away by the solar heat. This means that the "hot Jupiters" we find, which are mainly made of hydrogen and helium, must have formed further away from the star where it is cooler and then migrated inward.
Whilst the exoplanets detected so far have mostly been large gas-giants, we are just starting to detect smaller terrestrial planets like Earth and Mars, which no doubt exist in much larger numbers. Scientists are also trying to develop ways of checking if life could exist on those planets, such as looking for signs of oxygen in their atmosphere.<|endoftext|>
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Often not diagnosed quickly, osteoporosis is known as a silent disease as it sometimes goes unnoticed. Osteoporosis affects some 44 million women over the age of fifty in the United States according to the International Osteoporosis Foundation. Younger women and men may also develop this affliction. Osteoporosis literally mean porous bones and it develops as the body loses too much bone, makes too little bone or both. Often, osteoporosis does not display any obvious symptoms or discomfort. In other words, you don’t feel your bones weakening until you eventually experience bone fractures.
It is considered to be a major public health threat as it readily affects so many people. It is estimated that one in two women and up to one in four men over the age of 50 will break a bone due osteoporosis!
The most common symptoms of osteoporosis include:
- Osteoporotic bone breaks. Fractures and breaks most commonly occur in the hip, spine or wrist bones. They also affect the feet, knees and other parts of the body.
- Limited mobility, trouble getting around and difficulty completing everyday activities. Many elderly adults who break a bone will need to live long-term in nursing homes or will require assistance from an aid in their home.
- Bone pain, sometimes which is permanent and intense.
- Loss of height.
- Hunched or stooped posture. This occurs because the vertebrae, the bones of the spine, can become weaker.
Having osteoporosis can be quite emotionally devastating. One is at a loss due to lack of mobility and it can be life-threatening. The following natural handling of osteoporosis should be considered as you can strengthen the weakened areas. Osteoporosis should not be taken lightly, since weak and broken bones can be difficult to treat and cope with.
Treating osteoporosis naturally can be quite effective. This includes sufficient exercise, especially resistance training, preventing Vitamin D deficiencies and eat healthy and naturally. Diet plays a critical role in bone health. You must be taking in enough protein, vitamins and minerals, in particular, calcium, magnesium, phosphorous and manganese. These minerals are critical to bone development.
Here are five natural steps to help reduce excessive bone loss:
- Nourish your body with basic nutrients.
- Provide your body with specific bone-building nutrients.
- Eat an alkalizing diet.
- Generate stronger bone with some exercise.
- Minimize your stress.<|endoftext|>
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[Image above] Scientists at the Australian National University have developed a new spray-on material with a remarkable ability to repel water. Credit: Stuart Hay; ANU
One way to make materials and structures stand up better to environmental stress or chemical exposure is to engineer them from the outset to be resistant to these types of corrosive atmospheres.
But what about preserving and protecting existing materials and structures?
Enter: cutting-edge coatings.
From ice removal to water repelling to ultraviolet protection, coatings enable our vehicles to shed ice buildup, protect our structures from water damage, and keep our buildings cooler.
Protective coatings can help break the ice. Researchers at Rice University (Houston, Texas) developed a conductive graphene coating that offers a new tunable solution for surface ice shedding on aircraft, wind turbines, transmission lines, and other surfaces exposed to winter weather.
They can help keep it cool. A team of scientists from Johns Hopkins Applied Physics Laboratory (Laurel, Md.) created a glass-based paint that reflects light off metal surfaces to keep them cool and protects those surfaces from corrosion accelerated by the harsh rays of the sun.
Or they can have it made in the shade. A group of researchers at the Georgia Institute of Technology in Atlanta have developed a polymer coating for glass that can change the tint color instantly with a small, user-controlled electrical current—an innovation that could lead to major energy-saving potential for windows on buildings exposed to sunlight.
But when it comes to waterproofing, scientists at the Australian National University (ANU) say they’ve developed a new spray-on coating with a remarkable ability to repel water. And this new protective coating could eventually be used to waterproof cell phones, prevent ice formation on airplanes, and protect boat hulls against corrosion.
According to a university press release, this superhydrophobic coating is more robust than previous materials thanks to a novel combination of two plastics, one tough and one flexible.
“The surface is a layer of nanoparticles, which water slides off as if it’s on a hot barbecue,” explains William Wong, a Ph.D. student from the Nanotechnology Research Laboratory at the ANU Research School of Engineering. “It’s like two interwoven fishing nets, made of different materials.”
Credit: ANU TV; YouTube
The process to create the coating is relatively simple and more cost-efficient than what it takes to manufacture comparable coatings, the team says. To achieve this, the team developed two ways of creating the material.
“One method uses a flame to generate the nanoparticle constituents of the material. For lower temperature applications, the team dissolved the two components in a ‘sprayable’ form,” the release explains.
Plus, this water-repellent coating is also transparent and extremely resistant to ultraviolet radiation—key game-changing features of this type of material, says Antonio Tricoli, lead researcher and head of the ANU Nanotechnology Research Laboratory.
“The key innovation is that this transparent coating is able to stabilize very fragile nanomaterials resulting in ultra-durable nano-textures with numerous real-world applications,” explains Tricoli. “A lot of the functional coatings today are very weak, but we will be able to apply the same principles to make robust coatings that are, for example, anti-corrosive, self-cleaning, or oil-repellent.”
The research, published in ACS Applied Materials and Interfaces, is “Ultra-durable and transparent self-cleaning surfaces by large-scale self-assembly of hierarchical interpenetrated polymer networks” (DOI: 10.1021/acsami.6b03414).<|endoftext|>
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The Progressive Era was a period of social and political reformation that flourished under the leadership of President Woodrow Wilson. When the United States entered World War 1 on April 6, 1917, the whole nation was united under a moral cause. However, the war quickly busied Americans creating no time for Progressive movements and the events following the war put an end to the Era all together. During the beginning of the war, Wilson’s foreign policy was to remain on a neutral tide.
With his reelection, he even won against Charles Hughes under his slogan, “He kept us out of war”, that persuaded Americans that choosing Hughes would lead them into direct contact with the war. Americans united under this foreign policy trying to stay out of European affairs under the philosophy of isolationism. When Germany’s unauthorized submarines caused for the sinking of the RMS Lusitania and their efforts chose to continue unrestricted submarine warfare the United States was infuriated.
Only $13.90 / page
This is one factor that kept the United States busy with working to remain excluded from war. However, President Wilson addressed to Congress and declared he wanted “to make the world safe for democracy”, leading America into World War 1, under this moral cause. (Doc 2) Now that the United States was in the war, Americans started to become less concerned with social reforms and busier working, supporting, and fighting the war together. Women, for one, took over the jobs of all the men who went out to war, and worked to supply the belligerents, keeping women very busy.
The United States Food Administration, under Herbert Hoover, promoted rationing of food through “Meatless Mondays” and “Wheatless Wednesdays”. This helped to conserve food and save for those fighting overseas. (Doc 6) Trench warfare was how most of World War 1 was fought and living in these trenches was rather a tough condition. This ensured that supporting and working hard together in America for our men was required to aid their health and capability of being able to fight effectively. (Doc 3)
The war ended with a high number of casualties for America. Many of these casualties resulted from the influenza epidemic that attacked the world’s population and battle fights, such as those that occurred highly within the use of trenches. (Doc 3) This high death effected lives of Americans and their families, keeping them from being motivated to continue the Progressive movement. The Treaty of Versailles created at the Paris Peace Conference in 1919, put an end to World War 1, but did so with a lot of problems.
President Wilson, as devised in his 14 Point Peace Plan called for a creation of a League of Nations, an international peace organization. This organization caused for a split in American views. The United States, as a majority, voted not to join, as it took away the power of Congress to declare war and would ruin America’s self-determination. (Doc 9) This split in decision effected formally agreeing and creating more social reforms. The Progressive Era ended with World War 1, as many Americans were too busy with a plethora of events to help the United States fight the war.
When the United States entered the war, they were morally united under a new crusade. However, they had to put their further ideas of reformations on hold as the war demanded serious attention to be fought effectively. The further Americans became deeply involved into the war, the further attention was dragged away from the progressive movement. Even after the war was over, Americans were still busy caught up in dealing with the aftermath and the approach of another worldwide outbreak.<|endoftext|>
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# TRIGONOMETRIC RATIOS OF COMPOUND ANGLES
## About "Trigonometric ratios of compound angles"
Trigonometric ratios of compound angles :
An angle made up of the algebraic sum of two or more angles is called a compound angle.
Here, we are going see the formulas for trigonometric ratios of compound angles.
sin (A + B) = sinA cosB + cosA sinB
sin (A - B) = sinA cosB - cosA sinB
cos (A + B) = cosA cosB - sinA cosB
cos (A - B) = cosA cosB + sinA cosB
tan (A + B) = [tanA + tanB] / [1 - tanA tanB]
tan (A - B) = [tanA - tanB] / [1 + tanA tanB]
## Trigonometric ratio table
From the above table, we can get the values of trigonometric ratios for standard angles such as 0°, 30°, 45°, 60°, 90°
Now, let us look at some practice problems on "Trigonometric ratios of compound angles".
## Trigonometric ratios of compound angles - Practice problems
Example 1 :
Find the value of cos15°
Solution :
First, we have to write the given angle 15° in terms of sum or difference of two standard angles.
So, we have 15° = 45° - 30°
cos15° = cos (45° - 30°)
cos15° = cos45° cos30° + sin45° sin30°
Using the above trigonometric ratio table, we have
cos15° = (√2/2) x (√3/2) + (√2/2) x (1/2)
cos15° = (√6 / 4) + (√2/4)
cos15° = (√6 + √2) / 4
Hence, the value of cos15° is equal to (√6 + √2) / 4
Let us look at the next problem on "Trigonometric ratios of compound angles"
Example 2 :
Find the value of cos105°
Solution :
First, we have to write the given angle 105° in terms of sum or difference of two standard angles.
So, we have 105° = 60° + 45°
cos105° = cos (60° + 45°)
cos105° = cos60° cos45° - sin60° sin45°
Using the above trigonometric ratio table, we have
cos105° = (1/2) x (√2/2) - (√3/2) x (√2/2)
cos105° = (√2 / 4) - (√6/4)
cos15° = (√2 - √6) / 4
Hence, the value of cos15° is equal to (√2 - √6) / 4
Let us look at the next problem on "Trigonometric ratios of compound angles"
Example 3 :
Find the value of sin75°
Solution :
First, we have to write the given angle 75° in terms of sum or difference of two standard angles.
So, we have 75° = 45° + 30°
sin75° = sin (45° + 30°)
sin75° = sin45° cos30° + cos45° sin30°
Using the above trigonometric ratio table, we have
sin75° = (√2/2) x (√3/2) + (√2/2) x (1/2)
sin75° = (√6 / 4) + (√2/4)
sin75° = (√6 + √2) / 4
Hence, the value of sin75° is equal to (√6 + √2) / 4
Let us look at the next problem on "Trigonometric ratios of compound angles"
Example 4 :
Find the value of tan15°
Solution :
First, we have to write the given angle 15° in terms of sum or difference of two standard angles.
So, we have 15° = 45° - 30°
tan15° = tan (45° - 30°)
tan15° = [tan45° - tan30°] / [1 + tan45° tan30°]
Using the above trigonometric ratio table, we have
tan15° = [1 - 1/√3] / [1 + 1x1/√3]
tan15° = [(√3 - 1)/√3] / [(√3 + 1)/√3]
tan15° = [(√3 - 1)/√3] x [(√3/(√3 + 1)]
tan15° = (√3 - 1) / (√3 + 1)
By rationalizing the denominator, we get
tan15° = 2 - √3
Hence, the value of tan15° is equal to 2 - √3
Let us look at the next problem on "Trigonometric ratios of compound angles"
Example 5 :
If A and B are acute angles, sinA = 3/5, cosB = 12/13,
find cos (A+B)
Solution :
cos (A + B) = cosAcosB - sinAsinB
From the formula of cos (A + B), the value of sinA and cosA are given in the question.
We need the value of sinB and cosA.
We know that,
sin ² B = 1 - cos ² B
sin ² B = 1 - (12/13)²
sin ² B = 1 - 144/169
sin ² B = (169 - 144)/169
sin ² B = 25/169
sin ² B = (5/13)² --------> sinB = 5/13
cos ² A = 1 - sin ² A
cos ² A = 1 - (3/5)²
cos ² A = 1 - 9/25
cos ² A = (25 - 9)/25
cos ² A = 16 / 25
cos ² A = (4/5)² --------> cosA = 4 / 5
Plugging the values of cosA, cosB, sinA and sinB in the formula of cos (A+B), we get,
cos (A + B) = (4/5) x (12/13) - (3/5) x (5/13)
cos (A + B) = (4/5) x (12/13) - (3/5) x (5/13)
cos (A + B) = (48 / 65) - (15 / 65)
cos (A + B) = (48 - 15) / 65
cos (A + B) = 33 / 65
Hence, the value of cos(A+B) is 33/65.
## Compound angles without standard angles
As we have seen in the above examples, some angles can not be written in terms of sum or difference of two standard angles.
For example,
Let us consider sin 225°
Here, 225° can not be written in terms of sum or difference of two standard angles.
All that we can do is, 225° can be written in terms sum or difference of two angles where one of the angles will be quadrantal angles such as 0°, 90°, 180°, 270°
So, sin 225° can be written as
sin (180° + 45°) or sin (270° - 45°)
To evaluate sin (180° + 45°) or sin (270° - 45°), we have to know ASTC formula
ASTC formla has been explained clearly in the figure given below.
More clearly
In the first quadrant (0° to 90°), all trigonometric ratios are positive.
In the second quadrant (90° to 180°), sin and csc are positive and other trigonometric ratios are negative.
In the third quadrant (180° to 270°), tan and cot are positive and other trigonometric ratios are negative.
In the fourth quadrant (270° to 360°), cos and sec are positive and other trigonometric ratios are negative.
## Important conversions
When we have the angles 90° and 270° in the trigonometric ratios in the form of
(90° + θ)
(90° - θ)
(270° + θ)
(270° - θ)
We have to do the following conversions,
sin θ <------> cos θ
tan θ <------> cot θ
csc θ <------> sec θ
For example,
sin (270° + θ) = - cos θ
cos (90° - θ) = sin θ
For the angles 0° or 360° and 180°, we should not make the above conversions.
## Evaluation of trigonometric ratios using ASTC formula
Now, let us evaluate sin (180° + 45°)
To evaluate sin (90° + θ), we have to consider the following important points.
(i) (180° + 45°) will fall in the III rd quadrant.
(ii) When we have 180°, "sin" will not be changed.
(iii) In the III rd quadrant, the sign of "sin" is negative.
Considering the above points, we have
sin 225° = sin (180° + 45°)
sin 225° = - sin 45°
sin 225° = - √2 / 2
When we are not able to write the given angle in terms of sum or difference of two standard angles, we have to proceed the problem in this way.
After having gone through the stuff given above, we hope that the students would have understood "Trigonometric ratios of compound angles"
Apart from "Trigonometric ratios of 90 degree plus theta", if you need any other stuff in math, please use our google custom search here.
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# Video: Pack 3 • Paper 1 • Question 11
Pack 3 • Paper 1 • Question 11
04:05
### Video Transcript
There are some plain, some chocolate, and some raisin biscuits in a jar. There are exactly 16 chocolate biscuits in the jar. The ratio of the number of plain biscuits to the number of chocolate biscuits to the number of raisin biscuits is two to eight to five. Calculate the number of plain biscuits in the jar.
In the question, we’re given that the ratio of plain to chocolate to raisin biscuits is two to eight to five. And what this means is that for every eight chocolate biscuits, there has to be two plain biscuits and five raisin biscuits.
Now, the other piece of information given to us in the question is that there are exactly 16 chocolate biscuits in the jar. And so therefore, we could rewrite our ratio as something to 16 to something, where this first question mark represents the number of plain biscuits in the jar and the second question mark represents the number of raisin biscuits in the jar.
