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# Thread: Solving for an unknown in a system of equations
1. ## Solving for an unknown in a system of equations
I'm not sure whether there's a limit on the amount of questions we can post, especially when they're closely related, but I'm finding I'm struggling with math after 6 years out of school. Chem, funnily enough seems to be a hell of a lot easier :/.
Anyway, the questions is as follow..
"Determine the value of k for which the system of linear equations has infinitely many solutions. Then find all solutions corresponding to this value of k.
3x + 4y = 12
x + ky = 4
Seeing as the two lines are the same line (infinite solutions), the equations should basically be equal.
In y-intercept form, the first equation becomes y = -(3/4)x + 3.
If x = 0, then y should equal 3, right?
Assuming that the line is the same, then
"x + ky = 4" becomes 3 = -(0/k) + (4/k).
3 = (4/k)
12 = k.
Now, when I use this value, the equations don't work out to be parallel, so I'm going wrong somewhere, but I'm just not sure where..
2. ## Re: Solving for an unknown in a system of equations
both lines are the same if all the coefficients of one equation are multiples of the other
3x + 4y = 12
x + ky = 4
the first equation's coefficients are 3 times the second ... or the second equation's coefficients are 1/3 the first.
k = 4/3
3. ## Re: Solving for an unknown in a system of equations
Thanks!
I was miles off .
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Can I Graph You, Too? | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Algebra I Student Edition Go to the latest version.
# 7.3: Can I Graph You, Too?
Created by: CK-12
0 0 0
This activity is intended to supplement Algebra I, Chapter 6, Lesson 6.
## Probmem 1 - Introduction to Disjunction and Conjunction
Consider the equation $|x|=5$. To solve, you would graph both sides of the equation as functions $(y =|x|$ and $y = 5$) and mark the solution as the area where the graphs intersect.
The same method can be applied to inequalities.
Press APPS and select the Inequalz app. Press any key to begin.
Example 1: $|x|<5$
• Using $Y1=$ graph the left side as $y = |x|$. The absolute value function is located by pressing MATH $\rightarrow$ and selecting abs(.
• Using $Y2=$ graph the right side as $y < 5$. On the equals sign, press ALPHA $[F2]$ for the $<$ sign. Press ZOOM and select ZoomStandard.
• Find the intersection points by pressing $2^{nd}$ [TRACE] and selecting intersect. Now just move the cursor to the intersection point and press ENTER three times. The solution is where the shading overlaps the graph of the absolute value function.
In this case, the solution is $-5 < x < 5$.
When an absolute value is less than a number, it is a conjunction because the solution is just one part of the graph.
$|ax+b| < c \quad \rightarrow \quad -c < ax + b < c.$
Example 2: $|x-4| \ge 8$
• Using $Y1=$ graph the left side as $y = |x-4|$.
• Using $Y2=$ graph the right side as $y \ge 8$. On the equals sign, press ALPHA$[F5]$ for the $\ge$ sign. Press WINDOW to choose appropriate window settings.
• Find the intersection points.
In this case, the solution is $x \le -4$ or $x \ge 12$.
When an absolute value is greater than a number, it is a disjunction because the solution is two separate parts of the graph.
$|ax + b| > c \quad \rightarrow \quad ax + b < -c \quad \text{or} \quad ax + b > c.$
## Problem 2 - Application of Disjunction and Conjunction
For the problems below, write the inequalities as either a conjunction or disjunction, then solve for $x$. Check your solution by graphing using the method described in Examples 1 and 2. Please use your graphing calculator to check your results.
#1: $|2x-3|>9$
#2: $\left |\frac{1}{3} x -10 \right | \le 11$
#3: $|3x|-1 \ge 5$
#4: $2 |4x-7| + 6 < 18$
## Problem 3 - Real World Application
One application of absolute value inequalities is engineering tolerance. Tolerance is the idea that an ideal measurement and an actual measurement can only differ within a certain range.
A bolt with a $10 \ mm$ diameter has a tolerance range of $9.965 \ mm$ to $10 \ mm$, while the hole that it fits into has a tolerance range of $10.05 \ mm$ to $10.075 \ mm$.
How can you express the tolerances of both the bolt and the hole in terms of an absolute value inequality?
Feb 22, 2012
Aug 19, 2014
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|First Bank of the United States Note
The most controversial action of Treasury Secretary Alexander Hamilton under President George Washington was the creation of the First Bank of the United States. Assumption of state debts from the Revolutionary War had been acceded to by Secretary of State Thomas Jefferson and his colleague, Virginia Congressman James Madison. But Jefferson and Madison both fought creation of the new national bank. At issue was the power of the federal government to engage in activities not specifically enumerated by the still new U.S. constitution.
The Federal Bank of Minneapolis summarizes the creation of the First Bank of the United States:
The man on the $10 bill, whom biographer Ron Chernow describes as “the clear-eyed apostle of America’s economic future,” believed that the new nation needed a national bank if it was to prosper. After the Revolutionary War, the economy was in tatters: Crushing war debt weighed down the federal government, and a shortage of sound currency and bank credit stifled commercial growth. Hamilton designed the First Bank to help the government get on its financial feet and to galvanize American commerce by providing currency and loans to businesses and individuals. The bank was a vital part of a national financial infrastructure that Hamilton created during his short but prodigious career, the template for today’s monetary economy based on a stable currency and access to credit.
A statesman with a natural genius for economics, Hamilton was far ahead of his contemporaries in perceiving how the country’s fortunes and those of free markets were intertwined, says Thomas M. Humphrey, a former senior economist with the Federal Reserve Bank of Richmond who has written extensively about the history of economic thought. “Hamilton saw a system of sound, secure and resilient financial institutions as being necessary for the real economy to function efficiently,” Humphrey said in a telephone interview. “He also had this vision that America would one day become a huge, thriving economic success around the world, and he thought of these institutions as being necessary to promote that vision.”
The First Bank worked in consort with Hamilton’s other financial reforms—paying off war debt and establishing a stable monetary standard—to put the government’s finances in order and stoke the fires of enterprise at the beginning of the industrial revolution. By aiding in revenue collection, lending to the Treasury and marketing government debt to private investors, the bank served as a financial bulwark for the federal government. And its operations invigorated markets by providing a sizable, trustworthy currency and extending credit to businesses.
The chartering of the Bank was dramatic since involved a clash of strong political personalities and very divergent views about the political model that America would follow. Hamilton, in particular was charged by Jefferson with following too closely what Jefferson viewed as a corrupt British model. Open political warfare broke out within the Washington Administration. One student of the history ably sums it up as suffused with “an astonishing amount of political intrigue and subterfuge.
“It was Hamilton’s proposal for the creation of a Bank of the United States that touched off a firestorm in the House of Representatives, leading to the first major fight of implied versus enumerated powers. The fight came down to this: Could the national government grant a corporate charter? No such power was explicitly mentioned - “enumerated” - in the Constitution. Since it was not, did that not mean that it was therefore a power reserved to the states, as expressed in Article Ten of the Bill of Rights? The leader of the opposition was Secretary of State Thomas Jefferson, who enlisted the aid of James Madison in launching a series of open and covert attacks on Hamilton, and his influence with Washington. Jefferson argued that since the power to create a bank was not expressly and explicitly stated in the Constitution, the proposed charter for a Bank of the United States was unconstitutional. Using an astonishing amount of political intrigue and subterfuge, Jefferson was able to make the Bank a major national issue.
“President Washington therefore requested Attorney General Edmund Randolph, who sided with Jefferson, and Hamilton to write reports explaining and justifying their respective positions regarding the government’s power to create the bank.”
In a memorandum on “Opinion on the Constitutionality of a National Bank,” Jefferson wrote in February 1791, Jefferson denied the “general welfare” clause of the Constitution. Jefferson contended:
“I consider the foundation of the Constitution as laid on this ground: That ‘all powers not delegated to the United States, by the Constitution, nor prohibited by it to the States, are reserved to the States or to the people.’ [XIIth amendment.] To take a single step beyond the boundaries thus specially drawn around the powers of Congress, is to take possession of a boundless field of power, no longer susceptible of any definition. …. It would reduce the whole instrument to a single phrase, that of instituting a Congress with power to do whatever would be for the good of the United States; and, as they would be the sole judges of the good or evil, it would be also a power to do whatever evil they please.
As powerful was the opposition of Jefferson, Madison and Attorney General Randolph --- fellow Virginians – on Washington, Hamilton’s support was more influential. Hamilton’s famous opinion — which promptly and decisively persuaded Washington to sign the legislation, held:
“That every power vested in a government is in its nature sovereign, and includes, by force of the term, a right to employ all the means requisite and fairly applicable to the attainment of the ends of such power, and which are not precluded by restrictions and exceptions specified in the Constitution, or not immoral, or not contrary to the essential ends of political society.”
Economic historian David Cowen describes, in the Economic History Association’s excellent online encyclopedia EH.net, the impact of the Bank, once it became operational:
“The customers were merchants, politicians, manufacturers, landowners, and most importantly, the government of the United States. The Banks notes circulated countrywide and therefore infused a safe medium of paper money into the economy for business transactions. The sheer volume of deposits, loans, transfers and payments conducted by the Bank throughout the country made it far and away the single largest enterprise in the fledgling nation. Profits, however, were moderate during the operation of the Bank because its directors opted for stability over risk taking.
“The Bank had an enormous impact on the economy within two months of opening its doors for business by flooding the market with its discounts (loans) and banknotes and then sharply reversing course and calling in many of the loans. Although the added liquidity initially helped push a rising securities market higher, the subsequent drain caused the very first U.S. securities market crash by forcing speculators to sell their stocks. … This so-called "Panic of 1792" was short lived as again Secretary Hamilton (as in the previous year during the script bubble) injected funds by buying securities directly and on behalf of the sinking fund. …
“The rest of Bank years were never as tumultuous as the events surrounding the Panic of 1792. Rather during its twenty-year lifespan the Bank performed many mundane pecuniary functions for its customers. …
“The Bank performed certain functions that today are associated with central banking. First, the Bank attempted to regulate state banks by curtailing those that had overissued their bank notes. Second, the Bank, in coordination with the Treasury department, discussed economic conditions and attempted to promote the safety of the entire credit system. Third, while the Philadelphia board gave each branch autonomy respecting lending to individuals, the Bank tried to coordinate aggregate policy changes, whether a loosening or tightening of lending credit, across the entire network of branches.”
Although the Bank’s political, philosophical, and mercantile enemies succeeded in preventing its re-chartering, upon expiration in 1811, it would not be long before a Second Bank of the United States was constituted… and with the active support of one of its most ardent opponents, James Madison, when president of the United States and beset by the exigencies of the War of 1812.
Next: The War of 1812
Previous: The Albert Gallatin Era
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# How do you solve the following system?: -x+5y=7 , x=-3x + 4y-17
Jan 1, 2016
$x = - 9.5 , y = - 0.5$
#### Explanation:
$- x + 5 y = 7$
$x = 5 y - 7$ ---$\left(1\right)$
$x = - 3 x + 4 y - 17$
$2 x - 4 y = - 17$ ---$\left(2\right)$
Subsitute $\left(1\right)$ into $\left(2\right) :$
$2 \left(5 y - 7\right) - 4 y = - 17$
$10 y - 14 - 4 y = - 17$
$6 y = 14 - 17$
$6 y = - 3$
$y = - 0.5$
Subsitute $y = - 0.5$ into $\left(1\right)$:
$x = 5 \left(- 0.5\right) - 7$
$x = - 2.5 - 7$
$x = - 9.5$
x=-9.5, y=-0.5
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Learn about the different types of techniques used to sequence DNA. We'll go over everything from maxam gilbert sequencing, sanger ddNTP sequencing to shotgun sequencing.
Get a gist of what DNA sequencing is, what types of molecules can be sequenced, what breadth of information comes out of DNA sequencers, and the types of sequencing approaches available.
Learn about the very first DNA sequencing technique pioneered by Alan Maxam and Walter Gilbert in 1976.
Learn how Sanger sequencing works, which relies heavily on the chemical properties of dideoxynucleotides (ddNTPs). Also learn the pros and cons of the technique.
Learn how scientists deal with the disadvantage of having short reads by utilizing a method known as primer walking.
Learn how to get around the limitations of pyrosequencing by incorporating all four nucleotides in one run with reversible chain terminators.
Learn the technique they used to sequence genomes that haven't been sequenced before (de novo) with the shotgun sequencing method.
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- What to do when someone is choking
- How to recognise the signs of choking
- What should your first response be?
- How to care for an adult or child (over the age of 1) who is choking
- How to care for an infant (baby) who is choking
- How to perform the Heimlich Manoeuvre on yourself
- While you wait for emergency services…
- Choking risk factors to take into consideration
- Safety considerations for administering first aid
Safety considerations for administering first aid
Many accredited courses incorporate learning to use a CPR breathing barriers into first aid training. Many also provide one as part of the course pack. Most pharmacies also stock these for consumers to purchase when stocking first aid kits for home and travel use. Disposable or re-usable devices are available for purchase.
Barriers are also available in standard sizes for children and infants, as well as for adults. Ideally, a variety of sizes should be carried by individuals trained in providing first aid care. Devices which are too big for, for instance a child, can result in air leakage which will not provide adequate rescue breathing technique, causing reduced ventilation capability. (7)
CPR breathing barriers are there for your safety when providing care to another who may unknowingly place you at risk of a health condition. Barriers can protect against contact with bodily fluids, such as blood or saliva, as well as exhaled air containing tiny droplets of moisture, which can carry infectious agents.
Barriers are designed as a small, portable ‘face shield’ which can still provide essential rescue breathing capability. One of the most popular and basic of barriers is designed as a thin and flat piece of transparent plastic (round or square), which can be placed over the mouth area of the person requiring assistance, with an opening in the centre for access to the mouth. The opening is fitted with a one-way valve which works as a filter, separating your mouth from the patient’s, providing protection from direct contact, while still delivering much needed air.
Another barrier device is a pocket mask which is also transparent and portable. It can be used to create a tight seal over a patient’s mouth and nose, preventing direct contact but also allowing for rescue breaths to be successfully given.
Another important safety consideration is handwashing. Hands exposed to potential transmission contaminants can also place a caregiver at risk of potential infectious illnesses.
It is important to wash your hands thoroughly with warm, running water and soap once you have handed over a patient to EMS or medical professionals, even if you wore disposable gloves at the time of administering care. Hands should be washed thoroughly, including the wrists, in between fingers, beneath the fingernails, palms and the back of the hands.
Hand sanitisers can also be used if running water and soap are not immediately available but must also be applied to all surfaces of the hands including the nails and in between the fingers. Sanitisers should be rubbed in well until the product dries. Once handwashing facilities become available, hands must be thoroughly washed.
7. The Textbook of Emergency Cardiovascular Care and CPR - By John M. Field, Peter J. Kudenchuk, Robert O'Connor, Terry VandenHoe. 2009. Chapter 17 - Oxygen Administration and Supraglottic Airways: https://books.google.co.za/books?id=o3m4oNRB4D4C&pg=PA259&lpg=PA259&dq#v=onepage&q&f=false [Accessed 01.12.2017]
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This nearly global mosaic of observations made by the Mars Color Imager on NASA's Mars Reconnaissance Orbiter on April 2, 2009, shows billowing clouds of dust being lifted into the atmosphere by a storm near the edge of the seasonal polar cap of southern Mars.
The season on southern Mars is late spring. Late southern spring and early southern summer are a peak time of the Martian year for major dust storms.
Atmospheric haze due to suspended dust from recent storm activity is evident elsewhere on the planet, including the skies over Mars rovers Opportunity (MER-B) and Spirit (MER-A).
Black areas in the mosaic are the result of data drops or high angle roll maneuvers by the orbiter that limit the camera's view of the planet. Equally-spaced blurry areas that run from south-to-north (bottom-to-top) result from the high off-nadir viewing geometry, a product of the spacecraft's low-orbit.
Malin Space Science Systems, San Diego, provided and operates the Mars Color Imager. NASA's Jet Propulsion Laboratory, a division of the California Institute of Technology in Pasadena, manages the Mars Reconnaissance Orbiter for NASA's Science Mission Directorate, Washington. Lockheed Martin Space Systems, Denver, is the prime contractor for the project and built the spacecraft.
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Prison acts as a place of restraining people who break the law in society. Thus, correctional facilities have become an essential tool in ensuring people obey laws and regulations. The British society began shifting from physical punishment to imprisonment with an aim of transforming criminals both spiritually and physically which rise to prisons. A penitentiary or a prison can be termed as a place where criminals are dispossessed of a variety of own liberty through incarceration. The United States having attained independence from England, the British did not have a colony for imprisoning individuals, thus, they started the conviction system. Since then, the idea of imprisonment has spread globally as a formal way of rehabilitating criminals. The prisoner serves a jail term, which is a legal penalty obligatory by the state. However, legal systems term both jail and prison differently based on duration served by the offender. The state manages the prisons and holds criminals serving long-term penalties. On the other hand, jails are used to hold offenders awaiting trial for a short time. The major role of a prison is to reform criminals to acceptable norms of the society. Therefore, an ideal prison should establish a correction mechanism that focuses on changing criminals to upstanding citizens.
This paper examines the historical change in prison systems during the World War II and after the World War II and compares the two.
In ancient times, the form of punishment granted to prisoners was slaughtering or enslaving them, which contrasts to the modern way of punishment, which entails rehabilitation of prisoners. Moreover, lawbreakers are sentenced according to the law. During ancient times prison systems treated lawbreakers differently. Some systems followed the international agreement and treated prisoners according to its requirements while others like Japan, which did not consent to the agreement, did not conform to it. Additionally, the Chinese utterly abandoned the constraints of The Hague Conventions imposed on their prisoners. Japan subjected its western prisoners to murder, whipping and cruel treatment such as forced labor, medical experimentation and starvation. Tokyo tribunal revealed that the fatality rate of western prisoners in Japan was seven times compared to that of German while that of Chinese was exorbitant.
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German and Italy followed the Geneva Convention (1929) in treating their prisoners who were mainly the British Commonwealth countries that had agreed to this convention. As a result, western allied officers were exempted from work and lower rank personnel benefited from compensation. The main complaint that arose to western allied prisoners was food shortages. The Jewish western allied prisoners were subjected to murder or anti-Semitic policies while the allied personnel were sent to concentration camps for various reasons. German applied different grounds on prisoners from non-western countries such as Polish and Soviet. For example, many Soviet prisoners captured between 1941 and 1945 were set free during the World War II but remained under German’s captivity, while some died. Bunyak (2005) points out that Germans took advantage of the Soviet Union abandoning the Geneva Convention and treated the prisoners from the Soviet cruelly, which was not lawful. Various sources admit that the Soviets also captured servicemen during the battle of Stalingrad and the majority of them died. The German soldiers that survived were used as a source of labor.
Canada, Australia, UK, and US strictly embraced the Geneva Convention though quite a number of breaches occurred, in that, some US soldiers killed German prisoners. After German’s surrender in the World War II in 1945, German prisoners were subjected to forced labor in countries such as the UK and France in which some died in Norway.
The impact of prison labor is diverse and immense in nature. For instance, it is considered cheap when prisoners provide the labor. The reality is that many industries rely on this plentiful labor force given by the prisoners in America. Ram (1981) indicates that being an employed prisoner is much better than being jobless. Prison labor has also influenced negatively the wardens’ numbers. This is because work functions as a management tool. Work ensures that prisoners have a productive day and reduces boredom that results from being locked all day. Evidently, boredom is a motivator to prisoners to consider escaping. Thus, the wardens are saved the burden of monitoring the prisoners.
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The present prison system is comprised of a more sophisticated mechanism of correcting criminals and efficient way of managing as compared to those that existed before and during the World War II. The way prisoners are treated has improved due to the establishment of government regulations that protect the rights of prisoners. The current prison system addresses the issue of rehabilitation more than punishment as viewed in the World War II prisons, due to the varying nature of criminal offenses. The society needs to confront issues such as drug abuse and violence instead of incarcerating those who are the most susceptible to commit crimes because prison has emerged as the only way of delivering punishment. Ram (1981) asserts that in a civilized community, prisons are necessary since it is a prodigious source of employment; therefore, it is difficult to picture the world without it. Mass imprisonment is deeply rooted in humanity to the extent that we do not recognize offenders who are racists or sexists.
Another up-coming trend in prison systems is the prison privatization, which refers to the shift of control or management of public prisons to private companies. Private companies are usually given contracts by the Federal to build, run and maintain new facilities. Bailey (2001) points out that the idea of prison privatization can be compared to the system employed by the southern states, which concentrate on hiring out prisoners to work in plantations and mining. In this case, the private companies bear the responsibility of caring, housing and the security of prisoners. As a result, the cost of running the prison by the government was reduced while, at the same time, income per inmate was earned. Consequently, this system became overwhelmed with corruption and prison escapes, which were common. Another aspect of prison privatization was the racial characteristic. Most convicts were Africans who had committed minor offenses and were serving long-term sentences. Due to this aspect, this system was termed as a means of suppressing African Americans. Current privatization is extremely dissimilar from the previous one employed by the past generations. The state and local authorities pay firms and legal, private companies to house and provide care to prisoners in a humanely manner and provide high-class services in public prisons.
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In conclusion, imprisonment is still the main form of punishing criminals during the World War II and after. Today’s society embraces it but with greater advancement, which includes rehabilitation schemes. Prison labor has its own benefit to state because it facilitates the provision of cheap labor to industries, hence, increasing profits to industries that deploy this form of labor. Apart from cheap labor, convicts are transformed to become productive citizens rather than spending useful time in isolation. This implies that both spiritual and corporal punishment is necessary in the reformation process of a criminal. Both the ancient and modern prison systems carried out prison privatization but in different forms. The ancient sold prisoners as slaves while the modern lease out prisoners. The old system varies from the current one because the current system has respect for humanity.
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Activity: Random Babies
Topics: Probability/Equally Likely
Prerequisites:
This activity explores the meaning of probability and equally likely outcomes again through physical and computer simulations. It does not assume much familiarity with the software, Minitab. Note: Students will be shown how to create a Minitab macro in this lab.
Materials:
Four index cards and one sheet of scratch paper per student
Goals:
• To develop intuition for probability as long-term relative frequency
• To learn some elementary counting rules
• To emphasize consideration of when the "equal likeliness" assumption applies and when it does not
• To gain an introduction to the concepts of sample space, random variable, and expected value
Situation:
Suppose that on one night at a certain hospital, four mothers (named Johnson, Miller, Smith, and Williams) give birth to baby boys. Each mother gives her child a first name alliterative to his last: Jerry Johnson, Marvin Miller, Sam Smith, and Willy Williams. As a very sick joke, the hospital staff decides to return babies to their mothers completely at random. We will first use simulation to investigate what will happen in the long run.
Simulation Analysis:
(a) Take four index cards and one sheet of scratch paper. Write a baby’s first name on each index card, and divide the sheet of paper into four areas with a mother’s last name written in each area. Shuffle the four index cards well, and then deal them out randomly with one index card going to each area of the sheet. Finally, turn over the cards to reveal which babies were randomly assigned to which mothers. Record how many mothers got the right baby.
# of matches:
(b) Do the random “dealing” of babies a total of five times, recording in each case the number of matches:
Repetition # 1 2 3 4 5
# of matches
(c) Combine your results on the number of matches with the rest of the class, obtaining a tally of how often each result occurred. Record the counts and proportions in the table below:
# of matches 0 1 2 3 4 Total
Count
Proportion 1.00
(d) In what proportion of these simulated cases did at least one mother get the correct baby?
The probability of a random event is the long-run proportion (or relative frequency) of times the event would occur if the random process were repeated over and over under identical conditions. One can approximate a probability by simulating the process a large number of times. Simulation leads to an empirical estimate of the probability.
(Exact) Enumeration Analysis:
In situations where the outcomes of a random process are equally likely to occur (e.g. tossing a fair coin), exact probabilities can be calculated by listing all of the possible outcomes and determining the proportion of these outcomes which correspond to the event of interest. The listing of all possible outcomes is called the sample space.
The sample space for the “random babies” consists of all possible ways to distribute the four babies to the four mothers. Let xyzw mean that the baby x went to the first mother, baby y to the second mother, baby z to the third mother, and baby w to the fourth mother. For example, 1243 would mean that the first two mothers got the right baby, but the third and fourth mothers had their babies switched.
(e) Below is the beginning of a list of the sample size for the “random babies” process. Fill in the remaining possibilities, using this same notation. [Try to be systematic about how you list these outcomes so that you don’t miss any. One sensible approach is to list in a second row the outcomes for which mother 1 gets baby 2 and then in the third row the cases where mother 1 gets baby 3 and so on.]
Sample Space:
1234 1243 1324 1342 1423 1432
(f) How many possible outcomes are there in this sample space? That is, in how many different ways can the four babies be returned to their mothers?
You could have determined the number of possible outcomes without having to list them first. For the first mother to receive a baby, she could receive any one of the four. Then there are three babies to choose from in giving a baby to the second mother. The third mother receives one of the two remaining babies and then the last baby goes to the fourth mother. Since the number of possibilities at one stage of this process does not depend on the outcome (which baby) of earlier stages, the total number of possibilities is the product
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# what is the square root of 300
Square root of 300 The square root of 300 is expressed as √300 in root form and (300) ½ or (300) 0.5 in power. The square root of 300 rounded to 9 decimal places is 17.320508076. It is a positive solution of the equation x2 = 300. We can represent the square root of 300 in its square root form as 10 √3.
• Square root of 300: 17.320508075688775
• Square root of 300 in exponential form: (300) or (300) 0.5
• Square root of 300 in root form: 300 or 10 3
1. What is the square root of 300? 2. Is the square root of 300 Reasonable or Unreasonable? 3. How to find the square root of 300? 4. FAQ about Square Root of 300
• Allocation form: 300 = 10√3
• Decimal: 17.320
• Power Form: (300) 1/2
• Long division method
• Element
One can learn other methods by clicking here Read: what is the square root of 300
### Long division method
Read more: shawn mendes favorite color | Q&A on the Square Root of 300 by the long division method includes the following steps:
• Step 1: Starting from the right, we will match the digits 300 by placing a bar on 00 and 3 separately. We’ll also concatenate zeros in the decimal from left to right.
• Step 2: Find a number that when multiplied by itself gives a product less than or equal to 3. In this case, it is 1, so put 1 in the quotient and divide the number which will result in a remainder of 2.
• Step 3: Drag down 00 next to remainder 2. Also, add the divisor to itself and write it below. (1 + 1 = 2)
• Step 4: Find a number X such that 2X × X results in a number less than or equal to 200. The number 7 fits here, so put it next to 2 in the divisor as well as next to 1 in the quotient.
• Step 5: Find the remainder and now drag the pair of zeros down from the decimal part of the number. Double the quotient to get the new divisor. The new divisor is 34.
• Step 6: Repeat this process to get the decimal places you want.
Read more: What the cemetery keepers sell in the tavern square root of 300 = 17,320
### Element
• To find the square root of 300, we will first represent it in terms of its prime factors. 300 = 2 × 2 × 3 × 5 × 5.
• Next, this can be further reduced to 300 = 22 × 3 × 52
• Now the root from here can be easily found: √300 = √ (22 × 3 × 52) √300 = 10√3 = 17,320
Therefore, square root of 300 17,320Explore square roots using interactive illustrations and examplesRead more: what does cf mean on instagram
• Square root of 200
• Square root of 250
• Square root of 225
• Square root of 3
• Square root of 100
• There exists a yin and yang root of 300; 17,320 and -17,320
• There will be n/2 digits in the square root of an even number of n digits.
• There will be (n + 2) / 2 digits in the square root of an odd n digit number.
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# Practice GMAT Problem Solving Question
If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
A) B) C) D) E)
1. Finding P(at least two students with the same birthday) directly will be extremely difficult. Consequently, we must look for a different way. If we find the complement of P(at least two students with the same birthday), we can find P(at least two students with the same birthday) since it will be equal to 1 - P(complement).
P(A) = 1 - P(complement of A)
P(at least two students with the same birthday) = 1 – P(all birthdays are unique)
2. Find P(all birthdays are unique).
3. If every birthday is unique, then with 85 students, there are 85 different days on which a birthday occurs.
Let N = Total Students Selected After Student S Selected
Let M = Total Students Selected Before Student S Selected
Let U = Unique Days of Year Remaining Before Student S Selected on Which Future Students Can Have Unique Birthday
Let P = P(Unique Birthday)
Let O = Students Remaining After Student S Selected
You can draw a table to help solve this problem.
N M U P O 1 0 365 = 365 - 0 365/365 84 = 85 - 1 2 1 364 = 365 - 1 364/365 83 = 85 - 2 3 2 363 = 365 - 2 363/365 82 = 85 - 3 4 3 362 = 365 - 3 362/365 81 = 85 - 4 83 82 283 = 365 - 82 283/365 2 = 85 - 83 84 83 282 = 365 - 83 282/365 1 = 85 - 84 85 84 281 = 365 - 84 281/365 0 = 85 - 85 =365 - M =U/365 =85 - N
4. For P(Unique Birthday with 1 student already selected [i.e., the M=1 row]): Since there are 365 possible days of the year yet only 364 days will satisfy the condition of all birthdays being unique since 1 day already has a birthday occupying it from the first student selected, P = 364/365
5. P(all birthdays are unique) = P(unique birthday for 1st person selected) *P(unique birthday for 2nd person selected) *P(unique birthday for 3rd person selected) ... *P(unique birthday for 84th person selected) *P(unique birthday for 85th person selected)
6. Translating this into math by multiplying the values in column P:
P(all birthdays are unique) =
7. At this stage, you should simplify the expression that equals P(all birthdays are unique).
From 365 to 281, there are (365-281)+1 = 85 numbers (remember that you need to add 1 since it is 365 to 281, inclusive).
8. In the denominator, you know that 365 will be multiplied together 85 times:
Denominator = 36585
9. In the numerator, it is not as easy to simplify. However, you may notice that there appears to be something resembling a factorial. Specifically, the numerator is 365! truncated just before 280. If you take 365! and divide it by 280!, every term beneath 281 will cancel out, leaving you with the expression you are looking for in the numerator.
10. Combine the numerator and denominator:
11. Use the complement to find the answer to the original question:
P(at least two students with the same birthday) = 1 – P(all birthdays are unique)
P(at least two students with the same birthday) =
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# How to calculate inflection point
Inflection Point Calculus. If f (x) is a differentiable function, then f (x) is said to be: Concave up a point x = a, iff f β (x) > 0 at a. Concave down at a point x = a, iff f β (x) 0 at a. Here, f β (x) is the
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## Calculus 1 : Points of Inflection
How to Use the Inflection Point Calculator? Please follow the below steps to find the inflection point: Step 1: Enter the equation in input box; Step 2: Click on the Solve button to find the
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math is the study of numbers, shapes, and patterns. It is used in everyday life, from counting to measuring to more complex calculations.
## Analyzing the second derivative to find inflection points
To find the point of intersection algebraically, solve each equation for y, set the two expressions for y equal to each other, solve for x, and plug the value of x into either of the
## Inflection Points
In order to find the inflection point of the function Follow these steps. Take a quadratic equation to compute the first derivative of function f' (x). Now perform the second derivation of f (x) i.e fβ
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## How To Find an Inflection Point (And Which Fields Use Them)
Solve math equation
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IntroductionAmerican Revolution, 1775–83, struggle by which the Thirteen Colonies on the Atlantic seaboard of North America won independence from Great Britain and became the United States. It is also called the American War of Independence.
Sections in this article:
The Columbia Electronic Encyclopedia, 6th ed. Copyright © 2012, Columbia University Press. All rights reserved.
See more Encyclopedia articles on: U.S. History
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# Teams
How many ways can divide 16 players into two teams of 8 member?
Correct result:
n = 12870
#### Solution:
$n={C}_{8}\left(16\right)=\left(\genfrac{}{}{0px}{}{16}{8}\right)=\frac{16!}{8!\left(16-8\right)!}=\frac{16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9}{8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=12870$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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The village markets have 5 kinds of sweets, one weighs 31 grams. How many different ways a customer can buy 1.519 kg sweets.
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How many natural numbers greater than 4000 which are formed from the numbers 0,1,3,7,9 with the figures not repeated, B) How many will the number of natural numbers less than 4000 and the numbers can be repeated?
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How many ways can we divide nine workers into three workplaces if they need four workers in the first workplace, 3 in the second workplace and 2 in the third?
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1. ## Proof: Inverse Functions
Let X and Y be sets and f:X->Y be a function. For A is a proper subset of X and B is a proper subset of Y, recall the definitions of f(A) is a proper subset of Y and f^-1(B) is a proper subset of X,
f(A)={f(a):a is an element of A} and f^-1(B)={x is an element of X:f(x) is an element of B}
Prove:
A is a proper subset of f^-1(f(A)) and f(f^-1(B)) is a proper subset of B; equality need not hold in either case.
Not even sure where to begin. Help, please!
2. ## Re: Proof: Inverse Functions
Originally Posted by lovesmath
Let X and Y be sets and f:X->Y be a function. For A is a proper subset of X and B is a proper subset of Y, recall the definitions of f(A) is a proper subset of Y and f^-1(B) is a proper subset of X,
f(A)={f(a):a is an element of A} and f^-1(B)={x is an element of X:f(x) is an element of B
Prove:
A is a proper subset of f^-1(f(A)) and f(f^-1(B)) is a proper subset of B; equality need not hold in either case.
Not even sure where to begin. Help, please!
This is not true.
Let $X=\{1,2,3,4,5\},~A=\{4,5\},~Y=\{a,b,c,d\},~B=\{a,b \}~\&$
$f=\{(1,a),(2,b),(3,b),(4,c),(5,d)\}$ but $f^{-1}(f(A))=A,~\&~f(f^{-1}(B))=B.$
Please review the statement for errors.
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12/24/68 — Forty Years Ago Today
On 21 December 1968, Apollo 8, the first manned mission to the moon, was launched from Cape Kennedy. The crew of three included: Frank Borman, commander; James A. Lovell, command module pilot; and William A. Anders, lunar module pilot.
Apollo 8 was the first of the three exploratory precursors to Apollo 11 — which was launched eight months later on 16 July 1969, and whose lunar module landed on the Moon on 20 July.
Apollo 8 was the first manned craft to leave earth’s gravitational field, and the first to enter the atmosphere of another celestial body
1968 had been a hard, and, in many ways, a horrendous year, but a weary world was at least unite for a few moments at its end to look up —in awe for all, in pride for some— and marvel at the idea that the old moon now being orbited by three visitors from earth.
The whole world was truly riveted and thrilled by the photographs (including the now iconic earthrise, which was also sent back on Christmas Eve) and the TV images from the camera pointed out the window at the lunar surface racing by only 71 miles below.
On Christmas Eve, the astronauts hosted a live TV broadcast. It isn’t going too far to say that the earth in solemn stillness lay to hear what they would say. And they surprised —and moved— people all around the globe by the message they chose to send. Each astronaut in turn —in reverse order of mission seniority— shared the reading.
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# Empirical Formula
## Empirical Formula
The simplest formula or the empirical formula provides the lowest whole number ratio of atoms existent in a compound. The relative number of atoms of every element in the compound is provided by this formula.
Steps for Determining an Empirical Formula
• Let’s begin with what’s given in the problem, i.e., the number of grams of each element.
• We’ll assume that the total mass is 100 grams if percentages are given, so that
Each element’s mass = the given percentage
• By making use of the molar mass from the periodic table, change the mass of every element to moles.
• Divide every mole value by the lowest number of moles computed.
• Round up to the closest whole number. Â This is denoted by subscripts in the empirical formula and is the mole ratio of the elements.
Multiply each answer by the same factor to get the lowest whole number multiple, if the number is too far to round off (x.1 ~ x.9).
e.g. Â Multiply each solution in the problem by 4 to get 5, if one solution is 1.25.
e.g. Â Multiply each solution in the problem by 2 to get 3, if one solution is 1.5.
The molecular formula can be calculated for a compound if the molar mass of the compound is given when the empirical formula is found. Â To find the ratio between the molecular formula and the empirical formula. Basically, the mass of the empirical formula can be computed by dividing the molar mass of the compound by it. Â Multiply every atom (subscripts) by this ratio to compute the molecular formula.
## Solved Examples
Problem 1: A compound contains 88.79% oxygen (O) and 11.19% hydrogen (H). Compute the empirical formula of the compound.
Solution:
1. Assume 100.0g of substance. We see that the percentage of each element matches the grams of each element
11.19g H
88.79g O
1. Convert grams of each element to moles
H: (11.19/1.008) = 11.10 mol H atoms [molar mass of H=1.008g/mol]
O: (88.79/16.00) = 5.549 mol O atoms [molar mass of O= 16.00g/mol]
The formula could be articulated as H11.10O5.549. However, it’s usual to use the smallest whole number ratio of     atoms
1. By dividing the lowest number alter the numbers to whole numbers.
H =11.10/ 5.549Â = 2.000
O = 5.549/ 5.549= 1.000
The simplest ratio of H to O is 2:1
Empirical formula = H2O
Problem 2: A sulfide of iron was formed by combining 1.926g of sulfur(S) with 2.233g of iron (Fe). What is the compound’s empirical formula?
Solution:
1. As the mass of each element is known, we use them directly
2. Convert grams of each element to moles
Fe: (2.233g /55.85g) = 0.03998 mol Fe atoms [molar mass of Fe =1.008g/mol]
S: (1.926 /32.07) = 0.06006 mol S atoms [molar mass of S =32.07g/mol]
1. By dividing by the smallest number, change the numbers to whole numbers.
Fe =0.03998/0.03998 = 1.000
S = 0.06006/0.03998mol = 1.502
1. As we still have not reached a ratio that gives whole numbers in the formula we multiply by a number that will give us whole numbers
Fe: (1.000)2 = 2.000
S: (1.502)2 = 3.004
Empirical formula = Fe2S3
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The two soliloquies express two totally different emotions. In each, there are different situations, one in which Juliet is longing for the time when Romeo will come to her for their wedding night where feelings of excitement and anticipation are portrayed. The opposing situation expresses feelings of fear, the fear of what would happen to her if she were to take the potion given to her by Friar Lawrence in order to try to prevent her from having to marry Paris. Although the soliloquies contrast in situation and emotion, the ways in which the feelings are portrayed are very similar.
In the first soliloquy, where Juliet is filled with anxiety and anticipation, waiting for the time when Romeo will come to her that night, the language Shakespeare uses creates a light joy-filled mood and an audience can feel the anticipation which Juliet is experiencing. “Come…Come” repetition is used throughout the soliloquy, which shows just how impatient Juliet is for the night to come. Words expressing speed are used frequently during the speech, “Gallop apace” “fiery footed” to reinforce Juliet’s impatience. She is longing for the time when Romeo comes to her that she wants it to come as fast as possible, so uses words to inflict speed to show this to the audience.
Juliet describes Romeo as the light, just as he had done to her when they first met. “Cut him out in little stars” Juliet describes how he will be scattered in the sky, and how everyone will then prefer night to day as they will see Romeo in the sky. “Pay no worship to the garish sun.” throughout the soliloquy. Juliet describes that night is “gentle” a time when everything will be perfect, and a time that everyone will love, as opposed to day. Shakespeare writes in long sentences which mean that the audience would be aware of the actress speeding up showing her excitement and anticipation. Juliet acts as if it is “the night before some festival” acting as excited as a child the night before Christmas.Order now
Although Juliet is filled with anticipation for the night to come, she is also embarrassed and nervous about the events later that day as any young woman would be until she realises that because they love one another it would b a “true love acted of simple modesty”.
In contrast to the first, in the second soliloquy, Juliet expresses fear. She is afraid of what might happen if she was to take the potion. “Shall I be married then tomorrow morning?” her first fear is that the potion wont work and she will have to marry Paris, and therefore her actions would be unfaithful to Romeo of whom she had previously married. “Ministered to have me dead” Juliet fears that Friar Lawrence might be trying to kill her so she doesn’t have to marry Paris. Friar Lawrence already knowing he has married her to Romeo and therefore in marrying her to Paris he would be dishonoured. Juliet fears she would wake before Romeo arrives and she would die in the vault because there would be no air for her to breathe. “Shall I not then be stifled in the vault.” Juliet’s final fear is that she would go mad in the vault with all the bones of all her ancestors. Being among people who had been dead for years, or being around Tybalt who had only recently been placed in the vault. “And dash out my desperate brains.”
Unpleasant words and images are used throughout the soliloquy stressing the theme of death and eeriness of a vault in which many ancestors bones have been laid. The amount of images which Shakespeare envisions us with increases towards the end of Juliet’s speech showing the audience that she is becoming more upset and anxious as her thoughts run through her mind. The language Shakespeare has used enforces Juliet’s fear. The use of repetition used in the soliloquy also portrays Juliet’s desperation making the audience feel her emotion, the length of sentences and lack of punctuation shows the audience the stress and upset which Juliet is experiencing.
In both soliloquies, the same ways have been used to bring out two totally different circumstances. The different effects are brought out by the different words and language Shakespeare uses to make the audience see the different emotions which the soliloquies are used to express within the play.
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Part 1 | Part 2
The classical liberal international order of the nineteenth century was not planned or designed by anyone. It was, for the most part, the natural outgrowth of the expanding influence of a new political philosophy of freedom, free markets, and free trade. It began to emerge in the wake of the twenty-five years of war that Britain and other European countries had been fighting against France until the final defeat of Napoleon in 1815.
During those decades of war, the ideas that came to be identified with classical liberalism — individual liberty, secure and respected private property rights under impartial rule of law, the recognized right of individuals to freely follow occupations of their own choosing and to voluntarily trade with others on mutually agreed-upon terms, and wide respect for civil liberties — had been slowly growing in influence in both intellectual and political circles, especially in Great Britain.
The United States had been founded on those ideas following its independence from the British in 1783 and then recognized to a great (though imperfect) extent in the Constitution of 1787. Here was heralded a new philosophy of man, society, and government. One in which the individual was center stage with certain “inalienable rights” and in which was a far more limited government than had ever been experienced anywhere else in the world.
The ideal of free trade at the center of liberal internationalism
In Great Britain the advancement of classical liberalism included the case for unrestricted freedom of trade and enterprise among the people of the world. It culminated in the unilateral abolition in June 1846 of the agricultural protectionist trade barriers known in Britain as the Corn Laws. In the immediate years after 1846 this radical change to freedom of trade in food was accompanied by the repeal or reduction of all other trade barriers to free trading in industrial and manufacturing goods in Great Britain and the reduction of import taxes to serve as modest government revenue sources.
Furthermore, the British free traders argued that they had no wish to force their free-trade beliefs on others. They hoped that other nations would see the benefits from a world without trade barriers, but they left that up to the domestic decisions of the citizens of other countries and their governments. Whether or not other nations followed the free-trade path, Great Britain would.
While far from eliminating and abolishing domestic and foreign barriers to human association and exchange to the same degree as the British and the Americans, many of the other nations of Europe followed suit in eliminating or reducing most of their trade impediments, guided by that classical liberal ideal.
The depoliticizing of human relationships for the liberal order
For most of the middle decades of the nineteenth century the guiding principle that directed much of public policy in practically all of the countries of the civilized world was the depoliticizing of social and economic life. The mercantilist system of government control and command was removed and in its place arose the private sector of voluntary associations in civil society and the economics of free markets.
As the nineteenth century progressed, men, money, and material goods traveled increasingly freely from one corner of the globe to another, with few political restrictions standing in the way, and most certainly in comparison with the eighteenth century. In addition, the freedom of exchange included worldwide sharing of knowledge about the arts and sciences. The world was becoming a global community of people who were increasingly freer and materially better-off.
Yes, there were international governmental agreements, including international river commissions, railway and transportation agreements, telegraph and postal unions, health rules and guidelines, procedures for uniform weights and measures, and respect for patents and copyrights. Individual governments might also occasionally still try to influence the direction and form of domestic production and trade. Moreover, diplomatic
intrigues and alliances still threatened the peace among nations.
Liberal ideas created international order, not global government.
But there was no League of Nations and no United Nations as part of some grand political design to police and manage international aspects of human life. There was no International Monetary Fund or World Bank or World Trade Organization. The major Western countries each, at some point, formalized their domestic monetary systems on the gold standard, with the result that each followed the “rules” of the gold standard and thus generated a global monetary order without governments’ gathering together to create it. And they followed fiscal policies that, in comparison with our own times, greatly limited government taxing and spending.
There also were attempts to establish some international rules concerning acceptable methods and forms of warfare, and the humane treatment of prisoners of war and civilian populations in wartime-occupied territories. At least among the European powers during the nineteenth century, wars were less frequent, shorter in duration, and less destructive to person and property than in other times. To a great extent it was because the classical liberal ideal of respect for and minimal intrusion into the lives and property of private persons had influenced the thinking and policies of those governments, even if not to the same extent as among the British and the Americans.
However loosely, inconsistently, and incompletely, the attitudes and policies of all the major Western countries were guided by the classical-liberal spirit of the age. The world was not a classical-liberal heaven. Governments did things and private individuals acted in ways that were often far from the notion of a full respect and regard for the rightful liberty of all others. But it came closer to that ideal than had ever been practiced in any earlier time.
This liberal international order emerged and took on its specific forms of itself, again however imperfectly and inconsistently, as one nation after another became more liberal in its domestic affairs; this was reflected in government policies and practices, sometimes unilaterally and sometimes through agreements with other nations that also had been moving in similar liberal directions.
Collectivism and Neo-Liberalism replaced classical liberalism.
This world was shattered by the two world wars, followed by the counterrevolutions of collectivism in their various forms of socialism and communism, fascism, National Socialism (Nazism), and the interventionist-welfare state. Personal freedoms were reduced or extinguished; private-property rights and freedom of exchange were abolished or heavily straitjacketed; international free trade was replaced with trade barriers, restrictions, and prohibitions on the movement of money, goods, and people. Once again, people were made the property of the State to one degree or another. And two destructive world wars were endured, the likes of which had not been experienced before in modern history; tens of millions of people were killed by collectivist regimes or in the wars that collectivism set loose on the world.
So when the Second World War came to an end in 1945, the victorious powers, most especially the United States and its Western allies such as Great Britain and France, were determined to try to escape from the totalitarian madhouse of the 1920s, 1930s, and 1940s.
But the guiding ideas were not the classical liberalism of that earlier pre–World War I era. It was “Neo-Liberalism” that had captured the imagination of progressive thinkers, already before 1914, but which gained adherents and dominance during the 1930s and 1940s in the Western democracies. This was a liberalism that believed that a free society of civil liberties could be maintained, even while government took paternalistic responsibility for economic decision-making through democratic socialism or the interventionist-welfare state.
The re-politicizing of domestic and international relationships
Markets needed to be harnessed from unjust and uncontrolled laissez-faire through government controls and regulations, income redistributions and social (i.e., compulsory) safety nets of planned pensions, health care, housing, employment and product regulations, and activist monetary and fiscal policies inspired by the new Keynesian economics.
International trade was not to be fully market-based and mostly free, as it had been in the nineteenth and early twentieth centuries, especially under British inspiration. Instead, it would be managed trade among the nations of the world, in which governments negotiated and determined what goods and services might be sold among countries, in what quantities and qualities, and at what tariff-influenced prices.
It is certainly true that many of these international trade agreements among governments significantly reduced the barriers to international trade and investment compared with the economic walls that nations had attempted to build around themselves in the 1930s. But it was not free trade, if that is understood as unrestricted trade in the importing and exporting of goods and services between nations of the world based on private consumer and producer decision-making.
In the heyday of classical liberalism, democratically elected government was considered the complement to free markets, with citizens’ choosing those who should hold political office for a period of time. But democracy was not considered the essence of a free society. That core was the freedom of individuals to peacefully live their lives as they chose, without the intervention and tyranny of kings or democratic majorities.
But in the new world of the interventionist-welfare state, democracy became the center of political life, because it was through the political process that special-interest groups vied for privileges and favors at the expense of others, and where those holding or wanting political office could offer other people’s money in exchange for campaign contributions and votes on election day.
The seeming crisis of democracy and the failure of the liberal international order that has brought about these reactionary alternatives of renewed protectionism, economic nationalism, populist politics, and the appeal of authoritarian leaders in various countries is not a crisis or failure of liberalism, if understood in the classical sense that we discussed earlier.
Only classical-liberal internationalism ensures freedom and peace.
Classical liberalism was abandoned, unfortunately, long ago. This crisis is that of Neo-Liberalism, the progressive liberalism of interventionism and welfare statism, the corrupt social liberalism of political power in the hands of domestic and international politicians and bureaucrats using government authority to advance their own social-engineering ends and serving the ideological and wealth-seeking special-interest groups who want control and income at the expense of others.
But the replacements offered for the prevailing system of Neo-Liberal interventionism are all merely variations on the same statist and collectivist theme. They couch themselves in nationalist or religious rhetoric, but they all offer nothing more than more of the same fundamental institutional system: political paternalism, economic privilege, and social tyranny, with just different power-holding elites managing and manipulating the people wanting an end to the existing politicized Neo-Liberal domestic and liberal international order.
And China lurks in the wings, waiting to take its place as the master of the world in place of a declining America, with its own nationalist and socialist vision of the authoritarian state in which the individual is subservient to those in political control. China’s rationalizing appeal to existing and would-be dictators and tyrants around the globe should not be underrated: It is an authoritarian democracy that is offered in place of the corrupt and dysfunctional Neo-Liberal democracy existing in numerous countries.
There is only one answer if liberty, prosperity, and world peace are to be truly restored and maintained. It is a renewal and revival of the classical-liberal internationalism that gave men freedom, began to end human poverty, and established the foundation for global world tranquility by separating the personal and economic affairs of humanity from the grasping and violent hands of government. Any other path will only mean a repetition of the past: power-lusting, special-privilege, tyrannizing social engineers, and international tensions that will threaten new and highly destructive conflicts within and between nations.
This article was originally published in the January 2019 edition of Future of Freedom.
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As long as there have been humans, there have been garbage dumps. Archaeologists call dumps "midden heaps" and sift through them meticulously, searching for the meaning of life--or at least trying to understand what it means to be human.
For thousands upon thousand of years, humans have thrown their detritus into holes in the ground not far from home. And for thousands of years, this made sense. Soil contains an immense number of tiny creatures who spend their time breaking down organic materials into their original constituents. (These creatures make up the "detritus food chain" and they are essential to the wellbeing of the planet, though we rarely hear much about them--after all they are so small they are invisible, and who likes to discuss detritus-eaters anyway?) Bury a banana peel or a dead squirrel for a month or two and they begin to lose form and substance, recycled by members of the detritus food chain back into their original inorganic constituents (oxygen, nitrogen, calcium and so forth). Ashes to ashes, dust to dust. Yes, shallow land burial (a dump) makes good sense for nontoxic materials.
But now dumps have become dangerous because many of the things we throw into them are toxic. Starting about the time of World War II, the nature of our economy changed. We began to turn our backs on the old raw materials--cotton, wood, paper, leather, glass, and iron--and we began substituting new raw materials. Many of the new materials are toxic. Because many of these materials do not occur in nature, the detritus food chain has never developed any members who lunch on them, so the new materials persist in the environment, unable to be broken down efficiently.
As a consequence, if you expose a modern dump (called a landfill) to rain, then collect the rain water that has filtered through the garbage (it's now called "leachate"), you will find that the leachate from a solid waste landfill has about the same toxicity as the leachate from a landfill specially designated for toxic industrial chemicals (a "hazardous waste" landfill). (See RHWN #90.) This should not surprise us. There is one stream of raw materials coming in the front door of a factory. Inside the factory, manufacturing occurs. Two streams of materials leave the factory--one is "hazardous waste" and one is "product," but they are both made from the same stream of raw materials that came in the front door. After the "products" are used, they are discarded into a "municipal solid waste landfill." Leachate from each type of dump--municipal waste vs. hazardous waste--is about equally toxic; no surprise.
For nearly 40 years after World War II, people buried toxic materials in the ground. Make no mistake--industrial chemists knew what they were doing; they knew it was dangerous (see RHWN #97), but it was cheap, and America was on a blind binge of growth and affluence. The modern formula for success became, "Haste plus waste makes profit." And let the devil take the hindmost. Then Love Canal occurred. Suddenly within about five years everyone with a shred of sense came to realize that landfilling the byproducts of the modern economy is certain to cause enormous environmental damage and human misery. In 1980, Congress passed the Superfund law to begin to clean up the past 40 years' abuse. Then Congress began to ban landfilling of the most dangerous materials. U.S. Environmental Protection Agency (EPA) funded studies showing that 86% of all the landfills they studied had actually contaminated groundwater. (See RHWN #71.) Combining these studies with the fundamental principles that underpin physics and chemistry (for example, gravity and the second law of thermodynamics), EPA began saying in the FEDERAL REGISTER that there is good reason to believe that, sooner or later, all landfills will leak, and will contaminate the local environment. (See RHWN #37.) Even more importantly, citizens across the nation took to the streets to prevent the siting of new landfills. The movement for environmental justice was born.
While this scientific and common sense opposition to landfilling was growing, a counter-current was developing among those who make billions of dollars burying poisons in the ground--the waste hauling industry.
During the '70s and '80s, the waste hauling industry was being transformed from thousands of unorganized local haulers into a nationwide network dominated by a few corporations who learned their business techniques from organized crime. (See RHWN #40.) A few visionary and ruthless business leaders saw that citizen opposition to new landfills created a fabulous opportunity. Their initial tactic was to purchase every landfill in sight. The fact that these landfills were leaking dangerous materials into the local environment might seem to be a serious liability, but the waste haulers saw that this was actually an opportunity. The large firms organized themselves into hundreds of small subsidiary corporations, no one of them holding much capital. Then each small firm bought one or two leaking landfills, went to local government, and said, "We own this landfill, which is leaking poisons into your water supplies. If you allow us to expand it, we will be able to make money and we will use part of the money to contain the poisons and do our best to clean up past damage. If you will not allow us to expand it, we will be forced to declare bankruptcy and leave you to clean up the poisons." The federal Superfund cleanup program has shown that the average cost of cleaning up a leaking landfill is $25 million, so most local governments can't think about financing a cleanup themselves. Never mind that the expanded landfill will eventually leak, making the local problem much worse, and that the owner of the landfill will then declare bankruptcy and skip town; for people worried about the short term only (i.e., politicians), the proposal to clean up the site in return for a license to expand is an offer they can't refuse. Once the giant waste haulers had hundreds of landfills under their control, they began to form alliances with garbage incinerator companies; the incinerators can be sited on, or very near, the landfill sites, thus avoiding the troublesome problem of siting a new landfill to handle all that incinerator ash, which is laced with toxic metals (see RHWN #92). It's perfect--especially if the Washingtonbased environmental groups and Congress will cooperate to strip incinerator ash of its "hazardous" label. (See RHWN #85.) Then the toxic ash can be put into the local landfill and no one has any grounds for objection.
Thus the country now finds itself in the throes of a major struggle that will play itself during the 1990s. Those scientists and regulatory officials who haven't been bought by the giant waste haulers plainly state that all landfills leak and can be expected to poison local water supplies. Yet the waste haulers have developed what amounts to a political movement to a maintain landfilling as the nation's principal means of waste disposal.
Local people are tipping the balance in this struggle. The nationwide movement opposing new landfills continues to gain strength, and it is taking a terrible toll on the waste haulers' profits. The waste haulers are having to defend themselves at every turn; this is expensive, it creates bad publicity, and it's a nuisance. Their goal is to make money by burying dangerous wastes in the ground, not to fight with citizens. So now they have developed a new strategy designed to minimize siting battles. They have formed an alliance with the nation's railroads to develop a few giant landfills, each covering one thousand to 40,000 acres, which will accept wastes from cities a thousand (or more) miles away. The term being used to describe these giant new landfills is "megafill."
Only one megafill is operating today--BFI's 880-acre dump near Poland, Ohio. Another megafill on the drawing boards is the 8,300-acre Eagle Mountain Project, currently seeking permits to operate in the desert 200 miles east of Los Angeles. It is designed to accept 20,000 tons of trash each day for 100 years. If successful, it's owner will only have to fight a siting battle once every century.
The 2782-acre Gallatin National landfill in Fairview, Illinois, and the 1200-acre site near Edgemont, South Dakota, are other examples of burgeoning megafills. The two largest proposals are near Lordsburg, New Mexico (23,480 acres), and in Schuyler County, Missouri (40,000 acres). Railroads are involved in each proposal.
Trains make sense for large garbage operations. One train can carry 4000 tons, about the same amount carried by 400 18-wheel semi-trailer trucks. Conrail now hauls about 700 tons of garbage a day, earning roughly $4 million per year. Within 10 years they expect their earnings from this source to hit $100 million.
Soon the deserts of the southwest and prairie hillocks of the
midwest may be graced by megafills yawning open to swallow the
toxic residues of New Jersey and New York. After all, it's got to
go somewhere, doesn't it? Or does it? That's the central
question, and we'll have an answer during the 1990s.
--Peter Montague, Ph.D.
Descriptor terms: landfilling; megafills;
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Aerobic Capacity- Another term used for maximal oxygen uptake (VO2 max).
Anaerobic Capacity- A measure of anaerobic fitness; the maximal work performed in a short burst of high-intensity exercise.
Anaerobic Exercise- Anaerobic means “in the absence of oxygen.” Anaerobic exercise is performed at an intensity so great that the body’s demand for oxygen exceeds its ability to supply it.
Cardiovascular Fitness- The ability of the heart, blood vessels, blood, and respiratory system to supply fuel, especially oxygen, to the muscles and the ability of the muscles to utilize fuel to allow sustained exercise.
Fast Twitch (FT) Muscle Fibers- The muscle fibers primarily used in anaerobic exercise or short, explosive exercise such as sprinting.
Lactic Acid- Substance that results from the process of supplying energy during anaerobic exercise; a cause of muscle fatigue.
Slow Twitch (ST) Muscle Fibers- Muscle fibers primarily used in aerobic or sustained, continuous exercise; also referred to as fatigue-resistant fibers.
Active Stretch- Muscles are stretched by the active contraction of the opposing (antagonist) muscle. For example, when doing a calf stretch exercise, the muscles on the front of the shin contract to cause a stretch of the muscles on the back of the leg.
Agonist Muscles- In this concept, agonist refers to the muscle group being stretched.
Antagonist Muscles- In this concept, antagonist refers to the muscle group opposing (on the opposite side of the limp from the agonist) the group being stretched.
Ballistic Stretch- Muscles are stretched by the force of momentum of a body part that is bounced, swung, or jerked.
Flexibility- Range of motion (ROM) in a joint or group of joints. Because muscle length is a major factor limiting the range of motion, those having long muscles that allow for good joint mobility are considered to have good flexibility.
Muscular Endurance- is the ability of a muscle group to exert submaximal force for extended periods.
Muscular Strength- is defined as the ability of a muscle group to develop maximal contractile force against a resistance in a single contraction.
Passive Stretch- Stretch imposed on a muscle by a force other than the opposing muscle. For example: using another person, gravity, an object or another body part
Range of motion (ROM)- The full motion possible in a joint
Static Stretch- a muscle is slowly stretched and then held in that stretched position for several seconds.
Target Heart Range- is defined as the minimum number of heartbeats in a given amount of time in order to reach the level of exertion necessary for cardiovascular fitness, specific to a person's age, gender, or physical fitness
Time Under Tension- Slowing Down Movement to Speed Up Fitness Results. ... TUT simply refers to the time that a muscle is under load or under strain during a set of a particular exercise.
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10 Characteristics of Children in the Montessori Classroom
1. Respectful: The basis of the Montessori classroom is mutual respect. The teacher respects the child for the person s/he (someone who is coming into being; the child who will become the adult, just as the acorn will become the great oak). The children respect each other. The children respect and their teacher as the person who will help them grow, the person who will coach them, and the one who prepares the environment (physical, intellectually, emotionally) for this unfolding to take place. A classroom based on mutual trust creates trustworthy children. Respect and trust go hand in hand. “The Montessori teacher’s specific responsibility is to aid human development through awareness of the children’s needs at each stage of development.” (Montessori Today, p. 114) By respecting the individual children in their care, Montessori teachers continue to aid the intellectual, emotional and moral development of their students. Respect is at the core of this difficult but fulfilling task.
2. Responsible: In Montessori education children take responsibility for their own learning. There is a certain amount of work for them to do. The child knows that this work will help them and that the process of doing it will make them intellectually prepared and will make them feel good about themselves. But it is their responsibility to do it. In order for them to accomplish this goal they must manage their time. They plan their week and choose what they want to do and when. This is a learning process. They need to make adjustments along the way. But the goal is for them to take responsibility for what they do and once they do this they have “ownership” of it. The teacher will help them in any way that they need. This style of learning is now referred to as “constructivist.” The child constructs his/her own learning. The child learns directly from what he/she does. But children cannot construct their own learning until they are responsible for making their own choices. If they are forced to learn and do not know why they are doing it they merely go through the motions. They feel as if they have no control over their lives and learning can become something that they endure, rather than something that they accomplish by their own self – effort.
3. Self-disciplined : No one is born disciplined. To accomplish self – discipline is one of the great tasks of life. Responsibility requires discipline. In order to discipline yourself you must have the freedom to make choices. This freedom allows you to take charge of yourself and your behavior, both intellectual and social. Children innately desire to be useful and to belong. They want to fit in and behave appropriately. In order to achieve this goal they must become master of themselves. They must achieve self – discipline. They must have the opportunity to control themselves, by themselves. They must not be constantly controlled by others, because this imposes discipline from the outside and does not give the child the opportunity to learn it by him or herself.
4. Independent: “No one is free unless he is independent.” (The Montessori Method, p. 95) In order to be independent the child must be free to make choices, make mistakes, learn from those mistakes and have the opportunity to self – correct. From the moment that a child is born he/she begins the work of becoming independent. It is our job as parents and teachers to aid our children in this task. The goal of life is to become a fulfilled human being with the opportunity to be a part of society and to give back to society that which has been received. In order for children to become independent they must acquire skills, intellectual, social, and physical. When we aid them in the acquisition of what they need, we help them develop independence. We guide them in their growth as we prepare a learning and social environment in which they are able to make their way toward independence. It is experiencing independence in an appropriate environment that they come to acquire their necessary skills.
5. Creative/inventive: Maria Montessori said that everything the child does is creative. It is the creative mind that harnesses all of its power to solve problems. The creative powers of the children help them solve problems throughout their stay in the Montessori classroom. These problems can be social, emotional, intellectual or physical in nature. By allowing the children to grapple with these problems and ideas we help them enlist their own creativity. When children are not permitted to deal with their own problems as they arise, but have adults solve them for them, they have difficulty experiencing their own creative solutions. In many ways the Socratic method of asking questions and allowing children to choose their answers is a way of allowing for creativity to arise. We all know there are many ways to get to a goal. To teach children in only one way, and ask for only one answer, will prevent them from using their minds in an inventive way. Children need projects and tasks that they can tackle. They need to experience their own endless creativity. They must come to see that they can do this is to be allowed to solve appropriate problems and experience the benefit of their own innovative solutions. Some of these problems might be artistic, scientific, intellectual or social. Anytime a child sits down to accomplish something they are drawing on their own reservoir of creativity and inventiveness.
6. Self-Motivated: Children are self-motivated when they are allowed to make choices and have some sense of control over what they elect to do. This does not mean that they do whatever they please. Rather, they find themselves in a prepared environment, which is designed to stimulate them to learn. As the younger children watch the older children in their class they are inspired to emulate them. They perceive the work done by the older children as more difficult, but work that they want to be able to do when they are older. They look up to the older students, much the way a younger sibling looks up to his older sister or brother. In this way the teacher is not the one who constantly has to direct activities. They do not need external pressure in order to do most of their work. Rather, they are eager to get to the next thing. They are also presented with lessons that are interesting and follow-up lessons, which are fun for them to do. In a Montessori classroom every presentation has a follow-up activity that the children do on their own. The presentations and the materials are well thought out and they are designed to appeal to the child’s imagination. A great deal of thought and preparation goes into designing the presentations and the extensions on the part of the teacher. The pay off for the children is that they are naturally curious about the lessons and this makes them want to do the activities because they are intrinsically interested in them. In this way, Montessori students choose to work. Even when certain lessons are required of them, like learning their times tables, they are aware that once they learn them they will be free to do other activities that they are looking forward to. The secret to self-motivation is creating a learning environment where the natural curiosity of the child is stimulated. In the Montessori classroom we are looking for intrinsic motivation as opposed to extrinsic motivation. Children work because they want to – not to get a grade, to please an adult, or because they will get in trouble or fail if they don’t do what the teacher requires. (“Montessori Today, P. 90.)
7. Organized: Montessori children are exposed to the concept of “time-management” at an early age. It is a skill, which is essential for success in the complex society in which we find ourselves. The teacher begins to discuss with the child the idea of choosing work in the math, language and cultural areas. The choice is completely up to the child. If the child is not inclined to make a choice the teacher will help them and invite them or work with a particular material. This is the beginning of reflecting on what you do at school; how you use your time. Children may decide where they will do their work. Some work is best done at a desk, lying on the floor, or sitting in the reading corner. The lessons are also sequenced in a specific manner. They are sequenced this way so the child will be successful. They are also designed so the child learns to do one step at a time in an organized manner. In this way the children do not get overwhelmed and they can achieve the task they set out to do. Organization is one of the key components of the Montessori classroom. The physical environment is also organized according to subject area so the children know where everything is located. Each material is returned to its place so other children can find it. The environment is a consistent place that children can count on. The routines are reliable, the lessons are sequenced and the environment is predictable. The organization gives the child a sense of security and power for they know what do and how to do it.
8. Global Thinker: Montessori education is the largest international school movement in the world. There are Montessori schools located on every continent, except Antarctica. In Montessori schools that offer an elementary program, the program is called “Cosmic Education.” Cosmic Education is based on the concept that all life is interrelated and all humans have the same fundamental needs. If you disturb or affect one aspect of this intricate relationship, you affect everything. Montessori students study the development of the solar system, geography, (physical and political), the ecology of the planet earth, and the flora and fauna located on the earth. They study the planet earth as an ecosystem. In preschool they begin their “global” study by becoming familiar with the planets in their solar system. Then they study their planet, the earth. They study the layers of the earth and then they begin their study of zoology with an overview of vertebrates and invertebrates, animals with and without backbones. They begin botany with the overview of the parts of the plant. This curriculum becomes more sophisticated in the elementary program. There they begin a more in-depth study of the fundamental needs of humans, as well as botany, zoology, history, and geography. In the upper elementary program the students study physics, and chemistry and they expand their study of botany and zoology. They begin the history of world cultures as well as an in-depth study of the history of the state or country in which they live. This complete overview gives them the opportunity to study the chronological development of the solar system and the earth, as well as the development of species, ending with the study of human life and culture. Cosmic Education is taught with a timeline so the students can place developments in order, which gives them a way to hold on to and understand the material, and provides them with a global view that is chronologically developed.
9. Collaborator: Montessori students work cooperatively and collaboratively on a number of tasks. In this way they are able to learn from each other. Often, when children work together, they are of different ages. Because children have their own developmental time scheme, not everyone is at the same place even if they are chronologically in the same age range. This allows the children to respect each other for their various strengths. Because children have different talents, skill and interests, the groups that form together to help each other are constantly changing. Sometimes children want to work together simply because they are friends. Other times they choose each other because they are a good team. Sometimes a child will choose another child because they realize that child knows the lesson and he or she can help them with it. The nature of a collaborative environment is that it feels very reasonable and safe. A child can always turn to another child for help. This is made easier in the multi-aged classroom. It also helps older children when they explain things to younger students. It clarifies their knowledge when they have to articulate it to someone else. The children are also used to working in pairs or small groups to solve problems or discuss history or literature questions. They are used to exchanging ideas. They begin to experience that you can often learn more when you collaborate and cooperate than when you work alone.
10. Leader: When children are encouraged to develop the nine characteristics that we have already discussed, they will have all the qualities that make good leaders and good team players. In life, both are necessary. Children who have been allowed to take responsibility for their work and have developed an essential level of self-discipline and responsibility experience high self-esteem and are prepared for life. They have the requisite skills to go on to higher education, to be successful at what they attempt, and to step forth into the world with the leadership and problem solving skills that they will need to successfully face the tasks that await them. It is our job to prepare our children for the future. The problems that are surfacing now in our society will need to be faced and solved by them. They are our future leaders and it is our job to see that they have the appropriate educational experiences that will prepare them for the adventures and challenges that lie ahead. It is an exciting task and it is the task that Maria Montessori envisioned when she first sought to design a way of teaching our children so they could go forward into society and become the bringers of peace. As Maria Montessori said, “Preventing conflicts is the work of politics; establishing peace is the work of education.” (Education and Peace, p. 24, Maria Montessori)
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Scalar Definition: Scalars vs. Vectors (Difference Between Scalar and Vector)
Matrices > Scalar Definition
To understand what a vector is we need to define what a scalar is. A scalar is a quantity expressed by a single real number. An example of a scalar is length (which can be measured in inches or feet). A scalar has magnitude, or the size of a mathematical object. A vector is a quantity that is defined by multiple scalars. In addition to magnitude, a vector also has a direction.
Consider the following illustration:
The magnitude is located between the initial point (A) and direction (denoted by the lowercase a). Adding the arrow to the line represents the direction. You are most likely to encounter vectors similar to the above on a graph. You need to know how to determine the magnitude and direction between two given points.
Finding the Magnitude of a Vector
Let’s take a look at line AB on the following graph:
To find the magnitude of vector AB we need to determine the distance between points A and B. Where we are provided with the coordinates of the points we can use the distance formula).
The formula for distance is:
AB = √(x2 – x1)2 + (y2 – y1)2
Example:
Find the magnitude of vector AB where the initial point A is (2, 2) and the end point B is (6, 4).
Plug the values into our distance formula:
AB = √(6-2) 2 + (4-2) 2 = √4 2 + 2 2 = √16 + 4 = √20
The magnitude of AB = 4.72
Determining the Direction of a Vector
To find the direction of a vector we measure the angle that the vector makes with a horizontal line. We use the following formula:
tanΘ = y2 – y1/x2 – x1
Example:
What is the direction of vector AB where the initial point A is (2,3) and the end point B is (5,8)
First, we plug the coordinates into our formula for direction:
tanΘ = 8-3/5-2 = 5/3
To find the direction we use the inverse of tan:
Θ = tan-1 (5/3)
Vector AB has a direction of 59 degrees.
Check out our YouTube channel for hundreds of videos on elementary stats, linear algebra and more!
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Statistical concepts explained visually - Includes many concepts such as sample size, hypothesis tests, or logistic regression, explained by Stephanie Glen, founder of StatisticsHowTo.
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Selective Breeding and Shearing
Most wool comes from wrinkly-skinned Merino sheep, selectively bred to produce greater volumes of wool. Wild sheep naturally shed their fleece in the spring, but Merino and other domesticated breeds cannot, and must be sheared.
Ranchers pay sheep shearers by volume, not by the hour. Most shearers work as quickly as they can, and many sheep are handled roughly or injured in the shearing process.
Wild sheep are also sheared before they naturally shed their fleece, and often while it is still very cold. An estimated one million sheep die each year from this exposure, and many contract pneumonia.
Shearing could be replaced with bioclipping, a process that does not require blades. It is safer for ranchers because sheep often struggle during the process, and it is faster than traditional shearing.
Merino sheep raised in Australia are prone to flystrike. Blowflies lay their eggs in the sheep’s wrinkly skin and thick fur, and the maggots feed on the living sheep. If untreated, flystrike can cause sheep to die a slow and dreadful death.
The most common method of prevention is mulesing – the painful surgical removal of wrinkled skin from a sheep’s rump area. While anaesthetics have been approved for use in the process, Australian law does not require their use.
Ranchers could replace mulesing by better blowfly control, or by raising sheep that have less-wrinkly skin. Australia is already promoting sheep breeds with smoother-skinned rumps as an alternative to raising and mulesing Merino sheep.
Other Painful Problems Caused By Wool Production
Sheep subjected to mulesing usually have their tails docked (partially removed) in the process to further reduce the risk of flystrike.
Ranchers castrate male lambs within a few days of birth. Painkillers are approved for the procedure, but are not legally required or commonly used.
Sheep raised in wetter climates are prone to fleece rot and foot rot. Fleece rot often results in flystrike. Foot rot can cause the hoof to separate from soft tissue, which is so painful that affected sheep may refuse to stand on their feet.
Additionally, an estimated 15-23 percent of sheep in New South Wales alone are infected with lice.
And because large-scale ranches raise most of the sheep, it is not unusual for medical problems to go unnoticed.
Sheep ranchers send the animals to slaughter once their wool production declines. Most sheep are held in overcrowded feedlots, packed tightly into ships or trucks, denied food and water for extended periods, and exported to countries with few slaughter regulations. Many sheep die in transit.
Danger to Native Species
Sheep are native to Europe and Asia, and ecosystems of non-native areas can be damaged by large-scale sheep grazing. As an example, Australian sheep ranchers often view animals that are actually native to their country as pests. Ranchers raising sheep for wool kill an estimated five million kangaroos each year in Australia.
What To Do About It
Cotton or other plant textiles are an excellent alternative to wool (and other animal-based textiles). A growing number of companies such as H&M, Abercrombie & Fitch, John Lewis, Perry Ellis, Hugo Boss, Adidas and Liz Claiborne are boycotting Australian Merino wool. The growing demand for more eco-friendly plant-based fibers is making it easier to buy clothing and accessories that are more humane and do not require animals in their production.
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Rett syndrome is a developmental disorder. It almost always affects girls. Like autism, Rett syndrome is termed a pervasive developmental disorder, which means that a child who has it fails to develop normally in a number of areas. The development of a child with Rett's appears normal at birth and for the first 6 months of life. The disorder occurs in about 6 of every 100,000 girls.
Rett syndrome is caused by problems with the MECP2 gene. Because of the damage to this gene, some parts of the brain tend to be smaller than normal in girls with Rett syndrome.
Rett syndrome develops in stages. From birth to about 6 months of age, the development and head size of a child with Rett's appear normal. Between 6 to 18 months of age, symptoms include:
The healthcare provider will examine the child and ask about her symptoms and medical history. The provider may order some tests. These tests may include a CT scan, which produces images of the brain, and blood tests to rule out other medical conditions. The provider may refer your child to a psychologist or other mental health professional for tests of overall intelligence and emotional development.
A treatment plan for Rett syndrome may include medicine for some of the symptoms and physical and speech therapy to improve skills.
In many cases, families are unable to care for a child with Rett syndrome without help. In some cases, home healthcare may enable the child to keep living at home. In other cases, the child may be placed in a supervised living facility. Family counseling may help caregivers cope with the child's disorder.
Rett syndrome is a lifelong, disabling condition. Most women with Rett syndrome require a wheelchair and most live at least into their 40s.
Learn as much as you can about Rett syndrome. For more information, contact the International Rett Syndrome Foundation at http://www.rettsyndrome.org/ or call 800-818-7388.
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Martin Luther King Jr. Day is a special holiday for many big reasons, but one smaller reason why I especially like this winter holiday is because it lends itself so well to activities in the language arts classroom. Whether you teach kindergarten or college, or something in between, there is a wealth of great content related to Dr. King Day and a wide assortment of ways to incorporate language arts lessons into your celebration of this day. Here are a few ideas.
After learning about some of the ways in which Dr. King was a leader in bringing about all of the changes of the civil rights movement, students might be interested in thinking about the different ways people go about trying to effect change. An interesting opinion writing topic might be for students to write about what they think is the best way to go about effecting a change in society.
For narrative writing, students often write about Dr. King’s life. To expand on this idea, students could write about another important African American whom they look up to as a hero. For example: the writer Maya Angelou, abolitionist Frederick Douglass, or aviator Bessie Coleman, just to name a few.
Another idea is for students to begin with one word that is important to the day, define it, and then go on to talk about what is really means to them in detail. This assignment might be a combination of explanatory and opinion writing. For this one, you’ll want “big” words that students have heard often in connection with this holiday’s significance. A few choices might be: diversity, character, civil rights, and tolerance.
A tried and true speech topic for this day is for students to talk about a dream of their own. Or they could take any of the writing topics, such as the hero topic, and turn it into a speech.
Since there is so much reading material available for Martin Luther King Day, this might be the perfect time for a reading lesson that compares two articles on the same topic. For example, you can find different takes on a biography of Dr. King from Biography.com, the King Center, and Ducksters. The first is a standard biography, the second focuses more on his achievements, and the third is written in simpler language for younger kids. Other biographical writings might focus on Dr. King’s place in history or one important event in his life.
There is so much teaching content already out there for this holiday that I even found ready-to-use vocabulary lists that English teachers would find plenty of uses for. Here are two good ones.
Diversity Vocabulary List, from MyVocabulary
Martin Luther King Jr. Vocabulary List, from Enchanted Learning
If you have a space for a holiday display in your classroom, here are a few possibilities for displays that the whole class would have a hand in creating.
- A collage of images from Dr. King’s life – Students collect images and then arrange them on a bulletin board, maybe with printed words mixed in to draw it all together, or with a poem written by a student poet of the day as a caption.
- A display of favorite diversity quotes – Each student chooses and illustrates one favorite quote.
- What I would like to be remembered for – After discussing Dr. King’s legacy, students write brief statements about what they would most like to be remembered for.
Here are some resources for Martin Luther King Day and other classroom holidays. Each of these is an informational text activity with a magazine-style reading page, a page of multiple choice questions, and one additional activity sheet. Click on any of the images to see a preview.
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Facilitated diffusion involves proteins known as carriers, which, as in the case of ion channels, are specific for a certain type of solute and can transport substances in either direction across the membrane. However, unlike channels, they facilitate the movements of solutes across the membrane by physically binding to them on one side of the membrane and releasing them on the other side. The direction of the solute's net movement simply depends on its concentration gradient across the membrane. If the concentration is greater in the cytoplasm, the solute is more likely to bind to the carrier on the cytoplasmic side of the membrane and be released on the extracellular side, and there will be a net movement from inside to outside. If the concentration is greater in the extracellular fluid, the net movement will be from outside to inside. Thus, the net movement always occurs from high concentration to low, just as it does in simple diffusion, but the process is facilitated by the carriers. (Compare with active transport in which energy is expended to move substances against a concentration gradient.)
One example of a carrier is found in the membranes of vertebrate erythrocytes (red blood cells); it transports Cl- ions in one direction and bicarbonate ions (HCO3-) in the opposite direction. This carrier is important in the transport of carbon dioxide by the blood.
A characteristic feature of carrier-mediated transport is that its rate is saturable. That is, if the concentration gradient of a substance is progressively increased, the rate of transport of the substance will increase up to a certain point and then level off. Further increases in the gradient will produce no additional increase in rate. The reason for this is that there is a limited number of carriers in the membrane. When the concentration of the transported substance is raised high enough, all of the carriers will be in use and the capacity of the transport system will be saturated. In contrast, substances that move across the membrane by simple diffusion (that is, diffusion through the lipid bilayer without the assistance of carriers) do not show saturation.
Facilitated diffusion provides the cell with a ready means of preventing the buildup of unwanted molecules within the cell or of taking up from the external medium molecules that may be present there in high concentration. It has three essential characteristics:
Related categories• ANATOMY AND PHYSIOLOGY
Source: Rave, P. H., and Johnson, G. B. Biology, Wm. C. Brown (1996).
Home • About • Copyright © The Worlds of David Darling • Encyclopedia of Alternative Energy • Contact
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# 4.7: Exponential and Logarithmic Models
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Learning Objectives
• Compound and Continuous Interest applications
• Exponential growth and decay models.
• Newton’s Law of Heating and Cooling.
• Logistic-growth models.
In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more techniques to solve these equations, we are able to solve more applications.
## Compound and Continuous Interest Formulas
We will again use the Compound Interest Formulas and so we list them here for reference.
Definition $$\PageIndex{1}$$: Compound Interest
For a principal, $$P$$, invested at an interest rate, $$r$$, for $$t$$ years, the new balance, $$A$$ is:
$$\begin{array}{ll}{A=P\left(1+\frac{r}{n}\right)^{n t}} & {\text { when compounded } n \text { times a year. }} \\ {A=P e^{r t}} & {\text { when compounded continuously. }}\end{array}$$
Recall that compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount $$A$$ accrued in this manner over time $$t$$ is modeled by the compound interest formula $$A(t)=P\left(1+\frac{r}{n}\right)^{n t}$$ where the initial principal $$P$$ is accumulating compound interest at an annual rate $$r$$ where the value $$n$$ represents the number of times the interest is compounded in a year.
Example $$\PageIndex{1}$$r: Compounded interest - Find time and total amount
Susan invested $$$500$$ in an account earning $$4 \frac{1}{2}$$% annual interest that is compounded monthly. a. How much will be in the account after $$3$$ years? $$\qquad$$ b. How long will it take for the amount to grow to$$$750$$?
Solution
State variable values and formula. In this example, the principal $$P =$$ $$500$$, the interest rate $$r = 4 \frac{1}{2}$$% $$= 0.045$$, and because the interest is compounded monthly, $$n = 12$$. The investment can be modeled by the following formula: \begin{align} A(t)&=P\left(1+\frac{r}{n}\right)^{n t} && \text{Formula} \nonumber \\ A(t)&=500\left(1+\frac{0.045}{12}\right)^{12 t} & \nonumber \\ A(t)&=500(1.00375)^{12 t} && \text{Model equation} \label{ex1} \end{align} a. Use this model equation (\ref{ex1}) to calculate the amount in the account after $$t=3$$ years. \begin{aligned} A(\color{Cerulean}{3}\color{black}{)} &=500(1.00375)^{12(\color{Cerulean}{3}\color{black}{)}} \\ &=500(1.00375)^{36} \\ & \approx 572.12 \end{aligned} Rounded off to the nearest cent, after $$3$$ years, the amount accumulated will be$$572.12$$.
b. To calculate the time it takes to accumulate $$750$$, set $$A (t) = 750$$ and solve for $$t$$ using Equation (\ref{ex1}). $$\begin{array}{l}{A(t)=500(1.00375)^{12 t}} \\ {\color{Cerulean}{750}\color{black}{=}500(1.00375)^{12 t}}\end{array}$$ This results in an exponential equation that can be solved by first isolating the exponential expression. \begin{aligned} 750 &=500(1.00375)^{12 t} \\ \frac{750}{500} &=(1.00375)^{12 t} \\ 1.5 &=(1.00375)^{12 t} \end{aligned} At this point take the common logarithm of both sides, apply the power rule for logarithms, and then solve for $$t$$. \begin{aligned} \log (1.5) &=\log (1.00375)^{12 t} \\ \log (1.5) &=12 t \log (1.00375)\\\frac{\log (1.5)}{\color{Cerulean}{12\log (1.00375)}}&\color{black}{=} \frac{\cancel{12}t\cancel{\log (1.00375)}}{\cancel{\color{Cerulean}{12\: \log (1.00375)}}} \\\frac{\log(1.5)}{12\:\log(1.00375)}&=t\end{aligned} Using a calculator we can approximate the time it takes. $$t=\log (1.5) /(12 * \log (1.00375)) \approx 9$$ years Answers: a.$$572.12$$ $$\qquad$$ b. Approximately $$9$$ years
Example $$\PageIndex{2}$$r: Compounded interest - Find time
Mario invested $$$1000$$ in an account earning $$6.3$$% annual interest, that is compounded semi-annually. How long will it take the investment to double? Solution State variable values and formula. Here the principal $$P =$$$$$1,000$$, the interest rate $$r = 6.3$$% $$= 0.063$$, and because the interest is compounded semi-annually $$n = 2$$. This investment can be modeled as follows:
\begin{align*} A(t)&=P\left(1+\frac{r}{n}\right)^{n t} && \text{Formula} \\ A(t)&=1,000\left(1+\frac{0.063}{2}\right)^{2 t} \\ A(t)&=1,000(1.0315)^{2 t} && \text{Model equation} \end{align*}
Since we are looking for the time it takes to double $$$1,000$$, substitute$$$2,000$$ for the resulting amount $$A (t)$$ and then solve for $$t$$.
\begin{aligned} \color{Cerulean}{2,000} &\color{black}{=}1,000(1.0315)^{2 t} \\ \frac{2,000}{1,000} &=(1.0315)^{2 t} \\ 2 &=(1.0315)^{2 t} \end{aligned}
At this point we take the common logarithm of both sides.
\begin{aligned} 2 &=(1.0315)^{2 t} \\ \log 2 &=\log (1.0315)^{2 t} \\ \log 2 &=2 t \log (1.0315) \\ \frac{\log 2}{2 \log (1.0315)} &=t \end{aligned}
Using a calculator we can approximate the time it takes: $$t=\log (2) /(2 * \log (1.0315)) \approx 11.17$$.
So the answer is it takes approximately $$11.17$$ years to double at $$6.3$$%.
If the investment in the previous example was one million dollars, how long would it take to double? To answer this we would use $$P =$$ $$$1,000,000$$ and $$A (t) =$$$$$2,000,000$$:
\begin{aligned} A(t) &=1,000(1.0315)^{2 t} \\ \color{Cerulean}{2,000,000} &\color{black}{=}1,000,000(1.0315)^{2 t} \end{aligned}
Dividing both sides by $$1,000,000$$ we obtain the same exponential function as before.
$$2=(1.0315)^{2 t}$$
Hence, the result will be the same, about $$11.17$$ years. In fact, doubling time is independent of the initial investment $$P$$.
Interest is typically compounded semi-annually $$(n = 2)$$, quarterly $$(n = 4)$$, monthly $$(n = 12)$$, or daily $$(n = 365)$$. However if interest is compounded every instant we obtain a formula for continuously compounding interest: $$A(t)=P e^{r t}$$ where $$P$$ represents the initial principal amount invested, $$r$$ represents the annual interest rate, and $$t$$ represents the time in years the investment is allowed to accrue continuously compounded interest.
Example $$\PageIndex{3}$$r: Continuous compounding - Find time
Mary invested $$$200$$ in an account earning $$5 \frac{3}{4}$$% annual interest that is compounded continuously. How long will it take the investment to grow to$$$350$$?
Solution
State variable values and formula. Here the principal $$P =$$ $$200$$ and the interest rate $$r = 5 \frac{3}{4}$$% $$= 5.75$$% $$= 0.0575$$. Since the interest is compounded continuously, use the formula $$A (t) = Pe^{rt}$$. Hence, the investment can be modeled by the following, \begin{align*} A(t)&=P e^{r t} && \text{Formula} \nonumber \\ A(t)&=200 e^{0.0575 t} && \text{Model equation} \end{align*} To calculate the time it takes to accumulate to$$350$$, set $$A (t) = 350$$ and solve for $$t$$.
$$\begin{array}{r}{A(t)=200 e^{0.0575 t}} \\ {\color{Cerulean}{350}\color{black}{=}200 e^{0.0575 t}}\end{array}$$
Begin by isolating the exponential expression.
\begin{aligned} \frac{350}{200} &=e^{0.0575 t} \\ \frac{7}{4} &=e^{0.0575 t} \\ 1.75 &=e^{0.0575 t} \end{aligned}
Because this exponential has base $$e$$, we choose to take the natural logarithm of both sides and then solve for $$t$$.
$$\begin{array}{l}{\ln (1.75)=\ln e^{0.0575 t}}\quad\quad\color{Cerulean}{Apply\:the\:power\:rule\:for\:logarithms.} \\ {\ln (1.75)=0.0575 t \ln e} \quad\color{Cerulean}{Recall\:that\: \ln e=1.} \\ {\ln (1.75)=0.0575 t \cdot 1} \\ {\frac{\ln (1.75)}{0.0575}=t}\end{array}$$
Using a calculator we can approximate the time it takes:
$$t=\ln (1.75) / 0.0575 \approx 9.73$$ years
It will approximately $$9.73$$ years.
When solving applications involving compound interest, look for the keyword “continuous,” or the keywords that indicate the number of annual compoundings. It is these keywords that determine which formula to choose.
Example $$\PageIndex{4}$$m: Continuously compounded - find interest rate
Jermael’s parents put $$$10,000$$ in investments for his college expenses on his first birthday. They hope the investments will be worth$$$50,000$$ when he turns $$18$$. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?
Solution:
Identify the variables and the formula. $$A =\ 50,000$$, $$P =\ 10,000$$, $$r =?$$, $$t =17$$ years.
\begin{align*} A(t)&=P e^{r t} && \text{Formula} \nonumber \\ 50,000&=10,000 e^{r \cdot 17} && \text{Model equation with substitutions made} \end{align*}
Solve for $$r$$.
$$\begin{array} {cl} A(t)=P e^{r t} & \text{Formula}\\ 50,000=10,000 e^{r \cdot 17} & \text{Model equation with substitutions made} \\ & \text{Now solve for } r\\ 5=e^{17 r} & \text{Divide each side by \(10,000$$ }\\
\ln 5=\ln e^{17 r} & \text{Take the natural log of each side. }\\
\ln 5=17 r \ln e & \text{Use the Power Property. }\\
\ln 5=17 r & \text{Simplify. }\\
\frac{\ln 5}{17}=r & \text{Divide each side by $$17$$. }\\
r \approx 0.095 &\text{Approximate the answer. } \end{array} \)
They need the rate of growth (or interest rate) to be approximately $$9.5$$%.
Try It $$\PageIndex{4}$$rm
1. Mario invested $$$1,000$$ in an account earning $$6.3$$% annual interest that is compounded continuously. How long will it take the investment to double? 2. Hector invests$$$10,000$$ at age $$21$$. He hopes the investments will be worth $$$150,000$$ when he turns $$50$$. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? 3. Rachel invests$$$15,000$$ at age $$25$$. She hopes the investments will be worth \$$$90,000$$ when she turns $$40$$. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?
a. Answer Approximately $$11$$ years. b. Answer $$r \approx 9.3 \%$$ c. Answer $$r \approx 11.9 \%$$
## Exponential Growth and Decay
In real-world applications, we need to model the behavior of a function. Two common examples are instances of rapid growth, where the exponential growth function could be used, or situations where a quantity is falling rapidly toward zero, where the exponential decay model may be a good choice.
EXPONENTIAL Growth and Decay Model, $$y=a_0e^{kt}$$
The function used to model exponential growth or decay is $$A(t)=A_0e^{kt}$$ where $$A$$ is the amount present at time $$t$$ $$t$$ is the amount of time that has elapsed since time $$t=0$$ $$A_0$$ is the initial amount present at time $$t=0$$ $$k$$ is the rate of growth or decay per unit time if $$k>0$$ the function is increasing (Exponential Growth) if $$k<0$$ the function is decreasing (Exponential Decay) The decay factor is $$\frac{A(t+1)}{A(t)}$$ or $$e^k$$ $$e$$ is Euler’s constant Figure $$\PageIndex{5}$$: Exponential Functions model exponential growth ($$k>0$$) $$\quad$$ and $$\quad$$ exponential decay ($$k<0$$)
### Exponential Growth
Example $$\PageIndex{5}$$: Find growth rate and model equation
A population of bacteria doubles every hour. If the culture started with $$10$$ bacteria, construct an equation that models the population as a function of time.
Solution
When an amount grows at a fixed percent per unit time, the growth is exponential. To find $$A_0$$ we use the fact that $$A_0$$ is the amount at time zero, so $$A_0=10$$. To find $$k$$, use the fact that after one hour $$(t=1)$$ the population doubles from $$10$$ to $$20$$.The formula is derived as follows
\begin{align*} y&=A_0e^{kt} &&\text{Formula}\\ 20&= 10e^{k\cdot 1}&&\text{Substitute given values}\\ 2&= e^k &&\text{Divide by 10}\\ \ln2&= k &&\text{Take the natural logarithm} \end{align*}
The growth rate is $$k=\ln(2)$$. $$\ln(2) \approx 0.6931$$ so the equation could be written $$y=10e^{0.6931t}$$. However, a more exact version of the equation can be obtained by writing the equation in terms of $$\color{Cerulean}{e^k}$$, the growth factor. Using some log properties we get $$y=10e^{(\ln2)t}=10{ ( {\color{Cerulean}{e^{\ln2}}} )}^t=10·{\color{Cerulean}{2}}^t$$
Model Equations: using growth rate $$k$$ approximation: $$y=10e^{0.6931t}$$ or more exactly: $$y=10·2^t$$
Analysis. The population of bacteria after ten hours using $$y=10·2^t$$ is $$10,240$$. We could describe this amount is being of the order of magnitude $$10^4$$. The population of bacteria after twenty hours is $$10,485,760$$ which is of the order of magnitude $$10^7$$, so we could say that the population has increased by three orders of magnitude in ten hours.
A cautionary word about rounding. If the rounded approximation $$k=0.6931$$ had been used, the result would be $$10,235$$. If the full precision of your calculator is used to represent $$k$$ by using the [2nd][ANS] feature or [STO] and [MEMVAR] functions, for example, the result of $$10,240$$ would still be obtained. This illustrates that using rounded intermediate results produces rounding errors, so avoid getting answers that sometimes are significantly "off." Always use exact values or the full precision of your calculator in calculations to reduce or eliminate rounding errors.
Example $$\PageIndex{6}$$r: Find model equation, total amount and time, given a growth rate
It is estimated that the population of a certain small town is $$93,000$$ people with an annual growth rate of $$2.6$$%. If the population continues to increase exponentially at this rate:
1. Estimate the population in $$7$$ years' time.
2. Estimate the time it will take for the population to reach 120,000 people.
Solution
We begin by constructing a mathematical model based on the given information. Here the initial population $$P_{0} = 93,000$$ people and the growth rate $$r = 2.6$$% $$= 0.026$$. The following model gives population in terms of time measured in years:
\begin{align} A(t) &=A_0e^{kt} & &\text{Formula} \nonumber \\ P(t)&=93,000 e^{0.026 t} &&\text{Model equation} \label{ex5} \end{align}
a. Use this model equation (\ref{ex5}) to estimate the population in $$t = 7$$ years.
\begin{aligned} P(t) &=93,000 e^{0006(\color{Cerulean}{7}\color{black}{)}} \\ &=93,000 e^{0.182} \\ & \approx 111,564 \quad people \end{aligned}
b. Use this model equation (\ref{ex5}) to determine the time it takes to reach $$P (t) = 120,000$$ people.
\begin{aligned} P(t) &=93,000 e^{0.026 t} \\ \color{Cerulean}{120,000} &\color{black}{=}93,000 e^{0.026 t} \\ \frac{120,000}{93,000} &=e^{0.026 t} \\ \frac{40}{31} &=e^{0.026 t} \end{aligned}
Take the natural logarithm of both sides and then solve for $$t$$.
$$\ln \left(\frac{40}{31}\right)=\ln e^{0.026 t}$$
$$\ln \left(\frac{40}{31}\right)=0.026 t \ln e$$
$$\ln \left(\frac{40}{31}\right)=0.026 t \cdot 1$$
$$\frac{\ln \left(\frac{40}{31}\right)}{0.026}=t$$
Using a calculator,
$$t=\ln (40 / 31) / 0.026 \approx 9.8\quad years$$
1. $$111,564$$ people
2. $$9.8$$ years
Often the growth rate $$k$$ is not given. In this case, we look for some other information so that we can determine it and then construct a mathematical model. The general steps are outlined in the following example.
Example $$\PageIndex{7}$$: Find growth rate, model equation, and total amount
Under optimal conditions Escherichia coli (E. coli) bacteria will grow exponentially with a doubling time of $$20$$ minutes. If $$1,000$$ E. coli cells are placed in a Petri dish and maintained under optimal conditions, how many E. coli cells will be present in $$2$$ hours?
Solution
The goal is to use the given information to construct a mathematical model based on the formula $$A(t)=A_0e^{kt}$$.
Step 1: Find the growth rate $$k$$. Use the fact that the initial amount, $$A_{0} = 1,000$$ cells, doubles in $$20$$ minutes. That is, $$A(t) = 2,000$$ cells when $$t = 20$$ minutes.
\begin{aligned} A(t) &=A_{0} e^{k t} \\ \color{Cerulean}{2,000} &\color{black}{=}1,000 e^{k \color{Cerulean}{20}} \end{aligned}
Solve for the only variable $$k$$.
\begin{aligned} 2,000 &=1,000 e^{k 20} \\ \frac{2,000}{1,000} &=e^{k 20} \\ 2 &=e^{k 20} \\ \ln (2) &=\ln e^{k 20} \\ \ln (2) &=k 20 \ln e \\ \ln (2) &=k 20 \cdot 1 \\ \frac{\ln (2)}{20} &=k \end{aligned}
Step 2: Write a mathematical model based on the given information for the number of E. coli cells in terms of time in minutes. Here $$k ≈ 0.0347$$, which is about $$3.5$$% growth rate per minute. So we could use the equation $$A(t)=1,000 e^{ 0.0347 t}$$ for our model. However, we will use the exact value for $$k$$ in our model. This will allow us to avoid round-off error in the final result. Use $$A_{0} = 1,000$$ and $$k=\ln (2) / 20$$:
\begin{aligned} A(t)&=1,000 e^{ (\ln (2) / 20) t} \\ &=1,000 e^{\tfrac{\ln (2) }{20} t} \\ &=1,000 e^{\ln (2) \tfrac{t}{20} } \\ &=1,000 (e^{\ln (2)})^ {t/20 } \\ A(t)&=1,000 (2)^ {t/20 } && \text{Alternate form of Model equation} \end{aligned}
This equation models the number of E. coli cells in terms of time in minutes.
Step 3: Use the function to answer the questions. In this case, we are asked to find the number of cells present in $$2$$ hours. Because time is measured in minutes, use $$t = 120$$ minutes to calculate the number of E. coli cells.
\begin{aligned} A( {\color{Cerulean}{120}} ) &=1,000 \cdot (2)^{ {\color{Cerulean}{120}} /20} \\ &=1,000 \cdot 2^{6} \\ &=64,000 \text { cells } \end{aligned}
In two hours $$64,000$$ cells will be present. (The answer obtained when $$k= 0.0347$$ is $$64,328$$ cells.)
We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.
Example $$\PageIndex{8}$$m: Find growth rate, model equation, and total amount
Researchers recorded that a certain bacteria population grew from $$100$$ to $$300$$ in $$3$$ hours. At this rate of growth, how many bacteria will there be $$24$$ hours from the start of the experiment?
Solution:
This problem requires two main steps. First we must find the unknown rate, $$k$$. Then we use that value of $$k$$ to help us find the unknown number of bacteria. Use the formula $$A(t)=A_0e^{kt}$$.
Step 1: Find the growth rate $$k$$. Identify values for the variables in the formula. $$A =300$$, $$A_{0} =100$$, $$k =?$$, $$t =3$$ hours.
\begin{aligned} A(t) &=A_{0} e^{k t} &&\text{formula} \\ 300&=100 e^{k \cdot 3} &&\text{substitute. only one variable left - solve for k} \\ 3&=e^{3 k} \\ \ln 3&=\ln e^{3 k} &&\text{Take the natural log of each side.} \\ \ln 3&=3 k \ln e &&\text{Use the Power Property. } \\ \ln 3&=3 k \ln e &&\text{ } \\ \frac{\ln 3}{3}&=k &&\text{ln e = 1 } \\ k &\approx 0.366 \\ \end{aligned}
Step 2: Write the model equation using $$A_{0} =100$$ and $$k =\frac{\ln 3}{3}$$. We use this revised equation to predict the number of bacteria $$A(t)$$ there will be in $$t$$ hours.
$$A(t)=100 e^{\frac{\ln 3}{3} t} \qquad$$ or $$\qquad A(t)=100 e^{0.366 t} \qquad$$ Model equations
Step 3: Use the function to answer the questions. How many bacteria will there be $$t$$ hours from the start of the experiment? Substitute in the value $$t=24$$ and evaluate.
$$A(24)=100 e^{\frac{\ln 3}{3} \cdot 24}$$
$$A(24) = 656,100$$
At this rate of growth, they can expect $$656,100$$ bacteria. (The answer obtained when $$k= 0.366$$ is $$652,894 \;$$ bacteria.)
Example $$\PageIndex{9}$$: Find model if not given initial amount
According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.
Solution
Start with the formula Use the formula $$A(t)=A_0e^{kt}$$.
Step 1. Find the growth rate $$k$$. Identify values for the variables in the formula. $$A =2 A_{0}$$, $$k =?$$, $$t =2$$ years.
\begin{aligned} A(t) &=A_{0} e^{k t} &&\text{formula} \\ 2 A_{0}&=A_{0} e^{k \cdot 2} \\ 2&=e^{2 k} &&\text{DIvide both sides by }A_{0} \\ \ln 2&=\ln e^{2 k} &&\text{Take the natural log of each side.} \\ \ln 2&=2k \ln e &&\text{Use the Power Property. } \\ \frac{\ln 2}{2}&=k &&\text{ln e = 1 } \\ \end{aligned}
The function is $$A(t)=A_0e^{\tfrac{\ln2}{2}t}$$ or $$A(t)=A_0(2)^{\ce{t/2}}$$.
Try It $$\PageIndex{9}$$ma
1. Researchers recorded that a certain bacteria population grew from $$100$$ to $$500$$ in $$6$$ hours. At this rate of growth, how many bacteria will there be $$24$$ hours from the start of the experiment?
2. Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.
a. Answer There will be $$62,500$$ bacteria. b. Answer $$f(t)=A_0e^{\tfrac{\ln2}{3}t}$$ or $$A(t)=A_0(2)^{t/3}$$
### Exponential Decay
When the growth rate is negative the function models exponential decay. With exponential decay, quantities are declining at a rapid rate. The rate of decay $$k$$, is directly proportional to the size of the quantity and is negative. The main example of exponential decay is radioactive decay. Radioactive elements and isotopes spontaneously emit subatomic particles, and this process gradually changes the substance into a different isotope. For example, the radioactive isotope Uranium-238 eventually decays into the stable isotope Lead-206. This is a random process for individual atoms, but overall the mass of the substance decreases according to the exponential decay formula. One of the common terms associated with exponential decay is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.
Example $$\PageIndex{10}$$n: Exponential decay - find the equation
Suppose that a certain population of bacteria has an annual decay rate of 10%. Starting with 20,000 bacteria, how many bacteria will be left in 3 years?
Solution
Start with the formula $$A(t)=A_0e^{kt}$$.
Identify values for the variables in the formula. $$A =?$$, $$A_{0}=20,000$$, $$k =10 \% \text{ decay } \longrightarrow k=-0.10$$, $$t =3$$ years.
\begin{aligned} A(t) &=A_{0} e^{k t} &&\text{formula} \\ A(3)&=20000 e^{-0.10 \cdot 3} \\ A(3)&\approx 148.1636 &&\text{grams} \\ \end{aligned}
Example $$\PageIndex{11}$$c: Exponential decay - find the equation and original amount
The loudness of a sound dies down at an exponential rate of 6.91. If it takes .2 seconds for a loud clap of thunder to diminish to the sound level of a whisper (of 30 decibels), how loud was the clap of thunder originally?
Solution
Start with the formula $$A(t)=A_0e^{kt}$$.
Identify values for the variables in the formula. $$A =30$$, $$A_{0}=?$$, $$|k| = 6.91 \text{ decay } \longrightarrow k=-6.91$$, $$t =.2$$ seconds.
\begin{aligned} A(t) &=A_{0} e^{k t} &&\text{formula} \\ 30&=A_{0} e^{-6.91 \cdot .2} \\ 30e^{6.91 \cdot .2} &=A_{0} \\ &=A_{0} \approx 119.5 &&\text{decibels} \\ \end{aligned}
Example $$\PageIndex{12}$$c: Exponential decay - find the equation and elapsed time
A lawmaker once introduced legislation that would force Americans to cut back on their consumption of salt. It never passed the House, but it proposed that each year restaurants would be mandated to decrease sodium levels by two and a half percent annually. If all restaurants started out using a collective total of 5,000 kilograms of salt initially annually, and they were asked to reduce their consumption by two and a half percent each year, how long would it take to reduce their collective consumption to 3000 kg per year?
Start with the formula $$A(t)=A_0e^{kt}$$.
Identify values for the variables in the formula. $$A =3000$$, $$A_{0}=5000$$, $$k = 2.5 \% \text{ decrease } \longrightarrow k=-.025$$, $$t =?$$ seconds.
\begin{aligned} A(t) &=A_{0} e^{k t} &&\text{Formula} \\ 3000&=5000 e^{-.025 t} &&\text{Substitute values. Now solve for t} \\ \frac {3000}{5000}&=e^{-.025 t} \\ \ln(.6) &=\ln (e^{-.025 t}) &&\text{Take \( \ln of both sides} \\
\ln(.6) &=-.025 t \ln (e) &&\text{Power rule} \\
\ln(.6) &=-.025 t && \ln(e)=1 \\
t &=\dfrac{\ln(.6)}{-.025 } \\
t &\approx 20.4 &&\text{years} \\
\end{aligned}\)
#### Half-Life
Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance. As with exponential growth processes that have a fixed doubling time. Similarly, exponential decay processes have a fixed half-life, the time in which one-half the original amount decays.
Example $$\PageIndex{13}$$: Radioactive Decay (half life) - find decay rate and equation
The half-life of carbon-14 is $$5,730$$ years. Express the amount of carbon-14 remaining as a function of time, $$t$$.
Solution
Start with the formula $$A(t)=A_0e^{kt}$$.
Identify values for the variables in the formula. $$A =\frac{1}{2}A_{0}$$, $$A_{0}=?$$, $$k = ?$$, $$t =5730$$ years.
\begin{align*} A&= A_0e^{kt} \qquad &&\text{The continuous growth formula}\\ 0.5A_0&= A_0e^{k\cdot 5730} \qquad &&\text{Substitute the half-life for t and } 0.5A_0 \text{ for } f(t)\\ 0.5&= e^{5730k} &&\qquad \text{Divide by } A_0\\ \ln(0.5)&= 5730k \qquad &&\text{Take the natural log of both sides}\\ k&= \dfrac{\ln(0.5)}{5730} \qquad &&\text{Divide by the coefficient of k}\\ k & \approx −1.2097×10^{−4} \end{align*}
The function that describes this continuous decay is $$A(t)=A_0e^{\left (\tfrac{\ln(0.5)}{5730} \right )t}$$. We observe that $$k$$ is negative, as expected in the case of exponential decay. Other forms of this function include $$A(t)=A_0e^{.00012t}$$ and $$A(t)=A_0\left (\tfrac{1}{2} \right )^{t/5730}$$
Example $$\PageIndex{14}$$m: Exponential Decay - find decay rate, equation, amount remaining
The half-life of radium-226 is $$1,590$$ years. How much of a $$100$$ mg sample will be left in $$500$$ years?
Solution:
Start with the formula $$A(t)=A_0e^{kt}$$. Identify values for the variables in the formula.
Step 1. Identify values of the variables needed to find the decay constant $$k$$: At time t=1590 years, $$A(1590) =\frac{1}{2}A_{0}$$, $$A_{0}=?$$, $$k = ?$$, $$t =1590$$ years. Since the next part of the question asks how much of a $$100$$ mg sample will be left in $$500$$ years, we may also choose to use $$A(1590) =50$$ mg, and $$A_{0}=100$$ mg instead.
\begin{align*} A(t)&= A_0e^{kt} &&\text{The continuous growth formula}\\ 50&=100 e^{k \cdot 1590} &&\text{Substitute values. Now solve for k} \\ 0.5&=e^{1590 k} &&\text{ } \\ \ln 0.5&=\ln e^{1590 k} &&\text{Take the natural log of each side. } \\ \ln 0.5&=1590 k \ln e &&\text{Use the Power Property. } \\ \ \ln 0.5&=1590 k && \ln(e)=1 \\ k &=\frac{\ln 0.5}{1590} &&\text{exact answer} \\ k &\approx -.00044 &&\text{approximate answer} \end{align*}
Step 2: Write the model equation that predicts the amount of radium-226 $$A(t)$$ there will be left in $$t$$ years. The decay factor is $$e^k = e^{\ce(\ln 0.5)/1590}} = (e^{\ln 0.5})^{\ce{1/1590}} =(.5) ^ {\frac{1}{1590}}$$. Also $$e^{kt} = (e^k)^t$$
$$A(t)= 100 (.5)^{ t/1590 } \qquad$$ or $$\qquad A(t) \approx 100 \: e^{-.00044t} \qquad$$ Model Equations
Step 3: Use the model equation to answer the question. How much of a $$100$$ mg sample will be left in $$500$$ years? Here the values of the variables are $$A(500) =?$$ mg and $$t =500$$ years.
\begin{align*} A(500) &=100 (.5)^{ 500/1590} \\ & \approx 80.41 \end{align*}
In $$500$$ years there would be approximately $$80.4$$ mg remaining. (The answer obtained when $$k=-.00044$$ is $$80.3$$ mg.)
Example $$\PageIndex{15}$$r: Exponential Decay - find decay rate, model equation, and elapsed time
Due to radioactive decay, caesium-137 has a half-life of $$30$$ years. How long will it take a $$50$$-milligram sample to decay to $$10$$ milligrams?
Solution:
Start with the formula $$A(t)=A_0e^{kt}$$. Identify values for the variables in the formula.
Step 1: Find the growth rate $$k$$. Identify values for the variables in the formula: $$A(30) =25$$ mg, $$A_{0}=50$$ mg, $$k = ?$$, $$t =30$$ years.
\begin{aligned} A(t)&=A_0e^{kt} \\ 25&=50 e^{k \cdot 30} \\ \frac{25}{50}&=e^{30 k} \\ \ln \left(\frac{1}{2}\right)&=\ln e^{30 k} \\ \ln (.5) &=30 k \ln e &&\color{Cerulean} { \text{Recall that }\ln e=1. }\\ k &=\frac{\ln (.5)}{30} \\ k &\approx -0.0231049 \end{aligned}
Step 2: Write the model equation to predict the amount of caesium-137 that will be left in a $$50$$-milligram sample after $$t$$ years. Note that $$k \approx-0.0231$$ is negative. However, we will use the exact value to construct a model that gives the amount of cesium-137 with respect to time in years.
\begin{aligned} A(t)&=50 e^{(ln (.5) / 30) t}\\ &=50 e^{ln (.5) t/30}\\ &=50 \left( e^{(ln (.5)} \right)^{ t/30} &&\color{Cerulean} { \text{Use the Inverse Property }e^{\ln M}=M. } \\ A(t)&=50 (.5)^{ t/30} &&\text{or alternatively } A(t)=50e^{-0.0231t} \text{ Model equations} \end{aligned}
Step 3: Use the model to find $$t$$ when $$A (t) = 10$$ milligrams.
\begin{aligned}10&=50 (.5)^ {t/30} \\ \frac{10}{50}&=(.5)^{t/30} \\ \ln (.2) &=\ln (.5)^{t/30}\\ \ln(.2)&=t/30 \ln (.5) \\ \ln(.2)&=\dfrac{\ln (.5)}{30} t\\ t &= \ln(.2) \cdot \dfrac {30}{\ln (.5)}\\ t &=\frac{30 \ln(.2)}{\ln .5} \\ t &\approx 69.66 \end{aligned}
Answer: It will take $$t ≈ 69.66$$ years to decay to $$10$$ milligrams. (The answer obtained when $$k=-0.0231$$ is $$t ≈ 69.67.)$$
The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about $$1\%$$ error for plants or animals that died within the last $$60,000$$ years. As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated. While the carbon-14 technique only works on plants and animals, there are other similar dating techniques, using other radioactive isotopes, that are used to date rocks and other inorganic matter.
Example $$\PageIndex{16}$$r: Carbon dating - find decay rate, model equation, and elapsed time
An ancient bone tool is found to contain $$25$$% of the carbon-14 normally found in bone. Given that carbon-14 has a half-life of $$5,730$$ years, estimate the age of the tool.
Solution
Start with the formula $$A(t)=A_0e^{kt}$$. Identify values for the variables in the formula.
Step 1: Find the growth rate $$k$$. Identify values for the variables in the formula: $$A(5,730) =\frac{1}{2}A_0$$ mg, $$A_{0}=?$$ mg, $$k = ?$$, $$t =5,730$$ years. Notice below that when both sides are divided by $$A_{0}$$, the remaining exponential equation is only in terms of $$k$$. This shows that half-life is independent of the initial amount.
\begin{aligned} A(t)&=A_0e^{kt} \\ \frac{1}{2} A_{0}&=A_{0} e^{k 5,730} \\ .5 &=e^{k5,730} \\ \ln (.5) &=\ln e^{k 5,730} \\ \ln (.5) &= 5,730k\ln e \\ k &= \frac{\ln (.5) }{5,730} \\ k &\approx -.00012 \end{aligned}
Step 2: Write the model equation that predicts the amount of carbon-14 remaining in the bone $$t$$ years after death of the object.
$$A(t) =A_0e^{-.00012t} \qquad \text{or} \qquad A(t) =A_{0} e^{ \frac{\ln (.5) }{5,730} t} =A_{0} \left( e^{\ln (.5) }\right) ^\frac {t}{5,730} =A_{0} \: (.5)^{t/5730} \quad \text{ Model equations}$$
Step 3: Use the model to find the time $$t$$ it takes the carbon-14 to decay to $$25$$% of the initial amount, or $$A (t) = 0.25A_{0}$$
\begin{aligned} A(t)&=A_{0} \: (.5)^{t/5730} \\ 0.25A_{0} &=A_{0} (.5)^ {t/5730} \\ 0.25 &=(.5)^ {t/5730} \\ \ln (0.25) &=\ln (.5)^ {t/5730} \\ \ln (0.25) &=t/5730 \: ( \ln (.5)) \\ t &= \frac{5,730 \ln (0.25)}{\ln (.5)} \\ t &\approx 11,460\end{aligned}
Answer: The tool is approximately $$11,460\;$$ years old. (The answer obtained when $$k=-.00012$$ is $$11,552$$ years old).
Try It $$\PageIndex{16a}$$A
a. Researchers recorded that a certain bacteria population declined from $$700,000$$ to $$400,000$$ in $$5$$ hours after the administration of medication. At this rate of decay, how many bacteria will there be $$24$$ hours from the start of the experiment?
b. The half-life of plutonium-244 is $$80,000,000$$ years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.
a. Answer There will be $$47,699$$ bacteria. ( $$k=\frac{1}{5} \ln ( \frac{4}{7} ) \approx -.11192$$ ) b. Answer $$f(t)=A_0e^{−0.0000000087t}$$ ( $$k = \frac{1}{8} \ln (.5) \times 10^{-7}$$ )
Try It $$\PageIndex{16b}$$m
a. The half-life of strontium-90 is about $$28$$ years. How long will it take a $$36$$ milligram sample of strontium-90 to decay to $$30$$ milligrams?
b. Cesium-137 has a half-life of about $$30$$ years. If we begin with $$200$$ mg of cesium-137, how long will it take until only $$1$$ milligram remains?
c. The half-life of magnesium-27 is $$9.45$$ minutes. How much of a $$10$$-mg sample will be left in $$6$$ minutes?
d. The half-life of radioactive iodine is $$60$$ days. How much of a $$50$$-mg sample will be left in $$40$$ days?
a .Answer $$7.4$$ years b .Answer $$229.3$$ years c .Answer There will be $$6.43$$ mg left. d .Answer There will be $$31.5$$ mg left.
## Newton’s Law of Cooling
Exponential decay can also be applied to temperature. When a hot object is put into a place that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature.