In this part of the question, however, we’re only interested in finding the number of plain biscuits. So let’s focus on that. Now, we can compare the two ratios which we have. We see that to go from the first ratio to the second ratio, the number in the middle — so that’s the number that represents the number of chocolate biscuits — has doubled or multiplied by two. So therefore, in order to go from the first ratio to the second ratio, the other parts of the ratio must also double.
Therefore, we find that the number of plain biscuits in the jar will be two multiplied by two, giving us four biscuits. We’ve now answered the first part of the question since we found the number of plain biscuits in the jar to be four. Let’s move on to part b.
Lisa adds some more raisins biscuits to the jar. Given that the ratio of the number of plain biscuits to the number of chocolate biscuits to the number of raisin biscuits is now one to four to five, find the number of raisin biscuits that Lisa adds to the jar.
Let’s start by finding how many raisin biscuits were originally in the jar. Similarly to part a, we simply multiply the ratio by two to find the number of biscuits. So multiplying the five by two, we see that there were originally 10 raisin biscuits in the jar. Next, we need to find the number of raisin biscuits that are in the jar after Lisa added some more.
We know that the ratio of plain to chocolate to raisin biscuits is now one to four to five since this was given to us in the question. We also know that the number of plain biscuits and the number of chocolate biscuits in the jar from part a has not changed. So we still have four plain biscuits and 16 chocolate biscuits.
So we know that the current ratio of biscuits in the jar is four to 16 to something, where something is the number of raisin biscuits which are currently in the jar. Since this ratio must be equivalent to one to four to five, we just need to find what number we multiply one to four to five by in order to get this ratio of four to 16 to something.
Clearly, we can see that we multiply the one by four to get four and we multiply the four by four to get 16. So therefore, we must also multiply the five by four in order to find the number of raisin biscuits, which are currently in the jar. So five timesed by four gives us 20 raisin biscuits.
Now, we note that the question asked us to find the number of biscuits which Lisa added to the jar. So this will simply be the difference between the number of biscuits which are in the jar after she added them and the number of biscuits which were in the jar before she added them. And so this gives us 20 minus 10, telling us that Lisa has added 10 raisin biscuits to the jar. And this is our answer.<|endoftext|>
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# Difference between revisions of "2019 AIME I Problems/Problem 15"
## Problem
Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solution 1
$[asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); dot("O", O, S); dot("A", A, A); dot("B", B, B); dot("P", P, dir(70)); dot("Q", Q, dir(200)); dot("O_1", O1, SW); dot("O_2", O2, SE); dot("X", X, X); dot("Y", Y, Y); [/asy]$ Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$, respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$, so $\overline{OO_2}\parallel\overline{O_1P}$. Similarly, $\overline{OO_1}\parallel\overline{O_2P}$, so $OO_1PO_2$ is a parallelogram. Moreover, $$\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,$$whence $OO_1O_2Q$ is cyclic. However, $$OO_1=O_2P=O_2Q,$$so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$, $\overline{OQ}\perp\overline{XY}$, so $Q$ is the midpoint of $\overline{XY}$.
By Power of a Point, $PX\cdot PY=PA\cdot PB=15$. Since $PX+PY=XY=11$ and $XQ=11/2$, $$XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,$$ and the requested sum is $61+4=\boxed{065}$.
(Solution by TheUltimate123)
### Note
One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$. Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$. The most likely explanation for this is that $Q$ is the midpoint of $XY$, so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$. Then our answer is $m+n=61+4=\boxed{065}$. One can rigorously prove this using the methods above
## Solution 2
Let the tangents to $\omega$ at $A$ and $B$ intersect at $R$. Then, since $RA^2=RB^2$, $R$ lies on the radical axis of $\omega_1$ and $\omega_2$, which is $\overline{PQ}$. It follows that $$-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).$$ Let $Q'$ denote the midpoint of $\overline{XY}$. By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), $$RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,$$ whence $Q=Q'$. Like above, $XP=\tfrac{11-\sqrt{61}}2$. Since $XQ=\tfrac{11}2$, we establish that $PQ=\tfrac{\sqrt{61}}2$, from which $PQ^2=\tfrac{61}4$, and the requested sum is $61+4=\boxed{065}$.
(Solution by TheUltimate123)
## Solution 3
Firstly we need to notice that $Q$ is the middle point of $XY$. Assume the center of circle $w, w_1, w_2$ are $O, O_1, O_2$, respectively. Then $A, O_2, O$ are collinear and $O, O_1, B$ are collinear. Link $O_1P, O_2P, O_1Q, O_2Q$. Notice that, $\angle B=\angle A=\angle APO_2=\angle BPO_1$. As a result, $PO_1\parallel O_2O$ and $QO_1\parallel O_2P$. So we have parallelogram $PO_2O_1O$. So $\angle O_2PO_1=\angle O$ Notice that, $O_1O_2\bot PQ$ and $O_1O_2$ divide $PQ$ into two equal length pieces, So we have $\angle O_2PO_1=\angle O_2QO_1=\angle O$. As a result, $O_2, Q, O, O_1,$ lie on one circle. So $\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P$. Notice that $\angle O_1PQ+\angle O_2O_1P=90^{\circ}$, we have $\angle OQP=90^{\circ}$. As a result, $OQ\bot PQ$. So $Q$ is the middle point of $XY$.
Back to our problem. Assume $XP=x$, $PY=y$ and $x. Then we have $AP\cdot PB=XP\cdot PY$, that is, $xy=15$. Also, $XP+PY=x+y=XY=11$. Solve these above, we have $x=\frac{11-\sqrt{61}}{2}=XP$. As a result, we have $PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$. So, we have $PQ^2=\frac{61}{4}$. As a result, our answer is $m+n=61+4=\boxed{065}$.
Note that the tangents to the circles at $A$ and $B$ intersect at a point $Z$ on $XY$ by radical center. Then, since $\angle ZAB = \angle ZQA$ and $\angle ZBA = \angle ZQB$, we have $$\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},$$ so $ZAQB$ is cyclic. But if $O$ is the center of $\omega$, clearly $ZAOB$ is cyclic with diameter $ZO$, so $\angle ZQO = 90^{\circ} \implies Q$ is the midpoint of $XY$. Then, by Power of a Point, $PY \cdot PX = PA \cdot PB = 15$ and it is given that $PY+PX = 11$. Thus $PY, PX = \frac{11 \pm \sqrt{61}}{2}$ so $PQ = \frac{\sqrt{61}}{2} \implies PQ^2 = \frac{61}{4}$ and the answer is $61+4 = \boxed{065}$.<|endoftext|>
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# remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
Hence, $a^2+3a+4 = (7n+b)^2+3(7n+b)+4$ $= 49n^2 + 14nb + b^2 + 21n + 3b + 4$ $= 7(7n^2+2nb+3n) + (b^2+3b+4)$.
So, the remainder when $a^2+3a+4$ is divided by $7$ will be the same as the remainder when $b^2+3b+4$ is divided by $7$.
For the specific case when $b = 6$, we get that $a^2+3a+4 = 7(7n^2+12n+3n)+58$ $= 7(7n^2+12n+3n+8)+2$.
So the remainder when $a^2+3a+4$ is divided by $7$ is $2$.
The remainder of $a^2+3a+4$ divided by $7$ is sum of the remainder of each terms, modulo $7$.
So $a^2\equiv 1 \pmod{7}$ since $a=7k+6$ then $a^2=7l+1$; $\quad$ $3a\equiv 4 \pmod{7}$ since $3a=21k+18=21k+14+4$ and clearly $4\equiv 4 \pmod{7}$.
Finally $1+4+4 \equiv 2 \pmod{7}$ then the remainder is $2$.
• To obtain $a \equiv b \pmod{n}$, type a \equiv b \pmod{n} in math mode. Jul 28 '15 at 13:43
$a^2 + 3a + 4 \equiv a^2 - 4a + 4 \equiv (a-2)^2 \pmod 7$
If $a\equiv b \pmod 7$, then $a^2 + 3a + 4 \equiv (b-2)^2 \pmod 7$<|endoftext|>
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Giardiasis is a faeco-oral disease which results from infection of the small intestine by protozoa called Giardia intestinalis, also known as Giardia duodenalis. (Note: the same protozoa are called Giardia lamblia in older textbooks.). Like the parasitic amoebae described in the previous section, Giardia form resistant cysts in the person's intestines that pass out in the faeces. The cysts can be easily transmitted in water contaminated by faeces, from person-to-person through hand-to-mouth transmission and in food. They hatch out in the new person and the protozoa rapidly increase in number by cell division.
The commonest clinical manifestation of giardiasis is foul-smelling, pale, greasy diarrhoea, without blood or mucus (mucoid). The diarrhoea can be acute and resolve by itself within a few days, or it may be persistent (lasting for more than 14 days). Other symptoms of giardiasis include nausea, vomiting, abdominal cramps and abdominal distension (swelling).
You should suspect giardiasis in children if the diarrhoea is persistent, but not bloody or mucoid. For children with mild non-bloody or non-mucoid diarrhoea, the management does not require identification of the infectious agent; cases are managed with oral rehydration as already described for simple acute watery diarrhoea. If a child has persistent or severe diarrhoea, and giardiasis is one of the causes you suspect, treatment is carried out at a higher-level health facility on a case-by-case basis, taking into account the presence of other symptoms and/or malnutrition. Therefore, you should start ORS treatment and refer the child.
In adults, you should suspect a diagnosis of giardiasis in cases with acute or persistent, non-bloody or non-mucoid diarrhoea. However, as other diseases could also have similar manifestations, confirmation of the diagnosis is needed through detection of the parasite in laboratory examination of stool samples. Therefore, start all adult patients on rehydration with ORS and refer them for further diagnosis and treatment.
Amoebiasis and giardiasis both present with diarrhoea. What is the difference in the type of diarrhoea resulting from the two diseases?
The main difference in the type of diarrhoea is that in amoebiasis it is bloody or mucoid (contains mucus), whereas in giardiasis it is pale, greasy and foul smelling.<|endoftext|>
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 8.4: Alternating Series
Skills to Develop
In this section, we strive to understand the ideas generated by the following important questions:
• What is an alternating series?
• What does it mean for an alternating series to converge?
• Under what conditions does an alternating series converge? Why?
• How well does the nth partial sum of a convergent alternating series approximate the actual sum of the series? Why?
• What is the difference between absolute convergence and conditional convergence?
In our study of series so far, almost every series that we’ve considered has exclusively nonnegative terms. Of course, it is possible to consider series that have some negative terms. For instance, if we consider the geometric series
$2 − \dfrac{4}{3} + \dfrac{8}{9} − \ldots + 2 \left(\dfrac{−2}{3}\right)^n + \ldots ,$
which has $$a = 2$$ and $$r = − \dfrac{2}{3}$$, we see that not only does every other term alternate in sign, but also that this series converges to
$S = \dfrac{a}{1 − r} = \dfrac{2}{1 − \left(− \dfrac{2}{3}\right) } = \dfrac{6}{5}.$
In Preview Activity $$\PageIndex{1}$$ and our following discussion, we investigate the behavior of similar series where consecutive terms have opposite signs.
Preview Activity $$\PageIndex{1}$$
showed how we can approximate the number e with linear, quadratic, and other polynomial approximations. We use a similar approach in this activity to obtain linear and quadratic approximations to $$\ln(2)$$. Along the way, we encounter a type of series that is different than most of the ones we have seen so far. Throughout this activity, let $$f (x) = \ln(1 + x)$$.
a. Find the tangent line to $$f$$ at $$x = 0$$ and use this linearization to approximate $$\ln(2)$$. That is, find $$L(x)$$, the tangent line approximation to $$f (x)$$, and use the fact that $$L(1) \approx f (1)$$ to estimate $$\ln(2)$$.
b. The linearization of $$\ln(1 + x)$$ does not provide a very good approximation to $$\ln(2)$$ since 1 is not that close to 0. To obtain a better approximation, we alter our approach; instead of using a straight line to approximate $$\ln(2)$$, we use a quadratic function to account for the concavity of $$\ln(1 + x)$$ for $$x$$ close to 0. With the linearization, both the function’s value and slope agree with the linearization’s value and slope at $$x = 0$$. We will now make a quadratic approximation $$P_2(x) to f (x)$$ = $$\ln(1 + x)$$ centered at $$x = 0$$ with the property that $$P_2(0) = f (0), P'_2 (0) = f'(0)$$, and $$P''_2 (0) = f''(0)$$.
1. Let $$P_2(x) = x − \dfrac{x^2}{2}$$. Show that $$P_2(0) = f (0), P'_2 (0) = f'(0)$$, and $$P''2 (0) = f''(0)$$. Use $$P_2(x)$$ to approximate $$\ln(2)$$ by using the fact that $$P_2(1) \approx f (1)$$.
2. We can continue approximating $$\ln(2)$$ with polynomials of larger degree whose derivatives agree with those of $$f$$ at 0. This makes the polynomials fit the graph of $$f$$ better for more values of $$x$$ around 0. For example, let $$P_3(x) = x − \dfrac{x^2}{2} + \dfrac{x^3}{3}$$. Show that $$P_3(0) = f (0), P'_3 (0) = f'(0), P''_3 (0) = f''(0)$$, and $$P'''_3 (0) = f'''(0)$$. Taking a similar approach to preceding questions, use P3(x) to approximate $$\ln(2)$$.
3. If we used a degree 4 or degree 5 polynomial to approximate $$\ln(1 + x)$$, what approximations of $$\ln(2)$$ do you think would result? Use the preceding questions to conjecture a pattern that holds, and state the degree 4 and degree 5 approximation. .
Preview Activity $$\PageIndex{1}$$ gives us several approximations to $$\ln(2)$$, the linear approximation is 1 and the quadratic approximation is $$1 − \dfrac{1}{2} = \dfrac{1}{2}$$. If we continue this process we will obtain approximations from cubic, quartic (degree 4), quintic (degree 5), and higher degree polynomials giving us the following approximations to $$ln(2)$$:
Linear 1 1 Quadratric $$1- \dfrac{1}{2}$$ 0.5 Cubic $$1- \dfrac{1}{2}+ \dfrac{1}{3}$$ 0.83̅ Quartic $$1- \dfrac{1}{2}+ \dfrac{1}{3} - \dfrac{1}{4}$$ 0.583̅ Quintic $$1- \dfrac{1}{2}+ \dfrac{1}{3} - \dfrac{1}{4}+ \dfrac{1}{5}$$ 0.783̅
The pattern here shows the fact that the number $$ln(2)$$ can be approximated by the partial sums of the infinite series
$\sum_{k=1}^\infty (-1)^{k+1} \frac{1}{k} \tag{8.15} \label{8.15}$
where the alternating signs are determined by the factor $$(−1)^{k+1}$$. Using computational technology, we find that 0.6881721793 is the sum of the first 100 terms in this series. As a comparison, $$\ln(2) \approx$$ 0.6931471806. This shows that even though the series (Equation $$\ref{8.15}$$) converges to $$\ln(2)$$, it must do so quite slowly, since the sum of the first 100 terms isn’t particularly close to $$\ln(2)$$. We will investigate the issue of how quickly an alternating series converges later in this section. Again, note particularly that the series $$\ref{8.15}$$ is different from the series we have considered earlier in that some of the terms are negative. We call such a series an alternating series.
Definition: alternating series
An alternating series is a series of the form
$\sum_{k=0}^{\infty} (−1)^k a_k, \nonumber$
where $$a_k ≥ 0$$ for each $$k$$. We have some flexibility in how we write an alternating series; for example, the series
$\sum_{k=1}^{\infty} (−1)^{k+1} a_k, \nonumber$
whose index starts at $$k = 1$$, is also alternating. As we will soon see, there are several very nice results that hold about alternating series, while alternating series can also demonstrate some unusual behavior.
It is important to remember that most of the series tests we have seen in previous sections apply only to series with nonnegative terms. Thus, alternating series require a different test. To investigate this idea, we return to the example in Preview Activity $$\PageIndex{1}$$.