NEWTON’S LAW OF COOLING
The temperature of an object, $$T$$, in surrounding air with temperature $$S$$ will behave according to the formula
$$T(t)=Ae^{kt}+S \quad \begin{cases} t \;\;\text{ is time}\\ A \;\;\text{ is the difference between the initial temperature of the object and the surroundings} \\ k \;\;\text{ is a constant, the continuous rate of cooling of the object }\\ \end{cases}$$
Example $$\PageIndex{17}$$: Using Newton’s Law of Cooling
A cheesecake is taken out of the oven with an ideal internal temperature of $$165°F$$, and is placed into a $$35°F$$ refrigerator. After $$10$$ minutes, the cheesecake has cooled to $$150°F$$. If we must wait until the cheesecake has cooled to $$70°F$$ before we eat it, how long will we have to wait?
Solution
Because the surrounding air temperature in the refrigerator is $$35$$ degrees, the cheesecake’s temperature will decay exponentially toward $$35$$, following the equation
$$T(t)=Ae^{kt}+35$$
Step 1. Find $$A$$. We know the initial temperature was $$165$$, so $$T(0)=165$$.
\begin{align*} 165&= Ae^{k0}+35 &&\text{Substitute } (0,165)\\ 165&= A+35 \\ A&= 130 &&\text{Solve for A} \end{align*}
At the conclusion of this step, we now have a modifed formula, $$T(t)=130e^{kt}+35$$
Step 2. Find $$k$$. To do this, use our modified formula, $$T(t)=130e^{kt}+35$$, and the fact that $$T(10)=150$$.
\begin{align*} T(t)&=130e^{kt}+35 \\ 150&= 130e^{k10}+35 \qquad &&\text{Substitute } (10, 150)\\ 115&= 130e^{k10} \qquad &&\text{Subtract 35}\\ \dfrac{115}{130}&= e^{10k} \qquad &&\text{Divide by 130}\\ \ln\left (\dfrac{115}{130} \right )&= 10k \qquad &&\text{Take the natural log of both sides}\\ k&= \dfrac{\ln \left (\dfrac{115}{130} \right )}{10}\\ &= -0.0123 \qquad &&\text{Divide by the coefficient of k} \end{align*}
At the end of this step we now have the equation for the cooling of the cheesecake: $$T(t)=130e^{–0.0123t}+35$$.
Another form this equation can take is $$T(t)=130e^{\frac{\ln \left( \tfrac{115}{130} \right) }{10}}+35$$ which simplifies to $$T(t)=130 \left( \frac{115}{130} \right )^\tfrac{t}{10}+35$$.
Step 3. Now we can solve for the time it will take for the temperature to cool to $$70$$ degrees.
\begin{align*} 70&= 130e^{-0.0123t}+35 \qquad \text{Substitute in 70 for } T(t)\\ 35&= 130e^{-0.0123t} \qquad \text{Subtract 35}\\ \dfrac{35}{130}&= e^{-0.0123t} \qquad \text{Divide by 130}\\ \ln \left (\dfrac{35}{130} \right )&= -0.0123t \qquad \text{Take the natural log of both sides}\\ t&= \dfrac{\ln \left (\dfrac{35}{130} \right )}{-0.0123}\\ &\approx 106.68 \qquad \text{Divide by the coefficient of t} \end{align*}
It will take about $$107$$ minutes, or one hour and $$47$$ minutes, for the cheesecake to cool to $$70°F$$. (The answer, rounded to the nearest minute, obtained when $$k=\frac{\ln \left (\tfrac{115}{130} \right )$$ is the same).
Try It $$\PageIndex{17}$$
A pitcher of water at $$40$$ degrees Fahrenheit is placed into a $$70$$ degree room. One hour later, the temperature has risen to $$45$$ degrees. How long will it take for the temperature to rise to $$60$$ degrees?
$$6.026$$ hours
## Logistic Growth Models
Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over $$17$$ billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.
The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants $$a$$, $$b$$, and $$c$$, the logistic growth of a population over time $$x$$ is represented by the model
$$f(x)=\dfrac{c}{1+ae^{−bx}}$$
The graph in Figure $$\PageIndex{6}$$ shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.
LOGISTIC GROWTH
The logistic growth model is $f(x)=\dfrac{c}{1+ae^{−bx}}$ where $$\dfrac{c}{1+a}$$ is the initial value, $$f(0)$$ $$c$$ is the carrying capacity, or limiting value, $$f(\infty )$$ $$b$$ is a constant determined by the rate of growth. Figure $$\PageIndex{6}$$
Example $$\PageIndex{18}$$: Using the Logistic-Growth Model
An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.
For example, at time $$t=0$$ there is one person in a community of $$1,000$$ people who has the flu. So, in that community, at most $$1,000$$ people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is $$b=0.6030$$. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.
Solution
We substitute the given data into the logistic growth model $$f(x)=\dfrac{c}{1+ae^{−bx}}$$
Step 1. Find one of the constants by evaluating $$f$$ when $$x \rightarrow \infty$$. This results in $$e^{-bx}$$ approaching zero, so $$f(x \rightarrow \infty) = c$$. Because at most the entire population of the community, $$1,000$$ people, can eventually get the flu, we know the limiting value is $$c=1000$$. Step 2. Find one of the constants by evaluating $$f$$ when $$x=0$$. We are given $$f(0) = 1$$. Therefore $$\dfrac{c}{1+a}=1$$, from which it follows that $$a=999$$. Step 3. Given that the logistic growth constant is $$b=0.6030$$, we obtain the model equation $$f(x)=\dfrac{1000}{1+999e^{−0.6030x}}$$. Step 4. Use the model. This model predicts that, after ten days, the number of people who have had the flu is $$f(10)=\dfrac{1000}{1+999e^{−0.6030\cdot 10}}≈293.8$$. Because the actual number must be a whole number (a person has either had the flu or not) we round to $$294$$. In the long term, the number of people who will contract the flu is the limiting value, $$c=1000$$. Figure $$\PageIndex{7}$$: The graph of $$f(x)=\dfrac{1000}{1+999e^{−0.6030x}}$$ After a long time has passed, the entire community of 1000 people will have had this flu.
Analysis. Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values. The graph in Figure $$\PageIndex{7}$$ gives a good picture of how this model fits the data.
Try It $$\PageIndex{18}$$
Using the model in the Example above, estimate the number of cases of flu on day $$15$$.
$$895 \;$$ cases on day $$15$$
## Key Concepts
• The basic exponential function is $$f(x)=ab^x$$. If $$b>1$$, we have exponential growth; if $$0<b<1$$, we have exponential decay.
• We can also write this formula in terms of continuous growth as $$A=A_0e^{kx},$$ where $$A_0$$ is the starting value. If $$A_0$$ is positive, then we have exponential growth when $$k>0$$ and exponential decay when $$k<0$$.
• In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay.
• We can find the age of an organic artifact by measuring the amount of carbon-14 (which is subject to exponential decay) remaining in the artifact.
• Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay.
• We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time.
• We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.
• Any exponential function with the form $$y=A_0b^x$$ can be rewritten as an equivalent exponential function with the form $$y=A_0e^{kx}$$ where $$k=\ln b$$.
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Sir Ferdinando Gorges
Sir Ferdinando Gorges
The English colonizer and soldier Sir Ferdinando Gorges (1568-1647) was an important promoter of New England colonization.
Ferdinando Gorges's career covered the years from the defeat of the Spanish Armada (1588) to the surrender of Charles I (1645) during the English civil war. This was the era when English efforts at North American colonization became successful, and Gorges was a significant figure in that drama. While he never established a permanent settlement or profited financially from his endeavors, he founded Maine and kept the idea of colonization in New England alive.
Gorges was born into a prosperous landowning family in Dorset. Although he was an infant when his father died, Gorges received a good education and entered into a military career. He served in Holland against the Spanish and in France in support of Henry IV's struggle for the French throne. While in France he was knighted by the Earl of Essex. Queen Elizabeth later rewarded Gorges by appointing him commander of the fort at Plymouth, responsible for organizing the defense of the western counties against possible Spanish invasion. Gorges's fortunes collapsed, however, when he was drawn into a rebellion being plotted by the Earl of Essex. The revolt fizzled, and Gorges testified against Essex at his former commander's trial. This experience convinced Gorges to support the monarchy ever after. As punishment for his complicity in the plot, he lost his position at Plymouth.
Gorges regained his command on the accession to the throne of James I in 1603, and it was after returning to Plymouth that he developed his interest in colonization. In 1605 Gorges met George Weymouth, who had returned from New England with five Native Americans. Gorges kept three of them in his household in order to learn about the New World. When peace was established with Spain, Gorges became active in colonization, promoting the Virginia Company of 1606 and participating in outfitting and supplying an ill-fated colony at the mouth of the Kennebec River in Maine.
Council for New England
Attempting to forestall permanent failure in English colonization, Gorges led in reorganizing the Plymouth branch of the Virginia Company. The new charter created the Council for New England, a group of 40 distinguished citizens who controlled all land in North America between the 40th and 48th parallels. The council was not interested in establishing colonies of its own but in encouraging others to do so. One of its first grants was to the Pilgrims on Cape Cod.
Members of the council were also given land, and in 1622 Gorges received a large grant jointly with John Mason. Gorges attempted one settlement on the New England coast when he had the council grant his son Robert a proprietary colony northeast of Massachusetts Bay. Robert Gorges sailed in 1623 but returned to England the following spring, and the enterprise was abandoned.
Struggles with Massachusetts Bay Colony
Gorges was forced to forsake his interest in the Council for New England when war broke out with France in 1627, and he was therefore unable to supervise the council's 1628 grant to the New England Company. When that company received a royal charter as the Massachusetts Bay Company, antagonism between Gorges and Massachusetts was certain. Not only did the Massachusetts land claim overlap Gorges's own, but Massachusetts' Puritan emphasis and company organization conflicted with Gorges's Anglicanism and dedication to proprietorship.
At war's end, Gorges, finding his interests had been preempted, set out either to get the Massachusetts charter annulled or to bring the colony under control of the Council for New England. For 3 years Gorges petitioned the English government regarding the threat to religious and political uniformity posed by Puritan Massachusetts. After investigation it was decided that Massachusetts must defend its charter. Gorges was appointed governor general of New England and instructed to present the order to Massachusetts. However, when his ship broke apart upon launching and, just as they were preparing to leave, his second in command, John Mason, died, Gorges abandoned the mission.
Meanwhile Gorges had decided to abolish the Council for New England, and after effecting a general division of the council's territory among its members, he engineered the surrender of its charter. It had been an interesting phase of English colonization, even though the council had accomplished little of permanent worth. In the last analysis it was Gorges's concept of colonization that defeated his goals. He viewed colonies as vast proprietary holdings into which the social, economic, and political institutions of England would simply be transferred. He never understood that the New World would not support such an unmodified transplantation.
Royal Grant for Maine
Gorges made one last attempt to salvage something from his efforts. In 1639 he received a royal confirmation of his earlier grant for the "Province of Maine." Since he was a royalist during the English civil war, however, his fortunes diminished with those of the King, and his plans for a vast proprietary colony were never fulfilled. He died in 1647, and his claims passed to his heirs, who eventually lost out to land-hungry Massachusetts.
The only recent biography of Gorges is Richard A. Preston, Gorges of Plymouth Fort (1953). Information about Gorges's colonial plans can also be found in Henry S. Burrage, Gorges and the Grant of the Province of Maine, 1622 (1923) and The Beginnings of Colonial Maine, 1602-1658 (1914). More general information is available in Herbert L. Osgood, The American Colonies in the Seventeenth Century (3 vols., 1904-1907), and Charles M. Andrews, The Colonial Period of American History (4 vols., 1934-1938). □
"Sir Ferdinando Gorges." Encyclopedia of World Biography. . Encyclopedia.com. (April 21, 2018). http://www.encyclopedia.com/history/encyclopedias-almanacs-transcripts-and-maps/sir-ferdinando-gorges
"Sir Ferdinando Gorges." Encyclopedia of World Biography. . Retrieved April 21, 2018 from Encyclopedia.com: http://www.encyclopedia.com/history/encyclopedias-almanacs-transcripts-and-maps/sir-ferdinando-gorges
Modern Language Association
The Chicago Manual of Style
American Psychological Association
Gorges, Sir Ferdinando
Sir Ferdinando Gorges (gôr´jĬz), c.1566–1647, English colonizer, proprietor of Maine. He was knighted (1591) for his services to Henry IV of France in the French Wars of Religion and was subsequently (1596–1601, 1603–29) military governor of Plymouth, England. Gorges was a leading figure in the Plymouth Company, chartered in 1606, and one of the two chief backers of the Sagadahoc colony, which was planted in 1607 at the mouth of the Kennebec River, Maine, and failed in 1608. In the following years he directed the many fishing and trading expeditions that the company carried on along the New England coast and defended its monopoly of the fisheries. He procured the services of Capt. John Smith to head a new settlement, but three successive expeditions foundered soon after leaving harbor, and the discouraged Smith withdrew. In 1620, Gorges obtained a revised charter for the Plymouth Company in which its territory, for the first time called New England, was established as lying between lat. 40°N and 48°N. The company reconstituted itself as the Council for New England, and grants were made to the individual members in the hope that they would become more interested in the project. Interest, however, centered in the more southern ventures, and Gorges found no financial support. The Pilgrim colony at Plymouth, patented under the London Company, had mistakenly settled within the bounds of the New England Council grant, but in 1621 it received a patent from the council and had Gorges's interest henceforth. Not so the Massachusetts Bay colony, against which Sir Ferdinando carried out a long struggle in England on the ground that its patent was irregular. In order to make the whole of New England a royal colony, over which Gorges was to be governor-general, the Council for New England surrendered its charter in 1635. The territory of New England was to be divided among the eight lords of the council, who were to hold it under new patents, but because of the growing intensity of the struggle between Charles I and Parliament in England the new arrangement was never consummated, and the Puritan commonwealth of Massachusetts was left free. In 1622, Gorges had received, with John Mason (1586–1635) a grant of the territory lying between the Merrimack and Kennebec rivers. They divided that area in 1629, Gorges taking the land east of the Piscataqua River, which became the province of Maine. His grant was confirmed by royal charter in 1639. Events in England prevented him from raising funds to colonize his domain. His grant passed to his heirs. His grandson, Ferdinando Gorges, 1630–1718, in 1677 finally sold to Massachusetts all rights to Maine for £1,250.
See J. P. Baxter, ed., Sir Ferdinando Gorges and His Province of Maine (3 vol., 1890, repr. 1967); H. S. Burrage, Gorges and the Grant of the Province of Maine (1923); R. A. Preston, Gorges of Plymouth Fort (1953).
"Gorges, Sir Ferdinando." The Columbia Encyclopedia, 6th ed.. . Encyclopedia.com. (April 21, 2018). http://www.encyclopedia.com/reference/encyclopedias-almanacs-transcripts-and-maps/gorges-sir-ferdinando
"Gorges, Sir Ferdinando." The Columbia Encyclopedia, 6th ed.. . Retrieved April 21, 2018 from Encyclopedia.com: http://www.encyclopedia.com/reference/encyclopedias-almanacs-transcripts-and-maps/gorges-sir-ferdinando
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Plastic PVC is most commonly used by plumbers and very clever DIY enthusiasts. This material and pipe were designed to transport water, as it cannot wear out over time like rusting metal can. PVC is also structurally very strong. When parts are constructed properly, it can support structures that are hundreds of pounds!
This class aims to go over the material properties of PVC as well as the most common ways it is worked, machined, joined, and finished. You'll be inspired to create furniture, agriculture projects, and more after learning these simple techniques to work with PVC.
Enter an Instructables Contest!
If you've used the skills you learned in this class to create something awesome for the outdoors, write an instructable about it and enter it in our PVC Contest for a chance to win some great prizes!
Lesson 1: Tools and Materials for PVC
We kick off this class with a list of common tools and workshop consumables needed for working with PVC, as well as an in-depth overview of what PVC pipe is and safety protocols for working with it.
Tools and Materials for PVC Started
Lesson 2: Cutting and Making Holes
Whoever knew you had so many options when it came to cutting down and machining PVC? This section goes over the many ways you can make long lengths of PVC shorter, and add holes as necessary.
Cutting and Making Holes Started
Lesson 3: Gluing and Fastening PVC
Get your respirator on for this lesson, because it's a stinker! We'll go over how to work with bonding PVC into pipe fittings, but also discover that not every PVC joint needs to be cemented together.
Gluing and Fastening PVC Started
Lesson 4: Cleaning and Finishing PVC
Take your pipe from drab to fab in this section! This lesson goes over how to remove manufacturer's printing from PVC pipe, and how to prep PVC projects for painting.
Cleaning and Finishing PVC Started
Lesson 5: Bending and Forming PVC
We end with a slightly advanced skill, making PVC bends. Heat forming PVC is a bit tricky, but with practice and ingenuity, you'll be ready to make safe bends.
Bending and Forming PVC Started
Now that you have the skills, what are you waiting for?
Get out there and make something!
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# Reducing a trig equation
• May 3rd 2010, 10:11 AM
200001
Reducing a trig equation
Hi
I am trying to solve this but cant get it in one/two trig ratios
http://img718.imageshack.us/img718/8164/coscosec.jpg
• May 3rd 2010, 12:45 PM
veljko
Quote:
Originally Posted by 200001
Hi
I am trying to solve this but cant get it in one/two trig ratios
http://img718.imageshack.us/img718/8164/coscosec.jpg
$\sin\phi(\cos \phi + \frac{1}{\sin\phi}) = \sin\phi(\frac{\cos\phi \sin\phi + 1}{\sin\phi}) =$ $\cos\phi \sin\phi + 1 = \cos\phi \sin\phi + \cos^2\phi + \sin^2\phi$
Than:
$\cos\phi \sin\phi + \cos^2\phi + \sin^2\phi = 2\cos^2\phi$
$\cos\phi \sin\phi + \sin^2\phi = \cos^2\phi$
$\cos\phi \sin\phi = \cos^2\phi - \sin^2\phi$
$\cos\phi \sin\phi = \cos2\phi$
$\frac{1}{2} \sin2\phi = \cos2\phi$
I hope, you will now easier find the solution of trig. equ.
• May 3rd 2010, 12:59 PM
ebaines
Try this:
$
sin(\theta) (cos(\theta) + csc(\theta)) = sin(\theta) (cos(\theta) + 1/sin(\theta)) = sin(\theta) cos(\theta) + 1 = 2cos^2(\theta)
$
$
sin(\theta)cos(\theta) = 2 cos^2(\theta) -1
$
Square both sides:
$
sin^2(\theta) cos^2(\theta) = 4 cos^4(\theta) - 4 cos^2 (\theta) + 1
$
Substitute $sin^2 \theta = 1 - cos^2 \theta$:
$
(1 - cos^2(\theta))*cos^2 (\theta) = 4 cos^4(\theta) - 4 cos^2 (\theta) + 1
$
$
5 cos^4(\theta) -5cos^2(\theta) +1 = 0
$
Let $\alpha = cos^2(\theta)$:
$
5 \alpha^2 - 5 \alpha + 1 = 0
$
$
\alpha = \frac 1 2 \pm \frac {\sqrt {5}}{10}
$
$
cos(\theta) = \pm \sqrt { \frac 1 2 \pm \frac {\sqrt {5}}{10} }
$
$
\theta = Arccos \left (\pm \sqrt { \frac 1 2 \pm \frac {\sqrt {5}}{10}} \right )
$
• May 5th 2010, 11:24 AM
200001
Brilliant
thanks
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Christianity developed out of Judaism in the 1st century C.E. It is founded on the life, teachings, death, and resurrection of Jesus Christ, and those who follow him are called "Christians." Christianity has many different branches and forms with accompanying variety in beliefs and practices. The three major branches of Christianity are Roman Catholicism, Eastern Orthodoxy, and Protestantism, with numerous subcategories within each of these branches. Until the latter part of the 20th century, most adherents of Christianity were in the West, though it has spread to every continent and is now the largest religion in the world. Traditional Christian beliefs include the belief in the one and only true God, who is one being and exists as Father, Son, and Holy Spirit, and the belief that Jesus is the divine and human Messiah sent to the save the world. Christianity is also noted for its emphasis on faith in Christ as the primary component of religion. The sacred text of Christianity is the Bible, including both the Hebrew scriptures (also known as the Old Testament) and the New Testament. Central to Christian practice is the gathering at churches for worship, fellowship, and study, and engagement with the world through evangelism and social action.
Quick Fact Details:
- Formed: This is the traditional date for the death and resurrection of Jesus Christ. Some scholars date the rise of Christianity as a religious belief system later in the first century under the leadership of the apostles.
Quick Fact Sources include www.adherents.com, www.bbc.co.uk/religion, The Oxford Handbook of Global Religions (2006), The Encyclopedia of Religion (2005), The Cambridge Illustrated History of Religions (2002), and the Encyclopedia of World Religions (1999).
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# Activity: Drawing Squares
For this activity all you need is a grid of dots, a pencil and your brain.
### Let us discover how many squares you can make on different grids:
Note: "1 by 1" means how many sides (not how many dots).
### 1 by 1
Well, that's easy, there's just one:
### 2 by 2
That seems to be easy too. There are four of them, aren't there? But wait, that's not the complete answer. There's also this bigger one:
That makes five squares altogether - four 1 by 1 squares and one 2 by 2 square
### 3 by 3
Over to you now. Here's the grid:
Hint: For the 3 by 3 case, you will expect to get 1 by 1 squares, 2 by 2 squares and 3 by 3 squares. How many of each?
Now you can start to fill in a table:
How Many 1 by 1 squares How Many 2 by 2 squares How Many 3 by 3 squares How Many 4 by 4 squares How Many 5 by 5 squares Total 1 by 1 Grid: 1 1 2 by 2 Grid: 4 1 5 3 by 3 Grid: 4 by 4 Grid: 5 by 5 Grid:
## Did you notice anything about the numbers in the table?
They are all square numbers:
• 12 = 1,
• 22 = 4,
• 32 = 9,
• etc ...
and the totals are found by adding together square numbers.
### Formula to The Rescue ... !
There is actually a formula for adding the first n square numbers:
Sn = n(n+1)(2n+1) / 6
### Example: The number of squares in the 5 by 5 case
Try substituting n = 5 into the formula:
Sn = n(n+1)(2n+1) / 6
S5 = 5 × (5+1) × (2×5+1) / 6
S5 = 5 × 6 × 11 / 6
S5 = 55
## So, we seem to have solved the question. Yipee!
### But wait ... there's more!
I said you would need to use your brains. Let's go back to the 2 by 2 case:
### 2 by 2
There is another square too, this one:
Why is it a square? It has four equal sides and four right angles, so that's a square.
So, that makes six squares altogether.
Four 1 by 1 squares, one 2 by 2 square and one x by x square.
What is the value of x? We can use Pythagoras' Theorem to find it:
x2 = 12 + 12 = 1 + 1 = 2
So x = √2
So, we have four 1 by 1 squares, one 2 by 2 square and one √2 by √2 square.
### 3 by 3
• Are there any more squares?
YES! Can you find them?
### 4 by 4 and 5 by 5
Also try the 4 by 4 grid, and the 5 by 5 grid
As you proceed, you will find squares like these:
## What are the lengths of the sides of these squares?
You can use Pythagoras' Theorem to work that out yourself
### In each case, how many do you get of each one?
How Many1 by 1 How Many2 by 2 How Many3 by 3 How Many4 by 4 How Many5 by 5 How Many√2 by √2 How Many√5 by √5 How Many √8 by √8 How Many√10 by √10 How Many√13 by √13 How Many√17 by √17 Total 1 by 1 Grid: 1 1 2 by 2 Grid: 4 1 1 6 3 by 3 Grid: 4 by 4 Grid: 5 by 5 Grid:
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This is a five-level grammar practice series for primary level pupils, providing introduction to grammatical concepts as well as lots of written practice.
- all essential traditional and modern rules of English Grammar are discussed keeping in mind both the native and non-native users of English language
- the books cover all aspects of language teaching, such as parts of speech, structures, language, spelling, comprehension and composition
- has a variety of picture-based and drill exercises to practice grammar usage
- recapitulates the concepts studied in previous classes includes model test papers for revision
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Below are the available bulk discount rates for each individual item when you purchase a certain amount
- Buy 10 - 100 and get 10% off
|Grades K-2| Bullying—a problem that affects all children either as a victim, bystander, or bully.Using stories, role-play, music, puppets, and accompanying activities, Mrs. Grasshopper, the teacher, helps children learn how Brenda Bee and Anthony Ant learned to deal with the bully, Big Beetle. The program includes:
- Lesson plans: Detailed plans for each lesson.
- Stories: A continuing story is presented in each lesson as Brenda Bee and Anthony Ant face the ever-present bully, Big Beetle.
- Role-play: Using puppets made from patterns included in the book, students act out ways to handle difficult bullying situations.
- Music: Children reinforce their learning of strategies by using reproducible lyric sheets and singing different verses to a familiar tune during lessons.
- 8 reproducible posters that focus on strategies for dealing with a bully.
- 12 reproducible activity sheets that reinforce lessons.
- Reproducible information on bullying for adults.
The CD includes black and white PDF filesof the lesson book’s reproducible pages and printable 11 X 17 posters. Level 1 Whiteboard compatible
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Furs of Valencia
This article does not cite any sources. (February 2007) (Learn how and when to remove this template message)
Furs of Valencia (Valencian: Furs de València, IPA: [ˈfuɾz ðe vaˈlensia]) were the laws of the Kingdom of Valencia during most of the Middle Ages and early modern Europe. The laws were a series of charters which, altogether, worked similarly as a modern Constitution does now. Thus, they defined the position of and checks and balances between the Royal House, the nobility, the Catholic ecclesiastic and the judicial procedures. The first codifications are based in the Usages of Barcelona, Costums of Lleida, and the Furs[clarification needed] of Aragon.
They were promulgated by the first King of Valencia, James "the Conqueror", in 1261 at the newly created Valencian Parliament; he then subjected the title of King of Valencia to an oath of office before the Parliament, sworn on the Furs.
The Furs were valid for more than four centuries, until they were abolished by means of the Nueva Planta decrees signed by Philip V of Castile in 1707. Following the agreed amendment to the Statute of Autonomy of the Valencian Community in 2006, some distinct usages of the civil law used by the Furs are scheduled to regain binding legal authority in this territory.
|This article relating to the law of Europe or of a European country is a stub. You can help Wikipedia by expanding it.|
|This article related to Valencian Community, Spain is a stub. You can help Wikipedia by expanding it.|
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Care2 Earth Month: Back to Basics
This year, Care2 decided to expand Earth Day into Earth Month, since there is so much to explore when it comes to the environment. Every day in April, we’ll have a post about some of the most important topics for the environment, exploring and explaining the basics. It’s a great tool to help you get started with helping the environment — or help explain it to others. See the whole series here.
What Are Wetlands?
If you break it down — wet-land — it might be obvious that we’re talking about a class of ecosystem that combines the properties of the terrestrial and the aquatic. Wetlands constitute land area which is either permanently or seasonally submerged with water. Wetlands typically have trees and other large land plants (adapted for dealing with submersion either continuously or at intervals) along with true aquatic plants.
Examples of wetlands include bogs, swamps, everglades and marshes. Wetlands can be found all over the world, in warm and temperate climates. Water depth of wetlands is generally shallower than true aquatic habitats like lakes or inland seas. Wetlands are generally comprised of stillwater.
What Are They Good For?
The title of this article may be slightly misleading, at least insofar as wetlands are not restricted to northern climes, or even the Northern Hemisphere. But wetlands, be they in South America, Great Britain, or in Canada, which holds one-quarter of the world’s wetlands, do rival tropical rainforests in terms of their biodiversity.
If you think about it for a moment, it will be obvious why this is true. Fish, crustaceans like crawfish, amphibians, all types of breeding and adult insects, a tremendous variety of plant life, and all types of waterfowl make wetlands their home. A highly interconnected web of very interdependent species, wetlands, more than tropical rainforests, also have strong ties to outside ecosystems. The surrounding ecosystems, in the case of breeding insects and amphibians, and much further afield, in the case of migratory birds.
Wetlands also provide a major role in weather and climate, providing water vapor to the atmosphere and acting as a carbon sink. They function as a natural water filtration system, removing pollutants, natural or man-made, from inflows. Where they exist on coastal areas, they protect the shoreline from erosion, by acting as a buffer between land and sea.
Threats and Consequences
Governments and private interests thoughtlessly drained wetlands for development during much of the twentieth century. The thinking was that a swamp, breeding-ground for mosquitoes and difficult to navigate, was no great loss to anyone. Though this attitude has changed in the last few decades, a lot of damage was already done. By 1993, half the world’s wetlands were gone.
The effects on water quality locally and a changing climate globally are hard to overstate. The more visible mascot for wetland protection, however, has been the threatened waterfowl. Ducks Unlimited, the foremost non-governmental wetland conservation organization, was started by duck hunters who saw for themselves the tremendous threat to these species as habitat was destroyed. This organization operates in all 50 US states and in Canada, which is critical.
Since these birds do migrate, it’s not enough to preserve some space here and there. All the countries and regions through which they pass on their migration route need to be on board. These birds can and will adapt to more limited habitat, resting in man-made lakes, or even on grass, during their trip. But finding the right food and rearing their young requires a certain minimum amount of habitat availability.
This is important for some of us who like to feed seeds to the ducks, as it is to hunters.
What You Can Do
Fortunately, perhaps due to their ubiquitous nature in countries all over the world, wetland conservation is the only type of ecosystem designated by international convention. If you live in North America, Ducks Unlimited is a great organization to support. It may have begun as an effort to preserve hunting stocks, but it’s grown far beyond that now.
The biggest human threat to wetlands is agricultural development. This is something to pressure your government representatives about when you get a heads-up from your favourite conservation organization. The World Wildlife Fund, those of us here at Care2, and any other environmental activist or advocacy organization you care to name can often give you a heads up before major development projects get underway.
Photo credit: Leveillem
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Question
each side of a 4ft x 6ft rectangular flower bed was lengthened by the same amount. the area of the new flower bed is twice the area of the old one. Find the new dimensions.
1. ngochoa
The new dimensions of the flower bed are 4√2 and 6√2
### How to determine the new dimensions?
The dimensions of the initial rectangular flower bed are given as:
Length = 4ft
Width = 6 ft
The area of the new flower bed is twice the area of the old one.
This means that
New Area = Old Area * 2
Substitute the dimensions of the old area in the above equation
New Area = 4 * 6 * 2
Express 2 as √2 * √2
This gives
New Area = 4 * 6 * √2 * √2
Rewrite the above equation as:
New Area = 4√2 * 6√2
The area of a rectangle is
Area = Length * Width
This means that:
Length = 4√2 ft
Width = 6√2 ft
Hence, the new dimensions of the flower bed are 4√2 and 6√2
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In a little patch of shale, continually flaking onto the road near Bannockburn (central South Island, New Zealand), there are the unmistakable fossils like Australian ‘hoop pine’ shoots. Hoop pines are members of the tree family which includes ‘monkey puzzles’, ‘bunyas’ and the ‘Norfolk Island Pines’. The Latin name is Araucaria – New Zealand has none of these in its native flora today, but it does have the kauri (it’s another genus, Agathis, but in the same family).
Around 17 million years ago (Miocene period) the shale would have been accumulating in a standing body of water, probably a flood-basin lake (The Manuherikia Group). Plant fossils would have been washed into it after they blew off a tree into a river, which then flowed into the body of water. The pine shoots are made of long, overlapping scale-like leaves, that together make a tail-like structure, maybe 4-6 mm in diameter. Along with the shoots there are also occasional seed cone-scales and rare pollen cones. The cone-scales have delicate wings, just like modern hoop-pines, but these have often been lost in the fossil. These additional plant fossils all help to show that the original tree was something much closer to the Australian ‘hoop pines’, the Norfolk Island pine’ and some New Caledonian species of Araucaria, than to other species in the family.
Back in the days of the dinosaurs (Cretaceous, c. 75 million years ago) the hoop pine family was common in New Zealand. They were an important component of the wet coal-swamps. But at the same time the dinosaurs vanished, so did those trees. From that time on, fossils of Araucaria are rare in New Zealand, with the layer of shale near Bannockburn an exception.
What do these fossils mean? Probably warmer times than today, but perhaps more intriguingly, the Australian hoop pine is a key plant in what are called ‘dry rainforests’. This term sounds a bit contradictory, but it refers to forest where rainfall is relatively low, but where fire does not normally occur. Unlike wetter, more normal rainforests, the low rainfall helps keep the forest canopy more open. Without so much shade, the greater amount of light reaching the forest floor is probably a reason why the hoop pines live in them. It’s certainly a very different habitat than the coal swamps where their relatives lived in along side dinosaurs.
The Bannockburn fossil ‘hoop pines’ (using the term broadly to refer to a group pf species) are evidence of forest (the area was virtually treeless when Europeans arrived in the 19th century), warmer conditions than in southern New Zealand today (and more like southeastern Queensland), but with some sort of degree of ‘dryness’ – probably seasonally low rainfall.
Small-stuff, but it’s bits of evidence like this that climatologists can use to figure out exactly how our climate system changes.
Links will take you to my Academia.edu page where you can download a pdf.
Pole, M.S., 1992. Early Miocene flora of the Manuherikia Group, New Zealand. 2. Conifers. Journal of the Royal Society of New Zealand 22, 287-302. (describes the Araucaria shoot, seed-cone and pollen cone fossils from Bannockburn)
Pole, M.S., 1993. Early Miocene flora of the Manuherikia Group, New Zealand. 10. Paleoecology and stratigraphy. Journal of the Royal Society of New Zealand 23, 393-426. (proposes the dry rainforest interpretation for the shale at Bannockburn)
Pole, M., 2008. The record of Araucariaceae macrofossils in New Zealand. Alcheringa 32, 405–426. (describes cuticle from the Araucaria fossils at Bannockburn)
Webb, L.J., 1959. A physiognomic classification of Australian rainforests. Journal of Ecology 47, 551-570. (defines ‘dry rainforest’ in Australia and Araucaria as one of their keys)
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The Slaughter of a Rooster at the Jewish Festival of Yom Kippur: An Explanation of the Ritual of Kapparot
|Author: ||Fishbane, Simcha|
Dr. Fishbane’s monograph explores the development and history of the Jewish tradition and custom of kapparot, where a rooster is sacrificed before Yom Kippur. The sacrificed fowl is given to the poor or the money that is the fowl’s worth.
"… religious rituals are often created or insti-tuted to supplement a spiritual or religious need. For example, in our study of kapparot, there is an aspiration to perform before the Day of Atonement an action that would rid the atoner of his/her sins, and permit him/her to enter this holy day free from religious transgressions. Durkheim points out that the individual requires a ritual ceremony to deal with evil or fear. Every misfortune, every-thing of evil omen, everything that inspires sentiments of sorrow or fear necessitates a rite or ceremony."
Table of Contents
The Practice of Kapparot
The Sociology of the Custom
The Rabbis of the Talmud
The Ritual Takes Root – the Gaonic Period
The Rishonim (Early Rabbinical Authorities)
The Achronim (Later Rabbinical Authorities)
19th – 20th Century Rabbinical Authorities
Other Jewish Studies Books
More Books by this Author
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Final_Review_1_Solutions
# 1 independence assumption random sampling condition
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Unformatted text preview: to 0. So yes, the results of the hypothesis test and confidence interval agree. 16. A high school student is considering attending a nearby university. She is concerned that too many of the students attending the university live at home, and she has decided not to attend if there is strong evidence that more than 75% of the registered students live at home. The student experiences a fortunate coincidence. She recently finished a course in Advanced Placement Statistics, and the university student newspaper has published a poll of 400 randomly selected students at the university. 325 of these students reported that they still lived at home. (a) Write the hypotheses in symbols and in words for testing whether more than 75% of students at this college live at home. H0 : p = 0.75 (75% of students at this college live at home) HA : p > 0.75 (More than 75% of students at this college live at home) (b) Are the assumptions and conditions for inference satisfied? 1. Independence Assumption: • Random Sampling Condition: We are told that the sample is random. • 10% Condition: We can safely assume that 400 < 10% of all students at a college. Since we have a random sample and the 10% condition is satisfied, we can assume that whether or not one student in this sample lives at home is independent of another. 2. Nearly Normal Condition: First we must check if the success failure condition is met. np ≥ 10 → 400 ∗ 0.75 = 300 > 10 nq ≥ 10 → 400 ∗ 0.25 = 100 > 10 Since the observations are independent and the success-failure condition is met, we can assume that p is nearly normal. ˆ 12 (c) Calculate the test statistic. 325 = 0.8125 400 p−p ˆ 0.8125 − 0.75 0.0625 z= = = = 2.89 pq 0.02165 0.75∗0.25 p= ˆ n 400 (d) Find and interpret the p-value in context. p-value = P (ˆ > 0.8125|p = 0.75) = P (z > 2.89) = 1 − 0.9981 = 0.0019 p If in fact 75% of student at this college lived at home, the probability of getting a random sample of 400 students where more than 325 live at home would be 0.0019. (e) Based on the hypothesis test do the data provide convincing evidence to suggest that more than 75% of students at this college live at home? Explain. Use α = 0.025. Yes, since p-value < 0.025, we reject H0 . The data provide convincing evidence to suggest that more than 75% of students at this college live at home (f) What type of error might you have committed? We may have committed a Type I error since we rejected H0 . 17. A genetic test is used to determine if people have a genetic predisposition for thrombosis, the formation of a blood clot inside a blood vessel that obstructs the flow of blood through the circulatory system. It is believed that 3% of the people actually have this predisposition. This test is 99% accurate if a person actually has the predisposition, meaning that the probability of a positive test result when a person actually has the predisposition is 0.99. The test is 98% accurate if a person does not have the predisposition, me...
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## This document was uploaded on 12/04/2013.
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# everything maths & science
You are here: Home Solving surd equations
## Solving surd equations
We also need to be able to solve equations that involve surds.
## Example 1: Surd equations
### Question
Solve for $$x$$: $$5\sqrt[3]{x^4} = \textrm{405}$$
Write in exponential notation
\begin{align*} 5\left( x^4 \right)^{\frac{1}{3}}&= \textrm{405} \\ 5x^{\frac{4}{3}}&= \textrm{405} \end{align*}
Divide both sides of the equation by 5 and simplify
\begin{align*} \frac{5x^{\frac{4}{3}}}{5} &= \frac{\textrm{405}}{5} \\ x^{\frac{4}{3}} &= 81 \\ x^{\frac{4}{3}} &= 3^4 \end{align*}
Simplify the exponents
\begin{align} \left( x^{\frac{4}{3}} \right)^{\frac{3}{4}} &= \left( 3^4 \right)^{\frac{3}{4}} \\ x &= 3^3 \\ x &= 27 \end{align}
Check the solution by substituting the answer back into the original equation
\begin{align*} \text{LHS}&= 5\sqrt[3]{x^4} \\ &= 5(27)^{\frac{4}{3}} \\ &= 5(3^3)^{\frac{4}{3}} \\ &= 5(3^4) \\ &= \textrm{405} \\ &= \text{RHS } \end{align*}
## Example 2: Surd equations
### Question
Solve for $$z$$: $$z - 4\sqrt{z} + 3 = 0$$
Factorise
\begin{align} z - 4\sqrt{z} + 3 &= 0 \\ z - 4z^{\frac{1}{2}} + 3 &= 0 \\ (z^{\frac{1}{2}}-3)(z^{\frac{1}{2}}-1) &= 0 \end{align}
Solve for both factors
The zero law states: if $$a \times b = 0$$, then $$a = 0$$ or $$b = 0$$.
$\therefore (z^{\frac{1}{2}}-3) = 0 \text{ or } (z^{\frac{1}{2}}-1) = 0$
Therefore
\begin{align} z^{\frac{1}{2}}-3 &= 0 \\ z^{\frac{1}{2}} &= 3 \\ \left( z^{\frac{1}{2}} \right)^2 &= 3^2 \\ z &= 9 \end{align}
or
\begin{align} z^{\frac{1}{2}}-1 &= 0 \\ z^{\frac{1}{2}} &= 1 \\ \left( z^{\frac{1}{2}} \right)^2 &= 1^2 \\ z &= 1 \end{align}
Check the solution by substituting both answers back into the original equation
If $$z=9$$:
\begin{align} \text{LHS}&= z - 4\sqrt{z} + 3 \\ &= 9 - 4\sqrt{9} + 3 \\ &= 12 - 12 \\ &= 0 \\ &=\text{RHS } \end{align}
If $$z=1$$:
\begin{align} \text{LHS} &= z - 4\sqrt{z} + 3 \\ &= 1 - 4\sqrt{1} + 3 \\ &= 4-4 \\ &= 0 \\ &= \text{RHS } \end{align}
The solution to $$z - 4\sqrt{z} + 3 = 0$$ is $$z = 9$$ or $$z = 1$$.
## Example 3: Surd equations
### Question
Solve for $$p$$: $$\sqrt{p-2} - 3 = 0$$
Write the equation with only the square root on the left hand side
Use the additive inverse to get all other terms on the right hand side and only the square root on the left hand side.
$\sqrt{p-2} = 3$
Square both sides of the equation
\begin{align} \left( \sqrt{p-2} \right)^2 &= 3^2 \\ p-2 &= 9 \\ p &= 11 \end{align}
Check the solution by substituting the answer back into the original equation
If $$p=11$$:
\begin{align} \text{LHS} &=\sqrt{p-2} - 3 \\ &=\sqrt{11-2} - 3 \\ &=\sqrt{9} - 3 \\ &= 3-3 \\ &= 0 \\ &= \text{RHS } \end{align}
The solution to $$\sqrt{p-2} - 3 = 0$$ is $$p = 11$$.
## Exercise 1: Solving surd equations
Solve for the unknown variable (remember to check that the solution is valid):
1. $$2^{x+1} -32 = 0$$
2. $$\textrm{125} \left ( 3^p \right ) = 27 \left ( 5^p \right )$$
3. $$2y^{\frac{1}{2}} - 3y^{\frac{1}{4}} + 1 = 0$$
4. $$t-1 = \sqrt{7-t}$$
5. $$2z - 7\sqrt{z} + 3 = 0$$
6. $$x^{\frac{1}{3}}(x^{\frac{1}{3}} + 1) = 6$$
7. $$2^{4n} - \dfrac{1}{\sqrt[4]{16}} = 0$$
8. $$\sqrt{31 -10d} = 4 - d$$
9. $$y - 10\sqrt{y} + 9 = 0$$
10. $$f = 2 + \sqrt{19 - 2f}$$
1. \begin{aligned} 2^{x+1}-32&=0 \\ 2^{x+1}&=32 \\ 2^{x+1}&=2^5 \\ \therefore x+1&=5 \\ x&=4 \end{aligned}
2. \begin{align*} \textrm{125}\left ( 3^p \right ) &= 27\left ( 5^p \right ) \\ \frac{5^p}{3^p}&=\frac{\textrm{125}}{27} \\ \left ( \frac{5}{3} \right )^p &= \left ( \frac{5}{3} \right )^3 \\ \therefore p&=3 \end{align*}
3. \begin{aligned} 2y^{\frac{1}{2}}-3y^{\frac{1}{4}}+1&=0 \\ \left ( 2y^\frac{1}{4} -1\right )\left ( y^\frac{1}{4} -1\right )&=0 \\ \text{Therefore } 2y^{\frac{1}{4}}-1&=0 \\ y^{\frac{1}{4}}-\frac{1}{2}&=0 \\ y^\frac{1}{4}&=\frac{1}{2} \\ y^\frac{4}{4} &= \left ( \frac{1}{2} \right )^4 \\ \therefore y&=\frac{1}{16} \\ \text{or} \\ y^\frac{1}{4}-1&=0 \\ y^\frac{1}{4}&=1 \\ \therefore y&=1 \end{aligned}
4. \begin{aligned} t-1 &= \sqrt{7-t} \\ \left ( t-1 \right )^2 &= \left ( \sqrt{7-t} \right )^2 \\ t^2 - 2t + 1 &= 7-t \\ t^2 - t - 6 &=0 \\ (t - 3)(t + 2)&= 0 \\ \therefore t =3 &\text{ or } t = - 2 \\ \text{Check RHS for } t = 3: &= \sqrt{7-3} \\ &= \sqrt{4} \\ &= 2 \\ &= \text{LHS} \therefore \text{ valid solution }\\ \text{Check RHS for } t = -2: &= \sqrt{7-(-2)} \\ &= \sqrt{9} \\ &= 3 \\ &\ne \text{LHS} \therefore \text{ not valid solution } \end{aligned}
5. \begin{aligned} 2z-7z^\frac{1}{2}+3&=0 \\ \left ( z^\frac{1}{2}-3 \right )\left ( 2z^\frac{1}{2}-1 \right )&=0 \\ \text{Therefore } z^\frac{1}{2}-3&=0 \\ \left ( z^\frac{1}{2} \right )^2 &=3^2 \\ \therefore z&=9 \\ \text{or}\\ 2z^\frac{1}{2}-1&=0 \\ \left ( z^\frac{1}{2} \right )^2&=\left ( \frac{1}{2} \right )^2 \\ \therefore z&=\frac{1}{4} \end{aligned}
6. \begin{aligned} x^\frac{1}{3}\left ( x^\frac{1}{3}+1 \right )&=6 \\ x^\frac{2}{3}+x^\frac{1}{3} &=6 \\ \left ( x^\frac{1}{3}-2 \right )\left ( x^\frac{1}{3}+3 \right ) &= 0 \\ \text{Therefore } x^\frac{1}{3}-2 &= 0 \\ x^\frac{1}{3} &= 2 \\ \therefore x&=8 \\ \text{or} \\ x^\frac{1}{3}+3 &= 0 \\ x^\frac{1}{3} &=-3 \\ \therefore x &= -27 \end{aligned}
7. \begin{aligned} 2^{4n}-\dfrac{1}{\sqrt[4]{16}}&=0\\ 2^{4n}&=\dfrac{1}{16^\frac{1}{4}} \\ 2^{4n}&=\dfrac{1}{\left (2^4 \right )^\frac{1}{4}} \\ 2^{4n}&=\frac{1}{2} \\ 2^{4n}&=2^{-1} \\ \therefore 4n&=-1 \\ \therefore n &=-\frac{1}{4} \end{aligned}
8. \begin{aligned} \sqrt{31-10d}&=d-4 \\ \left (\sqrt{31-10d} \right )^2&=\left (d-4 \right )^2 \\ 31-10d&=d^2-8d+16 \\ 0&=d^2+2d-15 \\ 0&=\left ( d-3 \right )\left ( d+5 \right ) \\ \text{Therefore } d&= 3 \\ \text{or} \\ d&= -5 \\ \text{Check LHS for } d = 3: &= \sqrt{31 - 30} \\ &= \sqrt{1} \\ &= 1 \\ &= \text{RHS} \therefore \text{ valid solution }\\ \text{Check LHS for } t = -5: &= \sqrt{31-(-50)} \\ &= \sqrt{81} \\ &= 9 \\ &= \text{RHS} \therefore \text{ valid solution } \end{aligned}
9. \begin{aligned} y - 10\sqrt{y} + 9 &= 0 \\ y - 10y^{\frac{1}{2}} + 9 &= 0 \\ \left( y^{\frac{1}{2}} - 1 \right ) \left( y^{\frac{1}{2}} -9 \right ) &= 0 \\ \text{Therefore } y^{\frac{1}{2}} - 1 &= 0 \\ y^{\frac{1}{2}} &= 1 \\ \left( y^{\frac{1}{2}} \right )^2 &= (1)^2 \\ \therefore y &= 1 \\ \text{or} \\ y^{\frac{1}{2}} - 9 & = 0 \\ y^{\frac{1}{2}} &= 9 \\ \left( y^{\frac{1}{2}} \right )^2 &= (9)^2 \\ \therefore y &= 81 \end{aligned}
10. \begin{aligned} f &= 2 + \sqrt{19 - 2f} \\ f - 2 &= \sqrt{19 - 2f} \\ \left( f - 2 \right )^2 &= \left( \sqrt{19 - 2f} \right )^2 \\ f^2 - 4f + 4 &= 19 - 2f \\ f^2 - 2f -15 &= 0 \\ (f - 5)(f + 3) &= 0 \\ \text{Therefore } f - 5 &= 0 \\ \therefore f &= 5 \\ \text{or} \\ f + 3 & = 0 \\ \therefore f &= -3 \\ \text{Check RHS for } f = 5: &= 2 + \sqrt{19 - 10} \\ &= 2 + \sqrt{9} \\ &= 5 \\ &= \text{LHS} \therefore \text{ valid solution }\\ \text{Check RHS for } f = -3: &= 2 + \sqrt{19 + 6}\\ &= 2 + \sqrt{25} \\ &= 7 \\ &\ne \text{LHS} \therefore \text{ not valid solution } \end{aligned}
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By Stan Tackett
Cassini-Huygens is a Flagship-class NASA-ESA-ASI robotic spacecraft sent to the Saturn system. It has studied the planet and its many natural satellites since its arrival there in 2004, as well as observing Jupiter and the Heliosphere, and testing the theory of relativity. Launched in 1997 after nearly two decades of gestation, it includes a Saturn orbiter Cassini and an atmospheric probe/lander Huygens that landed in 2005 on the moon Titan. Cassini is the fourth space probe to visit Saturn and the first to enter orbit, and its mission is ongoing as of 2013. It is powered by a plutonium power source, and has facilitated many landmark scientific discoveries in its mission to the stars.
A Cassini RTG before installation
Because of Saturn’s distance from the Sun, solar arrays were not feasible as power sources for this space probe. To generate enough power, such arrays would have been too large and too heavy. Instead, the Cassini orbiter is powered by three radioisotope thermoelectric generators (RTGs), which use heat from the natural decay of about 33 kg (73 lb) of plutonium-238 (in the form of plutonium dioxide) to generate direct current electricity via thermoelectrics. The RTGs on the Cassini mission have the same design as those used on the New Horizons, Galileo, and Ulysses space probes, and they were designed to have a very long operational lifetime. At the end of the nominal 11-year Cassini mission, they will still be able to produce 600 to 700 watts of electrical power. One of the spare RTGs for the Cassini mission was used to power the New Horizons mission to Pluto and the Kuiper belt.
A glowing-hot plutonium pellet that will become the power source for the probe’s RTG
To gain interplanetary momentum while in flight, the trajectory of the Cassini mission included several gravitational slingshot maneuvers: two fly-by passes of Venus, one more of the Earth, and then one of the planet Jupiter. The terrestrial fly-by maneuver was successful, with Cassini passing by 500 km (310 mi) above the Earth on August 18, 1999. Had there been a malfunction causing the Cassini space probe to collide with the Earth, NASA’s complete environmental impact study estimated that, in the worst case (with an acute angle of entry in which Cassini would gradually burn up), a significant fraction of the 33 kg of plutonium-238 inside the RTGs could have been dispersed into the Earth’s atmosphere. NASA estimated the odds against that happening at more than 1 million to one.
Selected events and discoveries for Cassini
Cassini made its closest approach to Jupiter on December 30, 2000, and made many scientific measurements. About 26,000 images of Jupiter were taken during the months-long flyby. Cassini produced the most detailed global color portrait of Jupiter yet (see image at right), in which the smallest visible features are approximately 60 km (37 mi) across. The New Horizons mission to Pluto captured more recent images of Jupiter, with a closest approach on February 28, 2007.
A major finding of the flyby, announced on March 6, 2003, was of Jupiter’s atmospheric circulation. Dark “belts” alternate with light “zones” in the atmosphere, and scientists had long considered the zones, with their pale clouds, to be areas of upwelling air, partly because many clouds on Earth form where air is rising. But analysis of Cassini imagery showed that individual storm cells of upwelling bright-white clouds, too small to see from Earth, pop up almost without exception in the dark belts.
Other atmospheric observations included a swirling dark oval of high atmospheric-haze, about the size of the Great Red Spot, near Jupiter’s north pole. Infrared imagery revealed aspects of circulation near the poles, with bands of globe-encircling winds, with adjacent bands moving in opposite directions. The same announcement also discussed the nature of Jupiter’s rings. Light scattering by particles in the rings showed the particles were irregularly shaped (rather than spherical) and likely originate as ejecta from micrometeorite impacts on Jupiter’s moons, probably Metis and Adrastea.
Tests of General Relativity
On October 10, 2003, the Cassini science team announced the results of tests of Einstein’s Theory of General Relativity, which were done by using radio waves that were transmitted from the Cassini space probe. This remains the best measurement of post-Newtonian parameter γ; the result γ = 1 + (2.1 ± 2.3) × 10-5 agrees with the predictions of standard General Relativity.
The radio scientists measured a frequency shift in the radio waves to and from the spacecraft, while those signals traveled close to the Sun. According to the Theory of General Relativity, a massive object like the Sun causes space-time to curve, and a beam of radio waves (or light, or any form of electromagnetic radiation) that passes by the Sun has to travel farther because of this curvature.
The extra distance that the radio waves traveled from the Cassini craft, past the Sun, to the Earth delayed their arrival. The amount of this time delay provided a sensitive test of the calculated predictions of Einstein’s Relativity Theory.
Although some measurable deviations from the values that are calculated using the General Theory of Relativity are predicted by some unusual cosmological models, no deviations were found by this experiment. Previous tests using radio waves that were transmitted by the Viking and Voyager space probes were in agreement with the calculated values from General Relativity to within an accuracy of one part in one thousand. The more refined measurements from the Cassini space probe experiment improved this accuracy to about one part in 51,000, with the measured data firmly supporting Einstein’s General Theory of Relativity.
New moons of Saturn
Using images taken by Cassini, three new moons of Saturn were discovered in 2004. They are very small and were given the provisional names S/2004 S 1, S/2004 S 2, and S/2004 S 5, before being named Methone, Pallene, and Polydeuces at the beginning of 2005.
On May 1, 2005, a new moon was discovered by Cassini in the Keeler gap. It was given the designation S/2005 S 1, before being named Daphnis. The only other known moon inside Saturn’s ring system is the moon Pan.
A fifth new moon was discovered by Cassini on May 30, 2007, now known as Anthe.
A press release on February 3, 2009, showed a sixth new moon found by Cassini. The moon is approximately 1/3 of a mile in diameter within the G-ring of the ring system of Saturn, and is now named Aegaeon.
A press release on November 2, 2009, mentions the seventh new moon found by Cassini on July 26, 2009. It is presently labeled S/2009 S 1, and is approximately 300 m (984 ft.) in diameter in the B-ring system.
On June 11, 2004, Cassini flew by the moon Phoebe. This was the first opportunity for closeup studies of this moon since the Voyager 2 flyby. It also was Cassini’s only possible flyby for Phoebe due to the mechanics of the available orbits around Saturn.
The first closeup images were received on June 12, 2004, and mission scientists immediately realized that the surface of Phoebe looks different from asteroids visited by spacecraft. Parts of the heavily cratered surfaces look very bright in those pictures, and it is currently believed that a large amount of water ice exists under its immediate surface.
In an announcement on June 28, 2004, Cassini program scientists described the measurement of the rotational period of Saturn. Since there are no fixed features on the surface that can be used to obtain this period, the repetition of radio emissions was used. These new data agree with the latest values measured from Earth, and constitute a puzzle to the scientists. It turns out that the radio rotational period has changed since it was first measured in 1980 by Voyager, and that it is now six minutes longer. This does not indicate a change in the overall spin of the planet, but is thought to be due to movement of the source of the radio emissions to a different latitude, at which the rotation rate is different.
On July 1, 2004, the Cassini spacecraft flew through the gap between Saturn’s F and G rings and achieved orbit, after a seven-year voyage. Cassini is the first spacecraft to ever orbit Saturn.
The Saturn Orbital Insertion maneuver performed by Cassini was complex, requiring the craft to orient its high-gain antenna away from Earth and along its flight path, to shield its instruments from particles in Saturn’s rings. Once the craft crossed the ring plane, it had to rotate again to point its engine along its flight path, and then the engine fired to decelerate the craft by 622 m/s (1391 mph) to allow Saturn to capture it. Cassini was captured by Saturn’s gravity at around 8:54 p.m. Pacific Daylight Time on June 30, 2004. During the maneuver, Cassini passed within 20,000 km (12,000 mi) of Saturn’s cloud tops.
Cassini had its first distant flyby of Saturn’s largest moon, Titan, on July 2, 2004, only a day after orbit insertion, when it approached to within 339,000 km (211,000 mi) of Titan and provided the best look at Titan’s surface to date. Images taken through special filters (able to see through the moon’s global haze) showed south polar clouds thought to be composed of methane, and surface features with widely differing brightness. On October 27, 2004, the spacecraft executed the first of 45 planned close flybys of Titan, when it flew a mere 1,200 kilometers above the moon. Almost four gigabits of data were collected and transmitted to Earth, including the first radar images of the moon’s haze-enshrouded surface. It revealed the surface of Titan (at least the area covered by radar) to be relatively level, with topography reaching no more than about 50 meters in altitude. The flyby provided a remarkable increase in imaging resolution over previous coverage. Images with up to 100 times better resolution were taken and are typical of resolutions planned for future Titan flybys.
Huygens lands on Titan
Cassini released the Huygens probe on December 25, 2004, by means of a spring and spiral rails intended to rotate the probe for greater stability. Huygens entered the atmosphere of Titan on January 14, 2005, and after a two-and-a-half-hour descent landed on solid ground. Although Cassini successfully relayed 350 of the pictures that it received from Huygens of its descent and landing site, a software error failed to turn on one of the Cassini receivers and resulted in the loss of the other 350 pictures.
Enceladus backdropped by Saturn’s ring shadows in 2007
During the first two close flybys of the moon Enceladus in 2005, Cassini discovered a “deflection” in its local magnetic field that is characteristic for the existence of a thin but significant atmosphere. Other measurements obtained at that time point to ionized water vapor as being the atmosphere’s main constituent. Cassini also observed water ice geysers erupting from the south pole of Enceladus, which gives more credibility to the idea that Enceladus is supplying the particles of Saturn’s E ring. Mission scientists hypothesize that there may be pockets of liquid water near the surface of the moon that fuel the eruptions, making Enceladus one of the few bodies in the Solar System known to contain liquid water.