Activity $$\PageIndex{1}$$
Remember that, by definition, a series converges if and only if its corresponding sequence of partial sums converges.
Complete Table 8.7 by calculating the first few partial sums (to 10 decimal places) of the alternating series
$\sum_{k=1}^{\infty} (−1)^{k+1} \dfrac{1}{k}. \nonumber$
Plot the sequence of partial sums from part (a) in the plane. What do you notice about this sequence?
Activity $$\PageIndex{1}$$ exemplifies the general behavior that any convergent alternating series will demonstrate. In this example, we see that the partial sums of the alternating harmonic series oscillate around a fixed number that turns out to be the sum of the series.
$$\sum_{k=1}^{1} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{6} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{2} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{7} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{3} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{8} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{4} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{9} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{5} (−1)^{k+1} \dfrac{1}{k}$$= $$\sum_{k=1}^{10} (−1)^{k+1} \dfrac{1}{k}$$
Recall that if $$\lim_{k→∞} a^k \neq 0$$, then the series $$P a^k$$ diverges by the Divergence Test. From this point forward, we will thus only consider alternating series
$\sum_{k=1}^{\infty} (−1)^{k+1} a_k \nonumber$
in which the sequence $$a_k$$ consists of positive numbers that decrease to 0. For such a series, the $$n$$th partial sum $$S_n$$ satisfies
$S_n = \sum_{k=1}^n (-1)^{k+1} a_k. \nonumber$
Notice that
• $$S_1 = a_1$$
• $$S_2 = a_1 − a_2$$, and since $$a_1 > a_2$$ we have
$0 < S_2 < S_1. \nonumber$
• $$S_3 = S_2 + a_3$$ and so $$S_2 < S_3$$. But $$a_3 < a_2$$, so $$S_3 < S_1$$. Thus,
$0 < S_2 < S_3 < S_1 \nonumber$
• $$S_4 = S_3 − a_4$$ and so $$S_4 < S_3$$. But $$a_4 < a_3$$, so $$S_2 < S_4$$. Thus,
$0 < S_2 < S_4 < S_3 < S_1. \nonumber$
• $$S_5 = S_4 + a_5$$ and so $$S_4 < S_5$$. But $$a_5 < a_4$$, so $$S_5 < S_3$$. Thus,
$0 < S_2 < S_4 < S_5 < S_3 < S_1. \nonumber$
This pattern continues as illustrated in Figure $$\PageIndex{1}$$ (with $$n$$ odd) so that each partial sum lies between the previous two partial sums. Note further that the absolute value of the difference between the $$(n − 1)$$st partial sum $$S_n−1$$ and the nth partial sum $$S_n$$ is
$|S_n − S_n−1| = a_n. \nonumber$
Since the sequence {$$a_n$$} converges to 0, the distance between successive partial sums becomes as close to zero as we’d like, and thus the sequence of partial sums converges (even though we don’t know the exact value to which the sequence of partial sums converges). The preceding discussion has demonstrated the truth of the Alternating Series Test.
Figure $$\PageIndex{1}$$: Partial sums of an alternating series
Definition: The Alternating Series Test
Given an alternating series
$\sum (−1)^k a_k, \nonumber$
if the sequence $$\{a_k \}$$ of positive terms decreases to 0 as $$k \rightarrow \infty$$, then the alternating series converges.
Note particularly that if the limit of the sequence {$$a_k$$ } is not 0, then the alternating series diverges.
Activity $$\PageIndex{2}$$
Which series converge and which diverge? Justify your answers.
1. $$\sum_{k=1}^{\infty} \dfrac{(−1)^k}{k^2 + 2}$$
2. $$\sum_{k=1}^{\infty} \dfrac{(−1)^{k+1} 2k}{k + 5}$$
3. $$\sum_{k=2}^{\infty} \dfrac{(−1)^k}{\ln(k)}$$
The argument for the Alternating Series Test also provides us with a method to determine how close the $$n$$th partial sum $$S_n$$ is to the actual sum of a convergent alternating series. To see how this works, let $$S$$ be the sum of a convergent alternating series, so
$S = \sum_{k=1}^{\infty} (−1)^k a_k . \nonumber$
Recall that the sequence of partial sums oscillates around the sum $$S$$ so that
$|S − S_n| < |S_{n+1} − S_n| = a_{n+1}. \nonumber$
Therefore, the value of the term an+1 provides an error estimate for how well the partial sum $$S_n$$ approximates the actual sum $$S$$. We summarize this fact in the statement of the Alternating Series Estimation Theorem.
Alternating Series Estimation Theorem
If the alternating series
$\sum_{k=1}^{\infty} (−1)^{k+1} a_k \nonumber$
converges and has sum $$S$$, and
$S_n = \sum^n_{ k=1} (−1)^{ k+1} a_k \nonumber$
is the nth partial sum of the alternating series, then
$| \sum_{k=1}^{\infty} (−1)^{k+1} a_k − S_n | ≤ a_n+1. \nonumber$
Example $$\PageIndex{1}$$
Let’s determine how well the 100th partial sum $$S_{100}$$ of
$\sum_{k=1}^{\infty} \dfrac{(−1)^{k+1}}{k} \nonumber$
approximates the sum of the series.
Solution.
If we let S be the sum of the series P∞ k=1 (−1) k+1 k , then we know that
$|S100 − S| < a_{101}. \nonumber$
Now
$a_{101} = \dfrac{1}{101} \approx 0.0099, \nonumber$
so the 100th partial sum is within 0.0099 of the sum of the series. We have discussed the fact (and will later verify) that
$S = \sum_{k=1}^{\infty} \dfrac{(−1)^{k+1}}{k} = ln(2) , \nonumber$
and so $$S \approx 0.693147$$ while
$S_{100} =\sum_{k=1}^{\infty} \dfrac{(−1)^{k+1}}{k} \approx 0.6881721793. \nonumber$
We see that the actual difference between S and $$S_{100}$$ is approximately 0.0049750013, which is indeed less than 0.0099.
Activity $$\PageIndex{3}$$
Determine the number of terms it takes to approximate the sum of the convergent alternating series
$\sum_{k=1}^{\infty} \dfrac{(−1)^{ k+1}}{ k^ 4} \nonumber$
to within 0.0001.
### Absolute and Conditional Convergence
A series such as
$1 − \dfrac{1}{4} − \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{25} + \dfrac{1}{36} − \dfrac{1}{49} − \dfrac{1}{64} − \dfrac{1}{81} − \dfrac{1}{100} + \ldots \tag{8.16}\label{8.16}$
whose terms are neither all nonnegative nor alternating is different from any series that we have considered to date. The behavior of these series can be rather complicated, but there is an important connection between these arbitrary series that have some negative terms and series with all nonnegative terms that we illustrate with the next activity.
Activity $$\PageIndex{4}$$
a. Explain why the series
$1 − \dfrac{1}{4} − \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{ 25} + \dfrac{1} {36} − \dfrac{1}{49} − \dfrac{1}{64} − \dfrac{1}{81} − \dfrac{1}{100} + \ldots$
must have a sum that is less than the series
$\sum_{k=1}^{\infty} \dfrac{1}{ k^2}. \nonumber$
b. Explain why the series
$1 − \dfrac{1}{4} − \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{ 25} + \dfrac{1} {36} − \dfrac{1}{49} − \dfrac{1}{64} − \dfrac{1}{81} − \dfrac{1}{100} + \ldots \nonumber$
must have a sum that is greater than the series
$\sum_{k=1}^{\infty} - \dfrac{1}{ k^2}. \nonumber$
c. Given that the terms in the series
$1 − \dfrac{1}{4} − \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{ 25} + \dfrac{1} {36} − \dfrac{1}{49} − \dfrac{1}{64} − \dfrac{1}{81} − \dfrac{1}{100} + \ldots \nonumber$
converge to 0, what do you think the previous two results tell us about the convergence status of this series?
As the example in Activity $$\PageIndex{4}$$suggests, if we have a series $$\sum a_k$$, (some of whose terms may be negative) such that $$\sum |a_k |$$ converges, it turns out to always be the case that the original series, $$\sum a_k$$ , must also converge. That is, if $$\sum |a_k |$$ converges, then so must $$\sum a_k$$ .
As we just observed, this is the case for the series (8.16), since the corresponding series of the absolute values of its terms is the convergent p-series $$\sum d\frac{1}{k^2}$$. At the same time, there are series like the alternating harmonic series $$\sum (−1)^{k+1} \dfrac{1}{k}$$ that converge, while the corresponding series of absolute values, $$\sum \dfrac{1}{k}$$, diverges. We distinguish between these behaviors by introducing the following language.
Definition
Consider a series $$\sum a_k$$
1. The series $$\sum a_k$$ converges absolutely (or is absolutely convergent) provided that $$\sum |a_k |$$ converges.
2. The series $$\sum a_k$$ converges conditionally (or is conditionally convergent) provided that $$\sum |a_k |$$ diverges and $$\sum a_k$$ converges
In this terminology, the series (Equation \ref{8.16}) converges absolutely while the alternating harmonic series is conditionally convergent.
Activity $$\PageIndex{5}$$:
a. Consider the series $$\sum (-1)^k \dfrac{\ln(k)}{k}$$.
1. Does this series converge? Explain.
2. Does this series converge absolutely? Explain what test you use to determine your answer.
b. Consider the series $$\sum (-1)^k \dfrac{\ln(k)}{k^2}$$.
1. Does this series converge? Explain.
2. Does this series converge absolutely? Hint: Use the fact that $$\ln(k) < \sqrt{k}$$ for large values of $$k$$ and the compare to an appropriate $$p$$-series.
Conditionally convergent series turn out to be very interesting. If the sequence {$$a_n$$} decreases to 0, but the series $$\sum a_k$$ diverges, the conditionally convergent series $$\sum (−1)^k a_k$$ is right on the borderline of being a divergent series. As a result, any conditionally convergent series converges very slowly. Furthermore, some very strange things can happen with conditionally convergent series, as illustrated in some of the exercises.
#### Summary of Tests for Convergence of Series
We have discussed several tests for convergence/divergence of series in our sections and in exercises. We close this section of the text with a summary of all the tests we have encountered, followed by an activity that challenges you to decide which convergence test to apply to several different series.
Geometric Series
The geometric series $$\sum ar^k$$ with ratio $$r$$ converges for $$−1 < r < 1$$ and diverges for $$|r| ≥ 1$$.
The sum of the convergent geometric series $$\sum_{k=0}^\infty ar^k$$ is $$\dfrac{a}{1-r}$$
Divergence Test
If the sequence $$a_n$$ does not converge to 0, then the series $$\sum a_k$$ diverges.
This is the first test to apply because the conclusion is simple. However, if $$\lim_{n \rightarrow \infty} a_n = 0$$, no conclusion can be drawn
Integral Test
Let $$f$$ be a positive, decreasing function on an interval $$[c, \infty]$$ and let $$a_k = f (k)$$ for each positive integer $$k ≥ c$$.
• If $$\int_{c}^{\infty} f(t) dt$$ converges, then $$\sum a_k$$ converges.
• If $$\int_{c}^{\infty} f(t) dt$$ diverges, then $$\sum a_k$$ diverges.
Use this test when $$f (x)$$ is easy to integrate.
Direct Comparison Test
See Ex 4 in Section 8.3 .
Let $$0 ≤ a_k ≤ b_k$$ for each positive integer $$k$$.
• If $$\sum b_k$$ converges, then $$\sum a_k$$ converges.
• If $$\sum a_k$$ diverges, then $$\sum b_k$$ diverges
Use this test when you have a series with known behavior that you can compare to – this test can be difficult to apply.
Limit Comparison Test
Let an and bn be sequences of positive terms. If
$\lim_{k \rightarrow \infty} \dfrac{a_k}{b_k} = L \nonumber$
for some positive finite number $$L$$, then the two series $$\sum a_k$$ and $$\sum b_k$$ either both converge or both diverge.
Easier to apply in general than the comparison test, but you must have a series with known behavior to compare. Useful to apply to series of rational functions.
Ratio test
Let $$a_k \neq 0$$ for each $$k$$ and suppose
$\lim_{k \rightarrow \infty} \dfrac{ |ak+1|}{ |ak |} = r. \nonumber$
• If $$r < 1$$, then the series $$\sum a_k$$ converges absolutely.
• If $$r > 1$$, then the series $$\sum a_k$$ diverges.
• If $$r = 1$$, then test is inconclusive.
This test is useful when a series involves factorials and powers.
Root Test
Let $$a_k ≥ 0$$ for each $$k$$ and suppose
$\lim_{k \rightarrow \infty} \sqrt[k]{a_k}=r. \nonumber$
• If $$r < 1$$, then the series $$\sum a_k$$ converges.
• If $$r > 1$$, then the series $$\sum a_k$$ diverges.
• If $$r = 1$$, then test is inconclusive.
In general, the Ratio Test can usually be used in place of the Root Test. However, the Root Test can be quick to use when $$a_k$$ involves $$k$$th powers.
Alternating Series Test
If $$a_n$$ is a positive, decreasing sequence so that $$\lim_{k \rightarrow \infty} a_n = 0$$, then the alternating series $$\sum (−1)^{k+1} a_k$$ converges.
This test applies only to alternating series – we assume that the terms $$a_n$$ are all positive and that the sequence {$$a_n$$} is decreasing.
Alternating Series Estimation Theorem
Let $$S_n = \sum_{k=1}^n(-1)^{k+1}a_k$$ be the $$n$$th partial sum of the alternating series $$\sum_{k=1}^n(-1)^{k+1}a_k$$. Assume $$a_n > 0$$ for each positive integer $$n$$, the sequence an decreases to 0 and $$\lim_{n \rightarrow \infty} S_n = S$$. Then it follows that $$|S − S_n| < a_n+1\. This bound can be used to determine the accuracy of the partial sum \(S_n$$ as an approximation of the sum of a convergent alternating series.
Activity $$\PageIndex{6}$$:
For (a)-(j), use appropriate tests to determine the convergence or divergence of the following series. Throughout, if a series is a convergent geometric series, find its sum.
1. $$\sum_{k=3}^{\infty} \dfrac{2}{\sqrt{ k − 2}}$$
2. $$\sum_{k=1}^{\infty} \dfrac{k}{1 + 2k}$$
3. $$\sum_{k=0}^{\infty} \dfrac{2k^2 + 1}{k^3 + k + 1}$$
4. $$\sum_{k=0}^{\infty} \dfrac{100^k}{k!}$$
5. $$\sum_{k=1}^{\infty}\dfrac{2^k}{5^k}$$
6. $$\sum_{k=1}^{\infty} \dfrac{k^3 − 1}{k^5 + 1}$$
7. $$\sum_{k=2}^{\infty} \dfrac{3^{k−1}{7^k}$$
8. $$\sum_{k=2}^{\infty} \dfrac{1}{k^k}$$
9. $$\sum_{k=1}^{\infty} \dfrac{ (−1)^{k+1}}{\sqrt{k+1}}$$
10. $$\sum_{k=2}^{\infty} \dfrac{1}{k \ln(k)}$$
11. Determine a value of $$n$$ so that the nth partial sum $$S_n$$ of the alternating series $$\sum_{n=2}^\infty \dfrac{(-1)^n}{\ln(n)}$$ approximates the sum to within 0.001.
### Summary
In this section, we encountered the following important ideas:
• An alternating series is a series whose terms alternate in sign. In other words, an alternating series is a series of the form
$\sum (−1)^k a_k \nonumber$
where $$a_k$$ is a positive real number for each $$k$$.
• An alternating series $$\sum_{k=1}^\infty (−1)^k a_k$$. converges if and only if its sequence {$$S_n$$} of partial sums converges, where
$S_n = \sum_{k=1}^n (−1)^k a_k. \nonumber$
• The sequence of partial sums of a convergent alternating series oscillates around and converge to the sum of the series if the sequence of $$n$$th terms converges to 0. That is why the Alternating Series Test shows that the alternating series $$\sum_{k=1}^\infty (−1)^k a_k$$ converges whenever the sequence {$$a_n$$} of nth terms decreases to 0.
• The difference between the $$n$$ − 1st partial sum $$S_{n−1}$$ and the $$n$$th partial sum $$S_n$$ of a convergent alternating series $$\sum_{k=1}^\infty (−1)^k a_k$$ is $$|S_n − S_{n-1}| = a_n$$.. Since the partial sums oscillate around the sum S of the series, it follows that
$|S − S_n| < a_n. \nonumber$
• So the $$n$$th partial sum of a convergent alternating series $$\sum_{k=1}^\infty (−1)^k a_k$$ approximates the actual sum of the series to within $$a_n$$.
### Contributors
Matt Boelkins (Grand Valley State University), David Austin (Grand Valley State University), Steve Schlicker (Grand Valley State University)<|endoftext|>
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# Different kinds of numbers
Page 1 / 2
## Different kinds of numbers
CLASS ASSIGNMENT 1
• Discover the number system step by step....
1. General: Different kinds of numbers
Provide an example of each of the following numbers:
• Natural numbers N = {..................................................}
• Counting numbers N 0 = {..................................................}
• Integers Z+ = {..................................................}
Z- = {..................................................}
• Rational numbers Q = {..................................................}
• Irrational numbers Q’ = {..................................................}
• Real numbers R = {..................................................}
2. Natural numbers
Prime numbers ={..................................................} Compound numbers ={..................................................} Definition: ........................................................................................…………..................................................…………..................................................………….......... Definition: ................................................................…………..........................................................................…………..................................................…………..........
Prime numbers + Compound numbers = Natural numbers
3. Divisibility rules
Do you recall that
In each instance, select a number that is divisible by the given divisor and try to deduce a rule for each instance.
Number Divisor Divisibility rule 3.1 2 3.2 3 3.3 4 3.4 5 3.5 6 3.6 8 3.7 9 3.8 10 3.9 11
4. Determine by which numbers (1.3.1 - 1.3.9) 61 226 is divisible and provide a reason for each.
5. Explain what you understand the following terms to mean:
5.1 Multiple:
5.2 Factor:
5.3 Prime number:
5.4 Prime factor:
5.5 Even numbers and odd numbers:
• How do you determine the factors of a number? Look at the following.....e.g. F 24 = {1; 2; 3; 4; 6; 8; 12; 24} 1 x 24; 2 x 12; 3 x 8; 4 x 6
6. Determine the factors of 48.
7. Write out all the multiples of 6 between 23 and 56.
8. Determine the prime numbers between 17 and 78.
9 Determine all odd compound numbers between 16 and 50.
10 Write down all the factors of 50 that are prime numbers.
11. Write down all the factors of 50 that are compound numbers.
12. Explain: Cube numbers. Write down the first 6 cube numbers.
13. Explain : Square numbers. Write down the first 10 square numbers.
HOMEWORK ASSIGNMENT 1
1. Write the definition for each of the following:
1.1 Rational number:
1.2 Prime number:
1.3 Compound numbers:
1.4 Prime factors:
2. Select from {0; 1; 2; 3; 4; ... ; 36} and write down:
2.1 The first two compound numbers
2.2 Odd numbers that are not prime numbers
2.3 Multiples of 6
2.4 Factors of 12
2.5 Prime factors of 12
2.6 Factors of 36
3. Which of the following numbers - 9 / 3 ; 7 / 0 ; 0; 3; -9; 16; 2 1 / 3 are:
3.1 Integers?
3.2 Rational numbers?
3.3 Non-real numbers?
4. Tabulate the following:
4.1 Natural numbers<5 ...........................................................................
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how to synthesize TiO2 nanoparticles by chemical methods
Zubear
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!<|endoftext|>
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Related Search
Web Results
brainly.com/question/17686758
The prime factorization of fifty is 2 times 25. Step-by-step explanation: So, the prime factors of 50 are 2 × 5 × 5 or 2 × 52, where 2 and 5 are the prime numbers.
researchmaniacs.com/Calculator/PrimeFactors/Prime-Factorization-of-50.html
To find all the prime factors of 50, divide it by the lowest prime number possible. Then divide that result by the lowest prime number possible. Keep doing this until the result itself is a prime number. The prime factorization of 50 will be all the prime numbers you used to divide, in addition to the last result, which is a prime number.
The prime factorization of 50 is 2 × 5 × 5. To use a factor tree to find the prime factorization of a number, x, we use the following... See full answer below.
The prime factorization of 50 is 2x5x5 or 2 x 52. The prime factorization of 18 is 2x3x3 or 2 x 32. The prime factorization of 32 is 2x2x2x2x2 or 25.
Prime factorization is expressing a positive integer as product of its prime factors. Consider, for example the number 100. The factors of 100 are: 1, 2, 4, 5, 10, 20, 50 and 100.
brilliant.org/wiki/prime-factorization
A prime factor tree provides a pictorial representation of the prime factorization for a positive integer. Starting with the given integer N N N at the top of the tree, two branches are drawn toward two positive factors of N. N. N. The process is repeated for the numbers at the end of each branch that is drawn until each "leaf" is a prime number.
www.mathportal.org/calculators/numbers-calculators/prime-factorization...
Finding prime factorization and factor tree. Example: Find prime factorization of 60. Step 1: Start with any number that divides 60, in this we will use 10. So, $\color{blue}{60 = 6 \cdot 10}$.
www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization...
The prime factorization calculator can: Calculate the prime factorization of the number you type (Numbers above 10 million may or may not time out. Calculating the prime factorization of large numbers is not easy, but the calculator can handle pretty darn big ones!) Determine whether or not a number is prime; Create sieve of Erasthones for the ...
www.wikihow.com/Find-Prime-Factorization
Here's how to find the GCF of 30 and 36, using prime factorization: Find the prime factorizations of the two numbers. The prime factorization of 30 is 2 x 3 x 5. The prime factorization of 36 is 2 x 2 x 3 x 3. Find a number that appears on both prime factorizations. Cross it out once on each list and write it on a new line.
www.dummies.com/education/math/using-prime-factorizations
Using prime factorization to find the LCM. One method for finding the least common multiple (LCM) of a set of numbers is to use the prime factorizations of those numbers. Here’s how: List the prime factors of each number. Suppose you want to find the LCM of 18 and 24. List the prime factors of each number: 18 = 2 × 3 × 3. 24 = 2 × 2 × 2 × 3
Related Search<|endoftext|>
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Brittle stars are sea creatures with five limbs and no brain. Found on the seafloor across the world, they have no obvious front, unlike humans and most other animals. Now, a new study reports that the brainless creatures are nonetheless able to move in a coordinated way, by designating one limb as the “front-facing” limb, and using two others to propel forward.
“They are pushing forward with the front two limbs, like a turtle,” said Henry Astley, the study’s author and an evolutionary biologist at Brown University. “The back limbs aren’t highly involved.”
Mr. Astley reports his findings in The Journal of Experimental Biology.
Most animals, including humans, are bilaterally symmetrical. In other words, drawing a line down the center results in symmetrical halves. A few animals, brittle stars included, are radially symmetrical: They can be sliced in many different ways and still be symmetrical, giving them no clear “front.”
But while moving, the brittle star is able to designate a front, and act as if it is bilaterally symmetrical, Mr. Astley said. The brittle star is also able to switch its front-facing limb as needed, and this enables it to swiftly change direction, Mr. Astley said.
“If we turn when we’re walking, we have to change our direction,” he said. “They just decide another direction is front, and they’re off.”
Scientists have long thought that bilateral symmetry confers an evolutionary advantage, because it allows for directed movement when searching for food or avoiding predators. But the new study puts a twist on this notion, Mr. Astley said.Continue reading the main story
“You can get the benefits of bilateral symmetry without being bilaterally symmetrical,” he said. “You can become behaviorally bilaterally symmetrical.”Continue reading the main story<|endoftext|>
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## College Algebra (10th Edition)
$\displaystyle \frac{5x}{ (x-6)(x-1)(x+4)}$
Step 1: Factor each denominator For $x^{2}+bx+c$, we search for factors of c whose sum is b: $x^{2}-7x+6 = \quad$ ... -6 and -1... $=(x-6)(x-1)$ $x^{2}-2x-24 = \quad$ ... $-6$ and $+4$... $=(x-6)(x+4)$ Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. LCM = $(x-6)(x-1)(x+4)$ Step 3: Write each rational expression using the LCM as the denominator. $\displaystyle \frac{x}{ (x-6)(x-1)}-\frac{x}{ (x-6)(x+4)}$= $= \displaystyle \frac{x(x+4)}{ (x-6)(x-1)(x+4)}-\frac{x(x-1)}{ (x-6)(x+4)(x-1)}$= $=\displaystyle \frac{x^{2}+4x-(x^{2}-x)}{ (x-6)(x-1)(x+4)}$ $=\displaystyle \frac{5x}{ (x-6)(x-1)(x+4)}$<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.1: Lines and Angles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Define parallel lines, skew lines, and perpendicular planes.
• Understand the Parallel Line Postulate and the Perpendicular Line Postulate.
• Identify angles made by two lines and a transversal.
## Review Queue
1. What is the equation of a line with slope -2 and y\begin{align*}y-\end{align*}intercept 3?
2. What is the slope of the line that passes through (3, 2) and (5, -6)?
3. Find the y\begin{align*}y-\end{align*}intercept of the line from #2. Write the equation too.
4. Define parallel in your own words.
Know What? To the right is a partial map of Washington DC. The streets are designed on a grid system, where lettered streets, A through Z\begin{align*}Z\end{align*} run east to west and numbered streets, 1st\begin{align*}1^{st}\end{align*} to 30th\begin{align*}30^{th}\end{align*} run north to south. Every state also has its own street that runs diagonally through the city.
Which streets are parallel? Which streets are perpendicular? How do you know?
If you are having trouble viewing this map, look at the interactive map: http://www.travelguide.tv/washington/map.html
## Defining Parallel and Skew
Parallel: Two or more lines that lie in the same plane and never intersect.
To show that lines are parallel, arrows are used.
Label It Say It
AB||MN\begin{align*}\overleftrightarrow{AB} || \overleftrightarrow{MN}\end{align*} Line AB\begin{align*}AB\end{align*} is parallel to line MN\begin{align*}MN\end{align*}
l||m\begin{align*}l || m\end{align*} Line l\begin{align*}l\end{align*} is parallel to line m\begin{align*}m\end{align*}.
Lines must be marked parallel with the arrows in order to say they are parallel. Just because two lines LOOK parallel, does not mean that they are.
Recall the definition of perpendicular from Chapter 1. Two lines are perpendicular when they intersect to form a 90\begin{align*}90^\circ\end{align*} angle. Below lAB¯¯¯¯¯\begin{align*}l \perp \overline{AB}\end{align*}.
In the definitions of parallel and perpendicular, the word “line,” is used. Line segments, rays and planes can also be parallel or perpendicular.
The image to the left shows two parallel planes, with a third blue plane that is perpendicular to both of them.
An example of parallel planes could be the top of a table and the floor. The legs would be in perpendicular planes to the table top and the floor.
Skew lines: Lines that are in different planes and never intersect.
In the cube:
AB¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} and FH¯¯¯¯¯¯\begin{align*}\overline{FH}\end{align*} are skew
AC¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} and EF¯¯¯¯¯\begin{align*}\overline{EF}\end{align*} are skew
Example 1: Using the cube above, list:
(a) A pair of parallel planes
(b) A pair of perpendicular planes
(c) A pair of skew lines.
Solution: Remember, you only need to use three points to label a plane. Below are answers, but there are other possibilities too.
(a) Planes ABC\begin{align*}ABC\end{align*} and EFG\begin{align*}EFG\end{align*}
(b) Planes ABC\begin{align*}ABC\end{align*} and CDH\begin{align*}CDH\end{align*}
(c) BD¯¯¯¯¯¯\begin{align*}\overline{BD}\end{align*} and CG¯¯¯¯¯\begin{align*}\overline{CG}\end{align*}
## Parallel Line Postulate
Parallel Postulate: For any line and a point not on the line, there is one line parallel to this line through the point.
There are infinitely many lines that go through A\begin{align*}A\end{align*}, but only one that is parallel to l\begin{align*}l\end{align*}.
Investigation 3-1: Patty Paper and Parallel Lines
Tools Needed: Patty paper, pencil, ruler
1. Get a piece of patty paper (a translucent square piece of paper). Draw a line and a point above the line.
2. Fold up the paper so that the line is over the point. Crease the paper and unfold.
3. Are the lines parallel?
Yes. This investigation duplicates the line we drew in #1 over the point. This means that there is only one parallel line through this point.
## Perpendicular Line Postulate
Perpendicular Line Postulate: For any line and a point not on the line, there is one line perpendicular to this line passing through the point.
There are infinitely many lines that pass through A\begin{align*}A\end{align*}, but only one that is perpendicular to l\begin{align*}l\end{align*}.
Investigation 3-2: Perpendicular Line Construction; through a Point NOT on the Line
Tools Needed: Pencil, paper, ruler, compass
1. Draw a horizontal line and a point above that line. Label the line l\begin{align*}l\end{align*} and the point A\begin{align*}A\end{align*}.
2. Take the compass and put the pointer on A\begin{align*}A\end{align*}. Open the compass so that it reaches past line l\begin{align*}l\end{align*}. Draw an arc that intersects the line twice.
3. Move the pointer to one of the arc intersections. Widen the compass a little and draw an arc below the line. Repeat this on the other side so that the two arc marks intersect.
4. Take your straightedge and draw a line from point A\begin{align*}A\end{align*} to the arc intersections below the line. This line is perpendicular to l\begin{align*}l\end{align*} and passes through A\begin{align*}A\end{align*}.
To see a demonstration of this construction, go to:
Investigation 3-3: Perpendicular Line Construction; through a Point on the Line
Tools Needed: Pencil, paper, ruler, compass
1. Draw a horizontal line and a point on that line. Label the line l\begin{align*}l\end{align*} and the point A\begin{align*}A\end{align*}.
2. Take the compass and put the pointer on A\begin{align*}A\end{align*}. Open the compass so that it reaches out horizontally along the line. Draw two arcs that intersect the line on either side of the point.
3. Move the pointer to one of the arc intersections. Widen the compass a little and draw an arc above or below the line. Repeat this on the other side so that the two arc marks intersect.
4. Take your straightedge and draw a line from point A\begin{align*}A\end{align*} to the arc intersections above the line. This line is perpendicular to l\begin{align*}l\end{align*} and passes through A\begin{align*}A\end{align*}.
To see a demonstration of this construction, go to:
Example 2: Construct a perpendicular line through the point below.
Solution: Even though the point is below the line, the construction is the same as Investigation 3-2. However, draw the arc marks in step 3 above the line.
## Angles and Transversals
Transversal: A line that intersects two other lines.
The area between l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*} is the interior.
The area outside l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*} is the exterior.
Looking at t,l\begin{align*}t, l\end{align*} and m\begin{align*}m\end{align*}, there are 8 angles formed. They are labeled below.
There are 8 linear pairs and 4 vertical angle pairs.
An example of a linear pair would be 1\begin{align*}\angle 1\end{align*} and 2\begin{align*}\angle 2\end{align*}.
An example of vertical angles would be 5\begin{align*}\angle 5\end{align*} and 8\begin{align*}\angle 8\end{align*}.
Example 3: List all the other linear pairs and vertical angle pairs in the picture above.
Solution:
Linear Pairs: 2\begin{align*}\angle 2\end{align*} and 4, 3\begin{align*}\angle 4, \ \angle 3\end{align*} and 4, 1\begin{align*}\angle 4, \ \angle 1\end{align*} and 3, 5\begin{align*}\angle 3, \ \angle 5\end{align*} and 6, 6\begin{align*}\angle 6, \ \angle 6\end{align*} and 8, 7\begin{align*}\angle 8, \ \angle 7\end{align*} and 8, 5\begin{align*}\angle 8, \ \angle 5\end{align*} and 7\begin{align*}\angle 7\end{align*}
Vertical Angles: 1\begin{align*}\angle 1\end{align*} and 4, 2\begin{align*}\angle 4, \ \angle 2\end{align*} and 3, 6\begin{align*}\angle 3, \ \angle 6\end{align*} and 7\begin{align*}\angle 7\end{align*}
There are also 4 new angle relationships.