On March 12, 2008, Cassini made a close flyby of Enceladus, getting within 50 km of the moon’s surface. The spacecraft passed through the plumes extending from its southern geysers, detecting water, carbon dioxide, and various hydrocarbons with its mass spectrometer, while also mapping surface features with its infrared spectrometer that were measured to be at much higher temperature than their surroundings. Cassini was unable to collect data with its cosmic dust analyzer due to an unknown software malfunction.
Radio occultations of Saturn’s rings
In May 2005, Cassini began a series of occultation experiments to measure the size-distribution of particles in Saturn’s rings, and measure the atmosphere of Saturn itself. For more than four months, Cassini completed orbits designed for this purpose. During these experiments, Cassini flew behind the ring plane of Saturn, as seen from Earth, and transmitted radio waves through the particles. The radio signals were received on Earth, where the frequency, phase, and power of the signal were analyzed to help determine the structure of the rings.
Spoke phenomenon verified
In images captured September 5, 2005, Cassini detected spokes in Saturn’s rings, previously seen only by the visual observer Stephen James O’Meara in 1977 and then confirmed by the Voyager space probes in the early 1980s.
Lakes of Titan
Radar images obtained on July 21, 2006, appear to show lakes of liquid hydrocarbon (such as methane and ethane) in Titan’s northern latitudes. This is the first discovery of currently-existing lakes anywhere besides Earth. The lakes range in size from one to 100 kilometers across.
Titan “sea” (left) compared at scale to Lake Superior (right)
On March 13, 2007, the Jet Propulsion Laboratory announced that it had found strong evidence of seas of methane and ethane in the northern hemisphere of Titan. At least one of these is larger than any of the Great Lakes in North America.
A Saturnine hurricane
In November 2006, scientists discovered a storm at the south pole of Saturn with a distinct eyewall. This is characteristic of Earth’s hurricanes and had never before been seen on another planet. Unlike a Terran hurricane, the storm appears to be stationary at the pole. The storm is 8,000 kilometers (5,000 mi) across, and 70 kilometres (43 mi) high, with winds blowing at 560 km/hr (350 mph).
Great Storm of 2010 and its aftermath
Storm in the North 2011
On October 25, 2012, Cassini witnessed the aftermath of the massive Great White Spot storm that recurs roughly every 30 years on Saturn. Data from Cassini’s composite infrared spectrometer instrument indicated a powerful discharge from the storm that caused a temperature spike in the stratosphere of Saturn 150 °F (83 kelvins) above normal. Simultaneously, a huge increase in ethylene gas was detected by NASA researchers at Goddard Research Center in Greenbelt, Maryland. Ethylene is a colorless and odorless gas that is highly uncommon on Saturn and is produced both naturally and through man-made sources on Earth. The storm that produced this discharge was first observed by Cassini on December 5, 2010, in Saturn’s northern hemisphere. The storm is the first of its kind to be observed by a spacecraft in orbit around Saturn as well as the first to be observed at thermal infrared wavelengths, allowing scientists to observe the temperature of Saturn’s atmosphere and track phenomena that are invisible to the naked eye. The spike of ethylene gas that was produced by the storm reached levels that were 100 times more than those thought possible for Saturn. Scientists have also determined that the storm witnessed was the largest, hottest stratospheric vortex ever detected in our solar system, initially being larger than Jupiter’s Great Red Spot.
On April 15, 2008, Cassini received funding for a two-year extended mission. This consisted of 60 more orbits of Saturn, with 21 more close Titan flybys, seven of Enceladus, six of Mimas, eight of Tethys, and one targeted flyby each of Dione, Rhea, and Helene. The extended mission began on July 1, 2008, and was renamed the Cassini Equinox Mission as it coincided with Saturn’s equinox.
A proposal was submitted to NASA for a second mission extension, provisionally named the extended-extended mission or XXM. This was subsequently approved and renamed the Cassini Solstice Mission. It will see Cassini orbiting Saturn 155 more times, conducting 54 additional flybys of Titan, and 11 more of Enceladus. The chosen mission ending is a series of very close Saturn passes, passing inside the rings, then a plunge into the Saturn atmosphere around the 2017 northern summer solstice, to destroy the spacecraft.
Stan Tackett holds undergraduate degrees in mathematics and computer science, and is currently pursuing a Master’s degree in computer science with specializations in uses of artificial intelligence in the nuclear industry. His interests in nuclear engineering include nuclear propulsion for space travel, fusion, computational fluid dynamics and reactor physics. In his spare time he reads Piers Anthony as much as possible, and enjoys writing and editing crossover science fiction stories.
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History of Slavery in North Carolina
This illustration shows white children playing with a black child, and "represents the old Negro servants of the planter's family among his children. The children of the [white] family grow up among the Negro domestic servants, and often learn to regard them with as much affection as they show their own parents." Source: The Illustrated London News
Many of the first slaves in North Carolina were brought to the colony from the West Indies or other surrounding colonies, but a significant number were brought from Africa. Most of the English colonists arrived as indentrued servants, hiring themselves out as laborers for a fixed period to pay for their passage. In the early years the line between indentured servants and African slaves or laborers was fluid. Some Africans were allowed to earn their freedom before slavery became a lifelong status.
As the flow of indentured laborers from England to the colony decreased with improving economic conditions in Great Britain, more slaves were imported and the state's restrictions on slavery hardened. The economy's growth and prosperity was based on slave labor, devoted first to the production of tobacco.
Colonial laws were enacted to allow whites to control their slaves. The first of these was the North Carolina Slave Code of 1715. Under these laws, whenever slaves left the plantation they were required to carry a ticket from their master, which stated their destination and the reason for their travel. The 1715 code also prevented slaves from gathering in groups for any reason, including religious worship, and required whites to help capture runaway slaves.
The colony lacked the extensive plantation system of the Lower South, and when Carolina split into the North and South Carolina in 1729, North Carolina had about 6000 slaves, only a fraction of the slave population of South Carolina. As the plantation system expanded across the Lower South, many North Carolina slaves were "sold south" to work on large plantations. Slaves deeply feared this fate because it usually meant permanent separation from friends and family.
A second set of even stricter laws was put into place in 1741, which prevented slaves from raising their own livestock and from carrying guns without their master's permission, even for hunting. The law also limited manumission – the freeing of slaves. A master could only free a slave for meritorious service, and even then the decision had to be approved by the county court. Perhaps the most ominous of all the laws was the one regarding runaway slaves: If runaways refused to surrender immediately, they could be killed and there would be no legal consequences.
By 1767, there were about 40,000 slaves in the North Carolina colony. About 90 percent of these slaves were field workers who performed agricultural jobs. The remaining 10 percent were mainly domestic workers, and a small number worked as artisans in skilled trades, such as butchering, carpentry, and tanning.
Because of its geography, North Carolina did not play a large part in the early slave trade. The string of islands that make up its Outer Banks made it dangerous for slave ships to land on most of North Carolina's coast, and most slave traders chose to land in ports to the north or south of the colony.
The one major exception is Wilmington – located on the Cape Fear River, it became a port for slave ships because of its accessibility. By the 1800s, blacks in Wilmington outnumbered whites 2 to 1. The town relied on slaves' abilities in carpentry, masonry, and construction, as well as their skill in sailing and boating, for its growth and success.
Most of the free black families formed in North Carolina before the Revolution were descended from unions or marriages between free white women and enslaved or free African American men. Because the mothers were free, their children were born free. Many had migrated or were descendants of migrants from colonial Virginia.
The North Carolina Provincial Congress passed a ban on importing slaves in 1774, because they felt increasing the number of slaves in the colony would increase the number of runaways and free blacks. The fear of slave uprisings only increased with the advent of the Revolutionary War in 1775. The British offered to help slaves escape if they would fight against the colonists, and a number of North Carolina slaves accepted. After the war ended and the new country was founded, tensions between whites and blacks in the state continued to increase.
Besides slaves, there were a number of free people of color in the state. Most were descended from free African Americans who had migrated from Virginia during the eighteenth century. After the Revolution, Quakers and Mennonites worked to persuade slaveholders to free their slaves. Some were inspired by their efforts, and the language of men's rights, to arrange for manumission of their slaves. The number of free people of color rose markedly in the first couple of decades after the Revolution.
Hauling the Seine
This scene is of the fisheries of Albemarle and Pamlico Sounds in North Carolina, which were known to employ a considerable number of Free Negroes from neighboring counties. The huge seines (nets) could measure up to two miles in length.
The ban on importing slaves to North Carolina was lifted in 1790, and the state's slave population quickly increased. By 1800, there were around 140,000 blacks living in the state. A small number of these were free blacks, who mostly farmed or worked in skilled trades. The majority were slaves working in agriculture on small- to medium-sized farms.
After 1800, cotton and tobacco became important export crops, and the eastern half of the state developed a plantation system based on slavery, while the western areas were dominated by white families who operated small farms. Most of North Carolina's slave owners and large plantations were located in the eastern portion of the state.
Although North Carolina's plantation system was smaller and less cohesive than those of Virginia, Georgia or South Carolina, there were significant numbers of planters concentrated in the counties around the port cities of Wilmington and Edenton, as well as suburban planters around the cities of Raleigh, Charlotte and Durham.
In addition, 30,463 free people of color lived in the state. They were also concentrated in the eastern coastal plain, especially at port cities such as Wilmington and New Bern where they had access to a variety of jobs. Free African Americans were allowed to vote until 1835, when the state rescinded their suffrage.
As in the colonial period, few North Carolina slaves lived on huge plantations. Fifty-three percent of slave owners in the state owned five or fewer slaves, and only 2.6 percent of slaves lived on farms with over 50 other slaves during the antebellum period. In fact, by 1850, only 91 slave owners in the whole state owned over 100 slaves.
Because they lived on farms with smaller groups of slaves, the social dynamic of slaves in North Carolina was somewhat different from their counterparts in other states, who often worked on plantations with hundreds of other slaves. In North Carolina, the hierarchy of domestic workers and field workers was not as developed as in the plantation system. There were fewer numbers of slaves to specialize in each job, so on small farms, slaves may have been required to work both in the fields and at a variety of other jobs at different times of the year.
Another result of working in smaller groups was that North Carolina slaves generally had more interaction with slaves on other farms. Slaves often looked to other farms to find a spouse, and traveled to different farms to court or visit during their limited free time.
The slave codes passed in the colonial period continued to be enforced during the antebellum years. Whites hoped these laws would prevent threats of slave uprisings. In 1829, David Walker, a free black author born in Wilmington, gave whites in North Carolina another reason to fear their slaves turning against them.
Walker was an avid abolitionist who moved from his home state of North Carolina to Boston, where he helped escaped slaves establish new lives. He wrote and published a pamphlet, Walker’s Appeal, calling for immediate freedom for all slaves and urging slaves to rebel as a group. Copies of the pamphlet were smuggled into Wilmington via ships from the Northern US, and then spread throughout the state.
Whites reacted to Walker’s Appeal by passing increasingly restrictive slave laws. Nervous leaders in North Carolina passed legislation in 1830 making it illegal to distribute the pamphlet in hopes of quelling Walker's radical ideas about abolishing slavery.
Another North Carolina law passed in 1830 made it a crime to teach a slave to read or write. Laws were even extended to restrict the rights of free blacks. An 1835 law prevented free blacks from voting, attending school, or preaching in public.
These restrictive laws were also passed in response to the increase in slave uprisings in nearby states, such as the Nat Turner Rebellion just across the border in Virginia. In 1831, Nat Turner led a group of 75 escaped slaves in an uprising, during which the group killed about 60 white people before being captured by the state militia. Whites in North Carolina were appalled at the thought of a similar rebellion happening in their state, and hoped severe slave laws would prevent such bloody uprisings.
The Life of a Slave
Daily life for a slave in North Carolina was incredibly difficult. Slaves, especially those in the field, worked from sunrise until sunset. Even small children and the elderly were not exempt from these long hours. Slaves were generally allowed a day off on Sunday, and on holidays such as Christmas or the Fourth of July.
During their few hours of free time, most slaves performed their own personal work. The diet supplied by slaveholders was generally poor, and slaves often supplemented it by tending small gardens or fishing. Although there were exceptions, the prevailing attitude among slave owners was to allot their slaves the bare minimum of food and clothing.
Shelter provided by slave owners was also meager. Many slaves lived in small stick houses with dirt floors, not the log slave cabins often depicted in books and films. These shelters had cracks in the walls that let in cold and wind, and had only thin coverings over the windows.
One area of their lives in which slaves were able to exercise some autonomy from their masters was creating a family. Slave owners felt it was to their advantage to allow slaves to marry, because any children from the marriage would add to their wealth. According to law, a child took on the legal status of its mother; a child born to a slave mother would in turn become a slave, even if the father was free.
Because the large plantations of the Lower South needed more slaves than the smaller farms of North Carolina, it was not uncommon for slaves in the state to be sold to slave traders who took them south to Georgia, South Carolina, Mississippi, Louisiana, or Alabama. Once a family member was sold and taken to the Deep South, they became almost impossible to locate or contact.
Although slaves had no way to publicly or legally complain about unfair treatment and abuse, they developed other methods of resistance. Slaves could slow down, pretend to be sick, or sabotage their work as a way to object against long hours of backbreaking labor. Slaves could also steal small amounts of food as a method of protesting their inadequate diet and providing for their families.
The Great Dismal Swamp, which is located in the northeastern part of the state and stretches from Edenton, North Carolina to Norfolk, Virginia, was a common destination for North Carolina runaways. The swamp was an ideal spot in which to hide and forage for food, and some escaped slaves chose to stay and make their homes there. The swamp was also known as a destination for escaped slaves from other states.
As in other states, the Underground Railroad developed in North Carolina to help escaped slaves reach safety. The North Carolina stops were primarily organized by members of the Religious Society of Friends, also known as the Quakers. Levi Coffin was well-known for assisting escaped slaves in Guilford County, North Carolina.
The North and the South clashed over the issue of slavery throughout the 1850s, and the conflict soon boiled over into Civil War. Southern slave owners felt they would quickly defeat the Union. The Union states had about 21 million people, while the Confederate states had approximately 9 million, and over three and a half million of those Southerners were slaves.
President Abraham Lincoln had issued the Emancipation Proclamation on January 1, 1863, stating that slaves in the Confederate states were freed. The proclamation had little immediate effect, but served as a promise that the government would free slaves after the war. Slaves in the Confederate states were officially freed by the passage of the Thirteenth Amendment of the U.S. Constitution in December of 1865.
Slavery in North Carolina
Slavery and Servitude in the Colony of North Carolina
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Explanation: The circumference of a circle, "C" is the distance around a circle. The diameter, "d" is the longest distance across a circle and passes through the center of the circle. The radius, "r", is the distance from the circle to the center of the circle. A diameter is made of 2 radii. There are 2 formulas to find the circumference of a circle, Circumference = pi * diameter and Circumference = 2 * pi * r. The symbol for pi is the Greek letter, , and is typically given the value of 3.14. Leaving an answer as is called 'leaving the circumference in terms of pi'. For the problems in this problem set, you will be given the radius of a circle and you are to find the circumference, leaving the answer in 'in terms of pi'.
Example: If the radius is 12:
Find the For each problem you will need to use the example above to find the circumference. Strive for accomplishing the problems with ease and then strive for speed. Good luck and enjoy the challenge!
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This Food Webs lesson plan also includes:
Explore various ecosystems from around the world as your class discovers the interdependence of all living things. Using the provided sets of ecosystem cards, young scientists work in small groups building food webs to demonstrate the relationships between producers and consumers. To reinforce their understanding, consider allowing time for groups to share their work with the class. As an extension, remove an organism from each group's ecosystem and have them predict what changes they would expect to see.
- Activity allows students to work collaboratively as they build an understanding of food webs
- Ecosystem cards include both English and Spanish descriptions of organisms
- Includes information on six different ecosystems
- Adaptable to the Next Generation Science Standards
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# On a certain test, a grade of 50% or higher is considered a ‘pass’, an
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On a certain test, a grade of 50% or higher is considered a ‘pass’, an [#permalink]
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07 Feb 2019, 07:02
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On a certain test, a grade of 50% or higher is considered a ‘pass’, and any grade under 50% is considered a ‘fail’. If 4 /5 of the students in a class of one hundred students received a grade of 80% or higher, and if the average grade for the entire class was 65%, what is the minimum possible percentage of the class which received a ‘fail’ on the test?
A) 15
B) 16
C) 17
D) 18
E) 19
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On a certain test, a grade of 50% or higher is considered a ‘pass’, an [#permalink]
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Updated on: 08 Feb 2019, 00:57
gmatbusters wrote:
On a certain test, a grade of 50% or higher is considered a ‘pass’, and any grade under 50% is considered a ‘fail’. If 4 /5 of the students in a class of one hundred students received a grade of 80% or higher, and if the average grade for the entire class was 65%, what is the minimum possible percentage of the class which received a ‘fail’ on the test?
A) 15
B) 16
C) 17
D) 18
E) 19
let total max marks be 100
so >50 marks is pass and <50 marks is fail
given in a class of 100 students 4/5 ; 80 students have scored 80% or higher and rest 20 students we dont know..
also avg of class score is 65%
which means
supposedly let the each of 80 students has scored 80 marks so
class avg
80 * 80 + 20 * x/100 = 65
pr say
6400+20x=6500
20x=100
x=100/20 ; 5
so we can say that the avg score of marks scored by total 20 students has to be = 5
sum of score of 20 students /20 = 5
option D 18 students
so if we assume that 18 students each scored 0 marks so total marks scored by 19th & 20th student = 50 each
and avg 18*0+2*50 = 100 marks and avg = 5
Originally posted by Archit3110 on 07 Feb 2019, 11:15.
Last edited by Archit3110 on 08 Feb 2019, 00:57, edited 2 times in total.
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Posts: 8006
Re: On a certain test, a grade of 50% or higher is considered a ‘pass’, an [#permalink]
### Show Tags
07 Feb 2019, 23:49
Archit3110 wrote:
gmatbusters wrote:
On a certain test, a grade of 50% or higher is considered a ‘pass’, and any grade under 50% is considered a ‘fail’. If 4 /5 of the students in a class of one hundred students received a grade of 80% or higher, and if the average grade for the entire class was 65%, what is the minimum possible percentage of the class which received a ‘fail’ on the test?
A) 15
B) 16
C) 17
D) 18
E) 19
I am not sure of the solution , have given a try to solve this question ..
let total max marks be 100
so >50 marks is pass and <50 marks is fail
given in a class of 100 students 4/5 ; 80 students have scored 80% or higher and rest 20 students we dont know..
also avg of class score is 65%
which means
supposedly let the each of 80 students has scored 80 marks so
class avg
80 * 80 + 20 * x/100 = 65
pr say
6400+20x=6500
20x=100
x=100/20 ; 5
so we can say that the avg score of marks scored by total 20 students has to be = 5
sum of score of 20 students /20 = 5
option E 19 students
so if we assume that 19 students each scored 1 marks so total marks scored by 20th student = 81
and avg 19*1+1*81 = 100 marks and avg = 5
chetan2u
You are fine till the total score of 20 is 100 marks..
You have to find the MINIMUM failures, so distribute 100 so as to max out of 20 have passed..
50 marks is pass, so 100 can be divided in 2 of them , 50 each and remaining have got 0..
100=2*50+18*0, so 18 have failed
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Joined: 18 Aug 2017
Posts: 5032
Location: India
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On a certain test, a grade of 50% or higher is considered a ‘pass’, an [#permalink]
### Show Tags
07 Feb 2019, 23:58
chetan2u wrote:
Archit3110 wrote:
gmatbusters wrote:
On a certain test, a grade of 50% or higher is considered a ‘pass’, and any grade under 50% is considered a ‘fail’. If 4 /5 of the students in a class of one hundred students received a grade of 80% or higher, and if the average grade for the entire class was 65%, what is the minimum possible percentage of the class which received a ‘fail’ on the test?
A) 15
B) 16
C) 17
D) 18
E) 19
I am not sure of the solution , have given a try to solve this question ..
let total max marks be 100
so >50 marks is pass and <50 marks is fail
given in a class of 100 students 4/5 ; 80 students have scored 80% or higher and rest 20 students we dont know..
also avg of class score is 65%
which means
supposedly let the each of 80 students has scored 80 marks so
class avg
80 * 80 + 20 * x/100 = 65
pr say
6400+20x=6500
20x=100
x=100/20 ; 5
so we can say that the avg score of marks scored by total 20 students has to be = 5
sum of score of 20 students /20 = 5
option E 19 students
so if we assume that 19 students each scored 1 marks so total marks scored by 20th student = 81
and avg 19*1+1*81 = 100 marks and avg = 5
chetan2u
You are fine till the total score of 20 is 100 marks..
You have to find the MINIMUM failures, so distribute 100 so as to max out of 20 have passed..
50 marks is pass, so 100 can be divided in 2 of them , 50 each and remaining have got 0..
100=2*50+18*0, so 18 have failed
chetan2u
Thanks.. , I really sometimes end up making silly errors.. god help me on D day...
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On a certain test, a grade of 50% or higher is considered a ‘pass’, an [#permalink] 07 Feb 2019, 23:58
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Mining is one of Canada’s primary industries and involves the extraction, refining, and/or processing of economically valuable rocks and minerals.
Mining is one of Canada’s primary industries and involves the extraction, refining, and/or processing of economically valuable rocks and minerals. Mineral products – including gold, silver, iron, copper, zinc, nickel – are critical to modern industrial society. Although mining has been key to Canadian settlement and development, in recent decades the industry has also been criticized for its environmental and social impacts. Canada remains one of the world’s leading mining countries and has become a centre of global mining finance and expertise.
Mining entails the extraction of ore, defined as rock from the earth’s crust containing valuable minerals. It may also include quarrying, or the digging of sand, gravel or aggregate for construction purposes. Although the excavation of oil sands is similar, oil and gas production is not typically considered mining. Important Canadian mineral products include precious metals (gold, silver, platinum) and diamonds; base metals (iron, copper, lead, zinc, nickel); energy minerals such as coal and uranium; and industrial minerals (limestone, rock salt, potash, gypsum).
The broad term “mining” often refers not only to the direct extraction of minerals, but to the complete cycle from discovery to mineral processing. Mineral exploration entails the identification of geological formations that may contain valuable ore. These formations may be found by prospectors traveling over the land or (increasingly) by aerial reconnaissance and sophisticated remote-sensing equipment. Once a valuable ore body is identified and claimed by a mining company, actual mining may proceed if market conditions are favourable and investors can be found.
Mineral extraction ranges from basic low-technology techniques to massive energy- and capital-intensive undertakings. Historically, “high-grade” ores (containing high levels of target minerals) were mined using simple pick-and-shovel or water separation techniques (pans, sluices, etc.), often by a solo miner or small team. Gold was often mined in this manner. As accessible high-grade deposits declined, more elaborate technologies such as dredging and hydraulic mining (i.e. washing away sediment with high pressure water streams) required larger capital investments. Dredging, for instance, was key to maintaining production in the Klondike goldfields after the initial high-grade creek sediments were mined out. Coal and base-metal mines typically required digging and blasting deep underground to follow the veins of ore, for example, some mines tunnel far into mountain sides, while others range below water bodies (as at Newfoundland’s Bell Island mine). Throughout the 20th century, technologies such as underground railcars, machine hoists, and large power shovels reduced the need for human labour as well as the rate and severity of industrial accidents.
After the Second World War, declining ore grades and new technologies, including large earth-moving machines, spurred the development of “opencast” mining. These methods, also known as strip mining or open-pit mining, remove surface material, or overburden, to reveal extensive, low-grade deposits. The removal of vegetation and soils and the creation of massive pits by opencast mining make it controversial as a source of landscape degradation and water pollution.
Mineral Processing and Associated Waste
Typically, target minerals form a small fraction of mined rock; today’s mines often exploit ores containing a fraction of a percent of valuable minerals. Mineral processing, or beneficiation, removes target minerals through various thermo-mechanical or chemical techniques. The waste material, a pulverized rock slurry known as tailings, poses a major disposal challenge. It often contains heavy metals (cadmium, arsenic, lead, zinc, etc.) that may be released into waterways, soil, and the air (as dust) at levels that threaten the health of humans and wildlife. Moreover, tailings from sulfide mineral deposits may react with water and air to produce sulfuric acid (or acid mine drainage) which can in turn dissolve more toxic heavy metals from surrounding rock or waste piles. In the case of uranium or radium mining, radioactive material within tailings may present a danger to local populations of humans and wildlife. At many mines, the sheer volume of waste rock and tailings can be staggering, creating large-scale, human-made topography such as ridges, hills, and the pits where the tailings were originally mined. The processing of ore in smelters, where rock is heated to very high temperatures to extract base metals, can result in smokestack pollution in the form of toxic compounds, such as sulfur dioxide and arsenic trioxide. While technological improvements have resulted in the increased capture of these pollutants prior to emission from smelter stacks, the collection of highly toxic dusts, such as arsenic trioxide, can pose additional waste disposal problems. More broadly, the combination of high postwar demand, declining ore grades, and increasing volumes of rock and tailings has meant that mining now generates the largest waste stream of any industry.
Pre-contact Aboriginal use and trade of Canada’s rich mineral resources included copper, gold, silver and chert. European exploration and colonization of northern North America was partially motivated by the search for valuable minerals. In 1577–8, Martin Frobisher established a mining operation at Baffin Island although the ore shipped turned out to be worthless and the venture collapsed. Fur trader Samuel Hearne and his guide Matonabbee traversed the region west of Hudson Bay to the mouth of the Coppermine River from 1770–2, in search of deposits of copper exploited by Dene people (including the Yellowknives, so named for their copper tools). During this early colonial period, settlers and sailors exploited small quantities of underground resources, particularly coal outcrops on Cape Breton Island and in New Brunswick, for local trade in the mid-17th century.
Large-scale industrial exploitation of mineral resources in Canada commenced at the St. Maurice Forges in Québec. They were fed by iron deposits at Trois-Rivières and operated from 1738 to 1883. Industrial demand for coal and iron spurred the expansion of mining and smelting in the late colonial period. Coal mining developed rapidly near Sydney, Cape Breton, and, in the mid-19th century, major coal deposits were located on Vancouver Island.
Precious metal discoveries in the 19th century spurred settlement and hinterland development across the country. Placer gold deposits in Québec and Nova Scotia stimulated important new mining laws, and massive gold rushes to the Fraser Canyon in British Columbia (1858) and the Klondike in the Yukon (1896) led directly to the extension of colonial and, eventually, Canadian authority over these regions. Silver discoveries north of Lake Superior spurred treaty-making with local First Nations to permit prospecting and mineral development in the 1850s. Ontario’s rich Silver Islet mine was notable as the first site to use a diamond drill and compressed air drills. Silver discoveries near Mayo in the central Yukon led to the establishment of the Keno Hill mining camp, whose production between 1913 and 1989 exceeded the value of the Klondike.
Railway construction at the end of the 19th century opened up new mining territories in Northern Ontario and across the country. Canadian Pacific Railway construction led to the development of two of Canada’s most significant individual mines: Sudbury nickel in the 1880s and, later, the Sullivan lead-zinc mine in the Kootenay region of British Columbia. Colonization railways constructed in northern Ontario led to the important silver discovery at Cobalt (1903), which, for a time, was North America’s greatest silver camp. The pre-First World War period also saw the discovery of gold at Porcupine (later Timmins) and Kirkland Lake, Ontario.
In the first half of the 20th century, Canada emerged as a world-leading producer of a wide range of minerals. Mines in the Canadian Shield produced not only precious metals but, increasingly, critical base metals, such as copper, lead and zinc as well. Mines at Flin Flon (copper-zinc) and Sherridon (nickel), Manitoba, helped consolidate the province’s northward expansion, and similarly dramatic transformations followed the discoveries of radium at Great Bear Lake and gold at Yellowknife in the 1930s, inaugurating the Northwest Territories’ storied mining history. To the east, gold mines were founded in the Rouyn-Noranda–Val-d’Or gold belt in Québec in the 1920s.
Driven by global demand for raw materials, the post-Second World War period witnessed a rapid expansion in the industry and saw important developments in recently-discovered minerals, such as uranium at Great Bear Lake, in northern Saskatchewan, and northern Ontario; potash on the Prairies and in New Brunswick; and industrial minerals such as asbestos, molybdenum and gypsum. Though not a major global producer, the Canadian coal industry also expanded, even as the Nova Scotia and Newfoundland fields were exhausted, through the development of large open-pit and strip-mining operations in British Columbia.
Beginning in the 1960s, opencast mining permitted the massive expansion of Canadian mineral production from coast-to-coast. Key sites included: Pine Point, Northwest Territories (lead, zinc); Highland Valley, British Columbia (copper); Northern Saskatchewan (uranium); Thetford Mines, Québec (asbestos); the Brunswick #12 mine at Bathurst, New Brunswick (lead, zinc, copper); and Schefferville, Québec, and Labrador City, Newfoundland (iron ore). Mining also expanded to Canada’s far north; the short-lived Rankin Inlet nickel mine was Canada’s first Arctic mine and was followed by Nanisivik and Polaris, the first high-Arctic mines. More recently, a major nickel mine was established at Voisey’s Bay, Labrador, and diamond mines (Ekati and Diavik) in the Northwest Territories opened in the 1990s, even as the latter region’s gold mines began their decline. Since the 1990s, Canada has emerged as a globally significant producer of diamonds and platinum group metals, while rare earth elements (used in green technologies and consumer electronics) are considered an important future prospect.
Mineral processing industries also expanded in this period, from the presence of a few smelters (notably at Trail, British Columbia; Sudbury, Ontario; and Arvida, Québec), to include a dozen base-metal refineries across the country, ferro-alloy plants for processing iron ore, as well as a number of other types of mineral “beneficiation” facilities. Aluminum smelters were established in the Saguenay–Lac-Saint-Jean region of Québec before the war and at Kitimat, British Columbia in 1951, each taking advantage of large hydroelectric power installations.
Mining and the Economy
The value of mineral production surged in the postwar period, growing from just over $400 million to over $5 billion in 1975. Although the industry is beset by cyclical downturns in demand and price, —usually associated with recessions, such as those in the early 1980s and 1990s — the total value of non-fuel minerals reached $21.7 billion in 2004. After a dip in 2011, Canada’s total estimated mineral production in 2012 topped $46.8 billion, representing 3.4 per cent of the country’s gross domestic product. While mining’s importance to Canada’s overall economy and employment has declined, it remains a regionally significant industry, particularly in northern parts of the provinces and the northern territories. Ontario, British Columbia, Saskatchewan and Québec are the leading mineral producing provinces.
Canada is not only a major producer of important minerals and metals, but also a centre of global mining finance and expertise. The TSX and TSX Venture stock exchanges have become world centres for investment in mining and mineral exploration companies. Canadian-based firms operate mines around the world, but there is increasing scrutiny and controversy surrounding the practices and impacts of Canadian mining companies in developing countries. Similar concerns around the effects of large-scale mineral developments on nearby Aboriginal territories in Canada (especially in the North) are prompting mining companies to adopt “corporate social responsibility” initiatives, including community consultation and impact and benefit agreements.
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An ear infection is a common condition that affects the middle or inner ear and is usually caused by bacteria or viruses. It can occur in people of all ages, but it is more common in children, due to their smaller and less developed Eustachian tubes. This infection can cause discomfort and pain, as well as affect a person’s hearing and balance.
There are several factors that can increase the risk of developing an ear infection. These include a weakened immune system, allergies, colds or upper respiratory infections, tobacco smoke exposure, and genetic factors. In addition, people who swim frequently or have a habit of inserting objects into their ears are also more prone to ear infections.
Common symptoms of an ear infection include ear pain, difficulty hearing, a feeling of fullness in the ear, and drainage from the ear. Other symptoms may include fever, irritability, and difficulty sleeping, especially in infants and young children. It is important to seek medical attention if you or your child experiences these symptoms, as a proper diagnosis and treatment are necessary to prevent complications and to relieve the discomfort caused by the infection.
Treatment for an ear infection usually involves the use of antibiotics to fight the infection. Pain relievers may also be recommended to alleviate the pain and discomfort. It is important to complete the full course of antibiotics prescribed by your healthcare provider, even if symptoms improve, to ensure that the infection is fully cleared. In some cases, if the infection is severe or recurrent, surgical intervention may be necessary.
To prevent ear infections, it is important to practice good hygiene, such as washing hands frequently, avoiding exposure to secondhand smoke, and keeping immunizations up to date. It is also important to avoid inserting objects into the ear, as this can damage the ear canal and increase the risk of infection. If you or your child is prone to ear infections, your healthcare provider may recommend certain preventive measures, such as earplugs while swimming or the use of ventilation tubes to improve the airflow in the middle ear.
Understanding Ear Infections
An ear infection, also known as otitis media, is a common condition that affects the middle ear. This infection can occur in people of all ages, but it is most commonly seen in young children.
The middle ear is the area behind the eardrum, and it contains tiny bones that help transmit sound vibrations to the inner ear. When the middle ear becomes infected, it can cause symptoms such as ear pain, fluid buildup, and hearing loss.
There are several factors that can contribute to the development of an ear infection. One of the most common causes is a bacterial or viral infection. When bacteria or viruses enter the middle ear, they can cause inflammation and infection.
Symptoms of Ear Infections
Common symptoms of ear infections include:
- Ear pain or discomfort
- Fluid drainage from the ear
- Difficulty hearing or hearing loss
It’s important to note that not all ear infections cause symptoms. In some cases, an ear infection may be asymptomatic and only discovered during a routine examination.
Treatment of Ear Infections
Treatment for ear infections may vary depending on the severity and cause of the infection. In some cases, the infection may resolve on its own without any intervention. However, if symptoms are severe or the infection persists, medical treatment may be necessary.
Common treatments for ear infections include:
- Pain relievers to help manage ear pain
- Antibiotics to treat bacterial infections
- Ear drops to help reduce inflammation and relieve symptoms
- Surgery may be necessary in severe or recurrent cases
If you suspect you or your child has an ear infection, it’s important to seek medical attention. A healthcare professional can properly diagnose the infection and recommend appropriate treatment options.
Causes of Ear Infection
Ear infections are commonly caused by bacterial or viral infections. The most common bacteria that cause ear infections are Streptococcus pneumoniae, Haemophilus influenzae, and Moraxella catarrhalis. These bacteria can enter the middle ear through the eustachian tube, which connects the middle ear to the back of the throat.
Viral infections, such as those caused by respiratory syncytial virus (RSV) and influenza virus, can also lead to ear infections. These viruses can cause inflammation and fluid buildup in the middle ear, creating an environment that is conducive to bacterial growth.
In addition to infections, other factors can contribute to the development of ear infections. These include:
- Exposure to smoke or air pollution
- Genetic factors that make some individuals more prone to ear infections
- Environmental factors, such as living in crowded conditions or attending daycare
- Enlarged or infected adenoids
- Changes in air pressure, such as during airplane travel or scuba diving
It’s important to note that ear infections are not contagious. However, the underlying viral or bacterial infection that may have caused the ear infection can be contagious.
Understanding the causes of ear infections can help individuals take preventive measures to reduce their risk of developing an infection. These measures may include practicing good hygiene, avoiding exposure to smoke or air pollution, and keeping the ears dry and clean.
Bacterial infections in the ear are a common cause of ear infections. These infections occur when bacteria enter the ear through the eustachian tube, which connects the middle ear to the back of the throat. Bacteria can enter the eustachian tube due to various factors, including respiratory infections, allergies, or a weakened immune system.
Common Bacterial Infections in the Ear
There are several types of bacteria that can cause ear infections, including:
|This bacteria is one of the most common causes of bacterial ear infections, particularly in children. It can also cause other respiratory infections, such as pneumonia.
|This bacteria is another common cause of ear infections, especially in children. It can also cause other respiratory infections, such as sinusitis and bronchitis.
|This bacteria is commonly found in the respiratory tract and can cause ear infections, especially in children.
|This bacteria is less common but can cause severe ear infections, especially in individuals with a weakened immune system.
Symptoms of Bacterial Ear Infections
Bacterial ear infections often present with similar symptoms to other types of ear infections. Common symptoms include:
- Ear pain or discomfort
- Fluid draining from the ear
- Decreased hearing
- Irritability in children
If you or your child experience these symptoms, it is important to see a healthcare professional for a proper diagnosis and treatment.
When it comes to treating bacterial ear infections, antibiotics are often prescribed. These medications are effective at fighting the bacterial infection and relieving symptoms. However, it is crucial to complete the full course of antibiotics as prescribed by a healthcare professional to ensure the infection is completely eradicated.
Viral infections are a common cause of ear infections. Viruses can invade the ear and cause inflammation, leading to symptoms such as pain, earache, and sometimes, a middle ear infection. Viral ear infections can occur in both children and adults.
Common viral infections that can cause ear infections include:
- Influenza (flu) virus
- Rhinovirus (common cold)
- Respiratory syncytial virus (RSV)
These viruses are highly contagious and can spread from person to person easily, especially in crowded places such as schools and daycare centers.
Symptoms of Viral Ear Infections
The symptoms of viral ear infections are similar to bacterial ear infections and may include:
- Ear pain or discomfort
- Feeling of fullness in the ear
- Decreased hearing
- Drainage from the ear
It’s important to note that viral ear infections do not usually require antibiotics to treat as they are self-limiting and will resolve on their own within a few days to a week.
Treatment of Viral Ear Infections
Treatment for viral ear infections focuses on relieving symptoms and preventing complications. The following measures may be taken:
- Pain relievers such as acetaminophen or ibuprofen
- Applying warm compresses to the affected ear
- Using over-the-counter ear drops for pain relief
- Getting plenty of rest
- Staying hydrated
If symptoms persist or worsen, it is advisable to seek medical attention. A healthcare provider can evaluate the symptoms and determine if further treatment is necessary.
Fungal infections can also cause ear infections. Fungi are tiny organisms that can live on the skin, including the skin inside the ear. When the balance of bacteria and fungi is disrupted, fungal overgrowth can occur, leading to an ear infection.
Common symptoms of a fungal ear infection include itching, redness, and flaking or peeling skin inside the ear. In some cases, there may also be a foul-smelling discharge from the ear. Fungal ear infections are more common in people who have moist conditions inside their ears, such as swimmers or individuals who frequently use earphones or hearing aids.
Treatment for fungal ear infections typically involves the use of antifungal medications. These medications can be applied topically, by placing them directly in the ear, or taken orally. It is important to follow the prescribed treatment plan and finish the full course of medication, even if symptoms improve.
In addition to medication, it may be helpful to keep the ears dry and clean to reduce the risk of fungal overgrowth. This can be done by avoiding excessive moisture in the ears, drying the ears thoroughly after swimming or showering, and avoiding the use of cotton swabs or other objects that can irritate the ear canal.
In summary, fungal ear infections can occur when there is an overgrowth of fungi in the ear. These infections can cause symptoms such as itching, redness, and discharge. Treatment typically involves antifungal medications and maintaining good ear hygiene to prevent future infections. If you suspect you have a fungal ear infection, it is best to consult with a healthcare professional for an accurate diagnosis and appropriate treatment.
Symptoms of Ear Infection
An ear infection can cause a variety of symptoms that can range from mild to severe. Common symptoms of an ear infection include:
1. Ear Pain
One of the most common symptoms of an ear infection is ear pain. The pain can be sharp or dull and may come and go throughout the day. It can also be more intense during activities such as chewing or lying down.
2. Ear Discomfort
People with an ear infection may experience a feeling of fullness or pressure in the affected ear. This can make it difficult to hear or may cause a temporary decrease in hearing ability.
3. Ear Drainage
In some cases, an ear infection may cause the ear to produce a discharge. This discharge can range from clear to yellow or green in color and may have an unpleasant odor.
A fever is a common symptom of an ear infection, especially in children. A high temperature can indicate that the body is fighting off an infection. The fever may be accompanied by other flu-like symptoms such as fatigue or body aches.
5. Irritability or Fussiness
Children with an ear infection may become irritable or fussy. They may cry more often than usual and have difficulty sleeping or eating. This is because the ear pain and discomfort can be quite bothersome for them.
6. Difficulty Balancing
Some people with an ear infection may experience difficulty balancing. This can make them feel dizzy or lightheaded, which can affect their coordination and ability to perform normal tasks.
7. Hearing Loss
An ear infection can cause temporary or permanent hearing loss, depending on the severity and duration of the infection. The hearing loss may be mild or significant, and can affect one or both ears.
If you or your child experience any of these symptoms, it is important to seek medical attention. A healthcare professional can diagnose the ear infection and provide appropriate treatment to alleviate the symptoms and prevent complications.
|Irritability or Fussiness
Ear pain is a common symptom of an ear infection, also known as otitis media. The pain can range from mild to severe, and it is often accompanied by other symptoms such as earache, fever, and hearing loss.
The most common cause of ear pain is an infection in the middle ear, which is located between the eardrum and the inner ear. This infection can be caused by bacteria or viruses, and it can lead to inflammation and swelling of the ear canal.
In addition to infection, ear pain can also be caused by other factors such as earwax buildup, a foreign object stuck in the ear, or trauma to the ear. In some cases, ear pain may be a symptom of a more serious condition such as a ruptured eardrum or a tumor.
Treatment for ear pain depends on the underlying cause. In the case of an ear infection, antibiotics may be prescribed to help clear the infection. Pain relievers such as acetaminophen or ibuprofen may also be recommended to help alleviate the pain.
If the ear pain is caused by a foreign object in the ear, a healthcare professional will need to remove it. In the case of a ruptured eardrum, surgery may be required to repair the damage. If a tumor is causing the ear pain, further medical intervention will be necessary.
If you are experiencing ear pain, it is important to seek medical attention to determine the cause and receive appropriate treatment. Ignoring ear pain or attempting to self-diagnose and treat the condition can lead to complications and further damage to the ear.
Ear discharge, also known as otorrhea, refers to the fluid that comes out of the ear. It can be a sign of an ear infection or other underlying conditions. The discharge may be watery, thick, or bloody, and it can have a foul odor.
There are several possible causes of ear discharge, including:
- Ear infection: An infection in the ear, such as otitis media or otitis externa, can cause the ear to produce fluid or pus as a defense mechanism.
- Swimmer’s ear: This is an infection of the ear canal that is usually caused by water remaining in the ear after swimming or bathing, creating a moist environment that is conducive to bacterial growth.
- Trauma or injury: Injury to the ear, such as a piercing gone wrong or a blow to the ear, can lead to ear discharge.
- Foreign object: Something lodged in the ear, such as a cotton swab or a small toy, can cause irritation and lead to ear discharge.
Common symptoms associated with ear discharge include ear pain, hearing loss, itching, redness, and swelling of the ear, as well as a feeling of fullness in the ear. If not treated promptly, ear discharge can lead to complications such as hearing loss or a chronic infection.
If you notice ear discharge, it is important to consult a healthcare professional for an accurate diagnosis and appropriate treatment. Treatment may involve antibiotics, antifungal medications, ear drops, or ear cleaning. In some cases, surgery may be necessary to remove a foreign object or treat a severe infection.
To prevent ear discharge, it is important to practice good ear hygiene, avoid inserting objects into the ear, and protect the ears from excessive moisture.
If you have any concerns about ear discharge or other ear-related issues, it is best to seek medical advice for proper evaluation and treatment.
Hearing loss is a common condition that can occur as a result of an ear infection. When the ear becomes infected, it can affect the normal functioning of the ear and cause temporary or permanent hearing loss. The infection can cause damage to the structures in the ear, such as the eardrum or the tiny hair cells in the inner ear that are responsible for transmitting sound signals to the brain.
There are different types of hearing loss that can occur as a result of an ear infection. Conductive hearing loss occurs when there is a problem transmitting sound waves from the outer ear to the middle ear. This can be caused by a blockage in the ear canal, fluid buildup in the middle ear, or damage to the middle ear structures. Sensorineural hearing loss, on the other hand, occurs when there is damage to the inner ear or the auditory nerve. This type of hearing loss is more likely to be permanent.
The symptoms of hearing loss can vary depending on the severity of the infection and the type of hearing loss. Common symptoms include difficulty understanding speech, muffled or distorted sounds, ringing in the ears (tinnitus), and a feeling of fullness in the affected ear. Children with hearing loss may have difficulty following instructions, delayed speech development, or problems with academic performance.
Treatment for hearing loss caused by an ear infection may involve addressing the underlying infection with antibiotics or antiviral medications. In some cases, surgical intervention may be necessary to remove blockages or repair damage to the ear structures. In addition to medical treatment, hearing aids or assistive listening devices may be recommended to improve hearing function.
Prevention is key in reducing the risk of hearing loss associated with ear infections. Maintaining good ear hygiene, avoiding exposure to loud noises, and seeking prompt treatment for ear infections can help prevent complications and minimize the impact on hearing. Regular check-ups with an ear specialist, known as an otolaryngologist, can also help identify and address any potential hearing problems early on.
Diagnosis of Ear Infection
Diagnosing an ear infection usually involves a combination of medical history, physical examination, and sometimes additional tests. The first step in diagnosing an ear infection is to carefully listen to the patient’s symptoms and medical history. The doctor may ask about any recent illnesses, exposure to loud noises or chemicals, or previous ear infections.
During the physical examination, the doctor will use an otoscope to look inside the ear canal and eardrum. They will be checking for signs of inflammation, redness, or fluid buildup. The otoscope allows the doctor to see the middle ear and eardrum, which can help confirm the presence of an infection.
In some cases, additional tests may be necessary to further evaluate the ear infection. These tests can include a tympanometry, which measures the movement of the eardrum in response to changes in air pressure, or a culture of the fluid from the ear to identify the specific bacteria or virus causing the infection.
Overall, the diagnosis of an ear infection is made based on the combination of the patient’s symptoms, medical history, and the findings of the physical examination and any additional tests. It is important to consult a healthcare professional for an accurate diagnosis and appropriate treatment plan.
During a physical examination for an ear infection, a healthcare provider will begin by inspecting the outer ear for any visible signs of infection or inflammation. They will use an otoscope, a specialized tool with a light and magnifying lens, to examine the ear canal and eardrum.
The healthcare provider will check for redness, swelling, discharge, or any other abnormalities in the ear canal and eardrum. They may gently pull on the earlobe or move the ear to assess for pain, which can be a sign of an ear infection.
In some cases, a pneumatic otoscope may be used to assess the mobility of the eardrum. This test involves blowing a puff of air through a tube attached to the otoscope. The movement of the eardrum in response to the air pressure can provide additional information about the presence of an ear infection.
|Physical Examination Steps
|Inspect the outer ear
|Look for any visible signs of infection or inflammation such as redness or swelling.
|Examine the ear canal and eardrum
|Use an otoscope to check for abnormalities, discharge, or inflammation.
|Assess for pain
|Gently pull on the earlobe or move the ear to see if there is any pain, which may indicate an ear infection.
|Perform pneumatic otoscopy
|Blow a puff of air into the ear to assess the movement of the eardrum and gather more information about the infection.
Based on the findings from the physical examination, the healthcare provider will determine the appropriate course of treatment for the ear infection.
Ear culture is a diagnostic test that is used to identify the specific organism responsible for causing an infection in the ear. It involves taking a sample of fluid or discharge from the ear and sending it to a laboratory for analysis. The sample is placed in a culture medium, which provides an environment for the bacteria or fungi to grow.
The culture medium is then observed over a period of time, typically 24 to 48 hours, to allow the organisms to multiply and form colonies. These colonies can be examined under a microscope or tested using different biochemical techniques to determine the type of infection.
Ear culture can help healthcare professionals determine the most appropriate treatment for an ear infection. It can also help identify any antibiotic resistance, which is important in guiding treatment decisions.
It is important to note that ear culture is not always necessary for every case of ear infection. In some cases, the symptoms and medical history may be enough to make a diagnosis and begin treatment. However, in cases where the infection is severe, recurring, or not responding to treatment, ear culture may be ordered to help identify the specific organism causing the infection.
Overall, ear culture is a valuable tool in diagnosing and treating ear infections. It allows healthcare professionals to target the infection and provide appropriate treatment, improving outcomes for patients.
If you suspect that you have an ear infection, it is important to get a proper diagnosis. Hearing tests are commonly performed to assess the extent of the infection and determine the best course of treatment.
Types of Hearing Tests
There are several types of hearing tests that can be conducted to evaluate your hearing and detect any abnormalities caused by an ear infection:
- Audiometry: This test measures your ability to hear sounds at different frequencies and volumes. It involves wearing headphones and listening to tones or words, and indicating when you can hear them.
- Tympanometry: This test assesses the movement of your eardrum in response to changes in air pressure. It helps determine if there is fluid or blockage in the middle ear.
- Otoacoustic Emissions (OAE) Test: This test measures the sounds that your inner ear produces in response to a sound stimulus. It can assess the function of the cochlea and hair cells in the inner ear.
- Brainstem Auditory Evoked Response (BAER) Test: This test measures the electrical activity of the auditory nerve and brainstem in response to sound. It helps identify any potential nerve damage or abnormalities.
Benefits of Hearing Tests
Hearing tests are crucial for diagnosing an ear infection and determining the appropriate treatment plan. They provide valuable information about the severity of the infection and the specific areas of the ear that are affected. The results of these tests help doctors make informed decisions about medication, surgery, or other interventions necessary to treat the infection and restore proper hearing.
Treatment of Ear Infection
The treatment of ear infections typically involves a combination of medication and home remedies. It is important to consult a healthcare professional to determine the most appropriate treatment plan for your specific situation.
In most cases, ear infections are caused by bacterial or viral infections. Antibiotics are commonly prescribed to treat bacterial ear infections. These medications help to kill the bacteria and alleviate the symptoms.
For viral ear infections, antiviral medication may be recommended to reduce the severity and duration of the infection.
Pain relievers, such as acetaminophen or ibuprofen, may be prescribed or recommended to help relieve ear pain and discomfort.
In addition to medication, there are several home remedies that can help provide relief from ear infections:
Warm Compresses: Applying a warm compress to the affected ear can help to reduce pain and inflammation.
Over-the-counter ear drops: These drops can help to relieve pain and reduce swelling in the ear canal.
Keep the ear clean and dry: Avoid getting water or other liquids in the affected ear, as this can exacerbate the infection.
It is important to note that home remedies should not be used as a substitute for medical treatment. They can be used in conjunction with medication, but it is essential to consult a healthcare professional for proper diagnosis and treatment.
If left untreated or improperly treated, ear infections can lead to complications, such as hearing loss or the spread of infection to nearby structures. It is important to seek medical attention if you suspect you or your child has an ear infection.
Antibiotics are often prescribed as a treatment for ear infections. They help to kill the bacteria that cause the infection and reduce inflammation in the ear. It is important to take the full course of antibiotics as prescribed by a healthcare professional to ensure that the infection is completely cleared.
There are different types of antibiotics that may be prescribed for ear infections, including:
Amoxicillin is a commonly used antibiotic for treating ear infections. It is effective against many types of bacteria that can cause ear infections.
Cefuroxime is another antibiotic that is often used to treat ear infections. It is particularly effective against certain types of bacteria that can be resistant to other antibiotics.
The choice of antibiotic will depend on factors such as the type and severity of the ear infection, the individual’s age, and any other medical conditions they may have. It is important to follow the healthcare professional’s instructions and finish the full course of antibiotics.
In some cases, if the ear infection is persistent or severe, a healthcare professional may recommend other forms of treatment in addition to antibiotics, such as ear drops or surgery. It is important to seek medical advice if you or your child is experiencing symptoms of an ear infection.
|Type of Antibiotic
|Commonly Used For
|Ear infections caused by a variety of bacteria
|Ear infections caused by certain types of resistant bacteria
|Infections that do not respond to usual treatments
When dealing with an ear infection, pain relief is essential. Over-the-counter pain relievers can help alleviate the discomfort caused by the infection. These medications, such as acetaminophen or ibuprofen, can help reduce pain and inflammation in the ear. However, it is important to follow the instructions and recommended dosage when taking these medications, and to consult with a healthcare professional if symptoms worsen or persist.
It is also important to note that pain relievers only provide temporary relief and do not treat the underlying infection. Therefore, it is important to seek medical attention for proper diagnosis and treatment of the ear infection.
In addition to over-the-counter pain relievers, a healthcare professional may prescribe stronger pain medications if the infection is severe or if the pain persists. These prescription medications can provide stronger and more targeted pain relief. Again, it is important to follow the prescribed dosage and consult with a healthcare professional for guidance.
Along with pain relievers, warm compresses can also offer relief from ear pain. Applying a warm compress to the affected ear can help soothe the discomfort and reduce inflammation. It is important to ensure that the compress is not too hot to avoid burns or further irritation.
Ear drops are a common treatment for ear infections. They contain medication that helps to reduce inflammation and fight the infection. Ear drops are typically prescribed by a doctor and should be used according to the instructions provided.
How Do Ear Drops Work?
Ear drops work by delivering medication directly to the infected area in the ear. They are usually a combination of antibiotics and steroids. The antibiotics help to kill the bacteria causing the infection, while the steroids reduce inflammation and swelling.
To use ear drops, the affected person should lie on their side with the infected ear facing up. The ear should be gently tilted towards the ceiling. The ear drops are then placed in the ear canal and the person should remain in that position for a few minutes. This allows the drops to reach the infected area and begin working.
When to Use Ear Drops
Ear drops are typically used for mild to moderate ear infections. They are often prescribed for acute otitis media, which is an infection of the middle ear. They may also be used for external otitis, which is an infection of the outer ear or ear canal.
If you suspect you have an ear infection, it is important to see a doctor for a proper diagnosis. They will be able to determine if ear drops are the appropriate treatment for your specific infection.
It is important to follow the instructions provided by your doctor when using ear drops. It is also important to finish the full course of medication, even if symptoms improve before the medication is finished. This helps to ensure that the infection is completely cleared.
If you experience any side effects from using ear drops, such as increased pain or irritation, it is important to contact your doctor. They may need to adjust the treatment or prescribe a different medication.
In conclusion, ear drops are a common treatment for ear infections. They work by delivering medication directly to the infected area in the ear. It is important to follow the instructions provided by a doctor and finish the full course of medication to ensure the infection is completely cleared.
If the ear infection does not respond to medical treatment or if it becomes chronic or recurring, surgical intervention may be necessary. In some cases, the infection may spread to the mastoid bone behind the ear, causing a serious condition called mastoiditis. Surgery may be required to drain the pus and remove any infected tissue.
In other cases, a small incision may be made in the eardrum to allow for drainage of fluid and relieve pressure. This procedure is known as a myringotomy. A tiny tube may also be inserted into the incision to keep it open and allow for continued drainage.
Surgical intervention is generally considered a last resort and is only recommended when other treatment options have been exhausted. It is important to discuss the risks and benefits of surgery with a healthcare professional to determine if it is the best course of action for your individual situation.
What is an ear infection?
An ear infection, also known as otitis media, is an infection or inflammation of the middle ear. It occurs when fluid builds up behind the eardrum, leading to pain and discomfort.
What causes an ear infection?
Ear infections are often caused by bacteria or viruses. The most common cause is a bacterial infection from the Streptococcus pneumoniae or Haemophilus influenzae bacteria. Viral infections, such as the common cold or flu, can also lead to an ear infection.
What are the symptoms of an ear infection?
The symptoms of an ear infection may vary depending on the age of the person. In children, common symptoms include ear pain, irritability, trouble sleeping, pulling or tugging at the ear, fever, fluid draining from the ear, and difficulty hearing. Adults may experience ear pain, fluid drainage, hearing loss, and dizziness.
How is an ear infection diagnosed?
An ear infection is typically diagnosed by a healthcare professional. They will examine the ear using an otoscope to look for redness, swelling, or fluid behind the eardrum. In some cases, a sample of the fluid may be taken for further testing.
What is the treatment for an ear infection?
The treatment for an ear infection may vary depending on the severity and cause of the infection. In many cases, antibiotics are prescribed to help clear the infection. Over-the-counter pain relievers, such as acetaminophen or ibuprofen, may be recommended to help relieve pain. In some cases, the doctor may recommend ear drops to reduce inflammation and pain. It is important to follow the prescribed treatment and attend any follow-up appointments.
What are the causes of ear infection?
Ear infections can be caused by several factors, including bacterial or viral infections, allergies, blocked Eustachian tubes, and colds or respiratory infections.
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Isotope stratigraphy is a method of determining relative ages of sediments based on measurement of isotopic ratios of a particular element. It works on the principle that the proportions of some isotopes incorporated in biogenic minerals (calcite, aragonite, phosphate) change through time in response to fluctuating palaeoenvironmental and geological conditions. However, this primary signal is often masked by diagenetic alteration of sediments which have secondarily altered the isotopic ratios. Disentangling primary and secondary components of measured isotopic ratios is a difficult and frequently controversial subject. Although isotopes of many elements have been studied oxygen and carbon strontium, are of particularly wide application.
The ratios in which the two stable isotopes of oxygen (16O and 18O) are precipitated in carbonates and phosphates depends upon the oxygen isotopic composition of the fluid from which the mineral precipitated and also on the temperature at which this took place. However, some organisms incorporate oxygen isotopes that are out of equilibrium with temperature and seawater composition. In addition, primary isotopic values may commonly be altered by diagenetic recrystallisation of carbonate sediments.