Corresponding Angles: Two angles that are on the same side of the transversal and the two different lines. Imagine sliding the four angles formed with line l\begin{align*}l\end{align*} down to line m\begin{align*}m\end{align*}. The angles which match up are corresponding.
Above, 2\begin{align*}\angle 2\end{align*} and 6\begin{align*}\angle 6\end{align*} are corresponding angles.
Alternate Interior Angles: Two angles that are on the interior of l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*}, but on opposite sides of the transversal.
Above, 3\begin{align*}\angle 3\end{align*} and 5\begin{align*}\angle 5\end{align*} are alternate exterior angles.
Alternate Exterior Angles: Two angles that are on the exterior of l\begin{align*}l\end{align*} and m\begin{align*}m\end{align*}, but on opposite sides of the transversal.
Above, 2\begin{align*}\angle 2\end{align*} and 7\begin{align*}\angle 7\end{align*} are alternate exterior angles.
Same Side Interior Angles: Two angles that are on the same side of the transversal and on the interior of the two lines.
Above, 3\begin{align*}\angle 3\end{align*} and 5\begin{align*}\angle 5\end{align*} are same side interior angles.
Example 4: Using the picture above, list all the other pairs of each of the newly defined angle relationships.
Solution:
Corresponding Angles: 3\begin{align*}\angle 3\end{align*} and 7, 1\begin{align*}\angle 7, \ \angle 1\end{align*} and 5, 4\begin{align*}\angle 5, \ \angle 4\end{align*} and 8\begin{align*}\angle 8\end{align*}
Alternate Interior Angles: 4\begin{align*}\angle 4\end{align*} and 5\begin{align*}\angle 5\end{align*}
Alternate Exterior Angles: 2\begin{align*}\angle 2\end{align*} and 7\begin{align*}\angle 7\end{align*}
Same Side Interior Angles: 4\begin{align*}\angle 4\end{align*} and 6\begin{align*}\angle 6\end{align*}
Example 5: For the picture below, determine:
(a) A corresponding angle to 3\begin{align*}\angle 3\end{align*}?
(b) An alternate interior angle to 7\begin{align*}\angle 7\end{align*}?
(c) An alternate exterior angle to 4\begin{align*}\angle 4\end{align*}?
Solution:
(a) 1\begin{align*}\angle 1\end{align*}
(b) 2\begin{align*}\angle 2\end{align*}
(c) 5\begin{align*}\angle 5\end{align*}
Know What? Revisited For Washington DC, all of the lettered and numbered streets are parallel. The lettered streets are perpendicular to the numbered streets. We do not have enough information about the state-named streets to say if they are parallel or perpendicular.
## Review Questions
• Questions 1-3 use the definitions of parallel, perpendicular, and skew lines.
• Question 4 asks about the Parallel Line Postulate and the Perpendicular Line Postulate.
• Questions 5-9 use the definitions learned in this section and are similar to Example 1.
• Questions 10-20 are similar to Examples 4 and 5.
• Question 21 is similar to Example 2 and Investigation 3-2.
• Questions 22-30 are Algebra I review.
1. Which of the following is the best example of parallel lines?
2. Lamp Post and a Sidewalk
3. Longitude on a Globe
4. Stonehenge (the stone structure in Scotland)
2. Which of the following is the best example of perpendicular lines?
1. Latitude on a Globe
2. Opposite Sides of a Picture Frame
3. Fence Posts
4. Adjacent Sides of a Picture Frame
3. Which of the following is the best example of skew lines?
1. Roof of a Home
2. Northbound Freeway and an Eastbound Overpass
3. Longitude on a Globe
4. The Golden Gate Bridge
4. Writing What is the difference between the Parallel Line Postulate and the Perpendicular Line Postulate? How are they similar?
Use the figure below to answer questions 5-9. The two pentagons are parallel and all of the rectangular sides are perpendicular to both of them.
1. Find two pairs of skew lines.
2. List a pair of parallel lines.
3. List a pair of perpendicular lines.
4. For AB¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}, how many perpendicular lines would pass through point V\begin{align*}V\end{align*}? Name this line.
5. For XY¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}, how many parallel lines would pass through point D\begin{align*}D\end{align*}? Name this line.
For questions 10-16, use the picture below.
1. What is the corresponding angle to 4\begin{align*}\angle 4\end{align*}?
2. What is the alternate interior angle with 5\begin{align*}\angle 5\end{align*}?
3. What is the corresponding angle to 8\begin{align*}\angle 8\end{align*}?
4. What is the alternate exterior angle with 7\begin{align*}\angle 7\end{align*}?
5. What is the alternate interior angle with 4\begin{align*}\angle 4\end{align*}?
6. What is the same side interior angle with 3\begin{align*}\angle 3\end{align*}?
7. What is the corresponding angle to 1\begin{align*}\angle 1\end{align*}?
Use the picture below for questions 17-20.
1. If m2=55\begin{align*}m\angle 2 = 55^\circ\end{align*}, what other angles do you know?
2. If m5=123\begin{align*}m\angle 5 = 123^\circ\end{align*}, what other angles do you know?
3. If tl\begin{align*}t \perp l\end{align*}, is tm\begin{align*}t \perp m\end{align*}? Why or why not?
4. Is l||m\begin{align*}l || m\end{align*}? Why or why not?
5. Construction Draw a line and a point not on the line. Construct a perpendicular line to the one your drew.
Algebra Review Find the slope of the line between the two points, y2y1x2x1\begin{align*}\frac{y_2-y_1}{x_2-x_1}\end{align*}.
1. (-3, 2) and (-2, 1)
2. (5, -9) and (0, 1)
3. (2, -7) and (5, 2)
4. (8, 2) and (-1, 5)
5. Find the equation of the line from #22. Recall that the equation of a line is y=mx+b\begin{align*}y = mx + b\end{align*}, where m\begin{align*}m\end{align*} is the slope and b\begin{align*}b\end{align*} is the y\begin{align*}y-\end{align*}intercept.
6. Find the equation of the line from #23.
7. Find the equation of the line from #24.
8. Is the line y=x+3\begin{align*}y=-x+3\end{align*} parallel to the line in #26? How do you know?
9. Is the line y=x+3\begin{align*}y=-x+3\end{align*} perpendicular to the line in #26? How do you know?
1. y=2x+3\begin{align*}y = -2x + 3\end{align*}
2. m=6253=82=4\begin{align*}m = \frac{-6-2}{5-3} = \frac{-8}{2} = -4\end{align*}
3. 2=4(3)+b 2=12+b14=b y=4x+14\begin{align*}{\;} \ \ \ 2 = -4(3) + b\!\\ {\;} \ \ \ 2 = -12 + b\!\\ -14 = b\!\\ {\;} \ \ \ y = -4x + 14\end{align*}
4. Something like: Two lines that never touch or intersect and in the same plane. If we do not say “in the same plane,” this definition could include skew lines.
8 , 9 , 10
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Oct 28, 2015
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# Linear Equations in One Variable with Formula, Rules and Solved Examples
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Linear equations in one variable are the first basic topic that a candidate must know to solve questions of Algebra. Algebra is one of the most important topics in the Quantitative Aptitude section of any government examination. Hence, it is vital to be clear about the basics of linear equations before starting your preparations.
## What are Linear Equations in One Variable?
Linear equations in one variable is an algebraic equation of equality that involves only one variable, the greatest exponent of which is 1. There is a single root or solution for every linear equation in one variable. This root or solution is a value that when replaced with the variable makes the equation true.
Example: Consider the linear equation in one variable: $$4x+6=14$$. Can you tell what is the number that should replace ‘x’ so that the given equation results in a true statement?
The linear equation will result in a true statement when the value of ‘x’ is 2.
$$4*2+6=8+6=14$$
The linear equation will not result in a true statement if the value of ‘x’ is taken other than 2. Thus, the given equation 4x + 6 = 14 results in a true statement only when ‘x’ is replaced by 2. In this case, we call the number 2, the root or the solution of equation 4x + 6 = 14.
## Standard Form of Linear Equations in One Variable
Standard form of linear equations in one variable is written as ax+b=0, (a is not equal to zero) where x is the one variable while a and b are any real numbers.
Therefore, the formula of the linear equations in one variable is formulated as ax + b = 0, such that ‘a’ and ‘b’ are not equivalent to zero. When a graph is drawn for such an equation it seems to be a straight line either horizontally or vertically.
## Linear Equations in One Variable Formula
A linear equation is a straightforward approach to represent a mathematical statement. Moreover, solving a linear equation comprises a collection of simple methods. For this, we separate the variables on one side of the equation and the constants on another side of the given equation and conclude the final value of the unknown quantity/elements.
The important formulas of linear equations in one variable are:
Term Linear Equations in one Variable Formula Distributive Property x(y+z)=xy+xz Commutative Property x+y=y+x (Addition)x.y=y.x (Multiplication) Associative Property x+(y+z)=(x+y)+z (Addition)x.(y.z)=(x.y).z (Multiplication) Identity Property x+0=x (Additive)x.1=x (Multiplicative) Inverse Property x+(-x)=0 (Additive)x.(1/x)=1 (Multiplicative)
## How to Solve Linear Equations in One Variable?
We can solve linear equations in one variable by following the below steps:
• Step 1 – Clear the equation of fractions (or decimals) by multiplying both sides by the LCM of the denominators (or by a power of 10 in the case of decimals).
• Step 2 – Remove group symbols from both sides, if any, and simplify the equation.
• Step 3 – Bring all variable terms on one side and all the constants on the other. Combine the like terms.
• Step 4 – Divide the terms into both sides by the numerical coefficient of the variable.
Let us understand how to solve linear equations in one variable with the help of the following example:
Example: Consider the following equation and solve for ‘x’ $$\frac{x}{4}=28-\frac{x}{3}$$
Step 1. Clear the fraction by multiplying both sides by the LCM of denominators (4 and 3), which is 12.
$$12\frac{x}{4}=12*28-12\frac{x}{3}$$
Step 2. Simplify the equation
$$3x=336-4x$$
Step 3. Isolate all variable terms on one side and all the constants on the other and combine the like terms.
$$3x+4x=336$$
$$7x=336$$
Step 4. Divide the terms into both sides by the numerical coefficient of the variable, which is 7.
$$x=48$$
When you’ve finished with linear equations in one variable, you can learn about Linear Equations in Two variables
## Rules for Solving Linear Equations in One Variable
In these types of linear equations in one variable questions, a problem (or situation) will be given in verbal language involving certain numbers. You will have to find the solution to the problem by translating words into a linear equation.
There are some specific rules for solving linear equations in one variable which are mentioned as follows:
• Rule 1: Adding the same number (positive or negative) to both sides of an equation, then the resulting equation will be equivalent to the original equation.
• Rule 2: Multiplying or dividing both sides of an equation by the same non-zero number, then the resulting equation will be equivalent to the original equation.
• Rule 3: Transposing a term from one side to the other side of the equation, then the resulting equation will be equivalent to the original equation, but remember that the sign of the transposed term changes.
### Difference Between Linear Equation in One Variable and Non-Linear Equations
So far we have seen the linear equations in one variable definition and how to solve them. Now let us also learn about non-linear equations. As even non-linear equations have multiple applications in geometry, trigonometry, as well in calculus.
The following table shows the comparison of linear equations in one variable vs non-linear equations:
Linear Equations in One Variable Non-Linear Equations A linear equation in one variable is a single-degree equation that is represented by a line on a coordinate plane. A non-linear equation in contrast to a linear one is represented by a curve on the coordinate axis. Linear equation in one variable is of the form ax+b=0. A nonlinear equation is of a higher degree. The equation of curves such as parabola, circle, ellipse, hyperbola are examples of non-linear equations. Examples of linear equations in one variable:x = 8, 6x + 11 = 19, 2x + 4y = 22 Examples of non-linear equations:$$x^2 + y^2= 16$$, $$\frac{x^2}{8}+\frac{y^2}{25}=1$$
## Solved Examples of Linear Equations in One Variable
The following are a few examples of questions that can come in examinations, related to Linear Equations in one variable.
Example 1. The sum of the two numbers is 45 and their ratio is 2 : 3. Find the numbers.
Solution: Let x be one of the numbers. Then, the other number is 45-x.
Since the two numbers are in the ratio 2 : 3, we have
$$\frac{x}{45-x}=\frac{2}{3}$$
$$3x=2(45-x)$$
$$3x=90-2x$$
$$5x=90$$
$$x=\frac{90}{5}$$
$$x=18$$
Thus, one number is 18 and the other number is 45-18=27.
Example 2. Solve the following equations for $$x :$$ $$4(x+5)=14x+50$$
Solution: $$4x+20=14x+50$$ [Removing the brackets]
$$4x-14x=50-20$$ [Transposing like terms to same side]
$$-10x=30$$[Dividing both sides by -10]
$$x= -3$$
Example 3. Solve the following equations for $$x :$$ $$2(x-1)-3(x-2)=4(x-5)$$
Solution: $$2x-2-3x+6=4x-20$$
$$2x-3x-4x=2-6-20$$
$$-5x= -24$$
$$x=245$$
Example 4. Solve the following equations for $$x :$$ $$\frac{3x}{8}+\frac{1}{4}=\frac{x-3}{3}$$
Solution: $$24(\frac{3x}{8}+\frac{1}{4})=24\frac{x-3}{3}$$
$$9x+6=8x-24$$
$$9x-8x= -6-24$$
$$x = -30$$
Example 5. Solve the following equations for $$x :$$ $$\frac{5}{x}=\frac{9}{x-8}$$
Solution: Multiplying throughout by $$x(x – 8)$$, LCM of x and (x – 4), we get
$$5(x-8)=9x$$
$$5x-40=9x$$
$$5x-9x=40$$
$$-4x=40$$
$$x= -10$$
Example 6. Solve the following equations for $$x :$$ $$\frac{x+b}{a+b}=\frac{x-b}{a-b}$$
Solution: Multiplying throughout by $$(a + b)(a – b)$$, LCM of the denominators
$$x(a-b)+b(a-b)=a(x-b)+b(x-b)$$
$$xa-xb+ba-b^2=ax-ab+bx-b^2$$
$$ax-bx-ax-bx= -ab-b^2-ab+b^2$$
$$- 2bx= -2ab$$
$$x=a$$
Example 7. A woman is five times as old as her daughter. In two years’ time, she will be four times as old as her daughter. Find their present ages.
Solution: Let the present age of the daughter be $$x$$ years.
Then, the present age of woman = $$5x$$ years
After two years,
Daughter’s age = $$(x+2)$$ years
Woman’s age = $$(5x+ 2)$$years
According to the question,
$$5x+2=4(x+2)$$
$$5x+2=4x+8$$
$$5x-4x=8-2$$
$$x=6$$
Thus, the daughter’s present age is 6 years and the woman’s present age is (56) years, i.e., 30 years
Example 8. A number consists of two digits whose sum is 9. If 9 is subtracted from the number its digits are interchanged. Find the number.
Solution: Let the unit digit of the number be x.
Then, the tens digit =9-x
Therefore, Number =$$10(9-x)+x$$
$$=90-10x+x$$
$$=90-9x$$
Number with interchanged digits =10 x+9-x
$$=9x+9$$
According to the question,
$$(90-9x)-9=9x+9$$
$$81-9x=9x+9$$
$$9x+9x=81-9$$
$$18x=72$$
$$x=4$$
Therefore, the units digit $$=4$$ and the tens digit $$=9-4=5$$
Hence, the number is $$54$$.
Example 9. Rahul has 3 boxes of different vegetables. Box A weighs 2.50 kg more than box B and box C weighs 10.25 kg more than box B. The total weight of the three boxes is 48.75 kg. How much does box A weighs?
Solution:
Let the weight of box B be x kg.
Then,
• Weight of box A = $$(x+2.50)$$kg
• Weight of box C = $$(x+10.25)$$kg
According to the question,
$$(x+2.50)+x+(x+10.25)=48.75$$
$$x+x+x=48.75-2.50-10.25$$
$$3x=36$$
$$x=12$$
Therefore, Weight of box A =(x+2.50) kg=14.50 kg.
Example 10. The distance between the two stations is 340 km. two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/h. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.
Solution: Let the speed of the train starting from A be $$x$$ km/h.
Therefore, The speed of the second train starting from B will be $$(x + 5)$$ km/h.
The first train reaches C and the second train reaches D in 2 hours.
According to the question,
$$x2+30+(x+5)2=340$$
$$2x+30+2x+10=340$$
$$2x+2x=340-40$$
$$4x=300$$
$$x=75$$
Therefore, The speeds of the trains are 75 km/h and 80 km/h respectively.
We hope you found this article useful, informative and helpful in clearing all your doubts regarding linear equations in one variable and how to solve questions regarding the same.
If you are checking Linear Equations in One Variables article, check related maths articles: Linear Equations in Two Variable Pair of Linear Equations Solve Linear Differential Equation Linear Programming System of Equations Linear Functions
## FAQs on Linear Equations in One Variable
Q.1 What is a linear equation in one variable?
Ans.1 Linear equation in one variable is an algebraic equation of equality that involves only one variable, the greatest exponent of which is 1. It is represented as ax + b = 0, where a is not equal to 0.
Q.2 What is the formula for linear equations in one variable?
Ans.2 The formula for linear equations in one variable is ax + b = 0, where a is not equal to 0. It is also known as the standard form of linear equation in one variable.
Q.3 Can a linear equation include more than one variable?
Ans.3 A linear equation can contain more than one variable. Such equations are known as linear equations in two variables or linear equations in three variables depending on the number of variables.
Q.4 Can I get more than one answer for a linear equation in one variable?
Ans.4 No, there is only one solution (or root) to a linear equation in one variable.
Q.5 What is the power of the variable in a linear equation in one variable?
Ans.5 In a one-variable linear equation, the power of the variable is one.<|endoftext|>
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# Презентация
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```Mathematics & English
The 1st sign of equality of
triangles
Theorem on equality of triangles
If 2 sides and the angle between them
of the same triangle are accordingly equal to
2 sides and the angle between them of the
other triangle, then these triangles are equal.
B
A
C
B1
A1
C1
Given:
AB is equal to A1B1
AC is equal to A1C1
The angle A is equal to
the angle A1
To be to prove that:
The triangle ABC is
equal to the triangle
A1B1C1
B
A
C
B1
A1
C1
Proof
As the angle A is equal to the angle A1
then the triangle ABC can be put on
the triangle A1B1C1 so that the apex
A will be superposed with the apex
A1, and the sides AB and AC will be
accordingly put on the rays A1B1 and
A1C1. As far as, AB is equal to A1B1,
AC is equal to A1C1, then the side AB
will be superposed with the side A1B1
and the side AC with A1C1. In
particular the points B and B1, C and
C1 will be superposed. Therefore, the
sides BC and B1C1 will be superposed.
So the triangles ABC and A1B1C1 will
be completely superposed. That
means they are equal.
The theorem has been proved.
1. To proof the equality of the triangles AOB and
COD
D
Proof:
A
1) AO is equal to OC ( according to the condition)
2) BO is equal to OD (according to the condition)
3) The angle AOB is equal to the triangle COD ( as
vertical angles)
O
B
C
Therefore the triangle AOB is equal to the COD
(according to the first sing of triangles).
2. To proof the equality of the triangles AOB and
A
C
COD
K
B
O
D
P
Proof:
1)BA is equal to CD ( according to the condition )
2) BO is equal to OD ( according to the condition )
3)The angle ABО is equal to the angle CDO (as their
adjacent angles are equal )
Therefore the triangle AOB is equal to the triangle
COD ( according to the first sign of equality of
triangles).
• The presentation has been done by:
Mikhail Verbitskiy, Yana Karasyova,
7A form, school №511
• L.I. Kunaeva, Maths teacher
• V.V. Klyueva, English teacher
Moscow
2011
Библиография
1. Атанасян, Л.С. Геометрия: Учеб. для 7-9 кл. сред.
школы [Текст]/ Л.С. Атанасян, В.Ф. Бутузов, С.Б.
Кадомцев и др. – М.: Просвещение; ОАО «Моск.
учебн.» 2006. – 384с.
2. Рабинович, Е.М. Задачи и упражнения на готовых
чертежах. 7-9 классы. Геометрия [Текст]/ Е.М.
Рабинович. – М.:Илекса, 2006.-60с.
Подготовили:
• Ученики 7 «А» класса ГОУ СОШ № 511 Вербицкий Михаил и
Карасева Яна
• Учитель математики: Кунаева Л.И.
• Учитель английского языка: Клюева В.В.
```
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11 October 1995
The Royal Swedish Academy of Sciences has decided to award the 1995 Nobel Prize in Chemistry to
Professor Paul Crutzen, Max-Planck-Institute for Chemistry, Mainz, Germany (Dutch citizen),
Professor Mario Molina, Department of Earth, Atmospheric and Planetary Sciences and Department of Chemistry, MIT, Cambridge, MA, USA and
Professor F. Sherwood Rowland, Department of Chemistry, University of California, Irvine, CA, USA
for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone.
The ozone layer – The Achilles heel of the biosphere
The atmosphere surrounding the earth contains small quantities of ozone – a gas with molecules consisting of three oxygen atoms (O3). If all the ozone in the atmosphere were compressed to a pressure corresponding to that at the earth’s surface, the layer would be only 3 mm thick. But even though ozone occurs in such small quantities, it plays an exceptionally fundamental part in life on earth. This is because ozone, together with ordinary molecular oxygen (O2), is able to absorb the major part of the sun’s ultraviolet radiation and therefore prevent this dangerous radiation from reaching the surface. Without a protective ozone layer in the atmosphere, animals and plants could not exist, at least upon land. It is therefore of the greatest importance to understand the processes that regulate the atmosphere’s ozone content.
Paul Crutzen, Mario Molina and Sherwood Rowland have all made pioneering contributions to explaining how ozone is formed and decomposes through chemical processes in the atmosphere. Most importantly, they have in this way showed how sensitive the ozone layer is to the influence of anthropogenic emissions of certain compounds. The thin ozone layer has proved to be an Achilles heel that may be seriously injured by apparently moderate changes in the composition of the atmosphere. By explaining the chemical mechanisms that affect the thickness of the ozone layer, the three researchers have contributed to our salvation from a global environmental problem that could have catastrophic consequences.
How this knowledge evolved
Ozone is formed in the atmosphere through the splitting of ordinary oxygen molecules (O2) by ultra-violet radiation from the sun. The oxygen atoms thereby liberated react with the molecular oxygen according to:
O2+ uv-light -> 2O
O+O2+M -> O3+M
where M is a random air molecule (N2 or O2).
The English physicist Sidney Chapman formulated in 1930 the first photochemical theory for the formation and decomposition of ozone in the atmosphere. This theory, which describes how sunlight converts the various forms of oxygen from one to another, explains why the highest contents of ozone occur in the layer between 15 and 50 km, termed the ozone layer (Fig. 1). Later measurements, however, showed appreciable deviations from Chapman’s theory. The calculated ozone contents were considerably higher than the observed ones. Thus, there must be other chemical reactions contributing to the reduction of the ozone content. Some years later the Belgian Marcel Nicolet contributed important knowledge of how the decomposition of ozone was enhanced by the presence of the hydrogen radicals OH and HO2.
The scientist to take the next fundamental step towards a deeper understanding of the chemistry of the ozone layer was Paul Crutzen. In 1970 he showed that the nitrogen oxides NO and NO2 react catalytically (without themselves being consumed) with ozone, thus accelerating the rate of reduction of the ozone content.
NO + O3 -> NO2 + O2
NO2+O -> NO+O2
O3+uv-light -> O2+O
Net result: 2O3 -> 3O2
These nitrogen oxides are formed in the atmosphere through the decay of the chemically stable nitrous oxide N2O, which originates from microbiological transformations at the ground. The connection demonstrated by Crutzen between microorganisms in the soil and the thickness of the ozone layer is one of the motives for the recent rapid development of research on global biogeochemical cycles.
The first threat noted: supersonic aircraft
The power of nitrogen oxides to decompose ozone was also noted early by the American researcher Harold Johnston, who carried out extensive laboratory studies of the chemistry of nitrogen compounds. In 1971 he pointed out the possible threat to the ozone layer that the planned fleet of supersonic aircraft and supersonic travel (SST) might represent. These aircraft would be capable of releasing nitrogen oxides right in the middle of the ozone layer at altitudes of 20 km. Crutzen’s and Johnston’s work gave rise to a very intensive debate among researchers as well as among technologists and decision-makers. This was also the start of intensive research into the chemistry of the atmosphere which has made great progress during the past several years. (The subsequent cancellation of plans for a large SST fleet had other reasons than the environmental risks they involved.)
Spray cans and refrigerators damage the ozone layer
The next leap in our knowledge of ozone chemistry was in 1974, when Mario Molina and Sherwood Rowland published their widely noted Nature article on the threat to the ozone layer from chlorofluorocarbon (CFC) gases – “freons” – used in spray bottles, as the cooling medium in refrigerators and elsewhere and plastic foams. Molina and Rowland based their conclusions on two important contributions by other researchers:
– James Lovelock (England) had recently developed a highly sensitive device of measuring extremely low organic gas contents in the atmosphere, the electron capture detector. Using this he could now demonstrate that the exclusively man-made, chemically inert, CFC gases had already spread globally throughout the atmosphere.
– Richard Stolarski and Ralph Cicerone (USA) had shown that free chlorine atoms in the atmosphere can decompose ozone catalytically in similar ways as nitrogen oxides do.
Molina and Rowland realised that the chemically inert CFC could gradually be transported up to the ozone layer, there to be met by such intensive ultra-violet light that they would be separated into their constituents, notably chlorine atoms. They calculated that if human use of CFC gases was to continue at an unaltered rate the ozone layer would be depleted by many percent after some decades. Their prediction created an enormous attention. For the CFC gases were used in many technical processes and their very chemical stability and non-toxicity were thought to render them environmentally ideal. Many were critical of Molina’s and Rowland’s calculations but yet more were seriously concerned by the possibility of a depleted ozone layer. Today we know that they were right in all essentials. It was to turn out that they had even underestimated the risk.
Ozone content over Antarctica
Molina’s and Rowland’s report led to certain restrictions on CFC release during the late 1970s and early 1980s. Not until 1985, when the real shock came, was there any real urgency in the international negotiations on release restrictions. Then the Englishman Joseph Farman and his colleagues noted a drastic depletion of the ozone layer over the Antarctic, the “ozone hole” (Fig. 2). The depletion was, at least periodically, far greater than expected from earlier calculations of the CFC effect. The debate among researchers now intensified. Was this a natural climatic variation or was it chemical decomposition brought about by mankind? Thanks to pioneering research by many researchers, among them Crutzen, Molina and Rowland, as well as Susan Solomon and James Anderson, both from the USA, the picture has now cleared. The depletion is caused chiefly by ozone reacting chemically with chlorine and bromine from industrially manufactured gases.
|Fig 2. Thickness of the ozone layer (mean monthly value for October) over Halley Bay, Antarctica. Note the drastic depletion since the end of the 1970s.|
The surprisingly rapid depletion of the ozone layer over Antarctica could not be explained by transport processes or by gas phase chemical reactions. An alternative mechanism must exist which could accelerate the decomposition of ozone. Crutzen and colleagues identified this mechanism as chemical reactions on the surface of cloud particles in the stratosphere. Thus, the Antarctic ozone depletion appears to be connected with the extremely low prevailing temperatures, which lead to condensation of water and nitric acid to form “polar stratospheric clouds” (PSCs). The ozone-decomposing chemical reactions are greatly reinforced by the presence of cloud particles. This understanding has led to an exciting new branch of atmospheric chemistry: “heterogeneous” chemical reactions on particle surfaces.
The ozone layer and the climate
The ozone problem also has interesting connections with the issue of how mankind is affecting the climate. Ozone, like carbon dioxide and methane, is a greenhouse gas that contributes to high temperatures at the surface of the earth. (CFC gases have a similar effect). Model calculations have shown that the climate is specially sensitive to changes in the ozone content in the lower layers, the troposphere. Here the ozone content has increased markedly during the past century, chiefly because of the release of nitric oxide, carbon monoxide and gaseous hydrocarbons from vehicles and industrial processes and from the combustion of biomass in the tropics. The elevated ozone content in lower atmospheric layers is itself an environmental problem for the damage it can cause to crops and human health. Paul Crutzen has been the world’s leading researcher in mapping the chemical mechanisms that determine the ozone content at these levels.
What can we expect in the future?
Thanks to our good scientific understanding of the ozone problem – and very largely to Crutzen, Molina and Rowland – it has been possible to make far-reaching decisions on prohibiting the release of gases that destroy ozone. A protocol on the protection of the ozone layer was negotiated under the auspices of the United Nations and signed in Montreal, Canada, in 1987. Under the latest tightening-up of the Montreal Protocol, the most dangerous gases will be totally banned from 1996 (developing countries have a few years’ grace to introduce substitutes that do not harm the ozone layer). Since it takes some time for the ozone-destroying gases to reach the ozone layer we must expect the depletion, not only over Antarctica but also over parts of the Northern Hemisphere, to worsen for some years to come. Given compliance with the prohibitions, the ozone layer should gradually begin to heal after the turn of the century (Fig. 3). Yet it will take at least 100 years before it has fully recovered.
|Dotto, L. and Schiff, H. I., The ozone war, Doubleday, Garden City, New York, 1978.|
|Graedel, T. E. and Crutzen, P.J., Atmosphere, climate and change. Scientific American Library, 1995.|
|Gribbin, J., The hole in the sky, Corgi Books, London, 1988.|
|Ozonskiktet, Faktablad från Naturvårdsverket, Februari 1992.|
|Rodhe, H. och Bolin, B., Luftföroreningar, I Tidens miljöbok – En forskarantologi, Tidens förlag, 1992.|
|Rowland, F. S. and Molina, M. J., Ozone depletion: 20 years after the alarm, Chemical and Engineering News 72, 8-13, 1994.|
|Scientific assessment of ozone depletion 1994, WMO Report 37, World Meteorological Organization and United Nations Environment Programme, Geneva, 1995.|
|Toon, O. B. and Turco, R. P., Polar stratospheric clouds and ozone depletion, Scientific American 264, 68-74, 1991.|
|Wayne, R. P. Chemistry of Atmospheres, Oxford Science Publications, 1993.|
|WMO and the ozone issue, World Meteorological Organization, Report 778, 1992.|
Paul Crutzen was born in 1933 in Amsterdam. Dutch citizen. Doctor’s degree in meteorology, Stockholm University, 1973. Member of the Royal Swedish Academy of Sciences, the Royal Swedish Academy of Engineering Sciences and Academia Europaea.
Professor Paul Crutzen
Max-Planck-Institute for Chemistry
P.O. Box 3060
D-55020 Mainz, Germany.
Mario Molina was born in 1943 in Mexico City, Mexico. PhD in physical chemistry, University of California, Berkeley. Member of the US National Academy of Sciences.
Professor Mario Molina
Department of Earth, Atmospheric and Planetary Sciences
MIT 54 – 1312
Cambridge MA 02139, USA
F. Sherwood Rowland was born in Delaware, Ohio, USA, 1927. Doctor’s degree in chemistry, University of Chicago, 1952. Member of the American Academy of Arts and Sciences and of the US National Academy of Sciences, where he is currently Foreign Secretary.
Professor F. Sherwood Rowland
Department of Chemistry
University of California
Irvine CA 92717, USA<|endoftext|>
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Periodic table concept map. The periodic table is useful for understanding atomic properties that show periodic trends. Atomic structure and the periodic table 27 db most elements are metals. Choose from 201 different sets of periodic table concept map flashcards on quizlet. Periodic table and periodicity.