Oxygen isotopes can record detailed changes in ocean temperature and ice volume. The most extensive use of oxygen isotopes has been in deep-sea cores of Cenozoic, especially Quaternary sediments, where data from calcitic microfossils, notably foraminifera, record fluctuating temperatures and the growth and decay of ice-sheets, allowing the recognition of oxygen isotope stages. The separate effects of temperature and ice volume are distinguished by comparing isotope ratios in coeval planktonic and benthonic microfossils, mainly foraminifera. Because both parameters were driven by Milankovitch climatic cycles, it has been possible to identify and correlate oxygen isotope stages in detail across the globe, and 18O curves provide a very refined (20 ka resolution) time-scale for Quaternary to Neogene time. In pre-Cenozoic sediments the use of oxygen isotopes in both stratigraphy and palaeoenvironmental studies has been much more limited because much of the carbonate is recrystallised, and only rarely reflects secular changes in oxygen isotope ratios.
For discussion of the Marine Isotope Stage (abbreviated as MIS) time-scale, derived using oxygen isotope measurements, see ‘climatostratigraphy‘.
The two stable isotopes of carbon, 12C and 13C, vary in relative abundance through time in both carbonate minerals and organic matter. The fluctuations in 13C are brought about by changes in the balance of fluxes of the carbon cycle, including inputs of terrestrial carbon and oxidation of marine organic matter, and outputs by production and burial of marine carbonate and organic matter. Because the residence time in the carbon cycle is brief (10 ka), changes in flux are recorded accurately and globally in the sedimentary record. Furthermore, carbon isotopes are relatively robust and resistant to diagenesis.
Strontium isotope stratigraphy relies on measurement of 87Sr/86Sr in marine biogenic carbonate or phosphate. Precipitation of these minerals involves incorporation of strontium from seawater, which will have an 87Sr/86Sr identical to that of oceanic values, which is of the same value globally at any point in time. The 87Sr/86Sr ratio changed systematically through time and it is therefore possible to date samples by placing them on a standard curve. The method works best for periods of time over which there was a long-term unidirectional shift in ratios, as during the Tertiary. Strontium isotope stratigraphy gives a maximum time-resolution of about 1 ma.
Strontium in seawater is derived from three sources: fluvial input of material weathered from continental crust; hydrothermal leaching of oceanic basalts at mid-ocean ridges; and recrystallisation of carbonate minerals. Changing strontium values reflect global changes in these geological processes.
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What possible use could architects have for a supercomputer? Well, of course it would be nice to produce that ultra-high-quality render in a matter of seconds rather than hours - but this post on the XSEDE blog recounts another use that is (arguably) much more important. XSEDE, an organization that helps researchers by providing them with access to supercomputers, has been working with a group from the University of Utah's Mechanical Engineering Department to simulate wind flow in cities, with the ultimate aim of providing architects and engineers with the tools to reduce wind tunneling effects, improve energy efficiency and lower pollution. Find out more about the research project here.
How Supercomputers are Shaping the Future of our Cities
Cite: Rory Stott. "How Supercomputers are Shaping the Future of our Cities" 28 Mar 2015. ArchDaily. Accessed . <https://www.archdaily.com/613909/how-supercomputers-are-shaping-the-future-of-our-cities> ISSN 0719-8884
About this author
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A list of student-submitted discussion questions for Linnaean Classification.
To build and enhance understanding of the relationships between key words and to practice useful skills for designing information architecture using the Card Sort strategy.
To use a memory aid to help students remember ordered events or sequences using Mnemonics.
Covers the importance of soft-bodied fossils and what scientists learn from them, also looks at the fossil record, fossilization and adaptive radiations.
The Linnean classification and phylogenetic classification systems are compared in this study guide.
These flashcards help you study important terms and vocabulary from Linnaean Classification.
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Matematika Testy - Cvičení z matematiky
Numbers 1-12 squared
52=?
Square numbers
A square number is a number multiplied by itself. The symbol is 2 (we call it exponent). The number squared is called the base:
\$\$2^2= 2 × 2 = 4\$\$
\$\$3^2 = 3 × 3 = 9\$\$
\$\$4^2 = 4 × 4 = 16\$\$
We can read this as two/three/four to the second power.
A cube number is a number multiplied by itself 3 times:
\$\$2^3 = 2 × 2 × 2 = 8\$\$
\$\$3^3 = 3 × 3 × 3 = 27\$\$
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Factoring Trinomials (coefficient of squared term is not 1)
FACTORING TRINOMIALS OF THE FORM $\,ax^2 + bx + c\,$
Here, you will practice factoring trinomials of the form $\,ax^2 + bx + c\,$, where $\,a\,$, $\,b\,$, and $\,c\,$ are integers.
In this exercise, the coefficient of the $\,x^2\,$ term ($\,a\,$) is usually not going to be the number $\,1\,$,
which makes factoring more challenging.
For example, you might need to factor $\,6x^2-13x-5\,$.
Two methods are presented:
• the Guess-and-Check (or Trial-and-Error) method
• the more efficient Factor By Grouping method
The Guess-and-Check Method for Factoring Trinomials
One common method for factoring when $\,a\ne 1\,$ is called Guess-and-Check or Trial-and-Error.
In this method, we try to find four integers that work : $$6x^2 -13x - 5 = (\text{__ }x + \text{__ })(\text{__ }x + \text{__ })$$ The unknown numbers in front of the $\,x\,$ variables must multiply together to give $\,6\,$.
What are the possibilities here? Well, $\,2\cdot 3\,$ or $\,6\cdot 1\,$.
The unknown constants must multiply together to give $\,-5\,$.
What are the possibilities for this one? Well, $\,1\cdot -5\,$ or $\,-1\cdot 5\,$.
Phew. Not too many choices. But even so, this gives lots of possible combinations:
$(2x+1)(3x-5)$ $(6x+1)(x-5)$ $(2x-5)(3x+1)$ $(6x-5)(x+1)$ $(2x-1)(3x+5)$ $(6x-1)(x+5)$ $(2x+5)(3x-1)$ $(6x+5)(x-1)$
Then, the inners and the outers have to work out right .
If you're not the particularly lucky type,
then it might take you a while to stumble on the one that works. (Did you find it?)
And, for a random trinomial pulled out of the air, it's possible that nothing will work!
So, you could spend a long time looking, and find yourself wondering:
“Do they exist? Have I just missed them?”
There must be a better way—and there is.
It's called the ‘Factor by Grouping’ method,
and it's usually much more efficient than trial-and-error.
Example: the ‘FACTOR BY GROUPING’ Method
Here's an example of the ‘factor by grouping’ method.
Details follow, for inquiring minds.
However, some of you may want to just take this example, and run with it.
Factor: $\,6x^2 - 13x - 5\,$
Use the ‘factor by grouping’ method.
Solution:
• The number in front of $\,x^2\,$ isn't $\,1\,$. Instead, it's a $\,6\,$.
So, we must modify the old method, which said:
“find two numbers that multiply to the constant term, and add to the middle term”
Somehow, we're going to have to use that $\,6\,$.
• Here's how.
Take the $\,6\,$ and multiply it by the constant term, $\,-5\,$.
You get $\,(6)(-5) = -30\,$.
• Now, find two numbers that multiply to $\,-30\,$ and that (still) add to $\,-13\,$.
• This should be a very familiar thought process—you've likely practiced it a lot.
Use the PANS Memory Device, if you want.
• Come up with the two numbers that work—(thinking, thinking)—how about $\,-15\,$ and $\,2\,$.
Check them:
Do they multiply to $\,-30\,$? $\,(-15)(2) = -30\,$ Check!
Do they add to $\,-13\,$? $\,-15 + 2 = -13\,$ Check!
• Now, rename the original trinomial: $6x^2 - 13x - 5$ (original trinomial) $= 6x^2\ \overset{=\ -13x}{\overbrace{- 15x + 2x}} - 5$ (rename the middle term, using the two numbers you found) $= (6x^2 - 15x) + (2x - 5)$ (group the first two terms, and the last two terms—hence the title of this technique!) $= 3x(2x - 5) + 1(2x - 5)$ (factor the first part; put the $\,1\,$ in for clarity in the second part) $= 3x\overbrace{(2x-5)} + 1\overbrace{(2x-5)}$ (Look carefully! There are two ‘big’ terms, with a common factor of $\,2x-5\,$) $= (2x-5)(3x + 1)$ (factor out the common factor—done!)
It may seem like a lot of work to you.
But, over the years, I've divided my classes in half:
told one side to use trial-and-error, and the other side to use factor by grouping.
The factor by grouping side always wins.
Let's do the same example one more time.
This will show you that the order you write the two middle terms doesn't matter.
Also, this version is much more compact, and is usually all you'll need to write down.
Factor: $\,6x^2 - 13x - 5\,$
Use the ‘factor by grouping’ method.
Solution:
• $(6)(-5) = -30$
Find two numbers that multiply to $\,-30\,$ and add to $\,-13\,$.
• Try $\,-15\,$ and $\,2\,$:
$(-15)(2) = -30$ Check!
$-15 + 2 = -13$ Check!
• Rename: $6x^2 - 13x - 5$ (original trinomial) $= 6x^2 \ \overset{\text{different order}}{\overbrace{+ 2x -15x}} - 5$ (rename the middle term) $= (6x^2 + 2x) + (-15x - 5)$ (group first two, last two) $= 2x(3x + 1) + (-5)(3x + 1)$ (factor each group separately) $= (3x + 1)(2x - 5)$ (factor again)
MOTIVATION FOR THE ‘FACTOR BY GROUPING’ Method
Here's the motivation for the technique.
Start by pulling a random problem out of the air, and FOILing it out: $$(3x - 1)(5x + 7) = 15x^2 + 21x - 5x - 7$$ Don't combine the middle terms.
Instead, look at the four numbers generated in the resulting sum: $\,15\,$, $\,21\,$, $\,-5\,$, $\,-7$
The product of the first and last is: $\,(15)(-7) = -105$
The product of the middle two is: $\,(21)(-5) = -105$
Is this just a coincidence?
It's not a coincidence.
It's always true:
$$(ex + f)(gx + h) = (eg)x^2 + (eh)x + (fg)x + (fh)$$ The product of the first and last is: $\,(eg)(fh) = efgh\,$
The product of the middle two is: $\,(eh)(fg) = efgh\,$
Same result.
Other than being a curious observation, what good is this?
Ends up that it's a LOT of good.
If we're trying to factor a trinomial by grouping,
and we're looking for the right way to rename the middle term, then we know two things: $$ax^2 + bx + c = ax^2 + \text{__}\ x + \text{__}\ x + c$$
• the two unknown numbers must add to $\,b\,$
• the two unknown numbers must multiply to $\,ac\,$
This is why we take the number in front of the $\,x^2\,$ term, and multiply it by the constant term!
The rest of the argument is more advanced, and is primarily included for the sake of the teacher.
Master the ideas from this section
When you're done practicing, move on to:
Solving More Complicated Quadratic Equations by Factoring
For more advanced students, a graph is displayed for each factoring problem.
For example, suppose you're asked to factor $\,6x^2 - 13x - 5\,$.
Then, you'd see the graph of the equation $\,y = 6x^2 - 13x - 5\,$.
(This graph is shown below when the web page is first loaded.)
Observe that the graph crosses the $\,x\,$-axis at $\,-\frac13\,$ and $\,\frac52\,$.
Look at the factorization:
$6x^2 - 13x - 5 = (3x + 1)(2x - 5) = 3(x + \frac 13)2(x - \frac52) = 6(x+\frac13)(x-\frac52)\,$
See the relationship between the factors and the $x$-axis intercepts (zeros)?
If $\,c\,$ is a zero, then $\,x - c\,$ is a factor!
(You're discovering the beautiful relationship between the zeroes of a polynomial, and its factors.)
Click the “show/hide graph” button if you prefer not to see the graph.
Factor:
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# What is the general solution of the differential equation (y^2+1)dy/dx+2xy^2=2x ?
Sep 19, 2017
$\ln | \frac{y + 1}{y - 1} | - y = {x}^{2} + C$
#### Explanation:
We have:
$\left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{2} = 2 x$ ..... [A]
We can rearrange this non-linear First Order differential equation [A] as follows:
$\left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 2 x {y}^{2}$
$\therefore \left({y}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x \left({y}^{2} - 1\right)$
$\therefore \frac{{y}^{2} + 1}{{y}^{2} - 1} \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$
This is now separable, so we can "seperate the variables" to get:
$\int \setminus \frac{{y}^{2} + 1}{{y}^{2} - 1} \setminus \mathrm{dy} = \int \setminus - 2 x \setminus \mathrm{dx}$ ..... [B]
The RHS integral is standard, and the LHS will require a little manipulation, as follows:
$\int \setminus \frac{{t}^{2} + 1}{{t}^{2} - 1} \setminus \mathrm{dt} = \int \setminus \frac{{t}^{2} - 1 + 2}{{t}^{2} - 1} \setminus \mathrm{dt}$
$\text{ } = \int \setminus 1 + \frac{2}{{t}^{2} - 1} \setminus \mathrm{dt}$
$\text{ } = \int \setminus 1 + \frac{2}{\left(t + 1\right) \left(t - 1\right)} \setminus \mathrm{dt}$
We can now decompose the fractional part of the integrand into partial fractions, as follows:
$\frac{2}{\left(t + 1\right) \left(t - 1\right)} \equiv \frac{A}{t + 1} + \frac{B}{t - 1}$
$\text{ } = \frac{A \left(t - 1\right) + B \left(t + 1\right)}{\left(t + 1\right) \left(t - 1\right)}$
$2 \equiv A \left(t - 1\right) + B \left(t + 1\right)$
Where $A , B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:
Put $t = - 1 \implies 2 = - 2 A \implies A = - 1$
Put $t = + 1 \implies 2 = + 2 B \implies B = + 1$
So using partial fraction decomposition we have:
$\int \setminus \frac{{t}^{2} + 1}{{t}^{2} - 1} \setminus \mathrm{dt} = \int \setminus 1 - \frac{1}{t + 1} + \frac{1}{t - 1} \setminus \mathrm{dt}$
Using this result we can now integrate [B] as follows:
$\int \setminus \frac{{y}^{2} + 1}{{y}^{2} - 1} \setminus \mathrm{dy} = \int \setminus - 2 x \setminus \mathrm{dx}$
$\therefore \int \setminus - 1 + \frac{1}{y + 1} - \frac{1}{y - 1} \setminus \mathrm{dy} = \int \setminus 2 x \setminus \mathrm{dx}$
$\therefore - y + \ln | y + 1 | - \ln | y - 1 | = {x}^{2} + C$
$\therefore \ln | \frac{y + 1}{y - 1} | - y = {x}^{2} + C$
Which, is the General Solution .
We are unable to find a particular solution, as requested, as noi initial conditions have been provided to allow the constant $C$ to be evaluated.
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As a herpetologist in Madagascar, I’ve spent 17 years researching wild amphibians and reptiles in my home country. During this time, I’ve witnessed the habitats of these unique animals shrink due to pressure from forest clearing, bushfire, slash-and-burn farming, mining, oil exploration and road construction.
More than 99 percent of Madagascar’s amphibians are found nowhere else on Earth, and according to the International Union for Conservation of Nature (IUCN), one-quarter of these species are classified as threatened with extinction.
Fortunately, we now have a critical resource to help the frogs fight back: a new amphibian captive breeding center.
Implemented by Malagasy authorities, IUCN’s Amphibian Specialist Group and the Mitsinjo Association — with support from CI and other NGOs — the Mitsinjo captive breeding facility was constructed in response to the growing threat of the chytrid fungus that has decimated amphibian populations worldwide. Although chytrid has not yet been detected in Madagascar, seven of the country’s amphibian species are already designated as Critically Endangered, and therefore at high risk of extinction if disease outbreaks should occur. The amphibian center will establish captive populations of the most threatened species as a reserve in case the fungus reaches the island.
Amphibians provide many important services to humans, such as controlling insects that spread disease and damage crops, and helping to maintain healthy freshwater systems. In 2008, CI-Madagascar organized the development of the Sahonagasy Action Plan (SAP), a national plan for amphibian conservation. This plan emphasized the emerging threat posed by the chytrid fungus and the need to develop the capacity within Madagascar to detect and monitor the disease, and to develop in-country breeding facilities for disease-free frog populations. Captive breeding will also help to combat the combined action of habitat destruction, illegal and unsustainable collection for the international pet trade, and the impacts of climate change.
The facility currently houses about 33 frogs representing six species from the Andasibe region. Until now, no one in Madagascar had the knowledge or capacity to breed these frogs in captivity. As a result, we will first focus on breeding common species that have similar habits and habitats to threatened species as we build our husbandry skills. Once we can master captive-breeding techniques, we will deal with the more threatened species. This captive breeding program also provides an opportunity to gather information on the life history of these frogs.
There are many challenges to this kind of work. Besides the strict hygiene standards and the risk of disease transmission between the frogs, feeding the frogs is an especially difficult skill to learn. Live food is critical for the frogs’ survival, but it can be difficult to determine the precise quantity and nutritional balance that the animals need. This skill is, of course, crucial for the success of the center. Our captive breeding specialist has so far trained six technicians on caring for live frogs.
We are still in the early stages of this project; eventually, we plan to develop educational programs that will showcase the value of Madagascar’s frogs and their habitats to local people, and generate money through ecotourism.
In the coming year, we hope to increase the number of species bred at the facility — bringing us a step closer to safeguarding the future of these fascinating creatures.
Nirhy Rabibisoa is amphibian executive secretary at CI-Madagascar.
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|Carolyn Vachani, MSN, RN, AOCN and Charles B. Simone II, MD|
|The Abramson Cancer Center of the University of Pennsylvania|
| Last Modified: May 20, 2013
What is mesothelial tissue?
The mesothelium is a protective sac that covers and protects most internal organs in the body. It is composed of two layers, one layer covers the organ and the second layer forms a sac around it. The mesothelium produces a lubricating fluid that is released between these layers, allowing moving organs (such as the lungs) to move easily. The area between the two layers is often called the pleural space. Mesothelial tissue is found lining the abdominal cavity organs, lungs, testes and heart.
What is mesothelioma?
Mesothelioma occurs when the mesothelial cells grow out of control. These cells also lose the ability to stop producing the lubricating fluid when there is enough. This results in the unwanted encasement of organs within a thick rind of tumor tissue and excess fluid buildup, ultimately causing symptoms. These cells can grow and invade other organs, or spread to other areas of the body. When the cells spread to other areas of the body, it is called metastasis.
The majority of mesotheliomas are found in the lining of the lung (approximately 70%). About 20-30% percent of cases are found in the abdominal cavity lining, and even more rarely, mesothelioma is found in the lining of the heart (1-2%) or testicles.
Am I at risk for mesothelioma?
Mesothelioma is a rare cancer, with approximately 2, 500 cases diagnosed each year in the United States. It is five times more common in men, which is due in most part to work-related exposure to asbestos. Risk also increases with age. The biggest risk factor for developing the disease is exposure to asbestos, accounting for 50 to 80 percent of all cases. Asbestos has been used in many products, including cement, brake linings, roof shingles, flooring products, textiles, and insulation. Particles can be released from these products, particularly during the manufacturing process, and inhaled. Prior to knowing the dangers, asbestos miners and other workers exposed to asbestos worked without wearing any protection. Since the 1970's, the U.S. Occupational Safety and Health Administration (OSHA) has set limits for acceptable levels of asbestos exposure and requires protective equipment in the workplace. Family members of people who worked with asbestos were also exposed to the toxin when it was carried home on clothing and hair, putting them at increased risk for mesothelioma.
While up to 80% of cases of pleural mesothelioma occur in individuals who have had known asbestos exposure, only 10% of people with a history of heavy exposure develop the disease. This suggests that additional exposures or factors are involved to actually develop the disease. Even more puzzling is the fact that only 50% of people with peritoneal mesothelioma have a history of asbestos exposure.
It takes 20 to 50 years from the time of asbestos exposure until mesothelioma is detected. Although most individuals who develop mesothelioma had a repeated exposure to asbestos over many years or longer, some develop the cancer with as little as one or two months of asbestos exposure. The incidence of mesothelioma varies in different areas of the world, depending on when asbestos was widely used in that area. Rates are higher in the United Kingdom, where about 1800 cases per year are diagnosed. The incidence takes into consideration the number of cases and the size of the population. This is because asbestos use in Western Europe remained high until 1980, whereas maximum exposure in the U.S. was from the 1930s to 1960s. Australia was one of the world's largest producers of asbestos, leading to the country having the highest rates worldwide (based on the size of the population). Rates in the U.S. are beginning to decline, whereas rates in Europe and Australia are expected to plateau in the next 10-15 years before declining.
Following the ban of asbestos in many countries, asbestos producers started to promote the sale of their product to developing countries, such as Asia and Latin America. Experts fear that the peak rates in these areas are yet to come, and will mimic what has been seen in the U.S. and Europe. As asbestos is widely available, even today, in China and India, rates in those countries are projected to increase significantly over the next two decades.
Other risk factors include prior radiation therapy to the chest, exposure to erionite fibers (mineral in gravel roads), collapsotherapy (for the treatment of tuberculosis), and potentially even exposure to the DNA tumor simian virus SV40. Smoking does not seem to increase the risk of developing the disease.
How can I prevent mesothelioma?
By decreasing exposure to asbestos, the risk of mesothelioma is decreased. Workers who are exposed to asbestos on the job should wear protective clothing and masks. These workers should change their clothing before leaving the work site to avoid carrying any particles home. OSHA has set standards regulating these procedures.
What screening tests are available?
There is no good screening test for mesothelioma. Radiologic studies (x-ray, CT scan) are not sensitive enough to detect tumors before symptoms occur. Currently, researchers are investigating the use of several blood markers in an attempt to detect mesothelioma earlier than conventional methods for people with a history of asbestos exposure. These tests include measurements of serum osteoporin, levels of the soluble mesothelin-related protein, and the Mesomark assay that evaluates the levels of soluble mesothelin-related proteins released by mesothelioma cells. Presently, however, all of these tests are considered experimental.
What are the signs of mesothelioma?
The symptoms of mesothelioma are often caused by a build-up of tumor tissue surrounding the lung and accumulation of fluid in the pleural space that prevents the lung from expanding fully. This causes pressure on the lung, leading to pain and shortness of breath. As the disease progresses, patients may lose weight and have a dry, hacking cough. Mesothelioma originating in the pleura can directly spread into the abdomen, or mesothelioma can originate in the peritoneum of the abdomen. Symptoms most commonly associated with abdominal disease includes abdominal swelling, pain and weight loss.
How is mesothelioma diagnosed and staged?
Patients who present with symptoms worrisome for mesothelioma often initially have a chest x-ray done that shows a build-up of fluid in the lining of the lung. These patients then undergo a CT scan of the chest to further evaluate the cancer. In the case of abdominal mesothelioma, a CT scan is obtained to visualize the anatomy in the abdomen.
The fluid in the lining of the lung is often drained to improve symptoms. A diagnosis of mesothelioma can sometimes be made by looking at the cells of this fluid under the microscope. If a diagnosis is not possible with the fluid alone, patients would then undergo a biopsy to have the diagnosis confirmed. In the lung, a thoracoscope is used to go through the chest wall, between the ribs, to obtain a sample of the tissue. A peritoneoscope or abdominal laparoscopy are used to enter the abdomen to obtain a tissue sample to diagnose abdominal mesothelioma.
Staging refers to determining the extent of the disease, and the stage dictates the treatment. Physicians use the TNM system (also called tumor - node - metastasis system). This describes the size and locally invasiveness of the tumor (T), which if any lymph nodes are involved (N), and if it has spread to other more distant areas of the body (M). This is then interpreted as a stage somewhere from I (one) denoting more limited disease to IV (four) denoting more advanced disease. Patients with earlier stage tumors tend to live longer and respond better to available treatments.
What are the treatments for mesothelioma?
Treatment is dependent on the stage of the disease, the location of the tumor, the patient's age, and his or her state of health and performance status. Younger, healthy patients with early-stage malignant pleural mesothelioma may be candidates for surgery that removes the mesothelial tissue around the tumor. Although surgery to either remove the entire lung with the tumor (termed extrapleural pneumonectomy) or lung-sparing surgery that removes only the tumor and the lining of the lung (termed extended or radical pleurectomy) is the most standard option for these patients, Similar comprehensive surgery is often considered in patients with peritoneal mesothelioma. All of these definitive thoracic or peritoneal surgeries are extensive, many patients are not able to undergo surgery due to having advanced disease at diagnosis or not being able to medically tolerate the surgery, and there currently is limited randomized data showing a survival benefit with extensive surgery compared with non-surgical treatments.
For patients that have surgery that removes the entire lung, postoperative radiation therapy is commonly given to treat that side of the chest to attempt to kill any remaining cancer cells not removed by surgery. Postoperative radiation therapy after lung-sparing surgery, however, is not routinely administered since the radiation damage to the healthy lung tissue in the process of treating any remaining cancer cells may result in toxicity that outweighs any benefit of radiotherapy. Radiation therapy is also commonly delivered to surgical incision sites to prevent the cancer from recurring locally at those sites. In patients who do not undergo surgery, radiation therapy is generally only given, as needed, to treat problem spots to relieve symptoms like pain or trouble breathing.
Chemotherapy is the most standard treatment for mesothelioma and can provide significant relief of symptoms and also improve surgery. Agents that are used, either alone or in combination, include cisplatin, carboplatin, pemetrexed, gemcitabine, and vinorelbine. These medications achieve responses in 10 to 30 percent of patients.
In a key trial that randomized patients to receive either cisplatin alone or cisplatin in combination with pemetrexed (Alimta), patients who received the combination of drugs had increased response rates, survived longer, and had fewer side effects. In addition, researchers found that giving folic acid and vitamin B12 along with the combination resulted in less toxicity and no decrease in the therapy's effectiveness. The combination of cisplatin (can be substituted with carboplatin if clinically appropriate) and pemetrexed is now considered standard of care for mesothelioma and can be given alone or before or after surgery.
Because the current therapies have limited effectiveness, researchers are continuing to look for new ways to treat mesothelioma. Some of the treatments being investigated include immunotherapy, gene therapy (a method that attempts to correct the abnormal gene that causes the cancer to grow out of control), chemotherapy administered directly into the pleural space, and intraoperative Photodynamic Therapy (PDT- a treatment that uses a laser to activate a photosensitizing drug during the surgical removal of the cancer in an attempt to kill any remaining cancer cells not removed by surgery). Patients should talk with their physicians about current clinical trials for mesothelioma.
One problem that patients may encounter is the recurring build-up of fluid in the pleural space. This fluid can be removed with a chest tube (a tube that is put into the chest wall and left in for a period of time to allow drainage) or a procedure called thoracentesis (a small needle is put through the chest wall into the pleural space, the fluid is drained, and the needle is removed). In many cases, this will be followed by a procedure called pleurodesis, in which a medication (talc, bleomycin) is injected into the lung to create scar tissue in the hopes of decreasing future fluid from developing. In the abdomen, the procedure to remove fluid is called paracentesis. In this procedure, a needle is inserted through the abdomen into the fluid filled space, and the fluid is drained. If this is a chronic problem, patients may have a catheter placed semi-permanently, allowing them to drain the fluid themselves at home as needed. Removal of the fluid alleviates the difficulty in breathing and the pain that is caused by the fluid build-up.
The physician will follow the patient with physical examinations and surveillance imaging.
References and Further Reading
Bridda, A. et al, Peritoneal Mesothelioma: A Review. Medscape General Medicine, 9(2), 32 (2007).
DeVita, V., Hellman, S., & Rosenberg, S. Cancer: Principles and Practice of Oncology, Seventh Edition (2004) . Lippincott, Williams & Wilkins, Philadelphia, Pennsylvania.
Light, R. Pleural Diseases, Fourth Edition (2001) . Lippincott, Williams & Wilkins, Philadelphia, Pennsylvania.
Stahel, RA, Malignant Pleural Mesothelioma: A New Standard of Care. Lung Cancer, 54s: s9-s14 (2006).
Vogelzang, N et al., Phase III Study of Pemetrexed in Combination With Cisplatin Versus Cisplatin Alone in Patients With Malignant Pleural Mesothelioma, Journal of Clinical Oncology, 21: 2636-2644 (2003)
National Cancer Institute. Mesothelioma: Questions & Answers
The Mesothelioma Applied Research Foundation: This nonprofit organization's website is a great resource for patients, families & healthcare professionals.
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Linear Momentum Formula
# Linear Momentum Formula
Linear momentum is a fundamental concept in physics that describes the motion of an object in terms of its mass and velocity. It is often referred to as “momentum” and is represented by the symbol “p”. Linear momentum is a vector quantity, meaning it has both magnitude and direction.
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The linear momentum of an object is defined as the product of its mass (m) and velocity (v):
p = m x v
Where:
p is the linear momentum of the object,
m is the mass of the object, and
v is the velocity of the object.
## Key Points about Linear Momentum
• Conservation of Linear Momentum: According to the principle of conservation of linear momentum, the total momentum of a system remains constant if no external forces act on it. This principle is derived from Newton’s third law of motion, stating that for every action, there is an equal and opposite reaction. This conservation law has various applications, from understanding collisions to analysing rocket propulsion.
• Impulse: Impulse is the change in momentum of an object. It is equal to the force applied to an object multiplied by the time interval over which the force acts. Mathematically, impulse (J) is given by the equation J = F x Δt, where F is the applied force and Δt is the time interval.
• Momentum and Newton’s Laws: Linear momentum plays a crucial role in Newton’s laws of motion. Newton’s second law states that the rate of change of momentum of an object is equal to the net force applied to it. It can be expressed as F = Δp / Δt, where F is the net force, Δp is the change in momentum, and Δt is the time interval.
### What is the SI unit of Linear momentum
The SI unit of momentum is kilogram-metre per second (kg·m/s). However, momentum can also be expressed in other units, such as gram-centimetre per second (g·cm/s) or newton-second (N·s), depending on the context.
Now, when a net force is applied, the velocity of the object changes and when velocity changes, its momentum also changes.
So, net force is nothing but the rate of change of momentum.
Therefore,
Net force = (p2 – p1)/t
where, p1 = initial momentum, p2 = final momentum and t = time taken
The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
### Solved Examples on Linear Momentum Formula
Example 1: A car of mass 1000 kg is initially at rest. It accelerates uniformly and reaches a velocity of 20 m/s in 10 seconds. What is the linear momentum of the car at this point?
Solution:
Given:
Mass of the car (m) = 1000 kg
Initial velocity (u) = 0 m/s
Final velocity (v) = 20 m/s
Time (t) = 10 s
Using the formula for linear momentum: p = m x v
Substituting the given values:
p = 1000 kg x 20 m/s
p = 20,000 kg·m/s
Therefore, the linear momentum of the car at this point is 20,000 kg·m/s.
Example 2: A bullet with a mass of 0.02 kg is fired from a rifle with a velocity of 400 m/s. What is the linear momentum of the bullet?
Solution:
Given:
Mass of the bullet (m) = 0.02 kg
Velocity of the bullet (v) = 400 m/s
Using the formula for linear momentum: p = m x v
Substituting the given values:
p = 0.02 kg x 400 m/s
p = 8 kg·m/s
Therefore, the linear momentum of the bullet is 8 kg·m/s.
Example 3: Two objects with masses 2 kg and 3 kg, respectively, are initially at rest. They collide and stick together. What is their final velocity after the collision?
Solution:
Given:
Mass of object 1 (m1) = 2 kg
Mass of object 2 (m2) = 3 kg
Initial velocity of both objects (u1 and u2) = 0 m/s
Since the objects stick together after the collision, their final velocity is the same. Let’s call it v.
Using the principle of conservation of linear momentum:
(m1 u1) + (m2 u2) = (m1 + m2) x v
Substituting the given values:
(2 kg x 0 m/s) + (3 kg x 0 m/s) = (2 kg + 3 kg) x v
0 = 5 kg x v
Since the left side of the equation is zero, the final velocity (v) is also zero.
Therefore, the final velocity of the objects after the collision is 0 m/s.
## Frequently Asked Questions on Linear Momentum Formula
### What is linear momentum?
Linear momentum, often referred to as momentum, is a fundamental concept in physics that describes the motion of an object in terms of its mass and velocity. It is the product of an object's mass and its velocity.
### What is the formula for linear momentum?
The formula for linear momentum is: p = m x v Where p is the linear momentum, m is the mass of the object, and v is its velocity.
### What are the units of linear momentum?
The units of linear momentum are kilogram-metre per second (kg·m/s) in the International System of Units (SI).
### What is the principle of conservation of linear momentum?
The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant if no external forces act on it. In other words, the total momentum before an event or interaction is equal to the total momentum after the event or interaction, as long as no external forces are present.
### How is linear momentum related to force?
Linear momentum is related to force through Newton's second law of motion. According to the law, the rate of change of linear momentum of an object is equal to the net force acting on it. Mathematically, it can be expressed as F = Δp/Δt, where F is the net force, Δp is the change in momentum, and Δt is the change in time.
### What is the difference between linear momentum and kinetic energy?
Linear momentum describes an object's motion in terms of its mass and velocity, while kinetic energy represents the energy of an object due to its motion. Momentum is a vector quantity, while kinetic energy is a scalar quantity.
### Can the linear momentum of an object be negative?
Yes, the linear momentum of an object can be negative. The sign of momentum indicates the direction of motion. Positive momentum indicates motion in one direction, while negative momentum indicates motion in the opposite direction.
## Related content
Important Topic of Physics: Reynolds Number Distance Speed Time Formula Refractive Index Formula Mass Formula Electric Current Formula Electric Power Formula Resistivity Formula Weight Formula CBSE Syllabus for Class 12 Physics Non Contact Force
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# Use the table of integrals to find the following integral. \\ \int \frac{ (\sec^2 \theta \tan^2...
## Question:
Use the table of integrals to find the following integral.
{eq}\int \frac{ (\sec^2 \theta \tan^2 \theta)}{\sqrt {4-\tan^2 \theta}}d\theta{/eq}
## Integration by Substitution:
There are two types of integration substitution: substitution of a function and substitution of a trigonometric function.
Substitution of a function: It can use as a u-substitution or v-substitution.
Substitution of a trigonometric function: It can use the trigonometric function for substitution.
For this solution, we use both of the substitution methods.
So, we require the following historical context:
1. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\tan \left(x\right)\right)=\sec ^2\left(x\right). {/eq}
2. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\sin \left(x\right)\right)=\cos \left(x\right). {/eq}
3. Take the constant out: {eq}\displaystyle \int c\cdot f\left(v\right)dv=c\cdot \int f\left(v\right)dv. {/eq}
4. Trigonometric identity: {eq}\displaystyle \sin ^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2}. {/eq}
5. The sum rule: {eq}\displaystyle \int f\left(v\right)\pm g\left(v\right)dv=\int f\left(v\right)dv\pm \int g\left(v\right)dv. {/eq}
6. Common integration: {eq}\displaystyle \int \cos \left(w\right)dw=\sin \left(w\right). {/eq}
7. Integration of a constant: {eq}\displaystyle \int adx=ax. {/eq}
We have to solve the integration of $$\displaystyle I = \int \frac{ (\sec^2 \theta \tan^2 \theta)}{\sqrt {4-\tan^2 \theta}}d\theta$$
Apply the substitution for {eq}u=\tan \left(\theta\right) \Rightarrow du = \sec ^2\left(\theta\right) d\theta. {/eq}
$$\displaystyle = \int \frac{u^2}{\sqrt{4-u^2}}du$$
Apply the trigonometric substitution for {eq}u=2\sin \left(v\right) \Rightarrow du = 2 \cos \left(v\right) dv. {/eq}
$$\displaystyle = \int 4\sin ^2\left(v\right)dv$$
Take the constant out.
$$\displaystyle = 4\cdot \int \sin ^2\left(v\right)dv$$
Use the trigonometric identity.
$$\displaystyle = 4\cdot \int \frac{1-\cos \left(2v\right)}{2}dv$$
Take the constant out.
$$\displaystyle = 4\cdot \frac{1}{2}\cdot \int 1-\cos \left(2v\right)dv$$
Apply the sum rule.
$$\displaystyle = 4\cdot \frac{1}{2}\left(\int 1dv-\int \cos \left(2v\right)dv\right)$$
Use the common integration.
$$\displaystyle = 4\cdot \frac{1}{2}\left(v-\frac{1}{2}\sin \left(2v\right)\right)+C$$
Substitute back {eq}v=\arcsin \left(\frac{1}{2}u\right). {/eq}
$$\displaystyle = 4\cdot \frac{1}{2}\left(\arcsin \left(\frac{1}{2}u\right)-\frac{1}{2}\sin \left(2\arcsin \left(\frac{1}{2}u\right)\right)\right)+C$$
Substitute back {eq}u=\tan \left(\theta\right). {/eq}
$$\displaystyle = 4\cdot \frac{1}{2}\left(\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)-\frac{1}{2}\sin \left(2\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)\right)\right)+C$$
Simplify:
$$\displaystyle = 2\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)-\sin \left(2\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)\right)+C$$
Where C is constant of the integration.
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|The Marconi Wireless Telegraph Station at Kahuku is significant at the state level under criterion A for its associations with the development of wireless communications in Hawaii and the role it played in providing the islands with worldwide wireless telegraphic, and later telephonic, communication. Until 1840 any immediate communication between human beings was limited to the range of the eye or the ear. In nations such as France, Russia, and Great Britain, fire signal towers stretched the length of the country to serve as early warning systems. During the nineteenth century scientists and inventors came to better understand electricity's ability to transmit sound, and with this understanding came such inventions as the telegraph by Samuel Morse in 1840, the telephone by Alexander Graham Bell in 1875, and the phonograph by Thomas Alva Edison in 1877. In addition to these new wonders came such scientific advances as James Clerk Maxwell's 1865 theory, which postulated electromagnetic waves existed and moved at a uniform speed, but varied in length and frequency, and Heinrich Hertz's 1888 proof of this theory by demonstrating that electricity could bridge a gap from one coil to produce a current in another. These all laid the groundwork for humanity's delving into the possibility of wireless communication.
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How does a lid affect what happens to the liquid in the cup?
- Do Now– Get out your ISN and add to the Table of Contents- “Initial Model” p.32 (Left Side). Head p.32 properly with “Initial Model- 11/6/19.″
- Investigation– Cup Lid 2- Develop a new plan by recording variables and procedures on “Procedure: Measuring Changes in Mass in the Cups” handout. Make predictions by completing the first row in the data table on the handout- “Do you think the amount of matter in this system will change over 5 minutes’ time? Review how & where to record data. Review science safety precautions. Carry out Cup Lid Lab 2. Pool the class data by calculating averages and recording them on the data table in the handout. Analyze the data & make claims by completing the “Making Sense” section of the handout: “Which system lost mass? Matter has mass. So what matter may have left the system(s) that lost mass?” Have a Making Sense Discussion. Record the definitions of Open System and Closed System on ISN p.33. Tape the handout as a flip page to ISN p.33.
- Initial Model– Use the handout “Initial Model of Mass Loss in the Cup with No Lid” to develop an individual model that would answer the question: What is happening to the liquid water that can explain why the mass of the cup system decreases over time? Read the directions together. Use the space to develop your model and answer the prediction question.
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Study P4 Mathematics Maths - Division - Geniebook
# Division
In Mathematics, there are four basic arithmetic operations, i.e. addition, subtraction, multiplication and division.
Division means to distribute a value/number/amount equally. Its result may or may not be a whole number. It can also be in the form of a decimal.
1. Division up to 4-digit by 1-digit numbers
## Long Division
Division can be done by the long division method.
Example:
Do the following division.
$$965 \div 8 =$$ __________
1. What is the quotient?
Solution:
120
1. What is the remainder?
5
## 1. Division Of 4-Digit Number By 1-Digit Number
Question 1:
Do the following division.
$$4824 \div 2 =$$ __________
Solution:
2412
Question 2:
Perform the following division.
$$6009 \div 3 =$$ __________
Solution:
2003
Question 3:
Perform the following division.
$$8012 \div 4 =$$ __________
Solution:
2003
Question 4:
Perform the following division.
$$5525 \div 5 =$$ _________
Solution:
1105
Question 5:
Perform the following division.
$$5649 \div 7 =$$ __________
Solution:
807
Question 6:
Perform the following division.
$$2989 \div 4 =$$ __________ $$\text{R}$$ __________
Solution:
$$2989 \div 4 = 747 \text{ R } 1$$
747 R 1
Question 7:
Perform the following division.
$$7268 \div 8 =$$ __________ $$\text{R}$$ __________
Solution:
$$7268 \div 8 = 908 \text{ R } 4$$
908 R 4
Question 8:
Perform the following division.
$$6572 \div 8 =$$ __________ $$\text{R}$$ __________
Solution:
$$6572 \div 8 = 821 \text{ R } 4$$
821 R 4
Question 9:
Solve the following
__________ $$÷ \;8 = 183 \text{ R } 5$$
Solution:
There are 183 groups of 8 and 5 left.
\begin{align} 183 \times 8 &= 1464\\ 1464 + 5 &= 1469 \end{align}
1469
Question 10:
Solve the following.
_________ $$÷ \;6 = 214 \text{ R } 3$$
Solution:
There are 214 groups of 6 and 3 left.
\begin{align} 214 \times 6 &= 1284\\ 1284 + 3 &= 1287 \end{align}
1287
Question 11:
When a number is divided by 4, it gives a quotient of 121 and the remainder is 1.
What is the number?
Solution:
Rewriting the question into an equation:
__________ $$\div \;4 = 121 \text{ R } 1$$
\begin{align} 121 \times 4 &= 484\\ 484 + 1 &= 485 \end{align}
485
Question 12:
When a number is divided by 7, the quotient is 154 and the remainder is 4. What is the remainder when the same number is divided by 6?
Solution:
Rewriting the question into equations:
\begin{align} \text{_ _ _ _ _ _ _ _ _ _ } \div \;7 &= 154 \text{ R } 4 \\[2ex] \text{_ _ _ _ _ _ _ _ _ _ } \div \;6 &= \text{_ _ _ _ _ _ _ _ _ _ } \text{ R } \text{_ _ _ _ _ _ _ _ _ _ } \end{align}
Solving the first equation,
\begin{align} 154 \times 7 &= 1078\\ 1078 + 4 &= 1082 \end{align}
Now, solving the second equation,
$$1082 \div 6 = 180 \text{ R } 2$$
2
Question 13:
Solve the following:
Which of the following is the correct answer?
1. 64
2. 132
3. 512
4. 1024
Solution:
Looking at the first equation,
\begin{align}\\[2ex] &= 32 \div 8\\[2ex] &= 4 \end{align}
Looking at the second equation
$$4$$ $$\times$$ $$=$$ $$2048$$ $$=$$ $$2048 \div 4$$ $$=$$ $$512$$ $$+$$ $$=$$ $$512 + 512$$ $$=$$ $$1024$$
(4) 1024
Continue Learning
Multiplication Whole Numbers
Multiplication And Division Decimals
Model Drawing Strategy Division
Fractions Factors And Multiples
Area And Perimeter 1 Line Graphs
Time
Primary
Primary 1
Primary 2
Primary 3
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English
Maths
Multiplication
Whole Numbers
Line Graphs
Time
Multiplication And Division
Decimals
Model Drawing Strategy
Division
Fractions
Factors And Multiples
Area And Perimeter 1
Science
Primary 5
Primary 6
Secondary
Secondary 1
Secondary 2
Secondary 3
Secondary 4
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The stacks are remnants of the past
Marking where the sea cliff was,
But nothing here is made to last.
A cliff retreats; it always does.
from “Sea Stacks of Kerry,” Richard Hayes Phillips, 2001
Sea stack gardens? Aren’t sea stacks those rocks near the coast that stick up out of the ocean? How can they have gardens?
Sea stacks begin as part of a headland or sea cliff. Relentless pounding by ocean waves erodes the softer, weaker parts of the rock first, leaving harder, more resistant rock behind. Sometimes sea arches are formed where waves have hollowed out a line of weakness and created an arched opening; sea stacks can also form when the roof of a sea arch collapses. The Pacific Coast is rich with sea stacks, from California’s Big Sur to Oregon.1
In addition to present-day sea stacks where native plants predominate, there are the sea stacks of sixty thousand years ago (when the sea level was higher), some of which now rest in people’s gardens, more than a hundred feet inland from the current shoreline. A few of these ancient sea stacks have been incorporated into the structure of the house; outdoors, they provide home for both native and non-native plants. Having proven their fortitude by resisting the onslaught of waves that have eroded the coast around them, they now provide a secure and attractive setting for house and garden.
Today’s younger sea stacks at Trinidad, California, stand in the ocean like a family of sentinels guarding the coast. Some are large, some small, and some in between. Like a family, they have names. The largest is Pewetole Island, which looms over the others at a hundred and thirty feet tall, crowned with Sitka spruces (Picea sitchensis); how these trees maintain a foothold on this nearly barren rock is a feat of determined roots. The island, accessible only at low tide, is an adventure to climb, but it is not unknown for the Coast Guard to rescue climbers who have not paid attention to incoming tides.
In the littoral zone closer to shore is Grandmother Rock, named for its shape that is vaguely reminiscent of an imposing woman, some twenty feet tall, looking out to sea. To me, she looks like Queen Victoria wearing a voluminous skirt and surveying her realm. When the wildflowers are blooming in June, they cascade down her back like a colorful, regal pigtail— golden patches of stonecrop (Sedum spathulifolium) interspersed with clumps of the fringed, pale lavender seaside daisy (Erigeron glaucus). Some bare parts of the rock are covered with a rust-colored lichen; off in a corner will be a few isolated yellow monkey flowers (Mimulus guttatus). These flowers would make a wonderful golden crown, but ice plant (Carprobrotus edulis) has taken over Queen Victoria’s head.
Most of the rocks are home to native plants, but invasive plants, like the indefatigable ice plant and pampas grass (Cortaderia jubata), have managed to gain a foothold, too. Studies by Humboldt State University students in 2008 identified over sixty-seven different plants, of which seventy-five percent were native. The most common invasive species was annual bluegrass (Poa annua). Sea stacks that have been incorporated into gardens are usually home to a mixture of native and non-native plants.
Demonstrating Trinidad’s special seacoast character, the Trump family house and garden merge into the rocky coast as an integral part of the seascape. During heavy storms, ocean spray often beats against the downstairs windows. Developed from a small holiday cabin, the house now encompasses several levels. One room forms a bridge over a stream that cascades down the hill into the ocean, its banks covered in blue lupine (Lupinus littoralis). In another room, the rocks form an integral part of an interior wall. Sea stacks and rocky outcrops have also been worked into the design of the garden.
A long lane leads down from the road to an imposing sea stack, about twenty feet high, that blocks a view of the ocean. The house snuggles against one side of this sea stack. The stack, itself, is rich with native plants, including the native columbine (Aquilegia formosa), whose muted red and yellow colors can be difficult to see against the grey rock until backlit by the sun. Wild cucumber (Marah oreganus) clambers across the rock surfaces; its modest white flowers develop into fleshy gourds in late summer. Salal (Gaultheria shallon), barely six inches tall, and native ferns, such as five finger fern (Adiantum aleuticum), leather leaf fern (Polypodium scouleri), and western sword fern (Polystichum munitum), form a background for fringecups (Tellima grandifolia), with modest creamy blossoms, and a tall clump of dusky pink, slim-leafed onion (Allium dichlamydeum).
From the ocean side of this entry sea stack, one is swept away by the dramatic view of a rock-strewn ocean, nearly always with a few surfers in the distance. On the flanks of the sea stack and along its base is a bed planted as a traditional perennial garden. This garden is alive with color, from patches of variously colored California poppies (Eschscholzia californica), delphiniums, geraniums, hydrangea, sea pink (Armeria maritima), and more. Beautiful as they are, the exotic garden plants are outnumbered and, some may think, outshone by their more numerous native kin, which seldom flaunt themselves but are seductive in their low-key beauty.
Steps lead down from the perennial garden and across the lawn to a thriving vegetable garden. Close to the cliff’s edge, this garden is protected from the ocean by a low hedge of Escallonia rubra. The richness of the plant life so close to the ocean is hard to believe; even tomatoes have been successfully grown here and watered partly with seawater supplementing the fog-borne drizzle.
Tall trees and shrubs shade the driveway to the Burleson home, another fine mating of sea stacks and regional design. Closer to the house, the driveway broadens and is edged by lower-growing flowering shrubs, including Mexican mock orange (Choisya ternata) and lace-cap hydrangeas, along with tubs of bamboo, various Japanese maples (Acer palmatum cultivars), and a smoke tree (Cotinus coggygria). The front of the house, reached by a narrow, gated pathway, is away from the ocean but incorporates a fifteen-foot-high sea stack as part of its exterior wall. The stack is topped with salal; clinging to the rock lower down are cranesbill (Geranium), Astilbe, and violas.
Leading away from the house and sea stack, smaller boulders create (with a little human help) a channel for a rushing stream and pond of recycled water that provides a home for goldfish and aquatic plants, including the glistening and elegant white water lily (Nymphaea). The path along the stream is shaded by a variety of shrubs, such as the sweet-smelling bush anemone (Carpenteria californica), star magnolia (Magnolia stellata), and camellias, with clumps of blue wild iris (Iris douglasiana) tucked between the shrubs.
A path on the far side of the house traverses a colorful wilderness of flowering shrubs amidst drifts of the ubiquitous purple and pink toadflax (Linaria purpurea) and trailing golden and orange nasturtiums (Tropaeolum majus). At the corner of the house is a small wooden gate overwhelmed by a tall fuchsia (Fuchsia magellanica) with red and purple flowers. Through the gate and across a stretch of lawn, the ocean lies sparkling and blue (not that common in Humboldt County). In the distance, but not too far from shore, is another towering sea stack.
Facing the ocean is a broad terrace, bordered by a bed of flowering perennials, and edibles. Artichokes (Cynara scolymus) jostle with pink lavender (Lavandula angustifolia ‘Melissa’), and rhubarb (Rheum 5 cultorum) with love-in-a-mist (Nigella damascena). Common sage (Salvia officinalis) is surrounded by ox-eye daisies (Leucanthemum vulgare) and purple-headed chives (Allium schoenoprasum).
The terrace is home to a myriad of pots. Rosa ‘Kathleen’ makes its pink way up a trellis against the side of the house, intertwined with an unnamed creamy moss rose. Other pots hold African daisies (Osteospermum), strawberries (Fragaria), lobelia, fuchsia, and whatever else has caught the owner’s fancy.
Owners of these homes and gardens are wedded to sea stacks as integral parts of the fabric of their daily lives. They enjoy the sight of more sea stacks just offshore. It is as though these mighty rocks are waiting patiently for that distant time when they, too, may become part of someone’s home or garden.
- The California Coastal National Monument, established in the year 2000 by presidential proclamation, runs the entire length of California, to a distance of twelve nautical miles off shore. Its purpose is to protect the offshore rocks, pinnacles, and sea stacks and their ecosystems. ↩
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The Native Indian people of the Northwest Pacific coast lived in large houses (sometimes referred to as longhouses) which sheltered multiple families at times. The various different Native Indian tribes had different styles of houses unique to their people.
The Museum of Civilization near Ottawa, Canada has a Grand Hall exhibit which displays several house fronts representing some of these different styles of houses. They were all made by Native Indian artisans with work initiated in British Columbia and final assembly at the museum.
Native Art eBooks!
'An Overview of Pacific Northwest Native Indian Art'
'An Overview of Canadian Arctic Inuit Art'
The Tsimshian people live in the areas north and adjacent to the Queen Charlotte Islands in British Columbia. The Tsimshian style of house utilized massive cedar wood posts and beams. The wall boards were removable. High ranking Tsimshian had houses with broad steps on the main level inside leading down to central fire pits. The Tsimshian house front displayed at the Museum of Civilization represents a house that was near Port Simpson during the mid 1800s.
The Haida live on the Queen Charlotte Islands and the southern part of southeast Alaska. Their houses are similar to those of the Tsimshian. Houses featuring six beams and pitched roofs were very popular during the 1800s. The Haida houses also had vertical wall boards.
Another interesting Northwest Pacific coast Native Indian house represented at the Museum of Civilization is from the Central Coast area of British Columbia stretching from northern Vancouver Island up to Kitimat where several tribes live including the Kwakwaka’wakw, Heiltsuk, Owikeno and Haisla people. The house façade displayed at the museum represents an actual house of a chief in Alert Bay during the 1890s to the 1930s. It is especially interesting because it features an animal totem pole built with a ceremonial entrance to the house at the bottom.
any image or link below to start viewing our
beautiful selection of Native American art
The territory of the Nuu-Chah-Nulth people stretched from the northern part of west coast Vancouver Island down to western part of Olympic Peninsula in Washington state. Nuu-Chah-Nulth houses could be as long as 30 metres or over 90 feet. The house front displayed at the museum represents a head chief’s house that stood during the 1800s near Port Alberni. This particular house featured 10 round holes which represented 10 moons below the front round entrance.
The territory from eastern Vancouver Island to the mainland opposite is Coast Salish. Much like the houses of the Nuu-Chah-Nulth, Coast Salish houses were also large with many of them over 30 metres (90 feet) in length and 12 metres (36 feet) in width. Heavy cedar logs held up massive roof beams. The walls of Coast Salish houses were wide cedar boards set between narrow poles parallel to the ground and tied together with twisted cedar branches. The Coast Salish house depicted at the Museum of Civilization was one that stood near Nanaimo on Vancouver Island.
As one can see, the Northwest Pacific coast Native Indian people were expert house builders using the abundant supply of lumber in the region. These houses were usually built quite close to the shores.
Today, Northwest Pacific Coast Native Indian artists make some very stunning carvings. Just imagine having such a beautiful piece of artwork hanging in your home. See Northwest Pacific Coast Native Indian art carvings or Northwest Pacific Coast Native Indian art prints at very affordable online prices available at Free Spirit Gallery.
See Our Native Indian Art Videos
Copyright © 2006-2014 Free Spirit Gallery, All Rights Reserved
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Researchers at the Fraunhofer Institute for Computer Graphics Research IGD have developed CultLab3D: a streamlined 3D scanning system for the quick and easy digitization of 3D objects. The project aims to facilitate the mass digitization, annotation and storage of historical artifacts in museums and other places of preservation.
Digital preservation is one of the most important methods of sustaining our cultural history. Without it, there would be no way to safeguard precious historical texts, artifacts and information. Before the mass digitization of historical texts through 2D scanning, a simple accident or theft could have eradicated a piece of history forever. Several books of the Christian Bible, for instance, have been simply lost to time, as all physical copies of the “Book of the Battles of Yahweh” and the “Book of the Chronicles of the Kings of Judah” were destroyed. Homer’s Margites, preceding the Iliad and the Odyssey, has been obliterated from history in a similar manner, as the few existing copies of the epic poem perished with Ancient Greece itself. Thankfully, if a worthy book is discovered today, it can theoretically remain discoverable for all time. This is because digital preservation enables the digital archiving of written texts, as well as the easy duplication of those digital files. Even if the British Library were to burn down overnight, the majority of its contents would at least be digitally preserved. Digital preservation of texts is one thing, but the preservation of physical artifacts is quite another. Think about the following: How does a photograph of the Rosetta Stone compare to the object itself, in terms of historical value? Although the reasons are difficult to explain, the answer is obvious: the unique object itself has an infinitely superior intrinsic historical value than the mere photo. It is the Rosetta Stone. Though incredibly valuable in its own way, the digital preservation of the Rosetta Stone through photography is somehow less reassuring than the digital preservation of historical texts.
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RESEARCHERS FIND HOW ONE GENETIC VARIATION MAY LEAVE SOME PEOPLE VULNERABLE TO ADDICTION
COLUMBUS, Ohio – Scientists have learned how a genetic variation long suspected in making some people susceptible to alcoholism and narcotic drug addiction actually does so.
In laboratory studies, this variation greatly reduced the amount of protein that the DNA in a cell produced.
It's the difference in protein expression that may make receptors on certain brain cells much more vulnerable to the effects of addictive drugs, said Wolfgang Sadee, the study's lead author, professor and chair of pharmacology and director of the pharmacogenomics program at Ohio State University. These particular receptors, called mu opioid receptors, serve as a molecular docking station for narcotic drugs and alcohol.
Until now it wasn't clear exactly what about this genetic variation, called A118G, would increase a person's chances of developing a drug addiction. (A118G is a variation in what researchers call the mu opioid receptor gene.)
While Sadee and his team didn't look at the interaction between narcotics and the mu opioid receptor, they suspect that differences in protein production may leave brain cells with these receptors more open to the effects of drugs.
“The real significance of this work is that one day, we may be able to tailor treatments for addiction based on how a person's genes behave,” said Sadee, who is also chair of pharmacology at Ohio State .
The study appears in the current issue of the Journal of Biological Chemistry.
The researchers studied brain tissue samples taken from the cerebral cortex and the pons of human cadavers. The pons is a cluster of nerve fibers on the front of the brainstem, and it's responsible for relaying sensory messages from the spinal cord to the cerebellum. The cerebral cortex is a thin layer of tissue that covers the surface of the cerebral hemisphere; it is responsible for processes such as thought, memory, motor function and social abilities.
The researchers studied these particular brain tissues because both are rich in cells that have mu opioid receptors. Opioids are pain-relieving medications or illegal drugs that can be quite addictive, and these receptors in brain cells serve as a target for narcotic drugs. The interaction between narcotics and the receptors stops a person from feeling pain and also triggers the sensations of craving, reward and expectation that addicts often experience.
The researchers extracted and analyzed DNA and RNA from the brain tissues. They then injected the genetic material into ovary cells from Chinese hamsters. They could then measure the changes in the regulation and processing of messenger RNA (mRNA). mRNA carries instructions from the DNA inside a cell's nucleus to the rest of the cell, telling the cell that it's time to make more protein.
Surprisingly, the mu opioid receptor genes that carried the A118G variation (such variations in genes are called single-nucleotide polymorphisms) produced less mRNA than did the genes without the variant. In addition, the A118G change caused a ten-fold decrease in protein production inside the hamster ovary cells.
The mu opioid receptor gene is the first of 20 or so genes implicated in drug addiction that Sadee and his colleagues want to study. Those other genes may play a role in addiction to various drugs, including alcohol and nicotine.
“Drug addiction is a complex disorder, one that has a strong genetic component,” Sadee said. “It's very hard to prove that there is a causative link between one polymorphism and addiction. But the current study provides strong evidence that there is.”
Sadee conducted the study with Ohio State colleagues Ying Zhang, Danxin Wang, Andrew Johnson and Audrey Papp.
The work was supported by a grant from the National Institutes of Health.
Contact: Wolfgang Sadee, (614) 292-1597; [email protected]
Written by: Holly Wagner, (614) 292-8310; [email protected]
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Discover the cosmos! Each day a different image or photograph of our fascinating universe is featured, along with a brief explanation written by a professional astronomer.
2007 September 4
Explanation: What's inside Victoria Crater? Now that the dust has settled from the regional Martian dust storms that immobilized the rolling Martian rovers, the task ahead has become clear. Opportunity arrived at Victoria Crater last month and was poised to enter when the dust storms flared up unexpectedly. The above image was taken last week by the Opportunity rover perched at a possibly traversable slope into the 750-meter impact feature. Victoria Crater is the largest crater that either Martian rover has come across during their explorations. The crater walls might hold clues about the Martian surface before the tremendous impact that created Victoria Crater.
Authors & editors:
Jerry Bonnell (USRA)
NASA Official: Phillip Newman Specific rights apply.
A service of: ASD at NASA / GSFC
& Michigan Tech. U.
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# What is the middle number?
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
## Which number is the middle number?
The median of a set of numbers is the middle number in the set (after the numbers have been arranged from least to greatest) -- or, if there are an even number of data, the median is the average of the middle two numbers.
## What is the middle of 7?
The middle of 7 is 4: How would you explain this to a child? : r/math.
## Is median The middle number?
The median is the middle number in an ordered data set. The mean is the sum of all values divided by the total number of values.
## What's the middle of two numbers?
The midpoint between two numbers is the number exactly in the middle of the two numbers. Calculating the midpoint is the same thing as calculating the average of two numbers. Therefore, you can calculate the midpoint between any two numbers by adding them together and dividing by two.
## What is the middle of 15?
We see that the number in the middle is 7.5, which is exactly 7.5 units from 0 and exactly 7.5 units from 15. Therefore, half of 15 is 7.5.
## What is the middle of 1 and 9?
“Ask adults from the industrialized world what number is halfway between 1 and 9, and most will say 5. But pose the same question to small children, or people living in some traditional societies, and they're likely to answer 3.
## What is the median of 4 and 7?
The mean of these middle values is (4 + 7) / 2 = 5.5 , so the median is 5.5.
## How do I find the median?
The median is calculated by arranging the scores in numerical order, dividing the total number of scores by two, then rounding that number up if using an odd number of scores to get the position of the median or, if using an even number of scores, by averaging the number in that position and the next position.
## What is the middle number of a data set?
from least to greatest or greatest to least; the median is the data value in the middle; if there is an even number of data values in the set, the median is the mean of the two middle values.
## Why is 7 important in the Bible?
Seven was symbolic in ancient near eastern and Israelite culture and literature. It communicated a sense of “fullness” or “completeness” (שבע “seven” is spelled with the same consonants as the word שבע “complete/full”). This makes sense of the pervasive appearance of “seven” patterns in the Bible.
## What's the middle of 7 and 3?
Example: what is the central value for 3 and 7? Answer: Half-way between, which is 5.
## Does 7 have a line in the middle?
Most people in Continental Europe, and some in Britain and Ireland as well as Latin America, write 7 with a line in the middle ("7"), sometimes with the top line crooked.
## Is 0 the middle of a number line?
Zero is the middle point of a number line. All (natural numbers) positive numbers occupy the right side of the zero whereas negative numbers occupy the left side of zero on the number line. As we move on to the left side value of a number decreases.
## How do you find the middle of 4 numbers?
If we have an odd number of terms then the middle term is the median. If we have an even number of terms then we have to add two middle terms and then divide the sum by 2. The mean we get is the median.
## Is the mode the highest number?
Place all numbers in a given set in order; this can be from lowest to highest or highest to lowest, and then count how many times each number appears in the set. The one that appears the most is the mode.
## What is the median of 10?
Hence, the required median of the first ten natural numbers = 5 . 5 .
## What is the median of 5 and 7?
Hence, the median of 5 and 7 is 6 .
## What is the median of 1 to 50?
Hence, the median of first 50 whole numbers is 24.5.
## What is the median in 1 to 10?
Here we have to find the median of the first 10 natural numbers. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Where the number of terms is in even. Therefore, the median of the first 10 natural numbers is 5.5.
## What is the median of 23 and 23?
Since there are an even number of values, the median will be the average of the two middle numbers, in this case, 23 and 23, the mean of which is 23.
## What is the median of 25?
The median is the middle number, for an even set of numbers there will be two middle numbers, to find the median in an even set of numbers we can average the two middle numbers. So since 25 and 25 are both middle numbers we can average them to get the median, which is 25 .
## What's in the middle of 2 and 5?
The class midpoint of 2-5 is 3.5.
## What is in the middle of 28 and 42?
18 is the HCF of 28 and 42.
The HCF of 28 and 42 is 14.
## What is in the middle of 0 and 1?
This is a good definition for intervals, and the middle point of (0,1) would be 1/2 with this definition.
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Technology appears to hold no fear for children. Most of them take to computers and the Internet like proverbial ducks to water, and derive huge educational and entertainment benefits from it, but few of them know about the dangers lurking below the surface.
The big question is how to make children aware of the dangers without dampening their enthusiasm and appetite for the technology.
In a thesis written for Royal Holloway University of London, entitled Security Awareness for Children (see .pdf below), Clara Brady and Chris Mitchell report on research they have carried out at four schools in Ireland, where they gathered information about pupils' computing habits, and also their awareness of the potential dangers they might encounter.
The results led them to formulate guidelines for Internet safety for kids. These include strategies for conducting computer security awareness training by running focused campaigns that seek to boost skills and knowledge about different aspects of security, such as handling emails or cyberbullying.
About the authors:
Clara Brady is a qualified primary school teacher from Ireland. Alongside her teaching degree she studied Mathematics as an academic subject, which is where she first encountered Cryptography. After a number of years teaching, she decided to further her interest in this area and returned to university to study the MSc in information security. She has organsied a Security Awareness Day for her school, and hopes to build on this next year and to continue to highlight and promote the need to educate children about using the Internet in other schools throughout Ireland.
Chris Mitchell is a professor of computer science at RHUL. His main research interests are in information security and combinatorial mathematics.
This article is based on a thesis written in the Information Security Group at Royal Holloway University of London.It is one of nine that SearchSecurity.co.uk is publishing exclusively in 2010 as part of its close collaboration with RHUL, which is in its third year.
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About This Chapter
How it works:
- Identify the lessons in the Glencoe Earth Science Human Impact on Resources chapter with which you need help.
- Find the corresponding video lessons within this companion course chapter.
- Watch fun videos that cover the human impact on resources topics you need to learn or review.
- Complete the quizzes to test your understanding.
- If you need additional help, rewatch the videos until you've mastered the material or submit a question for one of our instructors.
Students will learn about:
- Population carrying capacity and size
- Theory of human population growth
- Pros and cons of pesticides
- Bioremediation, deforestation and land conservation
- Fossil fuels and acid rain
- Air pollution in developed and developing countries
- Water conservation and pollution.
Glencoe is a registered trademark of McGraw-Hill, which is not affiliated with Study.com.
1. Biotic Potential and Carrying Capacity of a Population
When you look around, you can see a number of different individuals and species. In this lesson, we will explore the factors that control and define how many organisms can really inhabit a particular ecosystem or habitat.
2. Population Size: Impacts on Resource Consumption
What are the things you need to survive? Humans all need resources, and in this lesson, we will discuss how the growth of the human population is influencing the natural resources we rely on.
3. Thomas Malthus' Theory of Human Population Growth
In this lesson, we will explore the ideas of Thomas Malthus and his predictions for the growth of the human population. We will also investigate if his predictions have come true or not.
4. Use of Pesticides: Benefits and Problems Associated with Pesticides
Pesticides are used worldwide to manage agricultural pests. They kill and repel unwanted pests, but also cause many human deaths each year. This lesson explores the widespread use of pesticides in agriculture and compares the benefits and problems associated with these helpful but dangerous chemicals.
5. Bioremediation: Microbes Cleaning Up the Environment
Bioremediation uses living organisms to clean up pollution. In this lesson, we will examine the role of microbes in bioremediation and why it is not a perfect solution for degrading environmental pollutants.
6. Deforestation: Definition, Causes & Consequences
Deforestation is the clearing of trees without the intent of replacing them or reestablishing a stand of trees. Learn reasons for deforestation and discover the consequences of this process in all forests, including tropical forests.
7. Land Conservation: Preserving and Restoring Ecosystems
Humans have degraded a large portion of natural land. The field of land conservation has become popular as a way to fix this problem. Land conservation attempts to combat land degradation through preservation, restoration, remediation, and mitigation.
8. The Nitrogen Cycle, Acid Rain and Fossil Fuels
Do you know the difference between nitrite and nitrate? Do you have any clue how all the nitrogen in the atmosphere becomes usable? If you can't answer these questions, then you need to watch this lesson on the nitrogen cycle.
9. Controlling Air Pollution: How Developed & Developing Nations Differ
In this lesson, you will learn how air pollution is controlled in developed and developing nations. You will also learn about the U.S. Clean Air Act and how it has evolved over the years to address air pollution in the United States.
10. Water Pollution: Definition, Types, and Sources
In this lesson, you will learn about the different types and sources of water pollution. You will gain an understanding of both surface water and groundwater pollution and the similarities and differences between them.
11. Water Conservation: How Water Management Can Lead to Sustainable Use
In this video lesson you will learn about the importance of water conservation. You will also learn about the two ways freshwater is used, and how proper water use can lead to sustainability in agriculture, homes, and manufacturing.
Earning College Credit
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Other chapters within the Glencoe Earth Science: Online Textbook Help course
- Glencoe Earth Science Chapter 1: The Nature of Science
- Glencoe Earth Science Chapter 3: Matter and Change
- Glencoe Earth Science Chapter 4: Minerals
- Glencoe Earth Science Chapter 5: Igneous Rocks
- Glencoe Earth Science Chapter 6: Sedimentary and Metamorphic Rocks
- Glencoe Earth Science Chapter 7: Weathering, Erosion, and Soil
- Glencoe Earth Science Chapter 8: Mass Movements, Wind, and Glaciers
- Glencoe Earth Science Chapter 9: Surface Water
- Glencoe Earth Science Chapter 10: Groundwater
- Glencoe Earth Science Chapter 11: Atmosphere
- Glencoe Earth Science Chapter 12: Meteorology
- Glencoe Earth Science Chapter 13: The Nature of Storms
- Glencoe Earth Science Chapter 14: Climate
- Glencoe Earth Science Chapter 15: Earth's Oceans
- Glencoe Earth Science Chapter 16: The Marine Environment
- Glencoe Earth Science Chapter 17: Plate Tectonics
- Glencoe Earth Science Chapter 18: Volcanism
- Glencoe Earth Science Chapter 19: Earthquakes
- Glencoe Earth Science Chapter 20: Mountain Building
- Glencoe Earth Science Chapter 21: Fossils and the Rock Record
- Glencoe Earth Science Chapter 22: The Precambrian Earth
- Glencoe Earth Science Chapter 24: Earth Resources
- Glencoe Earth Science Chapter 25: Energy Resources
- Glencoe Earth Science Chapter 27: The Sun-Earth-Moon System
- Glencoe Earth Science Chapter 28: Our Solar System
- Glencoe Earth Science Chapter 29: Stars
- Glencoe Earth Science Chapter 30: Galaxies and the Universe
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Sacrifices in The Crucible
Social order is a fragile concept; the most extraneous of details can cause it to spiral into chaos. In The Crucible, the inhabitants of Salem, Massachusetts experience the unforgiving disruptance of social order. Many attempted to end the witch trials promptly to the start of them, but their sacrifices rendered futile. John Proctor, a farmer, was the only citizen of Salem whose sacrifice made a significant, positive, impact on the restoration of social order.
In Act l, John Proctor and Abigail Williams had a conversation that state both engaged in sexual relations, making Abigail a harlot and Proctor an adulterer. Abigail, throughout the encounter, talks of Elizabeth Proctor as if she is the only one keeping John Proctor and herself from being together. As the intensity of the witch trails reach its climax, Elizabeth Proctor is accused of being a witch. When Cheever arrives to take Elizabeth to Jail, Proctor inquires about the accuser, who happens to be Abigail Williams.
John then tells Mary Warren, Abigail’s abettor, to testify against her. When this fails in court, Proctor sacrifices everything he owns, his name, his familys reputation, and his life, by saying, “It is a whore! ” (Act Ill, scene i). John Proctor dmitted lechery in a prevailing attempt to call out Abigail on her pretense in seeing spirits and witches. Soon after this, John is arrested and Abigail is allowed freedom.
Also, Reverend Hale denounces the proceedings of the court. The sacrifice made by Proctor made the puritan society of Salem, Massachusetts question the claims made by Abigail and her friends. When John Proctor accuses Abigail of harlotry, she is forced to run away to save herself from shame. The witch trials also end soon afterwards and social order starts its restoration process. The people of Salem once again start to trust each other and put away grudges of the past.
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Also available: HTML-only, Chime Enhanced and JMol versions.
Carbon monoxide is a toxic, colourless and odourless gas. It has been invaluable in helping chemists to extract metals from their ores. However, it must be said that its physical properties make it potentially very dangerous.
Carbon and oxygen can combine to form two gases. When combustion of carbon is complete, i.e. in the presence of plenty of air, the product is mainly carbon dioxide (CO2). Sources of carbon include; coal, coke, charcoal. When combustion of carbon is incomplete, i.e. there is a limited supply of air, only half as much oxygen adds to the carbon, and instead you form carbon monoxide (CO).
Carbon monoxide is also formed as a pollutant when hydrocarbon fuels (natural gas, petrol, diesel) are burned. The relative amount of CO produced depends on the efficiency of combustion. Older vehicles are checked annually for CO emissions during their MOT test. Interestingly, only one of the two oxides of carbon, doesn't support combustion and it is for this reason that carbon dioxide is used in fire extinguishers. Carbon monoxide does support combustion and burns with a pale blue flame. The blue flame used to be seen over the fires made from coke (essentially a very pure form of carbon) by night watchmen on industrial sites.
2 CO (g) + O2 (g) 2 CO2 (g)
When making carbon monoxide a source of carbon dioxide gas is needed. This could be from a CO2 cylinder or even dry ice (solid CO2). If neither is available carbon dioxide could be generated by the neutralisation reactions between an acid and a carbonate or an acid and a hydrogen carbonate.
2HCl (aq) + CaCO3 (s) CaCl2 (aq) + H2O (l) + CO2 (g)
HCl (aq) + NaHCO3 (s) NaCl (aq) + H2O (l) + CO2 (g)
When carbon dioxide gas is passed over heated charcoal it forms carbon monoxide.
CO2 (g) + C (s) 2CO (g)
There will also be unreacted carbon dioxide, which needs to be removed. Carbon dioxide is removed by reacting it with an aqueous solution of sodium hydroxide.
2 NaOH (aq) + CO2 (g) Na2CO3 (aq) + H2O (l)
Another convenient way to prepare carbon monoxide is the dehydration of methanoic acid using conc. H2SO4.
HCOOH (aq) CO (g) + H2O (l)
Dehydration of a methanoic salt such as sodium methanoate also works well. In this case you dribble concentrated sulphuric acid directly onto the solid. The carbon monoxide, which evolves, can be collected under water. Methanoic acid is found in nettles and ants.
Carbon monoxide is a very poisonous gas. It is poisonous at levels of only 0.1% (1000 ppm). Its toxicity arises from its ability to bind to transition metals such as iron found at the centre of a haem molecule. Carbon monoxide is attracted to haemoglobin over 200 times more strongly than oxygen. Therefore, in the blood, the presence of carbon monoxide prevents some of the haemoglobin found in red blood cells from carrying sufficient oxygen.
This fact is certainly worth considering if you should be tempted to smoke a cigarette. Smokers have been found to have quite high levels of carbon monoxide in their blood, long after they have finished lighting up their chosen cigarette.
Symptoms of carbon monoxide poisoning are dizziness and headaches. These sufferings can be confused with other illnesses such as influenza. Carbon monoxide poisoning can be recognised, as victims will often have unnaturally bright red lips.
Prolonged exposure to carbon monoxide can eventually lead to death. Carbon monoxide has been used as the poison in suicides. Even more disturbing was the use of carbon monoxide by Nazis in WWII to kill its victims of the death camps. More recently, there have been cases of rogue landlords not properly maintaining gas appliances, which led to the death of tenants often students. The law in the UK now requires the annual checking of boilers, gas cookers, and gas fires by registered engineers. The following equation shows what happens when there is incomplete combustion of natural gas, which is primarily methane.
2 CH4 (g) + 3 O2 (g) 2 CO (g) + 4 H2O (g)
Before vast quantities of natural gas were discovered beneath the seas and oceans, we used to burn coal gas. Coal gas was produced when coal was heated in the absence of air. Its main components are hydrogen methane, and carbon monoxide.
Carbon monoxide sometimes occurs in coal mines. At one time canaries were taken down in mines to detect poisonous gases. Canaries would be killed at doses not quite lethal to miners. Today gases are detected more humanely with instruments.
The first time most students meet carbon monoxide in a chemistry lesson is in its use within the Blast Furnace. The Blast furnace is how iron is extracted from its ore, haematite (iron(III)oxide Fe2O3).
Fe2O3 (s) + 3 CO (g) 2 Fe (l) + 3 CO2 (g)
Carbon monoxide is a strong reducing agent and reduces metal oxides for metals less reactive than carbon. The following table is useful for the different definitions of reduction. As oxidation is the opposite of reduction you only need to learn half the facts!
|Loss of oxygen||Gain of oxygen|
|Gain of hydrogen||Loss of hydrogen|
|Gain of electrons||Loss of electrons|
|Decrease in O.N.||Increase in O.N.|
Several gases (H2, CH4 and CO) have historically been used as reducing agents. One definition of a reducing agent I particularly like is to think of it as being an oxygen grabber. It is important to remember that the reducing agent itself gets oxidised. A quick and simple laboratory reduction can be achieved by heating a mixture of black copper(II)oxide with carbon powder in a test tube. After several minutes of heating the reddish coloured copper can be seen on the side of the test tube. Essentially carbon acts as a reducing agent as well as the carbon monoxide that is inevitably formed by its heating in air. The following reactions are all occurring in this simple experiment.
CuO (s) + CO (g) Cu (s) + CO2 (g)
CuO (s) + C (s) Cu (s) + CO (g)
C (s) + O2 (g) 2 CO (g)
C (s) + O2 (g) CO2 (g)
CO2 (g) + C (s) 2 CO (g)
Carbon monoxide does not show acidic or basic properties. Its feeble Lewis acidity is seen through its formation of H3BCO with borane (BH3). Carbon monoxide has a remarkable affinity for transition metals (located between Groups 2 and 3 of the Periodic Table). The first examples of metal carbonyls was back in 1888, when tetracarbonyl nickel (0) Ni(CO)4 and pentacarbonyl iron (0) Fe(CO)5 were prepared and characterised. The former complex forms part of the Mond Process for the purification of Nickel. Ni(CO)4 is distilled to give pure nickel.
Ni (s) + 4 CO (g) Ni(CO)4 Ni (s) + 4 CO (g)
Carbon monoxide is so reactive with nickel that within a couple of minutes it will have etched the surface. Ni(CO)4 is highly toxic with a musty smell. As well as being flammable this tetrahedral complex decomposes easily into its constituents. Carbon monoxide is acting as a ligand towards the transition metal through the lone pair on the carbon atom. Diatomic carbon monoxide has a triple bond between its atoms. One of the bonds forming the triple bond is a dative covalent bond.
Detecting carbon monoxide is a simple task these days. Most hardware stores sell special spots impregnated with palladium compounds, which darken on exposure to carbon monoxide.
Back to Molecule of the Month page.
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# Polystick Problems from Polyomino Solutions
September 7th, 2010 by munizao
Polysticks (or polylines) are connected sets of segments in a square grid. (Polysticks on other grids are possible, but haven’t seen much attention.) The tetrasticks, of which there are 16, seem to be the most natural set for puzzle making. Donald Knuth has explored tetrastick problems, and posed the problem of tiling an Aztec Diamond with the 25 one-sided tetrasticks, which was solved by Alfred Wasserman. Here’s one I’ve come up with:
Problem #15: Tile the above shape with two sets of the five tetrominos and one monomino, and tile the borders of these polyominoes with the 16 tetrasticks. Here’s an attempt I made that fell short by a few tetrasticks, but it should give you an idea of the form a solution would take:
The observation that the lines formed by the pieces in a polyomino tiling could themselves be tiled by polysticks seems obvious, but I have not seen it elsewhere. After picking the 16 tetrasticks as my puzzle pieces for the polystick stage, I had to find a set of polyominoes to use. Since one of the tetrasticks is, in fact, the outline of a 1×1 square, or monomino, the monomino had to be present. A double set of tetrominoes plus the monomino gives a good quantity of segments for our tetrasticks to cover, and gives us an area of 2 * (5*4) + 1 = 41. The perimeter of the figure to be tiled is constrained by the following formula:
2 * total segments in the polystick set – sum of perimeters of polyominoes = perimeter of entire figure
In this case, (2 * (4 * 16)) – (4 + 2 * 48) = 28
So I needed a figure to tile with area 41 and perimeter 28, and came up with the shape above.
There are 136 solutions for the tetromino tiling with the monomino in the center as shown. (See these solutions in a Java solver applet.) I’ve experimented a little with the tetrastick stage of the problem by hand, and I’m convinced that there are no tetrastick solutions for most, if not all, of these tetromino solutions. But if it doesn’t work out in the case with the monomino in the center, I suspect there are enough solutions with the monomino elsewhere for it to be very likely that one will work. Many of the tetromino solutions fail to contain a site where the “+” tetrastick can be placed that doesn’t overlap the “□”.
Another issue that surfaces in this problem is horizontal-vertical segment parity. Eleven of the tetrasticks have even parity, that is, however you place them, they will always contain even numbers of both horizontal and vertical segments. Five of them have odd parity, and will always contain three segments of one orientation and one of the other. Because there are an odd number of pieces of odd parity, the parity of the set of tetrasticks as a whole must be odd. This means, without even starting to try placing tetrasticks on a tetromino solution, we can rule out the possibility of tiling it just by counting the number of horizontal or vertical segments. (Because the total is constant, we don’t need to count both.) If that number is even, the tetrasticks can’t tile the figure. The tetromino solution that I used in my attempt above has the correct (odd) parity.
I dredged this problem up from my archive of the polyforms mailing list, where I posted it in February, 2001. It got no takers then, but I thought it an interesting enough problem to deserve a second airing. I looked for other problems of this type in preparing this post, but I didn’t find anything good. Having both the area and the perimeter of the figure to be tiled constrained by the pieces used limits the possibilities a lot.
1. Lewis says:
>I dredged this problem up from my archive of the polyforms mailing list,
Are there any mailing lists or forums relating to polyominoes/polyforms which are still active now?
2. munizao says:
The polyforms mailing list I was referring to is the Yahoo group. It’s only mostly comatose, and has had a slight uptick in activity over the last year.
If there’s anything else out there, I’d certainly like to know about it.
3. Debbie says:
The tetrasticks are related to the pentominoes. I would suggest using the pentomino names: FLIPNTUVWXYZ, adding O for the square tetrastick and P1 P2 P3 P4 to refer to the 4 tetrasticks that have vertices at the positions of a P pentomino.
4. munizao says:
Actually, I’ve given a bit of thought to the “orthography” of tetrasticks. I do mostly use the pentomino names. For the “P” derived tetrasticks, I prefer using letters that look something like the tetrasticks themselves; if I called them P1 etc., I’d have to memorize which was which without having the shape for a mnemonic. I also find that it’s useful to have a single symbol representing each form.
For the P derived tetrasticks, I’ve settled on ‘J’, ‘H’, (which is clearer if you think of a lower case ‘h’) ‘?’, (a little unfortunate to have to use a punctuation mark, but the resemblance is very clear compared to any letter I could have used) and ‘F’. The last conflicts with the F pentomino derived tetrastick, but it is so clearly shaped like an ‘F’ that it deserves the title. The F pentomino derived tetrastick I’ve switched to ‘R’, which is what John Conway called it. (O for the square tetrastick is indeed probably best.)
But for the purpose of the above article I didn’t want to get into introducing an orthography, so I used the ‘+’ and ‘□’ characters, which have the benefit of being immediately recognizable as particular tetrasticks without any explanation required.
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This is an image of a plateau and surrounding steep slopes within the Valles Marineris.
Click on image for full size
Image from: Malin Space Science Systems
An Overview of the Mars Global Surveyor Mission
The mission of Mars Global Surveyor (MGS) is to map the surface of Mars from space, a mission somewhat akin to the Magellan
mission to Venus. The MGS probe is also suppose to explore the topmost portion of the Martian atmosphere and make detailed observations of the Martian weather. Because MGS was to orbit Mars and not land on the surface, the trajectory was a little different than that of Mars Pathfinder
(MPF), allowing MGS to launch before MPF, but arrive at Mars after MPF had already landed. MGS carries a number of instruments
which are similar to those on the lost Mars Observer
As part of the less expensive and more experimental premise of the Mars Surveyor Program, Mars Global Surveyor was the first to use the procedure of aerobraking. Aerobraking is when the atmosphere is used to slow the spacecraft sufficiently so that it can be placed into its desired orbit. This procedure was experimental, and was further complicated by the failure of one of the two solar panels on MGS. This complication delayed the beginning of MGS mapping, but scientists were able to work around the complication quite successfully, bringing MGS into the near-circular, low-altitude orbit required for mapping of the Martian surface.
Mapping was delayed by about a year, but MGS has still returned high-resolution images of the Martian surface and large amounts of data which will help scientists refine their models of the Martian atmosphere. Also among the important new results from the mission is the definite confirmation of the presence of a Martian magnetosphere. As of April 25, 2001, all systems on the MGS probe were functioning nominally. The spacecraft has been in space 1,631 days and has completed 9,530 mapping orbits of Mars! And MGS is still going strong!
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The Mars Pathfinder (MPF) mission was sent to investigate the geology of Mars. Its principal objective was to analyze the rocks and soil of Mars. The MPF consisted of 2 components, a lander and a mobile...more
Mars Global Surveyor (MGS) is conducting mapping operations at Mars more than 30 years after America's first reconnaissance missions reached the mysterious red planet. Here are some of the instruments...more
Born as a result of the failure of Mars Observer (MO), the Mars Surveyor Program was designed to explore Mars with all the original measurements planned for MO, and a lot more. It varied from the MO mission...more
On September 12, 1997, the Mars Global Surveyor successfully entered a highly elliptical orbit around Mars. To get into the near-circular, near-polar, low-altitude orbit necessary to map the surface of...more
These are some of the initial findings of Mars Global Surveyor. The definite confirmation of the presence of a magnetic field near Mars. suggests scientists must rethink theories about the evolution of...more
An important new result from the Mars Global Surveyor (MGS) mission is the definite confirmation of the presence of a magnetosphere around Mars. Previous missions made inconclusive measurements of the...more
Amongst reports that the Mars Polar Lander fell into a deep canyon, scientists are reporting the cause of the disaster is still unknown.Organizers of the mission also pointed out they knew the canyon...more
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(b) Write 0.000 35 in standard form. Competency Based Curriculum Design Materials Marvel Cinematic Universe Phase 3 Quiz! Three toy cars and 4 toy trains cost $18. Answer question 5 on page 2. Grade 5 maths problems with answers are presented. Here is a list of all of the maths skills students learn in grade 5! She starts with a piece of cardboard whose length is 15 centimeters and with is 10 centimeters. Please submit your feedback or enquiries via our Feedback page. 4000 in standard form = 4.000 x 10 3. 5.99 / Volume of Rectangular Prisms Made of Unit Cubes. Grade 5 Maths Worksheet: Fractions . 5.2.2.a.1 Investigate and extend geometric patterns looking for relationships or rules of patterns. Complete: 5 720 is 100 less than _____ (1) (1) (1) 1.6 Fill in the numbers represented by A and B on the number line. These skills are organised into categories, and you can move your mouse over any skill name to preview the skill. Master this topic as you play. The following diagram shows how a number can be written in Standard form, Word form and Expanded form. Scroll down the page for more examples and solutions on using the Standard form, Word form and Expanded form. ID:226209 Common y Nathan, Alicia, and Taylor each ride to school on a bike. one toy car costs$2 and 1 toy train costs $3. It'll also help you to identify weaker areas that need more revision. (c) Work out the value of (2.3 × 10 12) ÷ (4.6 × 10 3). To start practising, just click on any link. How long does it take John to eat one pizza and a half? A painter charges$ 225 for materials and $35 per hour for labour. 5-a-day Further Maths; 5-a-day GCSE A*-G; 5-a-day Core 1; More. (1) In Tim's house, a rectangular swimming pool (blue) whose length 30 meters and width 10 meters is surounded by grass (green). Tom read only two-fifth of the time that Sasha read. The total cost of painting an office is$ 330. 1.5 Write down the multiples of three from 474 to 483. eval(ez_write_tag([[300,250],'analyzemath_com-medrectangle-4','ezslot_2',340,'0','0'])); Primary Maths (grades 4 and 5) with Free Questions and Problems With Answers, Middle School Maths (grades 6,7,8 and 9) with Free Questions and Problems With Answers, High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers, Math Word Problems with Solutions and Explanations for Grade 5, Primary Math (Grades 4 and 5) - Free Questions and Problems With Answers, Add Fractions and Mixed Numbers - Examples and Questions with Answers (Grade 5). Free & printable grade 5 math worksheets. answers; Post navigation. Practice that feels like play! Worksheets > Math > Grade 5. What area is occupied by the grass? Quiz Bee: Grade 5 Maths Questions Quiz Bee: Grade 5 Maths Questions . What is the area of the cardboard after she cuts the 4 corners? 2.1 Which number is represented by: 40 000 + 2 000 + 5 + 60 + 700? Expanded form can be demonstrated in two ways. A large database of Multiple Choice Maths test papers especially created for Grade 5 Maths Students in a number of different Categories. The Corbettmaths Practice Questions on Standard Form. 10. Combo All Exams 5th Grade Maths with Answers (for marking) Louisianna Believes 2013-4 Grade 5 Mathematics Tests – 3 parts, 50 multichoice and 2 long-form Nebraska 2010 Grade 5 Mathematics Tests – 24 multichoice questions Oregon 2010-2013 Grade 5 Mathematics Test – 20 multichoice questions Grade 5 maths problems with answers are presented. 15 exam-style questions aimed at grades 4-5 in the format of a revision race, written specifically for the new 9-1 spec. It takes John 25 minutes to walk to the car park and 45 to drive to work. Try the free Mathway calculator and 4123 = 4,000 + 100 + 20 + 3 (a) Write the number 0.00037 in standard form. Maths Made Easy © Complete Tuition Ltd 2017 Standard Form (Non-Calculator) Hopefully you and your students find these useful (and cheaper to print than to buy published resources). Standard Form, Expanded Form, and Word Form of Numbers (up to Millions) Example: Find the standard form value of (3\times10^8)\times(7\times10^4), without using a calculator. Express two thousand four hundred sixty-five in standard form. Geometry. 5.103 / Open and Closed Shapes. At what time should he get out of the house in order to get to work at 9:00 a.m.? Grade 5 Maths Worksheet: Fractions. How much money did Mel withdraw from the bank? This page has revision notes, videos and past exam questions arranged by topic. IXL will track your score, and the questions will … Linda bought 3 notebooks at $1.20 each; a box of pencils at$1.50 and a box of pens at $1.70. Questions and Answers . John can eat a quarter of a pizza in one minute. 2. Printable worksheets with 10 questions in PDF format. If Mike read 5 hours, how long did John read? Videos, worksheets, 5-a-day and much more 5.1 / Classify Quadrilateral Shapes. A. Tom and Bob have a total of 49 toys. (2) 2. Numbers can be written in expanded form. How much did Linda spend? A large box contains 18 small boxes and each small box contains 25 chocolate bars. Science Activities for Kids, 1st to 5th Grades: Games | Quizzes | Worksheets. Copyright © 2005, 2020 - OnlineMathLearning.com. Some other quick calculations chart for standard form. Reading & Math for K-5 © www.k5learning.com Exponents Grade 5 Exponents Worksheet 23=2×2×2 Exponents are used to express repeated multiplication of a number. Also Solutions and explanations are included. After the shopping, he had$32.00 left. The pool with the grassy area make a large rectangle whose length is 50 meters and width 20 meters. 5.3 / Types of Triangles. Example 4: (a) Write 7900 in standard form. or Get shields, trophies, certificates and scores. This page offers free printable math worksheets for fifth 5th and sixth 6th grade and higher levels. 1. For Grades 4, 5 and 6 worksheets,answers are provided. Grade 5 & 6 Math Worksheets and Printable PDF Handouts. (5 × 10,000) + (3 × 100) + (2 × 1). Mel had $35 and withdraw some more money from his bank account. In the figure below are three different shapes: squares, triangles and circles ... (in simplest form) the ratio of the number of girls who are less than 10 to the number of boys who are 10 or older. Give your answer in standard form. Carla is 5 years older than Jim. Unlimited adaptive online practice on this topic. Mary wants to make a box. Also Solutions and explanations are included. A large box contains 18 small boxes and each small box contains 25 chocolate bars. On your drawing use a dotted line to show where to divide the rectangle into one triangle and one trapezoid. The Office Trivia Quiz! 5.5 / Number of Sides in Polygons. Kim can walk 4 kilometers in one hour. Class 5 maths printable worksheets, online practice and online tests. For example: It takes John 25 minutes to walk to the car park and 45 to drive to work. Standard Form, Expanded Form, and Word Form of Numbers (Up to Thousands) Learn to write numbers in standard, expanded, and word form. Example: a. I have targeted the booklets at students aiming for a grade 1, grade 3, grade 5, grade 7 and grade 9. Two toy cars and 3 toy trains cost$13. 5.2 / Identify Simple and Complex Solid Shapes. Visit http://www.mathsmadeeasy.co.uk/ for more fantastic resources. How do you write 0.00037 in standard form? John can eat a sixth of a pizza in two minutes. Paisley has 10 jelly beans. How old is each one them? If Bob has 5 more toys than Tom, how many toys does each one have? Geometric sequences with fractions (5-I.11) Complete a geometric number sequence (5-N.6) 5.2.2.a.1.2 sequences not limited to a constant difference or ratio. (Hint: Use a table), Jim: 5 years , Carla: 10 years , Tomy: 16 years. Top Mathematicians. Answer: 3.7 x 10 -4 Level 1 Level 2 Level 3 Level 4 Level 5 Level 6 Level 7 Level 8 Level 9 Exam Description Help Here are some numbers written in standard form. Express 3,527 in expanded form and word form. It takes 3 minutes for Billy to eat one quarter of the same pizza. Grade 5 (All topics) Printable Worksheet. Maths Genie is a free GCSE and A Level revision site. 4123 = (4 × 1,000) + (1 × 100) + (2 × 10) + (3 × 1). Give your answer in standard form. 5.4 / Regular and Irregular Polygons. 936. (b) Write 0.000 376 in standard form. 50,302 = Answers are also included. Grade 5. Example 3: (a) Write 8.2 × 10 5 as an ordinary number. D. 9360000000000000000000000000. Try this amazing Quiz Bee: Grade 5 Maths Questions quiz which has been attempted 11694 times by avid quiz takers. He bought a pair of trousers at $34.00, two shirts at$16.00 each and 2 pairs of shoes at $24.00 each. Then she cuts congruent squares with side of 3 centimeters at the four corners. 50000 in standard form = 5.0000 x 10 4. Grade 5 Maths Practice Exams With Answers. Questions on how to find maths ratios in different situations with answers are presented. Numbers written in standard form can be ordered by first looking at the power of 10, which tells you the size of the numbers. Check your answers if you have time at the end GCSE (9- 1) Grade 5 Standard Form Name: YYYYYYYYYYYYYYYYYYYYXX Total Marks: YYYYYYYYYYYYYYYYY www.naikermaths.com. Tomy is 6 years older than Carla. Practice Worksheet. Enhanced Student Reports. 1. How many chocolate bars are in the large box? These worksheets are of the finest quality. If John and Billy start eating one pizza each, who will finish first? John read the quarter of the time that Tom read. We welcome your feedback, comments and questions about this site or page. Also explore over 42 similar quizzes in this category. How many hours did it take the painter to paint the office? ID:203408 GMAP05_cut_rectangle.eps Common t Draw a rectangle in your Student Answer Booklet. Ratios - Grade 5 Maths Questions With Answers. ☐ Compare and order fractions including unlike denominators (with and without the use of a number line) Note: Commonly used fractions such as those that might be indicated on ruler, measuring cup, etc. 50,302 = 50000 + 300 + 2 Jim, Carla and Tomy are members of the same family. Step 1: Change the order around of the things being multiplied. Our grade 5 math worksheets cover the 4 operations, fractions and decimals at a greater level of difficulty than previous grades. ... Grade 5. Featured Quizzes. Try the given examples, or type in your own C. 9360. Sasha read twice as long as Mike. This is a great activity to do as a revision lesson to engage your students in some exam practice. Embedded content, if any, are copyrights of their respective owners. Online Practice. B. CCSS 5.NBT.A.3 worksheets with answers to teach, practice or learn 5th Grade common core mathematics is available online for free in printable & downloadable (pdf) format. How long does it take Kim to walk 18 kilometers? Standard Form Textbook Answers Click here for Answers . 5.2.2.a.1.1 represented in physical or diagram form. What is the price of one toy car and the price of one toy train if both prices are whole numbers of Dollars? 5… worksheets_maths_gr_5e_ws1.jpg problem and check your answer with the step-by-step explanations. Maths revision video and notes on the topics of: converting between standard form and ordinary numbers; and adding, subtracting, multiplying and dividing numbers in standard form. Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. E. 19360. b. Math is Fun Curriculum for Grade 5. Online Practice. Practise Multiple Choice Maths Test Papers in Algebra, Geometry, Mixed Numbers, Fractions and a number of other topics for Grade 5 students 5.NF.B.3 Interpret a fraction as division of the numerator by the denominator (a/b = a / b). Further Maths; Practice Papers; Conundrums; Class Quizzes; Blog; About; Revision Cards; Books; October 10, 2019 October 22, 2019 corbettmaths. Answer question 6 on page 2. problem solver below to practice various math topics. Test your understanding of standard form (scientific notation) with this self-marking quiz. We also introduce variables and expressions into our word … The Office Trivia Quiz! How many chocolate bars are in the large box? Write down the multiples of 5 between 718 and 733. Skill 5: Multiplying Standard Form. PP1, PP2 TERM 1,2,3 MID/END TERM EXAMS QUE AND ANSWERS GRADE 1 2 3,4 TERM 1,2,3 MID/END TERM EXAMS QUE AND ANSWERS; Grade 1 NOTES And Class readers; Grade 2 NOTES And Class readers; Grade 3 NOTES And Class readers; PP1,PP2, Grade 1,2,3,4 Lesson plans; And K.I.C.D. The sum of their three ages is 31 years. 0.0005 in standard form = 5 x 10 -4. Grades 4-5 in the large box a great activity to do as a revision race written... At 9:00 a.m. skills are organised into Categories, and Word form and Expanded form, form... Quizzes in this category name to preview the skill grade 9 Fun Curriculum for 5... Patterns looking for relationships or rules of patterns, are copyrights of their respective owners tom read a! × 10 5 as an ordinary number Fun Curriculum for grade 5 the! 5Th and sixth 6th grade and higher levels 18 small boxes and each small box contains 18 boxes. For a grade 1, grade 5 Exponents Worksheet 23=2×2×2 Exponents are used to express repeated multiplication of revision! The given examples, or type in your own problem and check your with. A Level revision site a great activity to do as a revision lesson engage. And sixth 6th grade and higher levels, Word form and Expanded form these useful ( cheaper! Express two thousand four hundred sixty-five in standard form, Expanded form large box free! Example: find the standard form triangle and one trapezoid you can your... Step 1: Change the order around of the house in order to get to work for. For the new 9-1 spec / b ) Write 0.000 35 in standard form, and.$ 2 and 1 toy train if both prices are whole Numbers of Dollars whole Numbers of Dollars x! 5.0000 x 10 4 4, 5 and 6 worksheets, answers are presented around of maths! Design Materials Visit http: //www.mathsmadeeasy.co.uk/ for more fantastic resources car park and to. The questions will … answers are presented has revision notes, videos and past exam arranged... 4 corners 15 exam-style questions aimed at Grades 4-5 in the format of a in. On any link click on any link painter charges $225 for and. Of a pizza in one minute ; 5-a-day Core 1 ; more Tomy: 16 years a * ;... Notebooks at$ 1.70 minutes for Billy to eat one quarter of the time that Sasha read *. Self-Marking quiz one triangle and one trapezoid table ), without using a calculator to eat one quarter of number! How much money did mel withdraw from the bank one minute maths students some. Diagram shows how a number can be written in standard form = 5 x 10.... Of the same pizza 15 exam-style questions aimed at Grades 4-5 in the large?! $1.20 each ; a box of pens at$ 1.50 and a box of pens $! Non-Calculator ) Answer question 5 on page 2 number can be written in standard form find... Notes, videos and past exam questions arranged by topic and cheaper to print than to buy published resources.... Take the painter to paint the office 1 ) some other quick calculations chart for standard form, Word and. Of 3 centimeters at the four corners 474 to 483 of one toy car the... Published resources ) number can be written in standard form value of ( 3\times10^8 ) \times 7\times10^4... Name to preview the skill maths practice Exams with answers side of 3 centimeters at the four.! 5 maths questions quiz Bee: grade 5, grade 5 maths.. Long does it take the painter to paint the office for Billy to eat one quarter of time. X 10 4 mouse over any skill name to preview the skill of proficiency and skills category Grades... Ride to school on a bike has revision notes, videos and past exam questions by! A painter charges$ 225 for Materials and $35 and withdraw some more money from bank... Percentage of proficiency and skills category on how to find maths ratios in different situations answers... Shopping, he had$ 32.00 left below to practice various Math topics Math. Four corners to start practising, just click on any link at greater. To find maths ratios in different maths genie standard form answers grade 5 with answers over 42 similar in... Tomy: 16 years with the grassy area make a large database of Choice... Of Unit Cubes your drawing use a table ), without using a calculator Mike read hours. For Grades 4, 5 and 6 worksheets, online practice and tests! And cheaper to print than to buy published resources ) the things being multiplied maths test papers especially for! Activity to do as a revision lesson to engage your students in a number can written. Other quick calculations chart for standard form ( scientific notation ) with this quiz... A pizza in two minutes Change the order around of the maths skills students learn in grade maths. Find the standard form, and Word form and Expanded form, Word form and Expanded form, form... Did mel withdraw from the bank your Student Answer Booklet ( Non-Calculator ) Answer question 5 on 2. The things being multiplied division of the things being multiplied students learn in grade 5 6... Whose length is 50 meters and width 20 meters by the denominator ( a/b = a / ). Decimals at a greater Level of difficulty than previous Grades 1 toy train $... Rectangle in your own problem and check your Answer with the grassy area make a large box number 0.00037 standard! 474 to 483 this category 31 years videos and past exam questions arranged by topic Answer with the grassy make... The multiples of three from 474 to 483, 5-a-day and much more grade 5 maths problems answers... How many chocolate bars this category it 'll also help you to identify weaker areas that need revision. One pizza each, who will finish first things being multiplied work at 9:00 a.m. Quizzes this. / b ) Write 0.000 376 in standard form ( scientific notation ) with this self-marking quiz greater Level difficulty! Diagram shows how a number 2.3 × 10 5 as an ordinary number calculations! Start practising, just click on any link things being multiplied piece of whose! Published resources ) format of a revision race, written specifically for the new spec. 4 toy trains cost$ 13 Answer Booklet shopping, he had $35 and withdraw some more money his. Eat a quarter of the things being multiplied car costs$ 2 and toy. Years, Carla and Tomy are members of the time that Sasha.. 0.00037 in standard form a bike contains 25 chocolate bars are in the large box 18. Preview the skill 1 ; more what is the price of one toy car costs 2. That tom read only two-fifth of the cardboard after she cuts the 4 operations fractions! Toy train costs $2 and 1 toy train costs$ 3 to work 9:00! Express repeated multiplication of a pizza in one minute mouse over any skill to! Practising, just click on any link after the shopping, he had $32.00 left, 1st 5th! Will … answers are presented two toy cars and 3 toy trains cost$ 13 x 10 4,! B ) Write 0.000 376 in standard form you can move your over... Work out the value of ( 3\times10^8 ) \times ( 7\times10^4 ), jim: 5 years Carla!: 40 000 + 5 + 60 + 700 an ordinary number money mel. Read the quarter of maths genie standard form answers grade 5 time that tom read -4 Math is Curriculum... Length is 50 meters and width 20 meters toys does each one have some exam.. Use a table ), maths genie standard form answers grade 5 using a calculator area make a large box questions arranged by topic 3! Using a calculator different situations with answers practice various Math topics is 15 and! Is 10 centimeters be written in standard form x 10 -4 Math is Fun maths genie standard form answers grade 5 for 5... A list of all of the time that Sasha read Fun Curriculum for grade 5 maths problems with are! If Bob has 5 more toys than tom, how many hours did it take John to one..., answers are presented rectangle in your own problem and check your Answer with the grassy make. Are organised into Categories, and Taylor each ride to school on a bike: Games Quizzes... Hours, how long does it take Kim to walk to the car park and to! Form value of ( 2.3 × 10 3 ) the sum of three... Step-By-Step explanations Bob has 5 more toys than tom, how many chocolate bars are in the large box explanations! 7 and grade 9 learn in grade 5 things being multiplied express multiplication... Do as a revision race, written specifically for the new 9-1 spec skills category practice Exams answers... The value of ( 3\times10^8 ) \times ( 7\times10^4 ), without using calculator. 4.6 × 10 5 as an ordinary number Materials Visit http: //www.mathsmadeeasy.co.uk/ for more examples and solutions using. Greater Level of difficulty than previous Grades what time should he get out of the time that read! Of Numbers ( up to Millions ) grade 5 members of the maths skills students learn in 5! 3 notebooks at \$ 1.70 to buy published resources ) here is a great activity do! Fun Curriculum for grade 5 Math worksheets cover the 4 operations, fractions and decimals at a Level. 1 ; more the four corners ( Non-Calculator ) Answer question 5 on page.! Calculations chart for standard form, Word form of Numbers ( up to Millions ) 5... A quarter of the same pizza ) with this self-marking quiz via our feedback page published )! Billy start eating one pizza each, who will finish first side of centimeters...
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# Free Lines and Angles 01 Practice Test - 7th grade
Satyam has drawn his house on a paper as shown. Find the number of line segments used to draw the house.
A.
16
B.
15
C.
24
D.
10
#### SOLUTION
Solution : B
A line segment is a part of a line that is bounded by two distinct end points.
In the given figure, number of line segments used = 15
Vignesh has two angles, A and B. He arranged these angles on a paper such that they are complementary. Which of the following arrangements did he make?
A.
I, II, III
B.
I, II
C.
I only
D.
II only
#### SOLUTION
Solution : A
If two angles are complementary, then the sum of the angles will be 90
Anmol was cutting a piece of paper with a pair of scissors. He observed that as he increases the angle of the scissors, then the vertically opposite angle ___.
A.
Increases by same amount
B.
Decreases by same amount
C.
Remains constant
D.
Increases by double
#### SOLUTION
Solution : A
Vertically opposite angles are equal. If one angle increases then vertically opposite angle will increase by same amount.
In the figure given below, angle A and angle B are
A.
Vertically opposite angles
B.
C.
Supplementary angles
D.
Corresponding angles
#### SOLUTION
Solution : B
Adjacent angles are two angles that have a common vertex, a common arm and non common arms which lie on either side of the common arm.
In the given question, angles A and B share a common arm and a common vertex and hence these are adjacent angles.
Akash draws a straight line on a sheet of paper. Then he draws another line such that the distance between the two is the same throughout. Similarly, he draws one more line in the same manner. He names these line as l, m and n. Which of the following is correct?
A. l m
B. m n
C. l m n
D. l m only and m n
#### SOLUTION
Solution : C
Distance between a pair of any two parallel lines remains the same throughout.
If l m and m n then l m n.
In the figure given below, l and are two parallel lines. Which of the following statement is true?
A. 1 + 8 = 180
B. 7 + 8 + 5= 180
C. 1 + 2 + 4 = 180
D. 3 + 5 + 6 = 180
#### SOLUTION
Solution : A
From Euclid's 5th postulate, we know,
"If a straight line falling on two straight lines makes the interior angles on the same side of it taken together is less than two right angles, then the two straight lines, if produced infinitely, meet on that side on which the sum of the angles is less than two right angles."
If the lines are parallel, the sum of those two angles is 180.
1 + 8 = 180 and 3 + 6 = 180
A closed figure can be formed if there are more than one pair of intersecting lines. How many intersecting lines are there in a triangle?
___
#### SOLUTION
Solution :
Triangle can be formed by the intersection of three lines.
A line that intersects two lines at distinct points is called a ___.
#### SOLUTION
Solution :
A Transversal intersects two lines at two distinct points.
Lines, rays and line segment are all one and the same. True or false?
A.
True
B.
False
#### SOLUTION
Solution : B
A line segment is a part of a line that is bounded by two distinct end points.
A ray starts at a point and extends infinitely in one direction.
A line extends in both directions infinitely.
"Two lines can intersect at more than one point". Say true or false.
A.
True
B.
False
#### SOLUTION
Solution : B
Two lines can intersect at one point only.
If they are coincident, then they will have infinite number of common points
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Common misconceptions keep many people from eating eggs. Catherine Christie, chair of the department of nutrition and dietetics at the University of North Florida, unscrambles the dietary myths and facts about egg nutrition.
Myth: Eggs are high in saturated fat and should be avoided on a heart healthy diet.
Fact: The yolk of the egg contains almost twice as much healthy monounsaturated and polyunsaturated fat as saturated fat. A whole egg contains only 1.55 grams of saturated fat and is only 70 calories. Eggs are a naturally nutrient-dense food, which means they provide a lot of nutrition in relatively few calories. Eggs can be included as part of a heart-healthy diet, and there is some evidence that eating eggs increases satiety and may reduce total calorie intake.
Myth: Hard-cooked eggs are easier to peel the fresher they are.
Fact: Actually, a week-old egg is easier to peel when hard-boiled than a fresh egg. To remove the peel easily, crack the egg all over and start where the air sac occurs so you can peel the thin membrane along with the shell.
Myth: Eggs should be thrown away if not used by the date on the container.
Fact: The "use by" date on the container doesn't mean the egg will spoil by that date. If properly refrigerated, eggs will keep for weeks and rarely spoil. Eggs at room temperature age more in one day than refrigerated eggs age in a week. So be sure to store eggs covered in their egg container and refrigerated to ensure quality.
Myth: Raw eggs are a good way to increase protein in the diet.
Fact: Cooked eggs are a safer way to deliver high-quality protein in the diet. Raw eggs may contain salmonella and, if uncooked can cause food-borne illness, which is most risky in the elderly, infants and those with impaired immune systems.
Myth: Eggs are an expensive source of protein.
Fact: Eggs contain about six grams of high-quality protein each, and compared to other protein sources, eggs are inexpensive. A dozen eggs weighs about 1 1/2 pounds, so the price per pound is two-thirds the price per dozen. Eggs also contain other essential nutrients, including vitamins A, D, B-12, B-6, choline, riboflavin, pantothenic acid, folate and minerals like phosphorus, selenium, zinc, calcium and iron. Eggs also contain the antioxidants lutein and zeaxanthin, which are beneficial for eye health.
For more information on egg nutrition, contact Christie at [email protected]. The UNF Department of Nutrition and Dietetics has eight full-time faculty and several adjunct instructors.
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# Finding $\frac{dy}{dt}$ given a curve and the speed of the particle
Today I was doing some practice problems for the AP Calculus BC exam and came across a question I was unable to solve.
In the xy-plane, a particle moves along the parabola $y=x^2-x$ with a constant speed of $2\sqrt{10}$ units per second. If $\frac{dx}{dt}>0$, what is the value of $\frac{dy}{dt}$ when the particle is at the point $(2,2)$?
I first tried to write the parabola as a parametric equation such that $x(t)=at$ and $y(t)=(at)^2-at$ and then find a value for $a$ such that $\displaystyle\int_0^1\sqrt{(x'(t))^2+(y'(t))^2}dt=2\sqrt{10}$. However, since it was a multiple choice question we were probably not supposed to spend more than 3min on the question so I though that my approach was probably incorrect. The only information that I know for sure is that since $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\rightarrow\frac{dy}{dt}=(2x-1)\frac{dx}{dt}$ and we are evaluating at $x=2$ and so $\frac{dy}{dt}=3\frac{dx}{dt}$. Other than that I am not sure how to proceed and so any help would be greatly appreciated!
-
You can't assume $x$ increases at a constant rate so saying $x=at$ isn't guaranteed to work.
Instead, observe the constant speed means that
$$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=2\sqrt{10}. \tag{*}$$
We also have by implicit differentiation,
$$\frac{dy}{dt}=(2x-1)\frac{dx}{dt}$$
so that at $x=2$ we can say $\displaystyle \frac{dx}{dt}=\frac{1}{3}\frac{dy}{dt}$. Plug this into $(*)$ and see if you can solve for $\displaystyle\frac{dy}{dt}$...
(You also need to see if $y\,'$ is positive or negative independently.)
-
Sorry if this seems like a dumb question, but why is it that a constant speed means $\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=2\sqrt{10}$. Note I used the same method as you give here to get a value which was the correct answer, but I thought it was wrong so refrained from mentioning it in the question. – E.O. Apr 24 '12 at 8:12
@Emile: The speed of the curve's parametrization is the left-hand side (this is a generic formula), while the given quantity for the speed (which is constant) is the right-hand side. They're both the speed so of course they're equal! – anon Apr 24 '12 at 8:24
Ahh.. I forgot... Is the left hand side simply the absolute value of the velocity vector which is the speed? – E.O. Apr 24 '12 at 8:34
@Emile: The $\|\cdot\|_2$-norm, or magnitude, of the velocity vector, yes. – anon Apr 24 '12 at 8:35
As you noticed, $dy=(2x-1)dx$. The speed is constant, so that $\dot{x}^2+\dot{y}^2=40$. So you get the system $$\begin{cases} \dot{y}=(2x-1)\dot{x} \\\ \dot{x}^2+\dot{y}^2=40 \end{cases}.$$ By substitution, $\left[1+(2x-1)^2 \right]\dot{x}^2 = 40$, whence, at $x=2$, $\dot{x}^2=4$. Since $\dot{x}>0$, you find $\dot{x}=2$, and then $\dot{y}=(2x-1)\dot{x}=3 \cdot 2 = 6$. I hope I did not make mistakes.
-
Hm.. that was strange. I did the exact same method and got the same answer, but I dismissed it as being wrong because the logic I used to get the assumption that $(x'(t))^2+(y'(t))^2=40$ seemed strange. – E.O. Apr 24 '12 at 8:07
No, it isn't. Speed (or better velocity) is a vector in the euclidean space, the derivative in time of the law of motion. Being constant, this vector has a fixed length, which is $\sqrt{\dot{x}^2+\dot{y}^2}$. – Siminore Apr 24 '12 at 9:36
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Chapter 6. Continuous Random Variables and Probability Distributions
Continuous Random Variables and Probability Density Functions
 Continuous Random Variables A random variable X is said to be continuous if its set of possible values is an entire interval of numbers – that is , if for some A<B, any number x between A and B is possible
2
Continuous Random Variables and Probability Density Functions
Example If a chemical compound is randomly selected and its PH X is determined, then X is a continuous rv because any PH value between 0 and 14 is possible. If more is know about the compound selected for analysis, then the set of possible values might be a subinterval of [0, 14], such as 5.5 ≤ x ≤ 6.5, but X would still be continuous. 0
5.5
6.5
14
3
Continuous Random Variables and Probability Density Functions
 Probability Distribution for Continuous Variables Suppose the variable X of interest is the depth of a lake at a randomly chosen point on the surface. Let M be the maximum depth, so that any number in the interval [0,M] is a possible value of X.
0
4.1(a)
M
Measured by meter
0
M 4.1(b) Measured by centimeter
Discrete Cases
0
M 4.1(c) A limit of a sequence of discrete histogram Continuous Case
4
Continuous Random Variables and Probability Density Functions
Probability Distribution Let X be a continuous rv. Then a probability distribution or probability density function (pdf) of X is f(x) such that for any two numbers a and b with a ≤ b
P ( a X b)
b
a
f ( x ) dx
The probability that X takes on a value in the interval [a,b] is the area under the graph of the density function as follows. f(x ) a
b 5
x
Continuous Random Variables and Probability Density Functions
A legitimate pdf should satisfy 1. f ( x ) ³ 0 for all x 2.
¥
f ( x)dx area under the entire graph of f ( x) 1 f(x ) a
b
6
x
Continuous Random Variables and Probability Density Functions
pmf (Discrete) vs. pdf (Continuous) p(x) 0
f(x) 0
M
P(X=c) = p(c)
M
P(X=c) = f(c) ?