As mentioned earlier the higher the shell number the farther from the nucleus the electrons in that shell are likely to be. Physical science concept map. Choose from 213 different sets of concept map periodic table flashcards on quizlet. Quickly memorize the terms phrases and much more.
Alkaline earth metals 13. In other words the size of an atom is generally determined by the number of the valence electron shell. 66 revising the periodic table feedback for students a sample answer sheet for students revision summary for the periodic table is provided. Learn concept map periodic table with free interactive flashcards.
One such property is the atomic radius figure pageindex5. The periodic table can be reviewed and new elements can be added but only added after rigorous scientific examination. Periodic table by cp4 group2 1. The international union of pure and applied chemistry iupac verifies the additions of new elements and at the end of 2015 the 7th period of the periodic table of elements was completed.
Learn periodic table concept map with free interactive flashcards. Periodic table and periodicity.<|endoftext|>
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It was in 1952 in Bangladesh, when four students sacrificed their lives fighting the battle to recognize Bengali as the national language of Bangladesh. This instance led to declaring February 21st as the ‘Bengali Language Movement Day’ in Bangladesh. The initiative to recognize February 21st as International Mother Language Day was further passed by UNESCO General Assembly in 2000 to strengthen the linguistic roots and promote and encourage linguistic diversity.
Knowing one’s mother tongue helps preserve the culture and traditions of a community. Therefore, the theme for International Mother Language Day 2019 is:
‘Indigenous languages as a factor in development, peace, and reconciliation’.
Further, 2019 has been declared as the ‘International Year of Indigenous Languages’. One of the basic ideas to recognize and celebrate the International Mother Language Day is to empower the locals with their native language.
It helps make knowledge, education and learning accessible to the people who otherwise have no access to education as one of the biggest hindrances to access the information available is the barrier of language.
Initiatives taken in the same direction:
- There have been ample initiatives by the government of different countries and states to provide access to education and exposure to the outside world to the remote areas in the native language.
- Many large organizations like Google LLC are working towards developing and implementing initiatives like Google GLOW (Growing Local Language Content on Wikipedia).
- It will help avail and access information by the people around the world in their native language, keeping in mind their culture, ethnicity and traditional values.
- The launch of Digital Literacy Library by Facebook in six Indian languages-Hindi, Bengali, Telugu, Tamil, Kannada and Malayalam has also made information reach millions of people who are not well-versed with English, as initially English was the predominant language of Facebook.
- In many scientific researches and studies, the knowledge of the indigenous language and culture has helped develop and find new ways to cure many illnesses and to better understand the ecology of a place.
Here are some inputs about the languages of India:
- As per Schedule 8 of Constitution of India, there are 22 official languages of India. Though the exact number of languages spoken in the country is more in number and is still contradictory.
- Article 350A of Constitution of India, encourages the use of mother tongue for the mode of instruction for the primary school.
We should, therefore, pledge to keep alive our native language and culture in the following manner:
- Encouraging children to learn and speak their mother tongue with other languages.
- This would also help imbibe in the children receptiveness for linguistic diversity.
- Also make education accessible for all, irrespective of their regional and linguistic differences.
- Let language be a bridge to bring together people from all over and encourage harmony.<|endoftext|>
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Many science teachers perform a demonstration in which they immerse sealed cans of Coke and Diet Coke in a tank of water. The can of Coke immediately sinks, while the can of Diet Coke floats. Some instructors have adapted this demonstration as an experiment to teach students the principles of density and buoyancy. The experiments allow students to conduct measurements so that they can devise a scientific explanation for the observed behavior.
Density represents the relationship between a substance’s mass and the amount of space it occupies, i.e., its volume. Scientists normally express density in metric units of grams per milliliter, or g/mL. Pure water exhibits a density of 1.00 g/mL. Most other liquids exhibit densities between 0.8 and 1.3 g/mL.
Composition of Coke
Coca Cola consists mostly of water. But it also contains 39 g of sugars, primarily in the form of high-fructose corn syrup, or HFCS. The HFCS used by Coca-Cola and most other soft drink manufacturers exhibits a relative sweetness of 160 compared to the baseline sweetness of 100 exhibited by sucrose, or table sugar. This relative scale reflects perceived sweetness, which means that HFCS literally tastes 1.6 times sweeter than an equal amount of sucrose. As a result, a drink sweetened with HFCS requires less sweetener to achieve the desired level of sweetness than a drink sweetened with sucrose.
Composition of Diet Coke
Diet Coke contains an artificial sweetener: aspartame. Aspartame exhibits a perceived sweetness of 180 times that of sugar. Or, for a more direct comparison, aspartame exhibits a sweetness 113 times greater than high fructose corn syrup. As a consequence, Diet Coke only requires 0.35 grams of aspartame to achieve the same level of sweetness as the 39 g of HFCS in Coke.
Whether an object sinks or floats depends on its buoyancy. An object placed in water exerts a downward force on the water. The water, however, pushes back. Archimedes’ principle states that the buoyant force exerted by water or any other fluid on an object is equal to the weight of the water displaced by the object. If the weight of displaced water exceeds the weight of the object, the object floats. Otherwise, it sinks. This, in part, explains why metal ships float. It also explains why the can of Diet Coke sinks. Because the two cans exhibit identical shapes and sizes, they displace equal amounts of water when submerged. But the fact that the can of Coke sinks means it must weigh more than the amount of water it displaces, whereas the can of Diet Coke weighs less.
Both the Diet Coke and Coke cans contain 12 fluid oz., or 355 milliliters, of liquid. Both beverages consist primarily of water. The primary difference lies in the sweetener. Coke contains about 325 mL of water, with a density of 1.0 g/mL, and 39 g of sugar. The contents of the can therefore weigh 325 g + 39 g = 364 g. The can of Diet Coke, however, contains only 0.3 g of aspartame. It therefore consists almost entirely of water and the can’s contents therefore weigh about 355 g. This difference in weight makes the can of Diet Coke sufficiently buoyant to float. In terms of density, the density of Diet Coke is roughly 1.00 g/mL, the same as water. The Coke, however, exhibits a density of 1.03 g/mL.<|endoftext|>
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Everyone from popular-science communicators to politicians frequently emphasises the importance of mathematical education.
While this may seem like common knowledge in the abstract, many people (particularly mathematics students of all ages) cannot help but wonder specifically when and where, why and how, this is the case. After all, the average individual tends not to consciously perform numerical equations on the street or at the supermarket.
In truth, most everyday examples aren’t glaringly obvious, and yet mathematical patterns form the essential structure underlying all things.
Helping Students See Maths Everywhere
From networks of urban streets to several forms of drawing and painting to the growth of a tree, everything is governed by patterns. While numbers are crucial, the true soul of maths is patterns: In language terms, if numbers are the words and letters forming the fundamental language of reality, then patterns are the sentiments expressed.
Many students may think they’re uninterested in this rich field, and on the surface, a number of them are. However, people who have no areas of fascination at all are rare, if they even exist. The vast majority of students will respond positively to concrete examples of how a subject applies to their interests.
For instance, many people enjoy sports, music, or cooking. Explanations of how each operates mathematically may shine a light on the importance of maths.
Using Games and Music as Examples of Maths
In any given sport involving a ball or similar object, a series of mathematical patterns occur throughout a match. Each player occupies a carefully-coordinated section of the field or court to ensure maximum efficiency at his or her position. From an aerial view, the strategies enacted during a game can be depicted as shapes and line segments in two spatial dimensions and over a lapse of time.
People do not usually reflect on these ideas deliberately and in exact figures while engaging in the activity or even after the fact. Still, a consistent feature about patterns of these kinds is their continued existence, independent of human measurement.
In addition to the physical action taking place between athletes, calculating scores – at least in competitive contexts – involves a variable range of considerations whose complexity depends on the game. In some cases, adding up points and comparing is enough to determine the victor; other times, mean averages must be computed and perhaps even distributed accordingly.
In musical theory, patterns are arguably even more directly evident than in sports. Anyone who plays an instrument for long enough is surely familiar with studying wheel-like charts depicting the endless succession of scales between octaves. Time signatures also play their role: when two, three, or four beats fill each measure, a musician has already entered the realm of math via fractions.
When preparing a meal, fractions and proportions are critical. Adding four cups of salt or sugar instead of the inversion of that fraction, one-quarter cup could prove disastrous. Cooking time and temperature also bear consideration. Baking a loaf of bread for too long may produce a charred, inedible rock of a thing; not long enough leaves the risk of goopy pits of underdone dough beneath the outer crust – and possibly even health hazards.
These, along with an infinite quantity of other potential examples, might prove fruitful in persuading students of the importance of maths in daily life. Whether they all go on to become professional mathematicians is irrelevant, as they will have learned a set of invaluable skills for evaluating situations and making wise decisions.
Opportunities for Engaging Educators
Knowing how to engage more actively with students produces more well-rounded individuals and instils a love of learning, generally a lifelong one. Point to Point Education recruits the finest educators to teach in the overseas.
To learn more about the opportunities offered, contact us.
Every country has its own type of education system and its own methods of teaching. As an Australian teacher who wants to teach in the UK, you are probably wondering…Read More<|endoftext|>
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# What are the steps for simplifying radicals?
Jul 6, 2018
See if you can factor out a perfect square
#### Explanation:
In general, when we simplify radicals, we want to factor out a perfect square. For instance:
Let's say we're simplifying the radical $\sqrt{84}$:
Because of the radical law, we can rewrite a radical expression $\sqrt{a b}$ as $\sqrt{a} \cdot \sqrt{b}$.
In our example, we can rewrite $84$ as $4 \cdot 21$. We now have the radical
$\sqrt{4 \cdot 21} = \sqrt{4} \cdot \sqrt{21} = 2 \sqrt{21}$
Since $21$ has no perfect square factors, we cannot factor it any further.
The same goes if we had $\sqrt{54}$. We can rewrite $54$ as $9 \cdot 6$, which allows us to separate the radical as
$\sqrt{9} \cdot \sqrt{6} \implies 3 \sqrt{6}$
Once again, $6$ has no perfect square factors, so we are done.
Let's solidify this further with another example:
$\sqrt{162}$
We can rewrite $162$ as $81 \cdot 2$, which allows us to separate the radical as
$\sqrt{81} \cdot \sqrt{2} \implies 9 \sqrt{2}$
Hope this helps!<|endoftext|>
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B2.1 Cells and Cell Structures
All living things are made up of cells. The structures of different types of cells are related to their functions. To get into or out of cells, dissolved substances have to cross the cell membranes.
Cells • Cells are the smallest unit of life. • All living things are made of cells. • Most human cells, like most other animal cells, have the following parts: o nucleus o cytoplasm o cell membrane o mitochondria o ribosome
• Plant and algal cells also have: o cell wall o chloroplasts o permanent vacuole
What do these structures do? • Nucleus – controls the activities of the cell. • Cytoplasm – where most of the chemical reactions take place. • Cell membrane - controls the passage of substances in and out of the cell. • Mitochondria - where most energy is released in respiration. • Ribosome - where protein synthesis occurs. • Cell wall – made of cellulose and strengthens plant cells. • Chloroplasts - absorb light energy to make food in plant cells. • Permanent vacuole - filled with cell sap in plant cells.
Yeast • Yeast is a single-celled organism. • The cells have a nucleus, cytoplasm and a membrane surrounded by a cell wall.
Bacteria • Bacterium is a single-celled organism. • A bacterial cell consists of cytoplasm and a membrane surrounded by a cell wall. • The genes are not in a distinct nucleus.
• Cells may be specialised to carry out a particular function.
Movement into and out of cells
• To get into or out of cells, dissolved substances have to cross the cell membranes. • Solutes = particles in solution eg glucose, sodium ions, chloride ions. • Solvent = liquid in which the particles are dissolved eg water. • Solute and solvent molecules move around randomly. • Solutes can move into and out of cells by diffusion.
Diffusion • Diffusion is the spreading of the particles of a gas, or of any substance in solution, resulting in a net movement from a region where they are of a higher concentration. • Oxygen required for respiration passes through cell membranes by diffusion. • The greater the difference in concentration, the faster the rate of diffusion.
B2.1 Tissues, Organs and Organ Systems
The cells of multicellular organisms may differentiate and become adapted for specific functions. Tissues are aggregations of similar cells; organs are aggregations of tissues performing specific physiological functions. Organs are organised into organ systems, which work together to form organisms.
Multicellular organisms • Large multicellular organisms develop systems for exchanging materials. • During the development of a multicellular organism, cells differentiate so that they can perform different functions. • A tissue is a group of cells with similar structure and function. • Organs are made of tissues. • One organ may contain several tissues. • Organ systems are groups of organs that perform a particular function.
Examples of animal tissues include: • muscular tissue, which can contract to bring about movement • glandular tissue, which can produce substances such as enzymes and hormones • Epithelial tissue, which covers some parts of the body.
The stomach is an organ that contains: • muscular tissue, to churn the contents • glandular tissue, to produce digestive juices • Epithelial tissue, to cover the outside and the inside of the stomach.
The digestive system is one example of a system in which humans and other mammals exchange substances with the environment.
The digestive system includes: • glands, such as the pancreas and salivary glands, which produce digestive juices • the stomach and small intestine, where digestion occurs • the…<|endoftext|>
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The branch of mathematics concerned with the study of mathematical proofs and questions in the foundation of mathematics.
The idea of constructing a universal language for the whole of mathematics, and of the formalization of proofs on the basis of such a language, was suggested in the 17th century by G. Leibniz. But not until the middle of the 19th century did there appear the first scientific work on the algebraization of Aristotelean logic (G. Boole, 1847, A. de Morgan 1858). After G. Frege (1879) and C. Peirce (1885) put the logic of predicates, variables and quantifiers into the language of algebra, it became possible to apply this language to questions in the foundations of mathematics.
On the other hand, the creation of non-Euclidean geometry in the 19th century strongly shook the confidence of mathematicians in the absolute reliability of the geometrical intuition on which Euclidean geometry was founded. Doubts about the reliability of geometrical intuition were also promoted as a result of the fact that in the development of infinitesimal calculus, mathematicians came across unexpected examples of everywhere-continuous functions without derivatives. The need appeared to separate the notion of a real number from the vague notion of a "variable" , which was based on geometric intuition. This problem was solved in various ways by K. Weierstrass, R. Dedekind and G. Cantor. They showed the possibility of "arithmetizing" analysis and function theory, as a result of which the arithmetic of integers came to be considered as the foundation of the whole of classical mathematics. Subsequently, the axiomatization of arithmetic was undertaken (Dedekind, 1888, and G. Peano, 1891). In this connection, Peano created a more suitable symbolic representation for the language of logic. Afterwards, this language was perfected in the joint work of B. Russell and A. Whitehead, Principia Mathematica (1910), in which they attempted to reduce the whole of mathematics to logic. But this attempt was not crowned with success, since it turned out to be impossible to deduce the existence of infinite sets from purely logical axioms. Although the logistic program of Frege–Russell on the foundations of mathematics never achieved its major aim, the reduction of mathematics to logic, in their papers they created a rich logical apparatus without which the appearance of mathematical logic as a valuable mathematical discipline would have been impossible.
At the turn of the 19th century into the 20th century, antinomies (cf. Antinomy) related to the fundamental ideas of set theory were found. The strongest impression at that time was made by the Russell paradox. Let be the set containing exactly those sets which are not an element of itself. It is easy to convince oneself that is an element of itself if and only if is not an element of itself. Certainly one may circumvent this contradiction by stating that such a set cannot occur. However, if a set consisting precisely of all the elements satisfying some clearly defined condition, such as that given above in the definition of , need not exist, then where is the guarantee that in everyday work one will not meet with sets which also need not exist? And, in general, what conditions must the definition of a set satisfy so that the set does exist? One thing was clear: Cantor's theory of sets must somehow be restricted.