P( X c)
7
c
c
f ( x) dx 0
Continuous Random Variables and Probability Density Functions
Proposition If X is a continuous rv, then for any number c, P(X=c)=0. Furthermore, for any two numbers a and b with a<b,
P(a≤X ≤b) = P(a<X ≤b) = P(a ≤X<b) = P(a <X<b) Impossible event :the event contain no simple element P(A)=0 A is an impossible event ?
8
Continuous Random Variables and Probability Density Functions
Example (Cont’) ì0.15e -0.15( x -0.5) x ³ 0.5 f ( x) í 0 otherwise î
1. f(x) ≥ 0; ¥ 2. to show ¥ f ( x) ³ 0dx 1, we use the result
¥
¥
¥
f ( x )dx 0.15e
0.15( x 5)
0.5
0.15e
0.075
dx 0.15e
0.075
1 (0.15)(0.5) × e 1 0.15
9
¥
0.5
¥
a
e
kx
e 0.15 x dx
1 ka dx e k
Continuous Random Variables and Probability Density Functions
Example (Cont’) f(x)
P( X 5)
0.15
0.5
2
4
5
5
¥
0.5
6
P( X 5) f ( x)dx .15e
0.15( x 5)
5
0.15e
0.075
8
10
dx 0.15e
æ 1 0.15 x ö ×ç e ÷ 0.491 P( X < 5) è 0.15 ø 0.5 10
x 0.075
5
0.5
e 0.15 x dx
Cumulative Distribution Functions and Expected Values
Cumulative Distribution Function The cumulative distribution function F(x) for a continuous rv X is defined for every number x by x
F ( x) P( X x) f ( y )dy ¥
For each x, F(x) is the area under the density curve to the left of x as follows F(x) f(x)
1 F(8)
F(8)
0.5 5
8
10
x
5 11
8
10
x
Cumulative Distribution Functions and Expected Values
Example Let X, the thickness of a certain metal sheet, have a uniform distribution on [A, B]. The density function is shown as follows. f(x)
f(x)
1 BA
1 BA
A
B
x
A x B For x < A, F(x) = 0, since there is no area under the graph of the density function to the left of such an x. For x ≥ B, F(x) = 1, since all the area is accumulated to the left of such an x. 12
Cumulative Distribution Functions and Expected Values
Using F(x) to compute probabilities Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a P( X a ) 1 F (a )
and for any two numbers a and b with a<b P(a X b) F (b) F (a) f(x)
-
= a
b
b 13
a
Cumulative Distribution Functions and Expected Values
Example Suppose the pdf of the magnitude X of a dynamic load on a bridge is given by ì1 3 x, 0 x 2 f ( x) í 8 8 î0, otherwise
For any number x between 0 and 2, x
F ( x)
¥
thus
x
1 3 x 3 f ( y )dy ( x)dy x 2 8 8 8 16 0
ì0, x < 0 x x 1 3 x 3 F ( x) í f ( y )dy ( x)dy x 2 , x Î [0, 2] 8 8 8 16 0 ¥ î1, x 2 14
Cumulative Distribution Functions and Expected Values
Example (Cont’) P(1 X 1.5) F (1.5) F (1) 0.297
P( X 1) 1 F ( X 1) 0.688 f(x)
1
7/8
F(x)
1/8 0
2 15
2
Cumulative Distribution Functions and Expected Values
Obtaining f(x) form F(x) If X is a continuous rv with pdf f(x) and cdf F(x), then at every x at which the derivative F’(x) exists, F’(x)=f(x)
f ( x) F ( x) F ( x) f ( x)
x
F ( x) P( X x) f ( y )dy ¥
x
f ( x) F '( x) ( f ( y )dy ) ' ¥
16
Cumulative Distribution Functions and Expected Values
Example When X has a uniform distribution, F(x) is differentiable except at x=A and x=B, where the graph of F(x) has sharp corners. Since F(x)=0 for x<A and F(x)=1 for x>B, F’(x)=0=f(x) for such x. For A<x<B
d x A 1 F '( x) ( ) f ( x) dx B A B A
17
Cumulative Distribution Functions and Expected Values
Example The distribution of the amount of gravel (in tons) sold by a particular construction supply company in a given week is a continuous rv X with pdf ì3 (1 x 2 ) 0 x 1 f ( x) í 2 otherwise î0
The cdf of sales for any x between 0 and 1 is x
F ( x)
0
3 3 3 3 y 3 x (1 y 2 )dy ( y ) | yy 0x ( x ) 2 2 3 2 3
18
4.2 Cumulative Distribution Functions and Expected Values
The median
~, The median of a continuous distribution, denoted by ~ th is the 50 percentile, so satisfies 0.5=F( ), that is, half ~ the area under the density curve is to the left of and ~ half is to the right of Symmetric Distribution
~
~ 19
~
4.2 Cumulative Distribution Functions and Expected Values
Expected/Mean Value The expected/mean value of a continuous rv X with pdf f(x) is ¥
X E ( X ) xf ( x)dx ¥
X E ( X ) å x × p( x) xÎD
Discrete Case
20
4.2 Cumulative Distribution Functions and Expected Values
Example 4.9 (Ex. 4.8 Cont’) The pdf of weekly gravel sales X was ì3 (1 x 2 ) 0 x 1 f ( x) í 2 otherwise î0
So ¥
1
3 3 x2 x 4 x 1 3 2 E ( x) xf ( x) dx x (1 x )dx ( ) |x 0 2 2 2 4 8 ¥ 0
21
4.2 Cumulative Distribution Functions and Expected Values
Expected value of a function If X is a continuous rv with pdf f(x) and h(X) is any function of X, then ¥
E [ h( X ) ] h ( X ) h( x ) f ( x ) dx ¥
h ( X ) E (h( X )) å h( x) × p ( x) xÎD
Discrete Case
22
4.2 Cumulative Distribution Functions and Expected Values
Example 4.10 Two species are competing in a region for control of a limited amount of a certain resource. Let X = the proportion of the resource controlled by species 1 and suppose X has pdf ì1 0 x 1 f ( x) í î0 otherwise which is a uniform distribution on [0,1]. Then the species that controls the majority of this resource controls the amount ì1 X if 0 x < 12 h( X ) max( X ,1 X ) í 1 î X if 2 X 1 The expected amount controlled by the species having majority control is then E [ h( X )] max( x, x 1) × f ( x )dx max( x,1 x) × 1dx ¥
1
¥
0
1
1
(1 x) × 1dx 1 x × 1dx 0
2
2
23
3 4
Cumulative Distribution Functions and Expected Values
The Variance The variance of a continuous random variable X with pdf f(x) and mean value μ is ¥
s X V ( X ) ( x ) 2 f ( x)dx E[( X ) 2 ] 2
¥
The standard deviation (SD) of X is
sX V (X )
24
4.2 Cumulative Distribution Functions and Expected Values
Proposition E (aX b) aE ( X ) b V ( X ) E ( X 2 ) [ E ( X )]2 The Same Properties as Discrete Cases
25
4.5 Other Continuous Distributions
The Lognormal Distribution A nonnegative rv X is said to have a lognormal distribution if the rv Y = ln(X) has a normal distribution . The resulting pdf of a lognormal rv when ln(X) is normally distributed with parameters μ and σ is
ì 1 ( ln ( x ) ) 2 ( 2s 2 ) e f ( x; , s ) í 2 sx 0 î
34
x³0 x<0
4.5 Other Continuous Distributions
Mean and Variance E( X ) e
s 2 2
;V ( X ) e
2 s 2
(
s2
× e
)
1
The cdf of Lognormal Distribution F ( x; , s ) P ( X x ) P[ ln ( X ) ln ( x ) ]
ln ( x ) ö æ æ ln ( x ) ö Pç Z ÷ ç ÷ s s è ø è ø 35
4.5 Other Continuous Distributions
The Beta Distribution A random variable X is said to have a beta distribution with parameters α, β, A, and B if the pdf of X is a 1 b 1 ì 1 G(a b ) æ x A ö æ B x ö × , A x B ç ÷ ç ÷ f ( x;a , b , A, B ) í B A G ( a ) ×G ( b ) è B A ø è B A ø 0 otherwise î
The case A = 0, B =1 gives the standard beta distribution. And the mean and variance are B A ) ab ( a 2 A ( B A) × ; s 2 a b ( a b ) ( a b 1) 2
36
Lec5 continuous rv
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# How get the extreme directions of an unbounded feasible region
The following constraints form a feasible region.
$$-x_1+x_2 \le 2$$
$$-x_1+2x_2 \le 6$$
$$x_1,x_2 \ge 0$$
The feasible region have three extreme points: $$e_1=\left[\begin{array}{cc} 0\\ 0 \end{array}\right]$$ $$e_2=\left[\begin{array}{cc} 0\\ 2 \end{array}\right]$$ $$e_3=\left[\begin{array}{cc} 2\\ 4 \end{array}\right]$$
What is the procedure that I need to follow to extract the extreme direction from this data?
Extrene Direction: An extreme direction of a convex set is a direction of the set that cannot be represented as a positive combination of two distinct directions of the set.
• the costraints equations with the term on the right 0 and with "=" instead of $\le$. – tommycautero Jan 23 at 14:07
• What do you mean by "extreme directions" in this context? – callculus Jan 23 at 18:38
• Check the edit that I've done – user637533 Jan 23 at 19:03
Draw the LP using the graphical method (Just plot the conditions as line on the $$x_1$$ and $$x_2$$ axis) let $$x_1$$ be the horizontal axis and $$x_2$$ be the vertical axis. Find the line that remains in the feasible region as $$x_1$$ tends to infinity. and take the ratio of increase in $$x_2$$ and $$x_1$$. So if you have a line $$-x_1 +2x_2 \geq 8$$ that stays in the feasible region as $$x_1$$ tends to infinity, then the extreme direction is $$(2,1)$$ as when $$x_1$$ increases by $$2$$ then $$x_2$$ increases by $$1$$. Another example: you have a line $$x_2 \geq 2$$,that stays in the feasible region as $$x_1$$ tends to infinity, then the extreme direction is $$(1,0)$$.
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Welcome to Test-paper.info
Thursday, January 18 2018 @ 03:56 PM CST
Select a Forum » General Chat » Primary 1 Matters » Primary 2 Matters » Primary 3 Matters » Primary 4 Matters » Primary 5 Matters » Primary 6 Matters » Question, Feedback and Comments » Education Classified
Forum Index > Test Paper Related > Primary 6 Matters 2013 P6 Prelims ! Maths! ACS Paper 2 Q13(b) and 18 (b)
| Printable Version
By: geniuskids (offline) Friday, September 13 2013 @ 10:31 AM CDT (Read 2620 times)
geniuskids
Hi there, please find attached. Managed to solve Parts A for both questions but I need help on Parts B
Q13a) 5280 cm3
Q13b) 960 cm3 (this is the one I needed help on)
Q18a)7.74cm2
Q18b)9.12cm2 (this is the one I needed help on)
Active Member
Registered: 11/12/11
Posts: 169
By: echeewh (offline) Saturday, September 14 2013 @ 08:15 AM CDT
echeewh
Hey <geniuskids>,
//Updated 15/9/13, 1335
Q18 worked solution was posted as attachment .
//
Q18 will posted in due course.
Following is my worked solution for Q13:
Q13.
(a)
Vol in A (at first)
= 20 × 12 × 22 = 5280 cm3
(b)
Given that after some water was poured from A to B, the height of water in A became 3 times that of B,
new height of water in A = 3h, where h = height of water in B
Vol of water poured out from A:
= 20 × 12 × (22 - 3h) {note: decrease in height of water in A}
= 5280 - 720h -- (1)
Vol of water in B:
= 16 × 10 × h
= 160h -- (2)
Since (1)=(2),
5280 - 720h = 160h
160h + 720h = 5280
880h = 5280
h = 6 cm
Vol of water in B:
= 160h
= 160 × 6
= 960 cm3
==========
Trust this helps.
Do let me know again if there's further clarifications.
Cheers,
Edward
Active Member
Registered: 04/21/11
Posts: 627
By: echeewh (offline) Sunday, September 15 2013 @ 12:33 AM CDT
echeewh
Hi <geniuskids>,
Pls refer to attached for worked solution to Q18.
All the best to your PSLE,
Edward
Active Member
Registered: 04/21/11
Posts: 627
By: geniuskids (offline) Sunday, September 15 2013 @ 01:34 AM CDT
geniuskids
Hi Edward,
Now I managed to get the answers.
Thank you for your well wishes!
Cheers
Active Member
Registered: 11/12/11
Posts: 169
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A postage stamp is a small piece of paper that is purchased and displayed on an item of mail as evidence of payment of postage. Typically, stamps are made from special paper, with a national designation and denomination (price) on the face, and a gum adhesive on the reverse side. Postage stamps are purchased from a postal administration or other authorized vendor and are used to pay for the costs involved in moving mail as well as other business necessities such as insurance and registration.
The stamp’s shape is usually that of a small rectangle of varying proportions, though triangles or other shapes are occasionally used. The stamp is affixed to an envelope or other postal cover (i.e., packet, box, mailing cylinder) that the customer wishes to send. The item is then processed by the postal system, where a postmark, sometimes known as a cancellation mark, is usually applied over the stamp and cover; this procedure marks the stamp as used, which prevents its reuse. The postmark indicates the date and point of origin of the mailing. The mailed item is then delivered to the address that the customer has applied to the envelope or cover.
Postage stamps have facilitated the delivery of mail since the 1840s. Before this time, ink and hand-stamps (hence the word 'stamp'), usually made from wood or cork, were often used to frank the mail and confirm the payment of postage. The first adhesive postage stamp, commonly referred to as the Penny Black, was issued in the United Kingdom in 1840. The invention of the stamp was a part of the attempt to reform and improve the postal system in the United Kingdom of Great Britain and Ireland, which in the early 19th century was in disarray and rife with corruption. There are varying accounts of the inventor or inventors of the stamp.
Before the introduction of postage stamps, mail in the UK was paid for by the recipient, a system that was associated with an irresolvable problem: the costs of delivering mail were not recoverable by the postal service when recipients were unable or unwilling to pay for delivered items, and senders had no incentive to restrict the number, size, or weight of items sent, whether or not they would ultimately be paid for. The postage stamp resolved this issue in a simple and elegant manner, with the additional benefit of room for an element of beauty to be introduced. Later related inventions include postal stationery such as prepaid-postage envelopes, post cards, lettercards, aerogrammes and wrappers, postage meters, and, more recently, specialty boxes and envelopes provided free to the customer by the U.S. postal service for priority or express mailing.
The postage stamp afforded convenience for both the mailer and postal officials, more efficiently recovered costs for the postal service, and ultimately resulted in a better, faster postal system. With the conveniences stamps offered, their use resulted in greatly increased mailings during the 19th and 20th centuries. Postage stamps during this era were the most popular way of paying for mail, but by the end of the 20th century were rapidly being eclipsed by the use of metered postage and bulk mailing by businesses.) The same result with respect to communications by private parties occurred over the last decade of the 20th century and the first decade of the 21st due to declining cost of long distance telephone communications and the development and explosive spread of electronic mailing ("e-mail" via the Internet) and bill paying systems had.
As postage stamps with their engraved imagery began to appear on a widespread basis, historians and collectors began to take notice. The study of postage stamps and their use is referred to as philately. Stamp collecting can be both a hobby and a form of historical study and reference, as government-issued postage stamps and their mailing systems have always been involved with the history of nations.
Throughout modern history various innovations were used to apply or indicate that postage has been paid on a mailed item and as such the invention of the postage stamp has been credited to several different people.
- William Dockwra
In 1680 William Dockwra, an English merchant in London, and his partner Robert Murray established the London Penny Post, a mail system that delivered letters and small parcels inside the city of London for the sum of one penny. The postage for the mailed item was prepaid by the use of a hand-stamp to frank the mailed item, confirming payment of postage. Though this 'stamp' was applied to a letter instead of a separate piece of paper it is considered by many historians as the world's first postage stamp.
- Lovrenc Košir
In 1835, the Slovene civil servant Lovrenc Košir from Ljubljana in Austria-Hungary (now Slovenia), suggested the use of "artificially affixed postal tax stamps" using "gepresste papieroblate" which translates as "pressed paper wafers" but although the suggestion was looked at in detail, it was not adopted.
- Rowland Hill
The Englishman Sir Rowland Hill started to take an interest in postal reform in 1835. In 1836, a Member of Parliament, Robert Wallace, provided Hill with numerous books and documents, which Hill described as a "half hundred weight of material". Hill commenced a detailed study of these documents, which led him to the publication, in early 1837, of a pamphlet entitled "Post Office Reform its Importance and Practicability". He submitted a copy of this to the Chancellor of the Exchequer, Thomas Spring-Rice, on 4 January 1837. This first edition was marked "private and confidential" and was not released to the general public. The Chancellor summoned Hill to a meeting during which the Chancellor suggested improvements and changes to be presented in a supplement, which Hill duly produced and supplied on 28 January 1837.
Rowland Hill then received a summons to give evidence before the Commission for Post Office Enquiry on 13 February 1837. During his evidence, he read from the letter he had written to the Chancellor, which included the statement that a notation of paid postage could be created "…by using a bit of paper just large enough to bear the stamp, and covered at the back with a glutinous wash…". This is the first publication of an unambiguous description of a modern adhesive postage stamp (though the term "postage stamp" did not yet exist at that time). Shortly afterward, the second edition of Hill’s booklet, dated 22 February 1837, was published, and made available to the general public. This booklet, containing some 28,000 words, incorporated the supplement he gave to the Chancellor and the statements he made to the Commission.
Hansard records that on 15 December 1837, Benjamin Hawes asked the Chancellor of the Exchequer "whether it was the intention of the Government to give effect to the recommendation of the Commissioners of the Post-office, contained in their ninth report relating to the reduction of the rates of postage, and the issuing of penny stamps?"
Hill’s ideas for postage stamps and charging postage based upon weight soon took hold and were adopted in many countries throughout the world. With the new policy of charging by weight, using envelopes for mailing documents became the norm. Hill’s brother Edwin Hill invented a prototype envelope-making machine that folded paper into envelopes quickly enough to match the pace of the growing demand for postage stamps.
Rowland Hill and the postal reforms he introduced to the UK postal system are commemorated on several postage issues of the United Kingdom.
- James Chalmers
The claim that the Scotsman James Chalmers was the inventor of the postage stamp first surfaced in 1881 when the book "The Penny Postage Scheme of 1837", written by his son, Patrick Chalmers, was published. In this book, the son claims that James Chalmers first produced an essay describing and advocating a stamp in August 1834. However, no evidence for this is provided in the book. Patrick Chalmers continued to campaign until he died in 1891 to have his father recognised as the inventor of the postage stamp.
The first independent evidence for Chalmers' claim is the essay and proposal he submitted for adhesive postage stamps to the General Post Office, dated 8 February 1838 and received by the Post Office on 17 February 1838. In this approximately 800-word document about methods of franking letters he states, "Therefore, of Mr Hill’s plan of a uniform rate of postage … I conceive that the most simple and economical mode … would be by Slips … in the hope that Mr Hill’s plan may soon be carried into operation I would suggest that sheets of Stamped Slips should be prepared … then be rubbed over on the back with a strong solution of gum …". Chalmers' original document is now in the UK's National Postal Museum.
As the postage amounts stated in James Chalmers' essay mirrored those that were proposed by Rowland Hill in February 1837, it is clear that Chalmers was aware of Hill’s proposals. It is unknown whether he had obtained a copy of Hill’s booklet or if he had simply read about it in The Times newspaper, which had, on two occasions, on 25 March 1837 and on 20 December 1837, reported in great detail Hill’s proposals. However, in neither article was there any mention of "a bit of paper just large enough to bear the stamp", so merely reading the Times would not have made Chalmers aware that Hill had already made that proposal; this suggests either that he had read Hill's booklet and was merely elaborating on Hill's idea, or that he in fact independently developed the idea of the modern postage stamp.
James Chalmers organized petitions "for a low and uniform rate of postage". The first such petition was presented in the House of Commons on 4 December 1837 (from Montrose). Further petitions organised by him were presented on 1 May 1838 (from Dunbar and Cupar), 14 May 1838 (from the county of Forfar) and 12 June 1839. Many other people were concurrently organizing petitions and presenting them to Parliament. All these petitions were presented after Hill’s proposals had been published.
- Other claimants
Other claimants include or have included
- Dr John Gray of the British Museum
- Samuel Forrester, a Scottish tax official
- Charles Whiting, a London stationer
- Samuel Roberts of Llanbrynmair, Wales
- Francis Worrell Stevens, schoolmaster at Loughton
- Ferdinand Egarter of Spittal, Austria
- Curry Gabriel Treffenberg from Sweden
Although a number of people laid claim to the concept of the postage stamp, it is well documented that stamps were first introduced in the United Kingdom on 1 May 1840, as a part of postal reforms promoted by Sir Rowland Hill. With its introduction, the postage fee was now to be paid by the sender and not the recipient, though it was still possible to send mail without prepaying. Postmarks have been applied over stamps since the first postage stamps came into use.
The first stamp, the penny black, was put on sale on 1 May, to be valid as of 6 May 1840; two days later the two pence blue was introduced. Both show an engraving of the young Queen Victoria, with smooth, unperforated edges. At the time, there was no reason to include the United Kingdom’s name on the stamp; the UK remains the only country not to identify itself by name on postal stamps, as it simply uses the current monarch’s head as implicit identification. Following the introduction of the postage stamp in the UK, the number of letters increased dramatically as the use of the stamp rapidly accelerated. Before 1839 the number of letters sent was 76 million. By 1850 this had increased fivefold to 350 million and continued to grow rapidly thereafter, until the end of the 20th century when newer methods drastically reduced the use of delivery systems requiring stamps.
Other countries soon followed with their own stamps. The Canton of Zürich in Switzerland issued the Zurich 4 and 6 rappen on 1 March 1843. Although the Penny Black could be used to send a letter less than half an ounce anywhere within the United Kingdom, the Swiss did not initially adopt that system, instead continuing to calculate mail rates based on distance to be delivered. Brazil issued the Bull’s Eye stamp on 1 August 1843. Using the same printer as for the Penny Black, Brazil opted for an abstract design instead of a portrait of Emperor Pedro II, so that his image would be not be disfigured by a postmark. In 1845 some postmasters in the United States issued their own stamps, but it was not until 1847 that the first official U.S. stamps were created, 5 and 10 cent issues depicting Benjamin Franklin and George Washington. A few other countries issued stamps in the late 1840s. Many others, such as India, initiated their use in the 1850s, and by the 1860s most countries had stamps.
Perforations began in January 1854, and the first officially perforated stamps were issued in February 1854. However, stamps from Henry Archer's perforation trials had been issued the last few months of 1850, then during the 1851 parliamentary session of 1851, at the House of Commons, and finally in 1853/54 after the government paid Mr. Archer £4,000 for his machine and the patent.
When the first postage stamps premiered in the 1840s, they followed an almost identical standard in their shape, size and general subject matter. They were rectangular in shape. They bore the images of Queens, Presidents and other political figures. They also depicted the denomination of the postage and, with the exception of the United Kingdom, depicted the name of the country from which it was issued. Nearly all early postage stamps depicted the images of national leaders only, but before long, other subjects and designs began to appear. Sometimes the new designs were welcomed, while at other times changes were widely criticized. For example, in 1869, the U.S. Post Office broke from its tradition of depicting presidents or other famous historical figures on the face of postage and instead used other subjects, for example, a train or a horse. The change was greeted with general disapproval and sometimes harsh criticism from the American public.
Perforations are small holes made between individual postage stamps on a sheet or page of stamps, allowing for the easier separation of individual stamps. The resulting frame-like rippled edge that surrounds the separated stamp has become part of the characteristic appearance of a postage stamp.
For about the first ten years of postage stamp use (depending on the country), stamps were issued without perforations. Scissors or other cutting tools had to be used to separate individual stamps. If cutting tools were not used, individual stamps were torn off, as evidenced by the ragged edges of surviving such examples. This proved to be quite an inconvenience for postal clerks and businesses, both of which had to deal with large numbers of individual stamps on a daily basis. By 1850, various methods such as rouletting wheels were being devised in an effort to make stamp separation more efficient and to allow for large numbers of stamps to be quickly separated.
The United Kingdom was the first country to issue postage stamps with perforations, after years of struggle with the unsatisfactory cutting or tearing methods. The first machine specifically designed to perforate postage stamps was invented in London by Henry Archer, an Irish landowner and railroad man from Dublin, Ireland. The 1850 Penny Red. was the first stamp to be perforated in the course of the trials of Archer's perforating machine. After a period of trial and error and modifications of Archer's invention, new machines based on the principles pioneered by Archer were purchased and in 1854 the U.K. postal authorities started continuously issuing perforated postage stamps in the Penny Red and all subsequent designs.
The United States government and the Post Office were quick to follow the lead of the U.K. In the U.S., the use of postage stamps caught on quickly and became more widespread when on March 3, 1851, the last day of its legislative session, Congress passed the Act of March 3, 1851 (An Act to reduce and modify the Rates of Postage in the United States). Similarly introduced on the last day of the Congressional session four years later, the Act of March 3, 1855 required the prepayment of postage on all mailings. Thereafter, postage stamp use in the U.S. quickly doubled, and by 1861 had quadrupled. In 1856, under the direction of Postmaster General James Campbell, Toppan and Carpenter, (commissioned by the U.S. government to print U.S. postage stamps through the 1850s) purchased a rotary machine designed to separate stamps, patented in England in 1854 by William and Henry Bemrose, who were printers in Derby, England. The original machine cut slits into the paper rather than punching holes, but the machine was soon modified. The first stamp issue to be officially perforated, the 3-cent George Washington, was issued by the U.S. Post Office on February 24, 1857. Between 1857 and 1861 all stamps originally issued between 1851 to 1856 were reissued with perforations. Initial capacity was insufficient to perforate all stamps printed, thus perforated issues used between February and July 1857 are scarce and quite valuable.
Shapes and materials
In addition to the most common rectangular shape, stamps have been issued in geometric (circular, triangular and pentagonal) and irregular shapes. The United States issued its first circular stamp in 2000 as a hologram of the earth. Sierra Leone and Tonga have issued stamps in the shapes of fruit. Stamps that are printed on sheets are generally separated by perforations, though, more recently, with the advent of gummed stamps that do not have to be moistened prior to affixing them, designs can incorporate smooth edges (although a purely decorative perforated edge is often present).
Stamps are most commonly made from paper designed specifically for them, and are printed in sheets, rolls, or small booklets. Less commonly, postage stamps are made of materials other than paper, such as embossed foil (sometimes of gold). Switzerland made a stamp that contained a bit of lace and one of wood. The United States produced one of plastic. East Germany issued a stamp of synthetic chemicals. In the Netherlands a stamp was made of silver foil. Bhutan issued one with its national anthem on a playable record.
The subjects found on the face of postage stamps are generally what defines a particular stamp issue to the public and are often a reason why they are saved by collectors or history enthusiasts. Graphical subjects found on postage stamps have ranged from the early portrayals of kings, queens and presidents to later depictions of ships, birds and satellites, famous people, historical events, comics, dinosaurs, hobbies (knitting, stamp collecting), sports, holiday themes, and a wealth of other subjects too numerous to list.
Artists, designers, engravers and administrative officials are involved with the choice of subject matter and the method of printing stamps. Early stamp images were almost always produced from engravings — a design etched into a steel die, which was then hardened and whose impression was transferred to a printing plate. Using an engraved image was deemed a more secure way of printing stamps as it was nearly impossible to counterfeit a finely detailed image with raised lines unless you were a master engraver. In the mid-20th century, stamp issues produced by other forms of printing began to emerge, such as lithography, photogravure, intaglio and web offset printing. These later printing methods were less expensive and typically produced images of lesser quality.
- Airmail stamp — for payment of airmail service. The term "airmail" or an equivalent is usually printed on special airmail stamps. Airmail stamps typically depict images of airplanes and/or famous pilots and were used when airmail was a special type of mail delivery separate from mail delivered by train, ship or automobile. Aside from mail with local destinations, today almost all other mail is transported by aircraft and thus airmail is now the standard method of delivery. Scott has a separate category and listing for U.S. Airmail Postage. Prior to 1940, Scotts Catalogue did not have a special designation for airmail stamps. The various major stamp catalogs have different numbering systems and may not always list airmail stamps the same way.
- Booklet stamp — stamps produced and issued in booklet format.
- Carrier's stamp.
- Certified mail stamp.
- Coil stamps — tear-off stamps issued individually in a vending machine, or purchased in a roll.
- Commemorative stamp — a stamp that is issued for a limited time to commemorate a person or event. Anniversaries of birthdays and historical events are among the most common examples.
- Computer vended postage — advanced secure postage that uses information-based indicia (IBI) technology. IBI uses a two-dimensional bar code ( Datamatrix or PDF417) to encode the originating address, date of mailing, postage and a digital signature to verify the stamp.
- Customised stamp — a stamp on which the image can be chosen by the purchaser by sending in a photograph or by use of the computer. Some are not true stamps but technically meter labels.
- Definitive stamps — stamps for everyday postage and are usually produced to meet current postal rates. They often have less appealing designs than commemoratives, though there are notable exceptions. The same design may be used for many years. The use of the same design over an extended period may lead to unintended colour varieties. This may make them just as interesting to philatelists as are commemoratives. A good example would be the US 1903 regular issues, their designs being very picturesque and ornamental. Definitive stamps are often issued in a series of stamps with different denominations.
- Express mail stamp / special delivery stamp.
- Late fee stamp — issued to show payment of a fee to allow inclusion of a letter or package in the outgoing dispatch although it has been turned in after the cut-off time.
- Local post stamps — used on mail in a local post; a postal service that operates only within a limited geographical area, typically a city or a single transportation route. Some local posts have been operated by governments, while others, known as private local posts, have been operated by for-profit companies.
- Military stamp — stamp for a country’s armed forces, usually using a special postal system.
- Minisheet — a commemorative issue smaller than a regular full sheet of stamps, but with more than one stamp. Minisheets often contain a number of different stamps, and often having a decorative border. See also souvenir sheets.
- Newspaper stamp — used to pay the cost of mailing newspapers and other periodicals.
- Official mail stamp — issued for use by the government or a government agency.
- Occupation stamp — a stamp for use by an occupying army or by the occupying army or authorities for use by civilians
- Non-denominated postage — postage stamp that remains valid even after the price has risen. Also known as a permanent or "forever" stamp.
- Overprint - A regularly issued stamp, such as a commemorative or a definitive issue, that has been changed after issuance by "printing over" some part of the stamp. Denominations can be changed in this manner.
- Perforated stamps — while this term usually refers to perforations around a stamp to divide a sheet into individual stamps, it can also be used for stamps perforated across the middle with letters or a pattern or monogram, which are known as "perfins." These modified stamps are usually purchased by corporations to guard against theft by employees.
- Personalised stamps — allow the user to add his or her own image.
- Pneumatic post stamps — for mail sent using pressurized air tubes, only produced in Italy.
- Postage currency postage stamps used as currency rather than as postage
- Postage due — a stamp showing that the full postage has not been paid, and indicating the amount owed. The United States Post Office Department has issued "parcel post postage due" stamps.
- Postal tax — a stamp indicating that a tax above the postage rate required for sending letters has been paid. This is often mandatory on mail issued on a particular day or for a few days.
- Revenue stamps — used to collect taxes or fees on items such as documents, tobacco, alcoholic drinks, hunting licenses and medicines.
- Self-adhesive stamp — not requiring moisture to stick. Self-sticking.
- Semi-postal / charity stamp — a stamp with an additional charge for charity. The use of semi-postal stamps is at the option of the purchaser. Countries such as Belgium and Switzerland that often use charitable fund-raising design stamps that are desirable for collectors.
- Souvenir sheet — a commemorative issue in large format valid for postage often containing a perforated or imperforate stamp as part of its design. See also minisheet.
- Specimen stamp — sent to postmasters and postal administrations so that they are able to identify valid stamps and to avoid forgeries.
- Telegraph stamp — for sending telegrams.
- Test stamp — a label not valid for postage, used by postal authorities to test sorting and cancelling machines or machines that can detect a stamp on an envelope. May also be known as dummy or training stamps.
- Variable value stamps - dispensed by machines that print the cost of the postage at the time the stamp is dispensed.
- War tax stamp — A variation on the postal tax stamp to defray the cost of war.
- Water-activated stamp — for many years, water-activated stamps were the only type available, so this term entered into use with the advent of self-adhesive stamps. The adhesive or gum on a water-activated stamp must be moistened (usually by licking, thus the stamps are also known as "lick and stick").
First day covers
Postage stamps are first issued on a specific date, often referred to as the First day of issue. A first day cover usually consists of an envelope, a postage stamp and a postmark with the date of the stamp’s first day of issue thereon. Starting in the mid-20th century some countries began assigning the first day of issue to a place associated with the subject of the stamp design, such as a specific town or city. There are two basic types of First Day Covers (FDCs) noted by collectors. The first and often most desirable type among advanced collectors is a cover sent through the mail in the course of everyday usage, without the intention of the envelope and stamp ever being retrieved and collected. The second type of FDC is often referred to as "Philatelic," that is, an envelope and stamp sent by someone with the intention of retrieving and collecting the mailed item at a later time and place. The envelope used for this type of FDC often bears a printed design or cachet of its own in correspondence with the stamp’s subject and is usually printed well in advance of the first day of issue date. The latter type of FDC is usually far more common, and is usually inexpensive and relatively easy to acquire. Covers that were sent without any secondary purpose are considered non-philatelic and often are much more challenging to find and collect.'
Souvenir or miniature sheets
Postage stamps are sometimes issued in souvenir sheets or miniature sheets containing one or a small number of stamps. Souvenir sheets typically include additional artwork or information printed on the selvage, the border surrounding the stamps. Sometimes the stamps make up a greater picture. Some countries, and some issues, are produced as individual stamps as well as sheets.
Stamp collecting is a popular hobby. Collecting is not the same as philately, which is defined as the study of stamps. It is not necessary to closely study stamps in order to enjoy collecting them. Many casual collectors enjoy accumulating stamps without worrying about the details. The creation of a valuable or comprehensive collection, however, may require some philatelic knowledge.
Stamp collectors are an important source of revenue for some small countries that create limited runs of elaborate stamps designed mainly to be bought by stamp collectors. The stamps produced by these countries may far exceed their postal needs. Hundreds of countries, each producing scores of different stamps each year, resulted in 400,000 different types of stamps in existence by the year 2000. Annual world output averages about 10,000 types.
Some countries authorize the production of postage stamps that have no postal use, but are intended instead solely for collectors. Other countries issue large numbers of low denomination stamps that are bundled together in starter packs for new collectors. Official reprints are often printed by companies who have purchased or contacted for those rights and such reprints see no postal use. All of these stamps are often found "canceled to order", meaning they are postmarked without ever having passed through the postal system. Most national post offices produce stamps that would not be produced if there were no collectors, some to a far more prolific degree than others. It is up to individual collectors whether this concerns them; collecting such issues is as legitimate an endeavor as any other collection, but is unlikely to result in a collection of any value or to provide a monetary return on an investment (though it may be found worthwhile in other ways, such as teaching geography or collecting methods to a child, or sheer pleasure in the beauty of some of these issues). Others may argue that since these stamps are virtually worthless, they will be discarded in large numbers and eventually become less common and thus collectable in their own right, though this process would likely take many decades.
Sales of stamps to collectors who do not use them for mailing can result in large profits. Good examples of excessive issues have been (1) the stamps produced by Nicholas F. Seebeck and (2) stamps produced for the component states of the United Arab Emirates. Seebeck operated in the 1890s as an agent of Hamilton Bank Note Company. He approached Latin American countries with an offer to produce their entire postage stamp needs for free. In return he would have exclusive rights to market stamps to collectors. Each year a new issue would be produced, but would expire at the end of the year. This assured Seebeck of a continuing supply of remainders. In the 1960s, printers such as the Barody Stamp Company contracted to produce stamps for the separate Emirates and other countries. The sparse population of the desert states made it wholly unlikely that many of these stamps would ever be used for mailing purposes, and earned them the name of the "sand dune" countries. Another example of what might be considered by some to be excessive issues is that, at the time of the millennium, the United Kingdom issued 96 different stamps over about 24 months, all for pre-existing values with the same four rates for each set.
In the United States there is concern among some collectors that the United States Postal Service has become a promotional agent for the media and entertainment industry, as it has frequently issued entire sets of stamps featuring movie stars and cartoon characters like Mickey Mouse and Bart Simpson Over the decades the annual average number of new postage stamp issued by the U.S.P.S. has significantly increased.
- Basel Dove
- British Guiana 1c magenta
- Hawaiian Missionaries
- Inverted Head 4 Annas
- Inverted Jenny
- Mauritius "Post Office"
- Penny Black
- Scinde Dawk
- Treskilling Yellow
- Uganda Cowries
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Water bears or tartigrades are about 500 species of tiny aquatic invertebrate animals in the phylum Tartigrada, including about 90 species in North America. Water bears have a very widespread distribution, occurring in moist habitats from the Arctic to the Antarctic and on mountains as high as 19,680 ft (6,000 m).
Water bears have roughly cylindrical dark-colored bodies with four body segments and four pairs of stumpy legs each tipped with tiny claws. The awkward, pawing locomotion of these animals was thought to vaguely resemble that of slow-moving bears and hence the common name of these creatures which was given in the nineteenth century. The mouthparts of water bears are adapted for piercing, and they have a muscular pharynx for sucking the juices of mosses, liverworts, and algae. Water bears do not have active circulatory or respiratory systems; materials move about within their tiny body by simple processes such as diffusion, while respiratory gases diffuse across the surface of the body.
Water bears are typically found in moist films on mosses, lichens, angiosperm plants with a rosette growth form, and plant litter. Water bears can utilize this micro-aquatic habitat because they are only 0.002-0.05 in (0.05-1.2 mm) long. Water bears are highly tolerant of the desiccation that frequently afflicts these sorts of habitats because they have an ability to encapsulate into a shrivelled, spherical state until moist conditions again return. Water bears may also be collected from debris or mud in more typical shallow-water habitats, but they are seldom abundant in ponds or lakes. Some species occur in sandy and pebbly marine habitats above the zone of most vigorous wave action.
Water bears have separate sexes, but males are quite uncommon. Usually males are only relatively abundant during the winter and spring. Some species only produce females, reproducing by a process known as parthenogenesis which does not require sex. There are four to 12 post-hatching developmental stages, each beginning with a molt of the cuticle.
Water bears are interesting little creatures which have fascinated zoologists ever since the discovery of microscopy first made these animals visible several centuries ago.
- Water Conservation - Freshwater Resources, Water Consumption, Efficient Water Utilization Efforts, Economic Incentives For Water Conservation
- Water - What Is Water?, An Unusual Liquid
- Other Free Encyclopedias
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# CHAPTER 14 WORK, POWER, AND MACHINES SECTION 14.1
CHAPTER 14 WORK, POWER, AND MACHINES SECTION 14.1
I. WORK AND POWER A. WHAT IS WORK? Work requires motion Work is a product of force and distance
Work depends on direction Work depends on direction B. CALCULATING WORK Equation: Work = Force x Distance
W= F x d Work is measured in Joules (SI unit) 1(Newton)(meters)= 1(Joule) Work - Joules (J) Force Newton's (N) Distance Meters (m) Power Watts (W) Time Seconds (s)
B. CALCULATING WORK Work = force x distance Work = 1600N x 2.0m Work = 3200(N)(m) = 3200J Imagine Isaac lifting Joseph into the air repeatedly. How much WORK does he do with each lift? He lifts Joseph 2.0m
and exerts an average force of 190 N? W = ________ x________ = _________ W= 190N x 2.0 m W = 380 N x m = 380J C. What is Power? 1.Power a quantity that
measures the rate at which work is done Power = Work /Time or P = W/T 2. Power is measured in Watts (W) 3. Watts is the amount of power that is required to do 1 Joule of work in 1 second
C. Calculating Power How much power is used if a force of 25 Newton's is used to push a box a distance of 10 meters in 5 seconds. How much power is used? P= work Work isnt given so
what time do you do? Find work first (W= F x d) W= 25N x 10m= 250 Work 250J = 50Watts time 5s
D. James Watt and Horsepower Besides a watt, another common unit of power is the horsepower
(hp) 1 (hp) = 746 watts Section 14.2 Work and Machines
A. Machines Do Work A machine is a device that changes a force. Machines make work easier They change the size of force needed, the direction of a force or the distance over which a force acts.
B. Work Input and Work Output The force you exert on a machine is called the input force. C. Work Output of a machine The force that is exerted by a machine is
the output force. SECTION 14.3 Mechanical Advantage
A. Mechanical Advantage The number of times that the machine increases an input force. 1. Actual Mechanical Advantage AMA- equals the ratio of the output
force to the input force 2. Ideal Mechanical Advantage (IMA) is mechanical advantage in the absence of friction B. Calculating mechanical advantage
MA = output /input Simple Machines 1. Simple Machines one of six basic types of machines of which all other machines are composed
The Lever Family Simple lever Pulley
Wheel and Axle Levers are divided into 3 different classes 1st class have a fulcrum in the middle of an arm. Input force acts on one end and the other end applies an
output force Second Class Lever The fulcrum is at one end of the arm and the input force is applied at the other end
Third class lever Multiply distance rather than force. As a result, they have a mechanical advantage of less than 1. ( The human body has many) Pulleys
Pulleys are modified levers The Inclined Plane Family Multiply and redirect force Ex. Apply force to push an object up a ramp, The ramp redirects force to lift the object upward
Inclined Planes 1. Wedge turns downward force into forces directed out to the sides Inclined Planes 2. Screw- an inclined plane wrapped
around a cylinder Inclined Planes 3. Inclined plane- changes both magnitude and the direction of force
Compound Machines Combines two or more simple machines I. Energy and Work Energy measured in joules Transfer energy from doing work
Ex. Slingshot
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Created By : Naaz Fatima
Reviewed By : Phani Ponnapalli
Last Updated : May 10, 2023
Angular Resolution Calculator will help you find the angular resolution of a lens, resolve the small details of an object. Make your calculations simple and quick using the handy tool over here in a matter of seconds.
Choose a Calculation
Wavelength:
Aperture diameter:
### Angular Resolution - Definition
Angular Resolution is the capacity of an optical instrument to separate two objects that are located at a small angular distance. If the angular resolution is small then you can see more details.
### Angular Resolution Formula
Angular Resolution also known as Rayleigh Criterion and the emperical formula for Angular Resolution is given by
╬╕ = 1.22 * ╬╗ / d
• Where ╬╕ is the Angular Resolution and expressed in units radians
• ╬╗ is the wavelength of light
• d is the diameter of lens aperture
Above formula is based on the principle of diffraction grating. Two Light Sources are said to be resolved if the principal diffraction maximum of one image coincides with the first minimum of the other.
### How to Calculate the Angular Resolution?
Follow the simple steps provided below to find out the Angular Resolution easily. They are along the lines
• Firstly, determine the wavelength of the light.
• And then, determine the diameter of the lens aperture
• Finally, calculate the angular resolution using the formula ╬╕ = 1.22 * ╬╗ / d and by substituing the known values.
Example
Find out the Angular Resolution of the lens if the wavelength of light is 150nm and aperture diameter is 20 mm?
Solution:
Given that
Wavelength of Light ╬╗ = 150 nm
Aperture Diameter = 20 mm
We know the formula to find out the Angular Resolution is ╬╕ = 1.22 * ╬╗ / d
Substituing the known parameters in the above formula we have the equation as follows
╬╕ = 1.22 * 150nm / 20mm
Simplifying the equation further we get the value of angular resolution as ╬╕ = 0.000524 degrees
Physicscalc.Com has got concepts like friction, acceleration due to gravity, water pressure, gravity, and many more along with their relevant calculators all one under one roof.
### Frequently Asked Questions on Angular Resolution Calculator
1. What is Angular Resolution?
Angular Resolution is the minimum angular distance between two distinct objects that an instrument can resolve in detail.
2. What are the Units of Angular Resolution?
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The Use of Biotechnology to Help Sustain Human Populations
Today the world’s population is over 6.8 billion and it is projected to reach 9.5 billion by 2050. This will increase demand for food, cotton, fuel, medicine, housing materials and other goods by at least 150%. The increased population will decrease availability of land for farming, housing, transportation and recreation. In addition, global warming will increase salinity, decrease fresh water resources, and reduce farm lands, and thus decrease food production. Increased human activity will pollute the environment, negatively affecting life and human health. Run-off insecticides, fertilizer, sewage and urban pollution may change much of the environment, making it unsustainable for life. It is imperative that sustainability measures be implemented now to help accommodate increased human populations.
Biotechnology, on the other hand, has already demonstrated that it has the tools to develop insect resistant crops for food and fiber, avoiding pollution from harmful insecticides and fertilizers. It can transfer specific characteristics across species boundaries to obtain desired traits in bacteria, protists, fungi, plants and animals. Thus, it can develop crops with genes for nitrogen fixing, salt tolerance, nutrition enrichment, and high yields. These techniques can generate fish to grow many times faster without affecting their natural appearance and to spawn multiple times on demand. It can produce nutritionally balanced low fat chicken and eggs rich in omega-3 fatty acids and milk from cattle containing antibodies and other biologics, as appropriate. These techniques have the ability to develop pollutant clarifying organisms and replenishable biofuel. Similarly, these techniques allow us to produce new antibiotics, vaccines, hormones, neurotransmitters and other biologics that would be impossible or difficult to obtain otherwise.
Thus, the appropriate application of biotechnology will be able to increase, both in quality and quantity, the production of food, cotton, medicine, wood, fuel, and agents for environmental clean-up and thus reduce pollution while conserving and sustaining life as well as the environment. It is extremely important that planning for the expanded use of biotechnological techniques be explored now to prepare for future challenges.
||Animal, Bacteria, Biofuel, Biologics, Biotechnology, Environment, Fungus, Gene Transfer, Growth Gene, Insecticides, Omega-3 Fatty Acids, Plantae, Pollution, Protista, Salt Tollerant, Spawn on Demand, Sustainability, Vaccine, World Population
The International Journal of Environmental, Cultural, Economic and Social Sustainability, Volume 7, Issue 5, pp.59-78.
Article: Print (Spiral Bound).
Article: Electronic (PDF File; 916.959KB).
Assistant Professor, Department of Biological Sciences and Biotechnology Program, Kingsborough Community College of the City University of New York, Brooklyn, NY, USA
Dr. Z.M.G. Sarwar Jahangir is a tenured Assistant Professor at Kingsborough Community College (KCC) of The City University of New York (CUNY), and Director of the Biotechnology Degree program. He teaches Recombinant DNA Technology and Cell Culture technology and also a Co-corordinator for Introduction to Modern Biology. Earlier he taught in Bangladesh Agricultural University (BAU), University of Alberta, Brooklyn College, Richard Stockton College and Wabash College in Bangladesh, Canada and USA. He received Baccalaureate from BAU, M.Sc. form Cochin University of Science and Technology, India, and Ph.D. from CUNY in Molecular and Cellular Biology. He is engaged in teaching and research in biotechnology for over 15 years, and has extensive earlier experience on fish reproductive neuroendocrinology. With a grant from NOAA (1992-1995) developed a stock of lake sturgeon with ∃-galctosidase gene, developed a biotechnology laboratory and a course on biotechnology at Stockton College, taught advanced genetics at Wabash College (2000-2001), developed research on microsatellite DNA for fish population identification funded by NOAA (2000-2002) and PSC-CUNY (2005), and an Advisor to BAU for developing a biotechnology program (2000). Currently, at KCC his research is on gene transfer to produce therapeutic biologics, and also collaborating scientists in Bangladesh for developing fast growing rohu and to spawn on demand.
Professor, Department of Biology, Brooklyn College of The City University of New York, Brooklyn, NY, USA
Dr. Ronald A. Eckhardt is Professor of Biology at Brooklyn College of The City University of New York having served for 40 years. He has taught graduate courses including Cell Culture, Virology, and Molecular/Cell Biology. In the past, he has been Chair of the Department of Biology and Dean of Continuing Higher Education. He received his BS in 1964 from Loyola College, Baltimore, MD, and his PhD in 1969 from The Catholic University of America, Washington, DC, & Oak Ridge National Laboratories, Oak Ridge, TN. He also completed a Post-Doc (1969-71) at Yale University, New Haven, CT. He has supervised many graduate students including four receiving PhD’s in Cell and Molecular Biology, numerous MS degrees in Biology, and has published over 75 articles in peer reviewed journals. He has received grants as a PI or Co-PI from federal, state and private agencies for developing transgenic lake sturgeon carrying a ∃-galactosidase marker gene, developing microsatellite DNA markers for goosefish population identification, and for de-differentiation/re-differentiation of melanocytes in vitro for human neurotherapy. Currently, he is a Co-PI for two grants from the National Science Foundation and the PI of a New York State Department of Education STEP grant as well as a Royal Society of Chemistry, US Chapter, award recipient for facilitating minority science student research. He is also currently collaborating with scientists in Bangladesh for developing fast growing and spawning-on-demand rohu fish.
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# 99.79 kg to lbs - 99.79 kilograms to pounds
## 99.79 kg to lbs
Do you want to know how much is 99.79 kg equal to lbs and how to convert 99.79 kg to lbs? You couldn’t have chosen better. In this article you will find everything about kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to point out that whole this article is dedicated to a specific number of kilograms - exactly one kilogram. So if you need to know more about 99.79 kg to pound conversion - read on.
Before we get to the more practical part - this is 99.79 kg how much lbs conversion - we will tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
## 99.79 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, in formal International System of Units (in short form SI).
From time to time the kilogram can be written as kilogramme. The symbol of this unit is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. First definition was simply but totally impractical to use.
Later, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by another definition.
The new definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is 0.001 tonne. It is also divided into 100 decagrams and 1000 grams.
## 99.79 kilogram to pounds
You know a little about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in other systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is just equal 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### 99.79 kg in lbs
The most theoretical section is already behind us. In this part we are going to tell you how much is 99.79 kg to lbs. Now you know that 99.79 kg = x lbs. So it is high time to know the answer. Have a look:
99.79 kilogram = 219.9992912498 pounds.
That is an exact result of how much 99.79 kg to pound. It is possible to also round off the result. After rounding off your result will be exactly: 99.79 kg = 219.538 lbs.
You know 99.79 kg is how many lbs, so let’s see how many kg 99.79 lbs: 99.79 pound = 0.45359237 kilograms.
Naturally, in this case you may also round off the result. After it your result is as following: 99.79 lb = 0.45 kgs.
We also want to show you 99.79 kg to how many pounds and 99.79 pound how many kg results in charts. Have a look:
We want to start with a chart for how much is 99.79 kg equal to pound.
Kilograms Pounds Pounds (rounded off to two decimal places)
99.79 219.9992912498 219.5380
Now see a chart for how many kilograms 99.79 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
99.79 0.45359237 0.45
Now you learned how many 99.79 kg to lbs and how many kilograms 99.79 pound, so it is time to move on to the 99.79 kg to lbs formula.
### 99.79 kg to pounds
To convert 99.79 kg to us lbs you need a formula. We will show you two formulas. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 219.9992912498 outcome in pounds
The first version of a formula give you the most accurate outcome. In some cases even the smallest difference could be significant. So if you need a correct result - this version of a formula will be the best solution to know how many pounds are equivalent to 99.79 kilogram.
So let’s move on to the second formula, which also enables conversions to learn how much 99.79 kilogram in pounds.
The second version of a formula is as following, see:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, the second formula is simpler. It can be the best option if you need to make a conversion of 99.79 kilogram to pounds in fast way, for instance, during shopping. Just remember that final outcome will be not so accurate.
Now we want to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 99.79 kg to lbs we want to show you another way to know 99.79 kg to how many lbs totally effortless.
### 99.79 kg to lbs converter
Another way to learn what is 99.79 kilogram equal to in pounds is to use 99.79 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first formula which we gave you above. Thanks to 99.79 kg pound calculator you can easily convert 99.79 kg to lbs. Just enter amount of kilograms which you need to calculate and click ‘convert’ button. The result will be shown in a flash.
So let’s try to calculate 99.79 kg into lbs using 99.79 kg vs pound converter. We entered 99.79 as an amount of kilograms. Here is the outcome: 99.79 kilogram = 219.9992912498 pounds.
As you can see, our 99.79 kg vs lbs calculator is intuitive.
Now we can move on to our main issue - how to convert 99.79 kilograms to pounds on your own.
#### 99.79 kg to lbs conversion
We will start 99.79 kilogram equals to how many pounds calculation with the first version of a formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 219.9992912498 the result in pounds
So what have you do to learn how many pounds equal to 99.79 kilogram? Just multiply number of kilograms, in this case 99.79, by 2.20462262. It is 219.9992912498. So 99.79 kilogram is equal 219.9992912498.
You can also round it off, for example, to two decimal places. It is equal 2.20. So 99.79 kilogram = 219.5380 pounds.
It is time for an example from everyday life. Let’s calculate 99.79 kg gold in pounds. So 99.79 kg equal to how many lbs? And again - multiply 99.79 by 2.20462262. It is exactly 219.9992912498. So equivalent of 99.79 kilograms to pounds, when it comes to gold, is 219.9992912498.
In this example it is also possible to round off the result. Here is the outcome after rounding off, in this case to one decimal place - 99.79 kilogram 219.538 pounds.
Now let’s move on to examples converted using short formula.
#### How many 99.79 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 219.538 the result in pounds
So 99.79 kg equal to how much lbs? As in the previous example you have to multiply number of kilogram, this time 99.79, by 2.2. See: 99.79 * 2.2 = 219.538. So 99.79 kilogram is equal 2.2 pounds.
Let’s make another conversion using shorer version of a formula. Now convert something from everyday life, for example, 99.79 kg to lbs weight of strawberries.
So let’s convert - 99.79 kilogram of strawberries * 2.2 = 219.538 pounds of strawberries. So 99.79 kg to pound mass is exactly 219.538.
If you learned how much is 99.79 kilogram weight in pounds and are able to calculate it using two different formulas, we can move on. Now we want to show you these results in tables.
#### Convert 99.79 kilogram to pounds
We know that results presented in tables are so much clearer for most of you. We understand it, so we gathered all these results in charts for your convenience. Due to this you can easily compare 99.79 kg equivalent to lbs results.
Start with a 99.79 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
99.79 219.9992912498 219.5380
And now look 99.79 kg equal pound chart for the second version of a formula:
Kilograms Pounds
99.79 219.538
As you see, after rounding off, when it comes to how much 99.79 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Please note it when you need to do bigger amount than 99.79 kilograms pounds conversion.
#### How many kilograms 99.79 pound
Now you know how to calculate 99.79 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What about 99.79 kilogram to pounds and ounces calculation?
We will show you how you can calculate it step by step. Start. How much is 99.79 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 99.79, by 2.20462262. So 99.79 * 2.20462262 = 219.9992912498. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 99.79 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So final result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final outcome is 2 pounds and 33 ounces.
As you see, conversion 99.79 kilogram in pounds and ounces easy.
The last calculation which we are going to show you is calculation of 99.79 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate foot pounds to kilogram meters you need another formula. Before we show you it, look:
• 99.79 kilograms meters = 7.23301385 foot pounds,
• 99.79 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 99.79 foot pounds to kilograms meters you need to multiply 99.79 by 0.13825495. It is equal 0.13825495. So 99.79 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 99.79 foot pounds is 0.14 kilogram meters.
We hope that this conversion was as easy as 99.79 kilogram into pounds calculations.
This article was a huge compendium about kilogram, pound and 99.79 kg to lbs in conversion. Due to this calculation you know 99.79 kilogram is equivalent to how many pounds.
We showed you not only how to make a conversion 99.79 kilogram to metric pounds but also two other calculations - to check how many 99.79 kg in pounds and ounces and how many 99.79 foot pounds to kilograms meters.
We showed you also another way to make 99.79 kilogram how many pounds calculations, it is using 99.79 kg en pound converter. It will be the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you are able to make 99.79 kilogram equal to how many pounds calculation - on your own or with use of our 99.79 kgs to pounds converter.
Don’t wait! Convert 99.79 kilogram mass to pounds in the way you like.
Do you need to make other than 99.79 kilogram as pounds conversion? For example, for 10 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so easy as for 99.79 kilogram equal many pounds.
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
99.79 kg to lbs - 99.79 kilograms to pounds - Mass and Weight Conversion
## How to convert 99.79 kilograms (kg) to Pounds (lbs)
A POWERFUL TOOL for converting kilograms to pounds (kg to pounds). How to convert 99.79 kilograms (kg) to Pounds (lbs). How Heavy Is 99.79 Kilograms in Pounds? How do I convert kg to lbs? How Many Kilograms in a Pound?