L.E.J. Brouwer (1908) opposed the application of rules of classical logic to infinite sets. In his intuitionistic program it was suggested that the abstraction of actual infinity be removed from the discussion, that is, remove infinite sets as complete collections. While admitting the existence of arbitrarily large natural numbers, the intuitionist comes out against the consideration of the natural numbers as a complete set. They believe that in mathematics every proof concerning the existence of an object must be constructive, that is, must be accompanied by a construction of that object. If the premise that the object sought for does not exist results in a contradiction, then this, in the opinion of intuitionists, cannot be considered as a proof of its existence. Particular criticism on the part of intuitionists is aimed at the law of the excluded middle. Since the law was initially considered in association with finite sets, and taking into account the fact that many properties of finite sets are not satisfied by infinite sets (for example, that each proper part is less than the whole), intuitionists regard the application of this law to infinite sets as inadmissible. For example, in order to assert that Fermat's problem has a positive or a negative solution, the intuitionist must give the corresponding solution. So long as Fermat's problem is unsolved, this disjunction is regarded as illegitimate. The same requirement is imposed on the meaning of each disjunction. This requirement of intuitionists may create difficulties even in the consideration of problems connected with finite sets. Imagine that, with eyes closed, a ball is taken from an urn in which there are three black and three white balls, and that the ball is then put back. If no one sees the ball, then one cannot possibly know what colour it was. However, it is doubtful whether one can seriously dispute the certainty of the assertion that the ball was either black or white in colour.
Intuitionists have constructed their own mathematics, with interesting distinctive peculiarities, but it has turned out to be more complicated and cumbersome than classical mathematics. The positive contribution of intuitionists to the investigation of questions in the foundations of mathematics is that they have once more decisively stressed the distinction between the constructive and the non-constructive in mathematics; they have made a careful analysis of the many difficulties which have been encountered in the development of mathematics, and, by the same token, they have contributed to overcoming them.
D. Hilbert (see the Appendices VII–X in ) planned another way to overcome the difficulties arising in the foundations of mathematics at the turn of the 19th century into the 20th century. This route, based on the application of the axiomatic method in the discussion of formal models of interesting mathematics, and in the investigation of questions of consistency of such models by reliable finitary means, was given the name Hilbert's finitism in mathematics. Recognizing the unreliability of geometrical intuition, Hilbert first of all undertook a careful review of Euclidean geometry, liberating it from the appeal to intuition. As a result of this revision his Grundlagen der Geometrie (1910) appeared, .
Questions of consistency of various theories were essentially considered even before Hilbert. Thus the projective model of the non-Euclidean Lobachevskii geometry constructed by F. Klein (1871) reduced the question of consistency of Lobachevskii's geometry to the consistency of Euclidean geometry. The consistency of Euclidean geometry can similarly be reduced to the consistency of analysis, that is, the theory of real numbers. However, it was not clear how it would be possible to construct models of analysis and arithmetic for these consistency proofs. The merit of Hilbert is that he gave a direct way for the investigation of this question. Consistency of a given theory means that one cannot obtain a contradiction in it, that is, it is not possible to prove both an assertion and its negation . Hilbert suggested representing the theory under discussion as a formal axiomatic system, in which those and only those assertions are derivable which are theorems of that theory. Then for the proof of consistency it suffices to establish the non-derivability in the theory of certain assertions. Thus a mathematical theory whose consistency one wishes to prove becomes an object of study in a mathematical science which Hilbert called metamathematics, or proof theory.
Hilbert wrote that the paradoxes of set theory have arisen not from the law of the excluded middle but "rather that mathematicians have used inadmissible and meaningless formations of ideas which in my proof theory are excluded …. To remove from mathematicians the law of the excluded middle is the same as taking the telescope away from astronomers or forbidding a boxer to use his fists" (see ). Hilbert suggested distinguishing between "real" and "ideal" assumptions of classical mathematics. The first have a genuine meaning but the second need not. Assumptions corresponding to the use of the actual infinite are ideal. Ideal assumptions can be added to the real ones in order that the simple results of logic be applicable even to arguments about infinite sets. This essentially simplifies the structure of the whole theory in much the same way as the addition of the line at infinity intersecting any two parallel lines in the projective geometry of the plane.
The program suggested by Hilbert for the foundation of mathematics, and his enthusiasm for it, inspired his contemporaries into an intensive development of the axiomatic method. Thus the formation of mathematical logic as an independent mathematical discipline is linked with Hilbert's initiative, at the start of 20th century, and the subsequent development of proof theory based on the logical language developed by Frege, Peano and Russell.
The objective and fundamental branches of mathematical logic; relation to other areas of mathematics.
The objective of modern mathematical logic is diverse. First of all one must mention the investigation of logical and logico-mathematical calculi founded on classical predicate calculus. In 1930 K. Gödel proved the completeness theorem for predicate calculus, according to which the set of all valid purely logical assertions of mathematics coincides with the set of all derivable formulas in predicate calculus (see Gödel completeness theorem). This theorem showed that predicate calculus is a logical system on the basis of which mathematics can be formulated. Based on predicate calculus various logico-mathematical theories have been constructed (see Logico-mathematical calculus), representing the formalization of interesting mathematical theories: arithmetic, analysis, set theory, group theory, etc. Side-by-side with elementary theories (cf. Elementary theory), higher-order theories were also considered. In these one also admits quantifiers over predicates, predicates over predicates, etc. The traditional questions studied in these formal logical systems were the investigation of the structure of the deductions in the system, derivability of various formulas, and questions of consistency and completeness.
The Gödel incompleteness theorem on arithmetic, proved in 1931, destroyed the optimistic hopes of Hilbert for a complete solution of questions in the foundations of mathematics by the means mentioned above. According to this theorem, if a formal system containing arithmetic is consistent, then the assertion of its consistency expressed in the system cannot be proved by formalization within it. This means that with questions on foundations of mathematics the matter is not as simple as first desired or believed by Hilbert. But Gödel had already noted that the consistency of arithmetic could be proved by using fairly reliable constructive means, although still far from the means that are formalized in arithmetic. Similar proofs of the consistency of arithmetic were obtained by G. Gentzen (1936) and P.S. Novikov (1943) (cf. also Gentzen formal system).
As a result of the analysis of Cantor's set theory and the related paradoxes, various systems of axiomatic set theory were constructed with various restrictions on the formation of sets, to exclude the known inconsistencies. Within these axiomatic systems suitably extensive parts of mathematics could be developed. The consistency question for fairly rich axiomatic systems of set theory remains open. Of the most important results obtained in axiomatic set theory one must note Gödel's result on the consistency of the continuum hypothesis and the axiom of choice in the Bernays–Gödel system (1939), and the results of P. Cohen (1963) on the independence of these axioms from the Zermelo–Fraenkel axioms . One should note that these two axiom systems, and , are equiconsistent. For the proof of his result Gödel introduced the important idea of a constructible set (see Gödel constructive set) and proved the existence of a model of consisting of those sets. Gödel's method was used by Novikov for the proof of the consistency of certain other results in descriptive set theory (1951). For the construction of models of the set theory in which the negation of the continuum hypothesis or the axiom of choice holds, Cohen introduced the so-called forcing method, which subsequently became a fundamental method for constructing models of set theory with various properties (cf. also Model theory).
One of the most remarkable achievements of mathematical logic was the development of the notion of a general recursive function and the formulation of the Church thesis, asserting that the notion of a general recursive function makes precise the intuitive notion of an algorithm. Of the equivalent elaborations of the notion of an algorithm the most widely used are the idea of a Turing machine and a Markov normal algorithm. In essence all mathematics is connected with some algorithm or other. But the possibility of identifying undecidable algorithmic problems (cf. Algorithmic problem) in mathematics appeared only with the refinement of the notion of an algorithm. Undecidable algorithmic problems were discovered in many areas of mathematics (algebra, number theory, topology, probability theory, etc.). Moreover, it turned out that they could be connected with very widespread and fundamental ideas in mathematics. Research into algorithmic problems in various areas of mathematics, as a rule, is accompanied by the penetration of the ideas and methods of mathematical logic into the area, which then leads to the solution of other problems no longer of an algorithmic nature.
The development of a precise notion of an algorithm made it possible to refine the notion of effectiveness and to develop on that basis a refinement of the constructive directions in mathematics (see Constructive mathematics), which embodies certain features of intuitionism, but is essentially different from the latter. The foundations of constructive analysis, constructive topology, constructive probability theory, etc., were laid.
In the theory of algorithms itself it is possible to pick out research in the domain of recursive arithmetic, containing various classifications of recursive and recursively-enumerable sets, degrees of undecidability of recursively-enumerable sets, research into the complexity of description of algorithms and the complexity of algorithmic calculations (in time and extent, see Algorithm, computational complexity of an; Algorithm, complexity of description of an). An extensively developing area in the theory of algorithms is the theory of enumeration.
As noted above, the axiomatic method exerted a major influence on the development of many areas of mathematics. Of special significance was the penetration of this method into algebra. Thus at the junction of mathematical logic and algebra, the general theory of algebraic systems (cf. Algebraic system), or model theory, arose. The foundations of this theory were laid by A.I. Mal'tsev, A. Tarski and their followers. Here one should note research into the elementary theory of classes of models, in particular, decidability questions in these theories, axiomatizability of classes of models, isomorphism of models, and questions of categoricity and completeness of classes of models.
An important place in model theory is occupied by studies on non-standard models of arithmetic and analysis. Even at the dawn of differential calculus, in the work of Leibniz and I. Newton, infinitely-small and infinitely-large quantities were regarded as numbers. Later the notion of a variable quantity appeared, and mathematicians turned away from the use of infinitely-small numbers, the modulus of which was different from zero and less than any positive real number, since their use required the loss of the Archimedean axiom. Only after three centuries, as a result of the development of the methods of mathematical logic, was it possible to establish that (non-standard) analysis with infinitely-small and infinitely-large numbers is consistent relative to the usual (standard) analysis of real numbers (cf. Non-standard analysis).
One could not conclude without mentioning the influence of the axiomatic method on intuitionistic mathematics. Thus, as long ago as 1930, A. Heyting introduced formal systems of intuitionistic logic of propositions and predicates (constructive propositional and predicate calculi). Later, formal systems of intuitionistic analysis were introduced (see, for example, ). Much of the research in intuitionistic logic and mathematics is concerned with formal systems. Special study was made of so-called intermediate logics (cf. Intermediate logic; also called super-intuitionistic logics), that is, logics lying between classical and intuitionistic logics. The notion of Kleene realizability of formulas is an attempt to interpret the idea of intuitionistic truth from the point of view of classical mathematics. However, it turned out that not every realizable formula of propositional calculus was derivable in intuitionistic (constructive) propositional calculus.
Modal logic has also been formalized. However, in spite of the large number of papers on formal systems of modal logic and its semantics (Kripke models), this can still be said to be an accumulation of uncoordinated facts.
Mathematical logic has a more applied value too; with each year there is a deeper penetration of the ideas and methods of mathematical logic into cybernetics, computational mathematics and structural linguistics.
See also the references to the articles on the various branches of mathematical logic.
|||D. Hilbert, P. Bernays, "Grundlagen der Mathematik" , 1–2 , Springer (1968–1970)|
|||S.C. Kleene, "Introduction to metamathematics" , North-Holland (1951)|
|||E. Mendelson, "Introduction to mathematical logic" , v. Nostrand (1964)|
|||P.S. Novikov, "Elements of mathematical logic" , Oliver & Boyd (1964) (Translated from Russian)|
|||Yu.L. Ershov, E.A. Palyutin, "Mathematical logic" , Moscow (1979) (In Russian)|
|||J.R. Shoenfield, "Mathematical logic" , Addison-Wesley (1967)|
|||P.S. Novikov, "Constructive mathematical logic from a classical point of view" , Moscow (1977) (In Russian)|
|||S.C. Kleene, R.E. Vesley, "The foundations of intuitionistic mathematics: especially in relation to recursive functions" , North-Holland (1965)|
|||D. Hilbert, "Grundlagen der Geometrie" , Springer (1913) pp. Appendix VIII|
|||A. Levy, "Foundations of set theory" , North-Holland (1973)|
|||, Mathematics of the 19-th century. Mathematical logic. Algebra. Number theory. Probability theory , Moscow (1978) (In Russian)|
|||A. Mostowski, "Thirty years of foundational studies" , Foundational Studies: Selected Works , 1 , PWN & North-Holland (1979) pp. 1–176|
|[a1]||A. Heyting, "Intuitionism: an introduction" , North-Holland (1959)|
|[a2]||A. Robinson, "Non-standard analysis" , North-Holland (1971)|
|[a3]||Yu.I. Manin, "A course in mathematical logic" , Springer (1977) (Translated from Russian)|
|[a4]||L.E.J. Brouwer, "Collected works" , North-Holland (1975)|
|[a5]||N.A. [N.A. Shanin] Šanin, "On the constructive interpretation of mathematical judgements" Transl. Amer. Math. Soc. (2) , 23 (1963) pp. 109–190 Trudy Mat. Inst. Steklov. , 52 (1958) pp. 226–311|
|[a6]||A.S. Troelstra, D. van Dalen, "Constructivism in mathematics, an introduction" , 1–2 , North-Holland (1989)|
|[a7]||E.A. Bishop, "Foundations of constructive analysis" , McGraw-Hill (1967)|
|[a8]||J. Barwise (ed.) , Handbook of mathematical logic , North-Holland (1977) ((especially the article of D.A. Martin on Descriptive set theory))|
|[a9]||G. Müller (ed.) , Bibliography of mathematical logic , 1–4 , Springer (1988)|
|[a10]||A. Robinson, "Introduction to model theory and the metamathematics of algebra" , North-Holland (1963)|
Mathematical logic. S.I. Adyan (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Mathematical_logic&oldid=19187<|endoftext|>
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Bacteria living in sediments at the bottom of Aarhus Bay in Denmark can exchange electrons with other bacteria as far as 1 centimeter away—the equivalent of 12 miles if the bacteria were the size of people, Wired Science reported. The electrical signals are passed through long strands of the bacteria—a new species of the genus Desulfobulbaceae—that serve as transport cables for the community.
“To move electrons over these enormous distances in an entirely biological system would have been thought impossible,” co-author Moh El-Naggar, assistant professor of physics at the University of Southern California , said in a press release. But the bacteria seem built for the job. Cells line up end-to-end to form a single long filament. The outer surface of the cells are ribbed, and the researchers suspect these channels are what actually carry the electrons. Surrounding the channels is a membrane that may have insulating properites, like myelin sheaths of neurons.
The study, published yesterday (October 24) in Nature, adds to a growing body of literature on bacteria using electricity to communicate and survive. Earlier this year, Japanese researchers found that different species of microbes utilize minerals in the soil to transport electrons long distances. Other bacterial species have long extensions, known as nanowires, that also play a role in conducting electricity among the community. But the new Desulfobulbaceae species appears to be the first to physically link individuals together to create an ideal electrical cable.
The researchers suspect the Desulfobulbaceae filaments serve to help the bacteria respire and eat, with cells deeper in the sediment collecting and oxidizing hydrogen sulfide and cells closer to the surface gathering and using oxygen from the seafloor. “You have feeder cells on one end and breather cells on the other, allowing the whole living cable to survive,” El-Naggar said in the release. The bacteria in the filaments are so dependent on one another, they could be considered a single multicellular organism, co-author Lars Peter Nielsen of the Aarhus Department of Bioscience told Wired.<|endoftext|>
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# How can you evaluate 3/x+5/(2x)-11/(3x)?
Jul 17, 2015
You make sure everything has the same denominator ($6 x$).
#### Explanation:
$= \frac{3}{x} \cdot \frac{6}{6} + \frac{5}{2 x} \cdot \frac{3}{3} - \frac{11}{3 x} \cdot \frac{2}{2}$
$= \frac{18}{6 x} + \frac{15}{6 x} - \frac{22}{6 x}$
$= \frac{18 + 15 - 22}{6 x}$
$= \frac{11}{6 x} = \frac{11}{6} x$<|endoftext|>
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