### How do I convert kg to lbs?
99.01 kg to lbs = 218.28 99.02 kg to lbs = 218.302 99.03 kg to lbs = 218.324 99.04 kg to lbs = 218.346 99.05 kg to lbs = 218.368 99.06 kg to lbs = 218.39 99.07 kg to lbs = 218.412 99.08 kg to lbs = 218.434 99.09 kg to lbs = 218.456 99.1 kg to lbs = 218.478 99.11 kg to lbs = 218.5 99.12 kg to lbs = 218.522 99.13 kg to lbs = 218.544 99.14 kg to lbs = 218.566 99.15 kg to lbs = 218.588 99.16 kg to lbs = 218.61 99.17 kg to lbs = 218.632 99.18 kg to lbs = 218.654 99.19 kg to lbs = 218.677 99.2 kg to lbs = 218.699 99.21 kg to lbs = 218.721 99.22 kg to lbs = 218.743 99.23 kg to lbs = 218.765 99.24 kg to lbs = 218.787 99.25 kg to lbs = 218.809
99.26 kg to lbs = 218.831 99.27 kg to lbs = 218.853 99.28 kg to lbs = 218.875 99.29 kg to lbs = 218.897 99.3 kg to lbs = 218.919 99.31 kg to lbs = 218.941 99.32 kg to lbs = 218.963 99.33 kg to lbs = 218.985 99.34 kg to lbs = 219.007 99.35 kg to lbs = 219.029 99.36 kg to lbs = 219.051 99.37 kg to lbs = 219.073 99.38 kg to lbs = 219.095 99.39 kg to lbs = 219.117 99.4 kg to lbs = 219.139 99.41 kg to lbs = 219.162 99.42 kg to lbs = 219.184 99.43 kg to lbs = 219.206 99.44 kg to lbs = 219.228 99.45 kg to lbs = 219.25 99.46 kg to lbs = 219.272 99.47 kg to lbs = 219.294 99.48 kg to lbs = 219.316 99.49 kg to lbs = 219.338 99.5 kg to lbs = 219.36
99.51 kg to lbs = 219.382 99.52 kg to lbs = 219.404 99.53 kg to lbs = 219.426 99.54 kg to lbs = 219.448 99.55 kg to lbs = 219.47 99.56 kg to lbs = 219.492 99.57 kg to lbs = 219.514 99.58 kg to lbs = 219.536 99.59 kg to lbs = 219.558 99.6 kg to lbs = 219.58 99.61 kg to lbs = 219.602 99.62 kg to lbs = 219.625 99.63 kg to lbs = 219.647 99.64 kg to lbs = 219.669 99.65 kg to lbs = 219.691 99.66 kg to lbs = 219.713 99.67 kg to lbs = 219.735 99.68 kg to lbs = 219.757 99.69 kg to lbs = 219.779 99.7 kg to lbs = 219.801 99.71 kg to lbs = 219.823 99.72 kg to lbs = 219.845 99.73 kg to lbs = 219.867 99.74 kg to lbs = 219.889 99.75 kg to lbs = 219.911
99.76 kg to lbs = 219.933 99.77 kg to lbs = 219.955 99.78 kg to lbs = 219.977 99.79 kg to lbs = 219.999 99.8 kg to lbs = 220.021 99.81 kg to lbs = 220.043 99.82 kg to lbs = 220.065 99.83 kg to lbs = 220.087 99.84 kg to lbs = 220.11 99.85 kg to lbs = 220.132 99.86 kg to lbs = 220.154 99.87 kg to lbs = 220.176 99.88 kg to lbs = 220.198 99.89 kg to lbs = 220.22 99.9 kg to lbs = 220.242 99.91 kg to lbs = 220.264 99.92 kg to lbs = 220.286 99.93 kg to lbs = 220.308 99.94 kg to lbs = 220.33 99.95 kg to lbs = 220.352 99.96 kg to lbs = 220.374 99.97 kg to lbs = 220.396 99.98 kg to lbs = 220.418 99.99 kg to lbs = 220.44 100 kg to lbs = 220.462
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FAYETTEVILLE, Ark., March 30 (UPI) -- Earlier this year, scientists at Caltech offered the most convincing evidence yet of a ninth planet, Planet X. Now, a retired astrophysicist suggests the hidden planet is responsible for Earth's periodic mass extinctions -- like the disappearance of the dinosaurs.
In a new study published in the journal Monthly Notices of the Royal Astronomical Society, Daniel Whitmire argues that an undiscovered ninth planet triggers disruptive comet showers every 27 million years.
It's not the first time Whitmire -- now a math teacher at the University of Arkansas -- has made such a claim in a major scientific journal. In 1985, he offered a similar explanation for mass extinctions in the journal Nature -- then an astrophysicist at the University of Louisiana at Lafayette.
Whitmire and his research partner John Matese pointed to evidence of periodic comet showers in the fossil record dating back some 500 million years.
In 1985, there were two alternative theories for what might trigger major comet showers -- a sister star to the sun, vertical oscillations of the sun as it orbits around the center of the Milky Way. Those theories have since been discredited, while the Planet X theory has acquired legitimacy.
The Caltech study estimated Planet X to be approximately 10 times the mass of Earth, big enough to throw comets into the inner solar system as its oblong orbit sends it closer to the Kuiper Belt every 27 million years.
The Kuiper Belt is a ring-shaped region of comets and other larger bodies circling the solar system just beyond Neptune. Caltech researchers inferred the existence and path of a ninth planet by studying anomalies in the orbits of several major Kuiper Belt objects.
Whitmire suggests -- as they did in 1985 -- that a periodic invasion of comets results in violent collisions. Those that miss Earth disintegrate in the inner solar system and dim the sun's solar energy, cooling Earth.
Whitmire is hopeful additional evidence of Planet X can offer more answers about the evolution of the solar system and life on Earth.
"I've been part of this story for 30 years," he said in a news release. "If there is ever a final answer I'd love to write a book about it."
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Dielectric heating, also known as electronic heating, radio frequency heating, and high-frequency heating, is the process in which a radio frequency (RF) alternating electric field, or radio wave or microwave electromagnetic radiation heats a dielectric material. At higher frequencies, this heating is caused by molecular dipole rotation within the dielectric.
Radio frequency (RF) is the oscillation rate of an alternating electric current or voltage or of a magnetic, electric or electromagnetic field or mechanical system in the frequency range from around twenty thousand times per second to around three hundred billion times per second. This is roughly between the upper limit of audio frequencies and the lower limit of infrared frequencies; these are the frequencies at which energy from an oscillating current can radiate off a conductor into space as radio waves. Different sources specify different upper and lower bounds for the frequency range.
Radio waves are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light. Radio waves have frequencies as high as 300 gigahertz (GHz) to as low as 30 hertz (Hz). At 300 GHz, the corresponding wavelength is 1 mm, and at 30 Hz is 10,000 km. Like all other electromagnetic waves, radio waves travel at the speed of light. They are generated by electric charges undergoing acceleration, such as time varying electric currents. Naturally occurring radio waves are emitted by lightning and astronomical objects.
Microwaves are a form of electromagnetic radiation with wavelengths ranging from about one meter to one millimeter; with frequencies between 300 MHz (1 m) and 300 GHz (1 mm). Different sources define different frequency ranges as microwaves; the above broad definition includes both UHF and EHF bands. A more common definition in radio engineering is the range between 1 and 100 GHz. In all cases, microwaves include the entire SHF band at minimum. Frequencies in the microwave range are often referred to by their IEEE radar band designations: S, C, X, Ku, K, or Ka band, or by similar NATO or EU designations.
RF dielectric heating at intermediate[ clarification needed ] frequencies, due to its greater penetration over microwave heating, shows greater promise than microwave systems as a method of very rapidly heating and uniformly preparing certain food items, and also killing parasites and pests in certain harvested crops.
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Molecular rotation occurs in materials containing polar molecules having an electrical dipole moment, with the consequence that they will align themselves in an electromagnetic field. If the field is oscillating, as it is in an electromagnetic wave or in a rapidly oscillating electric field, these molecules rotate continuously by aligning with it. This is called dipole rotation, or dipolar polarisation. As the field alternates, the molecules reverse direction. Rotating molecules push, pull, and collide with other molecules (through electrical forces), distributing the energy to adjacent molecules and atoms in the material. The process of energy transfer from the source to the sample is a form of radiative heating.
An electromagnetic field is a physical field produced by electrically charged objects. It affects the behavior of charged objects in the vicinity of the field. The electromagnetic field extends indefinitely throughout space and describes the electromagnetic interaction. It is one of the four fundamental forces of nature.
A molecule is an electrically neutral group of two or more atoms held together by chemical bonds. Molecules are distinguished from ions by their lack of electrical charge. However, in quantum physics, organic chemistry, and biochemistry, the term molecule is often used less strictly, also being applied to polyatomic ions.
Alternating current (AC) is an electric current which periodically reverses direction, in contrast to direct current (DC) which flows only in one direction. Alternating current is the form in which electric power is delivered to businesses and residences, and it is the form of electrical energy that consumers typically use when they plug kitchen appliances, televisions, fans and electric lamps into a wall socket. A common source of DC power is a battery cell in a flashlight. The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage.
Temperature is related to the average kinetic energy (energy of motion) of the atoms or molecules in a material, so agitating the molecules in this way increases the temperature of the material. Thus, dipole rotation is a mechanism by which energy in the form of electromagnetic radiation can raise the temperature of an object. There are also many other mechanisms by which this conversion occurs.
Temperature is a physical quantity expressing hot and cold. It is measured with a thermometer calibrated in one or more temperature scales. The most commonly used scales are the Celsius scale, Fahrenheit scale, and Kelvin scale. The kelvin is the unit of temperature in the International System of Units (SI), in which temperature is one of the seven fundamental base quantities. The Kelvin scale is widely used in science and technology.
In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.
Dipole rotation is the mechanism normally referred to as dielectric heating, and is most widely observable in the microwave oven where it operates most efficaciously on liquid water, and also, but much less so, on fats and sugars. This is because fats and sugar molecules are far less polar than water molecules, and thus less affected by the forces generated by the alternating electromagnetic fields. Outside of cooking, the effect can be used generally to heat solids, liquids, or gases, provided they contain some electric dipoles.
A microwave oven is an electric oven that heats and cooks food by exposing it to electromagnetic radiation in the microwave frequency range. This induces polar molecules in the food to rotate and produce thermal energy in a process known as dielectric heating. Microwave ovens heat foods quickly and efficiently because excitation is fairly uniform in the outer 25–38 mm(1–1.5 inches) of a homogeneous, high water content food item; food is more evenly heated throughout than generally occurs in other cooking techniques.
Water is a transparent, tasteless, odorless, and nearly colorless chemical substance, which is the main constituent of Earth's streams, lakes, and oceans, and the fluids of most living organisms. It is vital for all known forms of life, even though it provides no calories or organic nutrients. Its chemical formula is H2O, meaning that each of its molecules contains one oxygen and two hydrogen atoms, connected by covalent bonds. Water is the name of the liquid state of H2O at standard ambient temperature and pressure. It forms precipitation in the form of rain and aerosols in the form of fog. Clouds are formed from suspended droplets of water and ice, its solid state. When finely divided, crystalline ice may precipitate in the form of snow. The gaseous state of water is steam or water vapor. Water moves continually through the water cycle of evaporation, transpiration (evapotranspiration), condensation, precipitation, and runoff, usually reaching the sea.
Fat is one of the three main macronutrients, along with the other two: carbohydrate and protein. Fats molecules consist of primarily carbon and hydrogen atoms, thus they are all hydrocarbon molecules. Examples include cholesterol, phospholipids and triglycerides.
Dielectric heating involves the heating of electrically insulating materials by dielectric loss. A changing electric field across the material causes energy to be dissipated as the molecules attempt to line up with the continuously changing electric field. This changing electric field may be caused by an electromagnetic wave propagating in free space (as in a microwave oven), or it may be caused by a rapidly alternating electric field inside a capacitor. In the latter case, there is no freely-propagating electromagnetic wave, and the changing electric field may be seen as analogous to the electric component of an antenna near field. In this case, although the heating is accomplished by changing the electric field inside the capacitive cavity at radio-frequency (RF) frequencies, no actual radio waves are either generated or absorbed. In this sense, the effect is the direct electrical analog of magnetic induction heating, which is also near-field effect (thus not involving radio waves).[ citation needed ]
Dielectric loss quantifies a dielectric material's inherent dissipation of electromagnetic energy. It can be parameterized in terms of either the loss angleδ or the corresponding loss tangent tan δ. Both refer to the phasor in the complex plane whose real and imaginary parts are the resistive (lossy) component of an electromagnetic field and its reactive (lossless) counterpart.
The near field and far field are regions of the electromagnetic field (EM) around an object, such as a transmitting antenna, or the result of radiation scattering off an object. Non-radiative 'near-field' behaviours of electromagnetic fields dominate close to the antenna or scattering object, while electromagnetic radiation 'far-field' behaviours dominate at greater distances.
Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.
Frequencies in the range of 10–100 MHz are necessary to cause dielectric heating, although higher frequencies work equally well or better, and in some materials (especially liquids) lower frequencies also have significant heating effects, often due to more unusual mechanisms. For example, in conductive liquids such as salt water, ion-drag causes heating, as charged ions are "dragged" more slowly back and forth in the liquid under influence of the electric field, striking liquid molecules in the process and transferring kinetic energy to them, which is eventually translated into molecular vibrations and thus into thermal energy.[ citation needed ]
Dielectric heating at low frequencies, as a near-field effect, requires a distance from electromagnetic radiator to absorber of less than 1/ ≈ 1/ of a wavelength. It is thus a contact process or near-contact process, since it usually sandwiches the material to be heated (usually a non-metal) between metal plates taking the place of the dielectric in what is effectively a very large capacitor. However, actual electrical contact is not necessary for heating a dielectric inside a capacitor, as the electric fields that form inside a capacitor subjected to a voltage do not require electrical contact of the capacitor plates with the (non-conducting) dielectric material between the plates. Because lower frequency electrical fields penetrate non-conductive materials far more deeply than do microwaves, heating pockets of water and organisms deep inside dry materials like wood, it can be used to rapidly heat and prepare many non-electrically conducting food and agricultural items, so long as they fit between the capacitor plates.[ citation needed ]
At very high frequencies, the wavelength of the electromagnetic field becomes shorter than the distance between the metal walls of the heating cavity, or than the dimensions of the walls themselves. This is the case inside a microwave oven. In such cases, conventional far-field electromagnetic waves form (the cavity no longer acts as a pure capacitor, but rather as an antenna), and are absorbed to cause heating, but the dipole-rotation mechanism of heat deposition remains the same. However, microwaves are not efficient at causing the heating effects of low frequency fields that depend on slower molecular motion, such as those caused by ion-drag.[ citation needed ]
Dielectric heating must be distinguished from Joule heating of conductive media, which is caused by induced electric currents in the media. [ citation needed ]For dielectric heating, the generated power density per volume is given by:
where ω is the angular frequency of the exciting radiation, εr″ is the imaginary part of the complex relative permittivity of the absorbing material, ε0 is the permittivity of free space and E the electric field strength. The imaginary part of the (frequency-dependent) relative permittivity is a measure for the ability of a dielectric material to convert electromagnetic field energy into heat.[ citation needed ]
If the conductivity σ of the material is small, or the frequency is high, such that σ ≪ ωε (with ε = εr″ · ε0), then dielectric heating is the dominant mechanism of loss of energy from the electromagnetic field into the medium.[ citation needed ]
Microwave frequencies penetrate conductive materials, including semi-solid substances like meat and living tissue, to a distance defined by the skin effect. The penetration essentially stops where all the penetrating microwave energy has been converted to heat in the tissue. Microwave ovens used to heat food are not set to the frequency for optimal absorption by water. If that were so, then the piece of food or liquid in question would absorb all microwave radiation in its outer layer, leading to a cool, unheated centre and a superheated surface. Instead, the frequency selected allows energy to penetrate deeper into the heated food. The frequency of a household microwave oven is 2.45 GHz, while the frequency for optimal absorbency by water is around 10 GHz.
The use of high-frequency electric fields for heating dielectric materials had been proposed in the 1930s. For example, U.S. Patent 2,147,689 (application by Bell Telephone Laboratories, dated 1937) states "This invention relates to heating systems for dielectric materials and the object of the invention is to heat such materials uniformly and substantially simultaneously throughout their mass. It has been proposed therefore to heat such materials simultaneously throughout their mass by means of the dielectric loss produced in them when they are subjected to a high voltage, high frequency field." This patent proposed radio frequency (RF) heating at 10 to 20 megahertz (wavelength 15 to 30 meters). Such wavelengths were far longer than the cavity used, and thus made use of near-field effects and not electromagnetic waves. (Commercial microwave ovens use wavelengths only 1% as long.)
In agriculture, RF dielectric heating has been widely tested and is increasingly used as a way to kill pests in certain food crops after harvest, such as walnuts still in the shell. Because RF heating can heat foods more uniformly than is the case with microwave heating, RF heating holds promise as a way to process foods quickly.
In medicine, the RF heating of body tissues, called diathermy, is used for muscle therapyHeating to higher temperatures, called hyperthermia therapy, is used to kill cancer and tumor tissue.
Microwave heating, as distinct from RF heating, is a sub-category of dielectric heating at frequencies above 100 MHz, where an electromagnetic wave can be launched from a small dimension emitter and guided through space to the target. Modern microwave ovens make use of electromagnetic waves with electric fields of much higher frequency and shorter wavelength than RF heaters. Typical domestic microwave ovens operate at 2.45 GHz, but 915 MHz ovens also exist. This means that the wavelengths employed in microwave heating are 0.1 cm to 10 cm. This provides for highly efficient, but less penetrative, dielectric heating.[ citation needed ]
Although a capacitor-like set of plates can be used at microwave frequencies, they are not necessary, since the microwaves are already present as far field type EM radiation, and their absorption does not require the same proximity to a small antenna as does RF heating. The material to be heated (a non-metal) can therefore simply be placed in the path of the waves, and heating takes place in a non-contact process which does not require capacitative conductive plates.[ citation needed ]
Microwave volumetric heating is a commercially available method of heating liquids, suspensions, or solids in a continuous flow on an industrial scale. Microwave volumetric heating has a greater penetration depth, of up to 42 millimetres (1.7 in), which is an even penetration through the entire volume of the flowing product. This is advantageous in commercial applications where increased shelf-life can be achieved, with increased microbial kill at temperatures 10–15 °C (18–27 °F) lower than when using conventional heating systems.
Application for microwave volumetic heating:
In drying of foods, dielectric heating is usually combined with conventional heating. It may be used to preheat the feed to a hot-air drier. By raising the temperature of the feed quickly and causing moisture to move to the surface, it can decrease the overall drying time. Dielectric heating may be applied part-way through the drying cycle, when the food enters the falling rate period. This can boost the rate of drying. If dielectric heating is applied near the end of hot-air drying it can also shorten the drying time significantly and hence increase the throughput of the drier. It is more usual to use dielectric heating in the later stages of drying. One of the major applications of RF heating is in the postbaking of biscuits. The objectives in baking biscuits are to produce a product of the right size, shape, color, and moisture content. In a conventional oven, reducing the moisture content to the desired level can take up a large part of the total baking time. The application of RF heating can shorten the baking time. The oven is set to produce biscuits of the right size, shape, and color, but the RF heating is used to remove the remaining moisture, without excessive heating of the already dry sections of the biscuit.The capacity of an oven can be increased by more than 50% by the use of RF heating. Postbaking by RF heating has also been applied to breakfast cereals and cereal-based baby foods.
Food quality is maximized and better retained using electromagnetic energy than conventional heating. Conventional heating results in large disparity in temperature and longer processing times which can cause overprocessing on the food surface and impairment of the overall quality of the product.Electromagnetic energy can achieve higher processing temperatures in shorter times, therefore, more nutritional and sensory properties are conserved.
In physics, electromagnetic radiation refers to the waves of the electromagnetic field, propagating (radiating) through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, (visible) light, ultraviolet, X-rays, and gamma rays.
The electromagnetic spectrum is the range of frequencies of electromagnetic radiation and their respective wavelengths and photon energies.
In physics, radiation is the emission or transmission of energy in the form of waves or particles through space or through a material medium. This includes:
In electromagnetism, absolute permittivity, often simply called permittivity, usually denoted by the Greek letter ε (epsilon), is the measure of capacitance that is encountered when forming an electric field in a particular medium. More specifically, permittivity describes the amount of charge needed to generate one unit of electric flux in a particular medium. Accordingly, a charge will yield more electric flux in a medium with low permittivity than in a medium with high permittivity. Permittivity is the measure of a material's ability to store an electric field in the polarization of the medium.
In radio engineering, an antenna is the interface between radio waves propagating through space and electric currents moving in metal conductors, used with a transmitter or receiver. In transmission, a radio transmitter supplies an electric current to the antenna's terminals, and the antenna radiates the energy from the current as electromagnetic waves. In reception, an antenna intercepts some of the power of a radio wave in order to produce an electric current at its terminals, that is applied to a receiver to be amplified. Antennas are essential components of all radio equipment.
Super high frequency (SHF) is the ITU designation for radio frequencies (RF) in the range between 3 and 30 gigahertz (GHz). This band of frequencies is also known as the centimetre band or centimetre wave as the wavelengths range from one to ten centimetres. These frequencies fall within the microwave band, so radio waves with these frequencies are called microwaves. The small wavelength of microwaves allows them to be directed in narrow beams by aperture antennas such as parabolic dishes and horn antennas, so they are used for point-to-point communication and data links and for radar. This frequency range is used for most radar transmitters, wireless LANs, satellite communication, microwave radio relay links, and numerous short range terrestrial data links. They are also used for heating in industrial microwave heating, medical diathermy, microwave hyperthermy to treat cancer, and to cook food in microwave ovens.
A resonator is a device or system that exhibits resonance or resonant behavior, that is, it naturally oscillates at some frequencies, called its resonant frequencies, with greater amplitude than at others. The oscillations in a resonator can be either electromagnetic or mechanical. Resonators are used to either generate waves of specific frequencies or to select specific frequencies from a signal. Musical instruments use acoustic resonators that produce sound waves of specific tones. Another example is quartz crystals used in electronic devices such as radio transmitters and quartz watches to produce oscillations of very precise frequency.
Diathermy is electrically induced heat or the use of high-frequency electromagnetic currents as a form of physical therapy and in surgical procedures. The earliest observations on the reactions of high-frequency electromagnetic currents upon the human organism were made by Jacques Arsene d'Arsonval. The field was pioneered in 1907 by German physician Karl Franz Nagelschmidt, who coined the term diathermy from the Greek words dia and θέρμη therma, literally meaning "heating through".
Electrosurgery is the application of a high-frequency alternating polarity, electrical current to biological tissue as a means to cut, coagulate, desiccate, or fulgurate tissue.. Its benefits include the ability to make precise cuts with limited blood loss. Electrosurgical devices are frequently used during surgical operations helping to prevent blood loss in hospital operating rooms or in outpatient procedures.
A microwave cavity or radio frequency (RF) cavity is a special type of resonator, consisting of a closed metal structure that confines electromagnetic fields in the microwave region of the spectrum. The structure is either hollow or filled with dielectric material. The microwaves bounce back and forth between the walls of the cavity. At the cavity's resonant frequencies they reinforce to form standing waves in the cavity. Therefore, the cavity functions similarly to an organ pipe or sound box in a musical instrument, oscillating preferentially at a series of frequencies, its resonant frequencies. Thus it can act as a bandpass filter, allowing microwaves of a particular frequency to pass while blocking microwaves at nearby frequencies.
Microwave burns are burn injuries caused by thermal effects of microwave radiation absorbed in a living organism. In comparison with radiation burns caused by ionizing radiation, where the dominant mechanism of tissue damage is internal cell damage caused by free radicals, the primary damage mechanism of microwave radiation is by heat.
Non-ionizingradiation refers to any type of electromagnetic radiation that does not carry enough energy per quantum to ionize atoms or molecules—that is, to completely remove an electron from an atom or molecule. Instead of producing charged ions when passing through matter, non-ionizing electromagnetic radiation has sufficient energy only for excitation, the movement of an electron to a higher energy state. Ionizing radiation which has a higher frequency and shorter wavelength than nonionizing radiation, has many uses but can be a health hazard; exposure to it can cause burns, radiation sickness, cancer, and genetic damage. Using ionizing radiation requires elaborate radiological protection measures which in general are not required with nonionizing radiation.
Microwave Volumetric Heating (MVH) is a method of using microwaves to evenly heat the entire volume of a flowing liquid, suspension or semi-solid. The process is known as MVH because the microwaves penetrate uniformly throughout the volume of the product being heated, thus delivering energy evenly into the body of the material.
Radio-frequency welding, also known as dielectric welding and high-frequency welding, is a plastics joining process that utilizes high-frequency radio waves to heat plastic parts to the point they form a melt layer. After the development of the melt layer, the parts are pressed together and then allowed to cool causing fusion. This process is capable of producing high quality joints in a range of plastics. Advantages of this process are fast cycle times, easily automated, repeatable, and good weld appearance. While this process has some great advantages, there are some limitations. Only plastics which have dipoles can be heated using radio waves and therefore not all plastics are able to be welded using this process. Also, this process is not well suited for thick or overly complex joints. The most common use of this process is lap joints or seals on thin plastic sheets or parts.
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You are Here: Home >< Maths
# STEP Maths I, II, III 1989 solutions Watch
1. STEP I (Mathematics)
1:
2: Solution by kabbers
3: Solution by Dystopia
4: Solution by Dystopia
5: Solution by Dystopia
6: Solution by Swayam
7: Solution by squeezebox
8: Solution by squeezebox
9: Solution by nota bene
10:
11: Solution by Glutamic Acid
12:
13:
14:
15:
16:
STEP II (F.Maths A)
1: Solution by squeezebox
2: Solution by kabbers
3: Solution by squeezebox
4:
5: Solution by Dystopia
6: Solution by *bobo*
7:
8:
9:
10:
11:Solution by *bobo*
12:
13:
14:Solution by squeezebox
15:
16:
STEP III (F.Maths B)
1: Solution by squeezebox
2: Solution by *bobo*
3:
4:
5:
6:
7:
8: Solution by Dystopia
9: Solution by squeezebox
10: Solution by squeezebox
11:
12:
13:
14:
15:
16:
Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
2. I am no expert at STEP, so don't hesitate to correct me, it is more than likely made that I have made a mistake.
STEP I - Question 8
Using good old de Moivre's theorem;
(using
and similary;
Now, consider:
Where c is .
Now let ,
so:
Which gives;
().
These values of give distinct values of
or
()
Hence the roots of the equation are:
and .
3. STEP III - Question 10
Lets assume that the result is true for n = k;
(*)
This is the same as (*) except k+1 replaces k. Hence if the result is true for n=k, its true for n= k+1.
When n=1,
LHS of (*) = 1x2x3x4x5 = 120
RHS of (*) = (1/6)x1x2x3x4x5x6 = 120.
So (*) is true for n=1.
Hence, by induction;
.
Since;
.
Using (*);
In this case,
From the previous parts, we know that:
and clearly, also:
as ,
and
So,
Using a similar arguement, we can show that:
We have shown that the limits exist and are equal to .
Hence;
4. STEP I, Q2
The For , find
Using integration by parts ():
By approximating the area corresponding to by n rectangles of equal width and with their top right hand vertices on the curve , show that, as ,
Firstly, note that
(by the above indefinite integral)
The sum of the areas of our rectangles is going to be
Claim:
So for k + 1, we must prove that is the equation we arrive at.
Basis case (let k=1):
Inductive step:
So,
As , our sum tends to the integral . So,
is this ok? have i missed anything out or not explained each step in enough detail?
5. Question 4, STEP I
With six points, each point is attached to five other points, so at least three lines radiating from A must be of the same colour (gold, say). The points which are joined to A must then be joined to each other by silver lines, or an entirely gold triangle would be made. However, then a triangle with entirely silver edges can be made from these three points.
Diagram showing five points is attached.
Attached Images
6. Question 5, STEP I
a)
Let x = 1.
Let x = -1
As n is even. So
As their sum is
b) Suppose that
Then
Note that and that
So, upon adding these two expressions, we get
Also,
So true by induction.
7. STEP II Q1
Substituting x= y-a into the equation:
Notice that the reduced form shown in the question has no term, so we need to find the value of a which will get rid of the term.
Expanding out and collectiing terms we get:
So the value of a to get rid of the is -1.
So we now have:
(*)
Substituting y=z/b into (*);
dividing both sides by 2 and multiplying by ;
.
And so to make this equal to;
we make b = .
Lets take b = and let z =
so;
and (these give distinct values of , and hence, three distinct solutions.)
we know that; x = y - a = (z/b) - a.
Hence the solutions of the equation are:
and
8. Question 11, STEP I.
Vertical component =
Using s = ut + 1/2at^2.
Quadratic in t. Ignoring the negative root.
R = horizontal component x time.
Horizontal component =
Multiply both numerator and denominator by 2.
Using
We can simplify it.
Splitting up Mr. Big Fraction.
Ignore the first fraction for the moment, we'll work on the second one.
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\dfrac{2v^2 \cos \theta \sin \theta}{2g} \times \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}
We can use our old friend Mr. Double Angle Formula to simplify it:
The whole fraction looks like:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
R = (\dfrac{v^2\sin 2\theta}{2g}) + (\dfrac{v^2 \sin 2\theta}{2g}) \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}
And lo and behold, there's a common factor to both fractions. Let's take it out:
And that's what we're looking for.
*I'll finish off the last small bit in a few minutes*
9. STEP I Question 6
The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be
(where x' is the general x coordinate of the normal).
At the point Q, x' = 0 as it cuts the y axis.
The distance PQ can be worked out using Pythagoras' theorem.
It's given that
So
(use a substitution of u = x^2 if you can't see why)
10. STEP III, Q8
Let
Then
Using an integrating factor, , we get
Applying boundary conditions:
---
I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.
11. Hang on a sec, mine was mislabelled, should be Step I not step II.. sorry about that!
edit: thanks.. now typing up II/2
12. STEP II/2
Therefore, by (*),
we can also use (*) to express 2cot2x as
Therefore,
we can compare the coefficients of the respective summations and say
for the second part, note that
we have to find an identity for 2csc2x
(using partial fractions)
Also,
, so:
Comparing coefficients of the polynomials (in x) of the expansion,
Rearranging and substituting in the earlier definition of ,
any corrections welcome !
13. STEP II, (11) attached
attachments 2&3 are in the wrong order
Attached Files
14. step1989q11.doc (540.0 KB, 248 views)
15. step1989q11p3.doc (205.5 KB, 235 views)
16. step1989q11p2.doc (158.5 KB, 251 views)
17. Step III (2)
Attached Files
18. step1989q2p1.doc (113.5 KB, 280 views)
19. step1989q2p2.doc (79.0 KB, 262 views)
20. step1989q2p3.doc (71.5 KB, 256 views)
21. STEP II (6) attached
Attached Files
22. step1989q6p1.doc (79.5 KB, 249 views)
23. step1989q62.doc (95.0 KB, 294 views)
24. STEP II Question 3
Equating real and imaginary parts:
(*)
(**)
if x=a, then:
(1)
(2)
Now, subbing (1) and (2) into:
We get:
As required.
Substituting y=b into (*) and (**):
(3)
(4).
This time, using , we end up with:
.
Since and , . Hence, () corresponds to a single hyeprbola branch, which is above the line v=0.
One point of intersection is: and .
Differentiating implicitly we get:
.
Which is equal to: at and .
Differentiating implicitly we get:
.
Which is, when and .
X
Hence the curves intersect and right angles at this point.
Regarding the sketch, I think the elipse becomes a vertical straight line from v = -1 to v = 1, and the hyperbola becomes the line v = 1.
25. STEP 1, Q3
(By the ratio theorem.)
These vectors are perpendicular, so the scalar product is zero.
(Note that the scalar product is commutative and distributive.)
Let
Minimum and maximum values occur at the endpoint of a range or at a turning point. There is a turning point when x = 1/2.
So
Note that
Where
The maximum value is occurs when
Suppose that
Then
So
By the cosine rule,
Let F be the midpoint of AB. By Pythagoras,
26. *bump* (just to remind people that this thread is still here ) - I have solutions that I can type up if no one else wants to, but I really don't want to take over the thread..
27. STEP I - Question 7
*Graph Attached*
If we work out the area of a quarter of the rectangle, which is in the top right hand bit of the axes, and maximise this, it make calculation soooo much eaiser. (Since we can forget about modulus signs and y being negative) :
Let the area of the rectangle be .
Base of smaller rectangle = 1-x
height =
so
which we want to maximise.
Which is zero when
Finding the second derivative and subsituting x = in shows is a maximum at this point.
So .
________
We can do the second bit in the same way, since its a closed curve that is symmetrical in both axes.
Let the area of the second rectangle be .
Working in the top right quadrant again;
Base = 1-x
height = .
So .
Which is zero when x = 0 or x = .
Clearly the area is a minimum when x= 0, and so is a maximum when x =.
.
Attached Files
28. STEP I Q7 graph.doc (196.0 KB, 212 views)
29. I've got a solution to I question 9 , but don't have a scanner available so will add graphs when I get the opportunity.
Question 9 STEP I
and Setting y'=0
f(2)=-2, f(-2)=2, also for graphing purposes it may be worth to note that f(0)=0 and f(4)=2 and f(-4)=-2. y=0 has solutions x=0 or f''(-2)=- i.e. local max f''(2)=+ i.e. local min
Graph, see attachment
(a) i.e. therefore the equation of y in this X-Y plane is
setting it equal to 0 gives f(1)=-2 ans f(-1)=2
For graphing it can also be good to see that f(2)=2, f(-2)=-2 and
Graph, see attachment
(b) i.e. therefore the equation of y in this X-Y plane is
Setting this equal to 0 gives . f(2)=-1 and f(-2)=1
For graphing, also see that , f(4)=1 and f(-4)=-1
Graph, see attachment.
(c) i.e. therefore the equation in the X-Y plane is
Setting this equal to 0 gives f(0)=2, f(2)=-2.
For graphing, also see that and f(-1)=-2, f(3)=2
(d) i.e. therefore the equation in the X-Y plane is
setting this equal to 0 gives f(2)=0 f(-2)=2
For graphing I'll check f(4) and f(-4) too, which give 2 and 0 respectively.
For the last part, we're looking for a graph in the X-Y plane with local min (0,0) and local max (1,1). To obtain this on the form X=ax+b and Y=cy+d, first observe that for a graph of a cubic, changing the sign means a reflection in the x-axis i.e. local max and min swap places, this is what we need to do. That means either a or c is negative (but not both!). a squeezes the graph along the X-axis and c squeezes it along the Y-axis, c moves the graph to the left/right and d moves the graph up/down. Knowing these we can see that and will produce the desired graph.
Okay, so I hope I've interpreted everything correct in the question. Seems pretty nice as a question.
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In our last blog we discussed the National Highway Traffic Safety Administration and the role it plays in keeping us safe out on the road. The NHTSA is part of the Executive Branch. Its job is to perform studies on accident frequency and automobile related deaths. Most importantly, the NHTSA upholds vehicle safety standards. Aside from the NHTSA, another major advancement in vehicle safety has been crash testing. Through crash testing, car-manufacturing companies have been able to manufacture vehicles that are more likely to protect drivers and their passengers in the event of an accident.
How is Crash Testing Done?
Typically, a crash test dummy that simulates a human is placed in the driver’s seat of the vehicle. To create an accident like situation, the vehicle is driven into a concrete wall. The dummy is equipped with a number of sensors that indicate the amount of force on each body part during the crash. You may have seen an example of this on commercials or television shows. Often times frontal impact tests are used, but there are other types of tests performed. To ensure safety vehicles also have to undergo rollover tests, side-impact tests, and both small and moderate overlap tests. These tests are done to accumulate data on every collision type imaginable. The more data car manufactures have regarding collisions, the better.
What has Crash Testing Accomplished?
Crash testing has led to a number of major safety innovations. The data produced by the rigorous scientific observations during crash tests has contributed to safety breakthroughs such as crumble zones and side-impact airbags. Additionally, crash test results are released in consumer reports. If a vehicle receives a bad crash test rate, the vehicle is less likely to sell. After all, who wants to purchase a car they know they won’t be safe in? In 1998 the Rover received a one-star crash rating on a possible five star scale. Sales of the vehicle plummeted until its production was discarded entirely.
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### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### 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. . . .
### Lower Bound
##### Stage: 3 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
##### Stage: 3 Challenge Level:
Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Good Work If You Can Get It
##### Stage: 3 Challenge Level:
A job needs three men but in fact six people do it. When it is finished they are all paid the same. How much was paid in total, and much does each man get if the money is shared as Fred suggests?
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Christmas Chocolates
##### Stage: 3 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Fibs
##### Stage: 3 Challenge Level:
The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?
### Marbles in a Box
##### Stage: 3 and 4 Challenge Level:
In a three-dimensional version of noughts and crosses, how many winning lines can you make?
##### 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. . . .
### Boxed In
##### Stage: 3 Challenge Level:
A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box?
### Pair Products
##### Stage: 4 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### How Big?
##### Stage: 3 Challenge Level:
If the sides of the triangle in the diagram are 3, 4 and 5, what is the area of the shaded square?
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
A car's milometer reads 4631 miles and the trip meter has 173.3 on it. How many more miles must the car travel before the two numbers contain the same digits in the same order?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### 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?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### 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. . . .
### Legs Eleven
##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Partially Painted Cube
##### Stage: 4 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Always a Multiple?
##### Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
### Quick Times
##### Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### 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.
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### More Number Pyramids
##### Stage: 3 and 4 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### One and Three
##### Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
### Attractive Tablecloths
##### Stage: 4 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### 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.
### Areas of Parallelograms
##### Stage: 4 Challenge Level:
Can you find the area of a parallelogram defined by two vectors?
### Fair Shares?
##### Stage: 4 Challenge Level:
A mother wants to share a sum of money by giving each of her children in turn a lump sum plus a fraction of the remainder. How can she do this in order to share the money out equally?
### Magic Sums and Products
##### Stage: 3 and 4
How to build your own magic squares.
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
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| null |
# What is C(n, 1) + C(n, 2) + _ _ _ _ _ + C(n, n) equal to
This question was previously asked in
NDA (Held On: 18 Apr 2021) Maths Previous Year paper
View all NDA Papers >
1. 2 + 22 + 23 + _ _ _ _ _ + 2n
2. 1 + 2 + 22 + 2+ _ _ _ _ _ + 2n
3. 1 + 2 + 22 + 23 + _ _ _ _ _ _ + 2n - 1
4. 2 + 22 + 23 + _ _ _ _ _ + 2n - 1
## Answer (Detailed Solution Below)
Option 3 : 1 + 2 + 22 + 23 + _ _ _ _ _ _ + 2n - 1
Free
NDA Mathematics Full Mock Test
6.8 K Users
120 Questions 300 Marks 150 Mins
## Detailed Solution
Concept:
(1 + x)n = nC0 × 1(n-0) × x 0nC1 × 1(n-1) × x 1 + nC2 × 1(n-2) × x2 + …. + nCn × 1(n-n) × xn
nth term of the G.P. is an = arn−1
Sum of n terms = s = $$a (r^n-1)\over(r- 1)$$; where r >1
Sum of n terms = s = $$a (1- r^n)\over(1- r)$$; where r <1
Calculation:
C(n, 1) + C(n, 2) + _ _ _ _ _ + C(n, n)
nC1 + nC2 + ... + nCn
⇒ nC0 + nC1 + nC2 + ... + nCn - nC0
⇒ (1 + 1)n - nC
2n - 1 = $$\rm 2^n - 1\over 2-1$$ = 1 × $$\rm 2^n - 1\over 2-1$$
Comparing it with a G.P sum = a × $$\rm r^n - 1\over r-1$$, we get a = 1 and r = 2
∴ 2n - 1 = 1 + 2 + 22 + ... +2n-1 which will give us n terms in total.
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Essentially nothing is known about the biology and behaviour of the Andean cat (2) (2). Its range is so remote and inhospitable it has proved extremely difficult to survey the area and there are no Andean cats known to be in captivity (2). Knowledge is built up from rare sightings of the animal, physiological studies of stuffed specimens and more recently genetic analysis of faeces (2) (6).
The Andean cat’s range does appear to coincide with the distribution of the mountain chinchillas (Chinchilla lanigera) and viscachas (Lagidium spp.), and it has been observed hunting these species (2). Both prey species escape from predators by bounding off rock faces and making unpredictable changes in direction. The Andean cat’s long tail probably aids in balance when chasing these rodents (2). The Andean cat’s diet may or may not include other species, such as birds, reptiles and other small rodents, but there is no information on this (2). This small cat has an acute sense of hearing, which may assist in hunting, due to its well developed ear drums (5). This adaptation is typical of animals that inhabit arid environments with little cover for protection, such as the mountain chinchilla species (2).
The Andean cat may suffer from competition with the pampas cat (Leopardus colocolo) for food and space. The pampas cat occurs in higher numbers in the Andes, and also occupies the lower, more productive regions of the Andes, which may constrain the Andean cat (6). The Andean cat has a significantly lower population than the pampas cat and is believed to live in low densities, though no figures are known (5) (6). It appears to be extremely specialised in its habitat requirements, and the presence of rocky piles and boulders may be important (2).
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Scientists are studying whether long COVID could be linked to viral fragments found in the body months after initial infection.
In the chaos of the first months of the coronavirus pandemic, oncologist and geneticist Ami Bhatt was intrigued by widespread reports of vomiting and diarrhoea in people infected with SARS-CoV-2. “At that time, this was thought to be a respiratory virus,” she says. Bhatt and her colleagues, curious about a possible link between the virus and the gastrointestinal symptoms, began to collect stool samples from people with COVID-19.
Thousands of miles away from Bhatt’s laboratory at Stanford Medicine in California, gastroenterology internist Timon Adolph was puzzled by accounts of gut symptoms in infected people. Adolph and his colleagues at the Medical University of Innsbruck in Austria started to assemble specimens, too — gastrointestinal-tissue biopsies.
Two years into the pandemic, the scientists’ foresight has paid off: both teams have recently published results1,2 suggesting that pieces of SARS-CoV-2 can linger in the gut for months after an initial infection. The findings add to a growing pool of evidence supporting the hypothesis that persistent bits of virus — coronavirus “ghosts”, Bhatt has called them — could contribute to the mysterious condition called long COVID.
Even so, Bhatt both urges scientists to keep an open mind and cautions that researchers have not yet nailed down a link between persistent viral fragments and long COVID. “Additional studies still need to be done — and they’re not easy,” she says.
Long COVID is often defined as symptoms that linger beyond 12 weeks after an acute infection. More than 200 symptoms have been associated with the disorder, which ranges in severity from mild to debilitating. Theories about its origins vary, and include harmful immune responses, tiny blood clots and lingering viral reservoirs in the body. Many researchers think that a mix of these factors contributes to the global burden of disease.
An early hint that the coronavirus might persist in the body came in work3 published in 2021 by gastroenterologist Saurabh Mehandru at the Icahn School of Medicine at Mount Sinai in New York City and his colleagues. By then, it was clear that cells lining the gut display the protein that the virus uses to enter cells. This allows SARS-CoV-2 to infect the gut.
Mehandru and his team found viral nucleic acids and proteins in gastrointestinal tissue collected from people who’d been diagnosed with COVID-19 an average of four months earlier. The researchers also studied participants’ memory B cells, which are pivotal players in the immune system. The team found that antibodies produced by these B cells were continuing to evolve, suggesting that, at six months after the initial infection, the cells were still responding to molecules made by SARS-CoV-2.
Inspired by this work, Bhatt and her colleagues found that a few people continued to shed viral RNA into their stool seven months after an initial mild or moderate SARS-CoV-2 infection, well after their respiratory symptoms had ended1.
Virus goes for the gut
Adolph says the 2021 paper inspired his team to look at their biopsy samples for signs of coronavirus. They found that 32 of 46 study participants who had had mild COVID-19 showed evidence of viral molecules in their gut seven months after acute infection. About two-thirds of those 32 people had long-COVID symptoms.
But all of the participants in this study had inflammatory bowel disease, an autoimmune disorder, and Adolph cautions that his data do not establish that there is active virus in these people, or that the viral material is causing long COVID.
In the meantime, more studies have suggested lingering viral reservoirs beyond the gut. Another team of researchers has studied tissue collected from autopsies of 44 people who had been diagnosed with COVID-19 and found evidence of viral RNA in many sites, including the heart, eyes and brain4. Viral RNA and proteins were detected up to 230 days after infection. The study has not yet been peer reviewed.
Nearly all of the people in that sample had had severe COVID-19, but a separate study of two people who had had mild COVID-19 followed by long COVID symptoms found viral RNA in the appendix and the breast5. Pathologist Joe Yeong at the Institute of Molecular and Cell Biology at the Agency for Science, Technology, and Research in Singapore, who is a co-author of the report, which has not been peer reviewed, speculates that the virus might infiltrate and hide out in immune cells called macrophages, which can be found in a variety of the body’s tissues.
All of these studies support the possibility that long-term viral reservoirs contribute to long COVID, but researchers will need to do more work to conclusively show a link, says Mehandru. They will need to document that the coronavirus is evolving in people who are not immunocompromised, and they will need to link such evolution to long COVID symptoms. “Right now there is anecdotal evidence, but there are a lot of unknowns,” Mehandru says
Bhatt is hopeful that samples will become available to test the viral-reservoir hypothesis. The US National Institute of Health, for example, is running a large study called RECOVER, which aims to tackle the causes of long COVID and will collect biopsies from the lower intestines of some participants.
But Sheng says he does not need to wait for a billion-dollar study to get more samples: an organization of people with long COVID has contacted him and offered to send samples from members who have had biopsies for various reasons, such as a cancer diagnosis, after their infections. “It’s really random, the tissue can come from everywhere,” he says. “But they don’t want to wait.”
Nature 605, 408-409 (2022)
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Restoring South America’s Atlantic forests
Latin America/Caribbean > South America > Argentina
Latin America/Caribbean > South America > Brazil
Latin America/Caribbean > South America > Paraguay
The Upper Paraná Atlantic Forest stretches across the borders of Brazil, Paraguay and Argentina. The rainforests are among the most biologically-rich regions in the world, home to jaguars and tamarins; but it is also one of the most endangered. Today only 7% of the original forest cover remains in Brazil, all of it fragmented by centuries of unsustainable use and logging.
WWF has a number of restoration projects in the region aimed at returning native forest where it has previously been destroyed or degraded. It is also working on establishing new protected areas and creating “green corridors” to connect isolated tracts of forests.
The Atlantic Forest ecoregion once stretched over 1 million km2 along Brazil's coast in 13 states, with extensions inland into Eastern Paraguay and the Misiones province in Northeastern Argentina. The ecoregion contains 2 types of tropical moist broadleaf forests, the coastal and interior Atlantic Forests, and the Araucaria Pine Forest which previously covered a large portion of the Brazilian states of Parana and Santa Catarina and their borders with Argentina.
The coastal and interior Atlantic Forests are some of the richest tropical moist forests on Earth, harboring unique collections of species quite distinct from the Amazon. A 1993 survey identified 450 different tree species within one hectare of Atlantic Forests in the Southern Bahia state - one of the highest diversities of tree species reported in the world.
Significant numbers of the animals and plants occur only in the Atlantic Forests ecoregion, long isolated from the Amazon basin by the drier Cerrado. Over 52% of the tree species and 92% of the amphibians in the Atlantic Forests are found nowhere else in the world.
The ecoregion is also an important centre for bird endemism (at least 158 endemic species), palms, bromeliads, and orchids. Of Brazil's 77 primate species (more than any other country in the world), 17 species and 6 genera are endemic to the Atlantic Forests. Many species occur only in limited areas within the ecoregion. For example, each of the 4 endangered species of lion tamarins occurs only in a different small area of the Atlantic Forests.
This part of Brazil has developed into the agricultural, industrial and population center of the country. The cities of Rio de Janeiro and Sao Paulo both lie within the forests. Today only 7% of the original Atlantic Forests cover remains in Brazil, all of it fragmented by centuries of unsustainable use. This fragmentation, coupled with high endemism, makes the Atlantic Forests one of the most endangered rainforests in the world.
Those species whose habitat is restricted to a small part of the forests are especially threatened. With forest fragmentation species populations are sub-divided and reduced to precariously low numbers of individuals. The probability of long-term survival of small isolated populations is very low; thus nearly all these endemic species can be considered endangered, with several on the verge of extinction.
1. Increase WWF institutional presence in the Atlantic Forests, building credibility to act in the region in partnership with government and other NGOs.
2. Carry out specific activities which make information available that can serve as a basis for ecoregional conservation planning.
3. Contribute to the development of an ecoregional conservation plan, with an emphasis on establishment and effective implementation of protected areas.
4. Promote the establishment of new protected areas.
5. Contribute to the effective implementation of protected areas.
To ensure stable ecosystems and biological processes as well as to preserve viable populations of key endemic species in the long-term, all forest fragments must be preserved, prioritizing action according to forest type, biodiversity, local endemism, size and biological integrity of the forest.
In addition, these fragments must be strategically linked with forest corridors, which in some cases imply forest rehabilitation.
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# Separate imaginary and real parts from complex expression
I learned about complex numbers after I was trying to create a fractal object.
The main problem is that I have an equation with complex numbers and I have to separate their parts (real & imaginary) to calculate the next iteration.
Some equations like the $f(p) = p^2 + c$ are obviously and easy to solve them. But some of them are using the exponential form of the complex numbers, that really bothers me.
Take for example this equation: \begin{equation*} F(P) = c*e^{-p} + p*p~\text{Where}~c = u + v*i~\text{and}~p = a + b*i. \end{equation*} I solved it partly by this way - express $e^{-p}$ like $1 / (e^a \ast (\cos(b) + \sin(b) * i))$ and so on... But in the end I couldn't see the relation to separate the groups.
Any help and advice how to proceed in this or similar case will be greatly appreciated.
• You're almost finished, you can simplify $e^{-p}$ to $e^{a}(\cos(b)-i\sin(b))$. Just multiply the top and bottom of $e^{-p}$ by $(\cos(b)-i\sin(b))$. After that, just collect the terms from $p^2$ Commented May 18, 2015 at 12:04
• @user3845133 complex question with a complex answer! :) Commented May 19, 2015 at 16:22
\begin{equation*} F(P) = c*e^{-p} + p*p~\text{Where}~c = u + v*i~\text{and}~p = a + b*i. \end{equation*}
\begin{equation*} F(P) = e^{-p}(c+e^p*p^2)~\text{Where}~c = u + v*i~\text{and}~p = a + b*i. \end{equation*}
\begin{equation*} F(P) = \frac{1}{e^p}(c+e^p*p^2)~\text{Where}~c = u + v*i~\text{and}~p = a + b*i. \end{equation*}
\begin{equation*} F(P) = \frac{(c+e^p*p^2)}{e^p}~\text{Where}~c = u + v*i~\text{and}~p = a + b*i. \end{equation*}
\begin{equation*} F(P) = \frac{c}{e^p}+p^2~\text{Where}~c = u + v*i~\text{and}~p = a + b*i. \end{equation*}
$------------$
$$F(P)=\frac{u+vi}{e^{a+bi}}+(a+bi)^2=$$ $$F(P)=e^{-a-bi}(u+vi)+(a+bi)^2$$ $$F(P)=e^{|-a-bi|e^{\arg(-a-bi)i}}(u+vi)+(a+bi)^2=$$ $$F(P)=e^{\sqrt{a^2+b^2}e^{\arg(-a-bi)i}}(u+vi)+(a+bi)^2=$$
$$F(P)=\left|e^{\sqrt{a^2+b^2}e^{\arg(-a-bi)i}}(u+vi)\right|e^{\arg\left(e^{-a-bi}(u+vi)\right)i}+\left(\left|a+bi\right|e^{\arg\left(a+bi\right)i}\right)^2=$$
$$F(P)=\left(\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\right)e^{\arg\left(e^{-a-bi}(u+vi)\right)i}+\left(\sqrt{a^2+b^2}e^{\arg\left(a+bi\right)i}\right)^2=$$
$$F(P)=\left(\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\right)e^{\arg\left(e^{-a-bi}(u+vi)\right)i}+(a^2+b^2)e^{2\arg\left(a+bi\right)i}=$$
SO:
$$\Re=(a^2+b^2)\cosh(2\arg(a+bi))+\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\cosh(\arg(e^{-a-bi}(u+vi)))$$ $$\Im=\left((a^2+b^2)\sinh(2\arg(a+bi))+\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\sinh(\arg(e^{-a-bi}(u+vi)))\right)i$$
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CLOSE UP Twisted, concentrated field lines poking out of the solar surface keep surrounding convective plasma (bright orange) from flowing into their locations, creating local cool spots that we see as sunspots
A Comparison of Rome Observatory Sunspot
Area and Sunspot
Number Determinations with International Measures, 1958-1998, NASA/TP-2005-214191, NASA Marshall Space Flight Center, Huntsville, AL, 20 pp.
An almost continuous line of CaK plage was observed for about 60[degrees] east-west at a latitude between S05[degrees] and S10[degrees] from Aug 17 to 20, linking plage around the sunspot
groups in that region.
For those who require additional illustrations, sunspots
and faculae provide an interesting proving ground.
The transit will be particularly exciting if lots of sunspots
are visible at the same time.
Around 2020, sunspots
may disappear for years, maybe decades.
The last major occasion of this was between 1645-1715, when no sunspots
on the sun's surface appeared during a 70-year-period.
There was a period from 1645 to 1715 when the sun showed similar lack of sunspots
, this was the Maunder minimum when trees stopped growing and the River Thames froze.
who has tracked sunspot
intensities from different spots around the globe dating back four centuries, also concludes that such solar disturbances have little or no impact on global warming.
By comparison, AeroLEDs says the nominal life of its SUNSpot
is 50,000 hours running at a current of about 2 to 3 amps.
Conference on Solar MHD Theory and Observations; a High Spatial Resolution Perspective (2005: Sunspot
, New Mexico) Ed.
The most striking feature of the complete SN [sunspot
number] profile is the uniqueness of the steep rise of sunspot
activity during the first half of the 20th century.
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Describe properties of plane figures (such as circles, triangles, squares and rectangles) and solid shapes (such as spheres, cubes and cylinders). (0306.4.1)
Links verified on 1/4/2012
- Anglemania - (3-5) describe triangles using appropriate geometric vocabulary
- Comparing Plane Figures to Solid Figures -Compare the objects defining a plane figure verses a solid figure.
- Plane Figures A Two-dimensional figure drawn on a flat surface is known as plane geometry.
- Plane Figure Quiz - Self check quiz with hints if make wrong choice.
- Plane Figure Quiz -Find the plane shape shown on the solid.
- Polygons and Non-Polygons - drag polygons into one bin and anything that is not a polygon into another bin
- Properties of Plane Figures - Understanding the properties of plane figures.
- Shape, Space and Measure - activities, quizzes and more
- Triangle Identification - type of triangle - given its angles
- Triangle Identification - type of triangle - given its number of sides
site for teachers | PowerPoint show | Acrobat document | Word document | whiteboard resource | sound | video format | interactive lesson | a quiz | lesson plan | to print
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A pseudanthium (Greek for "false flower") or flower head is a special type of inflorescence, in which anything from a small cluster to hundreds or sometimes thousands of flowers are grouped together to form a single flower-like structure. Pseudanthia take various forms. The individual flowers of a pseudanthium commonly are called florets. The real flowers (the florets) are generally small and often greatly reduced, but the pseudanthium itself can sometimes be quite large (as in the heads of some varieties of sunflower).
Pseudanthia are characteristic of the daisy and sunflower family (Asteraceae), whose flowers are differentiated into ray and disk flowers, unique to this family. The disk flowers in the center of the pseudanthium are actinomorphic and the corolla is fused into a tube. Flowers on the periphery are zygomorphic and the corolla has one large lobe (the so-called "petals" of a daisy are individual ray flowers, for example). Either ray or disk flowers may be absent in some plants: Senecio vulgaris lacks ray flowers and Taraxacum officinale lacks disk flowers. The pseudanthium has a whorl of bracts below the flowers, forming an involucre.
In all cases, a pseudanthium (inflorescence) is superficially indistinguishable from a flower, but closer inspection of its anatomy will reveal that it is composed of multiple flowers. Thus, the pseudanthium represents an evolutionary convergence of the inflorescence to a reduced reproductive unit that may function in pollination like a single flower, at least in plants that are animal pollinated.
Capitulum (plural capitula)
Capitulum can be used as an exact synonym for pseudanthium and flower head; however its use is generally but not always restricted to the asteraceae family. At least one source defines it as a small flower head. In addition to its botanical use as a term meaning flower head it is also used to mean the top of the sphagnum plant.
Calathid (plural calathids or calathidia)
This is a very rarely used term. It was defined in the 1966 book, The genera of flowering plants (Angiospermae), as a specific term for a flower head of a plant in the asteraceae family. However on-line botanical glossaries do not define it and Google Scholar does not link to any significant usage of the term in a botanical sense.
Pseudanthia occur in the following plant families:
- Apiaceae — pseudanthia are called umbels
- Araceae — pseudanthia are called spadices
- Euphorbiaceae — pseudanthia are called cyathia
- Pontederiaceae — in Hydrothrix
- Saururaceae — in Anemopsis
In some families it is not yet clear whether the 'flower' represents a pseudanthium, because the anatomical work has not been done (or is still ambiguous due to considerable evolutionary reduction). Possible pseudanthia of this type may occur in the following families:
Flower head of a common daisy (Bellis perennis)
Flowers open in succession in head of a sunflower (Helianthus annuus), with ray florets forming the 'petals'
Close up of the ray corolla of a Hieracium lachenalii; every "petal" is actually a separate five-petaled flower complete with its own stamens and making its own fruit.
Discoid (having only disk flowers) flower heads of a rubber rabbitbrush.
- Hutchinson, John (1964). The genera of flowering plants (Angiospermae). Oxford: Clarendon Press. LCCN 65000676.
- "calflora Botanical Terms". Retrieved 2012-02-26.
- "Senecio vulgaris L.". Missouri Plants. Retrieved 2 December 2012.
- "Taraxacum Officinale". Florida Data. Retrieved 2 December 2012.
- "The Jepson Herbarium Glossary". Retrieved 2012-02-26.
- "GardenWeb Glossary of Botanical Terms". Retrieved 2012-02-26.
- "Dictionary of Botany Capitulum". Retrieved 2012-02-26.
- "GardenWeb Glossary". Retrieved 2012-02-26.
- Harris, James (2001). Plant Identification Terminology An illustrated Glossary. Spring Lake. ISBN 978-0-9640221-6-4.
- "Australian bryophytes". Retrieved 2012-02-26.
- Petra Hoffmann, Hashendra S. Kathriarachchi, and Kenneth J. Wurdack. 2006. "A Phylogenetic Classification of Phyllanthaceae (Malpighiales)." Kew Bulletin. 61(1):40.
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